mm-360 === Subject: === Subject: Integer-Alterations On a Grid (fun)> It seems that this game's strategy would rely heavily upon the actual> integer-alteration rules being used for any particular round.> Knowing what you know about the general game, what would be some of> YOUR suggestions for integer-alteration rules which would make for> interesting play and strategy?> ;)Included among the rules can be those which give an output as a resultof anif/then question, such as: m = m+3 if m is a prime; m = m^2 otherwise. But I would discourage such rules which pretty much give the sameoutput no matter what is inputted, such as 'm = 4' or anything lessobviously of one-output. (Even such rules as m = floor(|m|^(1/|m|)),which only outputs 1 unless m is 0, should probably be discouraged immost cases.)Remember:Goal of game = FUN!Leroy Quet> Has anyone played the game?> Leroy Quet> This is a game I made up today.> It is unoriginal, in that, not only do I steal ideas from my own games I > have posted here in the past, but from other people's games also.> But, anyway, this game seems as if it would be SO fun to play that I HAD > to post it!> 2 players.> Played on n-by-n grid (as many of my games are). One player makes up n integer-alteration rules (see below for examples) > for the n rows, one rule per row.> And the other player makes up n rules for the n columns, one rule per > column.> Rule: a (meta)mathematical rule which changes any input integer into an > output integer.> At the game's beginning, after the rules are created, each player gets 3 > random integers, from 1 to n.> A player's first 2 random integer give the coordinate of the square the > player starts in, the last integer gives the player's starting integer.> And what player moves first is then chosen randomly.> (But no more randomnes is in the game once play begins.)> At each turn, a player can choose to apply the row-rule OR the > column-rule to his/her integer. (Each player has their own integer which > is altered.)> the square he/she is currently at.> The player can then choose to move either up, down, left, or right, > without regard to what integers (s)he must jump over, that number (m) > of grid-squares.> And the grid is considered topologically a torus, ie. going off the right > edge, for example, brings you back to the left side> (and the same (mod n)-topology for right-to-left or vertical moves).> (So, for example, a move from column 3 to the right 7 squares, on an n = > 6 grid, gets the player to column 2, which is one square to the *left* of > where (s)he started.)> But a player must only move to a square *without* an integer (written > there by either player) already in it.> And the game's winner is the last player able to move.> Notes:> *Be Creative in coming up with rules!> *The output integers' signs are, in practice, insignificant, since you > can move right OR left, up OR down. But an output of 0 is to be avoided!> *Start out with easy rules as you first learn how to play this game, or > if you simply are poor at math. It would be less fun to play this game if > you must spend an hour calculating each of your integers!> Example of rules:> (For n=6)> Rows:> A) m = m+1;> B) m = 2*m;> C) m = m + sum of distinct primes dividing all already-written > neighboring integers;> D) m = d(|m|)*phi(|m|) -m; (d(m) is number-of-divisors function, phi(m) > is Euler-phi function.)> E) m = (numerator)-(denominator) of reduced rational,> sum{k=1 to |m|} 1/k;> F) m = m + number of already-written integers in row F.> Columns:> 1) m = ceiling(sqrt(|m|))> 2) m = ceiling(secant(m)), m in radians;> 3) m = number of primes <= |m|;> 4) m = 2^|m| - m;> 5) m = number of divisors in product of all aready-written integers in > column 1;> 6) m = ceiling((|m|th Fibonacci number)/(|m|_th prime)).> (Yes, computers might be needed for some kinds of rules, and are allowed.)> And a grid to aid visualization:> ------------------> A! ! ! ! ! ! !> ------------------> B! ! ! ! ! ! !> ------------------> C! ! ! ! ! ! !> ------------------> D! ! ! ! ! ! !> ------------------> E! ! ! ! ! ! !> ------------------> F! ! ! ! ! ! !> ------------------> 1 2 3 4 5 6> Leroy QuetSubject: === Subject: zeta(r)^zeta(r)[I am replying here to the other reply of mine in this thread.]> Let a(1) = 1;> Let, for m >= 2,> a(m) = > (1/ln(m)) sum{p=primes} sum{k|m,k>=2} a(m/k) ln(k) H(c(p,k)),> where H(n) = 1+1/2+1/3+...+1/n, the n_th harmonic number,> and c(p,k) is a nonnegative integer where> p^c(p,k) is the highest power of the prime p which divides k. (And, oh yeah, H(0) =0.)> So, we have then:> sum{m=1 to oo} a(m)/m^r => zeta(r)^zeta(r),> unless I made a mistake.> (My math was not rigorous.)> But what I am wondering is,> what is a closed-form (non-recursive definition) for> a(m) ??>I might as well give the first few terms of this sequence:>a(m): 1, 1, 1, 2, 1, 3, 1, 7/2, 2,...>Is suspect that each term is a rational which depends only on the >exponents in the prime-factorization of that term's index, but I am not >absolutely certain.>If, for example, b(k) = a(p^k), p = prime, then:>sum{k=0 to oo} x^k b(k) =>(1/(1-x))^(1/(1-x)),>I believe.>And b(k)(k-1)! forms the sequence:>1, 2, 7, 35,...I gave the b-sequence from k = 1, of course.And b(k)*k! is:http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?Anum=A087761 >(Is this sequence, or anything related to the a-sequence, in the EIS??)Leroy QuetSubject: Bondy & Murty's book available on-line (was: === Subject: Eulerian path in infinite graph)> Just out of curiosity, where could I find more information about > infinite graphs? The material on infinite graphs covered in Wilson's > book is just approximately 5 pages long and the proofs could> certainly be improved (e.g., the proof of K.9anig's Infinity Lemma> presented there is not as careful as I like to read).While I am still interested in an (easy) reference on infinite graphs(or perhaps a survey), I did find something that left me quite happy.Prof. Adrian Bondy has put his *entire* book available on-line on hishomepage, which is something quite helpful, since the book seems to belong out-of-print and not all libraries have this book available(especially depending on where you live).His personal homepage is:. Hope this helps others in the newsgroups, Rog.8erio Brito.-- =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- =-=-=-= Rog.8erio Brito - rbrito@ime.usp.br - http://www.ime.usp.br/~rbrito=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=Subject: TI-89 questionSorry if this is the wrong group for this.On a TI-89, how do you input in parametric mode sin^2(theta)??When I use Mode>Parametric, it sets the F1 editor to xt1 and yt1. How doI get it to x theta 1 and y theta 1??StacySubject: === Subject: TI-89 question> Sorry if this is the wrong group for this.> On a TI-89, how do you input in parametric modesin^2(theta)??> When I use Mode>Parametric, it sets the F1 editor to xt1and yt1. How do> I get it to x theta 1 and y theta 1??******************************************************Why not use t instead of theta? You don't say whatparametric equations you are trying to graph, but if youwere trying to graph y = sin^2(x) parametrically, you coulduse xt1 = t and yt1 = sin^2(t). If you were trying tograph the polar coordinate equation r = sin^2(theta)parametrically, you could use xt1 = cos(t)sin^2(t) andyt1 = sin(t)sin^2(t)._______________________________________________ __________Eric J. Wingler (wingler@math.ysu.edu)Dept. of Mathematics and StatisticsYoungstown State UniversityOne University PlazaYoungstown, OH 44555-0001330-941-1817Subject: === Subject: area of a circle> Ignacio Larrosa Caestro>>Keith A. Lewis>given a circle with 2 chords cutting perpendicularly and then>bisecting the four angles with a 45 degree angle..thus giving 8>regions in a circle. turns out that the alternate regions areas sum>equals the other four regions areas. is there a nice proof for this.>seems like i should have learned this in my college geometry class/>thanks in advance.>>The integrals are messy, but cancel trivially ...>>Is there a elementary, better euclidean, proof?> I drew four lines through a point P, producing 8 equal angles, and then> a circle surrounding P. If we _rotate_ all the lines by dTheta around> P, one of the sum-areas changes by> (AA - BB + CC - ... - HH) dTheta/2> (where A,B,...,H are lengths of segments from P to the circle)>I think you mean here AA = A*A = A^2 where A = length of PA, one of the >two parts of chord AE.> which is zero by Euclid III.35 about two intersecting chords of a circle.>???? _ANY_ chords ???>Not shure... the proof should be detailed a little more...>If you draw 3 chords AB, CD, EF intersecting at 6 equal angles pi/3>with segments PA=a, PB=b...PF=f>(vertices A,C,E,B,D,F in that order on the circle) you get :>dArea = (a^2 - c^2 + e^2 - b^2 + d^2 - f^2) dTheta/2>IMHO not zro.>This is OK here because you have _eight_ equal angles, so the chords are >rectangular by pairs.>The hint is that for two _RECTANGULAR_ chords, with length>AB=PA+PB=a+b and CD=PC+PD=c+d>a^2 + b^2 + c^2 + d^2 = 4*R^2>(easy with some pythagoras with the center and the middles of chords)I came up with a pretty short proof based on the fact that when you havetwo intersecting chords, opposing arcs on the circle sum to twice theangle between the chords. Since the chords are perpindicular, we havethat the opposing arcs must be supplementary. Sliding these arcs andchords together, we see that the chords form a right triangle with thediameter of the circle as the hypotenuse. See http://www.whim.org/nebula/math/images/perpchord.gif>If you take two other _RECTANGULAR_ chords offset from the first set by >angle alpha, the sum of one over two regions satisfies :>dArea=(a^2 - a'^2 + b^2 - b'^2 + c^2 - c'^2 + d^2 - d'^2)*dTheta/2 = 0>because>a^2 + b^2 + c^2 + d^2 = a'^2 + b'^2 + c'^2 + d'^2 (= 4*R^2)>So this sum is constant when you rotate the whole pattern.>Of course you choose alpha = pi/4 to get equal areas by symetry.Rob Johnson I came up with a pretty short proof based on the fact that when you have> two intersecting chords, opposing arcs on the circle sum to twice the> angle between the chords. Since the chords are perpindicular, we have> that the opposing arcs must be supplementary. Sliding these arcs and> chords together, we see that the chords form a right triangle with the> diameter of the circle as the hypotenuse. See> http://www.whim.org/nebula/math/images/perpchord.gif[...]Fine !I had forgotten this general property of two intersecting chords.This improves the Pizza Theorem :Not only the areas are equal, but also the portions of the edge (where usually there is no toping...).-- philippe(chephip at free dot fr)Subject: === Subject: area of a circleEn el mensaje:404F0FEA.8010603@free.invalid,philippe 92 escribi.97:> [...] [Pizza theorem]> I came up with a pretty short proof based on the fact that when you> have two intersecting chords, opposing arcs on the circle sum to> twice the angle between the chords. Since the chords are> perpindicular, we have that the opposing arcs must be supplementary.> Sliding these arcs and chords together, we see that the chords form> a right triangle with the diameter of the circle as the hypotenuse.> See> http://www.whim.org/nebula/math/images/perpchord.gif> [...]> Fine !> I had forgotten this general property of two intersecting chords.> This improves the Pizza Theorem :> Not only the areas are equal, but also the portions of the edge (where> usually there is no toping...).Certainly, Rob Johnson's proof is much more easy and pretty.That didn't seems true for me is that is said in MathWorld:If a circular pizza is divided into 8, 12, 16, ...slices by making cuts atequal angles from an arbitrary point, then the sums of the areas ofalternate slices are equal.(http://mathworld.wolfram.com/PizzaTheorem.html)If the number n of slices is 4 multiple, but not 8 multiple, it is nottrue, or at least, it isn't supported by the mentionated proofs. If n = 12,by example, you can get 3 bunchs of 4 slices with equal sum of areas, butnot 2 bunch of six alternate slices verifying that.your first message,http://www.maths.unsw.edu.au/~mikeh/webpapers/paper57. pdf.(the 'crust' in the corollary)-- Best regards,Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.comSubject: === Subject: Points on a Line, InfinityDoes the acceptance of the Axiom of Choice imply the existence of amapping from N to R? If so does the acceptance of the Axiom of Choiceimply mappings between any two infinite sets?I posit that it does, the AC axiom. AC or Zorn's Lemma is often used to forestall certainimpredicabilities of considerations of orderings of the reals, a setcomprised of each real number.I think there are bijections from N to R of specific forms, and aswell interjections (in-, sur-, and bijections) among any two infinitesets, where an ordering can be distinguished upon any infinite setthat is not order-sensitive where an ordering makes it finite.I agree that the Cantor's (or rather, Cantorian as they have seendevelopment since his time) proofs of the inexistence of mappings fromset to powerset are quite strong results. Models without thoseresults or with constructions resolving those obstructions are notnecessarily inconsistent, variously with extensions to implicitinfinite elements of an infinite set, re-ordered function composition(deferral), or of the declaration of the set of all sets as its ownpowerset and extension of that statement.There is some utility in the consideration of qualities and quantitieswitihn infinite sets based upon their existence as numbers and thenumber-theortic properties of those numbers with respect to well-knownsets of numbers of which they are supersets and subsets. For example,the asymptotic density of the even integers within the integers is onehalf, selecting probabilistically with the set of integers as thesample sample oe value of the integers yields a probability that it isa multiple of two of one half.That is extended to the infinitesimal probabilistic case in terms ofitems being definite and indefinite in infinitistic scales relative toeach other, definitely for example in the case of a random integer (orreal, or number dense) being in a finitely sized interval or of ascalar multiple of that interval, and that indefinitely to the numberof primes or multiples of two, each an infinite set and indefinite,yet qualifiable, in terms of their relation to each other, and eachdefined in terms of their perspacity.It's possible to consider the existence of a set that has the samedensity of the prime numbers in the integers, but is disjoint theprime numbers, for example prime p plus five, or even of a completelydifferent construction as the prime numbers, yet still having thedensity of the primes (near zero) or even of the composites (nearone). The density of multiples of two, specifically in theintegers, is exactly the same as the density of the multiples offour unioned with the multiples of four, plus one.In terms of points on a line, the idea is to find ways to identifiycharacteristics of points on a line as they relate to the continuousline. Considering functions defined on one point on a line orregularly spaced points on a line, e.g. the integers, where a resultcan be taken in terms of a continuous region of points on a line inthe same coordinate system of a vector space offers the potential of abridge between quantum and continuous observation and analysis, andwhere in the analog continuous quantities are useful in engineeringand physics, the possibility of mathematical practice that providesthe correct answer across the infinite scales of the amount of allpoints on a line or separately of a segment of a line, and back, maybe a good thing.How many of the points on a unit line segment in the binary expansionof their value have equal numbers of one and zeros? Combinatoricsimplies near zero, measure theory near all, and I say half: half havehalf.Please discuss the tautologies with multiple meanings. Explain whythe empty set is not an infinite set.Ross F.Subject: Rolling a Dice for 1000 times. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2A2n6J16796;I have a question about rolling a dice. lets say I rolled a (Normal) dice, and I got 6 for the first time.rolled again and i got 6 again .. Obviously if i roll it again the probabilities of getting a 6 again is 1/6.Ok what if I roll this dice for 10 times or 100 times and i get nothing but 6? Do you think is something specious? Mathematically I can still roll the dice and get a 6 and the probability is 1/6, but why do I think something must be wrong, or the chance of having 6 is higher than the other numbers ?Subject: === Subject: Rolling a Dice for 1000 times.Behnam Mansouri wibbled:> I have a question about rolling a dice. > lets say I rolled a (Normal) dice, and I got 6 for the first time.> rolled again and i got 6 again .. Obviously if i roll it again the probabilities of getting a 6 again is 1/6.> Ok what if I roll this dice for 10 times or 100 times and i get nothing but 6? Do you think is something specious? > Mathematically I can still roll the dice and get a 6 and the probability is 1/6, but why do I think something must be wrong, or the chance of having 6 is higher than the other numbers ?Well, you've defined it as a normal die, so by your own definition there's a 1/6 chance of rolling any particular number at each roll.If you list every possible series of 1000 rolls, which of them is most likey? None - they're all the same. All 6s is no more or less likely than any other _particular_ series.But, if you hadn't defined it as a normal die, then you might approach your experiment as a test of the die's fairness, and conclude that there is a high probability that it isn't.Subject: === Subject: Rolling a Dice for 1000 times.In sci.math, Behnam Mansouri I have a question about rolling a dice. > lets say I rolled a (Normal) dice, and I got 6 for the first time.> rolled again and i got 6 again .. Obviously if i roll it again the> probabilities of getting a 6 again is 1/6.> Ok what if I roll this dice for 10 times or 100 times and i get> nothing but 6? Do you think is something specious? > Mathematically I can still roll the dice and get a 6 and the> probability is 1/6, but why do I think something must be wrong,> or the chance of having 6 is higher than the other numbers ?The probability of rolling a die and having it come up 6100 times in a row is 1/6^100, or about 1.53 * 10^-78.That's about the reciprocal of the number of atoms in theUniverse; if each of them were a die, and each of them wererolled 100 times simultaneously, then one of them -- somewhere --might have shown 100 6's in a row. (The actual probabilityis not certainty, as it turns out. It's about 1/e.)So if you do have a die that rolls 100 times in a row andcomes up 6, chances are it's loaded.-- #191, ewill3@earthlink.netIt's still legal to go .sigless.Subject: === Subject: Rolling a Dice for 1000 times.If you walk over and place the die on the table with the 6 up 100times, it's still a fair die. If you are a skilled dice setter inthe casino game of craps, and you set the straight sixes combinationand get an inordinate number of horn numbers (2,4 11, 12), the diceare still fair.Probability math presumes a probability universe and is not likely tobe correct in a physical universe with limits and conditions thatviolate the supposed mathematical nature of a probability universe. For example, a shuffle process is analyzed with probability math asallowing any combination, but no physical object that is beingshuffled can magically disappear and reappear somewhere else withoutbeing limited in movement by other physical elements involved in theshuffle process.A die has a set physical limit of where each number can appear, andthus physical conditions could cause that die to appear with the 6 upan inordinate number of times, and no amount of 'math' could make thatdie unfair.Very Respectfully,Ray Donald PrattSubject: === Subject: Rolling a Dice for 1000 times.> If you walk over and place the die on the table with the 6 up 100> times, it's still a fair die. If you are a skilled dice setter in> the casino game of craps, and you set the straight sixes combination> and get an inordinate number of horn numbers (2,4 11, 12), the dice> are still fair.> Probability math presumes a probability universe and is not likely to> be correct in a physical universe with limits and conditions that> violate the supposed mathematical nature of a probability universe. > For example, a shuffle process is analyzed with probability math as> allowing any combination, but no physical object that is being> shuffled can magically disappear and reappear somewhere else without> being limited in movement by other physical elements involved in the> shuffle process.> A die has a set physical limit of where each number can appear, and> thus physical conditions could cause that die to appear with the 6 up> an inordinate number of times, and no amount of 'math' could make that> die unfair. We know, since just a little bit of math called Casino Chips makes the die unfair. Hence ALL dice are loaded. > Very Respectfully,> Ray Donald PrattSubject: === Subject: Rolling a Dice for 1000 times.> I have a question about rolling a dice. It used to be that one rolled a die, and that dice was reserved strictly for the plural of die.> lets say I rolled a (Normal) dice, and I got 6 for the first time.> rolled again and i got 6 again .. Obviously if i roll it again the > probabilities of getting a 6 again is 1/6.> Ok what if I roll this dice for 10 times or 100 times and i get nothing but > 6? Do you think is something specious?The more times in succession that you roll only sixes, the more you might suspect that either the die or the rolling is not fair. However, no number of such rolls guarantees unfairness, since it could still occur by chance when fairly done.> Mathematically I can still roll the dice and get a 6 and the probability is > 1/6, but why do I think something must be wrong, or the chance of having 6 is > higher than the other numbers ?Subject: === Subject: Rolling a Dice for 1000 times.> I have a question about rolling a dice.> It used to be that one rolled a die, and that dice was reserved> strictly for the plural of die.> lets say I rolled a (Normal) dice, and I got 6 for the first time.> rolled again and i got 6 again .. Obviously if i roll it again the> probabilities of getting a 6 again is 1/6.> Ok what if I roll this dice for 10 times or 100 times and i get nothingbut> 6? Do you think is something specious?> The more times in succession that you roll only sixes, the more you> might suspect that either the die or the rolling is not fair.> However, no number of such rolls guarantees unfairness, since it could> still occur by chance when fairly done.Yes, indeed, although the probability becomes increasingly small.In the case of the O.P. a statistical approach could start by the hypothesisthat the die is fair. If a particular sequence of throws has a very smallprobability, say less than 1 %, given this hypothesis, then one rejects thehypothesis and concludes that the die is probably unfair. In this case,throwing three sixes out of three throws has a probability of about 0,5 %.So take any die and roll it exactly three times. If you get three sixes,throw the die away.If you repeat this experiment with 200 different dice, then there will beapproximately one false positive, i.e. a perfectly straight die that justhappened to roll some unlucky numbers.So you should ask yourself, how many false positives can you live with?-Michael.Subject: === Subject: Rolling a Dice for 1000 times.Michael Jrgensen wibbled:> If you repeat this experiment with 200 different dice, then there will be> approximately one false positive, i.e. a perfectly straight die that just> happened to roll some unlucky numbers.When I play Waddington's Formula 1, I typically find that one player rolls 3 or 11 (and thus spins off) much more often than the others.It looks weird, but it's just chance.Subject: === Subject: Rolling a Dice for 1000 times.>>lets say I rolled a (Normal) dice, and I got 6 for the first time.>>rolled again and i got 6 again .. Obviously if i roll it again the>>probabilities of getting a 6 again is 1/6.>>Ok what if I roll this dice for 10 times or 100 times and i get nothing>but>>6? Do you think is something specious?>The more times in succession that you roll only sixes, the more you>might suspect that either the die or the rolling is not fair.>In the case of the O.P. a statistical approach could start by the hypothesis>that the die is fair. If a particular sequence of throws has a very small>probability, say less than 1 %, given this hypothesis, then one rejects the>hypothesis and concludes that the die is probably unfair. In this case,>throwing three sixes out of three throws has a probability of about 0,5 %.>So take any die and roll it exactly three times. If you get three sixes,>throw the die away.>If you repeat this experiment with 200 different dice, then there will be>approximately one false positive, i.e. a perfectly straight die that just>happened to roll some unlucky numbers.>So you should ask yourself, how many false positives can you live with?> [Quote edited]The Bayesian approach models the OP's dilemma more closely. If the experimenter is absolutely certain that the original dice is fair (i.e., his prior on the die distribution is a point mass on the unbiased one), then the experimenter should have no reason to doubt it, even after many sixes in a row. On the other hand, if s/he leaves room for doubt berfore even starting the experiment (by a less concentrated prior), then s/he can determine the probability of the die's being tolerably close enough to fair via the posterior distribution.-- Stephen J. Herschkorn herschko@rutcor.rutgers.eduSubject: === Subject: Rolling a Dice for 1000 times.>lets say I rolled a (Normal) dice, and I got 6 for the first time.>rolled again and i got 6 again .. Obviously if i roll it again the>probabilities of getting a 6 again is 1/6.>Ok what if I roll this dice for 10 times or 100 times and i get nothing>but 6? Do you think is something specious?>The more times in succession that you roll only sixes, the more you>>might suspect that either the die or the rolling is not fair.>In the case of the O.P. a statistical approach could start by the hypothesis>that the die is fair. If a particular sequence of throws has a very small>probability, say less than 1 %, given this hypothesis, then one rejects the>hypothesis and concludes that the die is probably unfair. In this case,>throwing three sixes out of three throws has a probability of about 0,5 %.>So take any die and roll it exactly three times. If you get three sixes,>throw the die away.>If you repeat this experiment with 200 different dice, then there will be>approximately one false positive, i.e. a perfectly straight die that just>happened to roll some unlucky numbers.>So you should ask yourself, how many false positives can you live with?>[Quote edited]>The Bayesian approach models the OP's dilemma more closely. If the >experimenter is absolutely certain that the original dice is fair (i.e., >his prior on the die distribution is a point mass on the unbiased one), >then the experimenter should have no reason to doubt it, even after many >sixes in a row. On the other hand, if s/he leaves room for doubt >berfore even starting the experiment (by a less concentrated prior), >then s/he can determine the probability of the die's being tolerably >close enough to fair via the posterior distribution.For example...Suppose we are given ahead of time that the probability our die rollsonly 6 is p and the probability that our die is fair is 1-p. If we rolla 6 n times in a row, then the probability that our die is unfair is p u(p,n) = ------------- [1] p+(1-p)6^{-n}Thus, the probability that the next roll will be a 6 is u(p,n) + (1-u(p,n))/6 6p+(1-p)6^{-n} = ---------------- [2] 6(p+(1-p)6^{-n})If p = 0 (we are certain the die is fair), we get that the probabilitythat the next roll will be 6 is 1/6; whereas, if p = 1, (we are certainthe die is not fair), the probability is 1. Furthermore, for any p<>0,the probability that the next roll will be a 6 tends to 1 as n goes toinfinity.Therefore, depending on how certain we are that we have a fair die tostart with and how many consecutive rolls of 6 we have, the probabilitythat the next roll will be a 6 is between 1/6 and 1.Rob Johnson Therefore, depending on how certain we are that we have a fair die to> start with and how many consecutive rolls of 6 we have, the probability> that the next roll will be a 6 is between 1/6 and 1.How does the die know the state of your mind?Subject: === Subject: Rolling a Dice for 1000 times.> Therefore, depending on how certain we are that we have a fair die to> start with and how many consecutive rolls of 6 we have, the probability> that the next roll will be a 6 is between 1/6 and 1.>How does the die know the state of your mind?It doesn't. Depending on the nature of the die, it has a probability ofeither 1/6 or 1 of showing 6 next roll. What I should have said, forthe more literal readers, is that our _estimate_ of the probability thatthe next roll will be a 6 is between 1/6 and 1. Since we don't know thenature of the die, only its past behavior and the apriori probabilitythat it is fair, the best we can do is estimate the probability that thenext roll is a 6.Rob Johnson > Therefore, depending on how certain we are that we have a fair die to>> start with and how many consecutive rolls of 6 we have, the probability>> that the next roll will be a 6 is between 1/6 and 1.>How does the die know the state of your mind?> It doesn't. Depending on the nature of the die, it has a probability of> either 1/6 or 1 of showing 6 next roll. What I should have said, for> the more literal readers, is that our _estimate_ of the probability that> the next roll will be a 6 is between 1/6 and 1. Since we don't know the> nature of the die, only its past behavior and the apriori probability> that it is fair, the best we can do is estimate the probability that the> next roll is a 6.You must also recognize the Gambler's estimate: The probability ofrolling a 6 decreases with each successive consecutive 6.Subject: === Subject: convex problem.....>i know convex function.>example....f(x) = x^2 form....>f(x) = -x^2 is not convex.>because def of convex is>f((1-t)x + ty) <= (1-t)f(x) + t*f(y) , 0<= t <=1>and i know that f is convex <=> f''(x) >=0... assuming f is twice differentiable everywhere.>but...i think.......> /> / > / /> / />/ />this is graph...You can't always tell from ASCII art, but it doesn't lookvery differentiable at the corners.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2Subject: === Subject: Randomness is a notion.> Hi Servo,> You asked what the word ' random ' means.> First and foremost, randomness is a notion.> And the notion is this: Insufficient information.> For example:> Otherwise predicable weather is labeled as random > when one's information about it is insufficient.What is really interesting is that Chaos Theory was actually thebrainchild of a frustrated weatherman!Double-ASubject: === Subject: Randomness is a notion.[...]I won't entertain the notion notion again, and randomnessas insufficient information is already being debated inanother thread.However:> Because science is forever discovering more regularities> in nature, it's best to assume that> the future is perfectly fixed.> ( Even though the future is imperfectly known )Hawking apparently shared this view for a time, and even heldthat in the big crunch everything would just stop and thenreverse all events which had happenened in our universe sincethe Big Bang.Smashed dishes would come together whole again. Apples wouldfly from the ground and attach themselves to trees.The scattered molecules of Socrates would swoop from aroundthe world to eventually converge in Greece. He would awakenfrom death, sit up on his bed, and hemlock would spew fromhis mouth into a cup. The tears of his friends would run uptheir faces and into their tear ducts.I'm not sure he still holds this view, since changing beliefs onthe effects of expansion-contraction on entropy, and he seemsuncertain of time-symmetry issues. It would certainly be anextreme-determinist view.I believe there are so many reasons to reject this view, andso few to recommend it, that it incurs a huge burden of proof.One of the more obvious questions: What would make the universecontract if gravity were to start working in reverse?> This sticks in people's craws for the same> reason that evolution is so repulsive.> i.e. People would rather exploit unknowns> to confabulate delusion of grandeur.Believing in a universe in which we can actually make choices,where our actions and decisions can actually become new causes,and not in a deterministic 'everything-was-fated in an explosionof hot junk realm, is not grandeur. It boils down to: Are we whowe think we are? Are our lives and choices illusion, delusion, orare they as they seem? Do we have free will?Determinists are faced with the same dilemma as theists: Ifeverything had to be caused, what caused the causes? If thereare first causes which are exempted, why can't there besecond causes and third causes and so on?Assuming the future is perfectly fixed is without foundationin physical science. I believe the persistence of this idea hasmore to do with psychology--what better way to avoid responsibilityfor our own lives that to claim they were determined?Determinism is the ultimate claim to powerlessness and victimhood.The universe made me this way, goes the loudest possible whine,against the greatest bully of all, so don't hold me responsible!I'll pass. I'm assuming that we can shape our future, and I believethe evidence is on my side!If I'm wrong, well, then I couldn't help it, could I? The universemade me think this way.ServoSubject: Delusions of grandeur.Hi Servo, You erred in saying, Determinism is the ultimate claim to powerlessness and victimhood. .Can't give up your delusions of grandeur eh ?Can't stop taking credit for the work of others ?You're all too human, You reject physicalism for the same reason evolution gets rejected, you can't let go of your delusions of grandeur.You erred again by saying, Assuming ' the future is perfectly fixed ' is without foundation in physical science. .If that's not the best assumption then: Why do Hawking and Einstein think it is ? What do you think the best assumption is and why ?On this very topic, Hawking once said, Yet many people believe that science should be concerned only with the local laws which govern how the universe evolves in time. They would feel that the boundary conditions for the universe that determine how the universe began were a question for metaphysics or religion rather than science ... The only way to have a scientific theory is if the laws of physics hold everywhere including at the beginning of the universe. ... The Boundary Condition Of The Universe Is That It Has No Boundary. ... Because we can not see the whole of spacetime on account of black hole and cosmological event horizons, our observations are described by an ensemble of quantum states rather than by a single state. This introduces an extra level of unpredictability but it may also be why the universe appears classical. This would rescue Schroedinger's cat from being half alive and half dead. .You mentioned, What would make the universe contract if gravity were to start working in reverse ? .Wait a minute, What does Hawking's trust in physicalism ( i.e. the assumption that no place in nature is special ) have to do with the now discredited idea that the space-time in our universe is negatively curved ?Hawking goes with the observed data, and that data tells us that space-time has been undergoing a constant acceleration for the last 11 billion years or so. Our universe has always been cooling. The topology of space-time is hyperbolic, open. Yet space ( not space-time ) is perfectly flat. i.e. Every point in space finds itself at the center of the cosmos, similar to how every point on the surface of a balloon is the center. This strongly suggest that heat is a fifth dimension. Where the notion of heat is similar to the notion of insufficient information, i.e. it's prejudiced by our preferred scale.Subject: === Subject: Delusions of grandeur.> Hi Servo, You erred in saying,> Determinism is the ultimate claim> to powerlessness and victimhood. .> Can't give up your delusions of grandeur eh ?Answered.If you feel everything in the universe is determined, whyare you posting here? The thoughts of others are ordained,in this view.Is the universe making you do it?> Can't stop taking credit for the work of others ?Now you're just being a punk.[...]> This strongly suggest that heat is a fifth dimension.For some reason, I don't feel like debating with thinkingof this caliber.ServoSubject: Yet another jailhouse game.Hi Servo,My delusions of grandeur question was rhetorical, not a jibe directed at you personally. === Subject: How cosmic heat could be a fifth spatial dimension,You replied, For some reason, I don't feel like debating with thinking of this caliber. .Then call it Energy density , or The ' size ' of space-time .I call it heat because heat is related to randomness, i.e. degrees of coherence ... which is subjective. One must change his perspective to enter a singularity.You asked, If you feel everything in the universe is determined, why are you posting here ? Is the universe making you do it ? .We observe regularities and randomness, this give us choices . Only the more trivial the choice is, the more it looks like freedom to us. Usenet is just one of those trivial choices. Yet another jailhouse game.Subject: === Subject: Random pi digits and Mahler's theorem> ...................>But just how unlikely is this very same behavior with >a corresponding sequence of iid random variables? >In other words ...>If X is uniform on (0,1), then what is the value of P= >pr( for all integers p,q > 1, |3+X - p/q| > 1/q^42 )?> P < sum 2/q^42 < .5/10^12Herman, I think you did the wrong inequality: that would bea bound on pr{ for some integers p,q > 1, |3+X - p/q| < 1/q^42 }.So r.e.s.'s P > 1 - .5/10^12.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2Subject: === Subject: Integral of cyclotomic polynomial>If I enter the function>f(x) = (x^101 - 1)/(x-1) into Wolfram's Integrator, I get>x^101/101 + x^100/100 + X^99/99 + ... + x^2/2 + x>which is correct, being the term by term integration of>x^100 + x^99 + ... + x^2 + x + 1>but this is quite unwieldly. Is there a simpler expression?Maple would call it -ln(1-x) - LerchPhi(x,1,102) x^102.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2Subject: Math and Parallel UniversesI originally posted this alt.parallel.universes where they appear to beequation phobic; that is the explanation for the Warning, you will seesome equations!.You know all those wonderful theories of 20th century science (stringtheory, relativity, quantum mechanics). They all used math as a basisfor their existance, so until Parallel universes start getting some mathin them, we are just wasting our time. On that note, I will show you amath based model of Parallel universes. Warning, you will see someequations!pi/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ...This equation calculates the value of Pi (3.14159...), it is calledLiebniz' equation. There are many equations that calculate pi. Thisone is unique because it uses only numbers and four signs (+,-,=, and/). Other equations that calculate pi use arctangents or square rootsor other things. I point this out to you because real soon you aregoing to be questioning me about it.Now pi is the ratio between a circles circumference and its diameter.In Non-euclidian geometery, you can create geometeries where pi isgreater then 3.14159.. (Hyperbolic Geometery), and geometeries where piis less then 3.14159... (Elliptic Geometery), but all these geometerieshave a flaw where the smaller the space you measure the closer pi willget too 3.14159... Now if you have seen it I just presented you with aparadox. The paradox is, how can you be in a (let's say) a Hyperbolicgeometery and calculate twodifferent values of pi at the same time? One by the space you are inand the other by math. Where did math separate from space? Liebniz'equation above always calculates the same value of pi no matter whatspace you are in.So let us say we are in a space where the value of pi is different from3.14159..., and we want this space to also be shown in Liebnizequation. How would we go about doing this? I ask the question becauseI am going to give you an answer, an answer that involves Paralleluniverses. You will debate whether my answer is the correct one andsome of you may even come up with answers that don't involve paralleluniverses. Best of luck!I noted above that Liebniz' equation uses only numbers and four signs(+,-,=, and /). We can move the 4 to the other side of the equals signand get,pi = 4(1 - 1/3 + 1/5 - 1/7 + 1/9 - ...).As you look at this equation, you will see that the equation has aform to it. The odd numbers in the denominatior, the infinite numberof fractions, the alternating between plus signs and minus signs. Nowyou need to ask yourself, Do these same features also occur in aparallel universe? Does the form of the equation remain in paralleluniverses. We can answer that question in this way. Liebniz formulatedthe equation in the 18th century, so there have been 200+ years ofuniverses splitting off of our universe in those years and all of thoseuniverses will have the same form iof Liebniz' equation, but all ofthem will have different values of pi then ours. That still doesn'tanswer our main question as to what changes. So let us rewrite theequation with what we suspect doesn't change and see what we get,pi = 4 units (one unit - unit/3units + unit/5units - unit/7units + ...).We see that if we change the value of one(unit) we will arrive at adifferent value of pi. This shows that Liebniz' equation shows therelationship between pi and the natural numbers. There are other typesof numbers; imaginary, complex, irrational, etc. All I am proposing isthat each parallel universe has its own separate natural numbers,complex numbers, irrational numbers, imaginary numbers, etc.To read the paper that the above info is based on, go tohttp://www.efn.org/~janieg/para61.html. Warning, there will beequations in the paper.Email me and tell me what you think picrosser@yahoo.com.Jim AkerlundSubject: === Subject: Math and Parallel UniversesThis is heavily mistaken. Your main problem is in assuming thatmathematics is driven by physics; it is not. Mathematics isessentially the study of unreal things, and the reason physicsts useit is that certain real things look a lot like unreal things.The value of pi in Euclidean geometry is the value of pi that ariseswhen we accept certain axioms, namely Euclid's axioms of geometry. When we accept different axioms, it is possible to get other values ofpi, or even for pi to be a non-constant number (i.e. C/d increases ordecreases as the size of the circle changes.)Things in the real world do not have shapes, at least not in the samesense that geometric objects do. For example, a clock is not acircle, because a circle is a collection of points equidistant from agiven point, called the center. There are no such things as pointsin the real world, so there are no such things as circles in the realworld. However, we do have things in the real world that behave sortof like circles, in that the equations for the area and circumferenceof a circle are closely related to the vague real-world concepts wewould call the circumference of a clock (which basically means thenumber which we would read if we put a cloth ruler around the clock)and the area of a clock (by which we basically mean that if we coverit with paper and then cut the paper up to make something resembling asquare, the length of a side times itself).This is a very important concept to grasp -- supposing that tomorrow,gravity stopped working, and that all of a sudden objects which lookedlike squares all of a sudden changed to look like circles and viceversa, and the rest of the laws of physics went away, it would notchange mathematics at all. Mathematics says, If A is true, then B istrue, not A is true about the world in which we exist. This shouldbe easy to see if you consider a sufficiently abstract area ofmathematics -- category theory, say -- that never had, doesn't have,and most likely never will have any relation to the real world.Also to correct a couple minor points in your statement -- ParallelUniverse Theories have *always* involved mathematics; they wereoriginally posited as an explanation for quantum phenomena, andquantum mechanics, in case you haven't heard, uses some pretty heavymathematics. Also, the arctangent series reduces to a Taylor series;that is, a series involving only addition and multiplication.Also notice that the units in your proposed mechanism all cancel outwith each other, except for two, so you'd get the following value outof it:4 units (one unit - unit/3units + unit/5units - unit/7units + ...)= 4 units (unit - 1/3 + 1/5 - 1/7 + ...)= 4 units ((unit - 1) + 1 - 1/3 + 1/5 - 1/7 + ...)= 4 units ((unit - 1) + pi)= 4 units^2 + 4 units + piNow, unless the unit is a dimensionless constant, you have some bigtrouble here; your dimensions contradict each other. Just as youcannot add a square meter and a (linear) meter, you cannot add 4square units and 4 (linear) units. Furthermore, this equation doesn'tgenerate any useful equations for any reasonable choice ofdimensionless units.I'd like to stress that nobody is ignoring the Multiverse theory forlack of mathematical foundations. David Deutsch, a highly respectedand famous researcher into quantum mechanics, has been promoting thisview for quite a long time, and he's got plenty of math to back it up. I'd suggest reading some stuff about his work.> I originally posted this alt.parallel.universes where they appear to be> equation phobic; that is the explanation for the Warning, you will see> some equations!.> You know all those wonderful theories of 20th century science (string> theory, relativity, quantum mechanics). They all used math as a basis> for their existance, so until Parallel universes start getting some math> in them, we are just wasting our time. On that note, I will show you a> math based model of Parallel universes. Warning, you will see some> equations!> pi/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ...> This equation calculates the value of Pi (3.14159...), it is called> Liebniz' equation. There are many equations that calculate pi. This> one is unique because it uses only numbers and four signs (+,-,=, and> /). Other equations that calculate pi use arctangents or square roots> or other things. I point this out to you because real soon you are> going to be questioning me about it.> Now pi is the ratio between a circles circumference and its diameter.> In Non-euclidian geometery, you can create geometeries where pi is> greater then 3.14159.. (Hyperbolic Geometery), and geometeries where pi> is less then 3.14159... (Elliptic Geometery), but all these geometeries> have a flaw where the smaller the space you measure the closer pi will> get too 3.14159... Now if you have seen it I just presented you with a> paradox. The paradox is, how can you be in a (let's say) a Hyperbolic> geometery and calculate two> different values of pi at the same time? One by the space you are in> and the other by math. Where did math separate from space? Liebniz'> equation above always calculates the same value of pi no matter what> space you are in.> So let us say we are in a space where the value of pi is different from> 3.14159..., and we want this space to also be shown in Liebniz> equation. How would we go about doing this? I ask the question because> I am going to give you an answer, an answer that involves Parallel> universes. You will debate whether my answer is the correct one and> some of you may even come up with answers that don't involve parallel> universes. Best of luck!> I noted above that Liebniz' equation uses only numbers and four signs> (+,-,=, and /). We can move the 4 to the other side of the equals sign> and get,> pi = 4(1 - 1/3 + 1/5 - 1/7 + 1/9 - ...).> As you look at this equation, you will see that the equation has a> form to it. The odd numbers in the denominatior, the infinite number> of fractions, the alternating between plus signs and minus signs. Now> you need to ask yourself, Do these same features also occur in a> parallel universe? Does the form of the equation remain in parallel> universes. We can answer that question in this way. Liebniz formulated> the equation in the 18th century, so there have been 200+ years of> universes splitting off of our universe in those years and all of those> universes will have the same form iof Liebniz' equation, but all of> them will have different values of pi then ours. That still doesn't> answer our main question as to what changes. So let us rewrite the> equation with what we suspect doesn't change and see what we get,> pi = 4 units (one unit - unit/3units + unit/5units - unit/7units + ...).> We see that if we change the value of one(unit) we will arrive at a> different value of pi. This shows that Liebniz' equation shows the> relationship between pi and the natural numbers. There are other types> of numbers; imaginary, complex, irrational, etc. All I am proposing is> that each parallel universe has its own separate natural numbers,> complex numbers, irrational numbers, imaginary numbers, etc.> To read the paper that the above info is based on, go to> http://www.efn.org/~janieg/para61.html. Warning, there will be> equations in the paper.> Email me and tell me what you think picrosser@yahoo.com.> Jim AkerlundSubject: === Subject: Point in the right direction> Hello I was wondering if someone could point me in the right direction> when it comes to proving this proof. Thank you very much.> Question:> Let G be a group, and for any a E G let Ya : G -> G be the isomorphism> given by conjugation; i.e., Ya(x) = ax(a^−1). If g, h E G prove> that Yg = Yh if and only if g = hz for some element z E Z(G).May I presume that what appears on my screen as the sequence of characters following ax, namely ( a ^ & # 8 7 2 2 ; 1 ), though without the spaces, means the group inverse of a?Subject: === Subject: Point in the right direction> Hello I was wondering if someone could point me in the right direction> when it comes to proving this proof. Thank you very much.> Question:> Let G be a group, and for any a E G let Ya : G -> G be the isomorphism> given by conjugation; i.e., Ya(x) = ax(a^−1). If g, h E G prove> that Yg = Yh if and only if g = hz for some element z E Z(G).May I presume that what appears on my screen as the sequence of >characters following ax, namely ( a ^ & # 8 7 2 2 ; 1 ), though >without the spaces, means the group inverse of a?Yes, Unicode 0x2212 is the 'Minus Sign' in 'Mathematical Operators'.Therefore, what you have quoted should read: (a^-1)Rob Johnson > Hello I was wondering if someone could point me in the right direction>> when it comes to proving this proof. Thank you very much.>>> Question:>>> Let G be a group, and for any a E G let Ya : G -> G be the isomorphism>> given by conjugation; i.e., Ya(x) = ax(a^−1). If g, h E G prove>> that Yg = Yh if and only if g = hz for some element z E Z(G).>>May I presume that what appears on my screen as the sequence of >characters following ax, namely ( a ^ & # 8 7 2 2 ; 1 ), though >without the spaces, means the group inverse of a?> Yes, Unicode 0x2212 is the 'Minus Sign' in 'Mathematical Operators'.> Therefore, what you have quoted should read: (a^-1)> Rob Johnson take out the trash before replyingWouldn't it have been easier to use the ASCII dash, -? And that would have had the added advantage of wider readability, as well.Subject: Where is he?Where is he and what is he up to? Whatfantastic new breakthrough is he comingup with now? I wait with bated breath.Subject: === Subject: Where is he?>Where is he and what is he up to? Whe-e-e-e-ere is he? Whom I killfile not to see? Will I ever know the sweet Hell no that's meant for only me? Who can say where he may hide? Must I read groups far and wide Till I can behold the someone who I can mean trouble to? [...](with apologies to Lionel Bart.)Subject: === Subject: Where is he?> Where is he and what is he up to? What> fantastic new breakthrough is he coming> up with now? I wait with bated breath.He is alive and living in Paris...Subject: === Subject: Cantor's Diagonal Argument oo ____|mn / /_/ / _ / K-9/ /_/ - www.YeOldeCoffeeShoppe.com -/____/_____-------------->> This guarantees diagonal constructions can be applied to the list of>> computable numbers.>>> OK so far?> This construction only governs computable numbers. I suspect these>> are similar to algebraic numbers in that both are countable.> Are you specifying an explicit mapping (e.g., one can compute>> Computable(n) for all n > 0) or merely assuming one exists?>> The mapping I outlined should be explicit except for the> variable m (processing grain) and the particular UTM used.> What is m? Integer? Real? Random entity? Bellybutton lint? :-) UTMs exist in real life, I could actually start producing a binary> list from a UTM from a book.> I'm assuming UTM = 'von Neumann machine' here. (The classical UTM> is not useful here, although it can emulate all other Turing machines.> Note that a von Neumann machine does not have a tape, but it has> number of states. Of course, a large number of states are equivalent;good enough, the hardware is not important as long as different powered systemsget the same 'initial' list of computable numbers. Here initial means finite so we onlyneed an approximation to a TM (with a finite tape), i.e. a Von Neuman machine.the feature of UTMs is that the state transition component is finite and canpractically be used.> UTMs applied to a data input can emulate any computer function.> So in theory there exists a finite x for each UTM where> UTM(x,0) = 3. UTM(x,1) = 1 UTM(x,2) = 4> UTM(x,3) = 1 UTM(x,4) = 5 adding pi to the list of computables.> I'm trying to show the set of computable numbers is a larger class> than previously thought and includes all irrationals.> If it includes all irrationals that would certainly prove that> a bijective mapping between the reals and the natural numbers exists.> So go for it.> We need a definition of countable!?> Countable: can be associated with a bijective mapping to the natural> numbers {1, 2, ...}.> In short, the more or less usual definition.> The number could be fed on an input tape, in the more or less> usual form -- a sequence of digits, highest N-power first.> (For various interesting reasons N is not necessarily equal to 10.)To get a bijection I'd have to alter the definition of effectively computableto make sure added numbers are unique in the list of computable numbers.Cantors conjecture : can we find a number b that forall cEC, NotEqual(b, c)?Ready for inductive proof that NotEqual will never terminate?HercSubject: === Subject: Cantor's Diagonal ArgumentIn sci.math, |-|erc:> oo> ____|mn> / /_/ / _> / K-9/ /_/ - www.YeOldeCoffeeShoppe.com -> /____/_____> -------------->> This guarantees diagonal constructions can be applied to the list of>> computable numbers.>>> OK so far?> This construction only governs computable numbers. I suspect these>> are similar to algebraic numbers in that both are countable.> Are you specifying an explicit mapping (e.g., one can compute>> Computable(n) for all n > 0) or merely assuming one exists?>> The mapping I outlined should be explicit except for the> variable m (processing grain) and the particular UTM used.> What is m? Integer? Real? Random entity? Bellybutton lint? :-)> It allocates some m processing cyles to each parameter pair, then moves on to> ..> So this partially evaluates UTM(n, i) where (n, i) = (1,1),(1,1),(1,2),(1,1),(1,2),(2,1)...> m (stands for multi and) is used to get parallel execution from the UTM.> say UTM(1,1) takes 50 cycles to complete, a cycle would likely be> a state transition, and UTM(1,2) takes 5 cycles to complete.> set m = 20> Processing Sequence> (1,1) 20> (1,1) 20> (1,2) 5 halts> (1,1) 10 halts> (2,1) 20> ...> so m is a natural, if m was 1 it would be inefficient though.> There are alternate methods like :> 1 cycle for parameter set 1> 2 cycles for 1 2> 3 cycles for 1 2 3> 4 cycles for 1 2 3 4> which can dump previous computation, but still guarantess each well founded function> will halt.> UTMs exist in real life, I could actually start producing a binary> list from a UTM from a book.> I'm assuming UTM = 'von Neumann machine' here. (The classical UTM> is not useful here, although it can emulate all other Turing machines.> Note that a von Neumann machine does not have a tape, but it has> number of states. Of course, a large number of states are equivalent;> good enough, the hardware is not important as long as different> powered systems get the same 'initial' list of computable numbers.> Here initial means finite so we only need an approximation to a> TM (with a finite tape), i.e. a Von Neuman machine.> the feature of UTMs is that the state transition component is finite> and can practically be used.> UTMs applied to a data input can emulate any computer function.> So in theory there exists a finite x for each UTM where> UTM(x,0) = 3. UTM(x,1) = 1 UTM(x,2) = 4> UTM(x,3) = 1 UTM(x,4) = 5 adding pi to the list of computables.> I'm trying to show the set of computable numbers is a larger class> than previously thought and includes all irrationals.> If it includes all irrationals that would certainly prove that> a bijective mapping between the reals and the natural numbers exists.> So go for it.> We need a definition of countable!?> Countable: can be associated with a bijective mapping to the natural> numbers {1, 2, ...}.> In short, the more or less usual definition.> The number could be fed on an input tape, in the more or less> usual form -- a sequence of digits, highest N-power first.> (For various interesting reasons N is not necessarily equal to 10.)> To get a bijection I'd have to alter the definition of effectively computable> to make sure added numbers are unique in the list of computable numbers.> Cantors conjecture : can we find a number b that forall cEC, NotEqual(b, c)?Not quite. Cantor's idea is to explicitly construct a number b andthen show that for all c(j), NotEqual(b, c(j)) always terminates,with a result stating NotEqual.Your problem is slightly different; you need to show that for *all* b(within your problem domain, as we were initially assuming [0,1)at one point), that NotEqual(b, c(j)) will never terminate for some j,and that therefore b = c(j).Your UTM/diag timeslicer m is apparently but one step inthat argument, allowing the construction of a functionDiagNotEqual(b), which will terminate on all numbers c(j)if b is not equal to any of the numbers c(j). (There is ofcourse the problem that DiagNotEqual(b) won't terminateeither, as the number of j's are countably infinite,but to mathematicians that's a detail. :-) )> Ready for inductive proof that NotEqual will never terminate?Go for it.> Herc-- #191, ewill3@earthlink.netIt's still legal to go .sigless.Subject: === Subject: Cantor's Diagonal Argument oo ____|mn / /_/ / _ / K-9/ /_/ - www.YeOldeCoffeeShoppe.com -/____/_____-------------->> This guarantees diagonal constructions can be applied to the list of> computable numbers.> OK so far?> This construction only governs computable numbers. I suspect these> are similar to algebraic numbers in that both are countable.> Are you specifying an explicit mapping (e.g., one can compute> Computable(n) for all n > 0) or merely assuming one exists?>>> The mapping I outlined should be explicit except for the>> variable m (processing grain) and the particular UTM used.> What is m? Integer? Real? Random entity? Bellybutton lint? :-)> It allocates some m processing cyles to each parameter pair, then moves on to> ..> So this partially evaluates UTM(n, i) where (n, i) = (1,1),(1,1),(1,2),(1,1),(1,2),(2,1)...> m (stands for multi and) is used to get parallel execution from the UTM.> say UTM(1,1) takes 50 cycles to complete, a cycle would likely be> a state transition, and UTM(1,2) takes 5 cycles to complete.> set m = 20> Processing Sequence> (1,1) 20> (1,1) 20> (1,2) 5 halts> (1,1) 10 halts> (2,1) 20> ...> so m is a natural, if m was 1 it would be inefficient though.> There are alternate methods like :> 1 cycle for parameter set 1> 2 cycles for 1 2> 3 cycles for 1 2 3> 4 cycles for 1 2 3 4> which can dump previous computation, but still guarantess each well founded function> will halt.> UTMs exist in real life, I could actually start producing a binary>> list from a UTM from a book.> I'm assuming UTM = 'von Neumann machine' here. (The classical UTM>> is not useful here, although it can emulate all other Turing machines.>> Note that a von Neumann machine does not have a tape, but it has>> number of states. Of course, a large number of states are equivalent;>> good enough, the hardware is not important as long as different> powered systems get the same 'initial' list of computable numbers.> Here initial means finite so we only need an approximation to a> TM (with a finite tape), i.e. a Von Neuman machine.> the feature of UTMs is that the state transition component is finite> and can practically be used.>> UTMs applied to a data input can emulate any computer function.>> So in theory there exists a finite x for each UTM where>> UTM(x,0) = 3. UTM(x,1) = 1 UTM(x,2) = 4>> UTM(x,3) = 1 UTM(x,4) = 5 adding pi to the list of computables.>>> I'm trying to show the set of computable numbers is a larger class>> than previously thought and includes all irrationals.> If it includes all irrationals that would certainly prove that>> a bijective mapping between the reals and the natural numbers exists.>> So go for it.>> We need a definition of countable!?> Countable: can be associated with a bijective mapping to the natural>> numbers {1, 2, ...}.> In short, the more or less usual definition.> The number could be fed on an input tape, in the more or less>> usual form -- a sequence of digits, highest N-power first.> (For various interesting reasons N is not necessarily equal to 10.)>> To get a bijection I'd have to alter the definition of effectively computable> to make sure added numbers are unique in the list of computable numbers.> Cantors conjecture : can we find a number b that forall cEC, NotEqual(b, c)?> Not quite. Cantor's idea is to explicitly construct a number b and> then show that for all c(j), NotEqual(b, c(j)) always terminates,> with a result stating NotEqual.> Your problem is slightly different; you need to show that for *all* b> (within your problem domain, as we were initially assuming [0,1)> at one point), that NotEqual(b, c(j)) will never terminate for some j,> and that therefore b = c(j).OK, b takes the form 0. b_1 b_2 b_3 b_4 b_5...where b_n E {0,1,2,3,...8,9}Examine the 1st column of C.Cj i 1 2 3 4 51 32 43 74 35 6...This is an infinite sequence of digits 0..9, whatever value b_1 is, we can assigna probability that that digit is not on the list of about 1 in 10% * infinity.That is, probability(exists j, b_1 = c(j)_1) = 1.There are actually infinite number of values for j that satisfy the 1stdigit of b, occuring very roughly every 10 computable numbers.The second digit is similarly satisfied, probability(exists j, b_1 = c(j)_1 & b_2 = c(j)_2) = 1.This is obvious by filtering the list of computables to entries that10% of the entires match b_2 aswell.Similarly :Prob(exists j, for i = 1..n, b_i = c(j)_i)=1 -> Prob(exists j, for i = 1..n+1, b_i = c(j)_i)=1By induction on n :for all n, exists j, b_1 = c(j)_1, b_2 = c(j)_2, ... b_n = c(j)_nThis suggests NotEqual(b, c(j)) will not halt.> Your UTM/diag timeslicer m is apparently but one step in> that argument, allowing the construction of a function> DiagNotEqual(b), which will terminate on all numbers c(j)> if b is not equal to any of the numbers c(j). (There is of> course the problem that DiagNotEqual(b) won't terminate> either, as the number of j's are countably infinite,> but to mathematicians that's a detail. :-) )more a concern!HercSubject: === Subject: ZZCrap> In sci.math, ZZBunker> :>> to isn't a verb.>>> The last time I heard that to wasn't an intransitive>> verb was the last time that math wasn't spelled>> in the moronic fashion maths.>>> What's math?> Nobody knows, Since like we've been telling morons> for several thousand years, now: > Maths is philosophy not science.> See Plato for philosophy and festive pillar work.> Math is an abstract model, nothing more. Occasionally it's> useful; we've even harnessed the Queen of Uselessness,> number theory, into modern cryptography. :-) Who says> primes aren't useful? :-)> However, '1' cannot be captured in a butterfly net. 'pi',> 'e', and i=sqrt(-1) are even more elusive. Good luck> finding a Taylor or MacLaurin series running around in the wild. Nobody ever said that you can find *anything* related to moronic *Calculus* running around in the wild. We have merely stated that you can find *real* *Turing machines* running around in the wild.> There are examples of Nature taking advantage of Fibonacchi> numbers (e.g., sunflower seed spirals, plant stalk branchings) but > that doesn't mean Fibonacchi numbers are out there.> The main requirement for math is that it be self-consistent. The *only* requirement for math is that it exist in a non-existent vaccuum. Other than that it's the most trivial thing in the universe.Subject: === Subject: ZZCrapIn sci.math, ZZBunker:> In sci.math, ZZBunker> :>> to isn't a verb.>>> The last time I heard that to wasn't an intransitive>> verb was the last time that math wasn't spelled>> in the moronic fashion maths.>>> What's math?> Nobody knows, Since like we've been telling morons> for several thousand years, now: > Maths is philosophy not science.> See Plato for philosophy and festive pillar work.> Math is an abstract model, nothing more. Occasionally it's> useful; we've even harnessed the Queen of Uselessness,> number theory, into modern cryptography. :-) Who says> primes aren't useful? :-)> However, '1' cannot be captured in a butterfly net. 'pi',> 'e', and i=sqrt(-1) are even more elusive. Good luck> finding a Taylor or MacLaurin series running around in the wild.> Nobody ever said that you can find *anything* related> to moronic *Calculus* running around in the wild.> We have merely stated that you can find *real* > *Turing machines* running around in the wild.Models only, and imperfect ones at that. The ideal Turingmachine has an infinite length tape (most of which is blank).All modern machines have finite tapes -- if they have tapesat all.> There are examples of Nature taking advantage of Fibonacchi> numbers (e.g., sunflower seed spirals, plant stalk branchings) but > that doesn't mean Fibonacchi numbers are out there.> The main requirement for math is that it be self-consistent.> The *only* requirement for math is that it> exist in a non-existent vaccuum.> Other than that it's the most trivial> thing in the universe.Math does not exist except as a series of thought patternsand as ink on paper and bits on storage devices and travelingthrough cables and air -- and those are also imperfect models.In short, there's no there there. It is merely a way ofcoloring one's perceptions of the Universe; without math,we'd look at the Universe differently -- perhaps usingmysticism? -- and draw conclusions in a different fashion.-- #191, ewill3@earthlink.netIt's still legal to go .sigless.Subject: === Subject: P, NP and coNP <1oc2c.30834$6c5.19657@nwrdny01.gnilink.net> <55d84984.0403080955.6f53f664@posting.google.com> (Obvious counterexample: A = set of all integers, B = set of all even > integers).> Congratulations!> As far as I know, you are the first to spot that (with the possible> exception of the author of the paper/proof, who did not seem to know> what he/she was writing about).> The Crypto'03 committee and referees totally missed that. The '40> theorists' have said nothing about it. The tcne HTML page has been up> for quite some time and nobody, you are no exception, had made that an> issue. It is only after my OP made it explicit, you became the _FIRST_> one to realize it. How extraordinary! Have you ever wondered why all> those people never tried to make an issue out of it? I don't think it is extraordinary that noone pointed this out. When I first read over your paper and encountered that statement, I didn't bother to think about the cardinality part because showing that coNP is a subset of NP would be sufficient to show that coNP is equal to NP. I looked past the cardinality statement, as probably the rest of the people who knew was it would take to show P=NP also looked past it. Your statement about the set cardinalities was the first indication that you probably didn't know what you were talking about. Then looking up the proof that coNP is a subset of NP was a good certificate of the fact.JSubject: === Subject: P, NP and coNP> The Crypto'03 committee and referees totally missed that. The '40> theorists' have said nothing about it. The tcne HTML page has been up> for quite some time and nobody, you are no exception, had made that an> issue. It is only after my OP made it explicit, you became the _FIRST_> one to realize it. How extraordinary! Have you ever wondered why all> those people never tried to make an issue out of it?> I don't think it is extraordinary that noone pointed this out. When I I don't think I explicitly said that challenging Cantor isextraordinary, either. Should be a piece of cake. You did it also? Howearlier (than I) did you do it? Showing that you can do something inthis field adds so much prestige and 'first' is such a big deal inmath!You know what, not everybody is not in the crowd like you. Maybe, itis by Schrodinger's uncertainly or something else I do not claim tohave any understanding that all happen to look at it as a pAssiblemistake. But it does not sound right to me. Great minds are not borneveryday and great minds have struggled with the issue. Now if so manycan lecture me on how a triviality that is (without a lecture), shouldI believe (that they do know the issue)? I correct my grammaticalmistake. It should not be so much above the crowd. It perhaps shouldbe so much crowd above. It can not be right, simply can not! If youhear a crowd shout 1+1=2, you can never tell which amongst them areliars, even if it is a crowd consisted entirely of liars. How boringit would be if liars are crippled to only tell obvious lies! On theother hand, if you hear people in a certain regime say they believethat certain race is superior than others, you won't be able to tellwho truly holds the erroneous belief, either.> first read over your paper and encountered that statement, I didn't > bother to think about the cardinality part because showing that coNP is a> subset of NP would be sufficient to show that coNP is equal to NP. I > looked past the cardinality statement, as probably the rest of the people > who knew was it would take to show P=NP also looked past it.What did you say? An error that obviously shows that the author isignorant made you waste more time on the paper? Mathematical logic isbeautiful!Congratulations to you, too. You can find out, interested or not,which of you two was the first (to have done _THAT_), if 'first' issuch a big deal here as well. Better establish the priority now, lestothers may suddenly pop up like you, trying to join the rank as well,threatening the seniority.> Your statement about the set cardinalities was the first indication that > you probably didn't know what you were talking about. Then looking up the > proof that coNP is a subset of NP was a good certificate of the fact.> JI can be humble. You now stand a chance of making me see my ignorance.It is a very nice way to show off your math, logic and knowledge. Tellus how you can establish (coNP subsetof NP) ==> (coNP = NP) (first,and we can continue the conversation from there). We are makingprogress.If you put down something that is math, I can deal with it.If you dump crap like: You are just wrong. I can just repeat that mathcan never be degraded to that level. Show us the mathematics. If onecan not, but try in any way to pretend to be able to, he/she is aliar!But keep on dumping and do not write a single line of math if you area liar.So, simply show us the math. The counterexampler had a bit. I advisedto get the infinite size of natural number set straight (not that Iyou sound you can deliver so much better. Lecture me/us intounderstanding why (coNP subsetof NP) ==> (coNP = NP). I believethat there is much crap out there that cannot come to terms with thebeautiful fact you just pointed out.So, do not disappoint us! Show us the math! We know you can do it! Doit!Subject: === Subject: P, NP and coNP <1oc2c.30834$6c5.19657@nwrdny01.gnilink.net> <55d84984.0403080955.6f53f664@posting.google.com The Crypto'03 committee and referees totally missed that. The '40> theorists' have said nothing about it. The tcne HTML page has been up> for quite some time and nobody, you are no exception, had made that an> issue. It is only after my OP made it explicit, you became the _FIRST_> one to realize it. How extraordinary! Have you ever wondered why all> those people never tried to make an issue out of it?> I don't think it is extraordinary that noone pointed this out. When I > I don't think I explicitly said that challenging Cantor is> extraordinary, either. Should be a piece of cake. You did it also? How> earlier (than I) did you do it? Showing that you can do something in> this field adds so much prestige and 'first' is such a big deal in> math! When did I say anything about challenging Cantor? You said it was extraordinary that none of the reviewers pointed out that your sentence was wrong. I'm pointing out that they didn't simply because that part of the sentence was totally unnecessary.> first read over your paper and encountered that statement, I didn't > bother to think about the cardinality part because showing that coNP is a> subset of NP would be sufficient to show that coNP is equal to NP. I > looked past the cardinality statement, as probably the rest of the people > who knew was it would take to show P=NP also looked past it.> What did you say? An error that obviously shows that the author is> ignorant made you waste more time on the paper? Mathematical logic is> beautiful! The problem is that your paper lacked any Mathematical logic.> Congratulations to you, too. You can find out, interested or not,> which of you two was the first (to have done _THAT_), if 'first' is> such a big deal here as well. Better establish the priority now, lest> others may suddenly pop up like you, trying to join the rank as well,> threatening the seniority. I don't think anyone cares about who noticed your paper was flawed first.> I can be humble. You now stand a chance of making me see my ignorance.> It is a very nice way to show off your math, logic and knowledge. Tell> us how you can establish (coNP subsetof NP) ==> (coNP = NP) (first,> and we can continue the conversation from there). We are making> progress.> If you put down something that is math, I can deal with it. Look it up in any book. Any proof that coNP is a subset of NP can be essentially replicated to analogously show that Np is a subset of coNP, and you are done.> So, do not disappoint us! Show us the math! We know you can do it! Do> it! I hope you're happy with that. If not, refer to Papadimitriou's Complexity Theory, proposition 10.2: If a coNP-complete problem is in NP, then NP = coNP.Which can be found in countless places on the web.JSubject: === Subject: Strange Complex Variables Problem> Here is a problem I'm trying to figure out. I tried a bunch of> different things, but not sure what to use here -- the argument> principle?> I wonder if anyone can make heads or tails of it.> You have the space H of all analytic functions on a doubly-connected> domain D (it's a hilbert space under the inner product for functions,> they all have an L2 norm). Then consider the subset J of H consisting> of all functions which have integral 2PI*i going once around the hole> in the domain. Let F be the function mapping the domain onto an> annulus. Show that F'/F (logarithmic derivative) has the smallest L2> norm of all L2 norms of functions in J.> Try stating the problem precisely. What are you integrating over? What do > you mean the function F mapping the domain onto an annulus? Which > annulus? Etc, etc.I am integrating over the doubly connected domain D. The annulusapparently can be any annulus centered around the origin. The sizeof the annulus doesn't matter, as there can only be one ratio of thetwo radii in the annulus, and F'/F is not sensitive to that ratio(e.g. 2F'/2F the 2's cancel).-GregSubject: === Subject: Strange Complex Variables Problem> Here is a problem I'm trying to figure out. I tried a bunch of> different things, but not sure what to use here -- the argument> principle?> I wonder if anyone can make heads or tails of it.> You have the space H of all analytic functions on a doubly-connected> domain D (it's a hilbert space under the inner product for functions,> they all have an L2 norm). It's simply not true that all the analytic functions in D have an L2 norm; I'm surprised that Wade didn't mention this when heexpressed confusion. You actually mean that H is the set ofanalytic functions f in D _such that_ the L2 norm of f is finite,right?This then raises the question of what measure you're talkingabout when you take the L2 norm. One can only guess youmean Lebesgue measure (but for Lebesgue measure Ihave no idea what the solution is, while for another measureI do.)> Then consider the subset J of H consisting> of all functions which have integral 2PI*i going once around the hole> in the domain. Let F be the function mapping the domain onto an> annulus. Show that F'/F (logarithmic derivative) has the smallest L2> norm of all L2 norms of functions in J.> Try stating the problem precisely. What are you integrating over? What do > you mean the function F mapping the domain onto an annulus? Which > annulus? Etc, etc.>I am integrating over the doubly connected domain D. The annulus>apparently can be any annulus centered around the origin. The size>of the annulus doesn't matter, as there can only be one ratio of the>two radii in the annulus, and F'/F is not sensitive to that ratio>(e.g. 2F'/2F the 2's cancel).If we fix the annulus there is still more than one map from D ontothe annulus (even assuming that we're just talking about 1-1maps). They differ by composition with a conformal map ofD to itself. In fact the L2 norm of F'/F is not sensitive to thateither, but that's not so obvious, at least to me, I started to saythat it _was_ sensitive to that and caight myself at the last minute.(If phi is an automorphism of D then there must exist a with |a| = 1such that F o phi = a*F. At least I think so.)Another problem is that in general F'/F will not be in H,for example if D is a punctured disk then F(z) = z,and 1/z is not in L2(D). We need to assume that D issuch that the annulus has strictly positive inner radiusand finite outer radius; then a change of variablesshows that F'/F is in L2.Having said all that, I don't know how to show thatF'/F has minimal norm in J. Now, if we were talkingabout L2 norms with respect to a different measurethat would be different: If we were talking aboutthe measure such that mu(E) = m(F(E)), wherem is Lebesgue measure on the annlus, thenyou could assume without loss of generalitythat D _was_ an annulus, and then you haveto show that the identity function minimizesthe L2 norm in J; this is easy by looking atthe Laurent series (and noting that membershipin J is the same as saying that the coefficientof 1/z is 1; if there's another non-zero coefficientthat increases the L2 norm, while if there's notthen the function is the identity.)>-Greg************************Subject: === Subject: Strange Complex Variables Problem>> Here is a problem I'm trying to figure out. I tried a bunch of>> different things, but not sure what to use here -- the argument>> principle?>> I wonder if anyone can make heads or tails of it.>>> You have the space H of all analytic functions on a doubly-connected>> domain D (it's a hilbert space under the inner product for functions,>> they all have an L2 norm). >It's simply not true that all the analytic functions in D have an L2 >norm; I'm surprised that Wade didn't mention this when he>expressed confusion. You actually mean that H is the set of>analytic functions f in D _such that_ the L2 norm of f is finite,>right?>This then raises the question of what measure you're talking>about when you take the L2 norm. One can only guess you>mean Lebesgue measure (but for Lebesgue measure I>have no idea what the solution is, while for another measure>I do.)>> Then consider the subset J of H consisting>> of all functions which have integral 2PI*i going once around the hole>> in the domain. Let F be the function mapping the domain onto an>> annulus. Show that F'/F (logarithmic derivative) has the smallest L2>> norm of all L2 norms of functions in J.>Try stating the problem precisely. What are you integrating over? What do >you mean the function F mapping the domain onto an annulus? Which >annulus? Etc, etc.>I am integrating over the doubly connected domain D. The annulus>apparently can be any annulus centered around the origin. The size>of the annulus doesn't matter, as there can only be one ratio of the>two radii in the annulus, and F'/F is not sensitive to that ratio>(e.g. 2F'/2F the 2's cancel).>If we fix the annulus there is still more than one map from D onto>the annulus (even assuming that we're just talking about 1-1>maps). They differ by composition with a conformal map of>D to itself. In fact the L2 norm of F'/F is not sensitive to that>either, but that's not so obvious, at least to me, I started to say>that it _was_ sensitive to that and caight myself at the last minute.>(If phi is an automorphism of D then there must exist a with |a| = 1>such that F o phi = a*F. At least I think so.)>Another problem is that in general F'/F will not be in H,>for example if D is a punctured disk then F(z) = z,>and 1/z is not in L2(D). We need to assume that D is>such that the annulus has strictly positive inner radius>and finite outer radius; then a change of variables>shows that F'/F is in L2.>Having said all that, I don't know how to show that>F'/F has minimal norm in J. Now, if we were talking>about L2 norms with respect to a different measure>that would be different: If we were talking about>the measure such that mu(E) = m(F(E)), where>m is Lebesgue measure on the annlus, then>you could assume without loss of generality>that D _was_ an annulus, and then you have>to show that the identity function minimizes>the L2 norm in J; this is easy by looking at>the Laurent series (and noting that membership>in J is the same as saying that the coefficient>of 1/z is 1; if there's another non-zero coefficient>that increases the L2 norm, while if there's not>then the function is the identity.)And come to think of it, I have _serious_ doubtswhether what you say is _true_ if you're talkingabout Lebesgue measure on D:Say G:annulus -> D is the inverse of F. Definea measure nu on the annulus by |G'|^2 dx dy.Now the problem is to show that 1/z has minimalL2 norm wrt nu, among all functions on J suchthat the coefficient of 1/z is 1. This is the sameas saying that z^n is nu-orthogonal to 1/z for alln <> -1, and I can't imagine why that would beso, seems very implausible.So: exactly what measure on D are you talkingabout when you say L2(D)?>-Greg************************Subject: === Subject: Hypocritical Fortune Tellers (What are the odds?)> are you a mathematician or a numerologist?> is what you're posting anything more than a> glorified horoscope?I am not looking to the future but goink back tothe past, to the day the person was born. Godprovided the individual with his or her life andalso their name, back then. Horoscopes deal withastrology, can you see any discussions of planetsand stars in my postings? The mainstreamChristian churches embrace pagan holidays thatare linked to sun whoreship and merge them in withtheir Christian worship, you are likely a memberof one of these cults and now you hypocriticallycondemn me and my work as being astrology. Forexample, by December 25th the sun is visiblyreturning from the south, calling this pagansunwhoreshipping holiday Christmas is aviolation of God's Third Commandment, forit is not Christ's Mass but is instead a paganmass, it is the use of God's name in vain. Andfor example, the mainstream Christians embracethe pagan Easter holiday at a time based uponthe phase of the moon. And then look at all thepredictions that the mainstream mathematiciansattempt, you people are tossing coins and lookingfor the probababbility of them landing as headsor tails (while disavowing Satanic influences thatcan and do manipulate these coins). You mathe-maticians are looking for the probababbility ofevents on earth (while disavowing spiritualinfluences), and then have the audacity to callme a fortune teller. I am a mathematician whodeals with God and little primes, while themainstream mathematicians are Godlessnumerologists who salivate when they discovera prime that is large enough to fill a telephonebook. -Daryl S. Kabatoffand what does satan care about a coin toss result?if you have a problem with probablility theory,then why not tell us which parts exactly, and wecan discuss it...Satanic forces can and do make coins fall intopositions that practitioners of I Ching theninterpret and tell fortunes. Some laws allowfor tied elections to be determined my cointosses, but this is not a good idea for thecandidate that Satan and his dark angels mosthighly esteems will be the inevitable winner.yikes. so does God have any say in this? imean, doesnt God care that Satan is takingcontrol of all these coin tosses? if so,maybe we can conjecture that whenever thecoin ends up being tails its Satan, andwhenever it ends up being heads, its God.and my theory is that as the coin toss isrepeated, God will win the cointoss about50% of the time (if the coin isnt Satanic).can you not see how psychotic your theoryabout satan and his dark angels is about acointoss?Invisible forces surround you and the coins, when youthrow the coins with the expectation of a fortune beingtold to you, then Satanic agencies will manipulate thecoins and attempt to pass on some bad advice to theperson asking for the advice. People get caught up inI CHING because the coins actually do fall into aposition that coincides with the message that seemsappropriate to the person's mental and phyical stateat that moment in time. The invisible angels thatsurround you study you and seek entry into your soul,they are not human, they are superhuman beings thatcan and do manipulate the outcome of coin tosses.When it is to their benefit, Satanic agencies willallow the coins to fall randomly. I am not psychotic,and for you to throw out accusations of insanity whendefending your position (as opposed to logicalreasoning), means you are being abusive to me, andlikely also means you are not just a cultist, butsome kind of nazi as well. You are not just closedminded, but you are abusive as well. You confusethe argument by asking for my age, by daring me togo to some web site of your choice, by mocking me(these ploys are not being reproduced here inan attempt to keep with the issues at hand), andby throwing out accusations mental illness (youhave outright called me psychotic). Because youdo such things, reponding to your abusive unfairnessis hampering my abilty to do more constructive tasks,such as posting another example of God providing anindividual with his or her name in the math groups.-Daryl S. KabatoffSubject: === Subject: Hypocritical Fortune Tellers (What are the odds?)hey daryl, i'm waiting for a response.Subject: === Subject: Hypocritical Fortune Tellers (What are the odds?)> hey daryl, i'm waiting for a response.You called me psychotic and then you comepressuring me for a response to your verbalabusiveness, just who the hell are you?!!!You're just another abusive anonymousvulgar dog spewing out garbage andunwilling to back it up with your ownreal name. -Daryl Shawn KabatoffSubject: === Subject: Hypocritical Fortune Tellers (What are the odds?)> hey daryl, i'm waiting for a response.> You called me psychotic and then you come> pressuring me for a response to your verbal> abusiveness, just who the hell are you?!!!> You're just another abusive anonymous> vulgar dog spewing out garbage and> unwilling to back it up with your own> real name. -Daryl Shawn Kabatoffuhhhh, where did i call you psychotic? i'm just questioning yourmathematics. you're the one getting all defensive. i asked you to providelinks to the other forums that you supposedly post at. i got nothing.www.christianity.com is one of the biggest christian forums on the internet.if God told you to post on religious forums, then you would think that thissite would be one of your first stops.my name is Colin Dickson - October 27th, 1974. you can even do one of yourlittle readings on me if you want to. i didnt use my real name becausewhen it asked for a user name for the newsgroups, i put in the username thati use on all the other forums that i post at. happy now?Subject: === Subject: Hypocritical Fortune Tellers (What are the odds?)> hey daryl, i'm waiting for a response.> You called me psychotic and then you come> pressuring me for a response to your verbal> abusiveness, just who the hell are you?!!!> You're just another abusive anonymous> vulgar dog spewing out garbage and> unwilling to back it up with your own> real name. -Daryl Shawn Kabatoff> uhhhh, where did i call you psychotic? i'm just questioning your> mathematics. you're the one getting all defensive. i asked you to provide> links to the other forums that you supposedly post at. i got nothing.> www.christianity.com is one of the biggest christian forums on theinternet.> if God told you to post on religious forums, then you would think thatthis> site would be one of your first stops.> my name is Colin Dickson - October 27th, 1974. you can even do one ofyour> little readings on me if you want to. i didnt use my real name because> when it asked for a user name for the newsgroups, i put in the usernamethat> i use on all the other forums that i post at. happy now?I haven't head any instruction from God directing me topublish web sites on the internet, nor to contribute to otherpeople's web sites. I post on some religious forums on theusenet, not on the internet. Some people use the internetas a portal to access my usenet postings.You stated can you not see how psychotic your theory aboutsatan and his dark angels is about a cointoss? And by statingthat my beliefs are psychotic is little different than calling mepsychotic. God desires that I get off my butt and go for a walk,and try to make some one to one human contact, and show theindivual evidence that his or her name is a gift from God. Iwant the individuals middle name and also the names andbirthdays of the individual's family members. When you provideme with your common name and birthday, it's not unlikeshowing me a portion of a petal from a flower, rather than theentire flower. If you want to be used as an example to others,get me the missing stats, mail me photocopies of family ID,and send me some cash. I work full time on this, it's unfairthat you would request that I work for you and then ratherthan offer me payment, you throw abuse at me.-Daryl S. KabatoffSubject: === Subject: Hypocritical Fortune Tellers (What are the odds?)> You stated can you not see how psychotic your theory about> satan and his dark angels is about a cointoss? And by stating> that my beliefs are psychotic is little different than calling me> psychotic... Well, since you *are* psychotic, it makes no difference anyway.-- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, SilverlockSubject: === Subject: Hypocritical Fortune Tellers (What are the odds?)> Well, since you *are* psychotic, it makes no difference anyway.What is wrong with you people? What do you expect to gain by having a discussion with the lunatic? You are just wasting everyone's time and bandwidth. Please let him go back to posting his idiotic numerology, at least I already have kill filters set up for that! :-)meeroh-- If this message helped you, consider buying an itemfrom my wish list: Subject: === Subject: Hypocritical Fortune Tellers (What are the odds?)>Message-id: are you a mathematician or a numerologist?> is what you're posting anything more than a> glorified horoscope?>DARYL:> I am not looking to the future but goink back to> the past, to the day the person was born. God> provided the individual with his or her life and> also their name, back then. Horoscopes deal with> astrology, can you see any discussions of planets> and stars in my postings? The mainstream> Christian churches embrace pagan holidays that> are linked to sun whoreship and merge them in with> their Christian worship, you are likely a member> of one of these cults and now you hypocritically> condemn me and my work as being astrology. For> example, by December 25th the sun is visibly> returning from the south, calling this pagan> sunwhoreshipping holiday Christmas is a> violation of God's Third Commandment, for> it is not Christ's Mass but is instead a pagan> mass, it is the use of God's name in vain. And> for example, the mainstream Christians embrace> the pagan Easter holiday at a time based upon> the phase of the moon. And then look at all the> predictions that the mainstream mathematicians> attempt, you people are tossing coins and looking> for the probababbility of them landing as heads> or tails (while disavowing Satanic influences that> can and do manipulate these coins). You mathe-> maticians are looking for the probababbility of> events on earth (while disavowing spiritual> influences), and then have the audacity to call> me a fortune teller. I am a mathematician who> deals with God and little primes, while the> mainstream mathematicians are Godless> numerologists who salivate when they discover> a prime that is large enough to fill a telephone> book. -Daryl S. Kabatoff> i am not a member of any cult, but how typical of you to assume so.>good> job. you want to talk about being pagan and being part of cultish> practices? i would like to see you go over to the forums at> www.christianity.com and watch as they laugh you out of there.>DARYL:>When you accused my work of being a glorified horoscope,>was there not an implicit underlying message that my material>was cultish? And I don't really care if 100% of the churches>in the land teach things that are in opposition to Scripture,>and I don't care if 100% of their filthy members laugh at me,>what I care about is doing the things that God desires for me.>And He wants me to post this material in mathematical,>political and religious forums.You're wasting your time posting to these newsgroups thathardly anyone reads. If you really want to be effective, you should abandon these newsgroups and go where the action is: alt.test has more messages than all these newsgroups combined.>ZAPHOD:> and what does satan care about a coin toss result? if you> have a problem with probablility theory, then why not> tell us which parts exactly, and we can discuss it...>DARYL:>Satanic forces can and do make coins fall into positions that>practitioners of I Ching then interpret and tell fortunes.>Some laws allow for tied elections to be determined my coin>tosses, but this is not a good idea for the candidate that Satan>and his dark angels most highly esteems will be the inevitable>winner.>ZAPHOD:> your posts here are completely devoid of any kind of> mathematics. disagree? then describe your methods> and show your proofs.>DARYL:>I am giving example after example of mathematical harmony>that ties together people's names, birthdays and the numbers>of the Bible. The 29x29th chapter of the Bible contains 29>verses, why doesn't your analytical tools note this and>similar patterns?>ZAPHOD:> until then, you're simply a numerologist in denial who> uses his faith as a shield>DARYL:>I have faith that God provides us with our names and lives,>and that this God is the God of the Bible rather than the>god of the Koran or some other religious book. Note that>the 4th Book of the Bible is called Numbers, it seems that>God has an interest in numbers and that discussion of such>should be permitted in these religious and mathematical>forums. -Daryl S. Kabatoff--MensanatorAce of ClubsSubject: === Subject: Optimal Strategy for playing StockMarket VonNeumann game-theory; Sanofi drop Aventis and buy BMY> BMY should sell its cholesterol drug division and focus on its realm> of excellence of cancer drugs. Recently Imclone's erbitux was approved> and look at what the colon cancer drug of Genentech had done for> Genentech. So, BMY, why waste time on cholesterol drugs when in the> final analysis it is all phony medicine that gives license to people> to remain fat and eat fat.A good solution for BMY is to ask Sanofi to merge. I hear Sanofi istrying to acquire Aventis because it needs to market its new drug thatstops smoking and is a diet drug. So a merger between Sanofi and BMYwould make good sense. Besides, Aventis is fighting the Sanofiovertures.BMY has been in the doldrums for 5 years now and a merger with Sanofijust might lift both companies.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxiesSubject: geometry of a crystal vase that gives maximum sparkle and light luster??I think crystal glass comes closest to the sparkle of diamonds whenone cannot have a huge diamond of the size of a vase. I own 3 vases ofcrystal. Orrefors and several Czech and Poland.I see crystal vases as a diamond writ large only without the diamond.But I wonder what sort of geometry is needed for a crystal vase suchthat the sparkle and luster of light hitting it makes it look themaximum in dazzle and a sight to behold.Judging from my Orrefors vase, if you take a very thick crystal andnotch it so that you maximize the long edge lines achieves the goal ofsparkle and dazzle. Long edges running up and down the face of thecrystal vase maximizes the sparkle and luster-- or so I believe thisto be the case. I have another crystal vase that has diamond shapedslices running up and down which diminishes the luster. And the morepatterns makes the vase tend toward opaqueness.Summary: I suspect the geometry of a crystal vase that seeks tomaximize sparkle and light dazzle is a thick glass whose only patternare elongated vertical grooves that maximizes edges for it is thelength of edges and the number of edges that makes a crystal dazzlethe most.Anyone refute or support that hypothesis?Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxiesSubject: === Subject: geometry of a crystal vase that gives maximum sparkle and light luster?? charset=UTF-8 Archimedes Plutonium [CapitalEth][EDoubleDot][Micro] .b3 .b9[EDoubleDot].b9> I think crystal glass comes closest to the sparkle of diamonds when> one cannot have a huge diamond of the size of a vase. I own 3 vases of> crystal. Orrefors and several Czech and Poland.> Summary: I suspect the geometry of a crystal vase that seeks to> maximize sparkle and light dazzle is a thick glass whose only pattern> are elongated vertical grooves that maximizes edges for it is the> length of edges and the number of edges that makes a crystal dazzle> the most.> Anyone refute or support that hypothesis?Crystal and gemstone cutting has an extensive theory behind it. In the olddays it was mostly empirical, that's why many old gemstone cutters werevictims of cardiac problems.In short, optimal cutting of any crystal object depends on maximizing localprismatic apical angles, so that with the aid of internal reflections and byutilizing the refraction index of the material, one gets near optimaldispersion for the particular material's refraction index.The principle is very similar to the Littrow design with a half-mirroredprism, where light enters, gets refracted, reflected, refracted again andgoes back out the same edge it went in.As diamond has a real high refraction index (>2.1) its shapes are prettystandard and all apical angles in most common shapes conform to an exactmathematical relationship. For materials with such high refractive index,generally care is taken to not exceed certain angle values, otherwise thelight refracts and turns inwards, unable to escape.For materials with lower refraction index, more exotic shapes can bedesigned, but generally there is always an optimal cutting method. Thismethod is taken into account also when dividing the crude stone roughly soit can produce high quality parts. For diamonds this also includes cuttingalong _exactly_ the natural veins of the crude stone, so that the producedparts have no carbon specs or veins in them.If diamond did not have such a high refraction index, it would have beensuperceeded by artificial crystals, which can be manufactured relativelyinexpensively, such as LaSFN9, SFxx series and TiO2 or with crystals withhigh Pb content, like Swarovski. The later for example, started makingdecorative gems, but is now also in the business of advanced optics making,like high quality telescope achromats, etc.> Archimedes Plutonium> whole entire Universe is just one big atom where dots> of the electron-dot-cloud are galaxies--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------- -----------------------Eventually, _everything_ is understandableSubject: === Subject: geometry of a crystal vase that gives maximum sparkle and light luster??> ..I suspect the geometry of a crystal vase that seeks to> maximize sparkle and light dazzle is a thick glass whose only pattern> are elongated vertical grooves that maximizes edges for it is the> length of edges and the number of edges that makes a crystal dazzle> the most.Perhaps it is total internal reflection that traps and containsintense light inside of it making it bright and dazzling.> But I wonder what sort of geometry is needed for a crystal vase such> that the sparkle and luster of light hitting it ..Is there some theory to the lustrous diamond of standard tear shapedgeometry ?I had seen such dazzle in some polyhedral glass decoration items. Asolid crystal glass sphere ( several inches diameter, much bigger thanthe Cullinan in size) would give off dazzling rainbow colours insunlight at about 20 degrees subtended angle between sun,crystal andeye, by total internal reflection.> one cannot have a huge diamond of the size of a vase. By a ray tracing algorthm, one can even design a shape as well, usingrefraction and reflection of rays at all faces, to get a hugeartificial/fake but functional diamond. As the refractive index ofglass is less, bigger size is needed for same total internalreflection.You could post this in sci.optics also.Subject: === Subject: a digital game> The transcendental numbers form a dense G_delta (as is the complement> of any countable set).Ah, and hence the solution to the original puzzle (I think):(a) If the player who wants to force a transcendental is player #1, then he picks every other digit, and assigns 101001000100001... to those digits. Any infinitely-repeating pattern in the final number would imply an infinitely-repeating pattern in player #1's digits; since the latter doesn't exist, neither does the former.(b) If the player who wants to force a transcendental is player #2, then he considers X (the number that would result if he assigned all zeroes) and Y (the number that would result if he assigned all ones). Due to density, there is guaranteed to be a transcendental somewhere between X and Y; he chooses one, then chooses digits that match its expansion.Subject: === Subject: a digital game> The transcendental numbers form a dense G_delta (as is the complement> of any countable set).>Ah, and hence the solution to the original puzzle (I think):>(a) If the player who wants to force a transcendental is player #1,> then he picks every other digit, and assigns 101001000100001...> to those digits. Any infinitely-repeating pattern in the final> number would imply an infinitely-repeating pattern in player #1's> digits; since the latter doesn't exist, neither does the former.If there's no infinitely-repeating pattern, it's irrational, notnecessarily transcendental.>(b) If the player who wants to force a transcendental is player #2,> then he considers X (the number that would result if he assigned> all zeroes) and Y (the number that would result if he assigned> all ones). Due to density, there is guaranteed to be a> transcendental somewhere between X and Y; he chooses one, then> chooses digits that match its expansion.But he can't get all the numbers between X and Y. He's only choosingsome infinite subset of the digits. Nevertheless, by considering cardinality, all but countably many of the uncountably many choices player2 has available are transcendental. Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2Subject: === Subject: a digital gameath: meganewsservers.com>The transcendental numbers form a dense G_delta (as is the complement>of any countable set).>Ah, and hence the solution to the original puzzle (I think):>(a) If the player who wants to force a transcendental is player #1,> then he picks every other digit, and assigns 101001000100001...> to those digits. Any infinitely-repeating pattern in the final> number would imply an infinitely-repeating pattern in player #1's> digits; since the latter doesn't exist, neither does the former.>If there's no infinitely-repeating pattern, it's irrational, not>necessarily transcendental.After a very brief googling, I have come to the conclusion that it isvery difficult to prove that a number is transcendental.Here's a new game: player 1 picks a number that may or may notbe transcendental, player two wins if he can prove whether thenumber is transcendental or not.I think player 1 will win almost every time, but if player 2 wins,maybe he'll get a Nobel prize! >(b) If the player who wants to force a transcendental is player #2,> then he considers X (the number that would result if he assigned> all zeroes) and Y (the number that would result if he assigned> all ones). Due to density, there is guaranteed to be a> transcendental somewhere between X and Y; he chooses one, then> chooses digits that match its expansion.>But he can't get all the numbers between X and Y. He's only choosing>some infinite subset of the digits. Nevertheless, by considering >cardinality, all but countably many of the uncountably many choices player>2 has available are transcendental. Umm, aren't there uncountably many irrational numbers that aren'ttranscendental?>Robert Israel israel@math.ubc.ca>Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2GeorgeSubject: === Subject: a digital game charset=Windows-1252> Umm, aren't there uncountably many irrational numbers that aren't> transcendental?No, the set of algebraic (i.e. non-transcendental) reals is countably infinite, and the algebraic irrationals area subset of these.--r.e.s.Subject: === Subject: TopologieDave Rusin a .8ecrit:>>Let U & V be two connected open subsets of the plane R.>>If U & V have the same fondamental group, doe it mean than U & V are >>homeomophs ?>Yes, I think so. Here's one way to structure this question:> I would like to retract that statement until I've thought about it> a little more, for two reasons, one prompted by someone in email:>Using the natural complex structure on R^2, U becomes a (connected)>1-dimensional complex manifold. Its universal cover X is > I was being a little glib here. There are some technical conditions> which must be satisfied before one can make an identification like U = X/G;> usually one takes U to be locally connected and locally simply connected> or something like that, and the OP didn't impose any such constraints.> So at best my approach using covering spaces applies only to tame cases.I don't understand the argument.Every open subset odf R is locally simplmy connected and locally path connected and locally all you could imagine.>So your question becomes, if we have two groups G, H of automorphisms>of the complex plane or the unit disk, and the groups themselves are>isomorphic, are the quotient spaces X/G and X/H homeomorphic?> [...]Inn this particular case, our groups are free. It may be a little easier.>but I think I have seen this result in the two cases X=C, X=Disk> I'm going to have to track this down before I am willing to believe it> in generality. Even in a fairly tame case this is not so clear to me> right now. > If for example U and V are obtained from the plane by> removing a finite set of points, then each open set has a 1-simplex> as a deformation retract, from which it is clear that it has the > homotopy type of a bouquet of circles, and in particular, the fundamental> group is a free group on a number of generators equal to the number of> points removed; if the groups are isomorphic, this requires the numbers> of generators to be equal, so that the numbers of points removed from> the plane must be equal, making U and V homeomorphic.You can use Van Kampen directly.> But when the set of points is infinite, I would be tempted to compute> the fundamental group from a cover or something, leading to ( lim^1 ? )> computations and things which I have avoided whenever I see them coming.> For example, what if U = R^2 - {1, 2, 3, ...} Van Kampen> and > V = R^2 - {0, 1, 1/2, 1/3, ...} ? In that case U and V are obviously> homeomorphic (using inversion in C ). We have a family of inclusions> e.g. U -> U_n := R^2 - {1, 2, ..., n} and it looks to me like pi_1(U)> is the inverse limit (union) of the pi_1(U_n). Obviously the situation is> the same with the homeomorphic family of spaces V_n . But if instead we > start with V = R^2 - {1, 1/2, 1/3, ...}, then pi_1(V) must be larger than> the union of the pi(V_n), since there are paths like > f(t) = t cos(pi/t) (suitably completed to a closed loop) which are not> homotopic to a finite product of loops around single elements of R^2 - V.> I don't know for sure whether in this case pi_1(V) is isomorphic to> pi_1(U) (I suspect not) and I guess I also don't know for sure whether> V is homeomorphic to U. (This V is not locally simply-connected.)> I suppose you can make more interesting examples by removing other sets > of points from R^2, and I won't hazard a guess as to what happens if,> for example, you tell me to well-order R and then remove the first> uncountable subset of R from C ...I know the result, I've found it in a book. You can always deformation retract your open to a countable graph, so the fondamental group is free.I cannot explain it in a few words but I can give an example that genarilize( not trivially) to all cases.Let C be the Cantor set and U = RCU contain ]1/3,2/3[the rectangle ]1/3,2/3[x[-1,1] U {1/3,2/3}x([-1,1]{0}) deformation retract to{1/2}x]-1,1[ U ]1/3,2/3[x{-1,1}You can copy the retraction on every other rectangle rectangle.Thjis give you a rectaction to a graph. Yes this is a graph because the vertical lines cut the horizontal on a discrete set. So Pi1 is free with base N.So I know the fondamental group. But anyway, I dont see whyR/F is homeomorph to R/F whatever way the group acts.> So like I said, I'm back to not really knowing the answer to the> original question. > dave> PS -- the OP's original question may be adapted to ask, what if we know> that U and V have the same first integer homology group? But here> too it seems that we run into tricky issues regarding exactly which> homology theory is used and so on.I don't know well about homology. But I have a question here. Is it possible to make a triangulation of our open. Here is a similar trap I fell into once:> http://www.math-atlas.org/95/sum.vs.prodSubject: === Subject: Topologie> I don't understand the argument.> Every open subset odf R is locally simplmy connected and locally path > connected and locally all you could imagine.Not so. If you take the plane and remove, say, the Cantor set, whatis left is open and certain not locally connected nor locally simplyconnected. If you take two locally connected and locally simplyconnected open sets of the plane then the fundamental groups are goingto be free on the number of holes and then I believe it. Note it canstill be infinite; you can remove the integers.Just to focus attention on an example, is the plane with a convergentsequence and its limit deleted homeomorphic to the plane with twoconvergent sequences and their limits deleted? I think they have thesame fundamental group. The fundamental groups of each would seem tobe a direct limit of finitely generated free groups which is thecountably generated free group.Subject: === Subject: TopologieMichael Barr a .8ecrit:>I don't understand the argument.>Every open subset odf R is locally simplmy connected and locally path >connected and locally all you could imagine.> Not so. If you take the plane and remove, say, the Cantor set, what> is left is open and certain not locally connected nor locally simply> connected. Of course it is ! Because it is OPEN.>If you take two locally connected and locally simply> connected open sets of the plane then the fundamental groups are going> to be free on the number of holes and then I believe it. Note it can> still be infinite; you can remove the integers.> Just to focus attention on an example, is the plane with a convergent> sequence and its limit deleted homeomorphic to the plane with two> convergent sequences and their limits deleted? I think they have the> same fundamental group. Yes they have.>The fundamental groups of each would seem to> be a direct limit of finitely generated free groups which is the> countably generated free group.Yes it is so.And the real question is ...............Are they homeomorphic ?To answer yes or no is far from trivialSubject: === Subject: Topologie> I don't understand the argument.> Every open subset odf R2 is locally simplmy connected and locally path > connected and locally all you could imagine.>Not so. If you take the plane and remove, say, the Cantor set, what>is left is open and certain not locally connected nor locally simply>connected. Huh? The definitions of locally connected and locally simplyconnected that I (and apparently hubert) think are standard makeit both. That is, for every point x of the topological space U =(plane)(Cantor set), and every neighborhood V of x in U, there isa neighborhood W of x contained in V which is connected andsimply connected--for, indeed, there is (U being open) an opendisk W centered at x contained in V. The pathology (such as it is)of the embedding of the Cantor set in the plane creates no pathologyin its complement. The pathology is all at its ends >If you take two locally connected and locally simply>connected open sets of the plane then the fundamental groups are going>to be free on the number of holes and then I believe it. Note it can>still be infinite; you can remove the integers.>Just to focus attention on an example, is the plane with a convergent>sequence and its limit deleted homeomorphic to the plane with two>convergent sequences and their limits deleted? I think they have the>same fundamental group. The fundamental groups of each would seem to>be a direct limit of finitely generated free groups which is the>countably generated free group.I confess that I haven't though this through, quite, but doesn'tthe difference (plane){1, 1/2, 1/3, ..., 1/n, ..., 0} (deformation-)retract onto a Hawaiian ear-ring? The fundamental group of theHawaiian ear-ring (standard model: the union of the circles of radius1/n and center (1/n,0), for n a positive integer) is most certainly*not* a countable generated free group (as Sammy Eilenberg pointed out years ago, after the publication of a purported proof that itwas). So the fundamental group of (plane){1,1/2,1/3,...,1/n,...,0} wouldn't be free either.Lee RudolphSubject: === Subject: Topologie> I confess that I haven't though this through, quite, but doesn't> the difference (plane){1, 1/2, 1/3, ..., 1/n, ..., 0} (deformation-)> retract onto a Hawaiian ear-ring? The fundamental group of the> Hawaiian ear-ring (standard model: the union of the circles of radius> 1/n and center (1/n,0), for n a positive integer) No the deformation retract goes onto a collection of circle but not yours.The circles here have centers whose center formes a discrete set.So the fondamental group is countable free.Subject: === Subject: Topologie>Just to focus attention on an example, is the plane with a convergent>sequence and its limit deleted homeomorphic to the plane with two>convergent sequences and their limits deleted? I think they have the>same fundamental group. The fundamental groups of each would seem to>be a direct limit of finitely generated free groups which is the>countably generated free group.> I confess that I haven't though this through, quite, but doesn't> the difference (plane){1, 1/2, 1/3, ..., 1/n, ..., 0} (deformation-)> retract onto a Hawaiian ear-ring? The fundamental group of the> Hawaiian ear-ring (standard model: the union of the circles of radius> 1/n and center (1/n,0), for n a positive integer) is most certainly> *not* a countable generated free group (as Sammy Eilenberg pointed > out years ago, Yes but this example is compact so not good intuition.>after the publication of a purported proof that it> was). So the fundamental group of (plane){1,1/2,1/3,...,1/n,...,0} > wouldn't be free either.In fact it is. So is the complement of the Cantor set.And I explain how it deformation retract onto a graph arlier in this post.Subject: === Subject: Die Petry, die>> So what is Petry's mistake?>>> Petry started off asserting that observability matters, without>> consideration of what it means to observe. Actually, the question of what it means to observe is> exactly the issue I was addressing.> Was it?> But the point is that you failed to demonstrate the relevance> of your peculiar notions of seeing and observation> to mathematics.> Silly circular discussion.Thsnk you for summarizing your position in a nutshell.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_Subject: === Subject: Die Petry, die> So what is Petry's mistake?>>> Petry started off asserting that observability matters, without>> consideration of what it means to observe.> Actually, the question of what it means to observe is> exactly the issue I was addressing.> Was it? But the point is that you failed to demonstrate the relevance> of your peculiar notions of seeing and observation> to mathematics.> Silly circular discussion.You have the power to break that circle.> If you define mathematics to be the implications of set theory,> then indeed the notion of observability has no relevance to> mathematics.You seem to be the one determined to define it in that way. > But if we agree from the outset that mathematics> is intended to be a tool we can use to help us understand the> world around us That is only one of the uses, and the one of least interest to the pure mathematician, though of primary interest to the applied mathematican.- the world we observe - then the notion> of observability that I advocate is precisely the notion that> will keep mathematics from drifting off into the world> of make-believe.It is prfecisely that make-believe world that is of greatest interest to a large number of mathematicians, for example, G.H. Hardy.Sometimes the results of that make-believe world turn out to have profound effects on the real world, such as the effects of number theory results on the security of those electronic transactions by which billions of dollars are transferred every day.If Petry's limitations has been in effect, much of that number theory might still await discovery.Subject: === Subject: Die Petry, die>Sometimes the results of that make-believe world turn out to have >profound effects on the real world, such as the effects of number >theory results on the security of those electronic transactions by which >billions of dollars are transferred every day.I would like to believe that number theory has more meaning than justenabling people to buy crap on-line or letting governments spy ontheir people.-- I'm not interested in mathematics that might have anythingto do with reality. -- Russell Easterly, in sci.mathSubject: === Subject: How to solve this differention equation?> Assume g(x) is a a differentiable scalar function of x,> and the equation is> (g'(x))^2 = 1/(1+g^2(x)), where g'(x) is the first order> derivative of g(x) w.r.t x.> So given the above equation, how to get the> explicit form of g(x) or implicit value of g(x)> give a particular value x.In this case a parametric solution might be more convenient for somepurposes, and you can get one easily by taking g = (t^2 - 1)/(2t),so that dg/dx = 2t/(t^2 + 1) (absorbing the sign option into t)and dg/dt = (t^2 + 1)/(2.t^2), which together give:dx/dt = dg/dt / dg/dx = (t^2 + 1)^2 / (4.t^3) = t/4 + 1/(2.t) + 1/(4.t^3)giving: x = t^2/8 + log(t)/2 - 1/(2.t^2) + CCheersJohn RamsdenSubject: Want ro receive A+ paper?If yes, visit http://www.Bestessays.comand win A+ custom written essay!CindySubject: === Subject: Want ro receive A+ paper?> If yes, visit http://www.Bestessays.com> and win A+ custom written essay!These folks will even write a doctoral dissertation for you. Is this a great country or what?Subject: === Subject: Want ro receive A+ paper?> If yes, visit http://www.Bestessays.com> and win A+ custom written essay!> CindyHee-hee. Check out the slogan at the top of their website. letprofessionals take care of you papersNathanSubject: === Subject: Want ro receive A+ paper?> If yes, visit http://www.Bestessays.com> and win A+ custom written essay!> CindyI wonder if they do mathematics homework, too?Subject: === Subject: Want ro receive A+ paper?anniel@nym.alias.net.invalid:> If yes, visit http://www.Bestessays.com> and win A+ custom written essay! Cindy> I wonder if they do mathematics homework, too?I wonder if they will buy any of my work? A starving graduate student could always use some extra income to buy more books (and maybe food after I have enough books). I've gotten an 'A+' on a few papers. But then, how many people really care about inverse problems and regularization of gravity surveying? Much less 15 pages of it. - Tim-- Timothy M. BrauchGraduate StudentDepartment of MathematicsWake Forest Universityemail is:news (dot) post (at) tbrauch (dot) comSubject: === Subject: Data analysis software> It plots and analyses any x-y data for peak location, peakheight,> peak> width, semi-derivative, derivative, integral, semi-integral,> convolution,> deconvolution, curve fitting, and separating overlapped peaksand> background.> www.chemSoftware.comSubject: === Subject: Data analysis softwareAlthough I might be marginally interested in a data analysis tool forgambling inquiries related to complexity, synchronizations, etc., I'dhave to wonder about the quality of the software from somebroke-weenie who places his ads on free message boards.Very Respectfully,RaySubject: === Subject: easy problem....> f:R->R> f(x+y) = f(x) + f(y),> show that if> lim f(x+y) = L> x->0> then L = 0> ----------------------------> um.....it is right problem??*******************************************Hi:As it is given the above is not true: take f(x) = x, as counter example, sothat f(x + 1) --> f(1) = 1 when x --> 0...Good luck!TonioSubject: === Subject: Permutation Group Automorphisms> This is probably a trivial and/or basic question, but so far I> haven't been able to prove nor disprove it myself, nor find the> answer in any book I have on hand:> If G is a group with an (outer) automorphism that mixes 2> conjugacy classes of subgroups, are the images of the actions of> G on the cosets of the 2 classes necessarily permutation> isomorphic?writing the question down and posting it focused my thoughtsenough to see that the answer is `yes', and that it's indeedessentially trivial.But now I'm wondering about a partial converse: If G hasinequivalent faithful transitive actions whose images arepermutation isomorphic, does G necessarily have an (outer)automorphism linking the non-conjugate point stabilizers? (Maybeposting this will once again focus me enough to see the answermyself. :-) Note that the conjecture is easily seen to be false if therequirement that the actions be faithful is dropped: C_4 x C_2and the dihedral group of order 8 are counterexamples in degree2.-- Jim HeckmanSubject: === Subject: Permutation Group Automorphisms> This is probably a trivial and/or basic question, but so far I> haven't been able to prove nor disprove it myself, nor find the> answer in any book I have on hand:> If G is a group with an (outer) automorphism that mixes 2> conjugacy classes of subgroups, are the images of the actions of> G on the cosets of the 2 classes necessarily permutation> isomorphic?>writing the question down and posting it focused my thoughts>enough to see that the answer is `yes', and that it's indeed>essentially trivial.>But now I'm wondering about a partial converse: If G has>inequivalent faithful transitive actions whose images are>permutation isomorphic, does G necessarily have an (outer)>automorphism linking the non-conjugate point stabilizers? (Maybe>posting this will once again focus me enough to see the answer>myself. :-)> Note that the conjecture is easily seen to be false if the>requirement that the actions be faithful is dropped: C_4 x C_2>and the dihedral group of order 8 are counterexamples in degree>2.I think the answer is again yes. Let the actions on the cosets of subgroupsH1, H2 be f1: G -> I1, f2: G -> I2. Since they are permutation isomorphic,there is an isomorphism f: I1 -> I2, which maps a stabilizer f1(H1) in I1to a stabilizer in I2, which is a conjugate of f2(H2). Since the actions arefaithful, f1 and f2 are isomorphisms, and the appropriate composite is anautomorphism G -> G mapping H1 to a conjugate of H2.Derek Holt.to G, so f induces an automorphism of G mapping H1 to H2. Subject: LatLong inside polygonHere is the problem:I have a LatLong pair: upper-left corner and lower-right. The upper-rightand lower-left can ofcourse be extracted from these. Each side in thesquare can be several hundreds of kilometers apart.Now, how do I tell if a LatLong is inside this square? I find this rathertricky, especially if one of the poles is inside the square.Can anyone help?Subject: === Subject: LatLong inside polygon> Here is the problem:> I have a LatLong pair: upper-left corner and lower-right. The upper-right> and lower-left can ofcourse be extracted from these. Each side in the> square can be several hundreds of kilometers apart.> Now, how do I tell if a LatLong is inside this square? I find thisrather> tricky, especially if one of the poles is inside the square.What do you mean with square? Are all angles 90 degrees? Or do you meanthat all four sides have constant latitude or longitude?-Michael.Subject: === Subject: LatLong inside polygon <404f1491$0$1643$edfadb0f@dread14.news.tele.dk> Here is the problem:> I have a LatLong pair: upper-left corner and lower-right. The upper-right> and lower-left can ofcourse be extracted from these. Each side in the> square can be several hundreds of kilometers apart.> Now, how do I tell if a LatLong is inside this square? I find this>rather> tricky, especially if one of the poles is inside the square.>What do you mean with square? Are all angles 90 degrees? Or do you mean>that all four sides have constant latitude or longitude?The constant latitude or longitude shape would never contain a pole unlessit spanned more than 180 degrees of longitude, so that's probably not whathe means.The reasonable definition of a square on a spherical surface would be 4equal angles (approaching 90 as the size of the square decreases) and 4equal great-circle arcs. If you connected the vertices with straighttunnels, they would form a real square.To answer the question of whether a given point is in the interior orexterior, you might need to convert to rectangular coordinates. Project the surface point onto the plane containing the 4 vertices. If thedistance between the surface point and the projection point is greater thanthe radius of the earth, it's on the exterior (opposite side, in fact). Otherwise you can use vector methods to see if a trip around the squareencloses the projection point. If P is the projection point and ABCD is the square, check the vector crossproducts PAxPB, PBxPC, PCxPD, and PDxPA. If they all have the samedirection, it's in the interior. Opposite directions (should be 2 each way)mean it's on the exterior. <0,0,0> means colinear, which is either on theborder or exterior.--Keith Lewis klewis {at} mitre.orgThe above may not (yet) represent the opinions of my employer.Subject: [Newbie] What is the interest to use eigenvectors ?I don't see why to use the eigenvectors.Have you some example in geometry ?Thank youMarcSubject: === Subject: [Newbie] What is the interest to use eigenvectors ?> I don't see why to use the eigenvectors.> Have you some example in geometry ? Example 1. Eigenvectors are orthogonal to geometry. Example 2. Geometry is the dominant eigenvector of physics. Example 3. Euclid is perpendicular to Gauss. Example 4. Category Theory is an Eigenvector of Turing Machines.> Thank you> MarcSubject: === Subject: [Newbie] What is the interest to use eigenvectors ?En el mensaje:404f003f$0$5915$7a628cd7@news.club-internet.fr,marc escribi.97:> I don't see why to use the eigenvectors.> Have you some example in geometry ?The directions of the axis of a conic/cuadric are of the eigenvctors ofassociated matrix.-- Best regards,Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.comSubject: === Subject: [Newbie] What is the interest to use eigenvectors ?> I don't see why to use the eigenvectors.> Have you some example in geometry ?If you diagonalize a linear map (or matrix), the entries in the diagonalmatrix are just the eigenvalues - and the transformation matrix consists ofthe eigenvectors.Also, for matrices that are not diagonalizable, eigenvalues and vectors arecrucial: see Jordan normal form. Also, the connection eigenspaces todecomposition of finitely generated modules over principal rings gave mepersonally a lot of insight.-- I'm on warm milk and laxativesCherry-flavored antacidsreverse my forename for mail! - saibotSubject: === Subject: [Newbie] What is the interest to use eigenvectors ?>I don't see why to use the eigenvectors.>Have you some example in geometry ?For a rotation matrix, the axis of the rotation is parallel to theeigenvector on the plane.-- I'm not interested in mathematics that might have anythingto do with reality. -- Russell Easterly, in sci.mathSubject: === Subject: [Newbie] What is the interest to use eigenvectors ?>I don't see why to use the eigenvectors.>Have you some example in geometry ?>For a rotation matrix, the axis of the rotation is parallel to the>eigenvector on the plane.That's true only in 3D. And I don't know what on the plane means.Subject: === Subject: [Newbie] What is the interest to use eigenvectors ?Oh, no! motivationWasn't there just a thread on how bad itis to provide motivation for mathematical concepts?> I don't see why to use the eigenvectors.> Have you some example in geometry ?> Thank you> MarcSubject: === Subject: [Newbie] What is the interest to use eigenvectors ?: I don't see why to use the eigenvectors.: Have you some example in geometry ?Keep looking, they're used all over the place.-JustinSubject: === Subject: [Newbie] What is the interest to use eigenvectors ?>I don't see why to use the eigenvectors.>Have you some example in geometry ?Is it of interest to you to know that every rigid motion ofEuclidian 3-space R^3, fixing the origin and preserving theorientation of space, transforms the sphere {(x,y,z) in R^3 : x^2+y^2+z^2=1} by a rotation aboutsome axis? Certainly the simplest proof of that, whichI know, uses eigenvectors.Lee RudolphSubject: === Subject: Differential equation>>d d* ln(zz*) = 2pi delta(z)delta(z*)> Presumably delta denotes a Dirac delta, ie a point mass> at the origin. If so I have no idea what delta(z)delta(z*)> is supposed to be. In fact that Laplacian of ln(zz*)> is some multiple of _delta_.>delta(z)delta(z*) means the two-dimensional Delta distribution, which>is often written as a product of two one-dimensional ones.If you say so. I've never seen it written as such a product.In any case, saying delta(x)delta(y) might make sense,but delta(z) seems to me like it's _already_ a two-dimensionaldelta function!(If not: exactly what _does_ delta(z) mean? The integralof f(z) delta(z) is what, according to whatever definitionyou have in mind?)Anyway, now I think I know what you mean.>>Using ln(zz*) = lnz + lnz*, the divergence>>theorem and the residual theorem, I get>>Int(d^2z, ln(zz*)) = 4pi>>which gives a normalization of 2 for the delta distribution.>>I used d d* ln(zz*) = dd*(lnz + lnz*) = d*(1/z) + d(1/z*).>>On the other Hand,>>d d* ln(zz*) = d(1/z*) = d*(1/z). Integrating this gives the correct>>normalization Int(delta dist.) = 1.> Hard to say, since you didn't show the details of the> actual calculation.>Look at the quotation above. I used the divergence theorem, and the>theorem on residuals (hope this is the correct english term) in>complex analysis to first make a line integral out of the integral>over the complex plane and then calculate the residuals encircled by>the curve of the line integral.The correct English term is residue theorem. That's no problem,I assumed that that was what you meant.You can't integrate these things over the complex plane;they're not integrable. You can't apply the divergence theorem;things are not smooth.Or maybe you can, but most of the calculations belowdon't make any sense to me as they stand, theyneed to be justified carefully.> You say you used> d d* ln(zz*) = dd*(lnz + lnz*) = d*(1/z) + d(1/z*);> whether that's correct or not depends on exactly> what you mean, and exactly _where_ you> applied this equality. (It's not true in the> sense of distributions in the plane, for example...)>OK, I write the calculation in ASCII:>dd*ln(zz*) = 2pi delta^2(z,z*) <- two-dimensional Delta-Distribution>Now I am using ln(ab)=ln(a)+ln(b)This is not necessarily true for complex numbers; a complexnumber has many logs, and whether the log of a product isthe sum of the logs depends on exactly which ones youchoose. (There's no way to define log(z) for all complexz <> 0 in such a way as to make the formula right forall a, b.)>dd*(lnz + lnz*) = 2pi delta(z,z*)>d*(1/z) + d(1/z*) = 2pi delta(z,z*) Integrating over the complex plane>Int(dz, d*(1/z) + d(1/z*)) = 2pi Int(d^2z delta(z,z*))>The left Integral can be written as the line integral>i Int(dz*/z* - dz/z) = - 2*2pi i>with the curve encircling the origin. Calculating the Residue gives>the right hand side.>Now compare this with above:>2pi Int(d^2zdelta^2(z,z*)) = 2*2pi>this means that the Delta distribution has normalization two.>Going through the same calculation with dd*(lnzz*) = d*(1/z) = d(1/z*)>one gets the correct normalization one.>Ren.8e.************************Subject: === Subject: 0 and natural numbers> I'm not sure. But using my program, I have managed to show:> ALL(a):ALL(b):[a e n & b e n => EXIST(c):[c e z & (a,b) e c]> & ALL(c1):ALL(c2):[c1 e z & c2 e z & (a,b) e c1 & (a,b) e c2 => c1=c2]]> The equivalence relation between (a0,b0) and (a1,b1) can be expressed as> EXIST(c):[c e z & (a0,b0) e c & (a1,b1) e c]> The EXIST(c):[c e z & (a,b) e c] part you proved corresponds to the> reflexive property. The symmetric property is actually trivial with> this expression. (Other expressions of the equivalence relation would> have this property follow from [a+d=c+b <=> b+c=d+a].) And then your> proof of the last part probably looks a lot like a proof of the> transitive property.> In other words, for every pair of natural numbers a,b, there exists aunique> integer c such that (a,b) e c. I think this will allow me to construct> integer addition and multiplication based on addition and multiplicationfor> the natural numbers, which I have already established. Again, thisproject> is very much a work in progress.> Looking back at your definition of z:> ALL(i):[i e z <=> i e p & EXIST(a):EXIST(b):[a e n & b e n &> ALL(c):ALL(d):[c e n & d e n => (c,d) e i <=> a+d=c+b]]]> You might want to show the following:> ALL(i):ALL(a):ALL(b):ALL(c):ALL(d):> [a e n & b e n & c e n & d e n & i e z & (a,b) e i & (c,d) e i => a+d=c+b]I will soon complete a proof of something similar:ALL(i):ALL(j):ALL(a):ALL(b):ALL(c):ALL(d):[i e z & j e z & (a,b) e i & (c,d) e j => [i=j <=> a+d=c+b]]This will be a useful result for the development of integer addition in mysystem.DanFree download of DC Proof 1.0 at http://www.dcproof.comSubject: === Subject: 0 and natural numbers> I'm not sure. But using my program, I have managed to show:> ALL(a):ALL(b):[a e n & b e n => EXIST(c):[c e z & (a,b) e c]> & ALL(c1):ALL(c2):[c1 e z & c2 e z & (a,b) e c1 & (a,b) e c2 => c1=c2]]> In other words, for every pair of natural numbers a,b, there exists a unique> integer c such that (a,b) e c. I think this will allow me to construct> integer addition and multiplication based on addition and multiplication for> the natural numbers, which I have already established. Again, this project> is very much a work in progress.Using your program to show the following should be very useful for thedevelopment of integer addition and multiplication:ALL(z1):ALL(z2):[z1 e z & z2 e z => EXIST(z3):[z3 e z & ALL(a1):ALL(b1):ALL(a2):ALL(b2):[(a1,b1) e z1 & (a2,b2) e z2 = (a1+a2,b1+b2) e z3]]]ALL(z1):ALL(z2):[z1 e z & z2 e z => EXIST(z3):[z3 e z & ALL(a1):ALL(b1):ALL(a2):ALL(b2):[(a1,b1) e z1 & (a2,b2) e z2 = ((a1*a2)+(b1*b2),(a1*b2)+(b1*a2)) e z3]]]-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 WSubject: === Subject: 0 and natural numbers> I'm not sure. But using my program, I have managed to show:> ALL(a):ALL(b):[a e n & b e n => EXIST(c):[c e z & (a,b) e c]> & ALL(c1):ALL(c2):[c1 e z & c2 e z & (a,b) e c1 & (a,b) e c2 => c1=c2]]> In other words, for every pair of natural numbers a,b, there exists aunique> integer c such that (a,b) e c. I think this will allow me to construct> integer addition and multiplication based on addition and multiplicationfor> the natural numbers, which I have already established. Again, thisproject> is very much a work in progress.> Using your program to show the following should be very useful for the> development of integer addition and multiplication:> ALL(z1):ALL(z2):[z1 e z & z2 e z => EXIST(z3):[z3 e z &> ALL(a1):ALL(b1):ALL(a2):ALL(b2):[(a1,b1) e z1 & (a2,b2) e z2 => (a1+a2,b1+b2) e z3]]]> ALL(z1):ALL(z2):[z1 e z & z2 e z => EXIST(z3):[z3 e z &> ALL(a1):ALL(b1):ALL(a2):ALL(b2):[(a1,b1) e z1 & (a2,b2) e z2 => ((a1*a2)+(b1*b2),(a1*b2)+(b1*a2)) e z3]]]> -- that I find myself. A real torture test -- but fun!DanFree download of DC Proof 1.0 at http://www.dcproof.comSubject: === Subject: what is the z-transform of sinc function?> Can anybody tell me what is the z-transform of sinc function and> what is its region of convergence?> -Joenyim> Does this help?> sin(x) = x - (x^3)/3! + (x^5)/5! - .... (converges everywhere)> so sinx/x evaluates to:> 1 - (x^2)/3! + (x^4)/5! - ... > (still infinite converges everywhere, even at x=0)> with BZT applied as f(z) = f(s) where s=(z-1)/(z+1) this gives:> 1- ((z-1)/(z+1))^2/3! + ((z-1)/(z+1))^4/5! - ....> I'm not sure if this converges at z=-1.> I guess, it should, but I'm not able to see it from the formula.> BernhardI published my FOUR-BOX algorithm at http://www.wehner.org/euler/ .To create sin(x) you subtract my box 4 from my box 2.The convergence is FACTORIAL.Sin(x)/(x) therefore has the same convergence as sin(x), as youmultiply each term in the four-box structure by X and divide by thenext number up. Eventually, that next number overtakes argument x, andthe term converges.I call the standard McLaurin-Taylor formulae unilateral becausethere is a bilateral form, as described on that page.Charles Douglas WehnerSubject: quadratic formI think the following problem looks so easy such that I fell unsure about it. Consider the following quadratic form in R^n:f(x) = -(1/2)*x'Qx - c'xwhere- Q is symmetric- Q is indefinite- all n eigenvalues of Q are non-zeroSince all eigenvalues are non-zero Q is non-singular. Hence, inv(Q) exists. Then there is a unique saddle point x withx = -inv(Q)cfor the following reason:The gradient of f isdf(x) = -Qx - cSince the Hessian Q is indefinite, there exists no local extremum. Hence, x is a saddle point where eigenvectors of negative eigenvalues are directions of increasing f and eigenvectors of positive eigenvalues are directions of decreasing f.Is this argumentation correct?Subject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs>) It's a distinction without a difference. Whatever number is picked is>) a number you would have said (prior to the pick) would be impossible>) to be picked.>You can't go back in time and make me pick exactly that number.Doesn't matter.>(The fact that I won't be able to describe that number is beside the point.)Agreed.>If you pick two numbers randomly from [0..1], is it possible that those>two numbers are the same ?Yes. But it is not possible that either of those numbers is 1.5.-- Matthew T. Russotto mrussotto@speakeasy.netExtremism in defense of liberty is no vice, and moderation in pursuitof justice is no virtue. But extreme restriction of liberty in pursuit of a modicum of security is a very expensive vice.Subject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs)>If you pick two numbers randomly from [0..1], is it possible that those)>two numbers are the same ?) Yes. But it is not possible that either of those numbers is 1.5.So even though they won't be the same no matter how many times you try,you say that it's still possible ?SaSW, Willem-- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you !#EOTSubject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs>)>If you pick two numbers randomly from [0..1], is it possible that those>)>two numbers are the same ?>) Yes. But it is not possible that either of those numbers is 1.5.>So even though they won't be the same no matter how many times you try,>you say that it's still possible ?Who says they won't be the same no matter how many times I try? That'sthe way to bet, but it's not a certainty.-- Matthew T. Russotto mrussotto@speakeasy.netExtremism in defense of liberty is no vice, and moderation in pursuitof justice is no virtue. But extreme restriction of liberty in pursuit of a modicum of security is a very expensive vice.Subject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs>) Each rational number (and some irrational numbers) can be described in>) finite space and time.>Why do you guys keep repeating that ? At least I try to change my>statement each time to try and get the point across.>I'm not disagreeing with that statement, just saying that *given* finite>space and time, there are rational numbers you can *not* describe.>This is a different statement.It's the opposite of that statement, and it's false. Given finitespace and time, there is no rational number you cannot describe.-- Matthew T. Russotto mrussotto@speakeasy.netExtremism in defense of liberty is no vice, and moderation in pursuitof justice is no virtue. But extreme restriction of liberty in pursuit of a modicum of security is a very expensive vice.Subject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs)>I'm not disagreeing with that statement, just saying that *given* finite)>space and time, there are rational numbers you can *not* describe.)>This is a different statement.) It's the opposite of that statement, and it's false. Given finite) space and time, there is no rational number you cannot describe.I'll give you the finite space and time: the space is one piece of A4-sizepaper, the time is one hour. Now explain to me how you are going todescribe any one rational number.(If you want, I can make the space 'our planet' and the time 'one year'.)SaSW, Willem-- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you !#EOTSubject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs>)>I'm not disagreeing with that statement, just saying that *given* finite>)>space and time, there are rational numbers you can *not* describe.>)>This is a different statement.>) It's the opposite of that statement, and it's false. Given finite>) space and time, there is no rational number you cannot describe.>I'll give you the finite space and time: the space is one piece of A4-size>paper, the time is one hour. Now explain to me how you are going to>describe any one rational number.Oh, that's easy. I'm going to write p=, then I will write thedecimal form of the numerator in lowest terms. Then I'm going towrite q=, and write the decimal form of the denominator in lowestterms. I might have to write really really small and really reallyfast, of course.You got at what you were trying to say earlier -- that given aparticular finite space, a particular finite time, and a particularmethod of description, there are rationals which cannot be described.But that's not what you've been saying most of the time. Givenfinite space doesn't imply a particular amount of space.-- Matthew T. Russotto mrussotto@speakeasy.netExtremism in defense of liberty is no vice, and moderation in pursuitof justice is no virtue. But extreme restriction of liberty in pursuit of a modicum of security is a very expensive vice.Subject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs) You got at what you were trying to say earlier -- that given a) particular finite space, a particular finite time, and a particular) method of description, there are rationals which cannot be described.)) But that's not what you've been saying most of the time. Given) finite space doesn't imply a particular amount of space.To me it does, actually.Why does 'Given finite space' not imply to you that you have been given acertain finite space ?SaSW, Willem-- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you !#EOTSubject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs>) You got at what you were trying to say earlier -- that given a>) particular finite space, a particular finite time, and a particular>) method of description, there are rationals which cannot be described.>)>) But that's not what you've been saying most of the time. Given>) finite space doesn't imply a particular amount of space.>To me it does, actually.>Why does 'Given finite space' not imply to you that you have been given a>certain finite space ?Because that's not what the phrase has meant in any mathematics courseI have taken or text I have examined.-- Matthew T. Russotto mrussotto@speakeasy.netExtremism in defense of liberty is no vice, and moderation in pursuitof justice is no virtue. But extreme restriction of liberty in pursuit of a modicum of security is a very expensive vice.Subject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs)>To me it does, actually.)>Why does 'Given finite space' not imply to you that you have been given a)>certain finite space ?) Because that's not what the phrase has meant in any mathematics course) I have taken or text I have examined.Then what has it meant in those mathematics courses and texts ?SaSW, Willem-- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you !#EOTSubject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs )>To me it does, actually.> )>Why does 'Given finite space' not imply to you that you have been given a> )>certain finite space ?> ) Because that's not what the phrase has meant in any mathematics course> ) I have taken or text I have examined.> Then what has it meant in those mathematics courses and texts ? Usually, given X means, Somebody gives you something. Assumingit's X, go with it. So given finite space means given somespace -- doesn't matter how much, just assume it's finite. Finiteand less than some fixed amount K mean different things. Example: I can prove that any integer's decimal expansion fitsin finite space. IOW, one might say that given finite space, youcan represent any integer. I cannot prove (because it's false) that any integer's decimalexpansion fits in space bounded by some fixed K. IOW, one mightsay that given space K, some integer won't fit in it. Again, it's all about the positions of the quantifiers. Englishis really bad at expressing logical propositions, because we tendto be very free with the position in the sentence of quantifierslike for all, for any, and for some. When in doubt (as now), just use math-like notation. forall integers x, exists finite space K such that x fits in K. notexists finite space K such that forall integers x: x fits in K.-ArthurSubject: === Subject: puzzle: GCDs of Infinite Set of Integer PairsWe seem to be arguing very vague semantics of English here, which isnt'tquite that interesting, but hey, let's go with the flow ey ?)> Then what has it meant in those mathematics courses and texts ?) Usually, given X means, Somebody gives you something. Assuming) it's X, go with it. So given finite space means given some) space -- doesn't matter how much, just assume it's finite. Finite) and less than some fixed amount K mean different things.My train of thought is that 'Somebody gives you something' implies thatthat 'something' has thus been fixed, by the act of being given.) Again, it's all about the positions of the quantifiers. English) is really bad at expressing logical propositions, because we tend) to be very free with the position in the sentence of quantifiers) like for all, for any, and for some.) When in doubt (as now), just use math-like notation.Yeah, shoulda done that earlier.) forall integers x, exists finite space K such that x fits in K.)) notexists finite space K such that forall integers x: x fits in K.forall finite spaces K, sizeof ( all integers x: x fits in K ) is finite.Something like that ?SaSW, Willem-- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you !#EOTSubject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs> ) You got at what you were trying to say earlier -- that given a> ) particular finite space, a particular finite time, and a particular> ) method of description, there are rationals which cannot be described.> )> ) But that's not what you've been saying most of the time. Given> ) finite space doesn't imply a particular amount of space.> To me it does, actually.> Why does 'Given finite space' not imply to you that you have been given a> certain finite space ?I agree with Willem on this point. Where he keeps tripping up (as Isee it) is saying there are rationals which cannot be described -which wrongly implies there is a given rational which cannot bedescribed - when he really means one will consume the space describinga finite number of rationals and thus fail to describe an infinitenumber of other rationals.Subject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs) I agree with Willem on this point. Where he keeps tripping up (as I) see it) is saying there are rationals which cannot be described -) which wrongly implies there is a given rational which cannot be) described - when he really means one will consume the space describing) a finite number of rationals and thus fail to describe an infinite) number of other rationals.Well, no... If you choose the right rational, you'll consume all thespace trying to describe just that one number. Worse even, there are onlya finite number of rationals that *won't* consume all the space and thensome with their description.Of course, I'm implicitly assuming that there is some method of descriptionalready present, and you don't go around saying stuff like you'll use '42'to describe that number, because then there'd be *other* numbers that can'tbe described inside that space.SaSW, Willem-- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you !#EOTSubject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs> ) I agree with Willem on this point. Where he keeps tripping up (as I> ) see it) is saying there are rationals which cannot be described -> ) which wrongly implies there is a given rational which cannot be> ) described - when he really means one will consume the space describing> ) a finite number of rationals and thus fail to describe an infinite> ) number of other rationals.> Well, no... If you choose the right rational, you'll consume all the> space trying to describe just that one number. Worse even, there are only> a finite number of rationals that *won't* consume all the space and then> some with their description.This touches on the definition of given finite space, which you andothers have debated elewhere in the thread.If I understand you correctly, you are saying for each finite space,there is a rational number whose description does not fit in it.I am saying the following things:1) For each finite set of rational numbers, there is some finite space large enough to contain the description of all members simultaneously.2) For each infinite set of rational numbers, there is no finite space large enough to contain the description of all members simultaneously.and one which has just occurred to me:3) For some infinite sets of rational numbers, there is no finite space large enough to contain the description of each member one at a time. (In other words, for each finite space, the infinite set contains at least one member whose description doesn't fit into that space.)Subject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs)> Well, no... If you choose the right rational, you'll consume all the)> space trying to describe just that one number. Worse even, there are only)> a finite number of rationals that *won't* consume all the space and then)> some with their description.) This touches on the definition of given finite space, which you and) others have debated elewhere in the thread.True.) If I understand you correctly, you are saying for each finite space,) there is a rational number whose description does not fit in it.Also true. (Trivially, one might add.)) I am saying the following things:)) 1) For each finite set of rational numbers, there is some finite space) large enough to contain the description of all members simultaneously.Also true.) 2) For each infinite set of rational numbers, there is no finite space) large enough to contain the description of all members simultaneously.And also true.) and one which has just occurred to me:)) 3) For some infinite sets of rational numbers, there is no finite space) large enough to contain the description of each member one at a) time. (In other words, for each finite space, the infinite set) contains at least one member whose description doesn't fit into) that space.)I think that goes for all infinite sets, actually, and I also thinkthat implies what I was trying to say.SaSW, Willem-- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you !#EOTSubject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs)> I'm not disagreeing with that statement, just saying that *given* finite)> space and time, there are rational numbers you can *not* describe.)> This is a different statement.) It's a different statement, and it's a wrong statement. Each rational) number has a finite description.My statement is not wrong. I assume you're not reading it correctly.Let me try to state it in a more mathematical sense.For each finite space/time ST (and method of description M), there isa rational number x that can not be described in ST (using M).As a matter of fact, for each space/time ST and method M, there are only afinite amount of numbers that *can* be described in ST, using M.SaSW, Willem-- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you !#EOTSubject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs> )> I'm not disagreeing with that statement, just saying that *given* finite> )> space and time, there are rational numbers you can *not* describe.> )> This is a different statement.> ) It's a different statement, and it's a wrong statement. Each rational> ) number has a finite description.> My statement is not wrong. I assume you're not reading it correctly.> Let me try to state it in a more mathematical sense.> For each finite space/time ST (and method of description M), there is> a rational number x that can not be described in ST (using M).That is a different statement. Your first statement was wrong; this oneis merely uninteresting.> As a matter of fact, for each space/time ST and method M, there are only a> finite amount of numbers that *can* be described in ST, using M.Let q_1, q_2, ... be an enumeration of the rationals. Then yourstatement amounts to saying that for each n, the set { q_1, ..., q_n } isfinite. Huh? Why did you find it necessary to say that?-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. For each finite space/time ST (and method of description M), there is)> a rational number x that can not be described in ST (using M).) That is a different statement. Your first statement was wrong; this one) is merely uninteresting.To me, this is the same statement as my first statement.Tell me, where do you see the difference ?)> As a matter of fact, for each space/time ST and method M, there are only a)> finite amount of numbers that *can* be described in ST, using M.)) Let q_1, q_2, ... be an enumeration of the rationals. Then your) statement amounts to saying that for each n, the set { q_1, ..., q_n } is) finite. Huh? Why did you find it necessary to say that?Because that's what I've been saying all along, you just keptmisunderstanding me.The statementGiven finite space/time, you can only describe a finite amount of numbers.directly translates to the statement above.SaSW, Willem-- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you !#EOTSubject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs> )> For each finite space/time ST (and method of description M), there is> )> a rational number x that can not be described in ST (using M).> ) That is a different statement. Your first statement was wrong; this one> ) is merely uninteresting.> To me, this is the same statement as my first statement.> Tell me, where do you see the difference ?One statement mentions a specific upper bound and the other does not.> )> As a matter of fact, for each space/time ST and method M, there are only a> )> finite amount of numbers that *can* be described in ST, using M.> )> ) Let q_1, q_2, ... be an enumeration of the rationals. Then your> ) statement amounts to saying that for each n, the set { q_1, ..., q_n } is> ) finite. Huh? Why did you find it necessary to say that?> Because that's what I've been saying all along, you just kept> misunderstanding me.I did not misunderstand you. I also did not ask why you kept repeatingthe same statement. The question I asked, which you have not answered,is why you found it necessary to make your statement in the first place(before I entered the thread).> The statement> Given finite space/time, you can only describe a finite amount of numbers.> directly translates to the statement above.Your statement is ambiguous. I can think of at about three or four waysto translate it, using different placement of quantifiers.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. Because that's what I've been saying all along, you just kept)> misunderstanding me.)) I did not misunderstand you. I also did not ask why you kept repeating) the same statement. The question I asked, which you have not answered,) is why you found it necessary to make your statement in the first placeI kept repeating it because I felt you were misunderstanding it.I made it in the first place because it is a valid statement, andbecause it is a point one could use to argue what I was arguing.) (before I entered the thread).))> The statement))> Given finite space/time, you can only describe a finite amount of numbers.))> directly translates to the statement above.)) Your statement is ambiguous. I can think of at about three or four ways) to translate it, using different placement of quantifiers.If there is more than one way to interpret a statement, and all but one ofthose lead to the conclusion that it is trivially false, then why is it notobvious that the intended interpretation is the one that does *not* lead tosuch a trivial conclusion ?SaSW, Willem-- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you !#EOTSubject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs> )> The statement> )> )> Given finite space/time, you can only describe a finite amount of numbers.> )> )> directly translates to the statement above.> )> ) Your statement is ambiguous. I can think of at about three or four ways> ) to translate it, using different placement of quantifiers.> If there is more than one way to interpret a statement, and all but one of> those lead to the conclusion that it is trivially false, then why is it not> obvious that the intended interpretation is the one that does *not* lead to> such a trivial conclusion ?I can think of at least two interpretations of that statement that arenot trivially false. In fact, both are trivially true, but the meaningsare quite different.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. If there is more than one way to interpret a statement, and all but one of)> those lead to the conclusion that it is trivially false, then why is it not)> obvious that the intended interpretation is the one that does *not* lead to)> such a trivial conclusion ?) I can think of at least two interpretations of that statement that are) not trivially false. In fact, both are trivially true, but the meanings) are quite different.Well, there's the ambiguity about describing just one number, as opposedto multiple numbers, but that was the very first thing I clarified, IIRC.Or is there some other interpretation I'm missing here ?SaSW, Willem-- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you !#EOTSubject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs> ) Given finite space/time, you can only describe a finite amount of numbers.> )> )> If there is more than one way to interpret a statement, and all but one of> )> those lead to the conclusion that it is trivially false, then why is it not> )> obvious that the intended interpretation is the one that does *not* lead to> )> such a trivial conclusion ?> ) I can think of at least two interpretations of that statement that are> ) not trivially false. In fact, both are trivially true, but the meanings> ) are quite different.> Well, there's the ambiguity about describing just one number, as opposed> to multiple numbers, but that was the very first thing I clarified, IIRC.> Or is there some other interpretation I'm missing here ?There's also the difference between describing a set of numbers anddescribing all the members of the set. In fact, I thought that was theone you clarified.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. Well, there's the ambiguity about describing just one number, as opposed)> to multiple numbers, but that was the very first thing I clarified, IIRC.))> Or is there some other interpretation I'm missing here ?) There's also the difference between describing a set of numbers and) describing all the members of the set. In fact, I thought that was the) one you clarified.Oh, sorry. The other one I clarified to someone else. I'm havingdifficulty keeping track of all the different persons I'm discussingwith here.SaSW, Willem-- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you !#EOTSubject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs> For each space T, and method of description M, there are only finitely> many x in Q such that x can be described using method M in space T.Yes, that's fine. I think I've understood for a while what you'retrying to get at, namely it's possible to describe any one rationalnumber in finite time/space, but it's impossible to describe aninfinite number of rational numbers in finite time/space. Add arigorous rule that disqualifies phrases such as the set of allnumbers expressible as X/Y where X and Y are both integers and Yis non-zero, and you're good to go.> I'm not even saying there's a hole in it, I'm just asking what's so wrong> with saying that 'the probability is zero' is the same as 'it's impossible'> in everyday use. And I also happen to know a bit of probability theory,> I'm just not accepting that where mathematicians draw the line is the only> right way.I don't see anything wrong with that. My argument lies with yourideas of what things (have probability zero | are impossible) andwhat things (do not | are not).Subject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs charset=Windows-1252> There are real numbers that cannot be described, > but there is no such thing as a rational number > that cannot be described.If x is one of those reals that cannot be described,how would you describe the ratio between the firstand second term in the continued fraction of x? ;o)--r.e.s.Subject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs> There are real numbers that cannot be described, > but there is no such thing as a rational number > that cannot be described.> If x is one of those reals that cannot be described,> how would you describe the ratio between the first> and second term in the continued fraction of x? ;o)It's a ration of two integers and therefore describable.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.> There are real numbers that cannot be described, >> but there is no such thing as a rational number >> that cannot be described.> If x is one of those reals that cannot be described,> how would you describe the ratio between the first> and second term in the continued fraction of x? ;o)> It's a ration of two integers and therefore describable.Hmmm... If stating what it means to be a rational number is all it takes to describe one, then why isn't x described by simply stating the definition of a real number?--r.e.s.Subject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs>> There are real numbers that cannot be described, >> but there is no such thing as a rational number >> that cannot be described.> If x is one of those reals that cannot be described,> how would you describe the ratio between the first> and second term in the continued fraction of x? ;o)> It's a ration of two integers and therefore describable.> Hmmm... > If stating what it means to be a rational number is all > it takes to describe one, then why isn't x described by > simply stating the definition of a real number?Stating the definition of a rational number is not the same as describinga rational number. Stating what it means to be a rational number issufficient to establish that each rational number has a finitedescription, because each integer has a finite description.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.> There are real numbers that cannot be described, > but there is no such thing as a rational number > that cannot be described.>>> If x is one of those reals that cannot be described,>> how would you describe the ratio between the first>> and second term in the continued fraction of x? ;o)>>> It's a ration of two integers and therefore describable.> Hmmm... > If stating what it means to be a rational number is all > it takes to describe one, then why isn't x described by > simply stating the definition of a real number?> Stating the definition of a rational number is not the same as describing> a rational number. Stating what it means to be a rational number is> sufficient to establish that each rational number has a finite> description, because each integer has a finite description.As I'd already said in a follow-up posting, I was being less than to something I'd read in Minsky's old but excellent textbook (Computation: Finite and Infinite Machines), but may have taken too much to heart and/or misinterpreted: This intuitive notion [of a describable real number] is so vague that one may wonder whether it could have any useful precisely defined counterpart. At present this seems very unlikely, since no one has been able to point toward a way of excluding logical paradoxes, while preserving all or most of the descriptive freedom of natural language.Supposing that to be an accurate assessment, it was neverthelesswritten in the '60s, and the situation may now be quite different-- I admit to virtual ignorance on the subject, and do value yourcomments. --r.e.s. Subject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs charset=Windows-1252> Which reminds me of something tangentially related to where> this mess has now drifted: I've heard that you can well-order> the set of real numbers [such that every set of real numbers> will have a unique least member under this ordering]. How?> (Or is this one of those things that has been proven deductively,> but never by example?)That every set can be well-ordered is equivalent to the Axiom of Choice. It's my understanding that a w.o. of the reals can exist only in a non-constructive sense. Some links, in roughly decreasing order of accessibility:http://en.wikipedia.org/wiki/Axiom_of_choicehttp ://www.math.vanderbilt.edu/~schectex/ccc/choice.htmlhttp:// www.math.rutgers.edu/~ocone/choice.pdfhttp:// www.logic.univie.ac.at/~caicedo/wo-abridged.ps--r.e.s.Subject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs As I'd already said in a follow-up posting, I was being less than> to something I'd read in Minsky's old but excellent textbook> (Computation: Finite and Infinite Machines), but may have taken> too much to heart and/or misinterpreted:> This intuitive notion [of a describable real number] is> so vague that one may wonder whether it could have any> useful precisely defined counterpart. At present this> seems very unlikely, since no one has been able to point> toward a way of excluding logical paradoxes, while> preserving all or most of the descriptive freedom of> natural language.> Supposing that to be an accurate assessment, it was nevertheless> written in the '60s, and the situation may now be quite different> -- I admit to virtual ignorance on the subject, and do value your> comments. I suppose you get into paradoxes like What's the smallestinteger that cannot be described in twelve words or less? Of course, with the real numbers you can always point outthat the phrase the smallest real number that's not describableis probably meaningless -- maybe the set of non-describable reals*has* no smallest member! Which reminds me of something tangentially related to wherethis mess has now drifted: I've heard that you can well-orderthe set of real numbers [such that every set of real numberswill have a unique least member under this ordering]. How?(Or is this one of those things that has been proven deductively,but never by example?)-ArthurSubject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs> Which reminds me of something tangentially related to where> this mess has now drifted: I've heard that you can well-order> the set of real numbers [such that every set of real numbers> will have a unique least member under this ordering]. How?> (Or is this one of those things that has been proven deductively,> but never by example?)I think it uses the Axiom of Choice, and there is no finite way todescribe any such ordering.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 WSubject: === Subject: puzzle: GCDs of Infinite Set of Integer Pairs charset=Windows-1252>> There are real numbers that cannot be described, >> but there is no such thing as a rational number >> that cannot be described.> If x is one of those reals that cannot be described,> how would you describe the ratio between the first> and second term in the continued fraction of x? ;o)> ^^^^^> It's a ration of two integers and therefore describable.> Hmmm... > If stating what it means to be a rational number is all > it takes to describe one, then why isn't x described by > simply stating the definition of a real number?Oops, I forgot the smiley that time. ;o)Before this goes any further, I want to point out that what I've said above is in jest. I have no knowledge of the non-describable reals, except that I would suppose any computable real (and that certainly includesthe rationals) to not be among them.--r.e.s.Subject: === Subject: A Paradox of The Central Limit Theorem> I solved the paradox in my first post on the topic. > There are three n's in a certain expression, and you take > a limit letting two of them tend to infinity, leaving the > third just sit there.That is NOT quite true. If I had written lim_{n ->infinity} Pr( alpha*sigma/sqrt{n} + mu <=_n <= beta*sigma/sqrt{n} + mu) = Pr( mu <= _n <= mu), I wouldhave leftthird just sit there. <= beta*sigma/sqrt{n} + mu)= lim_{n ->infinity} P( mu <= _n <= mu).Thus, I am still letting the third tend to infinity. >You mentioned mistakes, but evey paradox is a kind of logical>mistakes. Don't you understand this truth?> Right. Like if I said 2 + 2 = 5 was a paradoxNon, 2 + 2 = 5 can't make a paradox, poor fellow. There is no logic in 2 + 2 = 5. That is to say, there are notherefores,no so thats etc. in 2 + 2 = 5.Subject: basis of the reals over the rationalsHi everybody!I just happened to notice a reference, a couple of monthsago, to an old posting of mine (Sept 1998). If it wasn'tclear in the early days of NNTP that those quick, informalmessages were indeed sub specie aeternitatis (and retro-actively too)... well, it was by 1998, and more so now.It is kind of you to show me confidence and quote me, soI'll say thanks by, eh, making it right!...(Baire Property, i.e. being = G delta M, G open, M meager)I should have added ... with nonempty G. That is, if H has BP, it cannot be comeager on an open set...the exact analogue of 'if H is measurable, it cannot have non-0 measure'. [At the time I had a reason for category to differ frommeasure here; of course it doesn't. H can be meager, e.g.a max. independent subset of Cantor's middle-thirds set.Seems to me there were other threads later, and I got tostate things properly... but this one I forgot.The topic can be deceptive. It seems plausible, byanalogy with other constructions of this type, thatH should be 'irregular. On more than one occasionpeople took for granted that H cannot be measurable, and were quite surprised when I told them otherwise.He, he... the category case has taken revenge.]The two properties have twin proofs. Sketch of proof:---------------If a set S violates either property, that is, if ithas BP with nonempty G [resp. is measurable with non-0measure] then the difference set S - S contains aninterval about the origin, i.e. for some c > 0 for any|t| < c there are s_1, s_2 in S with s_1 - s_2 = t{Hint: Find an interval I with small intersectionwith ~S, meaning: meager [resp. of measure < |I|/4], andconsider translates of I intersect S by x, for all sufficiently small x}Now if H - H contained an interval about the origin,pick an h in H and find n s.t. h/n is in the interval.Then h/n = difference of two members of H, violatinglinear independence.A mixed bag at best. The original question was Why can't Hbe Borel? Compared to five years ago the stuff above has amajor disadvantage, not settling the original question;... andonly one minor advantage (truth). Stock tricks are good for a quickie sometimes; e.g. a Borel Hhas a perfect subset, in effect a copy of Cantor space, whichmust be linearly independent... No cigar; such things do exist.Here is a reasonably simple approach.Background: Analytic set means continuous image of Borel set. Easy fact: A countable union of analytic sets is analytic. Luzin's Thm: X is Borel iff both X and ~X are analytic.[Analytic sets are measurable and have BP; but we don't need thisbelow. Current notation: analytic = Sigma_1_1 (boldface),co-analytic = Pi_1_1 (b.), both of the above = Delta_1_1 (b.) =Borel, proj of co-analytic = Sigma_1_2 (b), etc]For each finite nondescending sequence of non-0 rationals, say s = , let H_s be the set of corresponding reals:x is in H_s iff there exist h_0, h_1, h_2 {h_0, h_1, h_2 in H& ~ h_0=h_1 & ~ h_0=h_2 & ~ h_1=h_2 & x = q_0h_0 + q_1h_1 + q_2h_2}(with H_empty = {0}). Then {H_s} is a partition of R into countably many pieces. Suppose H is Borel.The existential quantifiers amount to projection (clearly acontinuous map) of the set in braces, which is Borel (in R^4).Hence H_s is analytic. So is its complement, being a countableunion of analytic sets; by Luzin, then, H_s is Borel.The rest is as before: one of the H_s must have non-0 measure,thus H_s - H_s contains an interval, hence some h/n, and thisviolates linear independence.The proof also works for H analytic (the set in braces is thenanalytic; proj of analytic = proj of proj of Borel = analytic).Credit for the theorem goes, I think, to Sierpinski; blame forthe proof to me. It's just routine application of elementary DST.ZFC can't do much more. Projective Determinacy implies all projective sets are measurable; with H projective the proofabove applies yet again. So ZFC + PD |- H is not projective. Onthe other hand ZFC + V=L |- H is projective (of level 2,using the Delta_1_2 wellordering of R.) IliasSubject: === Subject: Fun with axioms> .....................>> But then of course there's the ultimate reality check>> for mathematics: physics.>I believe you have that backwards. When physics is done ignoring the theoretical>coherence of mathematics you will find that a great many resources have been>unnecessarily directed toward an infinite regress of triangles.> There is a good example from physics. Some of you may have> heard about the Feynman integral. I am one of those who> looked at it (I have not published on this point), and it> is clear that Feynman assumed that there were mathematical> objects available which do not exist. I do not believe that> there is yet a reasonable fix which enables it to be used> as a direct means to evaluate quantum transitions. It is known that there is not. Since it's also known that the only Scientist less informed about the Uncertainty Principle in the entire 20th Century than Einstone, was Feynmann.Subject: === Subject: PDE (modified) Maximum Principle> I am struggling with where to start on the following problem...> Show the Maximum Principle for u_t = k(x)*u_xx on [0,L]x[0,T] with > 0 The Maximum Principle states that for the diffusion equation (u_t = > k*u_xx), the maximum (and minimum) must be reached on one of the> edges (either x=0, x=L or t=0) I've proven the Maximum Principle for u_t = k*u_xx with k a constant.> I've also proven the Comparison Principle (if u and v are solutions> to the diffusion equation with u<=v on the edges then u<=v> everywhere). I feel like this problem should be a consequence of> these two principles, but after a many futile attempts, I think I am> going the wrong direction. > Any suggestions on how to start?> The usual proof of the Maximum Principle for u_t = k * u_xx> should work also for variable k. Try that, and ask if you> get stuck somewhere.> HTH,> Michael.I think I have it, taking your advice. Since the function is bounded, the proof for the Maximum Principle seems to work in this case (using properties of derivatives). We'll just see if my professor agrees with me.Now another question, suppose k(x), or even k(x,t) is positive, but unbounded, or negative, my proof seems to fall apart. This is beyond the question I was working on, but I was just wondering is it still holds. Any thoughts? - Tim-- Timothy M. BrauchGraduate StudentDepartment of MathematicsWake Forest Universityemail is:news (dot) post (at) tbrauch (dot) comSubject: === Subject: How to solve a polynomial of degree 6 over a ring modulo a composite by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2AFSo901159;>Dear Research Community,>I have a generic polynomial equation..>f(t)= t^6 + a. t^5 + b. t^4 + c. t^3 + d. t^2 + e. t + g = 0>I need to solve the above polynomial of degree 6 over a ring modulo a>composite n (n is large of the order of 1024 bbits). Where the factors>of n is unknown and it is hard to find too. I want a solution that>does not amount to factoring n.>I have a finite but large enough set of {a,b,c,d,e,g} for which I need>to know at least one solution. > Can anybody please help?The answer to your last question is NO. Noone can help. If onecan factor/solve a polynomial modulo a composite, then one can factor that composite. e.g. look at solvingx^2 = 1 mod N. Finding the non-trivial roots immediately factors N.I would need to think about how to factor n given the roots mod n ofan irreducible sextic, but it should be doable.AFAIK the only way to solve your problem is to factor n, solve thesextic modulo each prime factor, [via any one of a number of fastalgorithms, e.g. Cantor-Zassenhaus], then use the Chinese Remainder Theorem to get a solution mod n.Subject: === Subject: Cantor Paradox by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2AFSoS01169;> Who cares if you and I don't know what's on the list? The list is> still a well-defined list (ignoring vagueness of natural language, of> course).>We can generate every finite string of ASCII characters like Nathan>suggested.>Determining if the string describes a real number could be problematic.> This is going in circles. I don't care to continue the point.> Computability is not a requirement for Cantor's diagonal argument.> Nathan's list is well-defined. The only issue is whether the diagonal> number thus constructed is expressible in the object language.>The diagonal number must be in the list.>For example, consider the string:>If the nth digit of the nth real in Nathan's list is 3 then the nth digit>of this number is 7 else the nth digit of this number is 3.>In fact, every possible definition of a diagonal number must be in the list.>The diagonal number must be on the list and it can't be on the list.>Russell>- 2 many 2 countRussell,I got tangled up with this same idea some time ago and posted a lot of incorrect stuff about it in this very newsgroup.I think the tricky aspect is that language is very slippery. Meanings change and multiply in a flash in a way that is difficult for us to keep track of. Take the string (let's call it String Q), The antidiagonal of the list of all real numbers derived from an alphabetical examination of all finite strings. This string does not specify a real number when we are attempting to build our list, but it can be taken to specify a real number after our list is built.This shift in meaning is what creates the paradoxical effect. Under the following assumptions the paradox disappears. (1) Any given string either specifies a real number or does not. (2) No string specifies more than one real number. (3) No string changes it meaning.Under these assumptions String Q causes us no trouble. If String Q does not specify a real number when we are building the list, String Q also does not specify a real number after our list is built. If String Q specifies a real number when we are building the list, then that number is included in the list and - despite how we might protest later - String Q cannot, therefore, actually specify the antidiagonal of our completed list.So under these assumptions the real numbers specifiable by finite strings are denumerable. And there is no Cantorian argument to show otherwise because the antidiagonal of any list of these numbers cannot be specified by a finite string (such as String Q) - as surprising as that fact at first seems. Danny PurvisSubject: === Subject: Cantor Paradox>.org...> Who cares if you and I don't know what's on the list? The list is>> still a well-defined list (ignoring vagueness of natural language, of>> course).>We can generate every finite string of ASCII characters like Nathan>suggested.>Determining if the string describes a real number could be problematic.>> This is going in circles. I don't care to continue the point.>> Computability is not a requirement for Cantor's diagonal argument.>> Nathan's list is well-defined. The only issue is whether the diagonal>> number thus constructed is expressible in the object language.>The diagonal number must be in the list.>For example, consider the string:>If the nth digit of the nth real in Nathan's list is 3 then the nth digit>of this number is 7 else the nth digit of this number is 3.>In fact, every possible definition of a diagonal number must be in the list.>The diagonal number must be on the list and it can't be on the list.>Russell>- 2 many 2 count> Russell,> I got tangled up with this same idea some time ago and posted a lot of > incorrect stuff about it in this very newsgroup.> I think the tricky aspect is that language is very slippery. Meanings change > and multiply in a flash in a way that is difficult for us to keep track of. > Take the string (let's call it String Q), The antidiagonal of the list of > all real numbers derived from > an alphabetical examination of all finite strings. This string does not > specify a real number when we are attempting to build our list, but it can be > taken to specify a real number after our list is built.> This shift in meaning is what creates the paradoxical effect. Under the > following assumptions the paradox disappears. (1) Any given string either > specifies a real number or does not. (2) No string specifies more than one > real number. (3) No string changes it meaning.> Under these assumptions String Q causes us no trouble. If String Q does not > specify a real number when we are building the list, String Q also does not > specify a real number after our list is built. If String Q specifies a real > number when we are building the list, then that number is included in the > list and - despite how we might protest later - String Q cannot, therefore, > actually specify the antidiagonal of our completed list.> So under these assumptions the real numbers specifiable by finite strings are > denumerable. And there is no Cantorian argument to show otherwise because > the antidiagonal of any list of these numbers cannot be specified by a finite > string (such as String Q) - as surprising as that fact at first seems. > Danny PurvisI.e., If a string's meaning is independent of the context in which it occurs, Nathan's proof fails. And if a string's meaning is dependent on the context in which it occurs, Nathan's proof fails. Neat.Subject: degrees of freedom in N dimensionsI asked about degrees of freedom...> degrees of freedom is not especially well defined.> But if you mean (something like) what is the dimension of the manifold of> possible isometric embeddings of a subset of R^n, then that's easy:you're> just asking for the dimension of the set of Euclidean motions, which> is a semidirect product of the group of translations (itself naturally> a copy of R^n) and the group of translations (dimension n(n-1)/2 ).> For example, there's a six-parameter family of ways to position an> object in 3-dimensional space.What could be a general formula for the number of degrees of freedom of anobject (not a point) in an n-dimensional (metric?) space?By degrees of freedom, I mean the number of independent ways of movingwithin the space. For example, simple translations, rotations, etc.My instinct tells me it is something like this:N = sum( d Mod(k), k, 1, d)N equals the sum of d Mod(k) as k goes from one to d. The only tricky thinghere, is that Mod differs from mod in that when x mod(k)= 0, x Mod(k)= k.where N is DOF and d is the dimensionality of the space.any challenges? thoughts? new ways of thinking about this? I have no ideaif there is a standard notation for this idea or what.Subject: === Subject: calculating expectation and optimal strategy for a two-choice game of chance> [helpful stuff]Subject: === Subject: Candidate looking for Employment> 1) Get a better handle. Nobody will consider The Lord of Chaos.Especially after pathetic Got speeding ticket thread.Subject: === Subject: Candidate looking for Employment> am looking for a job. Here's a copy of my resumenice one!> Name : Suresh kumar Devanathan> contact : mdsuresh ___@thing__ media __.thing__ mit __.thing__ edu> Experience : MIT Media Lab> Rutgers CAIP Labwhat were your duties?and why aren't you still there?> Awards : AP Scholar With Distinction> Other : Army Reserve> Worked for : Dr. Vishwani Agarwal( IEEE/ACM fellow) on VLSI ATPG> Dr. Edward Fredkin( CMU Distinguished professor) on> Cellular Automata/ SARS-TA> Public Projects:> http://www.sourceforge.net/projects/atpg> -----------------------------------------> Built this project, as a PROOF of CONCEPT for statistical> techniques, i developed> to test chips> http://www.sourceforge.net/projects/blitz> -----------------------------------------> Patch work for blitz, a highly successful C++ math library> in use, by more than> 15000 programmers worldwide> Job Skill : C++, C#, Unix, Cadence Design System, MATLAB, .NET> Education : 3 years of college in Electrical Engineering> Left college to jumpstart a partly successful> entrepreneurshiprumour is that those jobs are hard to find...Subject: === Subject: Candidate looking for EmploymentJust nitpicking...one of the clocks are incorrect.-Michael.Subject: === Subject: Candidate looking for Employment> Just nitpicking...> one of the clocks are incorrect.Uncle Al's computer's clock is set 58 seconds fast,http://www.time.gov/timezone.cgi?Pacific/d/-8/javaThen again, you know Uncle Al - always functioning 40 minutes into thefuture (twice as good as Max Headroom).-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net!Subject: === Subject: Candidate looking for Employment> ...>Education : 3 years of college in Electrical Engineering> Left college to jumpstart a partly successful>entrepreneurship> What's an entrepreneurship?Self-created business. Also known as working without pay :-)Subject: === Subject: Spectral sequences in knot theory> used to solve problems in knot theory. Any references ?>Vassiliev, VA, Complements of Discriminants of Smooth Maps; Topology and >Applications. Translated from the Russian by B. Goldfarb. Translations The idea is to calculate the cohomology of the space of all embeddings of a circle in R^3 - or some related space -using a spectral sequence. This idea gave rise to the theory of Vassiliev invariants, which quickly got taken over by quantumgroups, chord diagrams and the like. There's a decent quick intro with references here:http://mathworld.wolfram.com/VassilievInvariant.htmlbut for the details of how the spectral sequences are used,you'll have to go back to earlier work, like the monograph Aaron mentioned. Subject: Bourbaki's General TopologyHi all,I wonder if someone could please have look at Bourbaki's GeneralTopology (chapters 5-10) and tell whether or not chapter VII, section1.1 contains this proposition: if G is a discrete subgroup of (R^n,+),then there are linearly independent vectors v_1,...,v_m in R^n such thatG is the lattice generated by them.The reason I am asking this is because I would like to quote Bourbaki onthis in a text that I am writing in english. I only have access to theFrench 1974 edition (which contains that result at that place), but I'veread that the english version is a translation of an earlier frenchversion.Best regards,Jose Carlos SantosSubject: === Subject: Bourbaki's General Topology Adjunct Assistant Professor at the University of Montana.>Hi all,>I wonder if someone could please have look at Bourbaki's General>Topology (chapters 5-10) and tell whether or not chapter VII, section>1.1 contains this proposition: if G is a discrete subgroup of (R^n,+),>then there are linearly independent vectors v_1,...,v_m in R^n such that>G is the lattice generated by them.>The reason I am asking this is because I would like to quote Bourbaki on>this in a text that I am writing in english. I only have access to the>French 1974 edition (which contains that result at that place), but I've>read that the english version is a translation of an earlier french>version.I assume you mean this one, the last labeled result in the SpringerEnglish edition: THEOREM 1. Every discrete subgroup G of R^n, of rank equal to p, is generated by a free system of p points.Later, it says: Discrete subgroups of R^n which are of rank n are also called lattices in R^n.-- == It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)=== ===Arturo Magidinmagidin@math.berkeley.eduSubject: === Subject: Bourbaki's General TopologyArturo Magidin> Jos.8e_Carlos_Santos>I wonder if someone could please have look at Bourbaki's General>Topology (chapters 5-10) and tell whether or not chapter VII, section>1.1 contains this proposition: if G is a discrete subgroup of (R^n,+),>then there are linearly independent vectors v_1,...,v_m in R^n such that>G is the lattice generated by them.> I assume you mean this one, the last labeled result in the Springer> English edition:> THEOREM 1. Every discrete subgroup G of R^n, of rank equal to p, is> generated by a free system of p points.> Later, it says:> Discrete subgroups of R^n which are of rank n are also called> lattices in R^n.It says exactly the same in the 1966 English-language Hermann/Addison-Wesleyedition.LHSubject: Norm in infinite-dimensional separable inner product spaceDoes the squared norm of an element (i.e. the inner product of the elementwith itself) in an infinite-dimensional separable complex inner productspace have to be finite? If not, why? Please explain....MosheSubject: === Subject: Norm in infinite-dimensional separable inner product space>Does the squared norm of an element (i.e. the inner product of the element>with itself) in an infinite-dimensional separable complex inner product>space have to be finite? If not, why? All you need to do to answer this question is look up the_definitions_ of the terms you're using.>Please explain....>Moshe************************Subject: === Subject: Norm in infinite-dimensional separable inner product space> Does the squared norm of an element (i.e. the inner product of the element> with itself) in an infinite-dimensional separable complex inner product> space have to be finite? If not, why? Please explain....If f is real valued function, is f^2 a real valued function?Subject: === Subject: Norm in infinite-dimensional separable inner product spaceMoshe Adrian a .8ecrit:> Does the squared norm of an element (i.e. the inner product of the element> with itself) What is the definition of your inner product. If, as usual, it's a fonction form ExE to R then of course the result lie in R.If your fantasy allow your to, imagine your innner-product going toR U {+infinity} (why not), it's not so but it would be rather unual.>in an infinite-dimensionalDimension is irrelevant here.> separable irrelevant>complex inner product> space have to be finite? If not, why? Please explain....Sometimes, we replace the norm by what we calle semi-norm. It's different from a norm in two ways.Fist |x| = 0 does not mean x=0Second |x| can be infinite.I think it is that sort of thing you have seen.Subject: === Subject: Norm in infinite-dimensional separable inner product space> Does the squared norm of an element (i.e. the inner product of the element> with itself) in an infinite-dimensional separable complex inner product> space have to be finite? If not, why? Please explain....> MosheThat's the definition of in the space, right?...points with infinite norm are not in the space...-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/Subject: === Subject: New Randomness Test> If we accept your definition of random above, and if the digits >> of pi do pass his stated list of tests, then how can these digits >> not be random, according to your definition?> Because if you know how to measure it anywhere, you get the same > answer everywhere. To be more explicit, if it were random, you'd > never get the same answer twice in one place much less anywhere else.> ...my point was that > _whatever_ was given as the definition of random, anything that passes > the tests must be random. In other words, the details of how the > numbers were generated were irrelevant, unless addressed by the > definition (with its tests). I was thinking about measurement as a number generation process and its predictability (as used as a definition of nonrandom). My point is; if you compare two observables (whether in one location or two) that each express as apparently random (in the sense that they're internally unpredictable) digit sequences, but both sequences are found to be identical to whatever limits are available, are they still random? Explicitly, suppose you measure pi at one place and use that data to _predict_ what will be measured, either elsewhere or in that same location at some other time, (specifying the procedure of course) and find your prediction perfectly validated. More blatantly, if you take the first measurement to x places and the second to fewer, you can use the first to predict the rest of the second. Pi is thus predictable, but only from an external viewpoint. Do you see this as an example of two different kinds of randomness? Mark L. FergersonSubject: === Subject: New Randomness Test> But you are right that there must be breakeven points> where it is just as efficient to transmit the original> sequence as the formulaic version.> This does not alter the fact that the sequence is not> truly random, since it is entirely predictable given> the knowledge of the appropriate formula.> That fact is zero though. Since a formula is not> a prediction.> Really? Amazing. Do you expect to get different> digits everytime you use the same formula to> generate the digits of pi? Do you expect different> formulas that give the digits of pi to give different> results?> Hence pseudo-random numbers> are used because to compute the 2^3^5^7^11^13^17^19 th> digit of pi takes longer than the universe> has been around. Nevermind how long pi> formulas have been around.> Nevertheless, the 2^3^5^7^11^13^17^19th digit is given> by a formula, regardless of the computing time that> might be involved. Accessiblity is not a prerequisite> for existence in mathematics. But also regardless of idiotic pi formulii, e, sqrt(10), and probability theory are prerequistes for intelligence. If set theory retards haven't figured out the formula yet. Nobody but mathema-heads cares what's a prerequisite for existence in mathematics.> Also, what would one do with the 2^3^5^7^11^13^17^19th> digit of pi that one couldn't do with the 2^3^5^7th? Notice the pattern: Prime_n to the Prime_n+1 to the Prime_n+2 to Prime_n+3 to the dot dot dot.> Perhaps a given sequence might be thought of as> relatively random with respect to a given compression> method, sort of like numbers can be relatively prime?> But, perhaps not. Since you first> have to convince the mathema-philosophers> to develop a theory of relative that works> for longer than the expected lifetime of> tritium, which is not that long.> I have no idea what you're trying to say there. I didn't think you would, since tritium has something to do with stars rather than something to do with nowhere.Subject: === Subject: Determining the Equation of a Shadow Cast onto a Surface>Given a flat circular disk (high-gain antenna), a planar surface (a>solar array), and an arbitrary vector to a point light source (the>Sun), I would like to determine the equation of the resultant shadow>(which would be an ellipse) that would be cast onto the plane.I have posted an html version of my Maple worksheet for the shadow at:http://math.asu.edu/~kurtz/shadow.html--LynnSubject: === Subject: Determining the Equation of a Shadow Cast onto a Surface>Given a flat circular disk (high-gain antenna), a planar surface (a>solar array), and an arbitrary vector to a point light source (the>Sun), I would like to determine the equation of the resultant shadow>(which would be an ellipse) that would be cast onto the plane.If what you are really interested in is how much light is obstructed bythe antenna, what you really want is the area of the shadow. If wesimplify the problem by assuming your point source is at an infinitedistance (like the Sun), then the area of the shadow is simply 2 cos(a) pi r ------ [1] cos(s)where r is the radius of the antenna, a is the angle between the axisof the antenna and the direction to the sun, and s is the angle betweenthe normal to the solar array and the direction to the sun.The reasoning behind [1] is that the shadow is an elliptical cylinderwith axes r and r cos(a), and when such a cylinder meets a plane at anangle of s, its area is scaled by 1/cos(s).If instead, what you really do want is the shape of the shadow, that isa bit more complicated, but it can be done. If you wanted a finitedistance to the light source, then things get a lot more complicated.Rob Johnson Given a flat circular disk (high-gain antenna), a planar surface (a> solar array), and an arbitrary vector to a point light source (the> Sun), I would like to determine the equation of the resultant shadow> (which would be an ellipse) that would be cast onto the plane.Depending on whether source to antenna distance is [finite, finite,infinite] and antenna axis [is, is not, is not] aligned parallel tosource-antenna line, we have [circular conic, oblique conic, ellipticcylinder] intersections as ground plane shadows needing [5,7,6] shadowboundary points respectively to find the shadow equation. Inputsrequired to fully solve the problem are : aperture radius, (aperture /antenna- source distance) ratio , two angles of antenna orientation(not needed for normal incidence with circular conic sections) and thesource (or sun) incidence/inclination (beta) to ground plane.For low beta angles, even hyperbolic shadows could form.Subject: === Subject: Determining the Equation of a Shadow Cast onto a SurfaceIn sci.math, mikki:> Given a flat circular disk (high-gain antenna), a planar surface (a> solar array), and an arbitrary vector to a point light source (the> Sun), I would like to determine the equation of the resultant shadow> (which would be an ellipse) that would be cast onto the plane.No, it might not be an ellipse if the Sun is low enough. :-)Though admittedly it would require that the Sun be lower than thetop of the disc. :-)Anyway, you might have some work ahead of you. Define the diskas a point p (p_x, p_y, p_z), a radius r, and a normal n. Youcan either consider the disk as embedded in the plane defined byp and n, or just consider it as a disk with its boundarygiven by the point c(t) (domain: [0, 2pi), range: R^3)c_x(t) = p_x + r * cos(t) * v_x + r * sin(t) * w_xc_y(t) = p_y + r * cos(t) * v_y + r * sin(t) * w_yc_z(t) = p_z + r * cos(t) * v_z + r * sin(t) * w_zwhere v and w are unit vectors such that the triplet of vectorsv, w, n are all mutually perpendicular. For convenience, onecan take v to be parallel to the groundplane unless n isalready parallel to the groundplane, in which case one has aminor problem. I'm not sure which is better but the parametericboundary at least gives me an equation that looks solvable.The planar surface, of course, is just the Z=0 plane, though itdepends to some extent on what coordinate system you want to use.Slicing with an arbitrary plane is possible but more complicated.The idealized Sun is at s_x, s_y, s_z, s_z > 0. (s_z < 0means it's nighttime, presumably.)So now you have a large number of line segments, which willallow you to trace a parametric curve on the planar surfacegiven t, the point s, and the point c(t).I'll leave it to you to solve the equations but you should beable to get a nice parametric representation k_x(t), k_y(t)with any luck. If you actually want an equation in X and Yas opposed to a parameteric representation, you'll have somemore work to do. :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless.Subject: === Subject: Determining the Equation of a Shadow Cast onto a SurfaceAs additional information, the solar array is a group of maybe 10,000individual solar cells, and this analysis is to determine the poweroutput reduction over the entire array if a portion of those cells areshadowed from the Sun by the high-gain antenna. So, certainassumptions can simplify this analysis. Fringes can be ignored (a fewcells won't degrade the power output significantly). The Sun's disksize can be ignored.Given knowledge of the physical characteristics of this spacecraft(geometries and sizes), locations and orientations of the articulablesolar array and high-gain antenna, and solar system ephemerous, I wasplanning on using an orthoganal projection approach to model theshadow cast on the array.Yeah, I realize the shadow might project off to a parabola with somepoints of the disk at infinity if the Sun is low enough (I belive Icould use some sort of projective transformation).Subject: === Subject: Determining the Equation of a Shadow Cast onto a Surface> The idealized Sun is at s_x, s_y, s_z, s_z > 0. (s_z < 0> means it's nighttime, presumably.)It's probably relevant to assume that the sun is infinitely far away. Butwhat about the size of the sun's disk; can we neglect that? I doubt it.-Michael.Subject: === Subject: Determining the Equation of a Shadow Cast onto a Surface>Given a flat circular disk (high-gain antenna), a planar surface (a>solar array), and an arbitrary vector to a point light source (the>Sun), I would like to determine the equation of the resultant shadow>(which would be an ellipse) that would be cast onto the plane.Five points determine a conic, except in degenerate cases. So takefive points on the boundary of your antenna, project them onto theplane, and take the conic that goes through those five points by solvingfive linear equations in five unknowns, which can be taken in the form x_i^2 + a_1 x_i y_i + a_2 y_i^2 + a_3 x_i + a_4 y_i + a_5 = 0where your five points in the plane are (x_i, y_i), and the unknownsare a_1 to a_5. For a more numerically stable solution, you mighttake more than five points and solve the resulting system in theleast-squares sense.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israelUniversity of British ColumbiaVancouver, BC, Canada V6T 1Z2Subject: === Subject: Determining the Equation of a Shadow Cast onto a Surface>Given a flat circular disk (high-gain antenna), a planar surface (a>solar array), and an arbitrary vector to a point light source (the>Sun), I would like to determine the equation of the resultant shadow>(which would be an ellipse) that would be cast onto the plane.> ... So take five points on the boundary of your antenna, project them onto the plane ... how is it done ? ;)Subject: === Subject: Why are there 63360 inches in a mile?> Bill Vajk schrieb:>> If you want to know why there are 63360 inches in a mile, try this:>> http://math.ucr.edu/home/baez/inches.html>Cute.>However the change to metric is happening slowly and quietly.> But the metric system has certain disadvantages. In Orwells 1984, if I> remember correctly, there was that guy who complained about the metric> system. 1/2 liter of beer is just a bit too much, compared to the> pint...> Half a litre ain't enough, it doesn't satisfy. And a full litre's too> much. It sets my bladder running.> In quotes...should I recognize it?from 1984E could 'a drawed me off a pint,' grumbled the old man ashe settled down behind a glass. 'A 'alf litre ain't enough. Itdon't satisfy. And a 'ole litre's too much. It starts mybladder running. Let alone the price.'Rick> I was always proud of my ability to hold my beer. Until the day I went> to a urologist, and he told me my bladder had double the usual> capacity--I could hold TWO liters--and that was bad, because it was a> sure symptom of diabetes.> --Ron BruckSubject: === Subject: Why are there 63360 inches in a mile?> If you want to know why there are 63360 inches in a mile, try this:> http://math.ucr.edu/home/baez/inches.html>Cute.>However the change to metric is happening slowly and quietly.>For example, don't try taking an SAE (fractional inch) wrench>to your car manufactured in recent years.>There are plenty of other places as well.2 liter bottles of Coke. Volts and amps. 1 gallon engines are now called 3.8 liter engines. The U.S. military measures distances in kliks. Ever buy a lightbulb rated in BTU?-- A nice adaptation of conditions will make almost any hypothesis agreewith the phenomena. This will please the imagination but does not advanceour knowledge. -- J. Black, 1803.Subject: === Subject: Why are there 63360 inches in a mile?Gregory L. Hansen wibbled:> 1 gallon engines are now called > 3.8 liter engines. cc is much more useful for model aero enginesI have one downstairs sold as 0.049 cu in0.8 cc is just more imaginable'cept it's probably a .75, really.-- wanna 57Subject: === Subject: Why are there 63360 inches in a mile?> Gregory L. Hansen wibbled:> 1 gallon engines are now called > 3.8 liter engines. > cc is much more useful for model aero engines> I have one downstairs sold as 0.049 cu in> 0.8 cc is just more imaginable> 'cept it's probably a .75, really.My wife just bought some clamps here in Australia, a country that went metric some time ago. The packaging said they were 75mm clamps, but on the clamps themselves it says 3in. I think the clamps were made in China for the American market but somehow got diverted here.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)Subject: === Subject: Why are there 63360 inches in a mile?> lrudolph@panix.com (Lee Rudolph) schrieb:>Whoa! You mean all those spammers are covertly trying to destroy>the metric system when they keep sending me offers to increase the>length of my rod?> They always want to add inches to the rod. Is this a conspiracy to> change the units of length? No. Only mathema-heads *add* inches, people with brains multiply them. Only physics dorks multiply seconds, people with brains divide them. And even more particulary, only philosphers use rods, people with brain use guns.> ThomasSubject: === Subject: Why are there 63360 inches in a mile?> If you want to know why there are 63360 inches in a mile, try this:> http://math.ucr.edu/home/baez/inches.htmlBack in the old days I learned both systems at school (of course theyalso taught us to use slide-rules as well). I never gave it muchthought as to which was better until I did some work in fluidmechanics. I will never go back to imperial for this!Subject: === Subject: Why are there 63360 inches in a mile?Another interesting fact is that there are about 63241 AstronomicalUnits in a Light Year (if I calculated correctly). Since this is veryclose to the number of inches in a mile, it provides a pretty goodexample for helping people imagine distances in space.For example, if we can imagine the universe shrunk down to where thedistance from the Sun to the Earth is one inch (pretty easy to grasp)then the distance to Pluto is on average about 39 inches. It's theninteresting to ask people to speculate on how far away the nextnearest star would be. Since at that scale 1 light year would be 1mile away, Alpha Centauri would be over 4 miles away!- PaulSubject: === Subject: Why are there 63360 inches in a mile? The second point has a history that could be probably be traced> back to yokes and chariots etc. This can be traced to the> physical properties that made axles and wheels, etc. most> efficient w.r.t. weight bearing and maneuverability(sp?).> So, you'ld be right; it's a conspiracy of nature.>Standard American railroad gauge supposedly traces back to the distance>between thw wheels on Roman wagons.So why was that length a good length? It had to have somethingto do with other lengths that didn't work as well. This is thepoint I was trying to make but didn't express very well. The people who designed rolling, weight-bearing devices didn'twake up one morning and say Let each axle be n units of length.They had to have tried different designs, with all others breakingor not working or having some problem that wasn't worth gettingfixed. Note that these people also didn't have tarmac so travelcould be quite rocky. Cultivating fields certainly didn't havetarmac. Note that cultivating using yoked animals could notalso cultivate the flora that was supposed to produce./BAHSubtract a hundred and four for e-mail.Subject: === Subject: Why are there 63360 inches in a mile?> > The second point has a history that could be probably be traced>> back to yokes and chariots etc. This can be traced to the>> physical properties that made axles and wheels, etc. most>> efficient w.r.t. weight bearing and maneuverability(sp?).> So, you'ld be right; it's a conspiracy of nature.>Standard American railroad gauge supposedly traces back to the distance>between thw wheels on Roman wagons.> So why was that length a good length? It had to have something> to do with other lengths that didn't work as well. This is the> point I was trying to make but didn't express very well.> The people who designed rolling, weight-bearing devices didn't> wake up one morning and say Let each axle be n units of length.> They had to have tried different designs, with all others breaking> or not working or having some problem that wasn't worth getting> fixed. Note that these people also didn't have tarmac so travel> could be quite rocky. Cultivating fields certainly didn't have> tarmac. Note that cultivating using yoked animals could not> also cultivate the flora that was supposed to produce.The Romans built roads to handle their standard wagons,which had wheels set to accommodate the width of two horse'sbehinds, for obvious reasons.The roads had ruts of the same width, particularly onsteep hills where the ruts would help to keep thevehicles on-path.If you try to build a wagon with a different width and runit on the old roads, the wooden wagon will be destroyedin short order by the stone road. Thus the standard.Subject: === Subject: Why are there 63360 inches in a mile?> > The second point has a history that could be probably be traced>> back to yokes and chariots etc. This can be traced to the>> physical properties that made axles and wheels, etc. most>> efficient w.r.t. weight bearing and maneuverability(sp?).> So, you'ld be right; it's a conspiracy of nature.>Standard American railroad gauge supposedly traces back to the distance>between thw wheels on Roman wagons.> So why was that length a good length? It had to have something> to do with other lengths that didn't work as well. This is the> point I was trying to make but didn't express very well.> The people who designed rolling, weight-bearing devices didn't> wake up one morning and say Let each axle be n units of length.> They had to have tried different designs, with all others breaking> or not working or having some problem that wasn't worth getting> fixed. Note that these people also didn't have tarmac so travel> could be quite rocky. Cultivating fields certainly didn't have> tarmac. Note that cultivating using yoked animals could not> also cultivate the flora that was supposed to produce.> The Romans built roads to handle their standard wagons,> which had wheels set to accommodate the width of two horse's> behinds, for obvious reasons.> The roads had ruts of the same width, particularly on> steep hills where the ruts would help to keep the> vehicles on-path.Some existing Roman city strets show stepping stones placed to allowpedestrains to cross the street above the accumulated mud and animal manure.The gaps in the stones were set to allow passage of draft animals andstandard wheel/axle sets.> If you try to build a wagon with a different width and run> it on the old roads, the wooden wagon will be destroyed> in short order by the stone road. Thus the standard.Subject: David Ullrich's parachuteHello. I recently spoke with my good friend, the well respected Mr.Ullrich. He told me of a close call he experienced while parachutingout of an airplane. He asked me to share the story with our readers inthis forum. It was a lovely day at 30,000 ft MSL, and everything was goingwell as David Ullrich leapt out of the aircraft. Everything except onething, that is: Mr. Ullrich discovered, to his horror, that instead ofa parachute, his backpack contained 500 lbs. of sand. But our friend Mr. Ullrich was in luck, because he happened tohave a sand-making device in his pocket. Allow me to explain. You see, each grain of sand in Mr. Ullrich's backpack was beingacted on by 2 forces: a (relatively large) gravitational force pullingit towards Earth, and a (relatively small) gravitational force pullingit towards the Moon. (The moon, fortunately, happened to be at thezenith, so that these two forces acted in precisely oppositedirections) We can draw 2 real number lines, labelling the first Earth andthe second one Moon. Now for each grain of sand, we can mark a(relatively large) and equal section of the Earth line and a(relatively small) and equal section of the Moon. Doing this for eachgrain of sand, we see that much more force is pulling our friendtoward Earth than toward the moon. But here the sand-making device comes into play. As an avidCantorian, Mr. Ullrich knew that if he added arbitrarily large amountsof new sand to his backpack, the forces pulling him Earthward and theforces pulling him Moonward would remain locked in a bijection, so asthe amount of sand approached infinity, the difference in these forceswould even itself out. Working quickly, Mr. Ullrich used his sand-making device to addan infinite amount of new sand to his backpack. At infinity, theEarthward forces and the Moonward forces balanced eachother perfectly,and our friend remained suspended in mid-air until the Air Force couldrescue him. For those here who need further explanation, consider theEarthward and Moonward forces separately. For a single grain of sand,call the magnitude of the force pulling it earthward E. Call themagnitude pulling it moonward M. Now, obviously, E is far larger thanM: and (with a little reflection), indeed we find that by choosingappropriate integers a and b, we can make the equation aE=bMapproximately true to any arbitrary desired degree of accuracy. ButCantor's teachings assure us that in all of Z, the number of multiplesof a and the number of multiples of b, are in fact identical (namely,Alpeh_0). And this, you will realize at once, means that by addingthe infinitude of sand into his backpack, Ullrich was able to achieveperfect equilibrium between the forces. Mr. Ullrich now works for NASA, where he is working on developingnew methods of rocket propulsion based on filling the rockets withSubject: === Subject: David Ullrich's parachute Discussion, linux)> Working quickly, Mr. Ullrich used his sand-making device to add> an infinite amount of new sand to his backpack. At infinity, the> Earthward forces and the Moonward forces balanced eachother perfectly,> and our friend remained suspended in mid-air until the Air Force could> rescue him.I'm not a physicist, so you'll have to help me here. Does he achieveequilibrium before or after the black hole forms?We *are* creating an infinite mass in a very small volume to achievethis effect, right?-- But in our enthusiasm, we could not resist a radical overhaul of thesystem, in which all of its major weaknesses have been exposed,analyzed, and replaced with new weaknesses. -- Bruce Leverett (presumably with apologies to Ambrose Bierce)Subject: === Subject: Cantor Paradox by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2AIear25746;Russell,I got tangled up with this same idea some time ago and posted a lot of incorrect stuff about it in this very newsgroup.I think the tricky aspect is that language is very slippery. Meanings change and multiply in a flash in a way that is difficultfor us to keep track of. Take the string (let's call it StringQ), The antidiagonal of the list of all real numbers derived from an alphabetical examination of all finite strings. This string doesnot specify a real number when we are attempting to build our list,but it can be taken to specify a real number after our list is built.This shift in meaning is what creates the paradoxical effect. Underthe following assumptions the paradox disappears. (1) Any givenstring either specifies a real number or does not. (2) No stringspecifies more than one real number. (3) No string changes inmeaning.Under these assumptions String Q causes us no trouble. If String Qdoes not specify a real number when we are building the list, StringQ also does not specify a real number after our list is built. IfString Q specifies a real number when we are building the list, thenthat number is included in the list and - despite how we mightprotest later - String Q cannot, therefore, actually specify theantidiagonal of our completed list.So under these assumptions the real numbers specifiable by finitestrings are denumerable. And there is no Cantorian argument to showotherwise because the antidiagonal of any list of these numberscannot be specified by a finite string (such as String Q) - assurprising as that fact at first seems. Danny PurvisSubject: Determining Limits of an Arbitrarily Rotated EllispeAnyone have the derivation for the equations to determine the X andY-axis limits (mins and maxes) of an arbitrarily rotated ellipse in anX-Y plane? Currently given is the axis-alligned equation of theellipse and the angle of rotation to the X-axis. Conversly, thegeneralized ellipse equation can be derived, too.Subject: === Subject: Determining Limits of an Arbitrarily Rotated Ellispe> Anyone have the derivation for the equations to determine the X and> Y-axis limits (mins and maxes) of an arbitrarily rotated ellipse in an> X-Y plane? Currently given is the axis-alligned equation of the> ellipse and the angle of rotation to the X-axis. Conversly, the> generalized ellipse equation can be derived, too.A paper which derives the equations for determining these values for ageneralized ellipse can be found athttp://www.crbond.com/papers/ellipse.pdf .This paper was written to present a method for scan converting ellipses atarbitrary angles.--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.comSubject: === Subject: 1st-order Pigeonhole Challenge : : > : > Well, actually, he was full of . : : *plonk*Plonk is NOT a refutation. But what you CUT goes onto EXPLAIN EXACTLY WHY Ketland had nothing to say, giventhat he had no axiom-set to say it from.All I am saying is that I at least bother to postjustifications of my reactions, no matter how condemnatory.This is a skill that you personally are not likely to evermaster, at least not with THAT attitude. --- The history of our nation has demonstrated --- that separate is seldom, if ever, equal.-- Supreme Judicial Court of Massachusetts (4-3)Subject: Any Math for Determining Exact Sequence and Suit of Cards after X Perfect Shuffles?I'm interested in knowing the exact sequence and suit of each card ina deck after any given number of perfect shuffles, and I have shuffleda deck perfectly up through 26 shuffles and recorded the patterns ofeach shuffle. (At 26 shuffles, the cards are back in order, but thesequence of the four suits is changed from the original.)It occurred to me, however, that even though these are perfectlynecessary patterns based on perfect shuffles, I do not know of anymathematical technique that could delineate the exact order and suitof a deck of cards after any X number of perfect shuffles.Does anyone have any ideas?Very Respectfully,Ray Donald PrattSubject: === Subject: Any Math for Determining Exact Sequence and Suit of Cards after X Perfect Shuffles?> I'm interested in knowing the exact sequence and suit of each card in> a deck after any given number of perfect shuffles, and I have shuffled> a deck perfectly up through 26 shuffles and recorded the patterns of> each shuffle. (At 26 shuffles, the cards are back in order, but the> sequence of the four suits is changed from the original.)> It occurred to me, however, that even though these are perfectly> necessary patterns based on perfect shuffles, I do not know of any> mathematical technique that could delineate the exact order and suit> of a deck of cards after any X number of perfect shuffles.> Does anyone have any ideas?> Very Respectfully,> Ray Donald PrattDepends on your definition of a perfect shuffle. It must represent a permutation on the original order. If the original order of 8 cards is 12345678, what is the order after one perfect shuffle? As I see it, either 15263748 or 51627384 are possibilities.Subject: === Subject: Any Math for Determining Exact Sequence and Suit of Cards after X Perfect Shuffles?>I'm interested in knowing the exact sequence and suit of each card in>a deck after any given number of perfect shuffles, and I have shuffled>a deck perfectly up through 26 shuffles and recorded the patterns of>each shuffle. (At 26 shuffles, the cards are back in order, but the>sequence of the four suits is changed from the original.)>It occurred to me, however, that even though these are perfectly>necessary patterns based on perfect shuffles, I do not know of any>mathematical technique that could delineate the exact order and suit>of a deck of cards after any X number of perfect shuffles.>Does anyone have any ideas?>Very Respectfully,>Ray Donald PrattHanging and Other Mathematical Diversions by Martin Gardner. It originally appeared in Scientific American and is reprinted in this collection published in 1969.(Still in print. I checked with Amazon .com.-- To e-mail me get rid of the cats and dogs.Subject: === Subject: Pythagorean triples via trig half and double angle formulas> When I was their age, I was sitting down with Count de Vega's log tables> (no calculator available) trying to find an equal-tempered musical scale> that produced better harmony (53-step)> So what did you conclude? I've had the same idea, but not done much work on> it. Has anyone else considered such an approach? What does it sound like?> I haven't got the book with me, but Sir James Jeans, The Physics > of Music discusses such things near the end of the book. IIRC he says > that Helmholtz had a harmonium specially made in the 53-note scale. > Even if I've got some of the details wrong, this question of temperament > was certainly tackled seriously in the 19th century.> Ken Pledger. I got the details very wrong, so here's a correction FWIW. The book by Jeans is actually called Science and Music (1937). Helmholtz's harmonium is mentioned in another context, but it has nothing to do with this 53-step octave. I'd better type out what Jeans Ken Pledger. So far as is known, a 53-note scale was first proposed in Europe in the seventeenth century by Nicolas Mercator, Danish mathematician and astronomer, who, according to Yasser, found it mentioned in the writings of a Chinese theorist, King-Fang, of the second century B.C. In the middle of the last [i.e. 19th] century, two harmoniums with 53 notes to the octave were built, one for Mr R. H. M. Bosanquet of London, and one by Mr J. P. White of Springfield, Mass., but neither seems to have been regarded as more than a curiosity.Subject: === Subject: Want ro receive A+ paper? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2AJat332320;> If yes, visit http://www.Bestessays.com and win A+ custom written essay!> Cindy>Hee-hee. Check out the slogan at the top of their website. let>professionals take care of you papers>Nathanthere seems really professional.doing homework with money. :)Subject: === Subject: Proof that Multiplication Modulo n is Associative?> I'm trying to find an example of a proof > that multiplication modulo n is associative.> We want to show rem(rem(ab)*c) = rem(a*rem(bc)) > where rem(x) := remainder of dividing x by n.> Write ab = p n + r, 0 <= r < n> rc = q n + t, 0 <= t < n> bc = j n + s, 0 <= s < n> as = k n + u, 0 <= u < n> To show t = u> consider t-u = (rc-qn) - (as-kn)> = rc - as + (k-q)n> = (ab-pn)c-a(bc-jn) + (k-q)n> = (ab)c-a(bc)+(aj-pc + k-q)n |a r-ab| |a r|Simpler: 0 = | | = | | (mod n) |c s-cb| |c s| i.e. n| r-ab & n|s-bc => n|(r-ab)c - a(s-bc) = rc - asHence rc, as have equal residues (mod n). QEDAppended below are two of my prior sci.math posts whichprovide further related examples of (ring) congruences.http://google.com/groups?selm=y8zk7vhhrc6.fsf% 40nestle.ai.mit.edu> AHSME 1956 #48: If n is a positive integer then (3n+25)/(2n-5) > can be a positive integer iff> a. 3<=n; b. 3<=n<=35; c. n<=35; d. n=35; e. n=3 or n=35You can bound the divisors in your problems by eliminatingthe variable n from two known multiples, e.g. see below.Here I employ the standard notation x|y == x divides y.(1) 2n-5 | 3n+25 -> 2n-5 | 2(3n+25) - 3(2n-5) = 65 i.e. an+b | cn+d -> an+b | a(cn+d) - c(an+b) = ad-bc(2) n+9 | 10n -> n+9 | 10n - 10( n+9) = -90 via m(n+9) = 10n if (n+10)/(10n+m) = 1/m as below. So(1) 0 < 2n-5 <= 65 -> 3 <= n <= 35, i.e. ASHME choice (b).It proves insightful to examine related methods of solution.That an+b | ad-bc has a determinant representation as follows: | a an+b | | a b | cn+d = 0 (mod an+b) => 0 = | | = | | (mod an+b) | c cn+d | | c d |For n = -1 this is nothing but a |proof on congruence multiplication: +-- 2nd column = 0i.e. a=b, d=c => ad=bc (mod m) | hence det = 0as follows: | | a -a+b | | a b | c = d, a = b (mod m) => 0 = | | = | | => ad = bc | c -c+d | | c d |In fact we can directly derive our result by congruence multiplication: -cn = d, b = -an (mod an+b) | 3n = -25, 5 = 2n (mod 2n-5) |=> -acn = ad, bc = -anc | 6n = -50, 15 = 6n | thus ad = bc, i.e. ad-bc = 0 | so -50 = 15, i.e. 65 = 0This illustrates the power of congruence based problem-solving.The just-presented congruence approach allows us to use familiararithmetic operations to eliminate the variable n from the twolinear equations in n, thus deriving a purely numerical equation.Contrast this with the first approach, where we were forced tomanipulate cumbersome equations involving divisibility relations.Because the integers modulo m inherit their arithmetic laws fromthe integers, we can reuse all our knowledge about the integerswhen working in the integers modulo m, e.g. above we reused methods of elimination for solving linear equations. The commonstructure and laws shared between these two number systems isformalized when one studies abstract algebra at university.These two number systems are examples of a fundamental algebraicstructure known as a ring. Just as above, in general ringsone may compute modulo congruences, which may greatly simplifymatters. For example, parity arguments correspond to arithmeticof integers modulo 2, i.e. to arithmetic in the ring Z/(2).Here's an related congruence proof on Pythagorean triples:CLAIM x^2 + y^2 = z^2 => x or y is even, sinceMod 4: odd^2 + odd^2 = 2 isn't odd^2 (= 1) or even^2 (= 0).http://google.com/groups?selm=y8z1ya4tmcy.fsf% 40nestle.ai.mit.edu> Does A = a, B = b (mod n) => AB = ab (mod n) ?> ab - AB = (a-A)b + A(b-B)This has a pretty representation in determinant form: A = a | a-A A | | a A | (mod n) => 0 = | | = | | = ab - AB (mod n) B = b | B-b b | | B b | The same proof yields lim(AB) = lim(A) lim(B) in calculus, seehttp://google.com/groups?selm=y8zu1pepbho.fsf% 40nestle.ai.mit.eduIndeed Calculus is Algebra via NSA (NonStandard Analyis): the product rule for limits is just the product rule for congruences upon NSA algebraicization. For if I is the ideal of infinitesimals then lim x->0 F(x) = f iff F := F(i) = f (mod I), i in I andthen lim(FG) = lim(F)lim(G) = fg is just congruence multiplication F = f (mod I) FG = fg + (F-f)G + f(G-g) => G = g (mod I) = fg (mod I) since F-f, G-g in IThe analytic proofs use FG - fg = (F-f)G + f(G-g) e.g. see [**] below.In determinant speak: (see above) F = f | f-F F | | f F | (mod I) => 0 = | | = | | => FG = fg (mod I) G = g | G-g g | | G g |Spelled out in functional notation one sees the underlying identityas a product rule for differences f(x)g(x) - f(a)g(a) = [f(x)-f(a)] g(x) + f(a) [g(x)-g(a)] -> D(fg) = (Df) g + f (Dg)which becomes the Leibniz product rule for derivatives D afterdividing thru by x-a and passing to the limit as x -> a.In fact this approach yields an elegant algebraic proof of the Leibniz product rule for polynomials if one defines thederivative of a poly p(x) as p'(x) = Q(x,x) where p(x) - p(y) = (x - y) Q(x,y) in the poly ring R[x,y].-Bill Dubuque> Here is how I would do it.> |f(x)g(x) - fg| =< |f(x)g(x) - f(x)g| + |f(x)g - fg|> = |f(x)| |g(x) - g| + |f(x) - f| |g| [**]> < |f + 1| |g(x) - g| + |f(x) - f| |g + 1|> for x sufficiently close to c that |f(x) - f| < 1 and |g(x) - g| < 1.> Then given E > 0 and also < 1 (to enable the line above) and choosing > D sufficiently small that for 0 < |x - c| < D, both inequalities> |f(x) - f| < E/(2(g+1)) and |g(x) - g| < E/(2(f+1)) you get> the required inequality. > BTW, it is much easier with infinitesimals. For then the limit is the> ordinary part of f(x+h) for h a non-zero infinitesimal, assuming > that the ordinary part does not depend on h. Then for infinitesimal h,> there are infinitesimals i,j so that f(x+h)g(x+h) = (f + i)(g + j) = > f g + f j + i g + i j and the last three terms are infinitesimal. > All the nasty work is done once and henceforth hidden in the definition > of infinitesimal.Subject: === Subject: Proof that Multiplication Modulo n is Associative?Since somewhere in your proof you have to use the associativity of the original ring: if a(bc) = (ab)c then a(bc) == (ab)c mod n.Subject: === Subject: Proof that Multiplication Modulo n is Associative?it really does seem trivial,as the statement n | remainder((ab)c - a(bc)),since (real?) multiplication is associative, within the parentheses. especially in base one! > Now you want to prove that (ab)c = a(bc) (mod n); which means that you> need to show that n|[(ab)c - a(bc)], which is trivial since the> multiplication here is the usual multiplication for the integers.--Give Earth a Trickier Dick Cheeny -- out of office, after GIGA years.http://www.benfranklinbooks.com/http://www.rand.org/ publications/randreview/issues/rr.12.00/http:// members.tripod.com/~american_almanacSubject: === Subject: Proof that Multiplication Modulo n is Associative?> I'm trying to find an example of a proof that multiplication modulo n> is associative.> Every reference I can dig up on the web either deals with some other> problem and tells the reader that the associativity of modular> multiplication can be assumed, or else claims it trivially follows> from the associativity of regular multiplication.Can you give the precise definition of multiplication modulo n? The reason I ask it that associativity does follow trivially follow from the associativity of regular multiplication once the definition is understood.Subject: Rieman surfaceIve looked for a classification of Riemann non compact surfaces or topological 2-manifolds on the net but to no avail.Can someone direct me to the correct place ?Subject: === Subject: New KnowledgeHi Ha Ha Hanson, You asked, How, when and why does * New Knowledge * come to humankind ? .Given:1. Regularity and randomness in nature.2. A system with a birth, a death, and a consumption rate.3. A system that can remember and utilize that randomness and regularity in nature.Then New Knowledge occurs in that system as it utilizes that randomness and regularity in order to consume.But of course what is regularity vs. what is randomness is a subjective notion. ( i.e. A by-product of one's crippled information )Subject: === Subject: New Knowledge> How, when and why does *New Knowledge* come to humankind?Interesting question. I think there are three main elements. The first is the shoulders ofgiants. New knowledge usually is built on existing knowledge or anexisting framework.The second is, as some of the others have said - luck. Nothing canreplace a good bit of luck to send us down a chain of thought.I think the third is intuition. This is the most interesting. I donot mean intuition in a para-normal sense, but rather in a John Nashsense. I think that the brain is far more hardwired than we thinkand that in some people some of this wiring leads them to jump to aconclusion that just seems right (of course the hard part is fillingin the details so the rest of us can see that it is right).You need all three working together to get New Knowledge. Howeverbased on these three factors I think New Knowledge comes about by twomain methods.The first is through a pedantic method of assimilation. Person Apicks up on work done by Person B and Person C and sees that they arelinked or that they raise questions to be answered. This is the waymost knowledge is extended and is heavily based on the shoulders ofgiants elements.The second is through a genuine insight. A new idea that is way outin front of the field. This is heavily reliant on the intuitionelement.Based on the work of the insights the assimilators move in and buildup the surroundings parts and voila - a new body of knowledge.that I better go to bed.Good night,Yelling.Subject: === Subject: New Knowledge charset=iso-8859-1> How, when and why does *New Knowledge* come to humankind?> Interesting question.> I think there are three main elements. The first is the shoulders of> giants. New knowledge usually is built on existing knowledge or an> existing framework.> The second is, as some of the others have said - luck. Nothing can> replace a good bit of luck to send us down a chain of thought.> I think the third is intuition. This is the most interesting. I do> not mean intuition in a para-normal sense, but rather in a John Nash> sense. I think that the brain is far more hardwired than we think> and that in some people some of this wiring leads them to jump to a> conclusion that just seems right (of course the hard part is filling> in the details so the rest of us can see that it is right).> You need all three working together to get New Knowledge. However> based on these three factors I think New Knowledge comes about by two> main methods.> The first is through a pedantic method of assimilation. Person A> picks up on work done by Person B and Person C and sees that they are> linked or that they raise questions to be answered. This is the way> most knowledge is extended and is heavily based on the shoulders of> giants elements.> The second is through a genuine insight. A new idea that is way out> in front of the field. This is heavily reliant on the intuition> element.> Based on the work of the insights the assimilators move in and build> up the surroundings parts and voila - a new body of knowledge.> that I better go to bed.> Good night,> Yelling.Cool, dude!..........ahahahaha......after typing in fullconviction of certainty......a just once-over reflectionand these nasty ing doubts begin to bubble up...ahahahaha.......ahahahansonPS:Blessed are the little idiots for they have all the answersahahahahaha............ahahahahaha..........isn't that whatMel Gibson had in his original script for his passion?Subject: === Subject: New Knowledge(snip)>Despite my personal abhorrence, Bush's machine is more than Kerry's can>overcome. Another narrow loss will make Bush President for another 4>years. Does that count as new knowledge?You are posting to the wrong groups.Bruce-------------------------------------------------- ---------------------It was so much easier to blame it on Them. It was bleakly depressing to think that They were Us. If it was Them, then nothing was anyones fault. If it was Us, what did that make Me ? After all, Im one of Us. I must be. Ive certainly never thought of myself as one of Them. No-one ever thinks of themselves as one of Them. Were always one of Us. Its Them that do the bad things. <=> Terry Pratchett. Jingo.Subject: === Subject: my discovery of the truth of our originsGood stuff yeah ?Subject: === Subject: One Geometry algebra question>Reflect line segment ( 3 e1 - 2 e2 + 4 e3) in the mirror described by>e13 - e12.>As far as I know planes can be described in terms of the area spanned>by two directed line segments, we can refer to our mirror M as a>product of two perpendicular line segments. In other words, we could>write M = AB where A and B are perpendicular and have unit lengths.>Any pair of unit line segments work for A and B as long as they are in>the M plane and perpendicular. Lets chose A to be in the same>direction as the direction of the part of the line L that is parallel>to the mirror plane. More descriptive names will be assigned to A and>B for this case. A becomes mpar and B becomes m_{perp}.>The reflection of the line L in the mirror M is written as follows. >L' = m{par}^{-1} L m_{par}>I don't know how to find m_{par}I will assume that 3 e1 - 2 e2 + 4 e3 is a vector, not a line segment.I will also assume that e13 - e12 specifies any plane perpindicular toe2 + e3 (e13 = e1 x e3 = -e2 and e12 = e1 x e2 = e3).There are two ways to approach this problem. The first is to create areflection matrix as you attempted above. This matrix should preserveany component of a vector perpindicular to the normal of the mirror andreverse the component parallel to the normal. e1 and e2 - e3 are bothperpindicular to the normal of the plane, so we can use them and e2 + e3as a basis for R^3. We get the following 3 equations that we can writeas a matrix equation [ 1 0 0 ] [ 1 0 0 ] [ 0 1 -1 ] M = [ 0 1 -1 ] [ 0 1 1 ] [ 0 -1 -1 ]Therefore, [ 1 0 0 ]-1 [ 1 0 0 ] M = [ 0 1 -1 ] [ 0 1 -1 ] [ 0 1 1 ] [ 0 -1 -1 ] 1 [ 2 0 0 ] [ 1 0 0 ] = - [ 0 1 1 ] [ 0 1 -1 ] 2 [ 0 -1 1 ] [ 0 -1 -1 ] [ 1 0 0 ] = [ 0 0 -1 ] [ 0 -1 0 ]Applying this to 3 e1 - 2 e2 + 4 e3, we get [ 1 0 0 ] [ 3 -2 4 ] [ 0 0 -1 ] = [ 3 -4 2 ] [ 0 -1 0 ]Another approach is to isolate the part of the vector that is normal tothe plane of the mirror and reverse it. The part of [ 3 -2 4 ] that isnormal to the plane is <[ 3 -2 4 ],[ 0 1 1 ] ------------------------- [ 0 1 1 ] = [ 0 1 1 ] <[ 0 1 1 ],[ 0 1 1 ]To reverse this component, we must subtract it twice from the originalvector: [ 3 -2 4 ] - 2 [ 0 1 1 ] = [ 3 -4 2 ]Therefore, the reflection of the original vector is [ 3 -4 2 ].Rob Johnson Dear all,> I wonder under what condition can I say X=Y if I have AXA=AYA?> A is any matrix, not neccessarily square or non-singular... it can be> rectangular and singular...> Then can AXA=AYA reduce to X=Y?> -------------------------------------> I myself conceived that> I can write AXA=AYA to be A(X-Y)A=0, since A is arbitarily any matrix, the> only possible case is X=Y,> am I right?> -Joenyim-- Best regards,Alex.PS. To email me, remove loeschedies from the email address given.Subject: === Subject: matrix analysis: can I say matrix X=Y if I have AXA=AYA?> 1. the dimensions of A,X,Y should be compatible, in this case I think> they should be all square matrices.> 2. If you require that AXA=AYA for all matrices A, then it would> certainly hold for A=identity matrix.> 3. If A is some unknown matrix, i.e. then AXA=AYA tells you not much.> E.g. if A=[0]Hi!Where has my reasoning gone wrong here? Suppose all matrices are square,and that AXA = AYA is true. If I pre- and post-multiply both sides by A-inv,ISkipSubject: === Subject: matrix analysis: can I say matrix X=Y if I have AXA=AYA? Adjunct Assistant Professor at the University of Montana.> 1. the dimensions of A,X,Y should be compatible, in this case I think> they should be all square matrices.> 2. If you require that AXA=AYA for all matrices A, then it would> certainly hold for A=identity matrix.> 3. If A is some unknown matrix, i.e. then AXA=AYA tells you not much.> E.g. if A=[0]>Hi!>Where has my reasoning gone wrong here? Suppose all matrices are square,>and that AXA = AYA is true. If I pre- and post-multiply both sides by A-inv,Why are you assuming that A has an inverse? What if A is singular?If A has an inverse, then yes, you may conclude that X = Y. But whatif A does NOT have an inverse?-- == It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)=== ===Arturo Magidinmagidin@math.berkeley.eduSubject: === Subject: matrix analysis: can I say matrix X=Y if I have AXA=AYA?Originator: richard@cogsci.ed.ac.uk (Richard Tobin)>Where has my reasoning gone wrong here? Suppose all matrices are square,>and that AXA = AYA is true. If I pre- and post-multiply both sides by A-inv,What if A does not have an inverse?-- RichardSubject: === Subject: matrix analysis: can I say matrix X=Y if I have AXA=AYA? Adjunct Assistant Professor at the University of Montana.>1. the dimensions of A,X,Y should be compatible, in this case I think >they should be all square matrices.Not at all. For AXA to make sense, where X is n x m, we need A to be ? x n (so wecan multiply A by X), but it must also be m x ? (so we can multiply Xby A). Set A to be m x n, and everything works out. Clearly, then Yshould by n x m, same size as X. So X and Y should be the same size,but if X and Y are, say, 2 x 3 matrices, then you can let A be a 3 x 2matrix and all the operations are compatible.The result on both sides will be an m x n matrix.-- == It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)=== ===Arturo Magidinmagidin@math.berkeley.eduSubject: === Subject: matrix analysis: can I say matrix X=Y if I have AXA=AYA? Adjunct Assistant Professor at the University of Montana.>Dear all,>I wonder under what condition can I say X=Y if I have AXA=AYA?>A is any matrix, not neccessarily square or non-singular... it can be>rectangular and singular...>Then can AXA=AYA reduce to X=Y?No. For example, say A is the zero matrix.>------------------------------------->I myself conceived that>I can write AXA=AYA to be A(X-Y)A=0, since A is arbitarily any matrix, the>only possible case is X=Y,>am I right?Not unless by arbitrarily any matrix you mean for every matrix A. If all you know is that AXA = AYA for SOME matrix A, and you do notknow anything about A, then you cannot conclude anything. If you knowthat for EVERY matrix A we have AXA = AYA, then by setting A to beelementary matrices (matrices with all entries equal to zero exceptfor one equal to 1) you can systematically show that each entry of Xis equal to each entry of Y.(If X and Y are n x m, then A can be any m x n matrix and bothexpressions will make sense and yield an mxm matrix. If X and Y aresquare matrices, then A will be a square matrix, and then you canchoose A to be the identity to get X = Y).-- == It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)=== ===Arturo Magidinmagidin@math.berkeley.eduSubject: Is this travelling salesman problem version known?(was: How are these minimal paths called?)Since only a few knew what I'm talking about, I'll reformulate the problem and hope for a better understanding.Let V be be set of cities, E (containing some two-element sets of elements of V) is the set of connections between some cities, all these connectons have cost 1. Other connections (i.e. not in E) cost infinity. (I.e. the salesman goes only through the connections listed in E). Let W be a subset of V.Questions: (1) Find a shortest path which goes at least once through each city in W (and is allowed to visit cities in VW)? (2) What is the cost of such shortest path?-- Best regards,Alex.PS. To email me, remove loeschedies from the email address given.Subject: === Subject: Simple algebra question> Please, in advance, forgive this simplicity of this question and my > lack of understanding of fundamental concepts. I am not a mathematician.> Large polynomials of the type> x^n + x^(n-1) + x^(n-2) + ... + x^2 + x + 1 = 0> can be easily reduced to (x^(n+1)-1)/(x-1) by multiplying both sides by> (x-1). How would one know, a priori, to do that? Is this anything like> completing the square? Are there rules that let you know when seemingly> complex polynomials are easily reduced?This is a cyclotomic factorization. A web search will turn up much.More generally there's been some work on the problem of finding smallermultiples of a polynomial with integer coefficients, e.g. see this paperFilaseta, M; Robinson, M; Wheeler, FThe minimal Euclidean norm of an algebraic number is effectively computableJournal of Algorithms 16 (1994) 309-333. http://www.math.sc.edu/~filaseta/papers/polynnormpaper.ps> The new function is equivalent to the old one for all x but we have> introduced a point - i.e. x = 1 - where the function is not defined. Is> there a name for this - where one introduces a singularity? (Sorry about the> terminology!) If one wanted to evaluate the function at this point, or take> a derivative, or an integral, one would perhaps like to use the original,> which has no ill-defined point. However, the original polynomial is quite> unwieldy. So does one try to apply transformations to remove such points?Search on removable singularity.-Bill DubuqueSubject: === Subject: easy l'hospital question~> i know that l'hospital rule can apply form 00 / 00, and> i know that l'hospital rule can apply form ~ / 00 (regardless of numerator)> but if> lim (sin x) / x = 0 (trivial),> x->00> but if l'hospital apply this, cos / 1 => oscillation> why not same?? my thinking is wrong?? advice...please> L'Hopital's rule may ONLY apply when the original limit form is > indeterminant, .i.e., leads to 0/0 or to oo/oo situations.> Lim_{x -> oo} sin(x)/x is not indeterminant.> In fact, l'Hopital applications can be restricted to 0/0 cases with no > loss of generality, since oo/oo can always be recast as (1/oo)/(1/oo) > cases.Not true. The o.p. is correct in stating thatL'Hospital's rule for the form lim f/g, lim g = ooneeds no hypotheses on existence or value of lim fI append below a recent nontrivial example of such.http://google.com/groups?selm=y8zr80dib6o.fsf% 40nestle.ai.mit.edu> if f is differentiable on (0, infinite)> and lim [f(x) + f'(x)] = L (x->infinite)> show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite)>>|>>| f e^x (f + f') e^x>>| lim f + f' = L => lim f = lim ----- = lim ------------ = L>>| x->oo x->oo x->oo e^x x->oo e^x>Provided f e^x -> +-oo, in which case> 0 = L - L = lim f+f' - lim f = lim f'> No proviso is needed. This form of L'Hopital's rule needs only> that the denominator is infinite, not also the numerator [1][2].> So there's no need to treat this case specially as you do below.> Not even the existence of lim(x->oo) f(x) ?L'Hospital's rule for the form lim f/g, lim g = ooneeds no hypotheses on existence or value of lim f> [1] A. E. Taylor, L'Hospital's Rule> Amer. Math. Monthly, Vol. 59, No. 1 (Jan., 1952), pp. 20-24.> http://links.jstor.org/sici?sici=0002-9890(195201)59:3%3C20%3E > [2] A. M. Ostrowski, Note on the Bernoulli-L'Hospital Rule> Amer. Math. Monthly, Vol. 83, No. 4 (Apr., 1976), pp. 239-242.> http://links.jstor.org/sici?sici=0002-9890(197604)83:3%3C239% 3E-Bill DubuqueSubject: === Subject: SCHOENFELDS RANDOM THEOREM> As Uncle Al said:> Trivially disproven by example in Hofstadter's Godel, Escher, Bach. > It's only 777 pages long. If you look at each page for one second you> can find the table within 12 minutes.Uncle Al is wrong, and transistively, so are you. Try independent thought.JSSubject: === Subject: SCHOENFELDS RANDOM THEOREM>SCHOENFELDS RANDOM THEOREM:>With probability 1, all finite sequences of integers contained by>[n,m] occur infinitely in an infinite sequence of random integers>bound by [n,m].Well known, but I don't think it needs a new name. I don't know if it hasan old name but it's an obvious extension of every possible finite sequenceof digits occurs in pi (and most other irrational numbers), many times,which I learned in high school.>P(S occurs in R)> = 1 - lim j->+inf (1 - 1 / [(|m|-|n|) + 1]^|S|)^j> = 1 - 0> = 1> where> R : random sequence of integers bound by [n,m]> |R| = +inf> S : arbitrary sequence of integers> |S| < +inf>I also put forth that an infinite sequence S occurs within an infinite>random sequence R given |S| < |R|. Example: given an uncountably>infinite sequence of random reals, all countably infinite sequences of>integers occur with probability 1, countably infinite times.There's no such thing as an uncountably infinite sequence. If you can putit in a sequence, it's countable. --Keith Lewis klewis {at} mitre.orgThe above may not (yet) represent the opinions of my employer.Subject: === Subject: SCHOENFELDS RANDOM THEOREM13:36:06 -0800:>SCHOENFELDS RANDOM THEOREM:>With probability 1, all finite sequences of integers contained by>[n,m] occur infinitely in an infinite sequence of random integers>bound by [n,m].> Well known, but I don't think it needs a new name. I don't know if it has> an old name but it's an obvious extension of every possible finitesequence> of digits occurs in pi (and most other irrational numbers), many times,> which I learned in high school.Interesting that you learnt this about pi in school, when AFAIK it has neverbeen proved.Subject: === Subject: SCHOENFELDS RANDOM THEOREM> Well known, but I don't think it needs a new name. I don't know if it has> an old name but it's an obvious extension of every possible finite> sequence> of digits occurs in pi (and most other irrational numbers), many times,> which I learned in high school.> Interesting that you learnt this about pi in school, when AFAIK it has never> been proved.He must have gone to a better high school. Are you from California?SSubject: === Subject: SCHOENFELDS RANDOM THEOREMNo, I went to school in Australia. What makes you think he went to a betterhigh school?Two questions for you:1. Do you believe (and have evidence to support this belief) that everypossible finite sequence of digits occurs in pi, many times?2. Where did YOU go to high school?> Well known, but I don't think it needs a new name. I don't know if ithas> an old name but it's an obvious extension of every possible finite> sequence> of digits occurs in pi (and most other irrational numbers), manytimes,> which I learned in high school.> Interesting that you learnt this about pi in school, when AFAIK it hasnever> been proved.> He must have gone to a better high school. Are you from California?> SSubject: === Subject: SCHOENFELDS RANDOM THEOREM No, I went to school in Australia. What makes you think he went to a better>high school?I don't want to get into a high school war here. I went to 2 differentpublic high schools in the US and have a very high opinion of both. >Two questions for you:>1. Do you believe (and have evidence to support this belief) that every>possible finite sequence of digits occurs in pi, many times?If you are given the assumption that the digits of pi are unpredictable(which I accept as fact but have never seen a proof of), it's easy to provethat that the probability of the statement pi contains many copies of thesequence being true for any given sequence is 1.The OP wasn't talking about pi anyway; he was talking about truly randomsequences. --Keith Lewis klewis {at} mitre.orgThe above may not (yet) represent the opinions of my employer.Subject: === Subject: SCHOENFELDS RANDOM THEOREM22:27:49 +1100:>No, I went to school in Australia. What makes you think he went to abetter>high school?> I don't want to get into a high school war here. I went to 2 different> public high schools in the US and have a very high opinion of both.>Two questions for you:>1. Do you believe (and have evidence to support this belief) that every>possible finite sequence of digits occurs in pi, many times?> If you are given the assumption that the digits of pi are unpredictable> (which I accept as fact but have never seen a proof of), it's easy toprove> that that the probability of the statement pi contains many copies of the> sequence being true for any given sequence is 1.> The OP wasn't talking about pi anyway; he was talking about truly random> sequences.Any finite sequence of digits of pi is truly random with a uniformdistribution.Subject: === Subject: SCHOENFELDS RANDOM THEOREM> 22:27:49 +1100:>No, I went to school in Australia. What makes you think he went to a> better>high school?> I don't want to get into a high school war here. I went to 2 different> public high schools in the US and have a very high opinion of both.>Two questions for you:>1. Do you believe (and have evidence to support this belief) that every>possible finite sequence of digits occurs in pi, many times?> If you are given the assumption that the digits of pi are unpredictable> (which I accept as fact but have never seen a proof of), it's easy to> prove> that that the probability of the statement pi contains many copies ofthe> sequence being true for any given sequence is 1.> The OP wasn't talking about pi anyway; he was talking about truly random> sequences.> Any finite sequence of digits of pi is truly random with a uniform> distribution.Proof? URL? Anything?Subject: === Subject: SCHOENFELDS RANDOM THEOREM> 22:27:49 +1100:>No, I went to school in Australia. What makes you think he went to a> better>high school?> I don't want to get into a high school war here. I went to 2different> public high schools in the US and have a very high opinion of both.>Two questions for you:>1. Do you believe (and have evidence to support this belief) thatevery>possible finite sequence of digits occurs in pi, many times?> If you are given the assumption that the digits of pi areunpredictable> (which I accept as fact but have never seen a proof of), it's easy to> prove> that that the probability of the statement pi contains many copies of> the> sequence being true for any given sequence is 1.> The OP wasn't talking about pi anyway; he was talking about trulyrandom> sequences.> Any finite sequence of digits of pi is truly random with a uniform> distribution.> Proof? URL? Anything?http://mathworld.wolfram.com/PiDigits.htmlVery simple google search will give you results also.Subject: === Subject: SCHOENFELDS RANDOM THEOREMOriginator: richard@cogsci.ed.ac.uk (Richard Tobin)> Any finite sequence of digits of pi is truly random with a uniform> distribution.> Proof? URL? Anything?>http://mathworld.wolfram.com/PiDigits.htmlPlease quote the relevant part of that page.-- RichardSubject: === Subject: SCHOENFELDS RANDOM THEOREM>> Any finite sequence of digits of pi is truly random with a uniform>> distribution.>> Proof? URL? Anything?>http://mathworld.wolfram.com/PiDigits.html> Please quote the relevant part of that page.Quoting from the page GR_Learner@GR.grv referred you to http://mathworld.wolfram.com/PiDigits.htmlThe following distribution of decimal digits d is found for the first departure from a uniform distribution.Subject: === Subject: SCHOENFELDS RANDOM THEOREM> 22:27:49 +1100:>No, I went to school in Australia. What makes you think he went toa> better>high school?> I don't want to get into a high school war here. I went to 2> different> public high schools in the US and have a very high opinion of both.>Two questions for you:>1. Do you believe (and have evidence to support this belief) that> every>possible finite sequence of digits occurs in pi, many times?> If you are given the assumption that the digits of pi are> unpredictable> (which I accept as fact but have never seen a proof of), it's easyto> prove> that that the probability of the statement pi contains many copiesof> the> sequence being true for any given sequence is 1.> The OP wasn't talking about pi anyway; he was talking about truly> random> sequences.> Any finite sequence of digits of pi is truly random with a uniform> distribution.> Proof? URL? Anything?> http://mathworld.wolfram.com/PiDigits.html> Very simple google search will give you results also.You mean the bit on the wolfram page that states:It is not known if is normal (Wagon 1985, Bailey and Crandall 2001), ,which directly contradicts what you said?I did a Google search, and can find no site which states that Any finitesequence of digits of pi is random with uniform distribution. Indeed, everyrelevant site I could find (including Wolfram) states that this is anunsolved problem.begin 666 p2img2.gifSubject: === Subject: SCHOENFELDS RANDOM THEOREM> I did a Google search, and can find no site which states that Any finite> sequence of digits of pi is random with uniform distribution. Indeed, every> relevant site I could find (including Wolfram) states that this is an> unsolved problem.Quoting from the page GR_Learner@GR.grv referred you to http://mathworld.wolfram.com/PiDigits.htmlThe following distribution of decimal digits d is found for the first departure from a uniform distribution. More from Kanada http://www.cecm.sfu.ca/personal/jborwein/Kanada_50b.html ftp://www.cc.u-tokyo.ac.jp/README.our_latest_recordSubject: === Subject: SCHOENFELDS RANDOM THEOREM> I did a Google search, and can find no site which states that Anyfinite> sequence of digits of pi is random with uniform distribution. Indeed,every> relevant site I could find (including Wolfram) states that this is an> unsolved problem.> Quoting from the page GR_Learner@GR.grv referred you to> http://mathworld.wolfram.com/PiDigits.html> The following distribution of decimal digits d is found for the first> departure from a uniform distribution.> More from Kanada> http://www.cecm.sfu.ca/personal/jborwein/Kanada_50b.html> ftp://www.cc.u-tokyo.ac.jp/README.our_latest_recordExcellent! So something true for the first 6 billion cases is thus true forany finite number of cases.I am therefore pleased to be able to state as fact the following results:1. All zeroes of the Riemann zeta function have real value equal to 1/2.2. There are no odd perfect numbers.3. All even numbers can be expressed as the sum of two primes.4. There are only a finite number of twin primes ;)These all have considerably stronger proofs than the random distribution ofpi digits, as all have been investigated way past 6 billion.Of course, if I did claim in a maths newsgroup all even numbers could beexpressed as the sum of two primes, and then offered as proof a study thatshowed that this was only known to be true up to 6 billion, I would lookpretty stupid.Here's looking at you, kid.Subject: === Subject: SCHOENFELDS RANDOM THEOREM>Any finite sequence of digits of pi is truly random with a uniform>distribution.Even the sequence 1?-- I'm not interested in mathematics that might have anythingto do with reality. -- Russell Easterly, in sci.mathSubject: === Subject: SCHOENFELDS RANDOM THEOREM> Interesting that you learnt this about pi in school, when AFAIK it has never> been proved.See: http://mathworld.wolfram.com/Pi.htmlSubject: === Subject: SCHOENFELDS RANDOM THEOREMWhere in the page does it state that every possible finite sequence ofdigits occurs in the decimal expansion of pi??The only relevant thing I could find was the following statement:but it is not known if is normal to any base (Stoneham 1970). If the statement made by Keith Lewis thatevery possible finite sequence of digits occurs in pi many times WASN'Ttrue, then we would know that pi was not normal in the base 10^n (where n isthe length of the finite sequence of digits). But Wolfram doesn't make thisclaim.> Interesting that you learnt this about pi in school, when AFAIK it hasnever> been proved.> See: http://mathworld.wolfram.com/Pi.htmlbegin 666 p2img2.gifSubject: === Subject: SCHOENFELDS RANDOM THEOREM> Where in the page does it state that every possible finite sequence of> digits occurs in the decimal expansion of pi??> The only relevant thing I could find was the following statement:> but it is not known if is normal to any base (Stoneham 1970). > If the statement made by Keith Lewis that> every possible finite sequence of digits occurs in pi many times WASN'T> true, then we would know that pi was not normal in the base 10^n (where n is> the length of the finite sequence of digits). But Wolfram doesn't make this> claim.> Interesting that you learnt this about pi in school, when AFAIK it has> never> been proved.> See: http://mathworld.wolfram.com/Pi.html> [Image]Sorry--It is the references listed on the page that include the treasures.For example the reference made by Uncle Al, the conversation from Are QuantaReal? by J.M.Jauch -- as quoted in Douglas Hofstadter's G.9adel, Escher, Bachis quoted in David Blatner's The Joy of Pi.Niven, I., A simple proof of the irrationality of Pi, Bulletin of theAmerican Mathematical Society, Vol. 53 (1947), page 509.Are the digits of pi random? http://www.lbl.gov/Science-Articles/Archive/pi-random.html http://www.pnl.gov/energyscience/10-01/art3.htmPi is both irrational and its digits are random (at least in the firsttrillion+ digits), therefore, every possible finite sequence of digits occurs in pi many times is a reasonable assumption!Subject: === Subject: SCHOENFELDS RANDOM THEOREM>SCHOENFELDS RANDOM THEOREM:>With probability 1, all finite sequences of integers contained by>[n,m] occur infinitely in an infinite sequence of random integers>bound by [n,m].An easy corollary of the Borel-Cantelli Lemma or the Kolmogorov zero-one law, assuming your random integers are iid.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2Subject: === Subject: SCHOENFELDS RANDOM THEOREM>SCHOENFELDS RANDOM THEOREM:>With probability 1, all finite sequences of integers contained by>[n,m] occur infinitely in an infinite sequence of random integers>bound by [n,m].> An easy corollary of the Borel-Cantelli Lemma or the Kolmogorov zero-one > law, assuming your random integers are iid.A corollary it is not, since I provided an alternative, superior proof(which you forgot to quote in your response). > Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2JSSubject: === Subject: Die Petry, die: was === Subject: Die Cantor Die> I think what he means is that the undecidability of classical> predicates makes statements provable in classical logic using the> excluded middle pronciple unprovable in intuitionistic logic, as the> latter requires settling a statement before using it to prove another> statement. No it doesn't. That only pertain's to *Classical* intuitionist logic, where the term Valid is ALWAYS relevent. In modern intuitionistic logic, the term Valid is not ALWAYS relevent. Since the term Valid has it origins but where-else? The heads who invented the principle of *Exluded Middle*. Since the term *PROVABLE* has NOTHING to do with the term *VALID*. > Posted Via Usenet.com Premium Usenet Newsgroup Services> ----------------------------------------------------------> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **> ---------------------------------------------------------- > http://www.usenet.comSubject: === Subject: Die Petry, die: was Re: Die Cantor Dieyeah. :D but without knowing very much about it. the post was very helpful to me. thanks.Subject: Binary treeSomeone said: There are some binary trees that can not be constructed from theempty tree via successive insertions, it caused the average height ofa binary tree is approximately sqrt(n) Could you come up with an example ? Posted Via Usenet.com Premium Usenet Newsgroup Services------------------------------------------------------ ---- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **---------------------------------------------------------- http://www.usenet.comSubject: === Subject: Die Cantor Die> We don't lose any of our capacity to understand> the reality we live in if we believe that the purpose of all> abstract thinking is to relate the phenomena we observe in> the real world to phenomena we observe in the world of> computation.> Why would I think the purpose of all abstract thinking is to relate > observable phenomena to computational phenomena? The idea is that the mind can be understood by analogy withartificial intelligence software running on a computer (thebrain). Then it is clear that the mind itself is something thatlives in the world of computation. Then thinking is seen to bethe act of relating objects within the world of computation toobjects in the real world. When you are thinking about thinking,you are recognizing that the world of computation is itself a real world (though not part of the physical world).If that doesn't make sense to you, I suppose my whole thesiswill make no sense to you.Subject: === Subject: Die Cantor Die> here about 5 years ago]> The objection to Cantor's Theory (i.e. classical set> theory) comes from the constructivist view that mathematics> necessarily has something to do with computation. This > point of view is especially appealing to the younger> generation.If you are referring to Cantor's Theory as he stated it, the best objection is that it allowed for paradoxes. If you are referring to paradox-free versions, such as Zermelo-Fraenkel set theory, the rest of your arguments fail to address one simple thing:Which of the axioms do you object to?See http://mathworld.wolfram.com/Zermelo-FraenkelAxioms.html for a list of axioms.My guess is that you object to the Axiom of Infinity, which basicly says there is a set containing the natural numbers, and would prefer it to be that the natural numbers form a proper class. If so, then go ahead and develop a new set theory with an Axiom of Finitude or similar. Or it may already exist.-- Will Twentymanemail: wtwentyman at copper dot netSubject: === Subject: Die Cantor Die>>If mathematics, like religion, lived in the private domain,>>then the mathematicians, like the religionists, would be fully>>justified in closing their ears to objections from outsiders.>>But, of course, mathematics lives in the public domain.> And therefore mathematicians are obliged to listen to unmathematical>twaddle about mathematics?> I usually imagine that you're playing games and you know it,> but maybe you're genuinely lost and confused.> Here in the United States the intrusion of religion into> the public domain is a serious issue. The religionists would> love to have you on their side, at least if they were ever to> get into positions of power. To think that they could dismiss> any objections to their agenda as unreligious twaddle about> religion, which they're not obliged to listen to.There are a few problems with this statement. First, religion used to be in the public domain in the United States, starting with the Declaration of Independence. Second, what is a religionist? If by the term you refer to all Muslims, Christians, Hindus, Buddhists, Wiccans, etc, etc, etc, then it clearly makes no sense at all. These various groups of people don't agree with each other, much less have a united agenda.-- Will Twentymanemail: wtwentyman at copper dot netSubject: re:Die Cantor DiePetry, you'll have to choose your position - either computation shouldbe the foundation of mathematics or observation. Those are quiteantipodal concepts. We use different parts of the brain to performthose operations.Classically mathematics is the attempt to link them and make them worktogether. That's where set theory comes very helpful, as it turnssystems of equations to geometrical objects and geometrical objectsto formal concepts that can be fed into a computation.Without the language of set theory this can't be done anumbiguously. Posted Via Usenet.com Premium Usenet Newsgroup Services------------------------------------------------------ ---- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **---------------------------------------------------------- http://www.usenet.comSubject: === Subject: Die Cantor Die> To be sure, formal systems, and hence set theory, are themselves> objects that live in the world of computation. And hence they> are objects which a mathematician might study. But, classical> set theory is not a faithful representation of the universe> of mathematics (it contains ghosts), and is totally unworthy> of its current status as a foundation of mathematics.> So is this a suggestion that mathematics would be better off with no > foundation at all?> I can't believe I haven't made myself clear on this.> I'm advocating that a reality check (what I'm calling observability)> be incorporated into the foundations of mathematics. For lack of> such a reality check, mathematics has drifted off into a fantasy> world.******************************************************** **********************Hi all:Reality? Observability?? What does this have to do with maths AT ALL??! Andwho cares anyway? Mathematics has ALWAYS been an exquisite game, with its ownrules and in its own world. Wanna play? Accept the rules. Don't wanna play? Gobe a lawyer, and ingeneer or a politician, who cares. There hasn't been, there is not, and there won't ever be any reality check, whatever that could possibly mean (for example,attempt to define reality) in the realm of maths,lest it won't be anymore maths but something else. People all over the world care about maths and universities get money for it because it seems to give quite an important service to people...but that's people's call, not maths!Of course, and out of plain pragmatism, mathematics has been forced now and then to make clear its unbelievable and staggering contribution to lots ofthings we consider now as part of modern life in order to get that funding sonecessary in our universities, which by the way is much lower than other wellknown things like physics, chemistry or medicine, say. And because of that, maybe, is that we have received, several years now, that funny thing calledapplied mathematics.And yes, mathematics has its own marvelous fantasy world, and partly because of this reason is that it remains, in advanced and in NOT SO advancedcountries, the one and only subject taught and learned during ALL the 12 basic school years; apparently people, even those with harsh feelings towards mathsfor whatever reason, feel or know how important that stuff is for theindividual development of children...Good luck!TonioSubject: === Subject: Die Cantor Die> Mathematics has ALWAYS been an exquisite game, with its own> rules and in its own world. Wanna play? Accept the rules. In deed, that's seems to be the way the pure mathematiciansview it. To everyone else - those who actually apply it - mathematics is more than just a game. There is an underlyingreality to it, which is computation.Subject: === Subject: Die Cantor Die> If mathematics, like religion, lived in the private domain,> then the mathematicians, like the religionists, would be fully> justified in closing their ears to objections from outsiders.> But, of course, mathematics lives in the public domain.> And therefore mathematicians are obliged to listen to unmathematical> twaddle about mathematics?> I usually imagine that you're playing games and you know it,> but maybe you're genuinely lost and confused.Or perhaps it is Petry who is so confused he does not recognize his own confusion. Whiplash, among other things can have that effect.> Here in the United States the intrusion of religion into> the public domain is a serious issue. The religionists would> love to have you on their side, at least if they were ever to> get into positions of power. To think that they could dismiss> any objections to their agenda as unreligious twaddle about> religion, which they're not obliged to listen to.Then Petry is recommending the appointment of ministers, etc., be put in the hands of atheists. Yes?Subject: === Subject: Die Cantor Die> The fact is, I know people my age who> are _really_ into computers and programming, and they've never> expressed this opinion that mathematics should be all about> computation...> Try a little experiment.> Suggest to them a possible definition of mathematics> as the study of phenomena observable through computation.That would leave out all of classical Euclidean geometry, except through analytic geometry. > Point out, by way of example, that all of the axioms of> high school algebra make predictions about the results> of computation. Again, this ignores geometry.(For example, the distributive law tells> us that if we compute a*(b+c) for arbitrarily chosen> a,b,c, we will get the same result as when we compute> a*b + a*c.)> Then ask them if they see any possible deficiency in that> definition of mathematics.Again, this ignores geometry, among other things.Subject: === Subject: Die Cantor DieDavid Petry says...>I have no reason to value your opinion on anything related to>this topic, but you might change my mind if you could coherently>explain why you believe that set theory, with its wild and >speculative implications about the existence of a super-infinite>world having nothing to do the world we live in, is a better>foundation for mathematics than computation, especially in light>of your observation that mathematics can describe motion,>population growth, gravity, quantum mechanics, etc.*I'm* not trying to tell you what to base your mathematics on. Your theone who is being the totalitarian here.--Daryl McCulloughIthaca, NYSubject: === Subject: Die Cantor DieDavid Petry says...>Here in the United States the intrusion of religion into>the public domain is a serious issue. The religionists would>love to have you on their side, at least if they were ever to>get into positions of power. To think that they could dismiss>any objections to their agenda as unreligious twaddle about>religion, which they're not obliged to listen to.I think your position is closer to religious fanaticism thanthe position of most mathematicians. Mathematicians are justexploring their interests, they aren't furthering an agenda.Unlike you.--Daryl McCulloughIthaca, NYSubject: === Subject: Die Cantor Die> Mathematics itself can describe things having nothing to> do with computation: motion, population growth, gravity, quantum> mechanics, etc.> To make computation both a tool of mathematics and the *subject* matter> of mathematics is much too limiting. > To say that I'm trying to limit the subject matter of mathematics> to computation is akin to saying that set theory limits the subject> matter of mathematics to sets. It's a despicable lie.<25bac3c0.0403091254.58ee9dea@posting.google.com>,> I'm advocating that a reality check (what I'm calling observability)> be incorporated into the foundations of mathematics. So who is lying? Certainly not Daryl.Subject: === Subject: Die Cantor Die> Here in the United States the intrusion of religion into> the public domain is a serious issue. The religionists would> love to have you on their side, at least if they were ever to> get into positions of power. To think that they could dismiss> any objections to their agenda as unreligious twaddle about> religion, which they're not obliged to listen to. In so far as your comments have any relevant content, they merelyreiterate that you think it's awful how mathematics is promotingCantorian religion, and so on. So why should mathematicians either beobliged to, or be inclined to, pay attention to this twaddle?Subject: === Subject: Die Cantor Die> So why should mathematicians either be> obliged to, or be inclined to, pay attention to this twaddle?Should people in positions of power be obliged to listento the powerless?Subject: === Subject: Die Cantor Die> To be sure, formal systems, and hence set theory, are themselves> objects that live in the world of computation. And hence they> are objects which a mathematician might study. But, classical> set theory is not a faithful representation of the universe> of mathematics (it contains ghosts), and is totally unworthy> of its current status as a foundation of mathematics.> So is this a suggestion that mathematics would be better off with no > foundation at all?> I can't believe I haven't made myself clear on this.> I'm advocating that a reality check (what I'm calling observability)> be incorporated into the foundations of mathematics. For lack of> such a reality check, mathematics has drifted off into a fantasy> world.Then do you observe that the real world geometry is flat, positively curved(elliptic) or hyperbolic(negatively curved), just so that we know which geometries to throw out, you understand.Subject: === Subject: Die Cantor Die>But if>>we accept Cantor's interpretation of the diagonalization>>argument, as codified in classical set theory, we are led to>>believe in the existence of a super-infinite world that has>>no connection to the world of computation, other than the>>borrowing of terminology.>His argument simply states that if you construct a function f:N->[0,1], >then he can construct a number in [0,1] that is not in the codomain of >f. He concludes that there is no f that is an onto map. Conclusion: >[0,1] has more elements than N, based on a formalized notion of >comparing the number of elements in two sets. I'm not clear on why you >feel this causes a problem.> Small nit to be picked:> I hope you mean that there is a number not in the RANGE, or IMAGE of f, > since the 'codomain' of any f:A -> B is almost universally taken to be B.Sorry, IMAGE is the correct word.-- Will Twentymanemail: wtwentyman at copper dot netSubject: === Subject: Re : Birkhoff / Maclane Algebra - category theory Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.miscSubject: === Subject: Birkhoff / Maclane Algebra - category theory>i was wondering if Algebra by Birkhoff and Maclane is a good choice>for a self-study book in algebra. I have had previous courses in>abstact algebra, linear algebra, group theory (basic) and ring theory>(noetherian rings) at undergraduate level. >There are two such books; one is called simply Algebra. The other is>called A Brief Survey of Modern Algebra. The latter is much better>than the former.AFAIK they are aimed at different audiences. Also, the former isimportant in that it introduced the use of cathegorical concepts inthe *teaching* of algebra in courses at an undergraduate level...Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.miscSubject: === Subject: Birkhoff / Maclane Algebra - category theory> i was wondering if Algebra by Birkhoff and Maclane is a good choice> for a self-study book in algebra. I have had previous courses in> abstact algebra, linear algebra, group theory (basic) and ring theory> (noetherian rings) at undergraduate level. In fact, i kind of like> this book (at least the first chapter) and i know it is considered as> a good one but i am quite afraid of the fact that they use category> theory as a basis in the rest of the book.> Category theory seems a little bit rough to me, i don't want to rush> into something and spend a great amount of time on it if it is too> early in my learning process. I might miss quite a great amount of> things in the rest of the text if i don't have a strong grasp on basic> category theory.They do not develop pure category theory in the book. Categoricalnotions are introduced as far as it is helpful for understandingseveral constructions from other chapters.Which edition did you look at?In the 2nd and 3rd edition they consequently reordered the sequenceof the relevant chapters (quote from the introduction):| The treatment of many of the topics in the first edition has been| simplified---and clarified---in this second edition. The material on| universal constructions, formerly introduced at the end of the first| chapter, has now been assembled in Chapter IV, at a point where there| are at hand many more effective examples of these constructions.[...]So you will probably find the later editions more suitable for self-study.MarcSubject: === Subject: Birkhoff / Maclane Algebra - category theoryThank you for your comments,> i was wondering if Algebra by Mac Lane and Birkhoff is a good choice> for a self-study book in algebra.> Which edition did you look at?i looked at the 3rd edition, it is said to be completly different fromthe others with extra chapters along with clearer explanations.> In the 2nd and 3rd edition they consequently reordered the sequence> of the relevant chapters (quote from the introduction):> | The treatment of many of the topics in the first edition has been> | simplified---and clarified---in this second edition. The material on> | universal constructions, formerly introduced at the end of the first> | chapter, has now been assembled in Chapter IV, at a point where there> | are at hand many more effective examples of these constructions.[...]> So you will probably find the later editions more suitable for self-study.ok, this is the one i wanted to use anyway.All the comments seem to agree on the quality of (at least this edition)of this book even for self study and on its capability to give enoughintuitionof the subject to go further with confidence.CharlesSubject: FEM/CFD Job Cambridge UKResearch Associate : FEM/CFD : University of Cambridge http://www.jobsin.finite-element-method.infoSubject: AlgebroidsDear NG,I was skimming my notes in the past few years and I stumbled into algebroid function elements when I was studying advanced complex analysis. I wanted to know if I still have the correct notion of them. I wanted to know how other people in this newsgroup would define the algebroid. It is a general form of function (or functional, however it is called !?!) element that has to do with the Puisuex expansion in my definition. These algebroid can form Riemann surfaces and topology with interesting properties. Sincerely,Jose CapcoSubject: === Subject: Algebroids>Dear NG,>I was skimming my notes in the past few years and I stumbled into algebroid >function elements when I was studying advanced complex analysis. I wanted to >know if I still have the correct notion of them. I wanted to know how other >people in this newsgroup would define the algebroid. It is a general form of >function (or functional, however it is called !?!) element that has to do with >the Puisuex expansion in my definition. These algebroid can form Riemann >surfaces and topology with interesting properties. Yeah, that's what they mean to me, too. I believe you can definean algebroid branch (or algebroid function element) as a formalpower series in a formal nth root of the independent variable(a different n for any given branch, but we want to exclude, forinstance, a formal series like z^{1/2}+z{3/4}+...+z^{1-2^{-k}}+...;what is like algebraic about an algebroid branch is that the exponents of terms with non-zero coefficients all have a common denominator,which is a finiteness condition, which is like algebra ratherthan analysis, I guess). By applying the Puiseux expansion toan actual algebraic branch, one gets an algebroid branch which(when we started out over C, at least) is not merely formal, butactualyl has a nonzero radius of convergence. Of course notevery algebroid branch arises this way, so they're a truegeneralization. On the other hand, there's a comparison theorem(which I once somewhere called a remote ancestor of the comparison theorem of M. Artin, a theorem which I no longer am sure where to find, and certainly can't state) that says thatfor many purposes the generalization isn't too broad. Forinstance, a finite (and therefore convergent) initial segmentof the algebroid branch already contains a lot of Galois-type,singularity-theory-type information. Don't ask me what goes on when you aren't over C, though.Lee RudolphSubject: Finding the equation of a tilted parabolagiven its focus (2,-1) and the equation of the directrix 3x-5y+1=0I would be able to do this if the parabola was vertical or horizontal byfinding the focus and p through the equation y^2 = 4py. However theparabola is tilted and thus is a problem of another level.Subject: === Subject: Finding the equation of a tilted parabolaYou are given the focus and the directrix.Do you know their relation to the parabola?Use that relation to form an equation. Theequation will be between wo distances....You can lead a horse's ass to knowledge, but you can't make him think.Subject: multiplicative order of numbers which include powers of primesare there any proofs about how the multiplicative order of numberswhich include powers of primes is calculated?I put them in three categories:1. Numbers of the Form p^n where p is a prime and n a natural number 12. Numbers in general of the from a^m where a and m are naturalnumbers > 1 and both can be composite3. Numbers of the form b*a^m where all three can be natural numbers.I know the following two rules:Ord(x,m) divides Phi(m) (Ord(x,m) means Order x (mod m))and Ord(x,a*b) = least common multiple of [Ord(x,a) ; Ord(x,b)] if a and bshare no common divisor.By this it is almost clear, that the Order of Numbers of category 1.should look like (Ord(x,p) * p^(n-1) but I don't have a proof for this. If you have one, It would be reallynice if you could post it here, or post the address of it :o)****For 2. I did some computation, because at first I thought that I couldalso apply the fromula above to this numbers, but it seems that thisis not true.For example if you look at the orders of the powers of 6, 10 or 21:6^1 => 2 6^2 => 6 6^3 => 18 (base 5)10^1 => 4 10^2 => 20 10^3 => 100 (base 3)21^1 => 6 21^2 => 42 21^3 => 882 (base 2)The first line means, that Ord(5,6^1) = 2 and Ord(5,6^2)=6.The second line means, that Ord(3,10^1)=4 and so on...The most interesting thing about this is, that the orders of compositepowers doesn't follow the rule mentioned for prime powers. I guess therule here is something like:Ord(x,a))*(a / gcd(Ord(x,a);a))^(m-1)But please note that this is just a guess. I didn't check enoughnumbers nor do I have a proof. But maybe there is already a proofabout the acual form of the order for such numbers?!****For 3I think that depending on the actual form of the order of compositepowers (numbers of the second form) the form of this numbers can mostlikely be derived using the rule:Ord(x,a*b) = least common multiple of [Ord(x,a) ; Ord(x,b)]Here I have some examples (Ord(2,7*...)):7 => 3 7*3 => 6 => lcm [Ord(2,7) ; Ord(2,3)]7*3*3 => 6 => lcm [Ord(2,7) ; Ord(2,3*3)]7*3*3*3 => 187*3*3*3*3 => 543*7 => 6 (see above)3*7*7 => 423*7*7*7 => 294It again gets more interesting if the base is composite (here 10):10 => 4 => Ord(3,10)=45 => 4 => Ord(3,5)=42 => 1 => Ord(3,2)=17 => 6 => Ord(3,7)=67*10 => 12 => lcm [Ord(2,7) ; Ord(2, 10)]7*10*10 => 60 => lcm [Ord(2,7) ; Ord(2, 10*10)]7*10*10*10 => 300 => lcm [Ord(2,7) ; Ord(2, 10*10*10)]7*10*10*10 => 1500 => lcm [Ord(2,7) ; Ord(2, 10*10*10*10)]Note that the numbers don't increase by factor 10 but only by factor5.This behaviour could be explained by a rule like the one stated above{ Ord(x,a))*(a / gcd(Ord(x,a);a))^(m-1) }.Sorry if I maybe used inaccurate expressions in this posting.I need the form of the multiplicative order of this three for a proof.It would really be nice if you could forward me to sources about thistopic :o)Thank you in advanceJuergen BullingerSubject: === Subject: Cantor Paradox :-)I posted this somewhere in the web of emails that are children of theoriginal email ... but it's really in the wrong place ... so here it isagain, attached to the email to which I'm replying:A countable number of finite strings over a finite alphabet can describe anuncountable number of real numbers. Note:The set of real numbers> Hello :-)> I'm sure you're quite familiar with Cantor's proof that R is> uncountable: specifically, assume the opposite and construct an N<->R> bijection and find a diagonal r which is the image of nothing in N.> This proof is, of course, flawed: afterall, R is quite countable!> To illustrate the flaw, I shall prove that there exists an> uncountable number of finite-length strings of typable characters> (which, you'll acknowledge at once, is absurd)!> Theorem: The set of finite-length strings of typable characters> is uncountable.> Proof:> It will suffice to merely exhibit an uncountable subset, for how> can a countable set possibly contain an uncountable subset?> We claim that such a subset is the set of all finite-length> typable characters which, when read, define a real number. (An example> of such a string would be: The ratio of the circumference to the> diameter of the curve whose equation is x^2+y^2=1). Call this subset> S. Suppose, for sake of argument, that S is countable.> Then we can construct a countable list of all the distinct reals> defined by the strings in S.> Now if we pull a Cantor and go down the diagonal of this list,> adding 1 to each digit (mod 10), we end up with a real r which is not> in the list-- and which therefor cannot be defined using any finite> string of typable characters. But this is absurd: for, we have just> defined it, using a finite string of typable characters!> Our only option, then, is to discard our assumption: and conclude> that S, and therefor the larger set of ALL finite strings of typable> characters, are uncountable. Q.E.D.> As you see, we have to conclude that either Cantor's proofs are> worthless wastes of ink, or there exist an uncountable number of> finite-length strings of typable characters.> I'll let you decide. :-)> Your friend,> Nathan the Great> Age 11Subject: re:Cantor Paradox :-)What's the point of criticizing a correct proof? What's the point ofpicking out of all the mathematicians that used diagonal argumentsthe Jewish one and launching an assault on him?That's what cranks do nowadays. Posted Via Usenet.com Premium Usenet Newsgroup Services------------------------------------------------------ ---- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **---------------------------------------------------------- http://www.usenet.comSubject: === Subject: re:Cantor Paradox :-) Original post:_________________________________________________________ __Theorem: The set of finite-length strings of typable charactersis uncountable. Proof: It will suffice to merely exhibit an uncountable subset, for howcan a countable set possibly contain an uncountable subset? We claim that such a subset is the set of all finite-lengthtypable characters which, when read, define a real number. (An exampleof such a string would be: The ratio of the circumference to thediameter of the curve whose equation is x^2+y^2=1). Call this subsetS. Suppose, for sake of argument, that S is countable. Then we can construct a countable list of all the distinct realsdefined by the strings in S. Now if we pull a Cantor and go down the diagonal of this list,adding 1 to each digit (mod 10), we end up with a real r which is notin the list-- and which therefor cannot be defined using any finitestring of typable characters. But this is absurd: for, we have justdefined it, using a finite string of typable characters! Our only option, then, is to discard our assumption: and concludethat S, and therefor the larger set of ALL finite strings of typablecharacters, are uncountable. Q.E.D.________________________________________________________ ___A countable number of finite strings over a finite alphabet can describe anuncountable number of real numbers. Note:The set of real numbers> What's the point of criticizing a correct proof? What's the point of> picking out of all the mathematicians that used diagonal arguments> the Jewish one and launching an assault on him?> That's what cranks do nowadays.> Posted Via Usenet.com Premium Usenet Newsgroup Services> ----------------------------------------------------------> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **> ---------------------------------------------------------- > http://www.usenet.comSubject: === Subject: Cantor Paradox :-) Discussion, linux)> What's the point of criticizing a correct proof? What's the point of> picking out of all the mathematicians that used diagonal arguments> the Jewish one and launching an assault on him?Are you suggesting that cranks (and wags like Nathan) are motivated byanti-Semitism? Have you any evidence at all for this claim? I think that's evidence of a willingness to see racism and bigotry asthe source of dissent that has nothing to do with racism or(non-mathematical) bigotry. I don't know about Nathan and other, more serious dissenters, but thatCantor is Jewish is news to me. I've never noticed any comments abouthis religion or heritage. I find it very implausible that it playsany role.Do you suppose those that criticize Turing's notion of computabilityare homophobes?> That's what cranks do nowadays.-- Jesse F. HughesAnd I'm one of my own biggest skeptics as I had *YEARS* of wrongideas, and attempts that failed. Worse, for some of them it took*MONTHS* before I figured out where I screwed up. -- James HarrisSubject: === Subject: Cantor Paradox :-)> What's the point of criticizing a correct proof? What's the point of> picking out of all the mathematicians that used diagonal arguments> the Jewish one and launching an assault on him?> Are you suggesting that cranks (and wags like Nathan) are motivated by> anti-Semitism? Have you any evidence at all for this claim? > I think that's evidence of a willingness to see racism and bigotry as> the source of dissent that has nothing to do with racism or> (non-mathematical) bigotry. > I don't know about Nathan and other, more serious dissenters, but that> Cantor is Jewish is news to me. I've never noticed any comments about> his religion or heritage. I find it very implausible that it plays> any role.> Do you suppose those that criticize Turing's notion of computability> are homophobes? No. The only people who have criticized Turing's notion of computabilty have only crictized Turing's notion of halt, which is more stupid than Set Theory.> That's what cranks do nowadays.Subject: === Subject: Cantor Paradox :-) <87ishcgc0s.fsf@phiwumbda.org No. The only people who have criticized Turing's > notion of computabilty have only crictized > Turing's notion of halt, which is more stupid> than Set Theory.I'm telling you, you just don't have the zen poetry of yourpredecessor. You keep giving too much information which betrays anawareness of terminology that the last guy didn't have. Compare what Hume was profoundly stupid, since he predicted that he would be the last philosopher. His main mistake was assuming that he knew something about understanding, the rest of his errors in reasoning naturally followed just as the domino effect predicts all semi-intelligent, semi-coherent philosopher babblings will tumble.features, the crazy claims that come from nowhere, the bizarreveneration of engineering as the queen of human achievement (absent inthis post). You need to work on your technique. In private.-- Jesse F. Hughes[blah blah blah] which you are trying to dispute, but I've found thatyou've used ***definitions*** rather than mathematical logic, often inyour posts. -- James Harris on the insufficiency of math. definitions.Subject: === Subject: Cantor Paradox :-) > I don't know about Nathan and other, more serious dissenters, but that> Cantor is Jewish is news to me. I've never noticed any comments about> his religion or heritage.Oh but you have .... Petry accused him of introducing religiousmysticism into maths :-(-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_Subject: === Subject: Cantor Paradox :-)>I don't know about Nathan and other, more serious dissenters, but that>Cantor is Jewish is news to me. I've never noticed any comments about>his religion or heritage.> Oh but you have .... Petry accused him of introducing religious> mysticism into maths :-(Cantor introduced mathematics of the infinite into religious mysticism.-- Aatu Koskensilta (aatu.koskensilta@xortec.fi)Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-PhilosophicusSubject: === Subject: Cantor Paradox> Who cares if you and I don't know what's on the list? The list is> still a well-defined list (ignoring vagueness of natural language, of> course).We can generate every finite string of ASCII characters like Nathansuggested.Determining if the string describes a real number could be problematic.> This is going in circles. I don't care to continue the point.> Computability is not a requirement for Cantor's diagonal argument.> Nathan's list is well-defined. The only issue is whether the diagonal> number thus constructed is expressible in the object language.The diagonal number must be in the list.For example, consider the string:If the nth digit of the nth real in Nathan's list is 3 then the nth digitof this number is 7 else the nth digit of this number is 3.In fact, every possible definition of a diagonal number must be in the list.The diagonal number must be on the list and it can't be on the list.Russell- 2 many 2 countSubject: === Subject: Cantor Paradox <87ptbqvzhw.fsf@phiwumbda.org <4049e3e6$0$3954$afc38c87@news.optusnet.com.au <87n06uvwz1.fsf@phiwumbda.org <404a6bd0$0$22425$afc38c87@news.optusnet.com.au <87znatrqq8.fsf@phiwumbda.org> Who cares if you and I don't know what's on the list? The list is> still a well-defined list (ignoring vagueness of natural language, of> course).> We can generate every finite string of ASCII characters like Nathan> suggested.> Determining if the string describes a real number could be problematic.> This is going in circles. I don't care to continue the point.> Computability is not a requirement for Cantor's diagonal argument.> Nathan's list is well-defined. The only issue is whether the diagonal> number thus constructed is expressible in the object language.> The diagonal number must be in the list.> For example, consider the string:> If the nth digit of the nth real in Nathan's list is 3 then the nth digit> of this number is 7 else the nth digit of this number is 3.> In fact, every possible definition of a diagonal number must be in the list.> The diagonal number must be on the list and it can't be on the list.If the strings are strings in a natural language, then the fact thatthere is an evident paradox is not surprising and has no bearing onCantor's diagonal argument. Natural languages (which fail todistinguish between object and metalanguage) are fraught withparadoxes. It's just a feature of natural language, a subjectappropriate to linguists but not a subject with direct implicationsfor Cantor's theorem.On the other hand, if the strings are supposed to be strings in arepresentable by a formula in our language. In fact, Nathan's proofessentially yields an indirect proof that the relation the stringrepresented by n defines a number is not representable in the objectlanguage, as Rupert pointed out.Of course, maybe there's also a proof that the relation isrepresentable (say, if our basic mathematical foundations areinconsistent, but that's very unlikely). Go ahead and try to show itif you like. But I promise you, Russell, that such a proof (even ifit existed) is beyond your current background. You'll need a carefulreading of the issues surrounding Goedel's representation ofmetamathematic relations like x represents a proof of the formularepresented by y. In principle, you could master the material, but it would take time,patience and care. All this, only to discover that you cannot findthe proof that definability is representable. Oh yeah, unless, ofcourse, our foundations are inconsistent. Then you'll be the proudauthor of a proof of inconsistency and your name will go down inhistory.Bit of a long shot, though.-- It has been shown that no man can sit down to write without a very profounddesign. Thus to authors in general trouble is spared. A novelist, for example,need have no care of his moral. It is there -- that is to say, it is somewhere-- and the moral and the critics can take care of themselves. --E.A. PoeSubject: === Subject: Cantor Paradox> Who cares if you and I don't know what's on the list? The list is> still a well-defined list (ignoring vagueness of natural language, of> course).> We can generate every finite string of ASCII characters like Nathan> suggested.> Determining if the string describes a real number could be problematic.> This is going in circles. I don't care to continue the point.> Computability is not a requirement for Cantor's diagonal argument.> Nathan's list is well-defined. The only issue is whether the diagonal> number thus constructed is expressible in the object language.> The diagonal number must be in the list.> For example, consider the string:> If the nth digit of the nth real in Nathan's list is 3 then the nth digit> of this number is 7 else the nth digit of this number is 3.> In fact, every possible definition of a diagonal number must be in the list.> The diagonal number must be on the list and it can't be on the list.> Russell> - 2 many 2 countAs there are several proofs of The uncountability of the set of reals, what is the point of criticising just one proof?Subject: === Subject: Cantor Paradox> As there are several proofs of The uncountability of the set of reals,> what is the point of criticising just one proof?The diagonal argument is the best known and it has beenused to prove a lot of things besides the uncountability ofthe real numbers. It is important to know what it can proveand what it can't prove.Russell- 2 many 2 countSubject: === Subject: Cantor Paradox> As there are several proofs of The uncountability of the set of reals,> what is the point of criticising just one proof?> The diagonal argument is the best known and it has been> used to prove a lot of things besides the uncountability of> the real numbers. It is important to know what it can prove> and what it can't prove.> Russell> - 2 many 2 countWhat theorem, if any, has it been used to prove that has no other proof?If, as I suspect, none, then what is the point of criticizing it?Subject: === Subject: Cantor Paradox> The diagonal number must be in the list.> For example, consider the string:> If the nth digit of the nth real in Nathan's list is 3 then the nth digit> of this number is 7 else the nth digit of this number is 3.Does the following string also correspond to a number?The 1st digit of this number is 2 and the 1st digit of this number is8.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 WSubject: === Subject: Cantor Paradox> The diagonal number must be in the list.> For example, consider the string:> If the nth digit of the nth real in Nathan's list is 3 then the nthdigit> of this number is 7 else the nth digit of this number is 3.> Does the following string also correspond to a number?> The 1st digit of this number is 2 and the 1st digit of this number is> 8.No. (Well, it might be one of Cantor's missing numbers.)Any describable diagonal number must be on the list.After all, the describable numbers are countable, aren't they?Russell- 2 many 2 countSubject: === Subject: Help is needed here.>I need a little help with this problem. Thank you very much.>Statement:> Let G be a group and suppose that |G|=pq where p and q are primes. >Question:> Does it have to be the case that G is cyclic? If so prove. If not>give a counterexample.>thanks a lot everyoneCounterexample: The group of permutations S_3. This group is notcyclic, because it is not abelian, but |S_3|=6=2*3.HTH-- Benedikt Plittbeplitt@math.uni-muenster.deSubject: === Subject: Help is needed here.> I need a little help with this problem. Thank you very much.> Statement:> Let G be a group and suppose that |G|=pq where p and q are primes. > Question:> Does it have to be the case that G is cyclic? If so prove. If not> give a counterexample.The simplest countexample is KLEIN's 4 element group with themultiplication table (omitting the entries for 1) *| a b c -+----------- a| 1 c b b| c 1 a c| b a 1It appears in many mathematical contexts. For instance in linearalgebra as the group of 180deg rotations about the x- y- or z-axisincluding the identity mapping in an orthogonal coordinate systemusing the concatenation of rotations as multiplicationThis group may be represented by the 3x3 matrices1 0 0 1 0 0 -1 0 0 -1 0 00 1 0 0 -1 0 0 1 0 0 -1 00 0 1 0 0 -1 0 0 -1 0 0 1using the matrix multiplication or simply by the 3 vectors(1 1 1) (1 -1 -1) (-1 1 -1) (-1 -1 1)using the multiplication by element(x1 y1 z1) * (x2 y2 z2) := (x1x2 y1y2 z1z2)or as the set of the permutations(1,2,3,4) -> (1,2,3,4)(1,2,3,4) -> (2,1,4,3)(1,2,3,4) -> (4,3,2,1)(1,2,3,4) -> (3,4,1,2)of a 4 element sequence-- HorstSubject: === Subject: Help is needed here.Distribution: aus> I need a little help with this problem. Thank you very much.> Statement:> Let G be a group and suppose that |G|=pq where p and q are primes. > Question:> Does it have to be the case that G is cyclic? If so prove. If not> give a counterexample.> The simplest countexample is KLEIN's 4 element group [...]Another simple example which is slightly easier to describe is thesymmetric group on three elements; this group has order 6.Cheers,Geoff.----------------------------------------------- ------------------------------Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --ftww@maths.usyd.edu.au | Gameplayer by vocation.----------------------------------------------------- ------------------------Subject: Is this correct?Hello everyone I wondering if someone could look at this proof to seeif it sounds correct and if I need to add or change anything in it.Question: An affine map is a function f : R -> R of the form f(x) = ax + bwhere a and b are fixed real numbers with a /= 0. Prove that Aff, thesetof all affine maps, is a group under the operation of composition.Proof:If g : R -> R has the form g(x) =cx+d, where c /= 0, then fg(x)=f(g(x))=f(cx+d)=a(cx+d)+b=acx+(ad+b).Since ac /= 0, the composite fg is an affline map. The identityfunciton1 : R -> R is an affline map (set a=1 and b=0), while the inverse of Fis easliy seen to be x=> (a^-1)x - (a^-1)b.Again thank you for your help.Subject: === Subject: Is this correct?>Hello everyone I wondering if someone could look at this proof to see>if it sounds correct and if I need to add or change anything in it.>Question: An affine map is a function f : R -> R of the form> f(x) = ax + b>where a and b are fixed real numbers with a /= 0. Prove that Aff, the>set>of all affine maps, is a group under the operation of composition.>Proof:>If g : R -> R has the form g(x) =cx+d, where c /= 0, then> fg(x)=f(g(x))=f(cx+d)=a(cx+d)+b=acx+(ad+b).>Since ac /= 0, the composite fg is an affline map. The identity>funciton>1 : R -> R is an affline map (set a=1 and b=0), while the inverse of F>is easliy seen to be x=> (a^-1)x - (a^-1)b.>Again thank you for your help.Seems correct to me!Have a good time,Benedikt.-- Benedikt Plittbeplitt@math.uni-muenster.deSubject: === Subject: MedicationThe example given usses MANY assumptions in it's data..Does your situation have the same attributes?ie: half life of the medication?Do the calcs again ussing a half life of 2 weeks and you'll see it not levelout untill 2 weeks have pastjohnBah!! why isn't outlook putting preceeding >'s in the quoted part?The reason why I ask is because genuinely I am taking antidepressants, andthey say these take several weeks to kick in. I can't understand why.Brian> Well, this is pretty easy.> Lets assume that right after you take the first pill, your body> concentration is c grams/kilo or however you measure it.> Lets look at the concentration on day n, immediately after you take apill.> You get a concentration of:> c from the pill you just took> c/2 from yesterdays pill> c/4 from the pill 2 days ago> .> .> c/2^n from the pill n days ago, which was your very first pill.> So your concentration on day n right after the pill isc(1+1/2+1/4....1/2^n)> Your concentratiuon gets closer and closer to 2c immediately after youtake> a pill and c immediately before, without (in theory) ever reaching these> limits.> In practice, after a few days you are simply moving between 2c down to c24> hours later when you take another pill.> HTH> Peter Webb> I am really not very up with my maths, and I wonder whether anyone canhelp.> If I am taking medication at the rate of one pill per day, and thehalf-life> of the substance (concentration in specific location in the body) is one> day, how long does it take for the concentration to maximize? Assume some> sort of asymptotic curve, like y = k/(x +1) , where k is a constant, y is> the concentration in the body of one pill, and x is the time since taking> it. Assume maximum concentration at time of taking it.Subject: === Subject: How to solve this differention equation?> Dear All,> I encountered a differention equation problem as below.> Assume g(x) is a a differentiable scalar function of x,> and the equation is> (g'(x))^2 = 1/(1+g^2(x)), where g'(x) is the first order direvative of> g(x) w.r.t x.> So given the above equation, how to get the> explicit form of g(x) or implicit value of g(x) give a particular value x.> Fred> This method is probably in your differential equations textbook.> Let's write y=g(x), so we have (dy/dx)^2 = 1/(1+y^2).> take squareroot, we assume y increasing so dy/dx > 0, but> a similar solution can be done for y decreasing...> dy/dx = 1/sqrt(1+y^2)> so> sqrt(1+y^2) dy = dx> integrate> (asinh(y)+y*sqrt(1+y^2))/2 = x+C> so we get the implicit equation for g(x)> asinh(g(x))+g(x)*sqrt(1+g(x)^2) = 2*(x+C)You really shouldn't solve the lad's HW for him. How will he ever learnto do it on his own? It would be enough to point out that after takingthe square root of both sides the equation is separable. Let him do theintegral.-- Julian V. NobleProfessor Emeritus of Physicsjvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ God is not willing to do everything and thereby take away our free will and that share of glory that rightfully belongs to us. -- N. Machiavelli, The Prince.Subject: === Subject: How to solve this differention equation?>Assume g(x) is a a differentiable scalar function of x,>and the equation is>(g'(x))^2 = 1/(1+g^2(x)), where g'(x) is the first order direvative of>g(x) w.r.t x.>So given the above equation, how to get the>explicit form of g(x) or implicit value of g(x) give a particular value x.Maple's solution is g(x) / | 1 | - ----------------------- d_a - x - _C2 = 0, | 1/2 / (_C1 + 2 arctan(_a)) g(x) / | 1 | ----------------------- d_a - x - _C2 = 0 | 1/2 / (_C1 + 2 arctan(_a))Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2Subject: === Subject: How to solve this differention equation?>Assume g(x) is a a differentiable scalar function of x,>and the equation is>(g'(x))^2 = 1/(1+g^2(x)), where g'(x) is the first order direvative of>g(x) w.r.t x.>So given the above equation, how to get the>explicit form of g(x) or implicit value of g(x) give a particular value x.> Maple's solution is > g(x)> /> | 1> | - ----------------------- d_a - x - _C2 = 0,> | 1/2> / (_C1 + 2 arctan(_a))> g(x)> /> | 1> | ----------------------- d_a - x - _C2 = 0> | 1/2> / (_C1 + 2 arctan(_a))Did you use symmetry methods or the like?The usual command returns integral-free implicit solutions:> ode:=diff(g(x),x)^2=1/(1+g(x)^2):> sols:=dsolve(ode,g(x)); 2 1/2 sols := x - 1/2 g(x) (1 + g(x) ) - 1/2 arcsinh(g(x)) - _C1 = 0, 2 1/2 x + 1/2 g(x) (1 + g(x) ) + 1/2 arcsinh(g(x)) - _C1 = 0-- Thomas RichardMaple SupportScientific Computers GmbHhttp://www.scientific.deSubject: ----- help needed.. maths fundas in yarn image analysis ---hi guys can anyone tell me how can i find the best fit linei need to find the axis of the yarn body and for that i need to takesay 10 points and then i have to find the best fit linehow can i do thatcan u provide some functions / codes in MATLABone more thinghow can i use maths fundamentals to solve my problems see the link belowhttp://www.geocities.com/dhaval_patel_1/yarnprobs.htmlif anyone knows plz tell mewarm regardsDhavalSubject: How to solve a polynomial of degree 6 over a ring modulo a composite by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2A7ZYm14180;Dear Research Community,I have a generic polynomial equation..f(t)= t^6 + a. t^5 + b. t^4 + c. t^3 + d. t^2 + e. t + g = 0I need to solve the above polynomial of degree 6 over a ring modulo acomposite n (n is large of the order of 1024 bbits). Where the factorsof n is unknown and it is hard to find too. I want a solution thatdoes not amount to factoring n.I have a finite but large enough set of {a,b,c,d,e,g} for which I needto know at least one solution. Can anybody please help?thanks in advance..regards,MukeshSubject: Recommendation between two programs... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2AJatU32324;Hello: I am looking at getting a math software program for school. I amcurrently a college student, and I was just introduced to Maple 9 bymy Calculus instructor. WOW! I never knew such a product existed. Thinking I would do a little more research on Maple 9, I actuallydiscovered Mathematica, IDL, and Matlab (actually I was introduced toMatlab as well, but we didnt tool around with it). Here is a little about me: I am working toward a degree inBiochemistry; I want to take as many forms of higher maths as I can(Calc II, III, etc.); I have a little computer knowledge; pricedoesnt matter too much (Im not rich but I could justify spendingabout two hundred dollars if I could find a neat and useful softwareapplication); my only other math tools that Ive used were a TI-83+ SE(and a slide rule for fun). I am looking at buying this type ofsoftware as an aide in my studies and to double check my work. Right now, I am looking at either Mathematica or Maple 9. Any help oradvice you offer would be greatly appreciated. Thomas Styron from North Carolina Subject: === Subject: Recommendation between two programs...> Hello:> Right now, I am looking at either Mathematica or Maple 9. Any help or> advice you offer would be greatly appreciated.Consider also MathCad. It's a bit pricey, but is aworkhorse for data analysis and presentation. Itinterfaces reasonably well to Excel, too.Subject: === Subject: Recommendation between two programs...I like Maple better than Mathematica. I used Mathematica for a few years and then converted to Maple. I actually found Maple easier to use which I think is unusual as usually whatever you are most familiar with is what is easiest.If I understand correctly Maple comes with more libraries for the price so it would also seem to be better value for money.Russell.>Hello: >I am looking at getting a math software program for school. I am>currently a college student, and I was just introduced to Maple 9 by>my Calculus instructor. WOW! I never knew such a product existed. >Thinking I would do a little more research on Maple 9, I actually>discovered Mathematica, IDL, and Matlab (actually I was introduced to>Matlab as well, but we didnt tool around with it). >Here is a little about me: I am working toward a degree in>Biochemistry; I want to take as many forms of higher maths as I can>(Calc II, III, etc.); I have a little computer knowledge; price>doesnt matter too much (Im not rich but I could justify spending>about two hundred dollars if I could find a neat and useful software>application); my only other math tools that Ive used were a TI-83+ SE>(and a slide rule for fun). I am looking at buying this type of>software as an aide in my studies and to double check my work. >Right now, I am looking at either Mathematica or Maple 9. Any help or>advice you offer would be greatly appreciated. >Thomas Styron from North Carolina > Subject: === Subject: some questions about Hensel lifting> U = (Bk C) mod Gk mod pk (about mod Gk and mod Hk see below)> V = (Ak C) mod Hk mod pk> G2k = Gk + pk U> H2k = Hk + pk V> C' = D + Ak U + Bk V> A2k = Ak - pk C' Ak> B2k = Bk - pk C' BkIf my recollection is correct, it is possible to divide C' Akby Gk mod pk, and that will keep the degree of A2k < deg(G2k).This is compensated by the division of C' Bk and Hk.> Question 2: How are linear and quadratic combined in practice?In giac/xcas, I use quadratic lift for polynomial factorization,except for the last step if a linear lift is sufficient(in combination with true factor tries). Note that thereal situation is a little bit more complicated since you haveto lift a product of n terms instead of 2.You can have also a look in mupad source codein lib/POLYLIB/FACTOR/plift.muthe bound2list seems to find a best strategy in mixinglinear/quadratic lifts. But I'm not sure it is reallybetter than quadratic except last stepsince you need to compute the Ak and Bk.Subject: === Subject: some questions about Hensel lifting> Background: (please skip down to the good stuff if you wish)> In _Polynomials: An Algorithmic Approach_ authors Mignotte and Stefanescu > show how linear lifting and quadratic lifting work.> Now, one problem with working through this book is that there are typos > and worse on nearly every page, and the reader has to permute and/or > combine all conceivable corrections to these and determine the set of > corrections which, indeed, produces a valid proof of a given theorem.> Sometimes these corrections affect even the hypotheses, or even the > conclusions of the theorem. Sometimes these corrections in one theorem B > imply the inapplicability of an earlier theorem A that B depends on, so > you backtrack and try another set of corrections for theorem A, until you > find one which supplies the proper preparation for theorem B. So it's a > recursive procedure which tries all permutations/combinations and > sometimes returns to itself with a last didn't work so get next and call > me again return code.> OK. There were a couple of typos in the linear lifting theorem. And there > were quite a few typos and even conceptual errors in the quadratic > lifting theorem -- getting G and H backwards, confusing A and B, > introducing a polynomial E which isn't needed and getting it confused > with D, etc.> But here's the quadratic method which I constructed by merging the proofs > of the linear and quadratic cases:> (Good stuff:)> Let p be prime.> Let > F = Gk Hk mod pk> where F is a Z polynomial and Gk and Hk are Z/pk polynomials, where I'm > using pk to symbolize p^k. Let p2k similarly symbolize p^2k.> Assume> Ak Gk + Bk Hk = 1 mod pk> Then find G2k, H2k, A2k, B2k such that> G2k = Gk mod pk> H2k = Hk mod pk> F = G2k H2k mod p2k> A2k G2k + B2k H2k = 1 mod p2k> Algorithm:> Let:> F - Gk Fk = pk CThat should be F - Gk*Hk = ...> Ak Gk + Bk Hk = 1 + pk D> Then set:> U = (Bk C) mod Gk mod pk (about mod Gk and mod Hk see below)> V = (Ak C) mod Hk mod pk> G2k = Gk + pk U> H2k = Hk + pk V> C' = D + Ak U + Bk V> A2k = Ak - pk C' Ak> B2k = Bk - pk C' Bk> Now I am assuming lc(Gk) and lc(Hk) are not divisible by p, so they are > invertible and the mod operations cause no problem. If we started with G1 > and H1 as polynomials over F/p that is certainly true. Thus this > operation preserves the degrees and lcs of the Gks and Hks.> If we begin with deg(Ak) < deg(Hk) and deg(Bk) < deg(Gk), however, this > condition is not preserved by the construction as is. In fact, the > degrees of the Aks and Bks seem to double at each step.> MY QUESTIONS (finally):> Question (1): I'd like the degree condition preserved. So I need some > simple method for converting A2k and B2k to A' and B' such that> A' G2k + B' H2k = 1 mod p2k> deg(A') < deg(H2k)> deg(B') < deg(G2k)> How is this done in practice? Or in fact does no one bother, because > iterations of quadratic lifting are small in number and we just tolerate > the growing degrees?> Now if F is a 50th degree polynomial with 50 digit coeffs (probably > bigger than anything I might encounter) and p = 2 then the height of a > factor of F might be of order 2^50 * 2^150 = 2^200. That's 200 simple > linear lift steps, but only 8 or so quadratic lift steps.> Knuth discusses this in one of his problem solutions, warning that pure > quadratic lift could introduce huge integer multiplications (to which I > might add: huge degree polynomials), so maybe you'd mix linear and > quadratic.> Question 2: How are linear and quadratic combined in practice?For question (1) I would recommend you first have a look at adifferent reference. One that has the algorithm spelled out isJoachim von zur Gathen, Juergen Gerhard. Modern Computer Algebra.University Press, Cambridge. 1999.In particular have a look at pp. 419-420 and perhaps one or twosurrounding pages on either side. The basic version, on p. 418,looks at least similar to what you have. But the actual algorithmmaintains degree bounds. In particular, there is no issue of growingdegrees, and that nullifies your caveat to the remark in Knuth (which,you will agree, is a Good Thing).As for question (2)... in the past quadratic lifting was not generallyused at high order (say, once coefficients exceeded the size ofmachine integers). The simplicity of the linear lift, which involvesno modification to the {A,B} pairs (in the notation above), made itquite competitive.With nowadays heavy use of asymptotically fast integer multiplicationand univariate polynomial algebra, this is no longer the case:quadratic lifting performed in a reasonable way is, to my knowledge,not beatable. An example of a reasonable way would be the tree liftapproach due to Victor Shoup, presented on pp. 424-425 of theabove-cited text. A very different approach is based on Groebner basesover the integers. It is also a quadratic lift and tends to competewell in terms of speed. It can suffer from a bad prime problem butin practice that is not an issue.When one works over a different ring it may well be the case thatlinear lifting is preferable. Offhand I am not sure what cases mightbe readily optimized for quadratic lifting. I think lifts formultivariate polynomial algorithms are typically done in linearfashion.Daniel LichtblauWolfram ResearchSubject: Solution of Partial Differntial Equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2AKA5203774; I am looking for a way to solve the following Partial differentialequation (analytical, numerical-- anything) EI[NonBreakingSpace]4Y/[NonBreakingSpace]x4 + M [NonBreakingSpace]2/[NonBreakingSpace]t2=0> Note that '[NonBreakingSpace]' represents partial derivative, x and t are theindependent variables, Y is the dependent variable and 'E' and 'I'and'M' are constants. the conditions are : @t=0,Y =0 @x=0, Y=0 (where b is a constant) Please send your replies to gaurangkachhadiya@yahoo.co.in Thank you very much GAURANG.Subject: number pattern by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2A2n7n16806;can't find this pattern 1,8,27,_,_,_Subject: === Subject: number pattern>can't find this pattern> 1,8,27,_,_,_Cubes? ---- Paul J. GansSubject: === Subject: number pattern by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2ABU4X02937;>can't find this pattern> 1,8,27,_,_,_Hi Mary,maybe 1=1*1*1, 8=2*2*2, 27=3*3*3, 64=4*4*4, 125=5*5*5, 216=6*6*6 ?Best wishesTorsten.Subject: === Subject: number pattern> can't find this pattern> 1,8,27,_,_,_How's this: 1^3 2^3 3^3 4^3 5^3 ...OUPSubject: ------ best fit line equation ----hi guys can anyone tell me how can i find the best fit linei need to find the axis of the yarn body and for that i need to takesay 10 points and then i have to find the best fit linehow can i do thatcan u provide some functions / codes in MATLABone more thinghow can i use maths fundamentals to solve my problems see the link belowhttp://www.geocities.com/dhaval_patel_1/yarnprobs.htmlif anyone knows plz tell mewarm regardsDhavalSubject: === Subject: ------ best fit line equation ---->...i need to take say 10 points and then i have to find> the best fit line. how can i do thatTry my online curve fitting site http://zunzun.comand use the 2D Function Finder menu selection. James Phillips http://zunzun.comSubject: === Subject: ------ best fit line equation ---- >hi guys >can anyone tell me how can i find the best fit line >i need to find the axis of the yarn body and for that i need to take >say 10 points and then i have to find the best fit line >how can i do that >can u provide some functions / codes in MATLAB >one more thing >how can i use maths fundamentals to solve my problems >see the link below >http://www.geocities.com/dhaval_patel_1/yarnprobs.html >if anyone knows plz tell me >warm regards >Dhaval10 points for this body ? what to do with the outliers (the threads?)what you have is a body plus those threads. hence you could: take the two outer profiles, represent them by a spline and then, if you definitely want to represent this by a line, take the line which minimizes the maximum deviation from both splines. a least squares line will be heavily influenced by the outer threads. or -- neglect those, represent the profile of thebody by a large set of points (say 100 from the upper and the lower profile each)then use the laest squares line. in matlab:[a;b] = [ ones(N,1),X]Y;with X=column vector of x-coordinates Y=column vector of y (profile) coordinates N=number of pointsthe ordering of the points is arbitrary then. for the best fit in the sense of the maximum norm you could use the fseminf from the optimization toolbox (matlab)hthpeterSubject: netlib - Feb '04Attn: Scientific/engineering computing professionals Attendance (Feb): 4.738 million http://netlib.org -- sw repositories http://gams.nist.gov -- guide (index)For a quick intro on how to access/use netlib software resources, clickon toolboxes at the site below.--T.Silva----------------------------------------------- -------Applied Algorithms http://sdynamix.comSubject: A valid bubble sort algorithm?Using Visual Basic the BubbleLong sub below is my attempt at encoding thebubble sort algorithm given by program 6.4 in Algorithms in C by RobertSedegewick. The BubbleModLong sub below is from an unknown source and ismuch faster than the algorithm in program 6.4.However, the code in BubbleModLong looks suspect.Is it even a bubble sort?If not, anybody recognize the critter?Public Sub BubbleLong(a() As Long, L As Long, r As Long) ' Sedgewick 6.4 Dim i As Long Dim j As Long Dim temp As Long For i = L To r - 1 For j = r To i + 1 Step -1 If a(j - 1) > a(j) Then temp = a(j) a(j) = a(j - 1) a(j - 1) = temp End If Next j Next iEnd SubPublic Sub BubbleModLong(a() As Long, L As Long, r As Long) ' Sedgewick 6.4, modified Dim i As Long Dim j As Long Dim temp As Long For i = L To r - 1 For j = r To i + 1 Step -1 If a(i) > a(j) Then temp = a(j) a(j) = a(i) a(i) = temp End If Next j Next iEnd Sub-- http://www.standards.com/; See Howard Kaikow's web site.Subject: === Subject: A valid bubble sort algorithm?Howard, asking in comp.programming or comp.theory will be more fruitful ;)Martin-- Quidquid latine dictum sit, altum viditur.Subject: === Subject: A valid bubble sort algorithm?Thanx, I was not aware of those newsgroups.-- http://www.standards.com/; See Howard Kaikow's web site.> Howard, asking in comp.programming or comp.theory will be more> fruitful ;)> Martin> -- > Quidquid latine dictum sit, altum viditur.Subject: === Subject: Runge-Kutta Fourth Order Method in Maple codeIt's not a lot of help, but I did find something explaining the Runge-Kutta method using Maple (doesn't use the built-in function)http://www.mapleapps.com/powertools/des/html/unit16. htmlI have seen a wroksheet using the function you speak of on the mapleapps site previously, but I couldn't find it this time. IIRC it was an example of solving differential equations (possibly nonlinear circuit analysis) using the built-in runge-kutte method.The mapleapps site has been useful on several occassions for me. The Maple help is also quite good (in Maple 9, at least).Sorry I can't be more help.Russell.> I have to solve van der pols equation by the runge-kutta fourth>order method, by programming the method in maple, However I have been>told that maple has the fourth order method built in and to program it>in maple is trivial. I am awful at maple and really need to know if>any one has the code they could e-mail me PLEASE!! Also, it would be>very helpful if i could get the results in a table and be able to plot>them. PLEASE -I REALLY NEED THE CODE.> THANKS IN ADVANCE > Subject: Runge-Kutta and circuit simulationI'm trying to implement 4th-order Runge-Kutta in Matlab for a circuit simulator. The ODE looks something like this:Gx + Cx' = Wwhere W is the source vector and G and C are the modified nodal matrices of the time invariant and reactive components respectively.If am interpreting my reference material correctly then the function f in the Runge-Kutta method isf := x'=C^(-1) (W - Gx)The problem being that in this expression C is almost certainly singular and as a result it is not possible to solve for x'.Since it is quite common for circuit solvers to use Runge-Kutta methods, would someone mind explaining what my mistake is?Russell.Subject: === Subject: Runge-Kutta and circuit simulation> I'm trying to implement 4th-order Runge-Kutta in Matlab for a circuit> simulator. The ODE looks something like this:> Gx + Cx' = W> where W is the source vector and G and C are the modified nodal matrices> of the time invariant and reactive components respectively.> If am interpreting my reference material correctly then the function f> in the Runge-Kutta method is> f := x'=C^(-1) (W - Gx)> The problem being that in this expression C is almost certainly singular> and as a result it is not possible to solve for x'.> Since it is quite common for circuit solvers to use Runge-Kutta methods,> would someone mind explaining what my mistake is?If WC and GC are non singular then isn't x(t) = -(WC)^-1 +exp[-(GC)^-1t]x(0)?-- HTH,Gerry T.Subject: === Subject: Runge-Kutta and circuit simulationDistribution: inet> I'm trying to implement 4th-order Runge-Kutta in Matlab for a circuit> simulator. The ODE looks something like this:> Gx + Cx' = W> where W is the source vector and G and C are the modified nodal> matrices of the time invariant and reactive components respectively.> If am interpreting my reference material correctly then the function f> in the Runge-Kutta method is> f := x'=C^(-1) (W - Gx)> The problem being that in this expression C is almost certainly> singular and as a result it is not possible to solve for x'.> Since it is quite common for circuit solvers to use Runge-Kutta> methods, would someone mind explaining what my mistake is?> Russell.Probably you have some components in x which are dependant on others,try to move them out of you vector x may help. Actaully I believe incircuit analysis people will more likely solve DAE (DifferentialAlgebraic Equations) instead of ODE.-- Yongtao Yangemail: yangyongtao@yahoo.comSubject: conditions of nonsingular matrix inverses by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2AI4SV21049;If A is nonsingular, under what conditions is ||A^-1|| = ||A||^-1?thanksSubject: === Subject: conditions of nonsingular matrix inverses >If A is nonsingular, under what conditions is ||A^-1|| = ||A||^-1? >thanks looks much like a homework.hint: multiply by ||A||. what do you get? have you seen this before?hthpeterSubject: Cubic Spline with mixed boundary condition by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2AIuU127731;In the construction of cubic spline(for n-knots) on a segment [a,b],there are 4n unknowns with 4n-2 equations. So, two more conditionsto be put in 0order to find all the unknowns.Let S be the cubic spline approximation of a function f which is fourtimes continuously differenetiable.1. End conditions of 1st type: S'(a) = f'(a), S'(b) = f'(b)2. End conditions of 2nd type: S(a) = f(a), S(b) = f(b)There are two other conditions: Periodic Spline(3rd type)and not-a-Knot condition (4th type).It is also possible to construct spline with boundary mixedboundary condition from 1st type and 2nd type as S'(a) = f'(a)and S(a) = f(a). Is there any special name for such typeof boundary condition ? What is the Optimal error bounds forthe spline arising from mixed boundary condition ?Is the error bounds results of Hall and Meyer( Theorem-5, Journal ofApproximation theory,16,105-122,1976) or (Theorem 3.4,page-169, AnIntroductionto Numerical Analysis, K.E. Atkinson, 2nd edition, JohnWiley) are valid in this case ? Expecting favourable response fromspline experts.With best regards,Bedabrata ChandSubject: === Subject: Finite Volume Implementation!> Hi there!> implementation of the finite volume method is fully explained? Or> where can I find an example program with the appropriate> documentation?> GeorgeThere are some finite volume and finite difference programs in Fortran77 at http://www.sali.freeservers.com/engineering/cfd/cfd_codes.html . The codes are reasonably well documented, but you will probably needa separate guide to the theory.Subject: === Subject: Finite Volume Implementation! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2AEHYf23928;>Hi there!>implementation of the finite volume method is fully explained? Or>where can I find an example program with the appropriate>documentation?>GeorgeHi George, tryhttp://www.cfd-online.com/Books/show_book.php?book_id= 37Best wishesTorsten.Subject: === Subject: Finite Volume Implementation!>Hi there!>implementation of the finite volume method is fully explained? Or>where can I find an example program with the appropriate>documentation?>Georgetry thishttp://pauillac.inria.fr/cdrom/prog/unix/nsc2ke/eng.htm-- F.J.Subject: === Subject: Round off example: MATLAB ok, GCC not ok> BTW: I would like to check the assembly from MSVC as well,> but I cannot find a compiler switch to output the assembly > code.You can view the asm by running the program with Go (F5) and opening the disassembly view. In release mode, switch on mapfile generation in the linker options and use that to locate the code of interest in the disasm.Martin-- Quidquid latine dictum sit, altum viditur.Subject: === Subject: Round off example: MATLAB ok, GCC not okHi!Still no hope! I've tried to compile the C codewith -ffloat-store as well -march=pentium4 -mfpmath=sse.The first case still produces a stable output while thesecond one produces a core dump. I'm usinggcc (GCC) 3.2 20020927under Cygwin on a Pentium 4 computer running Windows XP.Yes, GCC seems to be VERY agressive on code optimization!Humberto.>> The Gnu compiler is pretty agressive about optimizing -- I could believe>> that it would see that x and y are equal, and therefore that z = y already,>> and basically optimize your whole loop to nothing.> I'm note sure what is happening. gcc is behaving strangely. > Let us try the C code:> #include int main()> {> double x = 2.1, y = 2.1, z;> int i; > for (i=0; i<40; i++) > {> z = 4.0*x - 3.0*y;> x = y;> y = z;> } > printf(x = %f,ty = %f,n,x,y);> }> I've been staring at this for while, and I think I know what's> happening:> `-ffloat-store'> Do not store floating point variables in registers, and inhibit> other options that might change whether a floating point value is> taken from a register or memory.> This option prevents undesirable excess precision on machines such> as the 68000 where the floating registers (of the 68881) keep more> precision than a `double' is supposed to have. Similarly for the> x86 architecture. For most programs, the excess precision does> only good, but a few programs rely on the precise definition of> IEEE floating point. Use `-ffloat-store' for such programs, after> modifying them to store all pertinent intermediate computations> into variables.> The Intel architecure has 80-bit floating point registers, instead of> the 64-bit specified for IEEE doubles. As the GCC text says, that's> usually good. I was able to reproduce the unstable results by> computing intermediate results for 4.0*x and 3.0*y, then subtracting.> The optimizer will kill those intermediate computations unless you> specify -ffloat-store.> The other choice is to use the SSE unit (on Pentium 4 processors),> which is faster for those computations it supports, and has only 64> bit registers. Using GCC, specify these options to use SSE math, and> original program will be unstable: -march=pentium4 -mfpmath=sse> So I think this all has to do with which compilers/environments > a) Store intermediate results in variables> b) Use SSE when supported> MATLAB falls into a), since the matlab function calls return values in> memory.> Sturla's other examples of Microsoft's GCC build and MSVC probably> have SSE math enabled by default, for speed.Subject: === Subject: Round off example: MATLAB ok, GCC not ok Hi!> Still no hope! I've tried to compile the C code> with -ffloat-store as well -march=pentium4 -mfpmath=sse.> The first case still produces a stable output while the> second one produces a core dump. I'm using> gcc (GCC) 3.2 20020927To use -ffloat-store, you need to store intermediate results invariables. So you can't say z = 4.0*x - 3.0*y, you have to saya = 4.0*x;b = 3.0*y;z = a - b;But there's gotta be a better way...-- Peter Boettcher Hi!> Still no hope! I've tried to compile the C code> with -ffloat-store as well -march=pentium4 -mfpmath=sse.> The first case still produces a stable output while the> second one produces a core dump. I'm using> gcc (GCC) 3.2 20020927>To use -ffloat-store, you need to store intermediate results in>variables. So you can't say z = 4.0*x - 3.0*y, you have to say>a = 4.0*x;>b = 3.0*y;>z = a - b;>But there's gotta be a better way...Here's a better way: A = [4 -3; 1 0] x = [2.1; 2.1] format long x = A*x x = A*x ... iterate using the up arrow key.The eigenvalues of A are 1 and 3. The initial x is an eigenvectorfor the eigenvalue at 1. If there is no perturbation, the iterationx = A*x is stationary. But any error is multiplied by 3 at each iteration. -- Cleve moler@mathworks.comSubject: === Subject: Round off example: MATLAB ok, GCC not ok> But there's gotta be a better way...There is a better way:#include P(trees), G|->{spanning trees of G} one-to-one?Originator: israel@math.ubc.ca (Robert Israel)Alex.Subject: === Subject: Spectral sequences in knot theoryOriginator: baez@math-cl-n03.math.ucr.edu (John Baez)Originator: israel@math.ubc.ca (Robert Israel)> used to solve problems in knot theory. Any references ?>Vassiliev, VA, Complements of Discriminants of Smooth Maps; Topology and >Applications. Translated from the Russian by B. Goldfarb. Translations The idea is to calculate the cohomology of the space of all embeddings of a circle in R^3 - or some related space -using a spectral sequence. This idea gave rise to the theory of Vassiliev invariants, which quickly got taken over by quantumgroups, chord diagrams and the like. There's a decent quick intro with references here:http://mathworld.wolfram.com/VassilievInvariant.htmlbut for the details of how the spectral sequences are used,you'll have to go back to earlier work, like the monograph Aaron mentioned. If you type Vassiliev invariant spectral sequence intoGoogle, you'll get some useful stuff.Subject: contract mathematicianOriginator: israel@math.ubc.ca (Robert Israel)I seek to hire a contract mathematician for about six hours of work. Thework would be perfect for a graduate student or graduate assistant. Ofcourse, faculty members are also invited to inquire. Please seewww.rekord.us for details.Subject: === Subject: Intersection theory in products of curvesOriginator: israel@math.ubc.ca (Robert Israel)> Let X be the product of n irreducible algebraic curves X_i. I am> looking for a reference for intersections of subvarieties (or cycles)> in X.> For any subset I of {1,..,n}, denote by p_I the projection from X onto> the coordinates listed in I, e.g. if I={i,j}, then p_I : X -> X_i x> X_j.> Let Y be an irreducible subvariety of X, and define the degree of Y,> deg(Y), to be the sup of the degrees of the projection maps p_I : Y -> p_I(Y), for all I for which these projections are finite (in these> cases #I = dim(Y)). If Y is not irreducible, we define deg(Y) to be> the sum of the degrees of the irreducible components of Y of maximal> dimension.> Does anybody know how to prove the following version of Bezout's> theorem, for this definition of degree: if Y_1 and Y_2 are two> subvarieties of X, then deg(Y_1 intersect with Y_2) <=> deg(Y_1).deg(Y_2) ?> This seems to be false.> Let's translate this into the language of cohomology. For simplicity> let's consider the special case where each X_i has genus zero. Then> the cohomology of X_i has one generator in degree 0, let's call this> 1, and one generator in degree 2, let's call this a_i. The> cohomology ring of the product of the X_i's is the tensor product of> the cohomology of the X_i's, which we can regard as the polynomial> ring Z[a_1,...,a_n] with the relations a_i^2=0 and a_ia_j=a_ja_i. > Every subvariety has some homology class, whose Poincare dual is an> element of this cohomology ring (in which the polynomial has> nonnegative coefficients). The coefficient of a monomial in this> polynomial is the degree of the projection to the product of all the> X_i's such that a_i is NOT a factor in this monomial. Under Poincare> duality, intersection of subvarieties (assuming they intersect> transversely) corresponds to multiplication of polynomials in the> cohomology ring.> Let's consider subvarieties in which all components have the same> dimension. Your definition of degree is to take the largest> coefficient in the corresponding polynomial in the cohomology ring. > So your conjecture in this case says that if p and q are two> polynomials with nonnegative coefficients, and if pq denotes their> product with a_i^2 set to zero, then the largest coefficient of pq is> less than or equal to the largest coefficient of p times the largest> coefficient of q. But it is easy to find counterexamples to this,> e.g. take n=2 and p=q=a_1+a_2.> In conclusion, degree is a useful concept when your cohomology ring> has only one generator in each dimension; more generally, you really> should take the whole cohomology ring into account.Ok, thanks a lot. What you have shown above in the case of where eachX_i has genus zero is that we still have deg(Y_1cap Y_2) <=2deg(Y_1)deg(Y_2), which follows from the characterization of theclasses [Y_1] in the intersection (or Chow, or cohomology) ring A(X)as polynomials in the a_i with coefficients giving the degrees ofvarious projection maps. A result of the form deg(Y_1cap Y_2) <=const.deg(Y_1)deg(Y_2) would still be good enough for my purposes, aslong as the constant depends only on the curves X_i.What happens in larger genus? Let's assume X_1=...=X_n are smoothprojective curves. Then we must study the intersection ring A(X_1^n).But what is this ring? I know that A(X_1) is generated by 1=[X_1] andthe elements of Pic(X_1), with multiplication rules ab=0 for a,b inPic(X_1). This ring is still managable. Furthermore, we have a ringhomomorphism from the n-fold tensor product of A(X_1) to A(X_1^n). ButI think that in general, this homomorphism is neither injective norsurjective. (It's an isomorphism in genus 0). What worries me is thatone probably needs it to be surjective if we are to describe theelements of A(X_1^n) in terms of behaviour under projections ontosubproducts X_1^m, m Hey> I have written a program to calculate the Determinant and it is giving> the answer as -2576. Similar to the answer you are getting(sign is> diff). It is a recursive program which calculates the determinants by> computing cofactors. Is there a problem in the 'regular' method of> calculating determinants when the Order of matrix increases.> SandeepI have also written a program to find determinants. I started by using theco-factors method, but noticed as the matrix got larger the time to completethe algorithm became excessively large.A better approach to finding determinants is to use 'elementary rowoperations' to out the matrix in 'upper triangular form' and using the traceof the UTF matrix along with a record of the row operations conducted, thenthe value of the determinant can be found.This is a far more efficient approach.Subject: HP49GAnybody know a good help site for this thing? This user manual sucks.LeviSubject: === Subject: HP49G> Anybody know a good help site for this thing? This user manual sucks.> LeviLook in comp.sys.hp48-- Will Twentymanemail: wtwentyman at copper dot netSubject: === Subject: Cubic polynomial> Can you show that there is some x for which f(x) > 0 and another for> which f(x) < 0?I managed to find an x for which f(x) > 0. However, finding an x such thatf(x) < 0 proved more tricky. I showed that the limit of the function as xapproaches minus infinity is minus infinity, and said that as the functionis continuous, then there must exist some x such that f(x) < 0.Is there a better way to find an x for which f(x) < 0?Subject: Algebra QuestionEvery subgroup of Z is expressed as nZ and it is normal subgroup of Z.Is it correct ?I think this is correct.When N is a subgroup of a additive group G,If g+n-g in N for all g in G, n in N ... (*)Then, Does this imply N is a normal subgroup of G ?My book only contains in case multiplicative group G,So I wonder what is NORMAL SUBGROUP TEST FOR (+)ve GROUP .Can I understand (*) as NORMAL SUBGROUP TEST FOR (+)ve GROUP ?Thank you.Subject: === Subject: Algebra QuestionContent-transfer-encoding: 8bit> Every subgroup of Z is expressed as nZ and it is normal subgroup of Z.> Is it correct ?> I think this is correct.> When N is a subgroup of a additive group G,> If g+n-g in N for all g in G, n in N ... (*)> Then, Does this imply N is a normal subgroup of G ?Yes> My book only contains in case multiplicative group G,> So I wonder what is NORMAL SUBGROUP TEST FOR (+)ve GROUP .The one above is OK or g + N = N + g for all g in G> Can I understand (*) as NORMAL SUBGROUP TEST FOR (+)ve GROUP ?Yes.-- Paul SperryColumbia, SC (USA)Subject: === Subject: Algebra Question Adjunct Assistant Professor at the University of Montana.>Every subgroup of Z is expressed as nZ and it is normal subgroup of Z.>Is it correct ?>I think this is correct.Yes, and yes.>When N is a subgroup of a additive group G,>If g+n-g in N for all g in G, n in N ... (*)If by additive you mean commutative, yes.>Then, Does this imply N is a normal subgroup of G ?Yes. The definition of normal subgroup for a group with operation * isA subgroup N is a normal subgroup if and only if for every n in N and every g inG, g*n*g^{-1} is in N.>My book only contains in case multiplicative group G,>So I wonder what is NORMAL SUBGROUP TEST FOR (+)ve GROUP .If by 'additive' you mean commutative', then note that for any g in Gand any n in N, you would haveg*n*g^{-1} = g*g^{-1}*n = e*n = nwhich lies in N. SoCOR: In a commutative group, every subgroup is normal.>Can I understand (*) as NORMAL SUBGROUP TEST FOR (+)ve GROUP ?If by additive you mean commutative, yes, but then it is silly,because every subgroup is normal. All you have to check is that it isa subgroup. If by additive you just mean the operation is written'+' and the inverse of a is written '-a', then yes, all you do istranslate the expression g*n*g^{-1} into additive notation:g*n*g^{-1} means: take g and n, apply the binary operation, and thenapply the binary operation to the result of that with the inverse ofg. If your binary operation is called + and the inverse of g iscalled -g, then this becomes g+n+(-g) which is usually writteng+n-g. -- == It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)=== ===Arturo Magidinmagidin@math.berkeley.eduSubject: Help in induction by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2AFktS03516;How do I solve this problem?...Show that a given set of n+1 positive integers, none exceeding 2n,there is at least one integer in this set that divides another integerin the set. Use mathematical induction to solve this problem.Subject: === Subject: Help in inductionContent-transfer-encoding: 8bit> How do I solve this problem?...> Show that a given set of n+1 positive integers, none exceeding 2n,> there is at least one integer in this set that divides another integer> in the set. Use mathematical induction to solve this problem.For the induction step take your set with n + 2 elements and remove anelement. If the rest are <= 2n you are done. If not one of the rest iseither 2n + 1 or 2n + 2.Remove the element you found above from the whole set and repeat. Younow have 2n + 1 and 2n + 2 in the whole set. Remove them both and add n + 1. Conclude that one of the originals divides n + 1 and hence 2n + 2.-- Paul SperryColumbia, SC (USA)Subject: === Subject: Help in induction> How do I solve this problem?...> Show that a given set of n+1 positive integers, none exceeding 2n,> there is at least one integer in this set that divides another integer> in the set. Use mathematical induction to solve this problem.Start with your base case.Since n=0 yields the statement a given set of 1 positive integers, none exceeding 0, ... You need to use n=1.So, show that a given set of of 1+1 positive integers, none exceeding 2(1), there is at least one integer in this set that divides another integer in the set.Next, move to the induction step:Assume that if you have a given set of n+1 positive integer, none exceeding 2n, there is at least one integer in this set that divides another integer in the set. Use that to prove the situation for n+1+1 and none exceeding 2(n+1).Hint: you will likely use a subset of the n+1+1 as part of your proof.-- Will Twentymanemail: wtwentyman at copper dot netSubject: matrix notationIn my diff eqs course, we are studying matrix exponentials, for exampledY/dt = AY, where Y is an n-dimensional vector, and A is an n*n coefficientmatrix. Our text using bold face for both the matrix and the vector. Whenwriting by hand, what is the accepted notation? I usually put the singleRich HuhnSubject: === Subject: matrix notation> In my diff eqs course, we are studying matrix exponentials, for example> dY/dt = AY, where Y is an n-dimensional vector, and A is an n*n coefficient> matrix. Our text using bold face for both the matrix and the vector. When> writing by hand, what is the accepted notation? I usually put the singleVector: underlineMatrix: double underline-- P.A.C. SmithThe vast majority of Iraqis want to live in a peaceful, free world.And we will find these people and we will bring them to justice.Subject: Recursive Help.. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2AKXrk06504;How do I define recursively the set of bit strings that have morezeros than ones.Subject: === Subject: Recursive Help..> How do I define recursively the set of bit strings that have more> zeros than ones.Let S(0) = { 0, 00, 000, ..., 000...0, ... }. Clearly every element of S(0) has more zeros than ones.Let S(n) = { s0t1u | stu in S(n-1) } U { s1t0u | stu in S(n-1) }In other words, you get more stings by inserting pairs of 0s and 1s into strings that already have more zeros than ones.Then S (the set of all such strings) us S(0) U S(1) U ... U S(i) U ...meeroh-- If this message helped you, consider buying an itemfrom my wish list: