>Can someone please tell me how to remove old posts. I still have post from The easiest way is probably to reformat the hard drive and then reinstall Windows. (Don't forget to back up anything you want to keep. Probably copying such things to something like C:Backup would be a good idea...) ************************ David C. Ullrich ==== >The easiest way is probably to reformat the hard drive and then >reinstall Windows. ^ +---- ... some other operating system besides ... ==== I seem to be unable to solve a probability problem exactly. I have written a computer program to simulate the problem, but I'd like to hear your opinions on a more general solution. The problem is as follows. I have two jars. Each jar contains 20 balls, 10 of which are blue and 10 of which are red. It is a fair play situation with no faked balls etc. The following moves are carried out 1000 times: i.e., if a red ball was chosen, a new red ball is added to the jar. That is, the number of balls in both jars remains the same, but the proportion of red and blue may change. What I'd like to calculate is the exact probability that, after the 1000th round has been completed, one of the jars contains only red balls and the other only blue balls, P(only red in left jar AND only blue in right jar) + P(only blue in left jar AND only red in right jar). Additionally, I'd like to have the number of balls in each jar (m) and the number of generations (n=1000 above) as parameters in such a formula, so that I would not have to modify my program accordingly for each run. There is surely a formula for this, but I seem to be unable to find it. Comments are highly appreciated, JR ==== > > I seem to be unable to solve a probability problem exactly. I have written a > computer program to simulate the problem, but I'd like to hear your > opinions on a more general solution. The problem is as follows. > > I have two jars. Each jar contains 20 balls, 10 of which are blue and 10 of > which are red. It is a fair play situation with no faked balls etc. > > The following moves are carried out 1000 times: > > i.e., if a red ball was chosen, a new red ball is added to the jar. > > That is, the number of balls in both jars remains the same, but the > proportion of red and blue may change. > > What I'd like to calculate is the exact probability that, after the 1000th > round has been completed, one of the jars contains only red balls and the > other only blue balls, P(only red in left jar AND only blue in right jar) + > P(only blue in left jar AND only red in right jar). Additionally, I'd like > to have the number of balls in each jar (m) and the number of generations > (n=1000 above) as parameters in such a formula, so that I would not have to > modify my program accordingly for each run. There is surely a formula for > this, but I seem to be unable to find it. > > Comments are highly appreciated, > > JR I'd like to hear more about the practical applications of the problem. To me, it looks very much like you're modelling the contingency of evolutionary processes -- the principle that, from the same initial conditions, widely divergent outcomes may be equally likely. Paul Epstein ==== >I seem to be unable to solve a probability problem exactly. I have written a >computer program to simulate the problem, Here's the answer to your problem: http://smjg.port5.com/faqs/usenet/xpost.html -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/ You find yourself amusing, Blackadder. I try not to fly in the face of public opinion. ==== I seem to be unable to solve a probability problem exactly. I have written a > computer program to simulate the problem, but I'd like to hear your > opinions on a more general solution. The problem is as follows. I have two jars. Each jar contains 20 balls, 10 of which are blue and 10 of > which are red. It is a fair play situation with no faked balls etc. The following moves are carried out 1000 times: > duplicated, > i.e., if a red ball was chosen, a new red ball is added to the jar. That is, the number of balls in both jars remains the same, but the > proportion of red and blue may change. What I'd like to calculate is the exact probability that, after the 1000th > round has been completed, one of the jars contains only red balls and the > other only blue balls, P(only red in left jar AND only blue in right jar) + > P(only blue in left jar AND only red in right jar). Additionally, I'd like > to have the number of balls in each jar (m) and the number of generations > (n=1000 above) as parameters in such a formula, so that I would not have to > modify my program accordingly for each run. There is surely a formula for > this, but I seem to be unable to find it. Comments are highly appreciated, JR Are you sure you need an exact answer? What is done in one jar does not affect the other, so let's concentrate on one jar. Let p(k, n) be the probability that there are k balls in the jar at stage n. Recursion equations for p(k, n): p(k, 0) = 10 if k=10, otherwise 0 p(0, n) = p(0, n-1) + p(1, n-1) * 1/20 p(1, n) = p(1, n-1)*18/20 + p(2, n-1) * 2/20 * 18/19 p(20, n) = p(20, n-1) + p(19, n-1) * 1/20 p(19, n) = p(19, n-1)*18/20 + p(18, n-1) * 2/20 * 18/19 and for 1 < k < 19: p(k, n) = p(k-1, n-1)*(k-1)(21-k)/(19*20) + p(k, n-1)[k(k-1)+(20-k)(19-k)]/(19*20) + p(k+1, n-1) *(k+1)(19-k)/(19*20) I don't think there will be simple closed formulas for these probabilities. Using the recursion formulas, I ran this on Mathematica, caching all the probabilities that occur at each stage. p(0, 1000) is the probability that all the balls are red at stage 1000. I am showing the numerator and denominator of this fraction separately below. Each has 2271 digits. For an approximation: p(0, 1000) ~= 0.496352. It's fairly clear that as n -> infinity, p(0,n) -> 1/2 and p(20,n) -> 1/2, and p(k,n) -> 0 for 0 < k < 20. The numerator of p(0,1000): 2314499037161895047364116820669633054057080251723471658267046296598395617607 6 5182792994683275215765102668492807990422476914130445849134341280239420354081 7 4220143744912195282188815594953334034212152104184002995472144361928412625050 9 6427813171232459314197073626635669847382753342321550477935817985431543744781 8 5243594200413159043544793305232542426155290785926006267267818816454758114438 4 7412643494159062183831120418373665312010997196359372518849024975535442541995 3 5277893727357707706189675870800568538506720255298514997441385256226419482823 8 0983501031134725765557616001232029932145151280135062598279276410472488301096 7 0215356927818023445620920121029027792444609054076663430593150718830393633393 8 3459397564507845473872102840528551301154808859950834363474604921739974835113 2 8938692202096759731371890617709305149890183711501965470415296820477457952733 2 4446105854788670938390551330671262307364930951609783258503673713157845597845 1 5203931269366677890841371642640469868878828517245002715161558802034012797017 5 2211537534040985084081079827870731215300797177439901855723151907738344447561 5 8870116913880307660726008153988357049959656156425740637492827683319510197844 9 0672952352586004267571659153483177635782126921770135543248958071108024125237 0 4422928883362448683429961592246581994194311921911045100834172116064295120846 1 5775472179218463810493804328205156609552732950804245286733860521642253803163 6 4427775236318818932766371066104554183377469639640312700240987617269964502780 5 6085814095817711507679190440430717103758287264164199453604918428211689050440 4 8576126571939540108432579150348075457945176433676266565575188039414253924730 0 1730943884778753985800578913100454840531727760440184620816615961607395419869 6 3058365631505521164694048003453790853814973136556685634361408418841369833513 5 4018904052817151237314099178134069649902268576610790715874771190773692950703 6 0815450044141063858742790868592586006819519423915021413282502396394432866615 2 6825235971398127135057093886186522892576674771801153143432613837871312928193 3 8558416319243176825155061110166482206678172223974196306385574025761785803697 9 9524389000115808219108092456315605197028293362209311925844277603548147790431 5 1879678385002562369967588742090052612848582217141637256082843513332804991530 1 717704239444134002021324031335464695799 The denominator of p(0, 1000): 4663021070652057592290832019889286084934075133849783154171286066597863569214 7 2511124227555789567116695839222947422128026678133707586830899602590877709022 8 1610737775229010673660142390985702014186575494300741958646865896313461672851 9 2982987560079019695688810801036565282198482101652789269817011984469805572406 6 9161147569517672386111449720275323548088248041289097391094998905936399257876 3 2380383843937389864921351629825907880664996967340474059737870939688501390775 1 7963551268274498878112128796135874293474789955010361192860348374295020998139 9 3834866624484867404017602344519630497914227793125318341970526606502958732309 9 0527732980925260178576167366047665366701721689458109341590061024411703497287 9 7369613329903084077319521722372056586492777331520528237594982091583840800659 3 7322360077051011177850657768451345645027046158444108068814214685194335118154 0 9888504668420921603601199627573149610144430376298676523552714383180274790689 0 1118560874820183353094832175704957293244622838735556043131459528014488016035 9 5678551182141129161679412728193745589997214700409787702196137511841994402045 3 7281766558584062074148057718724843634552742869304664185616622525966983731813 2 9139470308598774993716526275368710071250194782849925684483541657989479403542 9 8467208990188435636443026091209455479126955592187500000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 0000000000000000000000000000000000000000000000000000000000000000000000000000 0 000000000000000000000000000000000000000 ==== [omitted] In some sort of anti-communist frenzy I left out the word red. Corrected: Let p(k, n) be the probability that there are k red balls in the jar at stage n. ==== > > I seem to be unable to solve a probability problem exactly. I have written > a > computer program to simulate the problem, but I'd like to hear your > opinions on a more general solution. The problem is as follows. > > I have two jars. Each jar contains 20 balls, 10 of which are blue and 10 > of > which are red. It is a fair play situation with no faked balls etc. > > The following moves are carried out 1000 times: > duplicated, > i.e., if a red ball was chosen, a new red ball is added to the jar. > > That is, the number of balls in both jars remains the same, but the > proportion of red and blue may change. > > What I'd like to calculate is the exact probability that, after the 1000th > round has been completed, one of the jars contains only red balls and the > other only blue balls, P(only red in left jar AND only blue in right jar) > + > P(only blue in left jar AND only red in right jar). Additionally, I'd like > to have the number of balls in each jar (m) and the number of generations > (n=1000 above) as parameters in such a formula, so that I would not have > to > modify my program accordingly for each run. There is surely a formula for > this, but I seem to be unable to find it. > > Comments are highly appreciated, > > JR > > > > Are you sure you need an exact answer? > > What is done in one jar does not affect the other, so let's concentrate on > one jar. Let p(k, n) be the probability that there are k balls in the jar > at stage n. > > Recursion equations for p(k, n): > > p(k, 0) = 10 if k=10, otherwise 0 > p(0, n) = p(0, n-1) + p(1, n-1) * 1/20 > p(1, n) = p(1, n-1)*18/20 + p(2, n-1) * 2/20 * 18/19 > p(20, n) = p(20, n-1) + p(19, n-1) * 1/20 > p(19, n) = p(19, n-1)*18/20 + p(18, n-1) * 2/20 * 18/19 > > and for 1 < k < 19: > p(k, n) = p(k-1, n-1)*(k-1)(21-k)/(19*20) + > p(k, n-1)[k(k-1)+(20-k)(19-k)]/(19*20) + > p(k+1, n-1) *(k+1)(19-k)/(19*20) > > I don't think there will be simple closed formulas for these probabilities. > > Using the recursion formulas, I ran this on Mathematica, caching all the > probabilities that occur at each stage. I did a Matlab simulation after working out the transition probabilities. In your notation p(k+1,n | k,n-1) = k(20-k)/(20*19) p(k-1,n | k,n-1) = k(20-k)/(20*19) p(k,n | k,n-1) = [k(k-1)+(20-k)(19-k)]/(19*20) The probability of adding a red ball is exactly the same as the probability of subtracting a red ball on each step. As k approaches 0 or 20, these probabilites get smaller and the most common event is for nothing to change (the two balls picked are the same color). The symmetry is bothering my intuition a little. Because it's equally likely to add or subtract a red ball at each step, I'm not sure why there's this bias, that the random walk converges to one endpoint or the other once an initial choice is taken. Anyway it does converge rapidly. In 1000 test runs, the number that went to one endpoint or the other was about 99.5%, and they typically did so in about 150 iterations as I recall. - Randy ==== > > I did a Matlab simulation after working out the transition > probabilities. In your notation > > p(k+1,n | k,n-1) = k(20-k)/(20*19) > p(k-1,n | k,n-1) = k(20-k)/(20*19) > p(k,n | k,n-1) = [k(k-1)+(20-k)(19-k)]/(19*20) Will Self's exact formulas can be derived from these transition probabilities, of course. Given transition probabilities, we have the following exact formula: Writing pi(n,k) = probability of k blue balls at time n, we have: pi(n,k) = pi(n-1,k-1) * p(k,n | k-1,n-1) + pi(n-1,k) * p(k,n | k,n-1) + pi(n-1,k+1) * p(k,n | k+1,n-1) which gives: pi(n,k) = [ pi(n-1,k-1) * (k-1)*(m+1-k) + pi(n-1,k) * (k*(k-1)+(m-k)*(m-1-k) + pi(n-1,k+1) * (k+1)*(m-1-k) ] / (m*(m-1)) Except for notational differences, this is the same answer that Will got. Note that this formula is okay for all k=0 to 20 provided you use the convention pi(n,-1)=pi(n,m+1)=0 for all n. For computation, start with a vector pi[k] all zeroes except for pi[m/2]=1, and calculate a new vector pi'[k] using the above formula, and set pi=pi'. Now repeat n-1 more times. For n=1000 and m=20, I get a result vector: k pi[k] 0 0.496351829 1 0.000384018 2 0.000384018 3 0.000384018 4 0.000384018 5 0.000384018 6 0.000384018 7 0.000384018 8 0.000384018 9 0.000384018 10 0.000384018 11 0.000384018 12 0.000384018 13 0.000384018 14 0.000384018 15 0.000384018 16 0.000384018 17 0.000384018 18 0.000384018 19 0.000384018 20 0.496351829 and the first and last elements represent the probability of all red and all blue respectively. Note that those center elements aren't actually equal. What's happening is that the three largest eigenvalues of the transition matrix are 1, 1, and 1-2/(m*(m-1)). The eigenvectors associated with these values are (1,0,...,0,0), (0,0,...,0,1), and (-(m-1)/2,1,...,1,-(m-1)/2) respectively. We're seeing the decay of this third eigenvector. In another 1000 generations, it should be reduced by a factor of (1-2/(20*19))^1000 = 0.005107439 (and that's exactly what happens). Anyway, to answer the poster's original question, the probability that one urn goes to all blue and the other goes to all red in 1000 generations is: 2 * pi(1000,0) * pi(1000,20) = 2 * .496351829^2 = 0.4927303 > The symmetry is bothering my intuition a little. Because > it's equally likely to add or subtract a red ball at > each step, I'm not sure why there's this bias, that > the random walk converges to one endpoint or the other > once an initial choice is taken. Well, there's no problem with that. The states at each end are absorbing. The random walk is unbiased, but if it eventually hits either endpoint, it stays there forever. Thus, as we let n get larger and larger, the chance that our ever longer random walk eventually hits an absorbing state gets bigger and bigger, so the probability of being stuck at an endpoint gets higher and higher. It would be really bizarre if the probability of being in some other state (like equal number of blue and red balls) *didn't* go to zero. This would mean that, somehow, there's positive probability that your unbiased random walk, no matter how long it lasts, can somehow contrive to *avoid* the absorbing endpoints! -- Kevin ==== I am at present reading up about Generic Extensions and Boolean-Valued Methods in Set Theory. To place the following comments in context, and as an exposition of the notation which I am using, I will denote the ground transitive model for ZFC by V, and the complete Boolean algebra by B (so that B is an element of V, and Boolean addition and Boolean multiplication are binary functions on B, so that addition and multiplication, treated as subsets of (BxB)xB, are elements of V - similarly, the complement, treated as a subset of BxB, is an element of V, and any subset of B which is an element of V has a least upper bound and a greatest lower bound in B). The class V^B is defined as follows. V^B_0 is the empty set. For all ordinals alpha, V^B_{alpha+1} is the set of all functions x, which are elements of V, whose domain is a subset of V^B_{alpha} (i.e. dom(x) is a subset of V^B_{alpha} and dom(x) is an element of V) and whose range is a subset of B (i.e. range(x) is a subset of B and range(x) is an element of V). If alpha is a limit ordinal, then V^B_{alpha} is the union of { V^B_{beta} : beta < alpha }. The class V^B is the union of V^B_{alpha} over all ordinals alpha. Consequently, V^B is not an element of V. On the other hand, V^B_alpha is an element of V for all ordinals alpha. To each ZF formula phi(x_1,...,x_n) and for all values for x_1,...,x_n in V^B, there is a Boolean value in B, denoted ||phi(x_1,...,x_n)||. This is determined by induction on the formula by ||x in y|| = sup{ y(t).||x = t|| : t in dom(y) } for x, y in V^B, ||x a subset of y|| = inf{ x(t)'+||t in y|| : t in dom(x) } for x, y in V^B, ||x = y|| = ||x a subset of y||.||y a subset of x|| for x, y in V^B, ||not phi|| = ||phi||', ||phi and psi|| = ||phi||.||psi||, ||phi or psi|| = ||phi||+||psi||, ||phi implies psi|| = ||phi||'+||psi||, ||forall x phi(x)|| = inf{ ||phi(x)|| : x in V^B }, ||exists x phi(x)|| = sup{ ||phi(x)|| : x in V^B }. One part with which I felt uncomfortable with from the first time that I read these definitions is that there was no guarantee that { ||phi(x)|| : x in V^B } is a set (i.e. there is no guarantee that { ||phi(x)|| : x in V^B } is an element of V), and so there is no guarantee that ||forall x phi(x)|| or ||exists x phi(x)|| exists, and there was no indication in the text on how to demonstrate that ||forall x phi(x)|| and ||exist x phi(x)|| exist in general. After I had given it some thought, I think that I have hit on a proof. Could others please look at the proof and check its validity. Firstly, note that the equation b = ||phi(x)|| for b in B and x in V^B is expressible as a formula in ZF. More generally, let psi(x,b) be a formula in ZF such that for all x in V^B, there exists a unique element b of B such that psi(x,b) is true. For x in V^B, let F(x) be shorthand for that unique element of B such that psi(x,F(x)) is true, i.e. for b in B and x in V^B, b = F(x) just means psi(x,b). Theorem: There exists a subset A of V^B (i.e. A is a subclass of V^B and A is an element of V) such that F(x) <= sup{ F(y) : y in A } for all x in V^B. Here, <= denotes less than or equal to, so that for a, b in B, a <= b iff a.b = a iff a.b' = 0 iff a+b = b iff a'+b = 1, i.e. <= is the natural order on the Boolean algebra. Firstly, as a consequence of the Theorem, sup{ F(x) : x in V^B } exists, and sup{ F(x) : x in V^B } = sup{ F(x) : x in A }. Proof: Since the theorem is trivial if B is finite, then assume that B is infinite. Let aleph_{zeta} = aleph(|B|), where |B| is the cardinality of the Boolean algebra B, and aleph is Hartog's aleph, i.e. for transfinite cardinal m, aleph(m) is the cardinality of the set of ordinals which describe well-orderings of subsets of a set of cardinality m. Let omega_{zeta} be the initial ordinal for aleph_{zeta}, so that omega_{zeta} is the smallest ordinal which does not describe any well-ordering of any subset of B. For all ordinals alpha, define s(alpha) = sup{ F(x) : x in V^B_{alpha} }, so that s(0) = 0, and s(alpha) <= s(beta) if alpha < beta. Suppose that there is no set A such that F(x) <= sup{ F(y) : y in A } for all x in V^B, i.e. for all subclasses A of V^B such that A is an element of V, there exists x in V^B such that it is not true that F(x) <= sup{ F(y) : y in A}. Since V^B_{alpha} is an element of V for all ordinals alpha, then for all ordinals alpha, there exists an ordinal beta > alpha such that s_{beta} > s_{alpha}. By the definition of s(alpha) from F(x), and the fact that b = F(x) is a formula in ZF, then b = s(alpha) for b in B and for ordinals alpha is a formula in ZF. For ordinals alpha, define the ordinal f(alpha) by f(0) = 0, for all alpha, f(alpha+1) is the smallest ordinal larger than f(alpha) such that s(f(alpha)) < s(f(alpha+1)), if alpha is a limit ordinal, then f(alpha) = lim{ f(beta) : beta < alpha }, then beta = f(alpha) for ordinals alpha and beta, is a formula in ZF. Since for all ordinals alpha, there exists an ordinal beta > alpha such that s(beta) > s(alpha), then f(alpha) is well-defined for all ordinals alpha, and since b = s(alpha) for b in B and ordinal alpha is a formula in ZF, then beta = f(alpha) for ordinals alpha and beta is a formula in ZF. Also, f(alpha) is strictly increasing and continuous. As a result, the formula b = s(f(alpha)) for b in B and ordinal alpha is also a formula in ZF, and { s(f(alpha)) : alpha < omega_{zeta} } is a strictly increasing sequence in B, and thus there exists a bijection in V between the von Neumann ordinal omega_{zeta} and a subset of B, contradicting the choice of omega_{zeta}. Since the assumption that there is no subclass A of V^B such that A is an element of V and F(x) <= sup{ F(y) : y in A } for all x in V^B leads to a contradiction, it follows that such an A exists. David McAnally -------------- ==== I am using the same notation as previously, so I will not reiterate my introductory section. It struck me that perhaps { ||phi(x)|| : x in V^B } is a set, since x in V^B may be expressible as a formula in ZF, with parameters given by B and the Boolean opertaions, so that the set is given by { b in B : (exists x in V^B) b = ||phi(x)|| }. Could anybody confirm this? The result which I earlier stated, and think I proved [i.e. that Theorem: There exists a subset A of V^B (i.e. A is a subclass of V^B and A is an element of V) such that F(x) <= sup{ F(y) : y in A } for all x in V^B. Here, <= denotes less than or equal to, so that for a, b in B, a <= b iff a.b = a iff a.b' = 0 iff a+b = b iff a'+b = 1, i.e. <= is the natural order on the Boolean algebra. where psi(x,b) is a formula in ZF such that for all x in V^B, there exists a unique element b of B such that psi(x,b) is true, and for x in V^B, F(x) is shorthand for that unique element of B such that psi(x,F(x)) is true] is needed for the proof that the Axiom of Collection in valid, i.e. tha ||Axiom of Collection|| = 1. David McAnally -------------- ==== > > I am using the same notation as previously, so I will not reiterate my > introductory section. It struck me that perhaps { ||phi(x)|| : x in V^B } > is a set, since x in V^B may be expressible as a formula in ZF, with > parameters given by B and the Boolean opertaions, so that the set is > given by > > { b in B : (exists x in V^B) b = ||phi(x)|| }. > > Could anybody confirm this? Yes, certainly that is correct. Alan Stern ==== | It struck me that perhaps { ||phi(x)|| : x in V^B } | is a set, since x in V^B may be expressible as a formula in ZF, with | parameters given by B and the Boolean opertaions, so that the set is | given by | | { b in B : (exists x in V^B) b = ||phi(x)|| }. | | Could anybody confirm this? When I read your first post, I was going to ask why you didn't do it that way. I don't see anything wrong with it. Keith Ramsay ==== | It struck me that perhaps { ||phi(x)|| : x in V^B } | is a set, since x in V^B may be expressible as a formula in ZF, with | parameters given by B and the Boolean opertaions, so that the set is | given by | | { b in B : (exists x in V^B) b = ||phi(x)|| }. | | Could anybody confirm this? When I read your first posting, I was going to ask why you didn't do it this way. I don't see anything wrong with it. As a rule of thumb, the problem with classes that aren't sets is that they're too big, but this class is a subset of B. Keith Ramsay ==== >| It struck me that perhaps { ||phi(x)|| : x in V^B } >| is a set, since x in V^B may be expressible as a formula in ZF, with >| parameters given by B and the Boolean opertaions, so that the set is >| given by >| >| { b in B : (exists x in V^B) b = ||phi(x)|| }. >| >| Could anybody confirm this? >When I read your first posting, I was going to ask why you didn't >do it this way. I don't see anything wrong with it. As a rule of >thumb, the problem with classes that aren't sets is that they're >too big, but this class is a subset of B. The result that I proved (or that I believe that I proved) in my first posting is still used in the text that I am reading with the proof that ||Axiom of Collection|| = 1, so I don't consider the effort of stating the result and attempting to prove it to be a total waste of time. But perhaps there is a simpler proof of that result as well, now that it has been established that { b in B : (exists x in V^B) b = ||phi(x)|| } is a set. For each b in B for which such an x exists, the Axiom of Collection can be invoked to determine a set of x in V^B such that the equation holds, and the union of these sets over all such b in B could then be taken. That should produce a set A in V^B such that ||phi(x)|| <= sup{ ||phi(y)|| : y in A } for all x in V^B. >Keith Ramsay David McAnally -------------- ==== >> >> I am using the same notation as previously, so I will not reiterate my >> introductory section. It struck me that perhaps { ||phi(x)|| : x in V^B } >> is a set, since x in V^B may be expressible as a formula in ZF, with >> parameters given by B and the Boolean opertaions, so that the set is >> given by >> >> { b in B : (exists x in V^B) b = ||phi(x)|| }. >> >> Could anybody confirm this? >Yes, certainly that is correct. >Alan Stern David McAnally --------------- ==== I have a question about DAGs (Directed Acyclic Graphs). There are two kinds of properties a DAG might have. 1) There are (at least) two nodes so that there is no path from either to the other. 2) There is more than one node with a zero in-degree (i.e. there is no edge going into that node). Now 1 of course implies 2, but as 2 doesn't imply 1, these are different properties. Now my question is, what are they called? One of them could be called disconnected, but which? And what would the other be called? -- /-- 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! ------------/ It sure is cool having money and chicks. - Beavis and Butt-head ==== > 2) There is more than one node with a zero in-degree (i.e. there is no > edge going into that node). > nope Herc ==== Joona I Palaste scribbled the following: > Joona I Palaste scribbled the following: >> I have a question about DAGs (Directed Acyclic Graphs). There are two >> kinds of properties a DAG might have. >> 1) There are (at least) two nodes so that there is no path from either >> to the other. > This isn't what I meant. This kind of thing would be very common in > DAGs and does not imply the property below. > What I meant was: > 1) There are (at least) two nodes so that there are no paths from the > same node to both. I.e. in formal logic: > exists A: exists B: A!=B and not exists C: there is a path from C to > A and a path from C to B. Sorry, this isn't what I meant either. Sorry, it's late at night here. Here is the third and final try: 1) The graph can be partitioned into at least two parts so that there is no path from any node in one part to any node in the other part. >> 2) There is more than one node with a zero in-degree (i.e. there is no >> edge going into that node). >> Now 1 of course implies 2, but as 2 doesn't imply 1, these are >> different properties. Now my question is, what are they called? One of >> them could be called disconnected, but which? And what would the other >> be called? > Sorry for the confusion. -- /-- 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 Tolsa ==== >> I have a question about DAGs (Directed Acyclic Graphs). There are two >> kinds of properties a DAG might have. 1) The graph can be partitioned into at least two parts so that there > is no path from any node in one part to any node in the other part. >> 2) There is more than one node with a zero in-degree (i.e. there is no >> edge going into that node). >> Now 1 of course implies 2, but as 2 doesn't imply 1, these are >> different properties. Now my question is, what are they called? One of >> them could be called disconnected, but which? And what would the other >> be called? > How about strongly and weakly disconnected disconnected and directedly disconnected physically disconnected and disconnected Is 1 equivalent to the undirected version of the graph being disconnected? Then I'd call 1 undirected path disconnected 2 directed path disconnected ==== Joona I Palaste scribbled the following: > I have a question about DAGs (Directed Acyclic Graphs). There are two > kinds of properties a DAG might have. > 1) There are (at least) two nodes so that there is no path from either > to the other. This isn't what I meant. This kind of thing would be very common in DAGs and does not imply the property below. What I meant was: 1) There are (at least) two nodes so that there are no paths from the same node to both. I.e. in formal logic: exists A: exists B: A!=B and not exists C: there is a path from C to A and a path from C to B. > 2) There is more than one node with a zero in-degree (i.e. there is no > edge going into that node). > Now 1 of course implies 2, but as 2 doesn't imply 1, these are > different properties. Now my question is, what are they called? One of > them could be called disconnected, but which? And what would the other > be called? Sorry for the confusion. -- /-- 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! ------------/ When a man talks dirty to a woman, that's sexual harassment. When a woman talks dirty to a man, that's 14.99 per minute + local telephone charges! - Ruben Stiller ==== Same-generation predicate is defined as: sg(X,Y) :- flat(X,Y) sg(X,Y) :- up(X,U),sg(U,V),down(V,Y) Therefore, the case #1 is essentially same-generation with flat relation being identity. > Joona I Palaste scribbled the following: > I have a question about DAGs (Directed Acyclic Graphs). There are two > kinds of properties a DAG might have. > 1) There are (at least) two nodes so that there is no path from either > to the other. This isn't what I meant. This kind of thing would be very common in > DAGs and does not imply the property below. > What I meant was: > 1) There are (at least) two nodes so that there are no paths from the > same node to both. I.e. in formal logic: > exists A: exists B: A!=B and not exists C: there is a path from C to > A and a path from C to B. > 2) There is more than one node with a zero in-degree (i.e. there is no > edge going into that node). > Now 1 of course implies 2, but as 2 doesn't imply 1, these are > different properties. Now my question is, what are they called? One of > them could be called disconnected, but which? And what would the other > be called? Sorry for the confusion. -- > /-- 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! ------------/ > When a man talks dirty to a woman, that's sexual harassment. When a woman talks > dirty to a man, that's 14.99 per minute + local telephone charges! > - Ruben Stiller ==== > Same-generation predicate is defined as: sg(X,Y) :- flat(X,Y) > sg(X,Y) :- up(X,U),sg(U,V),down(V,Y) Therefore, the case #1 is essentially same-generation with flat relation > being identity. Some care must be taken to define up and down properly, of course. If up is defined as Graph Adjacency Relation, and down is just a reverse, then the paths would be constrained to be of equal lengths, of course. We have to add identity to both in order to prevent this from happening. ==== I was reading some logic notes and i got interested in the construction of the arithmetic of natural numbers from peano's axioms. Peano Axioms are: 1 is a natural number For every natural number x there exists another natural number x' called the successor of x. 1 # x' for every natural number x (x' being the successor of x) If x' = y' then x = y If Q is a property such that: 1 has the property Q if x has property Q then x' has property Q then the property Q holds for all natural numbers. I already proved that n' =/= n (i.e the succesor of n is different from n for all n). but then i'm stuck with the following problem: Prove n'= n + 1 where n' is the successor of n how can this be proven? i'm even stuck with the base case: for n=1 1'= 1 + 1 but we can't assume 1 + 1 = 2 can we? if someone knows how to prove this please let me know the steps of the proof i'm really interested in knowing how to show this thanks in advance Sincerely Carlos ==== > I was reading some logic notes and i got interested in the > construction > of the arithmetic of natural numbers from peano's axioms. Peano Axioms are: 1 is a natural number > For every natural number x there exists another natural number x' > called the successor of x. > 1 # x' for every natural number x (x' being the successor of x) > If x' = y' then x = y > If Q is a property such that: > 1 has the property Q > if x has property Q then x' has property Q > then the property Q holds for all natural numbers. I already proved that n' =/= n (i.e the succesor of n is different > from n for > all n). > but then i'm stuck with the following problem: > Prove n'= n + 1 where n' is the successor of n > how can this be proven? i'm even stuck with the base case: > for n=1 > 1'= 1 + 1 but we can't assume 1 + 1 = 2 can we? > if someone knows how to prove this please let me know the steps of the > proof > i'm really interested in knowing how to show this > thanks in advance After you axiomatize a number system, the next step would be to define common operations like addition, multiplication, an ordering relation, etc... The PA axioms are very primitive, you can't prove anything about higher level operations like addition based solely on the PA axioms because you haven't defined those things yet. You can define addition in terms of PA. For example, if instead of the counting numbers you started with the naturals (which gives us 0), you could say something like: There is exactly one function + in N such that (i) for each n in N, +<0, n> = n (ii) for all m and n in N, + = (+)' Which clearly establishes an identity element, commutativity, and the expected behavior of addition for any two numbers n and m; note that the last property is defined recursively between i and ii. Other properties of addition that we expect will emerge from this definition as well, I just listed the obvious ones. Now then, armed with a definition of addition in terms of PA, it is possible to prove things about addition. Later you might want to define multiplication in terms of addition and PA, and so on... I *think* there are other ways to define addition but I am not familiar with those alternatives, perhaps someone else will clear that up. > Sincerely > Carlos ==== > Prove n'= n + 1 where n' is the successor of n > how can this be proven? i'm even stuck with the base case: > for n=1 > 1'= 1 + 1 but we can't assume 1 + 1 = 2 can we? > You prove that using the defintion of + > You can define addition in terms of PA. For example, if instead of the > counting numbers you started with the naturals (which gives us 0), you > could say something like: There is exactly one function + in N such that (i) for each n in N, +<0, n> = n > (ii) for all m and n in N, + = (+)' Which clearly establishes an identity element, commutativity, and the > expected behavior of addition for any two numbers n and m; note that > the last property is defined recursively between i and ii. Other properties > of addition that we expect will emerge from this definition as well, I just > listed the obvious ones. > Note Carlos' system starts at 1. > I *think* there are other ways to define addition but I am not familiar > with those alternatives, perhaps someone else will clear that up. > Starting from 1 1+n = n'; m'+n = (m+n)' or n+1 = n'; m+n'= (m+n)' Starting from 0 0+n = n; m'+n = (m+n)' or n+0 = n; m+n'= (m+n)' So the difference in the two systems, starting from 0 and starting from 1, is in how 1 is interperted by the definition of +, and consequently by *. ==== In Kolmogorov and Fomin, Elementary Real Analysis, pg 54, problem 8b, the following question is posed: Let F be the Cantor set. Prove that the numbers of the form t1 + t2, where t1,t2 in F, fill the whole interval [0,2]. (I'll add that F is the standard Cantor middle third set on [0,1].) My question is, what does fill mean? Do they mean the set is dense in [0,2] or do they mean equality? I don't want any hints on the proof or anything like that. I _only_ would like to know what they mean by fill. Matthew T. Brenneman ==== > In Kolmogorov and Fomin, Elementary Real Analysis, pg 54, problem 8b, the > following question is posed: > Let F be the Cantor set. Prove that the numbers of the form t1 + t2, where > t1,t2 in F, fill the whole interval [0,2]. > (I'll add that F is the standard Cantor middle third set on [0,1].) > My question is, what does fill mean? Do they mean the set is dense in > [0,2] or do they mean equality? > I don't want any hints on the proof or anything like that. I _only_ would > like to know what they mean by fill. Fill means that for every number x in the interval [0, 2], there exist t1 and t2, belonging to F, such that x = t1+t2. ==== In Kolmogorov and Fomin, Elementary Real Analysis, pg 54, problem 8b, the >following question is posed: Let F be the Cantor set. Prove that the numbers of the form t1 + t2, where >t1,t2 in F, fill the whole interval [0,2]. >(I'll add that F is the standard Cantor middle third set on [0,1].) My question is, what does fill mean? Do they mean the set is dense in >[0,2] or do they mean equality? Equality. >I don't want any hints on the proof or anything like that. I _only_ would >like to know what they mean by fill. >Matthew T. Brenneman > ************************ David C. Ullrich ==== Let G = (finite, abelian, 2-generated by x and y) where order(x) = a and order(y) = b are the only relations imposed (so G is isomorphic to Z_a x Z_b). Let H be a cyclic subgroup of G where H = where f, g in Z. Now, given a group element n in G, let N be the coset of H in G containing n. That is, an uppercase letter is simply the coset containing the element of G that is the lowercase letter. Suppose the following facts are true: (1) G/H is cyclic with G/H = , i.e. G/H = { c(x - y) + H | c in Z }. (2) x = w(x - y) + h, y = (w - 1)(x - y) + h (for some w in Z, h in H). How do I present G/H in the form G/H = where r, s, t in Z ??? ==== > > Let G = (finite, abelian, 2-generated by x and y) where > order(x) = a and order(y) = b are the only relations imposed (so G is > isomorphic to Z_a x Z_b). Let H be a cyclic subgroup of G where H = > where f, g in Z. > > Now, given a group element n in G, let N be the coset of H in G > containing n. That is, an uppercase letter is simply the coset > containing the element of G that is the lowercase letter. Suppose the > following facts are true: > > (1) G/H is cyclic with G/H = , i.e. G/H = { c(x - y) + H | c in > Z }. > (2) x = w(x - y) + h, y = (w - 1)(x - y) + h (for some w in Z, h in > H). > > How do I present G/H in the form > > G/H = where r, s, t in Z ??? Let's see if I've got this straight. G is abelian, G = + with o(x) = a, o(y) = b and / = {0}. H is a cyclic subgroup of G and G/H is generated by (x - y) + H. (I guess that's possible - I haven't checked.) (2) Is no new information.The first equation follows from G/H = <(x - y) + H>. The second equation just says y = y. So, it boils down to G = + , H is a cyclic subgroup and G/H = <(x - y) + H> What I don't follow is what you want. Clearly G/H is generated by x + H and y + H. a*(x + H) = H = b*(y + H) = 0 -- Paul Sperry Columbia, SC (USA) ==== > Let G = (finite, abelian, 2-generated by x and y) where > order(x) = a and order(y) = b are the only relations imposed (so G is > isomorphic to Z_a x Z_b). Let H be a cyclic subgroup of G where H = > where f, g in Z. > Simply write X = x+H, Y = y+H or just use x+H, y+H. > (1) G/H is cyclic with G/H = , i.e. G/H = { c(x - y) + H | c in > Z }. > (2) x = w(x - y) + h, y = (w - 1)(x - y) + h (for some w in Z, h in > H). How do I present G/H in the form > G/H = where r, s, t in Z ??? > This is rather complex, however it does address some issues I raised in a similar post a few days ago. Your problem has eerie resemblance to the problem posed by Heather in 'Presentations of a cyclic group'. I refer you to my response in her thread as possible suggestion. Not having gone thru the details of your presentation, I'm jumping to conclusion in making the referral. If I've time later, I'll go thru the details; as for now I'm sensing some confusion in your presentation or lack of understanding upon my behalf for not having slugged thru the details. ==== I agree that a(x+H)=b(y+H)=0, but that doesn't mean that the order of x+H in G/H is a (and similarly for y+H). This would mean that |G/H|=|G|=ab, which obviously isn't correct. I think the original poster wanted something of the form G/H = , where u=ord(X) and b and c are chosen such that bY=cX is the first time that a multiple of Y can equal a multiple of X. ==== Investment on TAB horse racing. This software utilises Internet technology and expert system to track all the professional activity of the racing experts and actions of the professionals when they invest. www.racePredictor.com ==== I want to generate (pseudo) random points in the unit ball of R^n (uniform distribution). I have available a generator for random numbers in an interval with the uniform distribution, and a generator for random numbers with the normal(mu,sigma) distribution. The method of generating uniformly distributed points in the box [-1,1]^n and then keeping those with norm less than 1 has the drawback that for all but the smallest values of n, I throw away most of the trial points. Even for n=6 I keep only about 8% (i.e., volume of 6-ball divided by 2^6) of the trial points I generate. It occurred to me that I could generate a point p by having each coordinate of p be random(normal(0,1)). p would then be randomly located in R^n according to a multinormal distribution (right?). Then if I had a function f:[0,inf)->[0,1) to adjust the norm of p, I could get a point in the ball. (That is, replace p by at pictures, it appears that for n=1 and n=2 the same function f(x) = 1 - e^-x seems to be close, but not exactly right. Or if you have a different way altogether, I'd like to see it. ==== > I want to generate (pseudo) random points in the unit ball of R^n > (uniform distribution). I have available a generator for random > numbers in an interval with the uniform distribution, and a > generator for random numbers with the normal(mu,sigma) > distribution. > The method of generating uniformly distributed points in the box > [-1,1]^n and then keeping those with norm less than 1 has the > drawback that for all but the smallest values of n, I throw away > most of the trial points. Even for n=6 I keep only about 8% > (i.e., volume of 6-ball divided by 2^6) of the trial points I > generate. > It occurred to me that I could generate a point p by having each > coordinate of p be random(normal(0,1)). p would then be randomly > located in R^n according to a multinormal distribution (right?). > Then if I had a function f:[0,inf)->[0,1) to adjust the norm > of p, I could get a point in the ball. (That is, replace p by > at pictures, it appears that for n=1 and n=2 the same function > f(x) = 1 - e^-x seems to be close, but not exactly right. > Or if you have a different way altogether, I'd like to see it. I think you are on the right track with the normal distribution. How about the following: 1. Let p = (p_1, p_2, ..., p_n), where each p_i is normal(0,1)-distributed. 2. Choose d, uniformly distributed on (0,1). 3. Let the resulting point be d^(1/n) * p/||p||. That is, normalize p and then multiply the resulting unit vector by d^(1/n) to obtain a length with the proper distribution. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== >I want to generate (pseudo) random points in the unit ball of R^n >(uniform distribution). I have available a generator for random >numbers in an interval with the uniform distribution, and a >generator for random numbers with the normal(mu,sigma) >distribution. >It occurred to me that I could generate a point p by having each >coordinate of p be random(normal(0,1)). p would then be randomly >located in R^n according to a multinormal distribution (right?). >Then if I had a function f:[0,inf)->[0,1) to adjust the norm >of p, I could get a point in the ball. (That is, replace p by >at pictures, it appears that for n=1 and n=2 the same function >f(x) = 1 - e^-x seems to be close, but not exactly right. Lebesgue measure on R^n can be written as r^(n-1) dr dw where r is the radial coordinate and dw is a rotationally-invariant measure on the (n-1)-sphere. For your n-dimensional normal distribution, the measure is (2 pi)^(-n/2) exp(-r^2/2) r^(n-1) dr dw, and Prob(r I want to generate (pseudo) random points in the unit ball of R^n > (uniform distribution). I have available a generator for random > numbers in an interval with the uniform distribution, and a > generator for random numbers with the normal(mu,sigma) > distribution. > > The method of generating uniformly distributed points in the box > [-1,1]^n and then keeping those with norm less than 1 has the > drawback that for all but the smallest values of n, I throw away > most of the trial points. Even for n=6 I keep only about 8% > (i.e., volume of 6-ball divided by 2^6) of the trial points I > generate. > [snip] Some thoughts: First, you can generate a uniform point in [0,1]^n instead of [-1,1]^n, keep it if it has norm less than 1, and then choose n + or - signs. In this way you'll be throwing away less than 50% of your candidates, independently of n. There's still some waste, but not too bad, and that's probably the easiest way. Another option would be to work with (an n-dim generalization of) polar coordinates: (r, theta_1, theta_2, ..., theta_{n-1}). If you choose r and the theta_i uniformly in the appropriate ranges, I believe you get a uniform point in the ball. HTH - EM ==== > I want to generate (pseudo) random points in the unit ball of R^n > (uniform distribution). I have available a generator for random > numbers in an interval with the uniform distribution, and a > generator for random numbers with the normal(mu,sigma) > distribution. > > The method of generating uniformly distributed points in the box > [-1,1]^n and then keeping those with norm less than 1 has the > drawback that for all but the smallest values of n, I throw away > most of the trial points. Even for n=6 I keep only about 8% > (i.e., volume of 6-ball divided by 2^6) of the trial points I > generate. > > It occurred to me that I could generate a point p by having each > coordinate of p be random(normal(0,1)). p would then be randomly > located in R^n according to a multinormal distribution (right?). > Then if I had a function f:[0,inf)->[0,1) to adjust the norm > of p, I could get a point in the ball. (That is, replace p by > at pictures, it appears that for n=1 and n=2 the same function > f(x) = 1 - e^-x seems to be close, but not exactly right. > > Or if you have a different way altogether, I'd like to see it. Your p/|p| is a random point on the surface of the unit ball in R^n. The f you suggest would take |p| as an argument. Instead, I would define a random variable R in [0,1] as the distance from the origin to a randomly chosen point in the unit ball of R^n. Then, x = R * p/|p| would be a random point in the unit ball of R^n. I think the probability density function for R is: g(t):= n*t^{n-1}, for t in [0,1], g(t) := 0 if t not in [0,1] . So the cumulative density function for R is just: h(t) = t^n . There is a standard way to get a variable with c.d.f. t^n from a uniform distribution random variable on [0,1]. I might be wrong here, but if U is uniform [0,1], then U^{1/n} might have the required c.d.f. and p.d.f. (for R). David Bernier ==== > I want to generate (pseudo) random points in the unit ball of R^n > (uniform distribution). I have available a generator for random > numbers in an interval with the uniform distribution, and a > generator for random numbers with the normal(mu,sigma) > distribution. > > The method of generating uniformly distributed points in the box > [-1,1]^n and then keeping those with norm less than 1 has the > drawback that for all but the smallest values of n, I throw away > most of the trial points. Even for n=6 I keep only about 8% > (i.e., volume of 6-ball divided by 2^6) of the trial points I > generate. > > It occurred to me that I could generate a point p by having each > coordinate of p be random(normal(0,1)). p would then be randomly > located in R^n according to a multinormal distribution (right?). > Then if I had a function f:[0,inf)->[0,1) to adjust the norm > of p, I could get a point in the ball. (That is, replace p by > at pictures, it appears that for n=1 and n=2 the same function > f(x) = 1 - e^-x seems to be close, but not exactly right. > > Or if you have a different way altogether, I'd like to see it. A little googling around led to this (via Wolfram's site): http://www.math.niu.edu/~rusin/known-math/96/sph.rand. According to this, you can generate a uniform distribution on the nth dimensional unit sphere by selecting n uniformly distributed values, x_1 through x_n, and then dividing by the magnitude of the corresponding vector (i.e., normalizing) to land on the unit sphere. It seems to me that you could transform this into a uniform distribution on the unit ball by then selecting the radius with the appropriate distribution -- i.e., one where the likelihood of the radius is proportional to the size of the sphere with that radius. I'm not proposing this as any sort of concrete proof -- it's just an idea that occurs to me as I sit here playing around right before I head off to lullaby land. HTH. .... Bob P.S. The referring page on Wolfram's site is http://mathworld.wolfram.com/SpherePointPicking.html. Unfortunately, the ball point picking page is not yet written. ==== > I want to generate (pseudo) random points in the unit ball of R^n >> (uniform distribution). I have available a generator for random >> numbers in an interval with the uniform distribution, and a >> generator for random numbers with the normal(mu,sigma) >> distribution. >> >> The method of generating uniformly distributed points in the box >> [-1,1]^n and then keeping those with norm less than 1 has the >> drawback that for all but the smallest values of n, I throw away >> most of the trial points. Even for n=6 I keep only about 8% >> (i.e., volume of 6-ball divided by 2^6) of the trial points I >> generate. >> >> It occurred to me that I could generate a point p by having each >> coordinate of p be random(normal(0,1)). p would then be randomly >> located in R^n according to a multinormal distribution (right?). >> Then if I had a function f:[0,inf)->[0,1) to adjust the norm >> of p, I could get a point in the ball. (That is, replace p by >> at pictures, it appears that for n=1 and n=2 the same function >> f(x) = 1 - e^-x seems to be close, but not exactly right. >> >> Or if you have a different way altogether, I'd like to see it. A little googling around led to this (via Wolfram's site): >http://www.math.niu.edu/~rusin/known-math/96/sph.rand. According to this, >you can generate a uniform distribution on the nth dimensional unit sphere >by selecting n uniformly distributed values, x_1 through x_n, and then >dividing by the magnitude of the corresponding vector (i.e., normalizing) >to land on the unit sphere. Looking briefly at that page I don't see the method you describe. Which is a good thing, because it does _not_ give points uniformly distributed on the sphere! What you said would work (and also become a method I _do_ see on that page) if you changed the words n uniformly distributed values, x_1 through x_n to n normally distributed values, x_1 through x_n. >It seems to me that you could transform this into a uniform distribution on >the unit ball by then selecting the radius with the appropriate >distribution -- i.e., one where the likelihood of the radius is >proportional to the size of the sphere with that radius. I'm not proposing this as any sort of concrete proof -- it's just an idea >that occurs to me as I sit here playing around right before I head off to >lullaby land. HTH. .... Bob P.S. The referring page on Wolfram's site is >http://mathworld.wolfram.com/SpherePointPicking.html. Unfortunately, the >ball point picking page is not yet written. ************************ David C. Ullrich ==== >> I want to generate (pseudo) random points in the unit ball of R^n >> (uniform distribution). I have available a generator for random >> numbers in an interval with the uniform distribution, and a >> generator for random numbers with the normal(mu,sigma) >> distribution. >> >> The method of generating uniformly distributed points in the box >> [-1,1]^n and then keeping those with norm less than 1 has the >> drawback that for all but the smallest values of n, I throw away >> most of the trial points. Even for n=6 I keep only about 8% >> (i.e., volume of 6-ball divided by 2^6) of the trial points I >> generate. >> >[snip] >Some thoughts: First, you can generate a uniform point in [0,1]^n >instead of [-1,1]^n, keep it if it has norm less than 1, and then >choose n + or - signs. In this way you'll be throwing away less >than 50% of your candidates, independently of n. There's still some >waste, but not too bad, and that's probably the easiest way. No, the waste is the same. >Another option would be to work with (an n-dim generalization of) >polar coordinates: (r, theta_1, theta_2, ..., theta_{n-1}). If >you choose r and the theta_i uniformly in the appropriate ranges, >I believe you get a uniform point in the ball. HTH - EM ************************ David C. Ullrich ==== > > I want to generate (pseudo) random points in the unit ball of R^n > (uniform distribution). I have available a generator for random > numbers in an interval with the uniform distribution, and a > generator for random numbers with the normal(mu,sigma) > distribution. > > The method of generating uniformly distributed points in the box > [-1,1]^n and then keeping those with norm less than 1 has the > drawback that for all but the smallest values of n, I throw away > most of the trial points. Even for n=6 I keep only about 8% > (i.e., volume of 6-ball divided by 2^6) of the trial points I > generate. > > It occurred to me that I could generate a point p by having each > coordinate of p be random(normal(0,1)). p would then be randomly > located in R^n according to a multinormal distribution (right?). > Then if I had a function f:[0,inf)->[0,1) to adjust the norm > of p, I could get a point in the ball. (That is, replace p by > at pictures, it appears that for n=1 and n=2 the same function > f(x) = 1 - e^-x seems to be close, but not exactly right. > > Or if you have a different way altogether, I'd like to see it. >>A little googling around led to this (via Wolfram's site): >>http://www.math.niu.edu/~rusin/known-math/96/sph.rand. According to this, >>you can generate a uniform distribution on the nth dimensional unit sphere >>by selecting n uniformly distributed values, x_1 through x_n, and then >>dividing by the magnitude of the corresponding vector (i.e., normalizing) >>to land on the unit sphere. > > Looking briefly at that page I don't see the method you describe. > Which is a good thing, because it does _not_ give points uniformly > distributed on the sphere! Yes. It was a typo. However, I was tired. At least I gave an accurate reference. :-) > What you said would work (and also become a method I _do_ see > on that page) if you changed the words n uniformly distributed > values, x_1 through x_n to n normally distributed values, x_1 > through x_n. > >>It seems to me that you could transform this into a uniform distribution >>on the unit ball by then selecting the radius with the appropriate >>distribution -- i.e., one where the likelihood of the radius is >>proportional to the size of the sphere with that radius. >>I'm not proposing this as any sort of concrete proof -- it's just an idea >>that occurs to me as I sit here playing around right before I head off to >>lullaby land. >>HTH. >>.... Bob >>P.S. The referring page on Wolfram's site is >>http://mathworld.wolfram.com/SpherePointPicking.html. Unfortunately, the >>ball point picking page is not yet written. > > ************************ > > David C. Ullrich ==== > >> I want to generate (pseudo) random points in the unit ball of R^n >> (uniform distribution). I have available a generator for random >> numbers in an interval with the uniform distribution, and a >> generator for random numbers with the normal(mu,sigma) >> distribution. >> >> The method of generating uniformly distributed points in the box >> [-1,1]^n and then keeping those with norm less than 1 has the >> drawback that for all but the smallest values of n, I throw away >> most of the trial points. Even for n=6 I keep only about 8% >> (i.e., volume of 6-ball divided by 2^6) of the trial points I >> generate. >> >[snip] > >Some thoughts: First, you can generate a uniform point in [0,1]^n >instead of [-1,1]^n, keep it if it has norm less than 1, and then >choose n + or - signs. In this way you'll be throwing away less >than 50% of your candidates, independently of n. There's still some >waste, but not too bad, and that's probably the easiest way. > > No, the waste is the same. You're right, of course, and thanks for pointing out my silly mistake. Better leave it at that and not try to explain the source of my confusion :-) - EM ==== > I want to generate (pseudo) random points in the unit ball of R^n > (uniform distribution). I have available a generator for random > numbers in an interval with the uniform distribution, and a > generator for random numbers with the normal(mu,sigma) > distribution. The method of generating uniformly distributed points in the box > [-1,1]^n and then keeping those with norm less than 1 has the > drawback that for all but the smallest values of n, I throw away > most of the trial points. Even for n=6 I keep only about 8% > (i.e., volume of 6-ball divided by 2^6) of the trial points I > generate. It occurred to me that I could generate a point p by having each > coordinate of p be random(normal(0,1)). p would then be randomly > located in R^n according to a multinormal distribution (right?). > Then if I had a function f:[0,inf)->[0,1) to adjust the norm > of p, I could get a point in the ball. (That is, replace p by > at pictures, it appears that for n=1 and n=2 the same function > f(x) = 1 - e^-x seems to be close, but not exactly right. Or if you have a different way altogether, I'd like to see it. Fast methods for getting uniform random points on the n-sphere for n=2,3,4, using only uniform variates, are described in Ann. Math Stat,v 43, 645-646. For general n: Generate standard normal variates X_1,...,X_n and let R=1/sqrt(X_1^2+...+X_n^2). Then the point R*(X_1,...X_n) is uniformly distributed over the set {x_1^2+...+x_n^2 = 1} . (The converse is true, but only for n>2; see G.Marsaglia, Symposium on Statistics and Related Topics, A. K. Selah, Ed., Carleton Univ.,9.01--9.13, 1974 Math Review 55, #6633. ) Let G be a gamma-n variate, density cx^(n-1) and let B=G/(G+Y), with Y a standard exponential variate. Then the point B*R*(X_1,...,X_n) is uniformly distributed over the set {x_1^2+...+x_n^2 < 1} For very fast methods for getting normal X's, and exponential Y's, see the ziggurat method, J. Stat. Software, V. 5 Issue 8. For simple, fast gamma variates, see ACM Transactions Math Software, V 26, Issue 2. A uniform point inside the sphere can also be produced as R*(X_1,...,X_n)*U^(1/n), with U uniform in [0,1) but the resulting U, exp( ) and log( ) calls may be slower than using a fast G/(G+Y). Also, using (1-U)^(1/n) may avoid the 1-in-4billion chance that U will be zero with some RNGs, and exp(-log(U)/n) will crash. George Marsaglia ==== >I want to generate (pseudo) random points in the unit ball of [...] > Perhaps a simpler method computationally, although it requires > an extra random number, is to take r^(1/n) p/|p| where p is your > n-dimensional normal random variable and r is another independent > random variable that is uniform in [0,1]. ==== let cos(k Pi) = sqrt(5)/2-1/2. Is k a rational number? ==== Nicola Sottocornola escribi.97 en el let cos(k Pi) = sqrt(5)/2-1/2. Is k a rational number? No, but cos(2pi/5) = (sqrt(5) - 1)/4 -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== >let cos(k Pi) = sqrt(5)/2-1/2. Is k a rational number? No. The conjugate of sqrt(5)/2 - 1/2 is -sqrt(5)/2 - 1/2 whose absolute value is greater than 1. But if T_n is the Chebyshev polynomial of order n, then all the roots of T_n(x)-1 are cos(m/n Pi) for m even, and all the roots of T_n(x)+1 are cos(m/n Pi) for m odd. So all the conjugates of cos(m/n Pi) (i.e. all the other roots of its minimal polynomial over the rationals) have absolute value <= 1. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > cos(m/n Pi) for m odd. So all the conjugates of cos(m/n Pi) > (i.e. all the other roots of its minimal polynomial over the > rationals) have absolute value <= 1. > thank you. Does this minimal polynomial be a factor (in Q[x]) of T_n(x)+1 or T_N(x)-1? Nicola ==== >> cos(m/n Pi) for m odd. So all the conjugates of cos(m/n Pi) >> (i.e. all the other roots of its minimal polynomial over the >> rationals) have absolute value <= 1. >thank you. Does this minimal polynomial be a factor (in Q[x]) of >T_n(x)+1 or T_N(x)-1? Of course. The minimal polynomial of r (over Q) is a factor of any nonzero polynomial over Q which has r as a root. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > Your post is quite ambiguous. More background would help. > At this point, I can only take a stab at helping by pointing out that > one feature of a SYMMETRIC, REAL, matrix is that all its eigenvalues > are real. > > > David > > I would like to add that in general, if a matrix is Hermitian (over > Complex numbers) its all eigenvalues are real. > > But I honestly think the OP is just joking. Specially the second > question got me laughing :) Ok, ok, ok... What the topic is: I know a few sufficient conditions for a matrix to have real eigenvalues (like real and symmetric or complex and hermitian. But does anybody know a necessary condition? What follows if a matrix has real eigenvalues? Or more difficult: Is there a nice characterisations for those matrices wirth real eigenvalues? I hope you understand the problem and I hope you know some answers. ==== >I know a few sufficient conditions for a matrix to have real >eigenvalues (like real and symmetric or complex and hermitian. But >does anybody know a necessary condition? What follows if a matrix has >real eigenvalues? Or more difficult: Is there a nice characterisations >for those matrices wirth real eigenvalues? I doubt that there's a nicer characterisation than that the eigenvalues are real. Or you could say, these are the matrices of the form C J C^(-1) where C is invertible and J is in Jordan canonical form with real numbers on the diagonal. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== This is the most-general theorem of this type I have posted. Let {a(j,k)} be any 2-d sequence (ie. infinite array) defined for all positive integer indexes j and k. Let f(x) be a real -> real positive continuous monotonic (always either increasing or decreasing) function. Let g(x) be such that f(g(x)) = x for all x, and let f be such that g is continuous. Then, for m = any positive integer: m floor(f(k)) --- --- / / a(j,k) = --- --- k=1 j=1 For f monotonically INCREASING: floor(f(m)) m --- --- / / a(j,k) --- --- j=1 k=max(1,ceiling(g(j))) For f monotonically DECREASING: floor(f(1)) min(m,floor(g(j))) --- --- / / a(j,k) --- --- j=1 k=1 In linear-mode: sum{k=1 to m} sum{j=1 to floor(f(k))} a(j,k) = For f monotonically INCREASING: sum{j=1 to floor(f(m))} sum{k=max(1,ceiling(g(j))) to m} a(j,k) For f monotonically DECREASING: sum{j=1 to floor(f(1))} sum{k=1 to min(m,floor(g(j)))} a(j,k) - /0 /max(0,g(x)) For f monotonically DECREASING: /f(0) /min(m,g(x)) ! | ! | a(x,y) dy dx /0 /0 (all under the same constraints on f and g and m as above.) In linear-mode: integral{0 to m} integral{0 to f(y)} a(x,y) dx dy = For f monotonically INCREASING: integral{0 to f(m)} integral{max(0,g(x)) to m} a(x,y) dy dx For f monotonically DECREASING: integral{0 to f(0)} integral{0 to min(m,g(x))} a(x,y) dy dx This is actually unsophisticated and is most-probably well-known. But it is definitely worth mentioning and getting archived on sci.math. Leroy Quet ==== >The formulas from polar to rectangular and back are this x = cos(L)* cos(B) * R >y = sin(L) * cos(B) * R >z = sin(B) *R then >R = sqrt(x*x + y*y + z*z) >B = sin-1(z/R) >L = tan-1(y/x) Better is L = 2*atan(y/(x+sqrt(x*x+y*y))). This formula for L eliminates the uncertainty as to its quadrant. Rob Johnson take out the trash before replying ==== > Given that I've been debating some people about a logical flaw in > Wiles's approach I thought I'd toss out a link from, yet again, > Wolfram Research. http://mathworld.wolfram.com/Taniyama-ShimuraConjecture.html Intriguingly enough to me it outlines what I've been talking about as > it rather simply gives the assertion. If any mathematicians disagree with the Wolfram Research site that > would be useful information as I will reference it again in taking > down posters who argue with me about the flaw. James Harris What flaw? -- Every problem can be oversimplified to the point that no solution is possible. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== I know a fair bit of real analysis, but nothing much about complex analysis. I'm interested in seeing a proof of Liouville's boundedness theorem because the Fundamental Theorem of Algebra is such an easy corollary of it. Does anybody know of a good source that would take me by the most direct route possible to an understanding of the proof of the boundedness theorem? Any hints will be mucho appreciado. thanks ==== > I know a fair bit of real analysis, but nothing much about complex > analysis. I'm interested in seeing a proof of Liouville's boundedness > theorem because the Fundamental Theorem of Algebra is such an easy > corollary of it. Does anybody know of a good source that would take me by > the most direct route possible to an understanding of the proof of the > boundedness theorem? Any hints will be mucho appreciado. I would suggest Alan Beardon's textbook on complex analysis. This concentrates on the topological properties of holomorphic functions, exploiting thoroughly the notion of winding number of curves. With this formalism, FTA drops out instantly. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== >I know a fair bit of real analysis, but nothing much about complex analysis. I'm >interested in seeing a proof of Liouville's boundedness theorem because the >Fundamental Theorem of Algebra is such an easy corollary of it. Does anybody >know of a good source that would take me by the most direct route possible to an >understanding of the proof of the boundedness theorem? Any hints will be mucho >appreciado. Really, just about any book on complex analysis is going to include Liouville's theorem, and it's going to be by a fairly direct route: First one proves Cauchy's Theorem - that's basic. Once one knows Cauchy's Theorem there's always a spate of more or less immediate corollaries, one of which is Liouville's theorem. Now, Cauchy's Theorem comes in various flavors, with different hypotheses. Like first for a circle, then for closed curves in a disk, then eventually a sophisticated version where the hypotheses have to do with the index of a curve about points not in the domain. You might note that probably the first version of CT you find will suffice for Liouville - you can skip the fancy versions. >thanks > ************************ David C. Ullrich ==== > I know a fair bit of real analysis, but nothing much about complex analysis. > I'm > interested in seeing a proof of Liouville's boundedness theorem because the > Fundamental Theorem of Algebra is such an easy corollary of it. Does anybody > know of a good source that would take me by the most direct route possible to > an > understanding of the proof of the boundedness theorem? Any hints will be > mucho > appreciado. Edward Nelson's one paragraph proof of Liouville's theorem for bounded harmonic functions [Proc. Amer. Math. Soc. vol. 12 (1961) page 995] is hard to beat. The analytic function form of Liouville's theorem is a simple corollary. -- A. ==== > I know a fair bit of real analysis, but nothing much about complex analysis. > I'm > interested in seeing a proof of Liouville's boundedness theorem because the > Fundamental Theorem of Algebra is such an easy corollary of it. Does anybody > know of a good source that would take me by the most direct route possible to > an > understanding of the proof of the boundedness theorem? Any hints will be > mucho > appreciado. Suppose f(z) = a0 + a1z + a2z^2 + ... for all z in C and |f(z)| <= M for all z in C. Let's use some real analysis: If we integrate |f|^2 over the circle with radius r centered at 0, we get (2Pi)*M >= int [0, 2Pi] |f(rexp(it)|^2 dt = 2Pi*(|a0|^2 + |a1|^2 r^2 + |a2|^2 r^4 + ...) But the only way for the last line to be bounded above for r > 0 is to have 0 = a1 = a2 = ..., which implies f is constant. This proves Louiville's Theorem, if you accept the fact that an analytic function on the entire plane equals a power series there. But really, the best proof of FTA is the one that shows that for a polynomial p(z), |p| cannot acheive a minimum anywhere in C, unless f is constant. And that proof is essentially all basic real analysis. ==== > I'm > interested in seeing a proof of Liouville's boundedness theorem because the > Fundamental Theorem of Algebra is such an easy corollary of it. Suppose f(z) = a0 + a1z + a2z^2 + ... for all z in C and |f(z)| <= M for > all z in C. Let's use some real analysis: If we integrate |f|^2 over the > circle with radius r centered at 0, we get (2Pi)*M >= int [0, 2Pi] |f(rexp(it)|^2 dt > = 2Pi*(|a0|^2 + |a1|^2 r^2 + |a2|^2 r^4 + ...) > Any hint how to go about the integration? > But the only way for the last line to be bounded above for r > 0 is to have > 0 = a1 = a2 = ..., which implies f is constant. This proves Louiville's Theorem, if you accept the fact that an analytic > function on the entire plane equals a power series there. But really, the best proof of FTA is the one that shows that for a > polynomial p(z), |p| cannot acheive a minimum anywhere in C, unless f > is constant. And that proof is essentially all basic real analysis. > Is there some error? Let p(z) = z = x + yi |z|^2 = x^2 + y^2 |p| attains minimum at z = 0. -- random thoughts p(z) = sum(j=0..n) aj z^j; z^* = conjugate of z |p(z)|^2 = sum(j=0..n, k=0..n) aj (ak)^* z^j (z^*)^k ==== > But really, the best proof of FTA is the one that shows that for a > polynomial p(z), |p| cannot acheive a minimum anywhere in C, unless f > is constant. And that proof is essentially all basic real analysis. > Is there some error? > Let p(z) = z = x + yi > |z|^2 = x^2 + y^2 > |p| attains minimum at z = 0. It was stated too quickly. If p is a polynomial with no zeros in C, then |p| cannot achieve a minimum unless p is constant. > But really, the best proof of FTA is ... I agree. But algebraists probably don't. They want to use as little analysis as possible, and don't mind using some Galois theory to finish. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== > But really, the best proof of FTA is the one that shows that for a > polynomial p(z), |p| cannot acheive a minimum anywhere in C, unless f > is constant. And that proof is essentially all basic real analysis. > > Is there some error? > Let p(z) = z = x + yi > |z|^2 = x^2 + y^2 > |p| attains minimum at z = 0. It was stated too quickly. If p is a polynomial with no zeros in C, > then |p| cannot achieve a minimum unless p is constant. > The same start is used to apply Louivelle's theorem to 1/p. > But really, the best proof of FTA is ... I agree. But algebraists probably don't. They want to use as little > analysis as possible, and don't mind using some Galois theory to > finish. > I'm not bothered using the mean value theorem to show an odd degree polynomial in R[x] has a real root. Would you offer some hints how to show FTA by the 'prefered' real analysis method? Here's my rambling: Consider f(x,y) = |p| as a function of the two variables in z = x + iy ? First off don't I use theorem if f is a bounded below real continuous function of R^2 then f attains minimum m, since f -> oo as (x,y) -> oo ? Expand f(x,y) = |p(z)|^2 = p(z) p(z^*) and show no solution to f_x(x,y) = f_y(x,y) = 0 except f(x,y) = c ??? ==== I was wondering if anyone could give me a reference (or answer) to the following question: If I have a tournament that is being run with the swiss chess method, and the games have a range of outcomes larger than the usual Win/Draw/Loss ie there are degrees of victory and defeat will this help me resolve* the tournament faster (with less rounds) or not? How could I calculate how many rounds are required? *ie find the strongest player Has anyone done any/know of any work on this? rgds rob ==== >I was wondering if anyone could give me a reference >(or answer) to the following question: >If I have a tournament that is being run with the >swiss chess method, and the games have a range of >outcomes larger than the usual Win/Draw/Loss ie >there are degrees of victory and defeat will this >help me resolve* the tournament faster (with less >rounds) or not? How could I calculate how many >rounds are required? >*ie find the strongest player I think you'll have to be clearer about what assumptions you are making, if you want us to be able to answer your question. Let G be the graph with edges joining players who play a game. If you assume that the outcome of a game (on some continuous scale) is, say, the ratio of the strengths of the players, you can determine the strongest player as soon as G is connected, and that can happen after only two rounds: round 1 matching players 1-2, 3-4, 5-6 etc and round 2 matching 2-3, 4-5, 6-7 etc. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >I was wondering if anyone could give me a reference >(or answer) to the following question: >If I have a tournament that is being run with the >swiss chess method, and the games have a range of >outcomes larger than the usual Win/Draw/Loss ie >there are degrees of victory and defeat will this >help me resolve* the tournament faster (with less >rounds) or not? How could I calculate how many >rounds are required? >*ie find the strongest player I think you'll have to be clearer about what assumptions you are making, > if you want us to be able to answer your question. > Let G be the graph with edges joining players who play a game. > If you assume that the outcome of a game (on some continuous scale) is, > say, the ratio of the strengths of the players, But of course in chess, and most games of skill, you can't assume that it's always the case that a single game is won by the better player. I agree the OP needs to give more detail. It could maybe be interesting. you can determine the > strongest player as soon as G is connected, and that can happen after only > two rounds: round 1 matching players 1-2, 3-4, 5-6 etc and round 2 > matching 2-3, 4-5, 6-7 etc. Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 ==== > >I was wondering if anyone could give me a reference >(or answer) to the following question: > >If I have a tournament that is being run with the >swiss chess method, and the games have a range of >outcomes larger than the usual Win/Draw/Loss ie >there are degrees of victory and defeat will this >help me resolve* the tournament faster (with less >rounds) or not? How could I calculate how many >rounds are required? > >*ie find the strongest player > > I think you'll have to be clearer about what assumptions you are making, > if you want us to be able to answer your question. > Let G be the graph with edges joining players who play a game. > If you assume that the outcome of a game (on some continuous scale) is, > say, the ratio of the strengths of the players, you can determine the > strongest player as soon as G is connected, and that can happen after only > two rounds: round 1 matching players 1-2, 3-4, 5-6 etc and round 2 > matching 2-3, 4-5, 6-7 etc. But the OP asked about the swiss system, and this is not how a swiss system tournament is run. Rather, in each round people are paired, insofar as possible, against people who have acquired the same number of points. The question is how many rounds do you expect to have to play before you get to a situation where a single player leads the tournament and can reasonably be said to have done the best. For chess, there's a theorem - or maybe it's only a rule of thumb - that if there are n players you need sqrt(n) rounds. I regret that I cannot supply a reference. The OP question I take to be, what's the corresponding theorem/rule for a game that has a larger range of outcomes than chess? I regret that I don't know the answer. -- ==== >> I think you'll have to be clearer about what assumptions you are making, >> if you want us to be able to answer your question. >> Let G be the graph with edges joining players who play a game. >> If you assume that the outcome of a game (on some continuous scale) is, >> say, the ratio of the strengths of the players, you can determine the >> strongest player as soon as G is connected, and that can happen after only >> two rounds: round 1 matching players 1-2, 3-4, 5-6 etc and round 2 >> matching 2-3, 4-5, 6-7 etc. >But the OP asked about the swiss system, and this is not how a swiss >system tournament is run. Rather, in each round people are paired, >insofar as possible, against people who have acquired the same number >of points. The question is how many rounds do you expect to have to >play before you get to a situation where a single player leads the >tournament and can reasonably be said to have done the best. In the second round, the winner of 1 vs 2 plays the winner of 3 vs 4, the loser of 3 vs 4 plays the loser of 5 vs 6, etc. Is that not possible in a Swiss system? Assuming that the outcome of a game tells you the ratio of the strengths of the players, it gives you enough information to rank all players. >For chess, there's a theorem - or maybe it's only a rule of thumb - >that if there are n players you need sqrt(n) rounds. I regret that >I cannot supply a reference. That's a strange result. Without ties, if you start out with 2^k players then after just k rounds there will be exactly one player with a perfect record. What model are you using to determine when ties occur? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== thanks to all who replied so far. You asked for more detail, hopefully this will do. A single game does not give an absolute ranking of the relative strength of players. In the long run the stronger player will always have a higher average result though. Game outcomes can be either a simple Win/Draw/Loss like a chess tournament or they can be on a higher granularity scale eg Total victory/Narrow Victory/Draw/ Narrow Defeat/Total Defeat. In each case the same swiss chess pairing algorithm is used to match players for subsequent rounds. (ie ignoring the many variations of this algorthim). Assume that it is the popular one in chess ie for 2n players in a score group player 1 plays player n+1, ..., player n plays player 2n. Which system will converge quicker (take less rounds) to accurately* pick the strongest player? *I suppose that some definition of this would be required :-) Perhaps with greater than 50% chance of being correct? rgds rob ==== > thanks to all who replied so far. You asked for more > detail, hopefully this will do. > A single game does not give an absolute ranking of > the relative strength of players. In the long run > the stronger player will always have a higher > average result though. > > Game outcomes can be either a simple Win/Draw/Loss > like a chess tournament or they can be on a higher > granularity scale eg Total victory/Narrow Victory/Draw/ > Narrow Defeat/Total Defeat. > > In each case the same swiss chess pairing algorithm is > used to match players for subsequent rounds. (ie ignoring > the many variations of this algorthim). Assume that it is > the popular one in chess ie for 2n players in a score group > player 1 plays player n+1, ..., player n plays player 2n. > > Which system will converge quicker (take less rounds) to > accurately* pick the strongest player? > > *I suppose that some definition of this would be required :-) > Perhaps with greater than 50% chance of being correct? By details, I mean that in addition to specifying precisely the pairing algorithm, you would have to specify the probability distribution of the outcome of a game, given the abilities of the players. It's unlikely that any realistic model would have an exact solution, so results would probably have to come from simulations. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > >> I think you'll have to be clearer about what assumptions you are making, >> if you want us to be able to answer your question. >> Let G be the graph with edges joining players who play a game. >> If you assume that the outcome of a game (on some continuous scale) is, >> say, the ratio of the strengths of the players, you can determine the >> strongest player as soon as G is connected, and that can happen after only >> two rounds: round 1 matching players 1-2, 3-4, 5-6 etc and round 2 >> matching 2-3, 4-5, 6-7 etc. > >But the OP asked about the swiss system, and this is not how a swiss >system tournament is run. Rather, in each round people are paired, >insofar as possible, against people who have acquired the same number >of points. The question is how many rounds do you expect to have to >play before you get to a situation where a single player leads the >tournament and can reasonably be said to have done the best. > > In the second round, the winner of 1 vs 2 plays the winner of 3 vs 4, > the loser of 3 vs 4 plays the loser of 5 vs 6, etc. Is that not possible > in a Swiss system? It can happen that you have a connected graph after two rounds, although the way these things are run in practice I think it's more common to have several components after two rounds and a connected graph after three. > Assuming that the outcome of a game tells you > the ratio of the strengths of the players, it gives you enough > information to rank all players. The hypothesis is not fulfilled in practice. Spassky can beat Fischer from time to time, but Fischer will win the long match (sorry that my example is 30 years out of date). >For chess, there's a theorem - or maybe it's only a rule of thumb - >that if there are n players you need sqrt(n) rounds. I regret that >I cannot supply a reference. > > That's a strange result. Without ties, if you start out with 2^k players > then after just k rounds there will be exactly one player with a perfect > record. What model are you using to determine when ties occur? I'm not using any model, I'm just dredging up memories from the days when I used to play in weekend swiss-system tournaments, sometime last century. You might get better answers on one of the chess newsgroups, or maybe there's something at the US Chess Federation website. -- ==== I've been reading Introductory Real Analysis by Kolmogorov and Fomin so I can acquire an understanding before I actually take the course. There is a problem set on relations: Give an example of a binary relation which is reflexive, but neither symmetric nor transitive. I can't find a reflexive relation on the arithmetic algebra. Of course, we can define a set A = {1,2,3,4} and create a relation R = {(1,1) (2,2) (3,3) (4,4)} that demonstrates reflexivity, but there surely are other examples in arithmetic: Symmetry: a is not equal to b Transitivity: a is greater than b Reflexivity and Transitivity: a is divisible by b Reflexivity and Symmetry: a minus b is less than one How about an example that's just reflexive? Xevious ==== > I've been reading Introductory Real Analysis by Kolmogorov and Fomin > so I can acquire an understanding before I actually take the course. There is a problem set on relations: Give an example of a binary relation which is reflexive, but neither > symmetric nor transitive. I can't find a reflexive relation on the arithmetic algebra. Of course, we can define a set A = {1,2,3,4} and create a relation R = > {(1,1) (2,2) (3,3) (4,4)} that demonstrates reflexivity, but there > surely are other examples in arithmetic: Symmetry: a is not equal to b > Transitivity: a is greater than b > Reflexivity and Transitivity: a is divisible by b > Reflexivity and Symmetry: a minus b is less than one How about an example that's just reflexive? Xevious Notice that the example R that you gave is actually reflexive, symmetric, and transitive. You could summarize this by saying that equality is an equivalence relation. One example on your set A, or on some other set such as the set of integers, say, that is reflexive but neither symmetric nor transitive is 0 <= a-b <= 1. I'm sure contributors will send you many more :-) ==== > I've been reading Introductory Real Analysis by Kolmogorov and Fomin > so I can acquire an understanding before I actually take the course. > > There is a problem set on relations: > > Give an example of a binary relation which is reflexive, but neither > symmetric nor transitive. > > I can't find a reflexive relation on the arithmetic algebra. I do not know what is meant by the arithmetic algebra. > > Of course, we can define a set A = {1,2,3,4} and create a relation R = > {(1,1) (2,2) (3,3) (4,4)} that demonstrates reflexivity, but there > surely are other examples in arithmetic: > > Symmetry: a is not equal to b > Transitivity: a is greater than b > Reflexivity and Transitivity: a is divisible by b > Reflexivity and Symmetry: a minus b is less than one The last example is usually not symmetric. So you already found one. The symmetric variant would be : |a-b| less than 1. Marc ==== >Let a_n = 1/(1+z^n), where z is a complex number. I'm trying to find the region of convergence of the series sum a_n, >where the sum is from 1 to infinity. My work so far: Let r=|z|. If r<1 then a_n does not tend to 0 and so >the series doesn't converge. Also true, for the same reason, if r = 1. >If r>2, the series converges because >|1/(1+z^n)| <= 1/(r^n - 1) <= 1/(r-1)^n and then applying the >comparison with the geometric series. Or: If r > 1 then r^n -> infinity, hence r^n > 2 for n > N. For such n, 1 < r^n/2, so |1+z^n| >= r^n - 1 >= r^n/2. ************************ David C. Ullrich ==== I am a student and I have to do a random sampling as part of my research. I need a software that will do this for me, one that is freely downloadable, even a demo/student version is ok. Please, I would appreaciate any pointers to such software. Sun. ==== > > I am a student and I have to do a random sampling as part of my > research. I need a software that will do this for me, one that is > freely downloadable, even a demo/student version is ok. > > Please, I would appreaciate any pointers to such software. > > > Sun. What type of random sampling? There is a program called WinStat at http://math.exeter.edu/rparris/ that may do what you need. -- Will Twentyman ==== > Robert, it embarrasses me to no end to admit this... What is JimT talking about? He obviously suggests that I am wrong below. So why does he say that he is embarrassed. Perhaps he means to say that he is embarrassed to actually explore my example, despite going on and on about it for numerous posts. > ...I calculated profits on total financial capital for the a firm > using the Beta technique and the prices in Table 7 of your original > post. The profit rate was 75%. If anybody cares, Table 7 showed the higher wage but original price of iron. Beta consisted of processes A and D. Selecting the relevant bits, one has: TABLE 7: COSTS, WAGE 109/4040 BUSHELS PER PERSON-YEAR, PRICE OF IRON 55/1112 BUSHELS PER TON INDUSTRY PROCESS PROFITS Corn A 90.1% Iron D 23.4% JimT's incorrect calculation of the rate of return combined the above wage and price of iron with Table 5. TABLE 5: THE BETA TECHNIQUE PRODUCING CORN NET INPUTS Process D Process A Labor 3 304/357 Person-Years 1 2/3 Person-Years Iron 3 59/357 Tons Iron 3 1/3 Tons Iron Corn 0 Bushels Corn 2/3 Bushel Corn OUTPUTS 6 178/357 Tons Iron 1 2/3 Bushel Corn LABOR-INTENSITY: 5 185/357 Person-Years Per Bushel > You are seriously going to tell me > that your (competitive, profit-maximizing) firm responds to a wage > increase by equating profits rates in the two commodity streams to > obtain a lower overall profit rate (60%) according to some kind of > bizarre equilibrium consideration?!? I am seriously saying that JimT's calculation of the rate of return is incorrect. When one calculates the rate of return on the processes A and D seperately, one obtains rates of return of 90% and 23%, respectively. Obviously, only one number can be the rate of return. And it is neither 90% nor 23%. Bizarre equilibrium consideration is just JimT's way of registering incomprehension. Recall that the coefficients of production were: TABLE 1: INPUTS REQUIRED PER TON CORN PRODUCED Process A Process B 1 Person-Year 1 Person-Year 2 Tons Iron 1/2 Tons Iron 2/5 Bushels Corn 3/5 Bushels Corn TABLE 2: INPUTS REQUIRED PER TON IRON PRODUCED Process C Process D 1 Person-Year 275/464 Person-Years 1/10 Tons Iron 113/232 Tons Iron 1/40 Bushels Corn 0 Bushels Corn The Beta technique can be summarized by a row vector and a matrix: a0 = [ 275/464 1 ] A = [ 113/232 2 ] [ 0 2/5 ] (I have put the iron industry in the first column and the first row.) As I showed in my post of 3 July, the production of a bushel corn with the Beta technique requires only dated labor inputs, where the amount of labor per bushel corn i years before the current year is: l(i) = a0 A^i e2 where e2 is the second column vector in the 2x2 identity matrix. (I had a typographic error in my previous post.) So the rate of return when using the Beta technique for a given wage w is the value of r that equates past discounted costs and revenues (unity, since corn is numeraire): l0 w (1 + r) + l1 w (1 + r)^2 + ... = 1 There is nothing bizarre about the above definition of the rate of return to anybody that knows accounting. And, as I showed, one needn't sum up the infinite series. By the magic of mathematics, that series is equal to a0 ( I - A (1 + r) )^(-1) e2 w (1 + r) So one has the implicit definition of the rate of return for the Beta technique for a given w from the equation: a0 ( I - A (1 + r) )^(-1) e2 w (1 + r) = 1 For w = 109/4040 Bushels Per Person-Year, r is approximately 56.65%, not 75%. And one can express the Alpha technique as a sequence of dated labor inputs too. When one costs up those inputs with w = 109/4040, r = 56.65%, one finds that the Alpha technique is cheaper than the Beta technique. The cost of the Alpha technique is approximately 0.8743 Bushels Per Bushel. Now, last time I checked, the following was true: 0.8743 < 1 So a good manager (not JimT) would adopt the cheaper technique. And the rate of return in operating the Alpha technique is 60%. > A good manager would recognize iron production is necessary to remain > in the corn industry and seek the highest overall return on capital. Which is what I show. -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau ==== Robert Vienneau is an awfully condescending fellow for someone who knows so little. He might want to learn a little more about economics before commenting on its failings. Two things Robbie should think about: (1) Models of asset prices. (2) The difference between a static problem and a steady-state equilibrium to a dynamic problem. Robert should have paid more attention to poor Chris Auld. Let's look at Robert's problem. Robert defines profits as the ratio of the value of output and inventory at the end of period to the value of inventory and labour costs at the beginning of the period. Let w denote the wage rate, p the price of iron at the beginning of the period, and p* the price of iron at the end of the period. Profits r are then determined as follows r=(end period value of output + inventory)/(beginning period value of inventory +labour costs) The numerator can be written as a function of p*, say E(p*), while the denominator can be written as a function of p and w, say B(p,w). Profits are thus 1+r=E(p*)/B(p,w) What we now have can be used as pricing equation by the manager of the integrated manager to determine the beginning of period value of iron. Let r* be the market rate of return. The price of iron in any given application (technique alpha, corn production, etc.) is simply the value of p that solves the equation 1+r*=E(p*)/B(p,w) The manager uses the price to determine how to use the asset, placing the asset in the use with the highest value (i.e. value of p). Labour demand is found by taking p* as given, changing w, and re-evaluating p for each possible application. Again, the manager chooses the application with the highest value of p. Increasing w in this problem won't result in the firm increasing it's labour inputs, since doing so would reduce the value of the asset p. Poor Robert Vienneau In the static problem, p* is treated as exogeneous by the manager. Where does p* come from? Well, we can figure out where p* comes for the manager of the firm using backwards induction. We could solve the same problem in the next period, using the market rate of interest and wage for that period, and solve for p. Hence p* depends on next period's interest rate, next period's wage rate, and the value of iron at the end of the next period. However, for the purposes of the static problem, we can treat these all as given. Hence p* is given (and is based on the firm behaving optimally in the future). The static exercise has nothing to do with prices changing for all time. The long-run equilibrium simply asks how a firm optimally chooses inputs and outputs to a given set of input and output prices and the interest rate. Poor Robbie Vienneau What is Robert trying to do? One could consider a situation where the wage and interest rate forever assume a given interest rate. In this case, it seems reasonable to assume that the price of iron will be the same at the beginning and end of period. One could then set p=p* and solve the pricing equation for p for each technique. Again, the relevant steady state solution will be the one that maximizes p. Poor Robbie Vienneau Finally, just because the firm is integrated, it doesn't follow that there isn't a market price for iron. Presumably someone could buy the firm and use the assets as they pleased. In this case, the firm proceeds as before, calculated the value of p using the pricing equation. If the highest value of p is less than the market price, then the optimal action for the manager is to sell the firm. Poor Rob Vienneau ==== > The difference between a static problem and a steady-state > equilibrium to a dynamic problem. So JimT's position today seems to be that the intro textbook story of supply and demand doesn't apply to models in which capital goods are used. For the intro textbook story is static and doesn't distinguish goods by date. That is, the intro textbook story only applies in the special case in which production immediately converts unproduced inputs (land, labor) to consumer goods with no help from intermediate goods. I wish JimT good luck in finding a mainstream textbook that clearly states this. > Let's look at Robert's problem. Notice how JimT mostly ignores the posts to which he is pretending to respond. > Let w denote the wage rate, p the price of iron at the beginning of > the period, and p* the price of iron at the end of the period. This is the first time anybody has begun to introduce notation that one can use to explicitly discuss dynamic problems. Anyway, JimT's introduction of dynamics is merely a matter of changing the subject. (p* is bad notation, since * is often used to denote equilibrium quantities.) Notice that JimT has implicitly dropped his confused nonsense in previous posts about the invalidity of treating corn as numeraire. The price equations expressing cost-minimization for iron production are: [ a11(mu) pi(t) + a21(mu) pc(t) + a01(mu) w(t) ](1 + r(t)) >= pi(t + 1), mu = C, D The price equations for corn production are: [ a12(mu) pi(t) + a22(mu) pc(t) + a02(mu) w(t)](1 + r(t)) >= pc(t + 1), mu = A, B (I adopt notation that is standard in the literature.) r(t) is mathematically equal to the own rate of interest of the numeraire, which could be corn. One equality must occur in each display above if both iron and corn are to be produced for use in the next period. > Let r* be the market rate of return... One response of neoclassical economists to the Cambridge Capital Controversy was to insist that mainstream economics had no need of a central capital market. These economists said that price theory could be developed with completely disaggregated models which do not contain a market for financial capital and a price r. They put forward the Arrow-Debreu model as an exemplar of what they were talking about. I don't know why JimT rejects this mainstream approach. > The manager uses the price to determine how to use the asset, placing > the asset in the use with the highest value (i.e. value of p). Labour > demand is found by taking p* as given, changing w, and re-evaluating p > for each possible application. Again, the manager chooses the > application with the highest value of p. Increasing w in this problem > won't result in the firm increasing it's labour inputs, since doing so > would reduce the value of the asset p. Both JimT and I take available coefficients of production as given. JimT now sets pc(t) and pc(t + 1) to unity. He wants to take pi(t + 1) and r(t) as given. So the only unknowns in the above displays are w(t) and pi(t). Suppose the firm produces both iron and corn. Then one would end up with two equalities above. For each value of w(t), the system would overdetermine pi(t), as JimT describes it. So either JimT is describing a generally inconsistent system or else he is describing a firm that will not be willing to produce both iron and corn. In the latter case, firms will find they must deviate from JimT's solution in the next period. Some equilibrium. > The static exercise has nothing to do with prices changing for all > time. In the intro textbook model, the equilibrium price of a supply-and- demand model, e.g., of the labor market, is supposed to persist through time, given the data of the model. Anyways, the reduction to dated labor quantities is only a notional reduction. > The long-run equilibrium simply asks how a firm optimally > chooses inputs and outputs to a given set of input and output prices > and the interest rate. Suppose one adopts the convention that goods available at different dates are different commodities, with distinct prices. The problem of the firm, with this convention, is conveniently set in the Arrow-Debreu model or in a model of a sequence of temporary equilibrium. These models, in which initial endowments are given, are of (very) short-run equilibria. (Notice that in such a model, firms would change initial endowments at the start of an out-of-equilbrium path. Any time to approach a (flow) equilibrium path is too long in such models.) I have already stated how I think capital-reversing would be manifested in such models. One can find a counter-intuitive temporal sequence, I claim, with given technology, in which the wage and the quantity of labor firms want to hire increase together. I think one could also make this claim in certain models of cross-dual dynamics. In making this claim, I draw on relevant literature on economic dynamics, some of which I have previously referenced in this thread. > What is Robert trying to do? One could consider a situation where the > wage and interest rate forever assume a given interest rate. There are various interpretations of the setting of my mathematics. As I keep on citing Kieran Sharpe, one capable of rational discussion (in which one pays attention to what others are saying) would know that one possible setting is that of a stock equilibrium. Another setting would be of a limit point of short-run flow equilibria that JimT tries to consider. Turnpike theorems relate such flow equilibria to positions in which endowments have been adjusted until relative prices remain unchanged through time. Another setting would be of a limit point of cross-dual dynamics. Anyway, my claim has been that supply-and-demand stories about substitution and prices as relative scarcity indices do not apply to such models. This claim merely echoes the literature: It is possible that the outputs produced in an Arrow-Debreu economy in the far distant future are independent of its initial endowments. That would mean that in such an economy the relative scarcities prevailing now would have no influence on the relative prices and rentals in the distant future. This should be enough to persuade the critics that the theory is not committed to a relative scarcity theory of distribution, though they seem to believe it is and that often motivates them in their attacks. -- Frank Hahn, General Equilibrium Theory, in _The Crisis in Economic Theory_, (edited by Daniel Bell and Irving Kristol), 1981. Not that I have been using general equilibrium models to support this claim. One doesn't need to. > In this > case, it seems reasonable to assume that the price of iron will be the > same at the beginning and end of period. One could then set p=p* and > solve the pricing equation for p for each technique. Again, the > relevant steady state solution will be the one that maximizes p. JimT is confused. He presents the above as if he is contradicting my equations. But, if corn and iron are both to be produced and temporally constant relative prices are considered, he, too, would get prices satisfying the constraints in my cost-minimization linear program in my appendix to my original post on this thread. And one of each of the corn and iron inequalities would be satisfied as an equality. In short, nobody has yet pointed out a mistake in my original post in this thread. > Finally, just because the firm is integrated, it doesn't follow that > there isn't a market price for iron. Presumably someone could buy the > firm and use the assets as they pleased. Once again, JimT presents my point as if he is contradicting me. I have over several posts explained methods for determining an accounting price of iron in my example. And it's not my fault that poor Chris Auld could not even conduct a coherent conversation (e.g., one in which he answered questions, instead of changing the subject) with Keith Ramsay. Incidently, I found that over certain ranges of the pseudo-production function the technique that becomes elgible at a...lower real-wage rate...may be less labour- intensive (that is, may have a higher output per man employed) than that chosen at a higher wage rate, contrary to the rule of a 'well-behaved production function' in which a lower wage rate is always associated with a more labour-intensive technique. (I attributed this discovery to Ruth Cohen - a private joke.) Journal of Economics. May 1970. -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau ==== Poor Robert Vienneau apparently doesn't understand that the static equilibrium of a competitive to firm is to maximize profits. nothing to do with profit maximation. That the profit rates be the same in both endeavours is not a necessary condition for profit maximization, *given* the binding constraint that the firm regenerate its inventory. Rob has tried to slip it in into to replicate a multi-equilibrium model. Poor Rob has apparently never seen a corner solution. Poor Rob Vienneau. ==== > > JimT is confused. He presents the above as if he is contradicting > my equations. But, if corn and iron are both to be produced and > temporally constant relative prices are considered, he, too, > would get prices satisfying the constraints in my > cost-minimization linear program in my appendix to my original > post on this thread. And one of each of the corn and iron > inequalities would be satisfied as an equality. It's all in the intepretation. I wouldn't impose the requirement that both goods be produced at the firm level. I think that this is the contemporary way of thinking about this situation. In other words, the market produces the conditions that insure that both good are produced. So yes, in my concept of the firm equilibrium both goods are only produced for certain combinations of the market rate of return, the wage, and the price of corn (given that corn is the numeraire). The workings of the market produce the outcome whereby both goods are produced. Profit maximization thus depends only on the available prices and not on imposition of further constraints. Anyhow, Robert, I apologize for all the nasty comments that I made. I think our disagreements are largly conceptual in nature (whether it is appropriate to impose certain conditions at the level of the firm or the level of the industry/market). point in time in the future. JT ==== > > The difference between a static problem and a steady-state > equilibrium to a dynamic problem. > > So JimT's position today seems to be that the intro textbook > story of supply and demand doesn't apply to models in which > capital goods are used. For the intro textbook story is static > and doesn't distinguish goods by date. > > That is, the intro textbook story only applies in the special > case in which production immediately converts unproduced > inputs (land, labor) to consumer goods with no help from > intermediate goods. > > I wish JimT good luck in finding a mainstream textbook that > clearly states this. The assumption is simply that the price of these goods is established by markets and not the firm. > > One response of neoclassical economists to the Cambridge Capital > Controversy was to insist that mainstream economics had no need > of a central capital market. These economists said that price > theory could be developed with completely disaggregated models > which do not contain a market for financial capital and a > price r. They put forward the Arrow-Debreu model as an > exemplar of what they were talking about. > > I don't know why JimT rejects this mainstream approach. In the textbook approach, economic profits are defined as earnings above the market rate of return. If you want to discuss the textbook, you better provide a consistent example. > > The manager uses the price to determine how to use the asset, placing > the asset in the use with the highest value (i.e. value of p). Labour > demand is found by taking p* as given, changing w, and re-evaluating p > for each possible application. Again, the manager chooses the > application with the highest value of p. Increasing w in this problem > won't result in the firm increasing it's labour inputs, since doing so > would reduce the value of the asset p. > > Both JimT and I take available coefficients of production as > given. JimT now sets pc(t) and pc(t + 1) to unity. He wants to > take pi(t + 1) and r(t) as given. So the only unknowns in the > above displays are w(t) and pi(t). > > Suppose the firm produces both iron and corn. Then one would > end up with two equalities above. For each value of w(t), > the system would overdetermine pi(t), as JimT describes it. > > So either JimT is describing a generally inconsistent system > or else he is describing a firm that will not be willing to > produce both iron and corn. In the latter case, firms will > find they must deviate from JimT's solution in the next > period. Some equilibrium. One more comment here. Iron has no value on its own. Since there is no goods market for iron, we can rule out iron production for price sequences that result in iron specialization (these are essentially bubble solutions where profits are based on a perpetually growing capital stock). This suggests that the pricing equation should be p=max(0,p(int)) where p(int) is determined by running the integrated techniques through the pricing equation that I gave previously. Though I haven't tried this, I suspect that the technique that maximizes p(int) will not be the technique that you offer as an equilibrium for a given (real) wage rate. This nevertheless results in a less restrictive outcome that I presented earlier. I image that there is some critical value of w where the firm ceases to be an integrated firm and switches to pure corn production. I doubt there will be factor reversals with a fixed market rate of return, r*. Just as an aside, the iron inventories might have value to other kinds of firms in the economy. In this case, the zero in the above equation could be replaced with a scrap value p(scrap). I think the big problem with your example is that the textbook *does* fix the market rate of return when defining profits for the firm. Economic profits are profits above the market rate on assets. I can find you quotes to that extent in commonly used textbooks. Your factor reversals require extra degrees of freedom that are not present in the basic model. ==== > nothing to do with profit maximation. That the profit rates be the > same in both endeavours is not a necessary condition for profit > maximization, *given* the binding constraint that the firm regenerate > its inventory. Rob has tried to slip it in into to replicate a > multi-equilibrium model. JimT is confused. In my numerical example, I describe profit-maximizing/ cost-minimizing firms. I ask what should the price of iron be such that such profit maximizing firms will continue to want to produce, say, corn at the same level of output. There are two steps in this analysis, and JimT has confused the order of the steps. The correct analysis implies that the rate of profits will be the same in both corn and iron production, if both continue to be produced. As I show with the mathematics in the appendix in my first post on this thread. JimT apparently wants to impose the constraint that the firm regenerate its inventory in such a way that the firm is not profit-maximizing/cost-minimizing. I don't know why he rejects mainstream economics. > JimT is confused. He presents the above as if he is contradicting > my equations. But, if corn and iron are both to be produced and > temporally constant relative prices are considered, he, too, > would get prices satisfying the constraints in my > cost-minimization linear program in my appendix to my original > post on this thread. And one of each of the corn and iron > inequalities would be satisfied as an equality. > It's all in the intepretation. I wouldn't impose the requirement that > both goods be produced at the firm level. I think that this is the > contemporary way of thinking about this situation. In other words, > the market produces the conditions that insure that both good are > produced. If JimT were ever to follow the posts to which he is pretending to respond, he would know that he isn't arguing with anything I say. He still hasn't acknowledged that my analysis does not postulate that any market is in equilibrium, whether he wants to call certain conditions (a part of) market equilibrium conditions or not. > So yes, in my concept of the firm equilibrium Notice the word equilibrium doesn't appear in my quoted paragraph above. JimT's response, of course, is a non-sequitur. > both goods > are only produced for certain combinations of the market rate of > return, the wage, and the price of corn (given that corn is the > numeraire). I have continually pointed out that profit-maximizing firms would only produce both iron and corn for certain combinations of the price of corn, the price of iron, and the wage. An accounting rate of profits emerges from my analysis. I have no problem, of course, with the assertion that the firm would shut down when this rate falls below some rate for finance. There is a reason why I did not include an incoherent notion (the neoclassical ones) of the market for capital in my original post on this thread. It is also true that mainstream economists use the word equilibrium in different ways. Some use it to only mean the solution of a set of equations. This strictly syntactic meaning abandons the concerns addressed by the traditional notion, including as that traditional notion is used in intro textbooks. Notice JimT has yet to acknowledge that, in the textbook, a supply-and-demand equilibrium will result in equilibrium prices and quantities persisting unchanged as long as the data for that equilibrium persist unchanged. > The workings of the market produce the outcome whereby > both goods are produced. Profit maximization thus depends only on the > available prices and not on imposition of further constraints. Once again, JimT is not disagreeing with the profit-maximizing problem I posed in my original post on this thread. > > The difference between a static problem and a steady-state > equilibrium to a dynamic problem. > So JimT's position today seems to be that the intro textbook > story of supply and demand doesn't apply to models in which > capital goods are used. For the intro textbook story is static > and doesn't distinguish goods by date. > > That is, the intro textbook story only applies in the special > case in which production immediately converts unproduced > inputs (land, labor) to consumer goods with no help from > intermediate goods. > > I wish JimT good luck in finding a mainstream textbook that > clearly states this. > The assumption is simply that the price of these goods is established > by markets and not the firm. Which has nothing to do with my point. > One response of neoclassical economists to the Cambridge Capital > Controversy was to insist that mainstream economics had no need > of a central capital market. These economists said that price > theory could be developed with completely disaggregated models > which do not contain a market for financial capital and a > price r. They put forward the Arrow-Debreu model as an > exemplar of what they were talking about. > > I don't know why JimT rejects this mainstream approach. > In the textbook approach, economic profits are defined as earnings > above the market rate of return. If you want to discuss the textbook, > you better provide a consistent example. It is true that the out-of-date textbook approach was developed in the 1920s (and was also shown to mistaken then). I have already presented an exposition of the firm facing a given market rate of return. In my post of 8 July, I described a firm maximizing the function: max f(L, K) - w L - r K And I pointed out an error in the derivation of a downward-sloping labor demand curve that follows on from that maximization problem. It should be well-known that mainstream textbooks do not provide a coherent and consistent account of the return to capital. M. I. Naples and N. Aslanbeigui, What DOES Determine the Profit Rate? The Neoclassical Theories Presented in Introductory Textbooks, Cambridge Journal of Economics, V. 20, pp. 53-71, 1996. Of course, mainstream economists are socialized to ignore the errors in their theories pointed out by certain communities of economists. > The manager uses the price to determine how to use the asset, placing > the asset in the use with the highest value (i.e. value of p). > Labour > demand is found by taking p* as given, changing w, and re-evaluating > p > for each possible application. Again, the manager chooses the > application with the highest value of p. Increasing w in this > problem > won't result in the firm increasing it's labour inputs, since doing > so > would reduce the value of the asset p. > Both JimT and I take available coefficients of production as > given. JimT now sets pc(t) and pc(t + 1) to unity. He wants to > take pi(t + 1) and r(t) as given. So the only unknowns in the > above displays are w(t) and pi(t). > > Suppose the firm produces both iron and corn. Then one would > end up with two equalities above. For each value of w(t), > the system would overdetermine pi(t), as JimT describes it. > > So either JimT is describing a generally inconsistent system > or else he is describing a firm that will not be willing to > produce both iron and corn. In the latter case, firms will > find they must deviate from JimT's solution in the next > period. Some equilibrium. > [Confusion about cost-minimization and vertical integration - deleted] The inequalities I presented describe cost-minimization. I reaffirm that firms will deviate from JimT's solution in the next period. I suppose this could be said to be related to the Hahn problem. > I think the big problem with your example is that the textbook *does* > fix the market rate of return when defining profits for the firm. > Economic profits are profits above the market rate on assets. There's a reason why I spoke of accounting profits, not economic profits, in my first post on this thread. > I can > find you quotes to that extent in commonly used textbooks. Your > factor reversals require extra degrees of freedom that are not present > in the basic model. Notice JimT does not specify special case assumptions that (1) Rule out the technology in my example (2) Provide a degree of freedom that permit the consideration of different levels of wages with production persisting. I provided a correct analysis of which processes a profit-maximizing/ cost-minimizing firm would want to adopt at any set of given prices. (See the appendix to my original post in this thread.) JimT has yet to point out an error in that analysis. And the accounting rate of profits emerges from that analysis. To summarize: Intro textbooks present a supply-and-demand equilibrium for the labor market that has the following properties: o A supply-and-demand equilibrium will result in the equilibrium price and quantity persisting unchanged as long as the data for that equilibrium persist unchanged. o In drawing the labor demand curve, other prices (e.g. of iron) and the market rate of return to financial capital are taken as given. o Labor demand curves slope down. As is implied by the analysis of my first post on this thread, these conditions are generally inconsistent. -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau ==== > > JimT apparently wants to impose the constraint that the firm > regenerate its inventory in such a way that the firm is not > profit-maximizing/cost-minimizing. I don't know why he rejects > mainstream economics. Bob, I think you might both get along better if you avoided stuff like this last he rejects mainstream economics nonsense. You are both playing with a specific model, yours, and it's no more mainstream than anything else any*body* else dreamed up on Usenet some arbitrary morning. My opinion is that you guys are both really really confused in a bunch of elementary ways. Just to pick an easy one -- and I only pick on you rather than him because you're the latest one to have posted -- you above use the phrase profit-maximizing/cost-minimizing as though profit maximinzing and cost-minimizing were synonyms. (obviously the slashin the middle doesn't mean you're dividing one by the other.) Well that's nuts, both in the real world and in your model. In the real world you maximise profit by maximizing the difference between income and outgo, regardless of what the outgo is. Increasing the out go is very often the first step, or thenecessary continuous concomitant, of maximising profit. I'm not clear on the model -- but this is very largely, Robert, because you keep appealing to elementary textbook (here I assume you are trying to insult your interlocutor) types of equilibrium. If that is what you are uisng for your prices and returns, then you don't have a model: you have an argument about the real world, which argument is cluttered up by these broken fragments of model you have under your feet, around your neck, in your hands opponont on the head of for banging with. I don't think JT does much better. I have tried to follow what he of profit and a rate of profit (per period, per transaction, per sum of transactions, per anything) he sure as hell has not found a form of notation which gives anyone confidence that this is so -- let alone what the distinction is, if one charitably assumes that he has made it. Neither of you guys is functioning. In a discussion of energy storage in spinning rings the other day, I ran cross a guy writing about rate of change of angular velocity per cycle per rotation. You guys may not yet have levitated to that level of levity. But it's not clear that you're not trying hard. -dlj. ==== > > I'm not clear on the model -- but this is very largely, Robert, > because you keep appealing to elementary textbook (here I assume > you are trying to insult your interlocutor) types of equilibrium. It would help if Robert would ever actually provide a reference to the mysterious textbook case. The labour demand curve in the textbook on my desk is derived for an apple grower with an orchid of a fixed size (Mankiw, 3rd edition). The production function is given. One factor is fixed and the assumption of diminishing returns to labour is clearly stated. I don't know what relationship Robert's *dynamic* model is even suppose to have to this model. > > I don't think JT does much better. I have tried to follow what he > of profit and a rate of profit (per period, per transaction, per sum > of transactions, per anything) he sure as hell has not found a form > of notation which gives anyone confidence that this is so -- let > alone what the distinction is, if one charitably assumes that he has > made it. No doubt that I was sloppy. I was trying for a rate (economic profits measured as a rate of return above the investor's next best alternative), though I realize an amount is the standard usage in economics (total revenue less opportunity cost of inputs). > Neither of you guys is functioning. In a discussion of energy > storage in spinning rings the other day, I ran cross a guy writing > about rate of change of angular velocity per cycle per rotation. Dammit. I'm talking about profit rate above market rate per second above opportunity cost cubed. Or some such thing... > You guys may not yet have levitated to that level of levity. But > it's not clear that you're not trying hard. Heh...well at least someone else was following the thread. I hope it at least provided you with some entertainment value.... ==== Well, Robert, I don't expect you'll pay attention, but I'm going to tell you what the different between the short and long-run is in your model. The integrated firm is simultaneously involved in two activities: (1) Producing output for sale at the end of the period. and (2) Investing in inputs for the next period. The textbook definition of the long-run is a period of time over which all factors are variable. The short-run is A period of time over which some factors are fixed. Consider one period of your model. The short-run labour demand is found by determining how the firm responds to a change in the wage rate after the investment is made. So, for an expected future value of steel (see an earlier post for more explanation), the firm takes its current stock of steel, adds or subtracts to/from its corn stock through sales and purchases, and hires the appropriate amount of labour. The relationship between w and L is the labour demand curve. The firm can adjust both labour and corn inputs, but not the amount of the iron input. In the long-run, the firm is allowed to change it's investment decisions from the previous period so that they are optimal given the wage in the next period. Iron, corn and labour inputs are all adjusted by the firm in response to the wage change. There is no requirement that that wage change prevails for all time. You won't find any mention of wages changing for all time in any introductory textbooks that I'm familiar with. Long-run doesn't refer to a time frame in any temporal sense, but only to a situation where all outputs can be choosen optimally to maximize profits. Otherwise, I don't know why you would bother objecting to my pricing equation. If you plotted the value of iron against the wage rate in a steady-state, I suspect the two curves cross in a way that is entirely consistent with the steady-state outcome you describe...once the wage increases to some thresh-hold value the firm switches to what you describe as the labour intensive technique. Introducing a cost of capital or rate of return in an alternative venture will only change the crossing point (and maybe the shutdown conditions). Au revoir, JT ==== > It would help if Robert would ever actually provide a reference to the > mysterious textbook case. The labour demand curve in the textbook on > my desk is derived for an apple grower with an orchid of a fixed > size (Mankiw, 3rd edition). downward-sloping labor demand curve would result from the quantity of some other factor being given. > and the assumption of diminishing returns to labour is > clearly stated. My example has diminishing returns to labour. As I said in my first post on this thread, I present a discrete approximation: The production functions constructed in this manner exhibit properties typically assumed in neoclassical economics. In particular, they exhibit Constant Returns to Scale, and the marginal product, for each input, is a non-increasing step function. Many intro textbooks try to present their material in a manner suitable for a student that has not mastered calculus. So JimT shows that he is only pretending to follow this my story to the textbook case. > I don't know what relationship Robert's *dynamic* > model is even suppose to have to this model. Dynamic is JimT's word. Once again, JimT refuses to explain how the intro static textbook model can apply if capital goods are used. Does he tell his students the theory he presents only applies if production immediately converts unproduced inputs (land, labor) to consumer goods with no help from intermediate goods? > I was trying for a rate (economic profits > measured as a rate of return above the investor's next best > alternative), though I realize an amount is the standard usage in > economics (total revenue less opportunity cost of inputs). The rate of return I find is for the firm's best alternative, where alternatives are compared on the basis of net present value. Standard terminology. > I hope [the thread] at least provided...some > entertainment value.... I assume any mathematicians paying attention are amused by the irrationalism and anti-intellectualism manifested by the professional economists participating in this thread. -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau ==== > Well, Robert, I don't expect you'll pay attention, but I'm going to > tell you what the different between the short and long-run is in your > model. The integrated firm is simultaneously involved in two > activities: > > (1) Producing output for sale at the end of the period. > > and > > (2) Investing in inputs for the next period. > > The textbook definition of the long-run is a period of time over > which all factors are variable. The short-run is A period of time > over which some factors are fixed. Consider one period of your > model. JT, You're doing what I thought we both agreed yesterday was a no-no when Robert did it: quoting the textbook without identifying the textbook. The actual definitions of short and long term were laid down by David Lloyd-Jones, during his Dark Period (as a bank auditor.) I quote from his very words: The long term is when you can feel 1% eating away at you. The medium term is when nothing short of 3% makes a difference. At that point the elevator door opened to let in a bunch of brokers who somehow thought themselves fit to eat at our sky lounge, such are the effects of money upon society, so I missed what he said about short term.. by the testicles... seven percent... and other no doubt wise and informative remarks which were lost in the noise. -dlj. ==== > I assume any mathematicians paying attention are amused by > the irrationalism and anti-intellectualism manifested by the > professional economists participating in this thread. No, Robert, we are not. I am not in fact a professional mathematician, but I *do* have two gold medals, one from the AMA, and one from their local science fair, so I'm no totally out of the pool. Everybody here is really fed up with you, and with your wste of yourself. You are the most tremendously over-educated sophomore in economics on the planet. For two or three years now you have been boring everybody into catatonia with the same stupid exercise over and over again: Look, Ma! I can make up a bunch of whacko assumptions, and prove that anything anybody thinks on normal assumptions can be shown false under my scheme. Well, whoopdie-doo. Robert, everybody who actually goes to a school does that for a while, and then moves on. You are bright, you are well read. You are ready for school. Robert, it is time for you to go get into a graduate seminar where you get your head beat in good and thorough Monday, Wednesday, and Friday for a couple of years. After that, you settle down and do some work for a while, and after that you write it up and they give you a PhD. In the meantime, would you please just piss off and shut up? Your friend and supporter, -dlj. ==== > Well, Robert, I don't expect you'll pay attention, but I'm going to > tell you what the different between the short and long-run is in your > model. The integrated firm is simultaneously involved in two > activities: > > (1) Producing output for sale at the end of the period. > > and > > (2) Investing in inputs for the next period. > > The textbook definition of the long-run is a period of time over > which all factors are variable. The short-run is A period of time > over which some factors are fixed. Consider one period of your > model. > > JT, > > You're doing what I thought we both agreed yesterday was a no-no > when Robert did it: quoting the textbook without identifying the > textbook. Oops! The referee just flagged me for a foul. Before I rush off to the penalty box to sit the remainder of this game out, I'll mention that the definitions were lifted from Landsburg's Price Theory & Application (5th ed.). This may even be considered cheating, since the book is technically an intermediate text. I didn't have a elementary text on hand when I posted the last message, but I'm confident the definition will be the same. JT ==== Somebody who is neither a professional economist nor a professional mathematician responds to a meta comment of mine touching on professional economists and professional mathematicians. > Look, Ma! I can make up a bunch of whacko assumptions, and > prove that anything anybody thinks on normal assumptions can be > shown false under my scheme. The above is an example of irrationalism or, at least, a fallacy. The fallacy being illustrated is that of begging the question. No attempt has been made to correctly show how the assumptions in my example are whacko, which assumptions drive any interesting results, or how these assumptions contrast and compare with any so-called normal assumptions. Let me be explicit. I assert that the conclusions I draw - that Marshall's principle of substitution does not apply to comparisons of long run positions, that long run equilibrium prices are not scarcity indices - depend on any whacko assumptions. As a matter of fact, I am only echoing the advanced literature, The issue was settled in favour of Cambridge University when is taught', the 'paradoxes' are accepted, and their consequences for the concept of capital known. It is another matter that, on the basis of this criterion, many seats of learning in North America, as perhaps also elsewhere, do not teach informed economic theory. For instance, the 'new classicals', claiming otherwise to be meticulous theorists, keep on blithely using the Clarkian concept of capital in their production functions. The case for the implicit excuse that the absence of any easy alternative justifies such high-handedness has not so far been made in the literature. -- Syed Ahmad (1998) -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau ==== > The textbook definition of the long-run is a period of time over > which all factors are variable. The short-run is A period of time > over which some factors are fixed. Consider one period of your > model. Which, as I have mentioned, could be seen as part of a limit point of an Arrow-Debreu intertemporal equilibrium. Turnpoint theorems are important here. Anyways, conceived in this fashion, the relative endowments of corn and iron are treated as endogeneous, that is, as adjusted appropriately, in the limit. Considering such a limit point does not imply one is considering a wage change that prevails for all time. Obviously, what is taken as exogeneous can receive a shock. As I keep mentioning, there are other contexts in which one can think about my example. > The short-run... The firm can adjust both labour and > corn inputs, but not the amount of the iron input. JimT's claimed asymmetry between corn and iron arguably does not exist in my numerical example. Both corn and iron comprise circulating capital. (Not that the properties of a short run flow equilibrium matter for my numerical example.) > In the long-run, the firm is allowed to change it's investment > decisions from the previous period so that they are optimal given the > wage in the next period. Iron, corn and labour inputs are all > adjusted by the firm in response to the wage change. There is no > requirement that that wage change prevails for all time. Whatever. > You won't find any mention of wages changing for all time in any > introductory textbooks that I'm familiar with. JimT is being silly. You won't find the future path of wages anticipated for all time either in most intro textbooks. For example, the discussion of supply-and-demand equilibrium on pg. 56 of Baumol and Blinder, Economics: Principles and Policy, Third Edition, 1985 doesn't contain any dated commodities. It does say: An equilbrium is a situation in which there are no inherent forces that produce change; that is a situation that does not contain the seeds of its own destruction. So, according to the intro textbook, if the labor market were in a supply-and-demand equilibrium the equilibrium wage and employment would persist unchanged until the exogeneous data of the model changed. As I implied in my post kicking off this thread: Q: If the data under which the curve is derived remains unchanged (e.g., technology) and the wage and prices for a given point on the long run labor demand curve actually prevailed, would there be forces within firms moving firms away from that point. A: No. Which brings me to another point. JimT correctly says that the intro textbook story is one of (comparative) statics. And he seems to say my example is inherently dynamic. What would make it so is that I explicitly model the use of capital goods. Now what dynamics should mean to economists has long been contested. One mainstream approach to economic dynamics is provided by sequences of temporary equilibria. This short-run (first edition, 1939) [1] [2]. I find the following passage in the second edition of that book: Of course, people used to be able to content themselves with the static apparatus, only because they were imperfectly aware of its limitations. Thus they would often introduce into their static theory a 'factor of production' capital and its 'price' interest, supposing that capital could be treated like the static factors. (Cf. J. B. Clark's 'free capital' and Cassel's 'capital disposal'.) That some error was involved in this procedure would not have been denied; but the absence of a general dynamic theory, in which all quantities were properly dated, made it easy to underestimate how great the error was. Fundamental Principles of Economic Theory, 2nd edition, 1946, p. 116. treated in economic statics. So the question, again, is whether static models can handle And do intro texts say the opposite? I predict JimT will once again manifest himself incapable of rational discussion. He will say neither yea nor nay, even if he continues posting. > ...If you plotted the value of iron against the wage rate in a > steady-state, I suspect the two curves cross in a way that is entirely > consistent with the steady-state outcome you describe... JimT is, once again, just babbling. The value of iron against the wage rate in a steady state is one curve, not two. The price of iron versus the wage is a combination of the two plots on the second worksheet in the spreadsheet I gave in my first post on this thread: http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ChoiceOfTechnique. xls JimT's comment, I suspect, shows that he has just not been paying attention. One thing, however, is evident when we look at this stationary economy, which was not evident in the static theory when time was left out of account altogether. This is the dependence of the input-output relations (the production functions) on the quantity of intermediate products carried by the system. How will the quantity of intermediate products - the quantity of capital - be determined? It turns out to be determined through the rate of interest. A fall in the rate of interest would encourage the adoption of longer processes, requiring the use (at any moment) of larger quantities of intermediate products. (2nd edition, p. 118) value and capital. He thought it was not properly set in historical time. That is, he was convinced by Joan Robinson's independent argument. -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau ==== > Let me be explicit. I assert that the conclusions I draw - that > Marshall's principle of substitution does not apply to comparisons > of long run positions, that long run equilibrium prices are not > scarcity indices - depend on any whacko assumptions. Should be do NOT depend on any 'whacko' assumptions. -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau ==== > Somebody who is neither a professional economist nor a professional > mathematician responds to a meta comment of mine touching on > professional economists and professional mathematicians. > >> Look, Ma! I can make up a bunch of whacko assumptions, and >>prove that anything anybody thinks on normal assumptions can be >>shown false under my scheme. > > The above is an example of irrationalism or, at least, a fallacy. > The fallacy being illustrated is that of begging the question. No, Robert, it's a characterisation of all your model building and posting of the past couple of years. > No attempt has been made to correctly show how the assumptions in > my example are whacko, which assumptions drive any interesting > results, or how these assumptions contrast and compare with any > so-called normal assumptions. Well in normal life we don't usualy go around thinking that the entire economy runs on corn and ale, or whatever your model of the day happnes to output. -dlj. ==== > > >>Let me be explicit. I assert that the conclusions I draw - that >>Marshall's principle of substitution does not apply to comparisons >>of long run positions, that long run equilibrium prices are not >>scarcity indices - depend on any whacko assumptions. > > Should be do NOT depend on any 'whacko' assumptions. Yes, Robert, we noticed. The word any in there made it perfectly clear what you had intended to say, which is why I didn't even bother to write back My point exactly to your original post. :-) -dlj. ==== Well looks like I got bit on the factor of versus factors in common with controversy as a reply I just received from the chief editor of a German journal talks as if I'm saying that none of the a's has a factor of 5, that is 5. He also thought my lemma was a trivial result and he mentioned polynomial factors. Then he put me on a spam block, which I discovered when I tried to reply back to him. Hmmm...I guess I'll not get much leeway from Germans. In any event, if I'm facing real problems with the is factor of thing I'd just as soon change it, so I'll be evaluating my paper today to see about making changes to be more in line with common usage. I'll also consider the coprime part, which confused a mathematician at Cal-Tech, and I'm kind of sure there was at least one other that said so, which may mean that others were confused but never admitted it. It seems that mathematicians are kind of fragile on this terminology thing, so I'll try and get closer to standard usage, and then I'll find yet another journal. I say that as saying 3 has a factor of 12 is actually correct, as 3 has itself as a factor and 3 is a factor of 12, but mathematicians have gotten themselves stuck one way, and there's no point in fighting their slang. I'm curious about what will come up next. James Harris ==== > Well looks like I got bit on the factor of versus factors in common > with controversy as a reply I just received from the chief editor of > a German journal talks as if I'm saying that none of the a's has a > factor of 5, that is 5. He also thought my lemma was a trivial result > and he mentioned polynomial factors. Then he put me on a spam block, > which I discovered when I tried to reply back to him. :-) Hmmm...I guess I'll not get much leeway from Germans. In any event, if I'm facing real problems with the is factor of > thing I'd just as soon change it, so I'll be evaluating my paper today > to see about making changes to be more in line with common usage. > I'll also consider the coprime part, which confused a mathematician > at Cal-Tech, and I'm kind of sure there was at least one other that > said so, which may mean that others were confused but never admitted > it. It seems that mathematicians are kind of fragile on this terminology > thing, so I'll try and get closer to standard usage, and then I'll ... end up with a paper that describes how to add 1 and 1 to get 2? Dirk Vdm ==== >Well looks like I got bit on the factor of versus factors in common >with controversy as a reply I just received from the chief editor of >a German journal talks as if I'm saying that none of the a's has a >factor of 5, that is 5. The only reason there's any controversy is that you refuse to believe the evil people on sci.math who are actually trying to help you make sense. How about that, this editor says a has a factor of b means exactly what everyone has been saying it means, not what you've been insisting it _should_ mean. What a surprise. >He also thought my lemma was a trivial result >and he mentioned polynomial factors. Then he put me on a spam block, >which I discovered when I tried to reply back to him. Wonder why he'd do that? >Hmmm...I guess I'll not get much leeway from Germans. As opposed to the editors elsewhere, who have been enthusiastic about your work, right? >In any event, if I'm facing real problems with the is factor of >thing I'd just as soon change it, so I'll be evaluating my paper today >to see about making changes to be more in line with common usage. >I'll also consider the coprime part, which confused a mathematician >at Cal-Tech, and I'm kind of sure there was at least one other that >said so, which may mean that others were confused but never admitted >it. You're really determined to waste the time of half the mathematicians like this they have to either reply or be impolite. You really shouldn't do that, at least not until you have something that's both correct and interesting to say. (How could you tell whether you have something correct and interesting? Ask on sci.math. No, that doesn't work because they're all liars. Except that every time you've bothered a journal editor or a top mathematician it's turned out their opinion is the same. Huh.) >It seems that mathematicians are kind of fragile on this terminology >thing, so I'll try and get closer to standard usage, and then I'll >find yet another journal. I say that as saying 3 has a factor of 12 >is actually correct, as 3 has itself as a factor and 3 is a factor of >12, but mathematicians have gotten themselves stuck one way, and >there's no point in fighting their slang. I'm curious about what will come up next. not the only problem... >James Harris ************************ David C. Ullrich ==== > It seems that mathematicians are kind of fragile on this terminology > thing, so I'll try and get closer to standard usage, and then I'll > find yet another journal. Fragile? Quite the reverse. Mathematicians are rigorous as hell about terminology things, requiring strict definitions to be made and followed. Anything less leads to logical anarchy, as JSH frequently demonstrates. It is JSH who is fragile on the subject, tending to get his proofs fractured by every minor misapplication of terminology possible. ==== > It seems that mathematicians are kind of fragile on this terminology > thing, so I'll try and get closer to standard usage... and > there's no point in fighting their slang. If everybody in the world uses a phrase a certain way, and you use it a different way, a way unique to you, how is it that EVERYBODY ELSE is speaking slang and you're correct? - Randy ==== > It seems that mathematicians are kind of fragile on this terminology > thing, so I'll try and get closer to standard usage... and > there's no point in fighting their slang. If everybody in the world uses a phrase a certain way, and > you use it a different way, a way unique to you, how is it > that EVERYBODY ELSE is speaking slang and you're correct? - Randy http://www.webster.com/cgi-bin/dictionary?&va=autism | Main Entry: auátism | Pronunciation: 'o-ti-z&m | Function: noun | 1 : absorption in self-centered subjective mental activity | (as daydreams, fantasies, delusions, and hallucinations) | usually accompanied by marked withdrawal from reality | 2 : a mental disorder originating in infancy that is | characterized by self-absorption, inability to interact | socially, repetitive behavior, and language dysfunction | (as echolalia) | - auátisátic /o-'tis-tik/ adjective or noun | - auátisátiácalály /-ti-k(&-)lE/ adverb Dirk Vdm ==== > Well looks like I got bit on the factor of versus factors in common > with controversy as a reply I just received from the chief editor of > a German journal talks as if I'm saying that none of the a's has a > factor of 5, that is 5. He also thought my lemma was a trivial result > and he mentioned polynomial factors. Then he put me on a spam block, > which I discovered when I tried to reply back to him. > > Hmmm...I guess I'll not get much leeway from Germans. > > In any event, if I'm facing real problems with the is factor of > thing I'd just as soon change it, so I'll be evaluating my paper today > to see about making changes to be more in line with common usage. > I'll also consider the coprime part, which confused a mathematician > at Cal-Tech, and I'm kind of sure there was at least one other that > said so, which may mean that others were confused but never admitted > it. Well then I went back and reread my paper. I use coprime in the key places and not factor of which makes me wonder a bit about that editor, as he may have been following discussions on the newsgroup. > It seems that mathematicians are kind of fragile on this terminology > thing, so I'll try and get closer to standard usage, and then I'll > find yet another journal. I say that as saying 3 has a factor of 12 > is actually correct, as 3 has itself as a factor and 3 is a factor of > 12, but mathematicians have gotten themselves stuck one way, and > there's no point in fighting their slang. Well I'm liking my use of coprime in the paper more and more as it turns out there's nothing to change. And in fact the chief editor's comments seem rather odd now. And no, it's not a big deal to me that it was the chief editor. That was a new twist though, him adding me to a spam list. Guess he wasn't going to be bothered with an answer to his objections! Oh well, there are other countries, other math journals. Maybe I should try the Chinese again? No wait, I think it might be time to see about the Russians again. They're supposedly good at math. James Harris ==== >> Well looks like I got bit on the factor of versus factors in common >> with controversy as a reply I just received from the chief editor of >> a German journal talks as if I'm saying that none of the a's has a >> factor of 5, that is 5. He also thought my lemma was a trivial result >> and he mentioned polynomial factors. Then he put me on a spam block, >> which I discovered when I tried to reply back to him. >> >> Hmmm...I guess I'll not get much leeway from Germans. >> >> In any event, if I'm facing real problems with the is factor of >> thing I'd just as soon change it, so I'll be evaluating my paper today >> to see about making changes to be more in line with common usage. >> I'll also consider the coprime part, which confused a mathematician >> at Cal-Tech, and I'm kind of sure there was at least one other that >> said so, which may mean that others were confused but never admitted >> it. Well then I went back and reread my paper. I use coprime in the key >places and not factor of which makes me wonder a bit about that >editor, as he may have been following discussions on the newsgroup. Or he may be aware of what he have been telling you for a long time: that your use of both factor and coprime are incorrect. Your recent posts indicate that you think that coprime is a synonim for has no common factors other than units; that is not the case. The two conditions ->happen<- to be equivalent in the ring of all algebraic integers, but not in general. You are using a term whose meaning you do not know, to mean something different from what it truly means. In addition, the comments of the editor dove-tail perfectly with what others have said often: that your statements in your manuscript are imprecise are prone to confusion. You do not specify what sort of objects you are dealing with, you do not specify what sort of objects your factors are, or in what rings the divisibility statements take place. If and when you ever make the statement precise, you will discover one of two things: either the lemma is false once it is made precise, or else the result will be trivial but inapplicable in the situation you wish to apply it to. The editor clearly interpreted in its trivial sense, that in which all named elements and statements relate to the ring of polynomials with algebraic integer coefficients. In that setting, the lemma is not applicable to your FLT argument. ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > Oh well, there are other countries, other math journals. Maybe I > should try the Chinese again? > > No wait, I think it might be time to see about the Russians again. > They're supposedly good at math. > > > James Harris But then the Russians would be just whom you don't want! ==== Jimmy, You're out of your depth. You're not going to get a reputable journal to publish your paper. Your best bet would be to add some hints about cosmic implications and get some newage (ryhmes with 'sewage') rag to print it. -jcr ==== > Well looks like I got bit on the factor of versus factors in common > with controversy as a reply I just received from the chief editor of > a German journal talks as if I'm saying that none of the a's has a > factor of 5, that is 5. He also thought my lemma was a trivial result > and he mentioned polynomial factors. Then he put me on a spam block, > which I discovered when I tried to reply back to him. > > Hmmm...I guess I'll not get much leeway from Germans. This journal is eager to publish your work - http://www.jir.com/ ==== > Jimmy, > > You're out of your depth. You're not going to get a reputable journal > to publish your paper. Your best bet would be to add some hints about > cosmic implications and get some newage (ryhmes with 'sewage') rag to > print it. > > -jcr Well *supposedly* a correct and profound math paper can get published in a reputable journal which means that the journals I've faced so far may lose a lot of their luster once the full story comes out. As for being out of my depth my little paper so far has drawn comments and questions from someone as notable as Barry Mazur, though I hope he doesn't now regret replying to me about it, and has been looked at without claim of error by at least two of the editors of the New York Journal of Mathematics, which includes Andrew Granville who referred me up to the chief editor, and I hope he does't regret that either. Now then it seems to me that a lot of you may think of me as just some crank but that lot of you don't even register on the screen for a lot of the people I've already received comments from, as they don't know you and probably don't care to know you. But, of course, you can still think that your opinions about me matter. As if opinion is mathematics, as I fear that for many of you, opinion IS mathematics, which is why you don't register and shouldn't as you don't understand the point. Mathematics is about truth, not social opinion. James Harris ==== > >>Jimmy, >>You're out of your depth. You're not going to get a reputable journal >>to publish your paper. Your best bet would be to add some hints about >>cosmic implications and get some newage (ryhmes with 'sewage') rag to >>print it. >>-jcr > > > Well *supposedly* a correct and profound math paper can get published > in a reputable journal which means that the journals I've faced so > far may lose a lot of their luster once the full story comes out. > > As for being out of my depth my little paper so far has drawn comments > and questions from someone as notable as Barry Mazur, though I hope he > doesn't now regret replying to me about it, and has been looked at > without claim of error by at least two of the editors of the New York > Journal of Mathematics, which includes Andrew Granville who referred > me up to the chief editor, and I hope he does't regret that either. > You're wrong is a comment. What are you trying to say? is a question. > Now then it seems to me that a lot of you may think of me as just some > crank but that lot of you don't even register on the screen for a > lot of the people I've already received comments from, as they don't > know you and probably don't care to know you. > > But, of course, you can still think that your opinions about me > matter. As if opinion is mathematics, as I fear that for many of you, > opinion IS mathematics, which is why you don't register and shouldn't > as you don't understand the point. > > Mathematics is about truth, not social opinion. > Then why are you tossing around names like they will change the truth? > > James Harris -- Will Twentyman ==== > Jimmy, > > You're out of your depth. You're not going to get a reputable journal > to publish your paper. Your best bet would be to add some hints about > cosmic implications and get some newage (ryhmes with 'sewage') rag to > print it. > > -jcr He could always try Social Text... ==== James Harris burped forth in > Now then it seems to me that a lot of you may think of me as just some > crank REALLY???? You haven't figured that out YET???? The mere number of people that have commented on your delusions of grandeur hasn't cued you YET to the fact that I, along with numerous other posters, think that you're a CRANK??? GO FIGURE!!! > But, of course, you can still think that your opinions about me > matter. As if opinion is mathematics, as I fear that for many of you, > opinion IS mathematics, which is why you don't register and shouldn't > as you don't understand the point. Mathematics is about truth, not social opinion. > James Harris Sooooo, when people tell you the truth about your crock-of-shit 'mathematics' (notice usage of quotes!), that's not the truth?? The fact that mathematicians, in all these ng's, have consistently corrected your errors, that you have been directed where to get your answers, that even the editor of a German publication has added you to his block list, is NOT the truth, in your opinion, is laughable. As far as finding a periodical to publish your results, have you tried the Weekly World News?? They love 'scandals' like this one, and I'm sure they'll put you right next to the Page 5 Girl. Until then, get a life you self-absorbed kook. ~Bhuvan ==== > Jimmy, > > You're out of your depth. You're not going to get a reputable journal > to publish your paper. Your best bet would be to add some hints about > cosmic implications and get some newage (ryhmes with 'sewage') rag to > print it. > > -jcr Well *supposedly* a correct and profound math paper can get published > in a reputable journal which means that the journals I've faced so > far may lose a lot of their luster once the full story comes out. correct and profount? And what if paper impies non-standard meaning for some standard math terms, like factor and coprime? Should the reputable journal reject the paper on that basis or not? ==== > Jimmy, > > You're out of your depth. You're not going to get a reputable journal > to publish your paper. Your best bet would be to add some hints about > cosmic implications and get some newage (ryhmes with 'sewage') rag to > print it. > > -jcr He could always try Social Text... Alan Sokal called What the Social Text Affair Does and Does Not Prove. http://www.physics.nyu.edu/faculty/sokal/noretta.html At one point he says: <> How fitting to be lead to that quotation from a JSH thread. -- Clive Tooth http://www.clivetooth.dk ==== >> Jimmy, >> >> You're out of your depth. You're not going to get a reputable journal >> to publish your paper. Your best bet would be to add some hints about >> cosmic implications and get some newage (ryhmes with 'sewage') rag to >> print it. >> >> -jcr This one's a keeper: >Well *supposedly* a correct and profound math paper can get published >in a reputable journal which means that the journals I've faced so >far may lose a lot of their luster once the full story comes out. Now that was just standard JSH - no matter _what_ entity rejects his work it reflects more on the entity than on him. A journal declined to publish his paper, that's going to be Bad for the journal when the Truth finally comes out, all perfectly standard stuff. The good part starts here: >As for being out of my depth my little paper so far has drawn comments >and questions from someone as notable as Barry Mazur, Somehow the fact that it's drawn comments from a big shot proves something. Right. >though I hope he >doesn't now regret replying to me about it, and has been looked at >without claim of error by at least two of the editors of the New York >Journal of Mathematics, which includes Andrew Granville who referred >me up to the chief editor, and I hope he does't regret that either. Better yet: When the editor of a journal looks at something and then sends it to a different editor that proves something? Well actually maybe it does, but not what he thinks it does... heh-heh. >Now then it seems to me that a lot of you may think of me as just some >crank but that lot of you don't even register on the screen for a >lot of the people I've already received comments from, as they don't >know you and probably don't care to know you. Have any of these people agreed to _publish_ this paper that you're so proud that they've commented on? Getting someone to _comment on_ something is a big deal. Right. Actually many of us have had editors of journals comment on papers with words like this: Dr. Ullrich: I'm pleased to inform you that your paper [title] has been accepted for publication in [journal]. Let us know when someone comments on your work in words that remind you of that. >But, of course, you can still think that your opinions about me >matter. As if opinion is mathematics, as I fear that for many of you, >opinion IS mathematics, which is why you don't register and shouldn't >as you don't understand the point. Mathematics is about truth, not social opinion. Another goodie, although not as original as the stuff above: Math is about truth, not social opinion, _except_ that the fact that someone's commented on my paper proves something about it - when people are making explicitly disparaging comments is when math is not about social opinion, but when people are making comments that I'm too dense to recognize as polite brushoffs _those_ misunderstood opinions count for something. >James Harris ************************ David C. Ullrich ==== > Alan Sokal called What the Social Text Affair Does and Does Not Prove. > http://www.physics.nyu.edu/faculty/sokal/noretta.html Sokal deserves an award. Maybe the Nobel Prize for humour. Gib ==== >> Jimmy, >> >> You're out of your depth. You're not going to get a reputable journal >> to publish your paper. Your best bet would be to add some hints about >> cosmic implications and get some newage (ryhmes with 'sewage') rag to >> print it. >> >> -jcr Well *supposedly* a correct and profound math paper can get published >in a reputable journal which means that the journals I've faced so >far may lose a lot of their luster once the full story comes out. As for being out of my depth my little paper so far has drawn comments >and questions from someone as notable as Barry Mazur, though I hope he >doesn't now regret replying to me about it, and has been looked at >without claim of error by at least two of the editors of the New York >Journal of Mathematics, which includes Andrew Granville who referred >me up to the chief editor, and I hope he does't regret that either. Now then it seems to me that a lot of you may think of me as just some >crank but that lot of you don't even register on the screen for a >lot of the people I've already received comments from, as they don't >know you and probably don't care to know you. But, of course, you can still think that your opinions about me >matter. As if opinion is mathematics, as I fear that for many of you, >opinion IS mathematics, which is why you don't register and shouldn't >as you don't understand the point. Mathematics is about truth, not social opinion. > Well, you may have gotten letters from some editors, but when there was a period of time when my duties as a graduate assistant was to assist the editor of a journal, selecting which papers to send referrees to, etc. When a stange paper came in which was riduculous on the face of it, it was not sent to a referree, but returned with a non-commital letter. You might have gotten some of those. James Harris (this space unintentially left blank ..... ==== >Message-id: <3c65f87.0307160558.25b5c60a@posting.google.com >> Jimmy, >> >> You're out of your depth. You're not going to get a reputable journal >> to publish your paper. Your best bet would be to add some hints about >> cosmic implications and get some newage (ryhmes with 'sewage') rag to >> print it. >> >> -jcr Well *supposedly* a correct and profound math paper can get published >in a reputable journal which means that the journals I've faced so >far may lose a lot of their luster once the full story comes out. As for being out of my depth my little paper so far has drawn comments >and questions from someone as notable as Barry Mazur, though I hope he >doesn't now regret replying to me about it, and has been looked at >without claim of error by at least two of the editors of the New York >Journal of Mathematics, which includes Andrew Granville who referred >me up to the chief editor, and I hope he does't regret that either. Do you get more crank points when the big names ignore your work than when the little people on sc.math ignore your work? -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm ==== > Well looks like I got bit on the factor of versus factors in common > with controversy as a reply I just received from the chief editor of > a German journal talks as if I'm saying that none of the a's has a > factor of 5, that is 5. He also thought my lemma was a trivial result > and he mentioned polynomial factors. Then he put me on a spam block, > which I discovered when I tried to reply back to him. > > Hmmm...I guess I'll not get much leeway from Germans. > > This journal is eager to publish your work - > > http://www.jir.com/ JIR is still publishing? Excellent. I thought they had gone defunct or merged with some other publication or something. - Randy ==== > > Jimmy, > > You're out of your depth. You're not going to get a reputable journal > to publish your paper. Your best bet would be to add some hints about > cosmic implications and get some newage (ryhmes with 'sewage') rag to > print it. > > -jcr > > Well *supposedly* a correct and profound math paper can get published > in a reputable journal which means that the journals I've faced so > far may lose a lot of their luster once the full story comes out. If things were different, they'd be different. (That's what we call a tautology.) If you ever come up with a 'correct and profound' math paper, then give it another shot. In the meantime, you're just huffing and puffing and wasting editors' time. -jcr ==== > > Well looks like I got bit on the factor of versus factors in common > with controversy as a reply I just received from the chief editor of > a German journal talks as if I'm saying that none of the a's has a > factor of 5, that is 5. He also thought my lemma was a trivial result > and he mentioned polynomial factors. Then he put me on a spam block, > which I discovered when I tried to reply back to him. > > Hmmm...I guess I'll not get much leeway from Germans. > > This journal is eager to publish your work - > > http://www.jir.com/ Umm.. I think they're careful to limit their journal to people who *know* that they're writing jokes. -jcr ==== Mathematics is about truth, not social opinion. > Being an engineer.... I would say not entirely... Mathematics may point to the truth, but it depends entirely on what you consider to be the truth. Consider Stephen Wolfram. Many consider him a bit of a yahoo. I read his A New Kind of Science book. Quite profound. But yet no respect, even though probably in about 100 years people will look at him as well look at Newton. Convincing people is not about saying, Here is the answer you wanted. Read it and learn the truth. It is about convincing. Because sometimes we are wrong... Christian Gross ==== > > >Mathematics is about truth, not social opinion. > Being an engineer.... I would say not entirely... > > Mathematics may point to the truth, but it depends entirely on what > you consider to be the truth. Consider Stephen Wolfram. Many > consider him a bit of a yahoo. I read his A New Kind of Science > book. Quite profound. But yet no respect, even though probably in > about 100 years people will look at him as well look at Newton. It's about logic. A mathematical proof begins with a truth, and then proceeds by logical steps to a conclusion that then must be true. So it has nothing to do with what you consider to be the truth, except at the beginning, where if you screw up then you have GIGO. But in mathematics it's easier to begin with a truth than in other arenas, as for example, my major proof begins with the truth that x^2 + y^2 + vz^2 = x^2 + y^2 + vz^2 which is not debatable. That is, no reasonable human being could ever question such an identity, and no amount of social relativism or philosophical debate changes that reality. That's mathematics. Many people in looking over their own lives and the lives of others see complex situations where truth can be hard to determine, and may forget that part of the point of mathematics is that it's simplified in comparison to our lives. Mathematics is boiled down truths, where ambiguity is removed. It may get complex as you proceed through logical steps but the *beginning* is always an absolute truth, like my identity above, and more often than not, that absolute truth is not one with which rational people would argue. Now as for Stephen Wolfram's book, he's dealing with science, which is different in that it seeks to explain reality. That is, science begins with reality as a base, and comes up with theories to explain and predict. > Convincing people is not about saying, Here is the answer you wanted. > Read it and learn the truth. It is about convincing. Because > sometimes we are wrong... Being wrong is what gives the thrill of the chase for the discoverer. In looking for a proof you have to face down a tremendous amount of failure and yet continue, believing that you will succeed. Throughout history, discoverers face challenges, continue where others give up, yet somehow for reasons that at times don't become clear, they succeed. Today in our society people may find it easy to accept that mathematics is like any other area of human endeavor and that when there is disagreement between two parties the actual truth can be difficult or impossible to find. They may believe that arguments can be rational, where two sides disagree and cannot come to agreement despite the mathematical work being explained in detail. That is impossible. Given a mathematical argument, there is an answer because a math proof begins with a truth, and proceeds by logical steps to a conclusion which then must be true. I've sent the paper to math journals, and I've talked about it in detail on the web. The mathematial argument in that paper begins with a truth, and proceeds by logical steps to a conclusion which then must be true. All any mathematician, or *anyone* who wishes to challenge the paper need do, is break that chain. None can. James Harris ==== [snip] > The mathematial argument in that paper begins with a truth, and > proceeds by logical steps to a conclusion which then must be true. Only if the steps are error-free. Consider this: a = b a^2 = a*b a^2 - b^2 = a*b - b^2 (a+b)(a-b) = b(a-b) a+b = b 2a = a 2 = 1 How would a mathematician (or, indeed, any rational person) interpret the result that 2 = 1? Here's how: Certainly 2 does *not* equal 1. Therefore the argument must contain an error. And, indeed, it does. Similarly, you have concluded that the ring of algebraic integers is incomplete. But there is no such thing as an incomplete ring. Either the algebraic integers form a ring or they do not. Your conclusion is tantamount to claiming that the algebraic integers do *not* form a ring. But it is been proven that the *do* form a ring. Therefore your argument contains an error.Furthermore, if your conclusion leads to the assertion that the algebraic integers do not form a ring, then all your setup step which are based on operations within the ring of algebraic integers re, prima facie, contradictory. > All any mathematician, or *anyone* who wishes to challenge the paper > need do, is break that chain. None can. James Harris I just have! Along with several others. Ignoring or misrepresenting the refutations will not dispose of them -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com <3c65f87.0307160558.25b5c60a@posting.google.com> <3c65f87.0307200519.6a5e941b@posting.google.com> <3F1AA941.A19BF836@ix.netcom.com> ==== [snip] > The mathematial argument in that paper begins with a truth, and >> proceeds by logical steps to a conclusion which then must be true. Only if the steps are error-free. Consider this: a = b > a^2 = a*b > a^2 - b^2 = a*b - b^2 > (a+b)(a-b) = b(a-b) > a+b = b > 2a = a > 2 = 1 How would a mathematician (or, indeed, any rational person) > interpret the result that 2 = 1? Here's how: Certainly 2 does *not* equal 1. Therefore the argument must contain > an error. Apparently, I am not a rational person. I don't reason as you claim above. Either the theory in which the proof occurs is inconsistent or there is an error in the proof (or both). I do not have a priori knowledge of whether the theory is inconsistent or not[1], so I can't see how the mere fact that your proof concludes 2=1 indicates your proof is erroneous. > And, indeed, it does. Indeed it does. But, I ought to check the proof and not merely its conclusion before I claim that the proof has an error. Footnotes: [1] This claim may be controversial. There are some on the newsgroup who have claimed that it is obvious that Peano arithmetic is consistent, and hence that any proof of 2=1 in that theory is fallacious. -- If you see math knowledge as a tool--as a hammer--with which you can attack other people then ... you defeat rational discourse. I get to call my proof the Hammer. It's more powerful than *any* physical object. It is overwhelming force. -- Two JSH quotes <3c65f87.0307160558.25b5c60a@posting.google.com> <3c65f87.0307200519.6a5e941b@posting.google.com> ==== > [snip] > >> The mathematial argument in that paper begins with a truth, and >> proceeds by logical steps to a conclusion which then must be true. > > Only if the steps are error-free. Consider this: > > a = b > a^2 = a*b > a^2 - b^2 = a*b - b^2 > (a+b)(a-b) = b(a-b) > a+b = b > 2a = a > 2 = 1 > > How would a mathematician (or, indeed, any rational person) > interpret the result that 2 = 1? Here's how: > > Certainly 2 does *not* equal 1. Therefore the argument must contain > an error. Apparently, I am not a rational person. I don't reason as you claim above. Either the theory in which the > proof occurs is inconsistent or there is an error in the proof > (or both). I do not have a priori knowledge of whether the theory is > inconsistent or not[1], so I can't see how the mere fact that your > proof concludes 2=1 indicates your proof is erroneous. If you do not believe the theory in which the proof occurs is consistent, which is certainly your prerogative, then you will summarily reject the proof. Either way the conclusion fails. > And, indeed, it does. Indeed it does. But, I ought to check the proof and not merely its > conclusion before I claim that the proof has an error. Footnotes: > [1] This claim may be controversial. There are some on the > newsgroup who have claimed that it is obvious that Peano arithmetic is > consistent, and hence that any proof of 2=1 in that theory is > fallacious. -- > If you see math knowledge as a tool--as a hammer--with which > you can attack other people then ... you defeat rational discourse. > I get to call my proof the Hammer. It's more powerful than *any* > physical object. It is overwhelming force. -- Two JSH quotes -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com