mm-4789 === Subject: Re: e + P/N = +/-V Goo - wit David Harrison, stupid innumerate cracker, bullted: Stop trolling the math groups you ignorant son-of-a-bitch. We know why Shut the up, Goo. === Subject: Re: e + P/N = +/-V Stop trolling the math groups you ignorant son-of-a-bitch. We know why YOU don't want anyone to work up an equation >you dishonest misnomer hugging son of a bitch. And for that same >reason I DO want someone who has a clue about math to work up >an equation. Have you noticed that even the supposed math genius >eliminationist himself won't touch the challenge? It's BECAUSE >as always the reality works against you people. . . . >e + N = -V and e + P = +V Shut the up Even math people should be aware of that much Goober...even though they don't give a they should be aware of the huge differences between decent Animal Welfare and the gross misnomer animal rights. Since the objectives are completely different they should be thought of in completely different ways. AW wants to provide future animals with lives of positive value, while the misnomer wanst to prevent their existence. Completely different objectives, Goob, and we should point the differences out not shut the up about them. === Subject: Re: e + P/N = +/-V stupid pig-ing cracker, cockfighting specialist - woke up and said, How can I be even *more* stupid today than I was yesterday?, and so he lied: > Stop trolling the math groups you ignorant son-of-a-bitch. > We know why YOU don't want anyone to work up an equation > you dishonest misnomer hugging son of a bitch. And for that same > reason I DO want someone who has a clue about math to work up > an equation. Have you noticed that even the supposed math genius > eliminationist himself won't touch the challenge? It's BECAUSE > as always the reality works against you people. > . . . > e + N = -V and e + P = +V > Shut the up, Goo. Even math people should be aware of that much You do not know math, Goo, beyond simple elementary school arithmetic, and even that you don't know well. Shut the up, Goober, you stupid illiterate cracker. === Subject: Re: e + P/N = +/-V >Stop trolling the math groups you ignorant son-of-a-bitch. We know why YOU don't want anyone to work up an equation >you dishonest misnomer hugging son of a bitch. And for that same >reason I DO want someone who has a clue about math to work up >an equation. Have you noticed that even the supposed math genius >eliminationist himself won't touch the challenge? It's BECAUSE >as always the reality works against you people. >. . . >e + N = -V and e + P = +V Shut the up Even math people should be aware of that much Goober...even >though they don't give a they should be aware of the huge >differences between decent Animal Welfare and the gross misnomer >animal rights. You do not know math I know that even math people should be aware of the huge differences between the misnomer and decent AW, Goo... everyone should understand it Goober even if they don't give a , they should still understand. Why don't you want them to, Goo? >Since the objectives are completely different they >should be thought of in completely different ways. AW wants to >provide future animals with lives of positive value, while the misnomer >wanst to prevent their existence. Completely different objectives, Goob, >and we should point the differences out not shut the up about them. === Subject: Re: e + P/N = +/-V stupid pig-ing cracker, cockfighting specialist - woke up and said, How can I be even *more* stupid today than I was yesterday?, and so he lied: > > Stop trolling the math groups you ignorant son-of-a-bitch. > We know why YOU don't want anyone to work up an equation > you dishonest misnomer hugging son of a bitch. And for that same > reason I DO want someone who has a clue about math to work up > an equation. Have you noticed that even the supposed math genius > eliminationist himself won't touch the challenge? It's BECAUSE > as always the reality works against you people. > . . . > e + N = -V and e + P = +V > Shut the up > Even math people should be aware of that much Goober...even > though they don't give a they should be aware of the huge > differences between decent Animal Welfare and the gross misnomer > animal rights. > You do not know math I know that even math people You do not know math. You didn't even write any valid equations. You're an ignorant cracker who knows how to mop up vomit in a hole cracker beer bar. === Subject: Re: This Week's Finds in Mathematical Physics (Week 272) On 30-Nov-2008, baez@math.removethis.ucr.andthis.edu (John Baez) [...] > Another group theorist, Derek Holt, told Westbury about an intriguing > conjecture which he believes but has not had the energy to check. Take a finite simple group and choose two elements at random. What > is the probability that they generate the group? One cool fact is that this probability is never zero. Every finite > simple group can be generated by two elements! So, they're all > quotients of the free group on two generators. Another cool fact is that this probability tends to 1 as the order of > the group tends to infinity: 12) M. Liebeck and A. Shalev, The probability of generating a finite > simple group, Geometrica Dedicata 56 (1995), 103-111. This implies that this probability attains a *mininum value* for some > finite simple group. For A_6 - the group of even permutations of a > 6-element set - the probability is 53/90. Holt's conjecture is that > this is the minimum among all finite simple groups. Here are the probabilities for a few finite simple groups: PSL(2,4) = PSL(2,5) = A_5 19/30 PSL(2,7) = PSL(3,2) 19/28 PSL(2,8) 71/84 PSL(2,9) = A_6 53/90 PSL(2,11) 127/165 Here A_n is the group of even permutations of an n-element set, > while PSL(n,q) is the group of n x n matrices with determinant 1 > having entries in the field with q elements, mod the diagonal > matrices of this form. I wish someone would tell me the probabilities > for PSL(2,3) = A_4 It's 2/3. > and PSL(4,2) = A_9 As someone else noted, of course you mean A_8. But that one's a little too big for me to do quickly by hand. I suspect someone familiar with GAP could do it easily. > just so I'd have an excuse to list *all* the unexpected isomorphisms > between alternating groups (A_n's) and projective special linear > groups (PSL's). If you widen your horizon just slightly to include symmetric as well as alternating groups, you can also include PSL(2,2) ~= S_3, with probability 1/2 -- lower than the 53/90 of PSL(2,9) ~= A_6, but of course not a simple group. [...] -- Jim Heckman === Subject: How to win on Roulette 2 posting-account=utAcGQoAAACMlbVq2A0zImjAWJu5B3cS .NET CLR 2.0.50727; InfoPath.2; IEMB3),gzip(gfe),gzip(gfe) See a roulette table here: http://wizardofodds.com/play/roulette/ I noticed something. On the bottom of the numbers there is first third, second third, third third. In this long row of numbers in the third third there are 12 numbers, 8 red and 4 black numbers. This means that if we put a chip on the first third and second third, we cover most of the black numbers, except 4 black numbers. The same can be arranged with covering most of the red numbers, 36 numbers, 18 red and 18 black, meaning we cover 14 of them, and 12 would be two thirds, a little more black numbers or black numbers than we expect. This is something to pay attention to that by arranging two rows covering more red or more black numbers from underneeth the numbers (in the right end of the table). This means we can focus on red or black numbers. The center third on the bottom side of the table covers two more black numbers in the third upper row, and only two uncovered black numbers would remain them if we cover the black numbers in the first and second rows. The point is numbers are not arranged to carry equal red or black numbers in the rows if you analyze them, thus chances can tip between red and black here. Any ideas, send it to gmbajszar at yahoo dot com === Subject: Re: How to win on Roulette 2 Bet, say, $100 on black. If you win, repeat. If you lose, bet $200 on black If you win, repeat. If you lose, bet $400 on black Keep doubling your bet on the same colour until you win. This way, you will always make a profit - it pays all your debts. This is my favourite method === Subject: Re: How to win on Roulette 2 posting-account=utAcGQoAAACMlbVq2A0zImjAWJu5B3cS .NET CLR 2.0.50727; InfoPath.2; IEMB3),gzip(gfe),gzip(gfe) > See a roulette table here:http://wizardofodds.com/play/roulette/ I noticed something. On the bottom of the numbers there is first > third, second third, third third. In this long row of numbers in the third third there are 12 numbers, 8 > red and 4 black numbers. This means that if we put a chip on the first third and second third, > we cover most of the black > numbers, except 4 black numbers. The same can be arranged with covering most of the red numbers, 36 > numbers, 18 red and 18 black, meaning we cover 14 of them, and 12 > would be two thirds, a little more black numbers or black numbers than > we expect. This is something to pay attention to that by arranging two > rows covering more red or more black numbers from underneeth the > numbers (in the right end of the table). This means we can focus on > red or black numbers. The center third on the bottom side of the table > covers two more black numbers in the third upper row, and only two > uncovered black numbers would remain them if we cover the black > numbers in the first and second rows. The point is numbers are not > arranged to carry equal red or black numbers in the rows if you > analyze them, thus chances can tip between red and black here. Any ideas, send it to gmbajszar at yahoo dot com I figured it out. Place one on red, and one on the center row on the right of the table. === Subject: Re: How to win on Roulette 2 posting-account=utAcGQoAAACMlbVq2A0zImjAWJu5B3cS .NET CLR 2.0.50727; InfoPath.2; IEMB3),gzip(gfe),gzip(gfe) > See a roulette table here:http://wizardofodds.com/play/roulette/ I noticed something. On the bottom of the numbers there is first > third, second third, third third. In this long row of numbers in the third third there are 12 numbers, 8 > red and 4 black numbers. This means that if we put a chip on the first third and second third, > we cover most of the black > numbers, except 4 black numbers. The same can be arranged with covering most of the red numbers, 36 > numbers, 18 red and 18 black, meaning we cover 14 of them, and 12 > would be two thirds, a little more black numbers or black numbers than > we expect. This is something to pay attention to that by arranging two > rows covering more red or more black numbers from underneeth the > numbers (in the right end of the table). This means we can focus on > red or black numbers. The center third on the bottom side of the table > covers two more black numbers in the third upper row, and only two > uncovered black numbers would remain them if we cover the black > numbers in the first and second rows. The point is numbers are not > arranged to carry equal red or black numbers in the rows if you > analyze them, thus chances can tip between red and black here. Any ideas, send it to gmbajszar at yahoo dot com I figured it out. Place one on red, and one on the center row on the > right of the table. Sorry, once again I set you up for the opposite, the casino gains more advantage, not you. Pretty tricky invention. If you cheat, the casino wins even more. Let me think about it. === Subject: Re: How to win on Roulette 2 posting-account=utAcGQoAAACMlbVq2A0zImjAWJu5B3cS .NET CLR 2.0.50727; InfoPath.2; IEMB3),gzip(gfe),gzip(gfe) > See a roulette table here:http://wizardofodds.com/play/roulette/ I noticed something. On the bottom of the numbers there is first > third, second third, third third. In this long row of numbers in the third third there are 12 numbers, 8 > red and 4 black numbers. This means that if we put a chip on the first third and second third, > we cover most of the black > numbers, except 4 black numbers. The same can be arranged with covering most of the red numbers, 36 > numbers, 18 red and 18 black, meaning we cover 14 of them, and 12 > would be two thirds, a little more black numbers or black numbers than > we expect. This is something to pay attention to that by arranging two > rows covering more red or more black numbers from underneeth the > numbers (in the right end of the table). This means we can focus on > red or black numbers. The center third on the bottom side of the table > covers two more black numbers in the third upper row, and only two > uncovered black numbers would remain them if we cover the black > numbers in the first and second rows. The point is numbers are not > arranged to carry equal red or black numbers in the rows if you > analyze them, thus chances can tip between red and black here. Any ideas, send it to gmbajszar at yahoo dot com I figured it out. Place one on red, and one on the center row on the > right of the table. Sorry, once again I set you up for the opposite, the casino gains more > advantage, not you. Pretty tricky invention. If you cheat, the casino > wins even more. Let me think about it. Maybe I wasn't wrong. Calculate it: Reds win 16 while blacks loose 12. === Subject: Re: How to win on Roulette 2 posting-account=utAcGQoAAACMlbVq2A0zImjAWJu5B3cS .NET CLR 2.0.50727; InfoPath.2; IEMB3),gzip(gfe),gzip(gfe) See a roulette table here:http://wizardofodds.com/play/roulette/ I noticed something. On the bottom of the numbers there is first > third, second third, third third. In this long row of numbers in the third third there are 12 numbers, 8 > red and 4 black numbers. This means that if we put a chip on the first third and second third, > we cover most of the black > numbers, except 4 black numbers. The same can be arranged with covering most of the red numbers, 36 > numbers, 18 red and 18 black, meaning we cover 14 of them, and 12 > would be two thirds, a little more black numbers or black numbers than > we expect. This is something to pay attention to that by arranging two > rows covering more red or more black numbers from underneeth the > numbers (in the right end of the table). This means we can focus on > red or black numbers. The center third on the bottom side of the table > covers two more black numbers in the third upper row, and only two > uncovered black numbers would remain them if we cover the black > numbers in the first and second rows. The point is numbers are not > arranged to carry equal red or black numbers in the rows if you > analyze them, thus chances can tip between red and black here. Any ideas, send it to gmbajszar at yahoo dot com I figured it out. Place one on red, and one on the center row on the > right of the table. Sorry, once again I set you up for the opposite, the casino gains more > advantage, not you. Pretty tricky invention. If you cheat, the casino > wins even more. Let me think about it. Maybe I wasn't wrong. Calculate it: Reds win 16 while blacks loose 12. No. They balance out 12 wins and 12 losses. That was a waste of time. === Subject: Re: How to win on Roulette 2 See a roulette table here:http://wizardofodds.com/play/roulette/ I noticed something. On the bottom of the numbers there is first > third, second third, third third. In this long row of numbers in the third third there are 12 numbers, 8 > red and 4 black numbers. This means that if we put a chip on the first third and second third, > we cover most of the black > numbers, except 4 black numbers. The same can be arranged with covering most of the red numbers, 36 > numbers, 18 red and 18 black, meaning we cover 14 of them, and 12 > would be two thirds, a little more black numbers or black numbers than > we expect. This is something to pay attention to that by arranging two > rows covering more red or more black numbers from underneeth the > numbers (in the right end of the table). This means we can focus on > red or black numbers. The center third on the bottom side of the table > covers two more black numbers in the third upper row, and only two > uncovered black numbers would remain them if we cover the black > numbers in the first and second rows. The point is numbers are not > arranged to carry equal red or black numbers in the rows if you > analyze them, thus chances can tip between red and black here. Any ideas, send it to gmbajszar at yahoo dot com I figured it out. Place one on red, and one on the center row on the > right of the table. Sorry, once again I set you up for the opposite, the casino gains more > advantage, not you. Pretty tricky invention. If you cheat, the casino > wins even more. Let me think about it. Maybe I wasn't wrong. Calculate it: Reds win 16 while blacks loose 12. No. They balance out 12 wins and 12 losses. That was a waste of time. The casino breaks even on all bets when regular numbers hit, but collects on 0 and double-0. That's how they slowly take your money. Ever see those gigantic huge plush opulent casinos? They build those out of a few percent edge in mathematical expectation. And dumb tourists plied full of free drinks. Those are the most expensive free drinks you'll ever have. === Subject: Re: How to win on Roulette 2 posting-account=utAcGQoAAACMlbVq2A0zImjAWJu5B3cS .NET CLR 2.0.50727; InfoPath.2; IEMB3),gzip(gfe),gzip(gfe) > See a roulette table here:http://wizardofodds.com/play/roulette/ I noticed something. On the bottom of the numbers there is first > third, second third, third third. In this long row of numbers in the third third there are 12 numbers, 8 > red and 4 black numbers. This means that if we put a chip on the first third and second third, > we cover most of the black > numbers, except 4 black numbers. The same can be arranged with covering most of the red numbers, 36 > numbers, 18 red and 18 black, meaning we cover 14 of them, and 12 > would be two thirds, a little more black numbers or black numbers than > we expect. This is something to pay attention to that by arranging two > rows covering more red or more black numbers from underneeth the > numbers (in the right end of the table). This means we can focus on > red or black numbers. The center third on the bottom side of the table > covers two more black numbers in the third upper row, and only two > uncovered black numbers would remain them if we cover the black > numbers in the first and second rows. The point is numbers are not > arranged to carry equal red or black numbers in the rows if you > analyze them, thus chances can tip between red and black here. Any ideas, send it to gmbajszar at yahoo dot com I figured it out. Place one on red, and one on the center row on the > right of the table. Sorry, once again I set you up for the opposite, the casino gains more > advantage, not you. Pretty tricky invention. If you cheat, the casino > wins even more. Let me think about it. Maybe I wasn't wrong. Calculate it: Reds win 16 while blacks loose 12. No. They balance out 12 wins and 12 losses. That was a waste of time. The casino breaks even on all bets when regular numbers hit, but > collects on 0 and double-0. That's how they slowly take your money. Ever > see those gigantic huge plush opulent casinos? They build those out of a > few percent edge in mathematical expectation. And dumb tourists plied > full of free drinks. Those are the most expensive free drinks you'll > ever have. It is not true that that is how they take all the money through the 0 and 00. If one spreads all or most of the money, there is an unbalanced down trend. One looses money faster. If care to spend time with the math on what this means: Place all the money on three spots: 1-18, even, red. Eight possibilities can happen on the three spots: (0 looses, 1 wins) 000 001 010 011 100 101 110 111 000, all money lost 111, money doubled 001 99 dollars became 66 (bet spread in three spots) 010 99 dollars became 66 011 99 dollars became 132 100 99 dollars became 66 101 99 dollars became 132 110 99 dollars became 132 We find in the middle of all doubled or all lost 6 possibilities above where one looses 33 or wins 33 So we ignore for a second the 000 and 111. If we loose 33, and spread 66 remaining chips on three bets, we place 22,22,22 chips on 1-18,even,red. If it wins 22, 66+22=88. If on average there is one win and one loss in the long term, 99 dollars in two games on average moved from 100 to 88 dollars now. If 99 wins 33, and one has 132, then one plays 132 as 44,44,44 and looses two 44 but wins one, he again after a two game average has 88. The average if one plays most his money keeps dropping 25 percent more for the table's advantage every two spins. That's what happens if one keeps spreading most his money, the table wins more, but there are occasional 'all three wins' or 'all three looses' which seem to be the winning moments, but everything in between has a trend downward in the benefit of the table and the casino. === Subject: Re: How to win on Roulette 2 posting-account=utAcGQoAAACMlbVq2A0zImjAWJu5B3cS .NET CLR 2.0.50727; InfoPath.2; IEMB3),gzip(gfe),gzip(gfe) See a roulette table here:http://wizardofodds.com/play/roulette/ I noticed something. On the bottom of the numbers there is first > third, second third, third third. In this long row of numbers in the third third there are 12 numbers, 8 > red and 4 black numbers. This means that if we put a chip on the first third and second third, > we cover most of the black > numbers, except 4 black numbers. The same can be arranged with covering most of the red numbers, 36 > numbers, 18 red and 18 black, meaning we cover 14 of them, and 12 > would be two thirds, a little more black numbers or black numbers than > we expect. This is something to pay attention to that by arranging two > rows covering more red or more black numbers from underneeth the > numbers (in the right end of the table). This means we can focus on > red or black numbers. The center third on the bottom side of the table > covers two more black numbers in the third upper row, and only two > uncovered black numbers would remain them if we cover the black > numbers in the first and second rows. The point is numbers are not > arranged to carry equal red or black numbers in the rows if you > analyze them, thus chances can tip between red and black here. Any ideas, send it to gmbajszar at yahoo dot com I figured it out. Place one on red, and one on the center row on the > right of the table. Sorry, once again I set you up for the opposite, the casino gains more > advantage, not you. Pretty tricky invention. If you cheat, the casino > wins even more. Let me think about it. Maybe I wasn't wrong. Calculate it: Reds win 16 while blacks loose 12. No. They balance out 12 wins and 12 losses. That was a waste of time. The casino breaks even on all bets when regular numbers hit, but > collects on 0 and double-0. That's how they slowly take your money. Ever > see those gigantic huge plush opulent casinos? They build those out of a > few percent edge in mathematical expectation. And dumb tourists plied > full of free drinks. Those are the most expensive free drinks you'll > ever have. It is not true that that is how they take all the money through the 0 > and 00. If one spreads all or most of the money, there is an unbalanced down > trend. > One looses money faster. If care to spend time with the math on what this means: Place all the money on three spots: 1-18, even, red. Eight possibilities can happen on the three spots: (0 looses, 1 wins) 000 > 001 > 010 > 011 > 100 > 101 > 110 > 111 000, all money lost > 111, money doubled 001 99 dollars became 66 (bet spread in three spots) > 010 99 dollars became 66 > 011 99 dollars became 132 > 100 99 dollars became 66 > 101 99 dollars became 132 > 110 99 dollars became 132 We find in the middle of all doubled or all lost 6 possibilities > above where one looses 33 or wins 33 So we ignore for a second the 000 and 111. If we loose 33, and spread 66 remaining chips on three bets, > we place 22,22,22 chips on 1-18,even,red. If it wins 22, 66+22=88. If on average there is one win and one loss in > the long term, 99 dollars in two games on average > moved from 100 to 88 dollars now. If 99 wins 33, and one has 132, then one plays > 132 as 44,44,44 and looses two 44 but wins one, > he again after a two game average has 88. The average if one plays most his money keeps > dropping 25 percent more for the table's advantage > every two spins. That's what happens if one keeps > spreading most his money, the table wins more, > but there are occasional 'all three wins' or > 'all three looses' which seem to be the winning > moments, but everything in between has a trend > downward in the benefit of the table and the > casino. Small letters: I understand that the concept risk itself is attributed to a possibility of win (111) which attributes to the necessary downtrend in the middle of hitting the jackpot or loosing it all, I guess the greater the risk the more down trend one finds. Math is not fun this way, to understand and stuff. I will not come to a math forum with this because aspects are quickly tied to political and economic aspects, and not looking at the specific question which is I would like someone to tell me how I can win in Vegas. I didn't find such an email sitting in my email account and making my day. === Subject: Re: How to win on Roulette 2 STFU dork. EVERYONE ELSE KNOWS you cannot win at Roulette in the long run. Go play Kenno. === Subject: Re: How to win on Roulette 2 posting-account=utAcGQoAAACMlbVq2A0zImjAWJu5B3cS .NET CLR 2.0.50727; InfoPath.2; IEMB3),gzip(gfe),gzip(gfe) > STFU dork. > EVERYONE ELSE KNOWS you cannot win at Roulette in the long run. > Go play Kenno. Win really big. Quit gambling and start doing something constructive. Now that I got your attention homies who don't like messin with and are the same online as on the farm in a country of killers, here is what you do, and all you need is a cowboy head, Bush knows than it works, but I tell you only once: Play 1 token in the first and third third under the numbers on the right, and 2 on red. Now there is a third row accross in the middle that has black numbers, and there is the zero. Those are the only loosing numbers. If those hit, raise the bet from two ones on the right from ones to twos and the bet on red to four. If a red hits in the first and third row, return the bet to ones in the right in the first and third rows and a two on the red. Only if a black hits in the middle row again, you raise the bet then from two twos on the right and a 4 on the red to two threes in the right and a 6 on the red. Keep incresing, increasing it this way if hit the wrong black in the middle row or the zeros, and decreasing it back as -1 then on the right side and -2 on the red. Increase is +1 chip on the right positions if a black hits in the middle row or a zero and +2 for the red, decrease is -1 chip on the right side two thirds and -2 for the red. Else if other numbers hit that are not red but in the two rows where there are one chips on the right under those rows, or a red in the center row, keep playing the bet. === Subject: Re: How to win on Roulette 2 posting-account=utAcGQoAAACMlbVq2A0zImjAWJu5B3cS .NET CLR 2.0.50727; InfoPath.2; IEMB3),gzip(gfe),gzip(gfe) > STFU dork. > EVERYONE ELSE KNOWS you cannot win at Roulette in the long run. > Go play Kenno. Win really big. Quit gambling and start doing something constructive. Now that I got your attention homies who don't like messin with and > are the same online as on the farm in a country of killers, here is what you do, and all you need is a cowboy head, Bush knows > than > it works, but I tell you only once: Play 1 token in the first and third third under the numbers > on the right, and 2 on red. Now there is a third row accross in the middle that has black numbers, > and there is the zero. Those are the only loosing numbers. If those > hit, raise the bet from two ones on the right from ones to twos > and the bet on red to four. If a red hits in the first and third row, return the bet to > ones in the right in the first and third rows and a two on the red. Only if a black hits in the middle row again, you raise the bet > then from two twos on the right and a 4 on the red to two threes > in the right and a 6 on the red. Keep incresing, increasing it > this way if hit the wrong black in the middle row or the zeros, > and decreasing it back as -1 then on the right side and -2 on > the red. Increase is +1 chip on the right positions if a black > hits in the middle row or a zero and +2 for the red, decrease > is -1 chip on the right side two thirds and -2 for the red. Else if other numbers hit that are not red but in the two rows > where there are one chips on the right under those rows, > or a red in the center row, keep playing the bet. I appreciate 10 percent Catholic Church appreciation as I suffer from a genius problem, but if you got convinced otherwise, I appreciate 10 percent Catholic Church appreciation for the money you didn't spend in Vegas. gmbajszar @ yahoo dot com === Subject: Re: How to win on Roulette 2 posting-account=utAcGQoAAACMlbVq2A0zImjAWJu5B3cS .NET CLR 2.0.50727; InfoPath.2; IEMB3),gzip(gfe),gzip(gfe) STFU dork. > EVERYONE ELSE KNOWS you cannot win at Roulette in the long run. > Go play Kenno. Win really big. Quit gambling and start doing something constructive. Now that I got your attention homies who don't like messin with and > are the same online as on the farm in a country of killers, here is what you do, and all you need is a cowboy head, Bush knows > than > it works, but I tell you only once: Play 1 token in the first and third third under the numbers > on the right, and 2 on red. Now there is a third row accross in the middle that has black numbers, > and there is the zero. Those are the only loosing numbers. If those > hit, raise the bet from two ones on the right from ones to twos > and the bet on red to four. If a red hits in the first and third row, return the bet to > ones in the right in the first and third rows and a two on the red. Only if a black hits in the middle row again, you raise the bet > then from two twos on the right and a 4 on the red to two threes > in the right and a 6 on the red. Keep incresing, increasing it > this way if hit the wrong black in the middle row or the zeros, > and decreasing it back as -1 then on the right side and -2 on > the red. Increase is +1 chip on the right positions if a black > hits in the middle row or a zero and +2 for the red, decrease > is -1 chip on the right side two thirds and -2 for the red. Else if other numbers hit that are not red but in the two rows > where there are one chips on the right under those rows, > or a red in the center row, keep playing the bet. I appreciate 10 percent Catholic Church appreciation as I suffer > from a genius problem, but if you got convinced otherwise, > I appreciate 10 percent Catholic Church appreciation for the > money you didn't spend in Vegas. gmbajszar @ yahoo dot com So the rule once again is (and numbers are not organized in the same ways in Roulette tables in Vegas): The black numbers in the middle horizontal row and the zero(s) raise the next bet as I described, the red numbers in the outer horizontal rows reduce the bet as I described. Read through the puzzle presented in this thread and learn it. The money you send me I will waste in Vegas too. === Subject: Re: How to win on Roulette 2 posting-account=utAcGQoAAACMlbVq2A0zImjAWJu5B3cS .NET CLR 2.0.50727; InfoPath.2; IEMB3),gzip(gfe),gzip(gfe) > STFU dork. > EVERYONE ELSE KNOWS you cannot win at Roulette in the long run. > Go play Kenno. Win really big. Quit gambling and start doing something constructive. Now that I got your attention homies who don't like messin with and > are the same online as on the farm in a country of killers, here is what you do, and all you need is a cowboy head, Bush knows > than > it works, but I tell you only once: Play 1 token in the first and third third under the numbers > on the right, and 2 on red. Now there is a third row accross in the middle that has black numbers, > and there is the zero. Those are the only loosing numbers. If those > hit, raise the bet from two ones on the right from ones to twos > and the bet on red to four. If a red hits in the first and third row, return the bet to > ones in the right in the first and third rows and a two on the red. Only if a black hits in the middle row again, you raise the bet > then from two twos on the right and a 4 on the red to two threes > in the right and a 6 on the red. Keep incresing, increasing it > this way if hit the wrong black in the middle row or the zeros, > and decreasing it back as -1 then on the right side and -2 on > the red. Increase is +1 chip on the right positions if a black > hits in the middle row or a zero and +2 for the red, decrease > is -1 chip on the right side two thirds and -2 for the red. Else if other numbers hit that are not red but in the two rows > where there are one chips on the right under those rows, > or a red in the center row, keep playing the bet. I appreciate 10 percent Catholic Church appreciation as I suffer > from a genius problem, but if you got convinced otherwise, > I appreciate 10 percent Catholic Church appreciation for the > money you didn't spend in Vegas. gmbajszar @ yahoo dot com So the rule once again is (and numbers are not organized in the > same ways in Roulette tables in Vegas): The black numbers in the middle horizontal row and the > zero(s) raise the next bet as I described, the red numbers in > the outer horizontal rows reduce the bet as I described. Read > through the puzzle presented in this thread and learn it. The money you send me I will waste in Vegas too. It is not over. One can try learning putting the ones on the right under the first and second rows, and learn what happens. One wins three times more on the red than what one looses on the blacks. No loss or win on the reds in the first row, only four blacks and one or two zeros there. Bla bla. === Subject: Re: How to win on Roulette 2 posting-account=utAcGQoAAACMlbVq2A0zImjAWJu5B3cS .NET CLR 2.0.50727; InfoPath.2; IEMB3),gzip(gfe),gzip(gfe) STFU dork. > EVERYONE ELSE KNOWS you cannot win at Roulette in the long run. > Go play Kenno. Win really big. Quit gambling and start doing something constructive. Now that I got your attention homies who don't like messin with and > are the same online as on the farm in a country of killers, here is what you do, and all you need is a cowboy head, Bush knows > than > it works, but I tell you only once: Play 1 token in the first and third third under the numbers > on the right, and 2 on red. Now there is a third row accross in the middle that has black numbers, > and there is the zero. Those are the only loosing numbers. If those > hit, raise the bet from two ones on the right from ones to twos > and the bet on red to four. If a red hits in the first and third row, return the bet to > ones in the right in the first and third rows and a two on the red. Only if a black hits in the middle row again, you raise the bet > then from two twos on the right and a 4 on the red to two threes > in the right and a 6 on the red. Keep incresing, increasing it > this way if hit the wrong black in the middle row or the zeros, > and decreasing it back as -1 then on the right side and -2 on > the red. Increase is +1 chip on the right positions if a black > hits in the middle row or a zero and +2 for the red, decrease > is -1 chip on the right side two thirds and -2 for the red. Else if other numbers hit that are not red but in the two rows > where there are one chips on the right under those rows, > or a red in the center row, keep playing the bet. I appreciate 10 percent Catholic Church appreciation as I suffer > from a genius problem, but if you got convinced otherwise, > I appreciate 10 percent Catholic Church appreciation for the > money you didn't spend in Vegas. gmbajszar @ yahoo dot com So the rule once again is (and numbers are not organized in the > same ways in Roulette tables in Vegas): The black numbers in the middle horizontal row and the > zero(s) raise the next bet as I described, the red numbers in > the outer horizontal rows reduce the bet as I described. Read > through the puzzle presented in this thread and learn it. The money you send me I will waste in Vegas too. It is not over. One can try learning putting the ones on the right > under the first and second rows, and learn what happens. One wins > three times more on the red than what one looses on the blacks. > No loss or win on the reds in the first row, only four blacks > and one or two zeros there. Bla bla. Beating the odds a littlebit still means you have to be a millionaire, play a whole year day and night, then in the long term you win, but things are that 49 percent of the time you will loose. Nobody can take the stress of playing more than a few days, eyes sag and stuff. So one still needs millions to win. === Subject: Re: Execution and Expropriation of all Jews, effective as of 08.08.08 (www.grishenkoff.com) posting-account=xniw7woAAAAnt6ofCUUjWAXiVSXvo-X9 Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) Execution and Expropriation of all Jews, effective as of 08.08.08 (www.grishenkoff.com) === Subject: RFinancial Services Inter/Nationalization, effective as of 08.08.08 (www.grishenkoff.com/Finance.html) posting-account=PH0myQoAAAA7qQRYzdMqrJhCdRj4gqbi AppleWebKit/525.19 (KHTML, like Gecko) Chrome/0.4.154.29 Safari/525.19,gzip(gfe),gzip(gfe) Financial Services Inter/Nationalization, effective as of 08.08.08 (www.grishenkoff.com/Finance.html) === Subject: Real Estate Inter/Nationalization, effective as of 08.08.08 (www.grishenkoff.com/Realty.html) posting-account=gEwDVwoAAAAUxSDz0pccsve8dNzvQomL AppleWebKit/525.19 (KHTML, like Gecko) Chrome/0.4.154.29 Safari/525.19,gzip(gfe),gzip(gfe) Real Estate Inter/Nationalization, effective as of 08.08.08 (www.grishenkoff.com/Realty.html) === Subject: Re: multisection of poerwseries Am 07.12.2008 12:35 schrieb David Bernier: > Yes, I agree with you. It could be the same Somos who has a page > about the Busy Beaver problem with Turing machines, which is > a fine page. That motivated me to go looking based on clues in > Somos's page on q-series. David > Funny, I have also a Michael Somos in > http://www.research.att.com/~njas/sequences/A002119, > a sequence that appears in a little problem I'm looking > at these days (but completely unrelated with this > thread). I had a look at Somos' homepage. In late October, he put up draft > number 7 of an announcement about an eta-series identity, which > occupies one page and it's easy to understand what the identity > is all about. The one-page announcement can be found in > A Remarkable eta-product Identity > from the Web page: > http://cis.csuohio.edu/~somos/math.html#multiq I too have taken a look at this site. Here are some sentences: http://cis.csuohio.edu/~somos/multiq.pdf A Multisection of q-Series > Michael Somos 12 Aug 2008 > somos@cis.csuohio.edu > (draft version 30) 2 Trisection and Multisection of power series > This suggests that it might be interesting to split up the power series into parts > depending on the residue of the exponent modulo a fxed modulus. For modulus > two this is called bisection, and in the general case it is called multisection. It > is already interesting to try bisecting known series. For example, bisecting the > exponential power series gives the series for hyperbolic sine and cosine. Note > that there is an algebraic relation between the hyperbolic sine and cosine. Also > note that multisections can be expressed using n-th roots of unity. > For the series A = y(q), applying bisection gives nothing obvious, so we > trisect the series into three others and fnd that Gottfried === Subject: Re: multisection of poerwseries > Am 07.12.2008 12:35 schrieb David Bernier: > Yes, I agree with you. It could be the same Somos who has a page > about the Busy Beaver problem with Turing machines, which is > a fine page. That motivated me to go looking based on clues in > Somos's page on q-series. David > Funny, I have also a Michael Somos in > http://www.research.att.com/~njas/sequences/A002119, > a sequence that appears in a little problem I'm looking > at these days (but completely unrelated with this > thread). > I had a look at Somos' homepage. In late October, he put up draft > number 7 of an announcement about an eta-series identity, which > occupies one page and it's easy to understand what the identity > is all about. The one-page announcement can be found in > A Remarkable eta-product Identity > from the Web page: > http://cis.csuohio.edu/~somos/math.html#multiq I too have taken a look at this site. Here are some sentences: http://cis.csuohio.edu/~somos/multiq.pdf A Multisection of q-Series > Michael Somos 12 Aug 2008 > somos@cis.csuohio.edu > (draft version 30) 2 Trisection and Multisection of power series > This suggests that it might be interesting to split up the power series into parts > depending on the residue of the exponent modulo a fxed modulus. For modulus > two this is called bisection, and in the general case it is called multisection. It > is already interesting to try bisecting known series. For example, bisecting the > exponential power series gives the series for hyperbolic sine and cosine. Note > that there is an algebraic relation between the hyperbolic sine and cosine. Also > note that multisections can be expressed using n-th roots of unity. > For the series A = y(q), applying bisection gives nothing obvious, so we > trisect the series into three others and fnd that Gottfried Yes, I think it's interesting and I've explored some more. Euler did some work on the partition function. For example, 3 = 1 + 1 + 1 = 2 +1 = 3 so 3 can be partitioned 3 ways. So we write p(3) = 3. Cf.: http://www.research.att.com/~njas/sequences/A000041 (Sloane's OEIS). A generating function for p(n) is given by the reciprocal of what Wikipedia calls Euler's function: http://en.wikipedia.org/wiki/Partition_function_(number_theory)#Generating_f unction The Euler function can be given as an infinite sum or an infinite product. His pentagonal theorem gives the infinite sum version: http://en.wikipedia.org/wiki/Pentagonal_number_theorem Euler's function is often denoted phi(q): http://en.wikipedia.org/wiki/Euler_function (PDF,DVI,text,LaTeX) --> using the PDF file, it's clear that what Somos denotes as y(q) is Euler's phi(q). Section 5 is about a 7-section of y(q) ( or phi(q) ) into A_i(q) , for 0<=i<7 , where for example A_0(q) regroups the terms in the sum expansion of phi(q) where the terms in q^(7k) only appear. So y(q) = A_0(q) + ... + A_6(q). Using a computer algebra package, Somos can conjecture algebraic relations between some of these A_i . The next section is about the 7-section/sections and generalized Rogers-Ramanujan identities found by Andrews, Schilling and Warnaar. The section is quite brief, so I don't understand what the connection is. But I'm pretty sure the paper by Andrews et al is this one (preprint): http://xxx.lanl.gov/abs/math.QA/9807125 in Quantum Algebra and Combinatorics. Theorem 5.2 is on page 19, and gives three identities, and the connection to modulus 7 is obvious. They write that they didn't find an identity for the fourth character ... On page 10, they mention the W_3 and W_n algebras. These algebras are vertex operator algebras, (some kind of infinite dimensional Lie algebra, I believe). I thought that understanding what W_3 is would be feasible in terms of just a definition. But these objects are complicated, and I don't understand... Cf.: Peter Bouwknegt, Jim McCarthy, Krzysztof Pilch: ``The W_3 algebra: modules, semi-infinite cohomology and BV-algebras, [144 pages ] at http://lanl.arxiv.org/abs/hep-th/9509119v1 David Bernier === Subject: Re: multisection of poerwseries <22234218.1228606572186.JavaMail.jakarta@nitrogen.mathforum.org> <493ba439$0$941$ba4acef3@news.orange.fr> <6q2qajFaj92mU1@mid.dfncis.de> posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.4) Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > Am 07.12.2008 12:35 schrieb David Bernier: > Yes, I agree with you. It could be the same Somos who has a page > about the Busy Beaver problem with Turing machines, which is > a fine page. That motivated me to go looking based on clues in > Somos's page on q-series. David > Funny, I have also a Michael Somos in >http://www.research.att.com/~njas/sequences/A002119, > a sequence that appears in a little problem I'm looking > at these days (but completely unrelated with this > thread). I had a look at Somos' homepage. In late October, he put up draft > number 7 of an announcement about an eta-series identity, which > occupies one page and it's easy to understand what the identity > is all about. The one-page announcement can be found in > A Remarkable eta-product Identity > from the Web page: >http://cis.csuohio.edu/~somos/math.html#multiq I too have taken a look at this site. Here are some sentences: http://cis.csuohio.edu/~somos/multiq.pdf A Multisection of q-Series > Michael Somos 12 Aug 2008 > so...@cis.csuohio.edu > (draft version 30) 2 Trisection and Multisection of power series > This suggests that it might be interesting to split up the power series into parts > depending on the residue of the exponent modulo a xed modulus. For modulus > two this is called bisection, and in the general case it is called multisection. It > is already interesting to try bisecting known series. For example, bisecting the > exponential power series gives the series for hyperbolic sine and cosine. Note > that there is an algebraic relation between the hyperbolic sine and cosine. Also > note that multisections can be expressed using n-th roots of unity. > For the series A = y(q), applying bisection gives nothing obvious, so we > trisect the series into three others and nd that given a powerseries f oo --- j / a x --- j j=0 a simpson (m, n)-multisection of f which i often write |m | f(x) |n is the series --- j / a x --- j j = m (mod n) which can also be written 00 --- nj+m / a x --- nj+m j=0 this operator is a projection operator it is idempotent and all that and there is a classic formula of simpson which relates it to the roots of unity simpson multisection is the generalisation of decomposing functions into even and odd parts the concept of multisection may be extended to other bases of expansion and many of the properties carry over through the common projection properties though formula on the roots of unity changes so there are (m, n)-multisections of dirichlet series and (m, n)-multisections of fourier series and (m, n)-multisections of orthogonal polynomial series and ... and it is all very standard in the fields of combinatorial enumeration and special functions and there are many books that mention it the hyperbolic trigonometrics ch and sh are the (0, 2)- and (1, 2)-multisections of the classical exponential function the circular trigonometrics are ch(i x) and -i sh(i x) the generalised hyperbolic trigonometrics g (x) m n are (m, n)-multisections of the exponential and the circular generalisations are -m t (x) = w g (w x) m n 2n m n 2n where w 2n is exp(pi i / n) this simple extension of trigonometry very simple to describe and work with has an intricate algebraic structure that allows many extensions of classical theories including fourier analysis tchebyshef expansion many famous matrix expansions and reveals many additional generalisations that lead into deeper algebraic theories like generalised polynomials and their galois theory or generalised fractions and their intricate ring structure and recursive properties -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: multisection of poerwseries > Am 07.12.2008 12:35 schrieb David Bernier: > Yes, I agree with you. It could be the same Somos who has a page > about the Busy Beaver problem with Turing machines, which is > a fine page. That motivated me to go looking based on clues in > Somos's page on q-series. > David > Funny, I have also a Michael Somos in > http://www.research.att.com/~njas/sequences/A002119, > a sequence that appears in a little problem I'm looking > at these days (but completely unrelated with this > thread). > I had a look at Somos' homepage. In late October, he put up draft > number 7 of an announcement about an eta-series identity, which > occupies one page and it's easy to understand what the identity > is all about. > The one-page announcement can be found in > A Remarkable eta-product Identity > from the Web page: > http://cis.csuohio.edu/~somos/math.html#multiq > I too have taken a look at this site. Here are some sentences: > http://cis.csuohio.edu/~somos/multiq.pdf > A Multisection of q-Series > Michael Somos 12 Aug 2008 > so...@cis.csuohio.edu > (draft version 30) > 2 Trisection and Multisection of power series > This suggests that it might be interesting to split up the power series into parts > depending on the residue of the exponent modulo a xed modulus. For modulus > two this is called bisection, and in the general case it is called multisection. It > is already interesting to try bisecting known series. For example, bisecting the > exponential power series gives the series for hyperbolic sine and cosine. Note > that there is an algebraic relation between the hyperbolic sine and cosine. Also > note that multisections can be expressed using n-th roots of unity. > For the series A = y(q), applying bisection gives nothing obvious, so we > trisect the series into three others and nd that given a powerseries f oo > --- > j > / a x > --- j > j=0 a simpson (m, n)-multisection of f > which i often write |m > | f(x) > |n is the series --- > j > / a x > --- j > j = m (mod n) which can also be written 00 > --- > nj+m > / a x > --- nj+m > j=0 this operator is a projection operator > it is idempotent and all that > and there is a classic formula of simpson > which relates it to the roots of unity I'm not sure which Simpson or formula you're referring to. It would help if you could elaborate on this, preferably with a reference. David Bernier simpson multisection is the generalisation > of decomposing functions into even and odd parts the concept of multisection > may be extended to other bases of expansion > and many of the properties carry over > through the common projection properties > though formula on the roots of unity changes so > there are (m, n)-multisections of dirichlet series > and (m, n)-multisections of fourier series > and (m, n)-multisections of orthogonal polynomial series > and ... and it is all very standard > in the fields of combinatorial enumeration > and special functions > and there are many books that mention it the hyperbolic trigonometrics ch and sh > are the (0, 2)- and (1, 2)-multisections > of the classical exponential function the circular trigonometrics > are ch(i x) and -i sh(i x) the generalised hyperbolic trigonometrics > g (x) > m n > are (m, n)-multisections of the exponential > and the circular generalisations are -m > t (x) = w g (w x) > m n 2n m n 2n where w_2n is exp(pi i / n) this simple extension of trigonometry > very simple to describe and work with > has an intricate algebraic structure > that allows many extensions of classical theories > including fourier analysis > tchebyshef expansion > many famous matrix expansions > and reveals many additional generalisations > that lead into deeper algebraic theories > like generalised polynomials > and their galois theory > or generalised fractions > and their intricate ring structure > and recursive properties -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar === Subject: Re: multisection of poerwseries <22234218.1228606572186.JavaMail.jakarta@nitrogen.mathforum.org> <493ba439$0$941$ba4acef3@news.orange.fr> <6q2qajFaj92mU1@mid.dfncis.de> posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.4) Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > a simpson (m, n)-multisection of f > which i often write |m > | f(x) > |n is the series --- > j > / a x > --- j > j = m (mod n) which can also be written 00 > --- > nj+m > / a x > --- nj+m > j=0 this operator is a projection operator > it is idempotent and all that > and there is a classic formula of simpson > which relates it to the roots of unity I'm not sure which Simpson or formula > you're referring to. It would help if > you could elaborate on this, preferably > with a reference. i've put up the latest copy of a paper i am trying to find time for up here http://www.galathaea.org/multisectedHypergeometricDocs/hypergeometric.pdf it's got less priority than other things so it's not intended to be complete or even very correct since i haven't even reread any of it really but it's got some definitions if needed chapter 2 has the reference to simpson -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: multisection of poerwseries a simpson (m, n)-multisection of f > which i often write > |m > | f(x) > |n > is the series > --- > j > / a x > --- j > j = m (mod n) > which can also be written > 00 > --- > nj+m > / a x > --- nj+m > j=0 > this operator is a projection operator > it is idempotent and all that > and there is a classic formula of simpson > which relates it to the roots of unity > I'm not sure which Simpson or formula > you're referring to. It would help if > you could elaborate on this, preferably > with a reference. i've put up the latest copy > of a paper i am trying to find time for > up here http://www.galathaea.org/multisectedHypergeometricDocs/hypergeometric.pdf I understand your multisection notation. Concerning Thomas Simpson's formula, I've been thinking about the case f(x) = e^x and n = 3, so a trisection formula. But that will have to wait ... David Bernier > it's got less priority than other things > so it's not intended to be complete > or even very correct > since i haven't even reread any of it really > but it's got some definitions if needed chapter 2 has the reference to simpson -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar === Subject: Re: #27 Chapter 1, proof that all infinite sets have to be countably infinite; new book 2nd edition: Math a subset of Physics, AP-adics posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1 FOH:R051,gzip(gfe),gzip(gfe) so, can you place these countably infinite numbers into one-to-one correspondence with, ah, what?... is their a circular element to your proof, that they are now only countable WRT themselves? however, your essential notion about the kind of infinity is probably good, since its original usage was just an ideal of Leibniz to create calculus; perhaps, it is isomorphic to Universe in some sense, meaning that the notion of multiverses is simply an abrogation of the definition of the word, universal (or, as Cusa would say, catholic .-) OK; on to CHapter Two of Book Twenty-three & eight ninths, by the biggest proponent of physics on sci.math, who never does either! > Infinity comes in only one brand-- countable infinity. thus: no big deal. it's just that numbers, themselves don't have roots; an *equation* in one variable has roots or solutions, that are numbers!... well, x - pi = 0 has a root, two! set a quaternion in (w,x,y,z) with coefficients; doess that have roots? > possible roots of certain types of number. Maybe there is no --USA out of Darfur Cruizade! http://larouchepub.com/pr_lar/2008/lar_pac/081124anti_clintons.html --ROTC, your summer vacation in the Sahara Desert ( S u d a n ) ; presage the Draft for your middleschool class of '12 -- brought to you by Allstate (tm) and Oxford U.Press! http://larouchepub.com/pr/2008/080813moloch_brown.html http://wlym.com --Wikipedia deletes notice of nullification of preclearance rule of Voting Rights Act in LaRouche v. Fowler, March 27, 2000; is the VRAo1965 a dead letter? http://en.wikipedia.org/wiki/Voting_Rights_Act#Pre-Clearance_2 === Subject: Re: #27 Chapter 1, proof that all infinite sets have to be countably infinite; new book 2nd edition: Math a subset of Physics, AP-adics posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1 FOH:R051,gzip(gfe),gzip(gfe) also, is not infinity really the first ideal, before zero? --Cheeny & Zbiggy, fo'mo' years! http://tarpley.net http://larouchepub.com http://xplodeyourmlmbiz.com/?s1=Fuller+Brush&t=LSP01&gclid=CPiar6nispcCFQu-G godYm21kA http://www.rwgrayprojects.com/synergetics/plates/plates.html === Subject: Re: Negative quantities >Is there any valid negative real world quality? Yes, for example, the electrical charge of an electron. Dave Greene === Subject: Re: Negative quantities- Is there any valid negative real world quality? Yes, for example, the electrical charge of an electron. Dave Greene The electrical charge of an electron is negative by convention. Please mind the distinction between extensive and intensive quantities. Quantities like length, area, volume, mass, positive electrical charge, negative electrical charge are extensive quantities. Quantities like line density, surface density, volume density = specific mass are intensive quantities. Extensive quantities are represented by nature by positive numbers. Please resolve for yourself the paradox on positive as well as negative electrical charges: this involves physics (real sex; Richard Feynman), not only mathematics (masturbation). See http://en.wikipedia.org/wiki/Extensive_quantity for more. relatives. Good luck: Johan E. Mebius === Subject: Re: Negative quantities- > Is there any valid negative real world quality? Yes, for example, the electrical charge of an electron. The electrical charge of an electron is negative by convention. Yes, it could have been done the other way around. It could have been chosen that the electrical charge on an electron was positive. Then the negative quality would have been the charge on a proton. The point is that in spite of which way we want to call it one of them is the negative of the other. So, to sum it up, there are real world qualities that are negative to other real world qualities. Dave Greene === Subject: Re: Negative quantities- <9GW_k.338$c35.190@nwrddc02.gnilink.net> <493C80E6.3090408@xs4all.nl> <1x%_k.561$7I6.290@nwrddc01.gnilink.net> posting-account=PSw72goAAADTugQpI3-NCbn5m3k-KKO3 .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) Is there any valid negative real world quality? > Yes, for example, the electrical charge of an electron. The electrical charge of an electron is negative by convention. Yes, it could have been done the other way around. It > could have been chosen that the electrical charge on an > electron was positive. Then the negative quality would > have been the charge on a proton. The point is that in > spite of which way we want to call it one of them is the > negative of the other. So, to sum it up, there are real > world qualities that are negative to other real world > qualities. Dave Greene How do we know what pole of a magnetic is so called the north? Mitch Raemsch === Subject: Re: Negative quantities <9GW_k.338$c35.190@nwrddc02.gnilink.net> posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) >Is there any valid negative real world quality? Yes, for example, the electrical charge of an electron. Dave Greene Yeah, one would think that but negative charges don't really subtract from positive one's. If we sum up positive and negative magnetic fields it appears as if nothing remains but both fields are still present because the 2 overlapping fields do not originate from exactly the same point. You can observe the horrible consequences here: The flux pictures are photographs. Indeed it allows for self powered engines it's not beyond the scope of this topic. Electric charge we call a negative when it is below the background level. It works just like Cecilius and Kelvin. I did say negative quality but the real *trick* question is of course to show a negative quantity . We want a real world example of something like -10 * -10 I cant imagine finding anything in negative quantity. To be honest it sounds like the whole concept originates in banking. It doesn't lend it much credibility to say the least. Bankers omitting perpetual motion sounds much like things bankers would do. But ok, what if we have -10 people with -100 euro debt? Or say -100 euro worth of negative charge at 10 charge per euro. Does this really make sense or is it intended to be confusing? The negative charge travels from the - to the +. The north pole is the magnetic south pole. Howard Johnson only got a patent after elaborate demonstration of permanent magnets doing work. His prototype made a dime worth of energy per year and cost many thousands of dollars to build. He did however get his patent. But if physicists keep subtracting the fields.... yeah..... what shall remain to say about it? It is wrong... O o About the flux. Here you have the weapons application here is the most recent motor http://www.youtube.com/watch?v=xo6y2fwLZ Q&feature=PlayList&p=5D2EA02EABD01947&index=0&playnext=1 Here is a crude video about the Newman motor http://www.youtube.com/watch?v=ctWa kMnEqg Wesley Garry got his patent 3 October 1876 http://magnetmotor.go-here.nl/wesley-gary/ Credibility is not truth, the observations show the fields do not subtract. Only the observations are truth. The best part of this is where people observe it to be true then keep repeating how it cant be. Striking irony. It doesn't prove it is wrong, no it shows exactly why they didn't know it. Even if they do, no one would listen to them. God not real? How dare you! lol It goes without question people are willing to die to preserve their believe system. Economic death, ecological death.... War in Iraq for oil... you name it and they will die for it. I really don't know of any negative quanta if that makes me a crank then so be it! The question seems honest enough to me. What negative quanta? Where? Or lets turn the question around. Perhaps that will make it more obvious. Why do we only have one kind of negative quantity? It seems to me there can be any number of different negative amounts of something if there is any. If a banana is -1 apple then surely that means a pear is also -1 apple just like a kiwi. If not, why? === Subject: Re: Negative quantities >Is there any valid negative real world quality? Yes, for example, the electrical charge of an electron. Yeah, one would think that but negative charges don't really subtract >from positive one's. They can subtract form each other affecting the net electric charge. >If we sum up positive and negative magnetic fields...[snip crapola] nutcase alert ... >Electric charge we call a negative when it is below the background >level. It works just like Cecilius and Kelvin. No, it is unlike Kelvin, the reference is zero charge. There is no electrical charge equivalent to absolute zero. >I did say negative _quality_ but the real *trick* question is of >course to show a negative _quantity_. We want a real world >example of something like So now that you have been shown to be wrong you want to change the question? >Only the observations are truth. So, there were no microbes until the microscope was invented? I think gabydewilde is probably Mitch/BURT in drag. Dave Greene === Subject: Re: Negative quantities Is there any valid negative real world quality? Yes, for example, the electrical charge of an electron. Yeah, one would think that but negative charges don't really subtract >from positive one's. They can subtract form each other affecting the net electric charge. If we sum up positive and negative magnetic fields...[snip crapola] nutcase alert ... Electric charge we call a negative when it is below the background >level. It works just like Cecilius and Kelvin. No, it is unlike Kelvin, the reference is zero charge. There is no > electrical > charge equivalent to absolute zero. I did say negative _quality_ but the real *trick* question is of >course to show a negative _quantity_. We want a real world >example of something like So now that you have been shown to be wrong you want > to change the question? Only the observations are truth. So, there were no microbes until the microscope was invented? I think gabydewilde is probably Mitch/BURT in drag. A gdewilde instructed rec.bicycles.tech on how to use a turbine to harness the wind flowing past a bicycle and turn it into motive energy, but they were too benighted to understand. -- Michael Press === Subject: Re: Negative quantities posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY AppleWebKit/525.19 (KHTML, like Gecko) Chrome/0.3.154.9 Safari/525.19,gzip(gfe),gzip(gfe) >Is there any valid negative real world quality? > Yes, for example, the electrical charge of an electron. Yeah, one would think that but negative charges don't really subtract >from positive one's. They can subtract form each other affecting the net electric charge. If we sum up positive and negative magnetic fields...[snip crapola] nutcase alert ... Electric charge we call a negative when it is below the background >level. It works just like Cecilius and Kelvin. No, it is unlike Kelvin, the reference is zero charge. There is no > electrical > charge equivalent to absolute zero. I did say negative _quality_ but the real *trick* question is of >course to show a negative _quantity_. We want a real world >example of something like So now that you have been shown to be wrong you want > to change the question? Only the observations are truth. So, there were no microbes until the microscope was invented? I think gabydewilde is probably Mitch/BURT in drag. A gdewilde instructed rec.bicycles.tech on how > to use a turbine to harness the wind flowing past > a bicycle and turn it into motive energy, but > they were too benighted to understand. > -- > Michael Press The point of this idea is that people are to stupid to look into back emf generators, gravity engines, anti gravity, magnet motors, radiant energy. The subjects are to complicated to suspend disbelieves. Further than this I cant dumb it down: http://www.youtube.com/watch?v=-7M6S2brhFE So lets use the wind and the drag! Do feel free to forget about the turbine, we are going to use propellers now. You do understand propellers. There are no more excuses. First a few points: 1 - heavier than air flying machines can really fly. 2 - Sailboats can sail close to the wind. 3 - Windmills have only a few blades to prevent the wind from flowing around it. 4 - Wind energy grows at the cube of the speed. 5 - Vehicle drag grows at the cube of the speed. So this means we have 20 mph combined drag moving at 10 mph up against 10 mph wind. If you understand point (2) you know the boat may sail up the wind. If you understand point (3) you should be able to understand the blades are not moving backwards but they are moving sidewards. Sticking a windmill onto a car does not cause the blades to move backwards. Not moving backwards means the sailboat like blades can sail up the drag. The blades are not moving in the same direction as the vehicle but they are moving sidewards. I have more elaborate visual documentation but I put together this silly picture to keep this posting short and to the point. http://wind-car.go-here.nl/sci-math-wind-car-nl.jpg Please laugh at it first, then do the math and see the problems of windmills (3) are not there, of course the first thing that pops to mind is the drag but the drag is not there per point (4) The concept is very old in lots of different applications. The only thing that is missing is good math. We can use wind energy at zero mph, the idea it would be worse at 60 mph is fairly silly. http://wind-car.go-here.nl/ I also proposed to build giant drinking bird generators powered by e- vaporation. Drinking birds are no mere veneer of vanity, a conversation lawyering vestige of the vox populi, now vacant, vanished. However, this valorous physics visitation of a bygone perpetual motion vexation stands vivified to be vindicated, law violations of vaporisation vow to vanquish these ventures of venal virulent energy science vanguard vermin vouchsafing their violent vicious and voracious vocabular violation of volition. The capillarity vichyssoise, it's viral vaporization and violent virtue of vessel pressure verily. Votive verbiage, not in vain, for the value and veracity of such shall one day vindicate the vigilant and the virtuous. To veer the verbose from vacuum to vortical let me quote Victor: Implosion is no invention in the conventional sense, but rather the renaissance of ancient knowledge, lost over the course of time. This drinking bird visitation verily begs to vision vessel volume, velocity, vehicle validity and choice of vapor. Does it not? More than oh, ha-ha-ha the victims of the energy monopoly can not produce. You really deserve to be lied to by Don Lancaster, Eeyore, Hairy C., George Bush, Dick Cheney and MC Billo Rally. Look out a terrorist behind you! ROFLMSO !! === Subject: Re: Negative quantities posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) >Is there any valid negative real world quality? > Yes, for example, the electrical charge of an electron. Yeah, one would think that but negative charges don't really subtract >from positive one's. They can subtract form each other affecting the net electric charge. > This doesn't make the operator part of the number. You loose either way. Show me negative quality and negative quantity. Make it evident that -10 * -10 = 20 you poetically tried the truth by insult thing. Insult is always the last resort. Your statement doesn't prove that what you learn to be right, in stead it shows exactly why you never bothered to think about it. Show me negative quality and negative quantity. Stop telling me that eating 1 apple reduces the number of apples from 10 to 9. Show me -10 times 10 negative apples equals 100 apples. If we sum up positive and negative magnetic fields it appears as if nothing remains but both fields are still present because the 2 overlapping fields do not originate from exactly the same point. You can observe the consequences here: The flux pictures are photographs. IT ARE PHOTOGRAPHS DAVID GREENE. Talking to Galileo after looking at the moon the church censors suggested they should agree the picture of the moon was an illusion. The same thing happened around 1905 when Royal Rife invented a microscope. In the case of his microscope the truth was never accepted. I posted 20 videos in sci.physics.research and they rejected it because the microscope didn't follow established believe systems. Is it not obvious to you the observation overrules the text book? O MY GAWD!! Indeed it allows for self powered engines it's not beyond the scope of this topic. Electric charge we call negative when it is below the background level or they are 2 different entities. Never is the one a negative of the other. Things go mathmatically wrong that way. You seem not to be looking for the truth but repeating that what you want to be the truth. It's just like claiming Humans are the only life form in the galaxy. I don't object to any claim if you just show them to be true. Why would an opperator become part of a variable? > No, it is unlike Kelvin, the reference is zero charge. There is no > electrical charge equivalent to absolute zero. Either they are 2 different things or they are both part of the same scale. At no stage does it become a negative value. I did say negative quality but the real *trick* question is of course to show a negative quantity . We want a real world example of something like -10 * -10 > So now that you have been shown to be wrong you want > to change the question? No, now that I failed to explain what the point is I further elaborate the point in the hope you will now try to understand it. You know? Repetition? I cant imagine finding anything in negative quantity. It not just sounds dumb, in fact it really is nonsense until you prove it. Yes, I rather weird but it was not my claim. Sure, you learn this to be the truth. But why would you not question it before accepting it? To be honest it sounds like the whole concept originates in banking. It doesn't lend it much credibility to say the least. Bankers omitting perpetual motion sounds much like things bankers would do. But ok, what if we have -10 people with -100 euro debt? Or say -100 euro worth of negative charge at 10 charge per euro. Does this really make sense or is it intended to be confusing? The negative charge travels from the - to the +. The north pole is the magnetic south pole. Howard Johnson only got a patent after elaborate demonstration of permanent magnets doing work. His prototype made a dime worth of energy per year and cost many thousands of dollars to build. He did however get his patent. But if physicists keep subtracting the fields.... yeah..... what shall remain to say about it? Here you have the weapons application here is the most recent motor http://www.youtube.com/watch?v=xo6y2fwLZ Q&feature=PlayList&p=5D2EA02EABD01947&index=0&playnext=1 Here is a crude video about the Newman motor http://www.youtube.com/watch?v=ctWa kMnEqg Wesley Garry got his patent 3 October 1876 http://magnetmotor.go-here.nl/wesley-gary/ Credibility is not truth, the observations show the fields do not subtract. Only the observations are truth. The best part of this is where people observe it to be true then keep repeating how it cant be. Striking irony. It doesn't prove it is wrong, no it shows exactly why they didn't know it. Even if they do, no one would listen to them. God not real? How dare you! lol It goes without question people are willing to die to preserve their believe system. Economic death, ecological death.... War in Iraq for oil... you name it and they will die for it. I really don't know of any negative quanta if that makes me a crank then so be it! The question seems honest enough to me. What negative quanta? Where? Or lets turn the question around. Perhaps that will make it more obvious. Why do we only have one kind of negative quantity? It seems to me there can be any number of different negative amounts of something if there is any. If a banana is -1 apple then surely that means a pear is also -1 apple just like a kiwi. If not, why? === Subject: Re: Negative quantities >Is there any valid negative real world quality? > Yes, for example, the electrical charge of an electron. Yeah, one would think that but negative charges don't really subtract >from positive one's. They can subtract form each other affecting the net electric charge. This doesn't make the operator part of the number. 1) The subtraction operator is not part of the number. 2) The negative of the charge is not the operator. 3) QED >You loose either way. Talking to yourself again! >Show me negative quality and negative quantity. You asked for negative quality and I showed you. Then you admitted it, now you try to deny it. I never signed up to show a negative quantity. >Stop telling me that eating 1 apple reduces the >number of apples from 10 to 9. I never told you that. Why do you resort to lies? >If we sum up positive and negative magnetic fields it appears as if >nothing remains but both fields are still present because the 2 >overlapping fields do not originate from exactly the same point. Nutcase alert... Magnetic fields do not come in positive and negative. > No, it is unlike Kelvin, the reference is zero charge. There is no > electrical charge equivalent to absolute zero. Either they are 2 different things or they are both part of the same >scale. At no stage does it become a negative value. Here is your logical fallacy. You have assumed the conclusion - By implying same scale as a scale that cannot contain a true zero at the center. The only scale you allow is one such as Kelvin with zero on an endpoint. Well the scale of electric charges has a zero and not just because of mathematical shift as with the Celsius scale of temperature. Negative charges are on one side of the scale and positive charges are on the other. Dave Greene === Subject: Re: Negative quantities-- > Is there any valid negative real world quality? > Yes, for example, the electrical charge of an electron. > Yeah, one would think that but negative charges don't really subtract >from positive one's. They can subtract form each other affecting the net electric charge. > This doesn't make the operator part of the number. 1) The subtraction operator is not part of the number. > 2) The negative of the charge is not the operator. > 3) QED You loose either way. Talking to yourself again! Show me negative quality and negative quantity. You asked for negative quality and I showed you. Then you > admitted it, now you try to deny it. I never signed up to > show a negative quantity. Stop telling me that eating 1 apple reduces the > number of apples from 10 to 9. I never told you that. Why do you resort to lies? If we sum up positive and negative magnetic fields it appears as if > nothing remains but both fields are still present because the 2 > overlapping fields do not originate from exactly the same point. Nutcase alert... Magnetic fields do not come in positive and negative. No, it is unlike Kelvin, the reference is zero charge. There is no > electrical charge equivalent to absolute zero. > Either they are 2 different things or they are both part of the same > scale. At no stage does it become a negative value. Here is your logical fallacy. You have assumed the conclusion - By > implying same scale as a scale that cannot contain a true zero at the > center. The only scale you allow is one such as Kelvin with zero on > an endpoint. Well the scale of electric charges has a zero and not just > because of mathematical shift as with the Celsius scale of temperature. > Negative charges are on one side of the scale and positive charges > are on the other. Dave Greene Electrical charges happen to come in two kinds, commonly known as positive and negative. Let us name the two kinds of electrical charge in this newsgroup post just A and B. Each and any electrical charge is accompanied by an electrical field. Physical facts: Electrical fields of type A can be superimposed. Electrical fields of type B can be superimposed. Both types of electrical field are conveniently modelled as vector fields in 3D Euclidean space. These fields are free of singularities and nowhere zero in the exterior of the room taken by the electrical charges (assuming that the charges occupy a finite spatial extent). It is a physical fact, not a mathematical fact, that electrical fields of mixed types can be superimposed too. A most curious fact, given that on a subatomic scale electrical charges of different types cannot occupy the same space without exploding into pure energy. BTW, this is also true of charges of the same type. In the mathematical model the superposition of electrical fields of different types happens to be faithfully represented by pointwise vector subtraction. Assuming again a finite extent of the electrical charges, the general field is free of singularities in the exterior of the room taken by the charges, but may be zero at isolated points. The equilibrium at any zero-field point is unstable. So far the masturbation in Feynman's sense. The =real sex= always occurs when an electron the same spot and explode into pure light! The better part of this mini-essay is of course old hat (Michael Faraday, 1831, 1839, 1855; James Clerk Maxwell, 1873). Ciao: Johan E. Mebius === Subject: Re: Negative quantities <9GW_k.338$c35.190@nwrddc02.gnilink.net> posting-account=PSw72goAAADTugQpI3-NCbn5m3k-KKO3 .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) >Is there any valid negative real world quality? Yes, for example, the electrical charge of an electron. Dave Greene Dave? That is rediculous. What do you think the difference between a negative charge and a positive is? If you can define it for us it would be appreciated. Mitch Raemsch === Subject: Re: Negative quantities >Is there any valid negative real world quality? Yes, for example, the electrical charge of an electron. Dave? That is rediculous. What do you think the difference between a >negative charge and a positive is? If you can define it for us it would be appreciated. Who is us, Mitch? Is *anyone* on your side? Why don't you pick up a textbook? Criminy, you are the guy who thinks a spring cannot be compressed because there is a force against compression. Dave Greene === Subject: Re: Negative quantities posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) >Is there any valid negative real world quality? > Yes, for example, the electrical charge of an electron. Dave? That is rediculous. What do you think the difference between a >negative charge and a positive is? If you can define it for us it would be appreciated. Who is us, Mitch? Is *anyone* on your side? Yes, there really are people who want to know how the micro cosmos works. Antagonism isn't going to render such effort null. > Why don't you pick up a textbook? We want to learn how things work by figuring them out. The text book is not the real world. > Criminy, you > are the guy who thinks a spring cannot be compressed > because there is a force against compression. I'm sure you think fields behave like springs. It says so in the textbook so it has to be true. But let me tell you something. The text book is full of real discoveries from people outside academia. In fact your far most scientific heroes got all the opposition from academics their little flywheel effort could produce. Honest review was never part of the review. For 5 long years aeroplanes flew over the university. It was only then the phenomenon was accepted in the text book. This was because the Professors started to look like the lunatics they really are. Not because it was the truth, it was merely to safe face. Read all of this: If you don't have time see this: http://www.youtube.com/watch?v=wt3smrXkVpE (10:00) If you have an even shorter attention span - see this: http://www.youtube.com/watch?v=-7M6S2brhFE (1:21) What is the text book explanation for this one? Constant symmetry perhaps? That doesn't explain the observation at all, in fact the observations contradict it. When will it be in the text book if you will never look into it? Could that be Never? Well o well, what does that tell us about the text book? In stead you have censors and systems of surveillance. The truth is not an author, it's not a person, it's not a heap of paper, it is not some one's credentials. The truth is out there! No really! === Subject: Re: Negative quantities posting-account=PSw72goAAADTugQpI3-NCbn5m3k-KKO3 .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) >Is there any valid negative real world quality? > Yes, for example, the electrical charge of an electron. Dave? That is rediculous. What do you think the difference between a >negative charge and a positive is? If you can define it for us it would be appreciated. Who is us, Mitch? Is *anyone* on your side? > Why don't you pick up a textbook? Criminy, you > are the guy who thinks a spring cannot be compressed > because there is a force against compression. Dave Greene We think your point is moot. === Subject: Re: Strange function > I cannot call it entire, since analyticity fails at the grid of the > Gaussian integers, but the points where it is not analytic obviously has > measure 0. Err, nonsense. The set of non-analyticity has at least measure oo. The line > Re(w)=1 for example has infinite measure. Please skip the measure part on the question. -- > Ioannis That depends on the measure of course. > The exceptional set has Lebesgue measure 0. Sorry, let me clarify it further: the exceptional points are not only the Gaussian integers, but also the line segments that connect them, right and above of -1-i. Here's the set of exceptional points E described exactly: Call T = {-1,0} U N = {-1,0,1,2,3,...}. E = {win C: Re(w)in T OR Im(w)in T} For fixed zin C{0}, f(z,w) is analytic wrt w in CE. My measure theory is a bit rusty, and I was under the impression that a segment of positive length has Lebesgue measure |L|, so it doesn't look to me like E has Lebesgue measure 0, since it contains at least one line which extends to oo. If E has Lebesgue measure 0, I'd appreciate it if you could give me a rigorous argument for that, if possible. -- Ioannis === Subject: Re: Strange function <1228683677.259453@athprx04> posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > My measure theory is a bit rusty, and I was under the impression > that a segment of positive length has Lebesgue measure |L|, so > it doesn't look to me like E has Lebesgue measure 0, since it > contains at least one line which extends to oo. If E has Lebesgue measure 0, I'd appreciate it if you could > give me a rigorous argument for that, if possible. Robert Israel has pointed out that it has infinite linear measure and zero planar measure. In fact, it has infinite but sigma-finite linear measure (Hausdorff dimension 1), and consequently non-sigma-finite Hausdorff p-masure for any p < 1 and zero Hausdorff p-measure for any p > 1. [Sigma-finite *-measure means expressable as a countable union of sets each with finite *-measure. Non-sigma-finite *-measure is like being uncountable, except instead of being uncountable in the sense of not being expressable as a countable union of points, it's uncountable in the sense of not being expressable as a countable union of finite *-measure sets.] Indeed, this is still true when Hausdorff is replaced with packing, or even with upper Minkowski (the latter being called upper box in Falconer's books). The set is also nowhere dense in a strong way, being strongly lower porous -- even satisfying a fairly strong exterior cone and sphere conditions (in the sense used in potential theory) and being ball porous. Moreover, the closure of the set satisfies all these properties as well. The set is also countably 1-rectifiable in the sense of geometric measure theory. Personally, I'd just describe it as a countably family of uniformly isolated horizontal and vertical lines, or something fairly specific, unless there was some specific aspect of its size as a set having relevance elsewhere in your work that would lead you to want to emphasize that it has that size. This is because for just about any size measure of interest in which your set is small relative to that size measure, your set is so small relative to the size measure that it's a major understatment to say it's small relative to that size measure. For example, unless the fact that it has planar measure zero (or even zero measure in Hausdorff dimension 1.0001) has some signficiance later on, it's a bit silly to say that it has planar measure zero (because it's so much smaller in so many other ways as well). Dave L. Renfro === Subject: Re: Strange function > I cannot call it entire, since analyticity fails at the grid of the > Gaussian integers, but the points where it is not analytic obviously > has > measure 0. Err, nonsense. The set of non-analyticity has at least measure oo. The > line > Re(w)=1 for example has infinite measure. Please skip the measure part on the question. -- > Ioannis That depends on the measure of course. > The exceptional set has Lebesgue measure 0. Sorry, let me clarify it further: the exceptional points are not only the > Gaussian integers, but also the line segments that connect them, right and > above > of -1-i. Here's the set of exceptional points E described exactly: Call T = {-1,0} U N = {-1,0,1,2,3,...}. E = {win C: Re(w)in T OR Im(w)in T} For fixed zin C{0}, f(z,w) is analytic wrt w in CE. My measure theory is a bit rusty, and I was under the impression that a > segment > of positive length has Lebesgue measure |L|, so it doesn't look to me like > E has > Lebesgue measure 0, since it contains at least one line which extends to > oo. If E has Lebesgue measure 0, I'd appreciate it if you could give me a > rigorous > argument for that, if possible. You're thinking of Lebesgue measure on a line. Lebesgue measure in the plane corresponds to area, not length. Your set E has Lebesgue measure 0 because lines have area 0. On the other hand, its one-dimensional Hausdorff measure is infinity. -- === Subject: Re: Strange function >I cannot call it entire, since analyticity fails at the grid of the > Gaussian integers, but the points where it is not analytic obviously > has > measure 0. Err, nonsense. The set of non-analyticity has at least measure oo. The > line > Re(w)=1 for example has infinite measure. Please skip the measure part on the question. -- > Ioannis That depends on the measure of course. > The exceptional set has Lebesgue measure 0. Sorry, let me clarify it further: the exceptional points are not only the > Gaussian integers, but also the line segments that connect them, right and > above > of -1-i. Here's the set of exceptional points E described exactly: Call T = {-1,0} U N = {-1,0,1,2,3,...}. E = {win C: Re(w)in T OR Im(w)in T} For fixed zin C{0}, f(z,w) is analytic wrt w in CE. My measure theory is a bit rusty, and I was under the impression that a > segment > of positive length has Lebesgue measure |L|, so it doesn't look to me like > E has > Lebesgue measure 0, since it contains at least one line which extends to > oo. If E has Lebesgue measure 0, I'd appreciate it if you could give me a > rigorous > argument for that, if possible. You're thinking of Lebesgue measure on a line. Lebesgue measure in the plane > corresponds to area, not length. Your set E has Lebesgue measure 0 because > lines have area 0. On the other hand, its one-dimensional Hausdorff measure > is infinity. analytic almost everywhere on its domain of definition, like Klueless said. Over and out. -- Ioannis === Subject: Re: Strange function > I have a complex function f(z,w) at my disposal and I need to know how to > characterize it, in terms of analyticity, with respect to w. f(z,w) is defined for Re(w) > -1 and Im(w) > -1. And z where? > It is analytic away from the grid that connects the Gaussian integers. That > is, > it has a series expansion in w everywhere, as long as Re(w), Im(w) not in N U > {0,-1}. fixed z the above holds? What series expansions? Power series? > When Re(w) is in N U {0} f(z,w) is C^1, w.r.t. w. Do you mean wrt Im(w)? > Also, f(z,n) has series expansions in z. ??? > My question is, how can I characterize such a function? It appears to me the function that = 0 on the grid and = 1 elsewhere satisfies the conditions you've given. > I cannot call it entire, since analyticity fails at the grid of the > Gaussian > integers, but the points where it is not analytic obviously has measure 0. Can I call it quasi-analytic? Perhaps semi-analytic? Analytic everywhere except on a set of measure 0? > === Subject: Re: holomorphically convex Well - I went to the Cornell library and looked at the later book that variables. All this was fixed up in that book, which is loosely based on Gunning and Rossi. have the idea that if a domain is covered by an ascending chain of compact sets, an arbitrary compact set is contained in some member of the chain. The same idea is in theorem G11 in the first chapter, although it doesn't cause a problem with the proof. So perhaps his later book is the one to read instead. He seemed to have included a lot more detail. It's obscenely expensive, a 3-volume set and each book costs about $200. Maybe to pay rent in Princeton :( I like compressed books, because authors write in English and I don't think math in English. So it's more pleasant to figure things out than to read through the author's exposition. But if they are compressed with a lot of typos and errors, it can be frustrating. Laura === Subject: Re: holomorphically convex > Well - I went to the Cornell library and looked at the later book that > variables. All this was fixed up in that book, which is loosely based > on Gunning and Rossi. > have the idea that if > a domain is covered by an ascending chain of compact sets, an arbitrary > compact set is contained in some member of the chain. The same idea is in > theorem G11 in the first chapter, although it doesn't cause a problem with > the proof. > So perhaps his later book is the one to read instead. He seemed to have > included a lot more detail. It's obscenely expensive, a 3-volume set and > each book costs about $200. Maybe to pay rent in Princeton :( > I like compressed books, because authors write in English and I don't > think math in English. So it's more pleasant to figure things out than > to read through the author's exposition. But if they are compressed > with a lot of typos and errors, it can be frustrating. > Laura Many MA and PhD thesises are available online for free. There must be dozens and dozens of thesises on several complex variables. Some thesises have quite a lot of background material... David Bernier === Subject: Re: linear > Reading Shilov, Linear Algebra, which has a series of problems, ... > 1. For matrices A,B, trace(AB) = trace(BA) > 2. AB-BA != E (identity matrix), and a note pointing out that this is > only true for finite-dimensional spaces. > 3. An example of two linear operators (in the space of polynomials p(t)) > A, B, where AB-BA = E > One of these operators (p(t) -> tp(t)) appears to require an > infinite-dimensional space. It's not to me whether my 'proof' of (1)(taking traces of AB & BA as > double sums, and reversing the order of summation in one of these) is > limited to finite dimensions, or if it's that suitable operators for (3) > require infinite dimensions. Or something else. I'd appreciate a (hopefully) simple elaboration on this, or a pointer to > somewhere to read more. If my statement is unclear, I am not a > mathmetician. But, I can try to clarify. The trace of a linear operator on an infinite dimensional space is in general not defined: it requires an infinite sum, which will not converge in general. For bounded linear operators on a Banach space, it is still true that AB - BA can't be E. You can't use the trace to prove that, though. -- === Subject: Re: linear > Reading Shilov, Linear Algebra, which has a series of problems, ... > 1. For matrices A,B, trace(AB) = trace(BA) > 2. AB-BA != E (identity matrix), and a note pointing out that this is > only true for finite-dimensional spaces. > 3. An example of two linear operators (in the space of polynomials p(t)) > A, B, where AB-BA = E > One of these operators (p(t) -> tp(t)) appears to require an > infinite-dimensional space. It's not to me whether my 'proof' of (1)(taking traces of AB & BA as > double sums, and reversing the order of summation in one of these) is > limited to finite dimensions, or if it's that suitable operators for (3) > require infinite dimensions. Or something else. I'd appreciate a (hopefully) simple elaboration on this, or a pointer to > somewhere to read more. If my statement is unclear, I am not a > mathmetician. But, I can try to clarify. The trace of a linear operator on an infinite dimensional space is in general >not defined: it requires an infinite sum, which will not converge in general. No problem there (I think). But, it's not clear why that means you can't take the difference between two such sums, when they can be identified on a term-by-term basis ... which is what I think happens in trace(AB)-trace(BA). G === Subject: Re: Basic introduction to Nevanlinna theory posting-account=5dWnDwkAAAAJX56s4Iiwi0owHzqwvK-v CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Can anyone recommend a good basic introduction to > Nevanlinnna theory in English? [...] It is OK if > is also deals with recent work, such as Vojta's > analogies with the problem of integral solutions of > Diophantine equations. But my main interest is in > understanding Nevanlinna's original groundbreaking > work in 1923-25. Are there some lecture notes or > semi-popular accounts of this early creative work? To answer my own question, Paul Vojta's four lectures on diophantine approximation and Nevanlinna theory (19 pages) start out with a short introduction to Nevanlinna theory (Picard's theorem and the two main of theorems of Nevanlinna theory): http://math.berkeley.edu/~vojta/cime/cimebeamer.pdf There is is also an expanded version on Vojta's lecture notes (122 pages): Diophantine Approximation and Nevanlinna Theory http://math.berkeley.edu/~vojta/cime/cime.pdf Both of these PDF files are available at his home page: http://math.berkeley.edu/~vojta/ === Subject: Help with packings of convex sets on the plane Hi I took as a challenge to understand the proof of Theorem 1.1.1 in http://assets.cambridge.org/97805218/01577/excerpt/9780521801577_excerpt.pdf but I got dissappointed since I got stuck when trying to obtain equation (1.5) on page 6. So I have given up but I would like somebody to help me to derive that equation. === Subject: Re: f(x) = f(3x) + f(3x+1) + f(3x+2) posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008111317 Ubuntu/8.04 (hardy) Firefox/3.0.4,gzip(gfe),gzip(gfe) > hello this is probably the dumbest question ever but anyways : f(1) = log(3) > f(x) = f(3x) + f(3x+1) + f(3x+2) i think a solution exists expressed with rational functions and their inverses and logaritms. its clear that we can solve it by using a taylor series and the binomium (3x + a) ^ n and find the taylor coefficients that way. how many solutions ? f: N -> R or f: R->R continuous or ... ? === Subject: Re: f(x) = f(3x) + f(3x+1) + f(3x+2) posting-account=HJnNYQoAAACtdMyfjpc6YoWX6U0epxWM FunWebProducts; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > hello this is probably the dumbest question ever but anyways : f(1) = log(3) > f(x) = f(3x) + f(3x+1) + f(3x+2) i think a solution exists expressed with rational functions and their inverses and logaritms. its clear that we can solve it by using a taylor series and the binomium (3x + a) ^ n and find the taylor coefficients that way. how many solutions ? kisses amy666 Well first, replace x with -x, 1 with -1, and then log(3) with n. You get f(-1) = n. Then try f(0) = p and f(1) = q. There should be a two- dimensional family of solutions depending upon your choices for p and q. Now that we've got that settled you, I, or someone else can then figure out the details. === Subject: Re: Gravity Flows Energy In Freefall posting-account=eClhIgoAAACIxAUKGXHwqJ9Am4Bt2i0u .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Matter jumps to speed when dropped. Then it begins to weightlessly > accelerate. This of course always remains finite. There is nothing > more to say. Then the changed momentum is evenly distributed. At the macro level, > acceleration appears smooth. Double-A jump to a higher speed by a quantum leap.? At the microscopic level, all objects are like globs of jello: and transmitting momentum by exchanges of photons. Such changes in motion intermediated by photons must be quantum leaps. We don't know how gravity imparts motion, but if it is by gravitons or quantized photon exchanges, all quantum leaps. When an object is propelled by a rocket engine, each molecule combusting must sent a wave of photon increased paricle motion, and in increased motion of the rocket as a whole. Also a curious wave nature is associated with the increased motion (glob of jello shaking), though this is probably not the same as its quantum wave nature. But this is why the macro concept of a force is hard to pin down. Gravity acts in a much smoother manner, in a quantum leap manner. > If i hold out my arm and drop a rock, and one second > later it hits the ground, it's going 32 feet/sec when it > hits the ground, isn't it? Only by macro averaged out approximation. Double-A > happy days and... > starry starry nights! -- > Indelibly yours, > Paine Ellsworth P.S.: I think on-stage nudity is disgusting, shameful > and damaging to all things American. But if I > were 22 with a great body, it would be artistic, > tasteful, patriotic and a progressive religious > experience. > Shelley Winters P.P.S.: http://yummycake.secretsgolden.com > http://garden-of-ebooks.blogspot.com > http://painellsworth.net === Subject: Re: Gravity Flows Energy In Freefall posting-account=PSw72goAAADTugQpI3-NCbn5m3k-KKO3 .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) > Matter jumps to speed when dropped. Then it begins to weightlessly > accelerate. This of course always remains finite. There is nothing > more to say. Then the changed momentum is evenly distributed. At the macro level, > acceleration appears smooth. Double-A jump to a higher speed by a quantum leap.? At the microscopic level, all objects are like globs of jello: > and transmitting momentum by exchanges of photons. Such changes in > motion intermediated by photons must be quantum leaps. We don't know > how gravity imparts motion, but if it is by gravitons or quantized > photon exchanges, all quantum leaps. When an object is propelled by a > rocket engine, each molecule combusting must sent a wave of photon > increased paricle motion, and in increased motion of the rocket as a > whole. Also a curious wave nature is associated with the increased > motion (glob of jello shaking), though this is probably not the same > as its quantum wave nature. But this is why the macro concept of a > force is hard to pin down. Gravity acts in a much smoother manner, > in a quantum leap manner. If i hold out my arm and drop a rock, and one second > later it hits the ground, it's going 32 feet/sec when it > hits the ground, isn't it? Only by macro averaged out approximation. Double-A happy days and... > starry starry nights! -- > Indelibly yours, > Paine Ellsworth P.S.: I think on-stage nudity is disgusting, shameful > and damaging to all things American. But if I > were 22 with a great body, it would be artistic, > tasteful, patriotic and a progressive religious > experience. > Shelley Winters P.P.S.: http://yummycake.secretsgolden.com > http://garden-of-ebooks.blogspot.com > http://painellsworth.net- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - Electic and magnetic can cause matter to flow just as gravity can. Mitch Raemsch === Subject: Re: Gravity Flows Energy In Freefall > Matter jumps to speed when dropped. Then it begins to weightlessly > accelerate. This of course always remains finite. There is nothing > more to say. Then the changed momentum is evenly distributed. At the macro level, > acceleration appears smooth. Double-A jump to a higher speed by a quantum leap.? At the microscopic level, all objects are like globs of jello: and transmitting momentum by exchanges of photons. That would be quite profound if you defined photon. http://www.androcles01.pwp.blueyonder.co.uk/AC/spin.gif How does a magnet move another? Such changes in motion intermediated by photons must be quantum leaps. We don't know how gravity imparts motion, but if it is by gravitons That's going overboard. or quantized is likely quantum in nature. All velocities are relative. ALL velocities are RELATIVE. ALL VELOCITIES ARE RELATIVE. ***** ALL VELOCITIES ARE RELATIVE! ***** photon exchanges, all quantum leaps. When an object is propelled by a rocket engine, each molecule combusting must sent a wave of photon increased paricle motion, and in increased motion of the rocket as a whole. The rocket as a whole includes it's exhaust.When a rocket accelerates part of it goes the other way. The total momentum remains zero. Also a curious wave nature is associated with the increased motion (glob of jello shaking), though this is probably not the same as its quantum wave nature. But this is why the macro concept of a force is hard to pin down. Forces act between TWO objects. It is action at a distance that is perplexing, even at the microscopic level. If the Moon is gravitationally attracted to the Earth then the neutron is gravitationally attracted to the proton. Only the scale has changed, the inverse square law still applies. Gravity acts in a much smoother manner, Smoother than what? An electron falling in an electrostatic field (such as in a CRT) accelerates smoothly. You are applying macroscopic (analogue) thinking and mixing it with quantum (digital) ideas. in a quantum leap manner. > If i hold out my arm and drop a rock, and one second > later it hits the ground, it's going 32 feet/sec when it > hits the ground, isn't it? Only by macro averaged out approximation. Double-A > happy days and... > starry starry nights! -- > Indelibly yours, > Paine Ellsworth P.S.: I think on-stage nudity is disgusting, shameful > and damaging to all things American. But if I > were 22 with a great body, it would be artistic, > tasteful, patriotic and a progressive religious > experience. > Shelley Winters P.P.S.: http://yummycake.secretsgolden.com > http://garden-of-ebooks.blogspot.com > http://painellsworth.net === Subject: Re: Gravity Flows Energy In Freefall posting-account=eClhIgoAAACIxAUKGXHwqJ9Am4Bt2i0u .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Matter jumps to speed when dropped. Then it begins to weightlessly > accelerate. This of course always remains finite. There is nothing > more to say. Then the changed momentum is evenly distributed. At the macro level, > acceleration appears smooth. Double-A jump to a higher speed by a quantum leap.? At the microscopic level, all objects are like globs of jello: > and transmitting momentum by exchanges of photons. That would be quite profound if you defined photon. See QED - Feynman. > http://www.androcles01.pwp.blueyonder.co.uk/AC/spin.gif How does a magnet move another? > Such changes in > motion intermediated by photons must be quantum leaps. We don't know > how gravity imparts motion, but if it is by gravitons > That's going overboard. or quantized > is likely quantum in nature. > All velocities are relative. Yes. > ALL velocities are RELATIVE. Yes. > ALL VELOCITIES ARE RELATIVE. Yes. > ***** ALL VELOCITIES ARE RELATIVE! ***** Yes. Therefore also all momentum and kinetic energy. photon exchanges, all quantum leaps. When an object is propelled by a > rocket engine, each molecule combusting must sent a wave of photon > increased paricle motion, and in increased motion of the rocket as a > whole. > The rocket as a whole includes it's exhaust.When a rocket accelerates > part of it goes the other way. The total momentum remains zero. Yes. Also a curious wave nature is associated with the increased > motion (glob of jello shaking), though this is probably not the same > as its quantum wave nature. But this is why the macro concept of a > force is hard to pin down. > Forces act between TWO objects. It is action at a distance that > is perplexing, even at the microscopic level. All actions are at a distance. > If the Moon is gravitationally > attracted to the Earth then the neutron is gravitationally attracted to > the proton. Only the scale has changed, the inverse square law still > applies. Yes. Gravity acts in a much smoother manner, > Smoother than what? A rocket exhaust. > An electron falling in an electrostatic field > (such as in a CRT) accelerates smoothly. At some level it is not smooth. >You are applying > macroscopic (analogue) thinking and mixing it with quantum (digital) > ideas. There is only one unified reality. in a quantum leap manner. If i hold out my arm and drop a rock, and one second > later it hits the ground, it's going 32 feet/sec when it > hits the ground, isn't it? Only by macro averaged out approximation. Double-A happy days and... > starry starry nights! -- > Indelibly yours, > Paine Ellsworth P.S.: I think on-stage nudity is disgusting, shameful > and damaging to all things American. But if I > were 22 with a great body, it would be artistic, > tasteful, patriotic and a progressive religious > experience. > Shelley Winters P.P.S.:http://yummycake.secretsgolden.com >http://garden-of-ebooks.blogspot.com >http://painellsworth.net Double-A === Subject: Re: Gravity Flows Energy In Freefall > Matter jumps to speed when dropped. Then it begins to weightlessly > accelerate. This of course always remains finite. There is nothing > more to say. Then the changed momentum is evenly distributed. At the macro level, > acceleration appears smooth. Double-A jump to a higher speed by a quantum leap.? At the microscopic level, all objects are like globs of jello: > and transmitting momentum by exchanges of photons. That would be quite profound if you defined photon. See QED - Feynman. No, I'm not asking Feynman, he's been dead for 20 years. If he were alive I would. YOU answer my question. > http://www.androcles01.pwp.blueyonder.co.uk/AC/spin.gif How does a magnet move another? === Subject: Re: Gravity Flows Energy In Freefall posting-account=eClhIgoAAACIxAUKGXHwqJ9Am4Bt2i0u .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Matter jumps to speed when dropped. Then it begins to weightlessly > accelerate. This of course always remains finite. There is nothing > more to say. Then the changed momentum is evenly distributed. At the macro level, > acceleration appears smooth. Double-A jump to a higher speed by a quantum leap.? At the microscopic level, all objects are like globs of jello: > and transmitting momentum by exchanges of photons. That would be quite profound if you defined photon. See QED - Feynman. > No, I'm not asking Feynman, he's been dead for 20 years. > If he were alive I would. YOU answer my question. Double-A http://www.androcles01.pwp.blueyonder.co.uk/AC/spin.gif How does a magnet move another? === Subject: Re: Gravity Flows Energy In Freefall > Matter jumps to speed when dropped. Then it begins to weightlessly > accelerate. This of course always remains finite. There is nothing > more to say. leap. > Then the changed momentum is evenly distributed. At the macro level, > acceleration appears smooth. Double-A jump to a higher speed by a quantum leap.? At the microscopic level, all objects are like globs of jello: > and transmitting momentum by exchanges of photons. That would be quite profound if you defined photon. See QED - Feynman. > No, I'm not asking Feynman, he's been dead for 20 years. > If he were alive I would. YOU answer my question. Double-A That would be quite profound if you defined photon. You should take a course in English reading comprehension. Let me try again with very simple language and slightly different words. How the does a magnet move another, you drooling moron? >http://www.androcles01.pwp.blueyonder.co.uk/AC/spin.gif How does a magnet move another? === Subject: Re: Gravity Flows Energy In Freefall <39i%k.2949$Fx1.2551@newsfe30.ams2> posting-account=eClhIgoAAACIxAUKGXHwqJ9Am4Bt2i0u .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Matter jumps to speed when dropped. Then it begins to weightlessly > accelerate. This of course always remains finite. There is nothing > more to say. leap. > Then the changed momentum is evenly distributed. At the macro level, > acceleration appears smooth. Double-A jump to a higher speed by a quantum leap.? At the microscopic level, all objects are like globs of jello: > and transmitting momentum by exchanges of photons. That would be quite profound if you defined photon. See QED - Feynman. > No, I'm not asking Feynman, he's been dead for 20 years. > If he were alive I would. YOU answer my question. > Double-A That would be quite profound if you defined photon. You should take a course in English reading comprehension. > Let me try again with very simple language and slightly different words. > How the does a magnet move another, you drooling moron? http://www.androcles01.pwp.blueyonder.co.uk/AC/spin.gif How does a magnet move another? Photons. Double-A === Subject: Re: Gravity Flows Energy In Freefall > Matter jumps to speed when dropped. Then it begins to > weightlessly > accelerate. This of course always remains finite. There is > nothing > more to say. leap. > Then the changed momentum is evenly distributed. At the macro > level, > acceleration appears smooth. Double-A jump to a higher speed by a quantum leap.? At the microscopic level, all objects are like globs of jello: > and transmitting momentum by exchanges of photons. That would be quite profound if you defined photon. See QED - Feynman. > No, I'm not asking Feynman, he's been dead for 20 years. > If he were alive I would. YOU answer my question. > Double-A That would be quite profound if you defined photon. You should take a course in English reading comprehension. > Let me try again with very simple language and slightly different words. > How the does a magnet move another, you drooling moron? http://www.androcles01.pwp.blueyonder.co.uk/AC/spin.gif How does a magnet move another? Photons. Double-A Enough! *plonk* === Subject: Re: Gravity Flows Energy In Freefall posting-account=PSw72goAAADTugQpI3-NCbn5m3k-KKO3 .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) > Matter jumps to speed when dropped. Then it begins to weightlessly > accelerate. This of course always remains finite. There is nothing > more to say. What do you mean by Matter jumps to speed when > dropped.? If i hold out my arm and drop a rock, and one second > later it hits the ground, it's going 32 feet/sec when it > hits the ground, isn't it? No. It is less than a second though. It starts off instantaneously at a speed close to 15 feet persecond. happy days and... > starry starry nights! -- > Indelibly yours, > Paine Ellsworth P.S.: I think on-stage nudity is disgusting, shameful > and damaging to all things American. But if I > were 22 with a great body, it would be artistic, > tasteful, patriotic and a progressive religious > experience. > Shelley Winters P.P.S.: http://yummycake.secretsgolden.com > http://garden-of-ebooks.blogspot.com > http://painellsworth.net === Subject: Re: Is max(x,y) f separable? > On 7 d=E9c, 03:47, Robert Israel a Good morning on this site, Let us consider a smooth =3DA0 fonction g(x,y) on R^2 =3DA0and > a continuous function f =3DA0which =3DA0separates > variables =3DA0such as : f(g(x,y) =3D3D m(x) + h(y) =3DA0, > m and h =3DA0both real smooth functions. In this case f(g(x,y) =3D3D m(x) + m(y) =3DA0, > =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 > ==3DA0g(= > x,y) =3D3D max(x,y) > =3D=3D3Dmax(y=3D > ,x) =3DA0 =3DA0, > I'm not quite sure, but I think you're asking whether there > exist functions f and m such that f(max(x,y)) =3D3D m(x) + m(y). > The answer is that all solutions are trivial > (m and f are constant). =3DA0Suppose a < b. f(b) =3D3D f(max(a,b)) =3D3D m(a) + m(b) > f(b) =3D3D f(max(b,b)) =3D3D m(b) + m(b) > so m(a) =3D3D m(b). > -- > Robert Israel =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 > =3DA0isr...@math.MyUniversitysInitial=3D > s.ca > Department of Mathematics =3DA0 =3DA0 =3DA0 > =3DA0http://www.math.ubc.= > ca/~israel > University of British Columbia =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 > =3DA0Van= > couver, BC, > Cana=3D > da Bonjour Robert, Since they do not depend so much upon the values of their > variables ,I am not very interested about constant functions. > I am working on separation problems , > Ex: g(x,y) =3D3D (x -cos(y)) / (x + cos(y)) > f(g(x,y)) =3D3D m(x) + h(y) , > And searching harder cases , more elaborate too, I believe there are links towards implicit functions, > functional equations , iteration ... More generally, if f(g(x,y)) =3D m(x) + h(y), then > m(x) =3D m(0) + f(g(x,y)) - f(g(0,y)) (which must not depend on y) > h(y) =3D h(0) + f(g(x,y)) - f(g(x,0)) (which must not depend on x) > f(g(x,y)) =3D =A0f(g(x,0)) + f(g(0,y)) - f(g(0,0)) In your example above, the last equation becomes f((x-cos(y))/(x+cos(y))) =3D f((x-1)/(x+1)) Exercise: this implies that f is constant. > -- > Robert Israel =A0 =A0 =A0 =A0 =A0 =A0 > =A0isr...@math.MyUniversitysInitial= > s.ca > Department of Mathematics =A0 =A0 =A0 =A0http://www.math.ubc.ca/~israel > University of British Columbia =A0 =A0 =A0 =A0 =A0 =A0Vancouver, BC, > Cana= > da- Masquer le texte des messages pr=E9c=E9dents - - Afficher le texte des messages pr=E9c=E9dents - > f((x-cos(y))/(x+cos(y))) =3D f((x-1)/(x+1)) > implies that f is constant. > For x =3Dcos(y) <> -1 , f(0) =3D f((cos(y) -1)/(cos(y)+1)) > So f doesn't depend upon its argument. The case g(x,y) =3D (x -cos(y)) / (x + cos(y)) has got a > separation function f(u)=3Dln((u-1)/(u+1)) > Or ln((g(x,y)-1)/(g(x,y) +1)) =3D ln(cos(y)) - ln(x) , > Well, the intervals of definition of g and f(g) must > be considered and work out, Yes, I should have been more careful about domains. Suppose you have intervals (a,b) and (c,d) such that g(x,y), f(g(x,y)), m(x) and h(y) are all defined for x in (a,b) and y in (c,d) and f(g(x,y)) = m(x) + h(y) there. Take p in [a,b] and q in [c,d]. Then m(x) = m(p) + f(g(x,y)) - f(g(p,y)) h(y) = h(q) + f(g(x,y)) - f(g(x,q)) f(g(x,y)) = f(g(p,y)) + f(g(x,q)) - f(g(p,q)) Now x = 0 is a very special case in your example because g(0,y) = -1 for all y, so the intervals must be chosen to exclude that. And indeed f(u) = ln((1-u)/(u+1)) is undefined at u=-1. -- === Subject: Re: Is max(x,y) f separable? posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) On 7 d.8ec, 23:51, Robert Israel a Good morning on this site, Let us consider a smooth =3DA0 fonction g(x,y) on R^2 =3DA0and > a continuous function f =3DA0which =3DA0separates > variables =3DA0such as : f(g(x,y) =3D3D m(x) + h(y) =3DA0, > m and h =3DA0both real smooth functions. In this case f(g(x,y) =3D3D m(x) + m(y) =3DA0, > =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 > ==3DA0g(= > x,y) =3D3D max(x,y) > =3D=3D3Dmax(y=3D > ,x) =3DA0 =3DA0, > I'm not quite sure, but I think you're asking whether there > exist functions f and m such that f(max(x,y)) =3D3D m(x) + m(y). > The answer is that all solutions are trivial > (m and f are constant). =3DA0Suppose a < b. f(b) =3D3D f(max(a,b)) =3D3D m(a) + m(b) > f(b) =3D3D f(max(b,b)) =3D3D m(b) + m(b) > so m(a) =3D3D m(b). > -- > Robert Israel =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 > =3DA0isr...@math.MyUniversitysInitial=3D > s.ca > Department of Mathematics =3DA0 =3DA0 =3DA0 > =3DA0http://www.math.ubc.= > ca/~israel > University of British Columbia =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 > =3DA0Van= > couver, BC, > Cana=3D > da Bonjour Robert, Since they do not depend so much upon the values of their > variables ,I am not very interested about constant functions. > I am working on separation problems , > Ex: g(x,y) =3D3D (x -cos(y)) / (x + cos(y)) > f(g(x,y)) =3D3D m(x) + h(y) , > And searching harder cases , more elaborate too, I believe there are links towards implicit functions, > functional equations , iteration ... More generally, if f(g(x,y)) =3D m(x) + h(y), then > m(x) =3D m(0) + f(g(x,y)) - f(g(0,y)) (which must not depend on y) > h(y) =3D h(0) + f(g(x,y)) - f(g(x,0)) (which must not depend on x) > f(g(x,y)) =3D =A0f(g(x,0)) + f(g(0,y)) - f(g(0,0)) In your example above, the last equation becomes f((x-cos(y))/(x+cos(y))) =3D f((x-1)/(x+1)) Exercise: this implies that f is constant. > -- > Robert Israel =A0 =A0 =A0 =A0 =A0 =A0 > =A0isr...@math.MyUniversitysInitial= > s.ca > Department of Mathematics =A0 =A0 =A0 =A0http://www.math.ubc.ca/~israel > University of British Columbia =A0 =A0 =A0 =A0 =A0 =A0Vancouver, BC, > Cana= > da- Masquer le texte des messages pr=E9c=E9dents - - Afficher le texte des messages pr=E9c=E9dents - > f((x-cos(y))/(x+cos(y))) =3D f((x-1)/(x+1)) > implies that f is constant. > For x =3Dcos(y) <> -1 , f(0) =3D f((cos(y) -1)/(cos(y)+1)) > So f doesn't depend upon its argument. The case g(x,y) =3D (x -cos(y)) / (x + cos(y)) has got a > separation function f(u)=3Dln((u-1)/(u+1)) > Or ln((g(x,y)-1)/(g(x,y) +1)) =3D ln(cos(y)) - ln(x) , > Well, the intervals of definition of g and f(g) must > be considered and work out, Yes, I should have been more careful about domains. > Suppose you have intervals (a,b) and (c,d) such that > g(x,y), f(g(x,y)), m(x) and h(y) are all defined for > x in (a,b) and y in (c,d) > and f(g(x,y)) = m(x) + h(y) there. Take p in [a,b] and q in [c,d]. > Then m(x) = m(p) + f(g(x,y)) - f(g(p,y)) > h(y) = h(q) + f(g(x,y)) - f(g(x,q)) > f(g(x,y)) = f(g(p,y)) + f(g(x,q)) - f(g(p,q)) Now x = 0 is a very special case in your example > because g(0,y) = -1 for all y, so the intervals must be chosen > to exclude that. And indeed f(u) = ln((1-u)/(u+1)) is undefined > at u=-1. > -- > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada- Masquer le texte des messages pr.8ec.8edents - - Afficher le texte des messages pr.8ec.8edents - I've done mistakes too: ln((1 -g(x,y))/(1+g(x,y))) = ln|cos(y)| - ln|x| . With separation of variables I tried to understand unexplained tricks of expert mathematicians. I do believe there a link between separable functions and iteration, Amicalement, Alain === Subject: Country Genealogy & Family History Studies: posting-account=6qVsEAkAAAAP4-CBBRBdYxtxzWaP0v1H 1.1.4322; .NET CLR 2.0.50727; yplus 5.1.04b),gzip(gfe),gzip(gfe) Country Genealogy & Family History Studies: Comprehensive Indexing of Family Genealogy & History Internet Education Directory BEGIN at: Ancestor Roots Information: A-Z http://www.academic-genealogy.com/ancestorrootsinformationdatabases.htm sample country is repeated for every other country in the world, throughout the Ancestor Roots Information site. Every available site in the world has been placed online, with information related to civil records, culture, genealogy, libraries, museums, printed country family history outlines, country studies groups, individual identity information, indigenous peoples, etc. Country Selected: ITALY, at: http://www.academic-genealogy.com/ancestorrootsinformationdatabases.htm#Ital y First Link goes to ITALY, at: http://www.academic-genealogy.com/regionalgenealogy.htm#Italy (2) Second entry is country profile - Italia (3) Third entry is Wikipedia Portal:Italy (4) Fourth entry is Library of Congress Portals to the World: Italy (5) Fifth entry is Genealogy and Population: sites on ITALY (6) Sixth entry is: General sites, where available, re: ITALY Second Link goes to ITALY: U.S. Department of State Country Documents http://travel.state.gov/visa/frvi/reciprocity/reciprocity_3589.html#docs Third Link goes to ITALY research, that Includes hundreds of published sources. http://www.encyclopedia.com/topic/Italy.aspx Fourth Link goes to ITALY Encyclopedia reference, with data on Libraries & Museums. http://www.encyclopedia.com/doc/1G2-2586700275.html#LIBRARIES_AND_MUSEUMS Fifth Link goes to ITALY Wikipedia Universities. http://www.academic-genealogy.com/schoolscollegesuniversities.htm#ITALY Reference, or other related education resource, enhanced with links and references to top 500 ranked universities (& sub links) worldwide. 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An Introduction. http://academic-genealogy.com/documentfilefamilyhistory.htm News, Media and Travel http://academic-genealogy.com/newsmediatravel.htm Worldwide information for genealogists, family historians and other research specialists involved in travel, tours and accommodations. Regional Genealogy and Local History Research: Local History & Genealogy Portals to the World: http://academic-genealogy.com/regionalgenealogy.htm areas, countries, directories, ethnic group populations, organizations, local ancestry & local history studies. SEARCH DIRECTORY: using Google - Pico - Yahoo!. http://academic-genealogy.com/searchthissite.htm What's New in Genealogy & Family History Resources? http://academic-genealogy.com/whatsnewingenealogyfamilyhistory.htm MISSION STATEMENT: To promote scholarly educational access to all key worldwide Internet genealogical and family history databases and resources: scholarly mega site map of world wide Internet resources. === Subject: Re: JSH: Normalization issue? posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081029 Firefox/2.0.0.18,gzip(gfe),gzip(gfe) [...] Questions. Is 7 there in 1? No, Why not? 1/7 * 7 = 1 so why can't 7 be there in 1? - William Hughes If 7x = 14, then x=2. Get it yet? I gave that example in my previous reply as well, and you > deleted it out. With 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) where b_1(x) = a_1(x) and b_2(x) = a_2(x) + 1, which is the normalization, where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0 you can verify that the 7 in 5b_1(x) + 7 is actually from the 7 on the > left hand side, so you have that 7*1= 7. My point is that if the 2 were multiplied by 7--even on the complex > plane--or by ANY factors in common with 7, then it wouldn't be 2, now > would it? multiplied by 7 either. Because if it were multiplied by 7, then it > wouldn't be 7, now would it? It would be 49. On the other hand you > could multiply 1 by 7 to get 7. Just like you could multiply 2/7 by 7 > to get 2. So how do you conclude that the 7 in 5b_1 + 7 is from the 7 > on the other side, but the 2 in 5b_2 + 2 is not? By the distributive property verified at x=0. If a*(f(x) + b) = a*f(x) + a*b, and f(0)=0, doesn't the DP still work? > It works just fine. Notice that in the distributive property, there is just one factor on the outside of the parenthesis. But you want to use it to factor 7 out of (5b_1(x) + 7)*(5b_2(x) + 2). There are two parentheses here, not just 1 as in the DP. So you want to write 7*(5c_1(x) + 1)*(5b_2(x) + 2) as your proposed way of factoring out 7, where c_1(x) = b_1(x)/7. Not a problem in the complex plane. But suppose 7 is written as 7 = A * B. Then you could also write 7 * (5b_1(x)/A + 7/A)*(5b_2(x)/B + 7/B). Which is also perfectly valid in the complex plane. No problem so far. But you want to make a statement about divisibility of b_1(x) = a_1(x) by 7 in other rings. So you say: OK, a_1(0) = 0, and therefore even if I factor 7 as 7 = A * B, I must have that A = 7, and then of course B = 1. And this must hold for all x, since A and B are constants. But there is where you make a hidden assumption: that A and B are constants. What if they are functions of x ? Say, A(x) and B(x). Of course you can still say A(0) = 7. But what about other values of x? What forces A(x) to ALWAYS equal 7? Not the distributive property. It just applies to single instances. My example shows explicitly, that A(1) is not divisible by 7 and B(1) is not coprime to 7 in the ring of algebraic integers. You say, this cannot be! I have a proof that it cannot be! But then there is just this little problem. I gave A(1) and B(1) explicitly. You can check that they work. It is just a matter of arithmetic. It is just like 2 + 2 = 4. So what you are saying is: 2 + 2 is not equal to 4. You have a proof. Can't be true that 2 + 2 = 4. But it is. Just arithmetic. You keep running away from it. The problem is not the distributive law. That's fine. The problem is, assuming that A(x) and B(x) are constant functions. They're not. You have pulled the wool over your own eyes. And you cannot face up to it! > Now with 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) x=0, reveals that 7(2) = (5(0) + 7)(5(0) + 2) = (7)(2) and you expect me or anyone else to believe that you *cannot* tell > that the 7 distributed through one factor? > Looks just fine for x = 0. But I showed that for x = 1, it is not right. Just plain arithmetic. You cannot refute it. All this mindless repetition of your old argument, without facing up to the arithmetic, is getting you nowhere. The arithmetic says you are wrong. The arithmetic doesn't care. It just tromps you in the dirt like a dung beetle. > To make it explicit where nothing is hidden by functions I use the > example: 7(x^2 + 3x + 2) = (7x + 7)(x+2) where the only change is the mathematical beastie as you have a > linear function now, which reveals the 7, in such a way that you can't > debate but you don't want to debate that anyway because you like that > result, right? > A simplistic analogy which has nothing to do with your situation. First, you are not dealing with a polynomial with constant coefficients. As you love to say, you are dealing with nonpolynomial factorization. In your real problem, a_1(x) and a_2(x) are not integers, are not even constants like in your baby example above. Secondly, your polynomial is actually a polynomial in the number 5. You introduced this years ago and it has confused you over and over again a hundred times. But that fact is your basis for showing that a_1(x) and a_2(x) are roots of an associated polynomial, a^2 - (7x - 1) + (49x^2 - 14x). You have probably long since forgotten this. You would be better off in terms of clarity if you replace 5 with y, and thought of your function as a polynomial in y. > The math you like you do not want argue about. But you do NOT like the result that the 7 cannot get a mind of its own > with 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2). > Get a mind of its own ??? Is that supposed be a mathematical argument? That numbers have minds of their own? What you are really saying here is: All I understand is factoring by inspection. I cannot possibly conceive of anything else. Therefore that is the only possible way to factor. > If I am wrong, give an example ***in the complex plane*** where when > x=0 shows one thing with a factor distributing, x=1 or some other > value shows something else. > The example I gave is in the ring of algebraic integers. But that ring is a subring of the complex plane. Therefore my example is also an example in the complex plane. So that is exactly what you asked for. But I bet you are not happy with it. It is NOT SUFFICIENT for you to consider only the complex plane, not if you want to conclude that one of a_1(x) and a_2(x) is divisible by 7, while the other has no factors in common with 7. In the complex plane, 7 divides every- thing. Further, if you COULD get the conclusion you want regarding 7 in the complex plane, you would also get it for every other nonzero number. All numbers (except 0) are factors of all other numbers in the complex plane. All you can get out of that is a triviality. Once more: through all of the last several days you have FAILED to show anything wrong with my example. It's arithmetic. It contradicts your claims. It proves you are wrong. There is a flaw in your argument. I have not only found a counterexample to your conclusion (which is sufficient to show your argument must be wrong) but I have also identified where your logic breaks down. OK, I concede that you are incapable of understanding the logic part. But you are NOT incapable of arithmetic, and that's sufficient. > You keep arguing because you do not like the mathematical result that > is true. It's like claiming there is no sun because you don't like sunshine. > Shoe. Other. Foot. Deal with it. Marcus. > James Harris === Subject: Re: JSH: Normalization issue? posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > If I am wrong, give an example ***in the complex plane*** where when > x=0 shows one thing with a factor distributing, x=1 or some other > value shows something else. f(x) = 7^x g(x) = 7^(1-x) f(x)*g(x)=7 For x=0, 7 divides only the second factor. For x=1/2, 7 does not divide either factor but sqrt(7) does. - William Hughes === Subject: Re: JSH: Normalization issue? posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > If I am wrong, give an example ***in the complex plane*** where when > x=0 shows one thing with a factor distributing, x=1 or some other > value shows something else. f(x) = 7^x > g(x) = 7^(1-x) f(x)*g(x)=7 There is no distribution here. Do you remember that thing called the > distributive property? Apparently you prefer the normalized form. h(x) = (7^x -1) i(x) =(7^(1-x) -7) f(x) = h(x) + 1 g(x) = i(x) + 7 f(x)*g(x)=7 For x=0, 7 divides only the second factor. For x=1/2, 7 does not divide either factor but sqrt(7) does. - William Hughes === Subject: Re: JSH: Normalization issue? posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > If I am wrong, give an example ***in the complex plane*** where when > x=0 shows one thing with a factor distributing, x=1 or some other > value shows something else. f(x) = 7^x > g(x) = 7^(1-x) f(x)*g(x)=7 There is no distribution here. Do you remember that thing called the > distributive property? Apparently you prefer the normalized form. h(x) = (7^x -1) > i(x) =(7^(1-x) -7) f(x) = h(x) + 1 > g(x) = i(x) + 7 f(x)*g(x)=7 For x=0, 7 divides only the second factor. For x=1/2, 7 does not > divide either factor but sqrt(7) does. There is no 7 on the left hand side!!! Making the 7 on the LHS explicit h(x) = (7^x -1) j(x) =(7^(1-x) -7)/7 f(x) = h(x) + 1 g(x) = i(x) + 7 7 * 1 = 7 * (h(x) + 1) * (j(x) + 1) At x=0 (h(x) + 1) does not contain factors of 7. At x=1 it does. -William Hughes === Subject: Re: JSH: Normalization issue? posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > If you know that 7 is there in 14, > then you know it was multiplied by 2. Questions. Is 7 there in 1? No, Why not? 1/7 * 7 = 1 so why can't 7 be there in 1? - William Hughes If 7x = 14, then x=2. So 7 can be there in 14. > If 7x = 1 then x=1/7. So why can't 7 be there in 1? - William Hughes If you divide 1 by 7 then you have 1/7. Please answer the question: Why can't 7 be there in 1? - William Hughes 7 can have multiplied to produce 1 as a product if it multiplied times > 1/7. e.g. 7(1/7) = 1. Understand yet? Nope. You have said exactly nothing about why 7 can't be there in 1. In fact from what you have said we would conclude 7 can be there in 1. Please answer the question Why can't 7 be there in 1? > NOTICE that if you know that 1 is the product and 7 > was multiplied times *something* that you know that something was > 1/7. Now then I answered your question, go back in context to the problem > at hand, talk from the equations for my special construction and > explain from there ***in the complex plane*** please. In the complex plane you have the factor (5b_2(x) + 2). This is a multiple of 7, in particular 7 * (5b_2(x)/7 + 2/7). 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Can we infer the summation of the principle eigenvector's entries? I have obtained a lower bound for sum(v)/sqrt(L), but it seems too loose. This bound is: B1=sqrt(L)*min(sum(A))/lamda, where lamda is the largest eigenvalue of A, which can be inferred by some matrix norms somewhat tightly. However, B1 is very loose compared with the numerical result. For an example, when L=5, B1 is approximately 0.4, while sum(v)/sqrt(L) is 0.9944! Can we obtain a tighter lower-bound for sum(v)/sqrt(L)? === Subject: Re: Direct sum of finite Noetherian A-modules posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.4) Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > The direct sum of finite Noetherian A-modules is a Noetherian A- > module, according to my notes. Is this true for any Noetherian A- > modules or is there some sort of restriction, such as the A-modules > giving an ascending chain? I.e. assume we are considering the direct > sum B + C + D, must we have that C is a Noetherian A-submodule of B > and that D is a Noetherian A-submodule of C, or can these be picked in > any way? for two modules n1, n2 if b = n1 (+) n2 then b / (n1 (+) 0) iso= n2 so looking at the direct sum is basically asking if given two modules b and c/b are noetherian show c is noetherian one way to show this is to take an ascending chain of submodules of c call them c1 <= c2 <= ... corresponding to this chain there are ascending chains in b and c/b c1 / b <= c2 / b <= ... and (c1 + b)/b <= (c2 + b)/b <= ... for each of these (which are chains in noetherian modules) there is some m at which they become constant so in particular (c m + b)/b = (c j + b)/b forAll j >= m or c m + b = c j + b forAll j >= m so c j = c j / (c j + b) = c j / (c m + b) = c m + (c j / b) = c m + (c m / b) = c m and so c j also eventually becomes finite and c is noetherian too now you have it for the direct sum of two you can do it for the direct sum of any finite number -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: bessel function asymptotics In sneddon's Fourier Transforms p 58 he examines the asymptotic representation for I = h/(p^2-m^2) [ p*J_m(m*h)J_p+1(p*h) - m*J_p(p*h)*J_m+1(m*h) - 1/(p+m)*J_m(m*h)*J_p(p*h) (1) which he states is I = 2/(pi*sqrt[m*p] ) * { sin[(p-m)*(h-pi/2)]/(p-m) - sin[(p+m)*(h-pi/2)]/(p+m) } + P/h (2) where P remains finite as h->oo. I don't see how he gets this however: I start with the fact that J_n(z) ~ sqrt[2/(pi*z)] * cos[ z - pi/4 -pi*n/2 ] so (1) becomes I = h/(p^2-m^2) { p*sqrt[2/(pi*m*h)] * cos[ m*h- pi/4 -pi*m/2 ]*sqrt[2/(pi*p*h)] * cos[ p*h - pi/4 -pi*(p+1)/2 ] - m* sqrt[2/(pi*p*h)] * cos[ p*h - pi/4 -pi*p/2 ]* sqrt[2/(pi*m*h)] * cos[ m*h - pi/4 -pi*(m+1)/2 ] } + P/h I = 2/(pi*(p^2-m^2)*sqrt[m*p]) * { p* cos[ m*h- pi/4 -pi*m/2 ]* sin[ p*h - pi/4 -pi*p/2 ] - m*cos[ p*h - pi/4 -pi*p/2 ]*sin[ m*h - pi/4 -pi*m/2 ] } + P/h Now p*sin[u]*cos[v]-m*cos[u]*sin[v] = 1/2*{ (p+m)*sin[u-v] - (p-m)*sin[u+v] here u= p*h - pi/4 -pi*p/2 and v = m*h- pi/4 -pi*m/2 so i get I = 1/(pi*(p^2-m^2)*sqrt[m*p]) * { (p+m)*sin[(p-m)*(h-pi/2)] -(p-m)*sin[(p+m)*(h-pi/2)-pi/2] } So I = 1/(pi*sqrt[m*p]) * { sin[(p-m)*(h-pi/2)]/(p-m) + cos[(p+m)*(h-pi/2)] / (p+m)} (3) which is almost (2) except the last sin is a cos. He then goes on to use (2) in some dirichlet type integrals proving a particular inverse transform theorem. Given his form in (2) his uses Dirichlet type integrals, namley 1/2(f(x+0) + f(x-0)) = lim (h->oo) 1/pi int(-oo,oo) f(t) sin(h*(t-x))/(t-x) dt But wouldn't this be screwed due to the fact that the last sin in (2) should be a cos i.e. (3) - unless of course i have fudged it completely. does lim (h->oo) 1/pi int(-oo,oo) f(t) cos(h*(t+x))/(t+x) dt have any meaning? I mean by the lebesgue-riemann lemma it should be zero except possibly for t=-x where the integrand is divergent. === Subject: Re: Putnam 2008 QUESTIONS posting-account=wNa2rQkAAADD36N8RX63z_c2LsxPan45 Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) Hey Genius: Please be more careful about posting Putnam problems before all people have taken the exam. While most people took the 2008 test on Saturday, some did take it on Sunday, which means they could have had early access to the problems. === Subject: Math Forum. posting-account=HJnNYQoAAACtdMyfjpc6YoWX6U0epxWM Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) Is there a place where I can discuss higher level maths (Set theory, Recursion theory, Model theory, Proof theory) that ISN'T usenet? === Subject: Re: Math Forum. > Is there a place where I can discuss higher level maths (Set theory, > Recursion theory, Model theory, Proof theory) that ISN'T usenet? Google Groups. You should have added a smiley... Google groups is the worse Web interface to most USENET groups, but has also other specific groups, inside Google. (good luck in scanning the 483 math groups in English language...) In fact these are not discussion groups, but merely communities. Most of them don't talk directly about math. Try also Yahoo Groups ? -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Re: Math Forum. posting-account=HJnNYQoAAACtdMyfjpc6YoWX6U0epxWM Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > Is there a place where I can discuss higher level maths (Set theory, > Recursion theory, Model theory, Proof theory) that ISN'T usenet? Google Groups. You should have added a smiley... Google groups is the worse Web interface to most USENET groups, > but has also other specific groups, inside Google. > (good luck in scanning the 483 math groups in English language...) > In fact these are not discussion groups, but merely communities. > Most of them don't talk directly about math. Try also Yahoo Groups ? > Philippe Ch., mail : chephip+n...@free.fr > site :http://mathafou.free.fr/ (recreational mathematics) It looks good, but ultimately I'm looking for people into things in mathematical logic such as matroid theory, knot theory, and of course the Big 4. === Subject: Re: Math Forum. --------------------------------------------------------------------- > Is there a place where I can discuss higher level maths (Set theory, > Recursion theory, Model theory, Proof theory) that ISN'T usenet? Google Groups. You should have added a smiley... Google groups is the worse Web interface to most USENET groups, > but has also other specific groups, inside Google. > (good luck in scanning the 483 math groups in English language...) > In fact these are not discussion groups, but merely communities. > Most of them don't talk directly about math. Try also Yahoo Groups ? > Is there a place where I can discuss higher level maths (Set theory, > Recursion theory, Model theory, Proof theory) that ISN'T usenet? A University math department! === Subject: Re: Math Forum. posting-account=HJnNYQoAAACtdMyfjpc6YoWX6U0epxWM Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) Is there a place where I can discuss higher level maths (Set theory, > Recursion theory, Model theory, Proof theory) that ISN'T usenet? A University math department! I need something that's free! === Subject: An asteroid ring posting-account=PSw72goAAADTugQpI3-NCbn5m3k-KKO3 .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) Why didn't it become a planet? Why isn't there a partially formed planet there at that orbital interval? Gravitational order designed the solar system. Mitch Raemsch === Subject: Re: An asteroid ring posting-account=qfTKWQoAAADrbu8lwB1KpYv95yNmrUOu Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > Why didn't it become a planet? Why isn't there a partially formed > planet there at that orbital interval? Gravitational order designed the solar system. Mitch Raemsch The fifth planet was destroyed, the Asteroid belt is an ancient planet that exploded long time ago... In the Bible is written that Satan did it to prevent the birth of the Messiah... Also is written in the genesis, Cain attacked his brother Abel and killed him and God installed a belt above our heads... now our Asteroid Belt. Let say that planets are born out solar eruptions ... but that is another tail.. === Subject: Re: An asteroid ring > Why didn't it become a planet? Why isn't there a partially formed > planet there at that orbital interval? Gravitational order designed the solar system. Mitch Raemsch The fifth planet was destroyed, the Asteroid belt is an ancient planet > that exploded long time ago... How do you know there was a planet? > In the Bible is written that Satan did > it to prevent the birth of the Messiah... Also is written in the > genesis, Cain attacked his brother Abel and killed him and God > installed a belt above our heads... now our Asteroid Belt. Hah! > Let say that planets are born out solar eruptions ... but that is > another tail.. A foolish one. -- Timberwoof http://www.timberwoof.com Most of the universe is extremely hostile to life as we know it. It seems obvious that it was all designed by some creature that hates life... And here you are, trying to attract its attention. === Subject: Re: An asteroid ring posting-account=qfTKWQoAAADrbu8lwB1KpYv95yNmrUOu Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) Why didn't it become a planet? Why isn't there a partially formed > planet there at that orbital interval? Gravitational order designed the solar system. Mitch Raemsch The fifth planet was destroyed, the Asteroid belt is an ancient planet > that exploded long time ago... How do you know there was a planet? > Asteroid Belt is the evidence which once existed as a planet similar to your own. > In the Bible is written that Satan did > it to prevent the birth of the Messiah... Also is written in the > genesis, Cain attacked his brother Abel and killed him and God > installed a belt above our heads... now our Asteroid Belt. Hah! > Christmas with a Capital C http://www.youtube.com/watch?v=IAckfn8yiAQ > Let say that planets are born out solar eruptions ... but that is > another tail.. A foolish one. -- The Sun birthed our Earth, just like all planets... Not so foolish as you may think... The Truth is that Planets are born out of solar eruptions and become Sun on their turn if not destroyed by Mad man. === Subject: Re: An asteroid ring Why didn't it become a planet? Why isn't there a partially formed > planet there at that orbital interval? Gravitational order designed the solar system. Mitch Raemsch The fifth planet was destroyed, the Asteroid belt is an ancient planet > that exploded long time ago... How do ?you know there was a planet? > Asteroid Belt is the evidence which once existed as a planet similar > to your own. Bzzzt. Your 15 minutes are up. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: An asteroid ring > Why didn't it become a planet? Why isn't there a partially formed > planet there at that orbital interval? Gravitational order designed the solar system. Mitch Raemsch The fifth planet was destroyed, the Asteroid belt is an ancient planet that exploded long time ago... In the Bible is written that Satan did it to prevent the birth of the Messiah... Also is written in the genesis, Cain attacked his brother Abel and killed him and God installed a belt above our heads... now our Asteroid Belt. Let say that planets are born out solar eruptions ... but that is another tail.. ------ Actually god created the asteroid belt to pummel every person that that believes in the bible. There is one asteroid for each person and on an unknown date they will all rain down on earth. God created a bible to seperate out the morons from the ones that can think for themselfs. God knows the only true religion is mathematics because he created it and made the universe follow it. It knows that any intelligent creature would be able to understand that and he made many false gods to lead ignorant humans astray. He knows that only intelligence matters. He is not a forgiving god as the bible says. He shows us this every day by giving us hints incase we are teatering on the edge of stupidity. === Subject: Re: An asteroid ring > (snip) God created a bible to seperate out the morons from the ones that can think >for themselfs. God knows the only true religion is mathematics because he >created it and made the universe follow it. It knows that any intelligent >creature would be able to understand that and he made many false gods to >lead ignorant humans astray. He knows that only intelligence matters. He is >not a forgiving god as the bible says. He shows us this every day by giving >us hints incase we are teatering on the edge of stupidity. > Jon, God presented the information that has been documented as the Bible, in a format that all people of all ages could interpret, equally. That is, there is no advantage to any society or person because of increased knowledge or understanding of the universe. Those Biblical allegories, parables, similes, etc., are easily interpreted, then accepted or rejected, by all persons of all times. No one has or had an advantage. God does not present Himself to us with overwhelming, obvious presence that no one could reject. Doing so would destroy any form of individual sovereignty. The balance here is set up such that anyone can make their own choices and go their own way. God allowed Lucifer/Satan to run his agenda, and, in the process, expose all of us to the long-term effects. It was/is God's intention that we will learn enough during this mortal phase of our existence to insure that none of us will ever want to go back and further explore those ideas that intrigued Lucifer/Satan. This on-going school of hard knocks is the only way we could have learned enough about this. Once we've been granted immortality and absolute sovereignty, we will do God's will with absolute reliability, not by reason of being constrained or intimidated into compliance. We will do God's will because our individual absolute sovereign wills shall then be exactly and precisely the same as God's will. Those false gods you mention are an example of Lucifer/Satan's on-going project and we are expected to observe and learn from these exhibits of Lucifer/Satan's program. Can you, or can anyone outline a process by which we could have been brought to a level of competence before God, such that He could justify granting us immortality and absolute sovereignty? Restraining us as pre-programmed puppets would not have resulted in any REAL level of sovereignty. On the other hand, absolute sovereignty without a firm, thorough, hands-on experience level of understanding of those things God wants us to turn away from would surely have led to many, or maybe all of us falling back into this mess, the same way Lucifer did originally. Gordon === Subject: Re: An asteroid ring (snip) God created a bible to seperate out the morons from the ones that can >think >for themselfs. God knows the only true religion is mathematics because he >created it and made the universe follow it. It knows that any intelligent >creature would be able to understand that and he made many false gods to >lead ignorant humans astray. He knows that only intelligence matters. He >is >not a forgiving god as the bible says. He shows us this every day by >giving >us hints incase we are teatering on the edge of stupidity. > Jon, God presented the information that has been documented as the > Bible, in a format that all people of all ages could interpret, > equally. That is, there is no advantage to any society or person > because of increased knowledge or understanding of the universe. Those > Biblical allegories, parables, similes, etc., are easily interpreted, > then accepted or rejected, by all persons of all times. No one has or > had an advantage. > So I guess those people that existed before the bible are out of luck? Or those people where the bible hasn't been translated into? So do you except all the bible or only parts of it that you agree with? If it's the word of god then surely you must except all of it? Oh I get it.... you have to learn from a book about morality and ethics rather than develop your own understanding? You have to be told Thou shall not kill instead of understand that it is good or bad? So what your telling me is that the bible was created by god for idiots that couldn't reason about morality and ethics? Ok, that explains a lot then! === Subject: Re: An asteroid ring > (snip) God created a bible to seperate out the morons from the ones that can >think >for themselfs. God knows the only true religion is mathematics because he >created it and made the universe follow it. It knows that any intelligent >creature would be able to understand that and he made many false gods to >lead ignorant humans astray. He knows that only intelligence matters. He >is >not a forgiving god as the bible says. He shows us this every day by >giving >us hints incase we are teatering on the edge of stupidity. > Jon, God presented the information that has been documented as the > Bible, in a format that all people of all ages could interpret, > equally. That is, there is no advantage to any society or person > because of increased knowledge or understanding of the universe. Those > Biblical allegories, parables, similes, etc., are easily interpreted, > then accepted or rejected, by all persons of all times. No one has or > had an advantage. > > So I guess those people that existed before the bible are out of > luck? Or those people where the bible hasn't been translated into? So do you except all the bible or only parts of it that you agree with? If > it's the word of god then surely you must except all of it? Oh I get it.... you have to learn from a book about morality and ethics > rather than develop your own understanding? You have to be told Thou > shall not kill instead of understand that it is good or bad? So what your telling me is that the bible was created by god for idiots > that couldn't reason about morality and ethics? Ok, that explains a lot > then! Jon, It looks like the Bible is Gods version of the popular Idiots Guide to ... series of books; sort of an Idiots Guide to Christianity. The writer (under his/her pen name God, AKA other names such as The Creator, The Great Spirit, etc.), after much debate with the publisher's marketing staff, decided to follow their advice and write many variations of this book in an effort to capture a greater market share; King James Bible, Mormon Bible - Book of Mormon, etc.. The debate is on going and has lasted many centuries leading to long delays between revisions. God was wise to follow the advice of this most experienced counsel; look at the number of books sold and you can see the marketing genius of the publisher. Some of his/her other works are also available in print but many are only available in closed circles and only available, in full context, via the spoken word - consult your local Medicine Man for details. Don't worry about, ... those people that existed before the bible ..., the Mormons are doing their part to clean up that mess with their Baptism for the dead. Chris === Subject: Re: An asteroid ring > Why didn't it become a planet? Why isn't there a partially formed > planet there at that orbital interval? Gravitational order designed the solar system. Mitch Raemsch It would be very difficult for cool matter to condence and form a solid substance... in fact it would be virtually impossible. The matter must be hot enough to be partially liquified so that the gravitational effects would be well distributed over the material collection. Think of mercury. The substance collects to itself because of surface tension. if you poke it with something it doesn't effect it much and usually stays together. If that material is now salt then poking it or pushing it aside seperates it into groups. So suppose that some collection of small asteroids were grouped together. Any perbutation could easily destroy it(such as another asteroid being attracted to it). But if it is liquified then it is much easier for it to absorb in incoming material. In fact it is almost a necessariy condition. For the matter to condense and grow they must attract other matter and it must be a somewhat stable configuration. If it is hard material, then it is a non-elastic collision and the energy gets distrubted throughout decreasing the stability. If it is elastic then they tend to clump together forming larger pieces and once it reaches a critical size it easily accepts harder material. It might be possible in some rare circumstance for the gravitational forces to be so great that the material is fused together and resulting in a stable state but this is highly unlikely. (of course once one gets a planetoid like structure and it's gravitational attraction is strong enough then it can hold cool material) So, as others have misunderstood, it mainly is depends on temperature. Some of the other sayings might be true in specific instances but not in general. We can surmise that the asteroid belts are left over material from collisions or planet forming that cooled down too much and were to far away from any large structure for significant attraction. Once they reached a critical state the probability of them forming a planet is virtually impossible. Take mercury, if it is too cool then it will not combine with other pieces. But at room temperature you can combine them. That is the basic idea and also has to do with atomic physics, chemestry, pool, etc... (it's basically a manifestation of atomic physics at a very large scale) === Subject: Re: An asteroid ring Why didn't it become a planet? Why isn't there a partially formed > planet there at that orbital interval? Gravitational order designed the solar system. Mitch Raemsch It would be very difficult for cool matter to condence and form a solid > substance... in fact it would be virtually impossible. The matter must be hot enough to be partially liquified so that the > gravitational effects would be well distributed over the material > collection. Think of mercury. The substance collects to itself because of surface > tension. if you poke it with something it doesn't effect it much and > usually stays together. If that material is now salt then poking it or > pushing it aside seperates it into groups. So suppose that some collection of small asteroids were grouped together. > Any perbutation could easily destroy it(such as another asteroid being > attracted to it). But if it is liquified then it is much easier for it to > absorb in incoming material. In fact it is almost a necessariy condition. For the matter to condense > and grow they must attract other matter and it must be a somewhat stable > configuration. If it is hard material, then it is a non-elastic collision > and the energy gets distrubted throughout decreasing the stability. If it > is elastic then they tend to clump together forming larger pieces and > once it reaches a critical size it easily accepts harder material. It might be possible in some rare circumstance for the gravitational > forces to be so great that the material is fused together and resulting in > a stable state but this is highly unlikely. (of course once one gets a > planetoid like structure and it's gravitational attraction is strong > enough then it can hold cool material) So, as others have misunderstood, it mainly is depends on temperature. > Some of the other sayings might be true in specific instances but not in > general. We can surmise that the asteroid belts are left over material from > collisions or planet forming that cooled down too much and were to far > away from any large structure for significant attraction. Once they > reached a critical state the probability of them forming a planet is > virtually impossible. Take mercury, if it is too cool then it will not combine with other > pieces. But at room temperature you can combine them. That is the basic > idea and also has to do with atomic physics, chemestry, pool, etc... (it's > basically a manifestation of atomic physics at a very large scale) The temperature idea you have is sound and makes sense, Jon. And it supports the mainstream idea that Timberwoof already covered, which is the fact that the gas giant, planet Jupiter, continues to this day to jostle the asteroids in the belt, occasionally sending one on a wild ride that has infrequently brought the asteroid a little too close to Earth for comfort. If we note the image and details on the Wiki page... http://en.wikipedia.org/wiki/Asteroid_belt ...we see some interesting facts about how if one of the asteroids enters an orbit around the Sun that is in orbital resonance with Jupiter, the asteroid is subject to the pertubations that can send it flying out of the belt in any one of an infinite number of directions. These resonance areas are called Kirkwood gaps, and there is a graph further down the page that depicts where the gaps are found today. That Wiki page also tells us that more than half the mass of the main belt is contained in the four largest objects: Ceres, 4 Vesta, 2 Pallas, and 10 Hygiea. All of these have mean diameters of more than 400 km, while Ceres, the main belt's only dwarf planet, is about 950 km in diameter. The remaining bodies range down to the size material since the Solar system formed that there's only about enough mass left to form an object about 0.1% of the Earth's mass (1/1,000th). Most people seem to harbor the idea that the asteroid belt is densely packed with rocks and dust. However if we were able to stand on just about any given asteroid, it would be very unlikely that we could even see another asteroid nearby. However in spite of this, there are still asteroid encounters and collisions on infrequent occasion. What appears to have happened is that sometime during the cooling and accreting stage, the matter in that area, perhaps at the time enough to form a planet as large as Earth, continued to be purturbed by the growing planet Jupiter. As Jupiter and its tremendous gravitational field resonances, would have swept across the asteroid belt, dynamically exciting the region's population and also increasing their velocities relative to each other. That was during perhaps the first 10 million years of the Solar system's formative period. The planetisimals in the asteroid belt were never able to form into very large bits of matter. So they continued circling the Sun as small asteroids, evolving into what we see today. happy days and... starry starry nights! -- Indelibly yours, Paine Ellsworth P.S.: I think on-stage nudity is disgusting, shameful and damaging to all things American. But if I were 22 with a great body, it would be artistic, tasteful, patriotic and a progressive religious experience. > Shelley Winters P.P.S.: http://yummycake.secretsgolden.com http://garden-of-ebooks.blogspot.com http://painellsworth.net === Subject: Re: An asteroid ring > Why didn't it become a planet? Why isn't there a partially formed > planet there at that orbital interval? Gravitational order designed the solar system. Mitch Raemsch There could have been two planets there that kept apart from billions of and annihilating each other... There could have been a gaseous band there at the formation of the Solar System, that never had what it takes to become a planetoid... There could have been a double planet there, whose two reciprocals finally destroyed one another... A single planet may have been annihilated by a rogue asteroid. Take your pick. === Subject: Re: An asteroid ring Why didn't it become a planet? Why isn't there a partially formed >planet there at that orbital interval? Gravitational order designed the solar system. Mitch Raemsch > There could have been two planets there that kept apart from billions of > and annihilating each other... There could have been a gaseous band there at the formation of the Solar > System, that never had what it takes to become a planetoid... There could have been a double planet there, whose two reciprocals finally > destroyed one another... A single planet may have been annihilated by a rogue asteroid. Take your pick. Mitch can't be bothered learning physics, so why speculate about this? === Subject: Re: An asteroid ring Why didn't it become a planet? Why isn't there a partially formed >planet there at that orbital interval? Gravitational order designed the solar system. Mitch Raemsch > There could have been two planets there that kept apart from billions of > and annihilating each other... There could have been a gaseous band there at the formation of the Solar > System, that never had what it takes to become a planetoid... There could have been a double planet there, whose two reciprocals finally > destroyed one another... A single planet may have been annihilated by a rogue asteroid. Take your pick. Mitch can't be bothered learning physics, so why > speculate about this? It's more fun to make up than be burdened with having to think about what really happened. Nevertheless, it's more fun to pretend it's real than to acknowledge that it's just foolish flights of fancy. -- Timberwoof http://www.timberwoof.com Most of the universe is extremely hostile to life as we know it. It seems obvious that it was all designed by some creature that hates life... And here you are, trying to attract its attention. === Subject: Re: An asteroid ring Why didn't it become a planet? Why isn't there a partially formed >planet there at that orbital interval? Gravitational order designed the solar system. Mitch Raemsch > There could have been two planets there that kept apart from billions > of > other > and annihilating each other... There could have been a gaseous band there at the formation of the > Solar > System, that never had what it takes to become a planetoid... There could have been a double planet there, whose two reciprocals > finally > destroyed one another... A single planet may have been annihilated by a rogue asteroid. Take your pick. Mitch can't be bothered learning physics, so why > speculate about this? It's more fun to make up than be burdened with having to think > about what really happened. Nevertheless, it's more fun to pretend it's > real than to acknowledge that it's just foolish flights of fancy. A little more imagination and a little less following of the holy traditions handed down by our forefathers, and we may even discover some of the secrets of the universe. === Subject: Re: An asteroid ring posting-account=vpv29woAAAAjASPtep1bSfPEIqZVH03h 4334.5000; Windows NT 5.0; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) spider-mtc-ta10.proxy.aol.com[400C700A] (Prism/1.2.1), HTTP/1.1 cache-mtc-aa06.proxy.aol.com[400C740A] (Traffic-Server/6.1.5 [uScM]) > Why didn't it become a planet? Why isn't there a partially formed > planet there at that orbital interval? Which orbital interval? Are you looking at the asteroid belt, or what? Maybe there once was a planet in that orbit but it got whacked by another chunk of rock. Wait a few million years and see if it does get back together. Try to be coherent. Doug Chandler === Subject: Re: An asteroid ring > Why didn't it become a planet? Why isn't there a partially formed > planet there at that orbital interval? Jupiter's gravity apparently causes too many perturbations to allow a planet to form. > Gravitational order designed the solar system. I'm not sure what that means. -- Timberwoof http://www.timberwoof.com Most of the universe is extremely hostile to life as we know it. It seems obvious that it was all designed by some creature that hates life... And here you are, trying to attract its attention. === Subject: Re: An asteroid ring posting-account=PSw72goAAADTugQpI3-NCbn5m3k-KKO3 .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) Why didn't it become a planet? Why isn't there a partially formed > planet there at that orbital interval? Jupiter's gravity apparently causes too many perturbations to allow a > planet to form. Gravitational order designed the solar system. I'm not sure what that means. -- > Timberwoof http://www.timberwoof.com > Most of the universe is extremely hostile to life as we know it. It seems > obvious that it was all designed by some creature that hates life... And here > you are, trying to attract its attention. No. God will sustain any world with intelligent life. He doesn't care about the rest. Maybe you do? Mitch Raemsch === Subject: Re: An asteroid ring Why didn't it become a planet? Why isn't there a partially formed > planet there at that orbital interval? Jupiter's gravity apparently causes too many perturbations to allow a > planet to form. Gravitational order designed the solar system. I'm not sure what that means. -- > Timberwoof http://www.timberwoof.com > Most of the universe is extremely hostile to life as we know it. It seems > obvious that it was all designed by some creature that hates life... And > here > you are, trying to attract its attention. No. God What's that? > will sustain any world with intelligent life. He doesn't care > about the rest. How do you know? > Maybe you do? Mitch Raemsch -- Timberwoof http://www.timberwoof.com People who can't spell get kicked out of Hogwarts. === Subject: Re: An asteroid ring posting-account=rsfFvAoAAADqHCqmunbjtmaKjfuP-pwP .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) Why didn't it become a planet? Why isn't there a partially formed > planet there at that orbital interval? Jupiter's gravity apparently causes too many perturbations to allow a > planet to form. Gravitational order designed the solar system. I'm not sure what that means. -- > Timberwoof http://www.timberwoof.com > Most of the universe is extremely hostile to life as we know it. It seems > obvious that it was all designed by some creature that hates life... And > here > you are, trying to attract its attention. No. God What's that? will sustain any world with intelligent life. He doesn't care > about the rest. How do you know? Maybe you do? Mitch Raemsch -- > Timberwoof http://www.timberwoof.com > People who can't spell get kicked out of Hogwarts. A soprano's voice causes a glass to crack. Could the asteroid belt be the remnants of a planet, hit by a non-dumbed down physics in a silly cosmic war? Read 'Dark Mission: the Secret History of NASA' - by Hoagland and Bara. -- foolsrushin. === Subject: Re: An asteroid ring Why didn't it become a planet? Why isn't there a partially formed > planet there at that orbital interval? Jupiter's gravity apparently causes too many perturbations to allow a > planet to form. Gravitational order designed the solar system. I'm not sure what that means. -- > Timberwoof http://www.timberwoof.com > Most of the universe is extremely hostile to life as we know it. It > seems > obvious that it was all designed by some creature that hates life... > And > here > you are, trying to attract its attention. No. God What's that? will sustain any world with intelligent life. He doesn't care > about the rest. How do you know? Maybe you do? Mitch Raemsch -- > Timberwoof http://www.timberwoof.com > People who can't spell get kicked out of Hogwarts. A soprano's voice causes a glass to crack. Could the asteroid belt be > the remnants of a planet, hit by a non-dumbed down physics in a silly > cosmic war? Given the amount of foolishness being written daily in this newsgroup, I'm surprised that it hasn't caused an epidemic of failed hard drives in NNTP servers all over the Internet. > Read 'Dark Mission: the Secret History of NASA' - by Hoagland and > Bara. This one? Product Details * Paperback: 550 pages * Publisher: Feral House; 1 edition (October 16, 2007) * Language: English * ISBN-10: 1932595260 * ISBN-13: 978-1932595260 * Product Dimensions: 8.8 x 6 x 1.3 inches * Shipping Weight: 1.8 pounds (View shipping rates and policies) * Average Customer Review: 109 Reviews 5 star: 49% (54) 4 star: 11% (13) 3 star: 10% (11) 2 star: 7% (8) 1 star: 21% (23) See all 109 customer reviews... 3.6 out of 5 stars See all reviews (109 customer reviews) * Amazon.com Sales Rank: #5,583 in Books (See Bestsellers in Books) Popular in these categories: (What's this?) #6 in Books > Science Fiction & Fantasy > Fantasy > Historical 20 of 26 people found the following review helpful: 1.0 out of 5 stars Beam me up, Scotty, August 31, 2008 By Stephen Pletko Uncle Stevie (London, Ontario, Canada) - See all my reviews (TOP 500 REVIEWER) XXXXX Consider the following: (1) The Face on planet Mars is an artificial structure and the face is part of a city on Cydonia Planitia (a region of Mars) . (2) Numerous objects surrounding the landing sites of the Mars Exploratory Rovers (called Spirit and Opportunity that are still exploring Mars) are, in fact, pieces of Martian machinery. (3) There are large semi-transparent structures constructed of glass on the moon's surface. (4) The twelve moon-walkers have had their memories selectively edited so that they no longer remember seeing evidence of a lunar civilization. (5) The evidence of a advanced civilization that once existed on the moon and Mars is being suppressed by NASA. The above five claims are just some of the tamer claims (there are many, some of which really (I mean really) become outlandish) that you will find in this book, authored by Richard Hoagland (perhaps best known as principal investigator of the Internet site called the Enterprise Mission, an organization that examines NASA data for the possibility of archaeological ruins on Mars and the Moon) and Michael Bara (an aerospace structural engineer). ... As far as the pictures go, I had trouble seeing what the authors saw unless I used a heavy dose of imagination. These pictures reminded me of a Rorschach test (after the psychiatrist who had this last name) or more generally an inkblot test where the person looking at the blot states freely what he or she sees. I predict that hardcore science fiction readers and/or those that cannot distinguish between science and science fiction will adore reading this book. I also noticed that all the images were manipulated (enhanced, filtered, etc.) in some way. If you manipulate photos enough, you're bound to come up with something interesting. With respect to the main narrative, I noticed that it was filled with assumptions, beliefs, a jumping to conclusions, etc. Conspiracy theories abound in this book. Thus, conspiracy addicts should also enjoy reading it. ... In conclusion, if you are either a conspiracy addict and/or a hardcore science fiction reader, then you should thoroughly enjoy this book. ----- So why should I read that? -- Timberwoof http://www.timberwoof.com Most of the universe is extremely hostile to life as we know it. It seems obvious that it was all designed by some creature that hates life... And here you are, trying to attract its attention. === Subject: Re: Theorem You may know it, but what I know is that a homomorphism between connected Lie groups is a covering if and only if lie groups have *isomorphic* Lie algebras. yes you are right. sorry , but the difinition of an infinite dimensional manifold is the same for finite dimensional and charts are from manifold to a t.v.s . === Subject: Re: 2 equations posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > i believe this was already posted by alain verghote but i cant find that thread again :s its about these 2 equations , and what possible solutions might exist. in particular the solutions that are not moebius transforms. f(-x) + 1 = inv f(x) Something of the form (x - 1/2)^2 + (f(x) + 1/2)^2 = r^2 (i.e. a circle centred at (1/2, -1/2)) kind of works but is not single-valued -- you need to choose the right values. However, by chopping parts out of this I think you should be able to get a function uniquely defined, and with inverse uniquely defined, over an appropriate range, though it won't be continuous everywhere. There will be infinitely many similar solutions, not necessarily circles (basically any old shape defined over one quadrant and then appropriately replicated will do I think). === Subject: Re: 2 equations posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) i believe this was already posted by alain verghote but i cant find that thread again :s its about these 2 equations , and what possible solutions might exist. in particular the solutions that are not moebius transforms. f(-x) + 1 = inv f(x) Something of the form (x - 1/2)^2 + (f(x) + 1/2)^2 = r^2 (i.e. a > circle centred at (1/2, -1/2)) kind of works but is not single-valued > -- you need to choose the right values. However, by chopping parts > out of this I think you should be able to get a function uniquely > defined, and with inverse uniquely defined, over an appropriate range, > though it won't be continuous everywhere. There will be infinitely many similar solutions, not necessarily > circles (basically any old shape defined over one quadrant and then > appropriately replicated will do I think). Bonjour Tommy, Be cool and carefully explore the given equations : Inside f(-x) + 1 = f^[-|] (x) we've got very simple planar movings : f(-x) a symmetry, + 1 a translation of function, f^[-|] (x) an other symmetry . Before bringing into big tools such as Moebius transforms, Chaos theory and even derivation , quietly analyze the case in front of you , Amicalement, Alain === Subject: Re: 2 equations posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) On 8 d.8ec, 10:10, alainvergh...@gmail.com circle centred at (1/2, -1/2)) kind of works but is not single-valued > -- you need to choose the right values. However, by chopping parts > out of this I think you should be able to get a function uniquely > defined, and with inverse uniquely defined, over an appropriate range, > though it won't be continuous everywhere. There will be infinitely many similar solutions, not necessarily > circles (basically any old shape defined over one quadrant and then > appropriately replicated will do I think). Bonjour Tommy, > Be cool and carefully explore the given equations : > Inside f(-x) + 1 = f^[-|] (x) we've got very simple > planar movings : > f(-x) a symmetry, > + 1 a translation of function, > f^[-|] (x) an other symmetry . Before bringing into big tools such as Moebius transforms, > Chaos theory and even derivation , quietly analyze the case in front > of you , Amicalement, > Alain- Masquer le texte des messages pr.8ec.8edents - - Afficher le texte des messages pr.8ec.8edents - Bon, with iterates ... First one <=> (-f(x))^[2] = 1 -x (a minus sign!) But what about real solutions... Second one <=> (-f(x))^[2] = x - 1 -f(x) = x -1/2 f(x) = 1/2 -x Notice (f(x))^[2] = x Alain === Subject: Re: 2 equations <32429507.1228772176271.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) On 8 d.8ec, 10:10, alainvergh...@gmail.com > On Dec 7, 10:24pm, amy666 i believe this was already posted by alain > verghote but i cant find that thread again :s its about these 2 equations , and what possible > solutions might exist. in particular the solutions that are not > moebius transforms. f(-x) + 1 = inv f(x) Something of the form (x - 1/2)^2 + (f(x) + > 1/2)^2 = r^2 (i.e. a > circle centred at (1/2, -1/2)) kind of works but > is not single-valued > -- you need to choose the right values. > However, by chopping parts > out of this I think you should be able to get a > function uniquely > defined, and with inverse uniquely defined, over > an appropriate range, > though it won't be continuous everywhere. There will be infinitely many similar solutions, > not necessarily > circles (basically any old shape defined over one > quadrant and then > appropriately replicated will do I think). Bonjour Tommy, > Be cool and carefully explore the given equations : > Inside f(-x) + 1 = f^[-|] (x) we've got very > simple > planar movings : > f(-x) a symmetry, > + 1 a translation of function, > f^[-|] (x) an other symmetry . i know ... i dont immediatly see how it helps ... Suppose that f(x) is defined at x = a such that f(a) = b, represented by the point (a, b) on the graph of f(x) versus x. Then f(-x) + 1 = inv f(x) => f(b + 1) = -a, which gives another point (b + 1, -a). As a consequence of said symmetries/translations it transpires that (b + 1, -a) is (a, b) rotated clockwise through 90 degrees about the point (1/2, -1/2). So, the graph of f(x) versus x must be unchanged by such a rotation. Any definition of f(x) that has this property will work (the simplest of which is the circle or subset thereof that I mentioned), subject of course to the consideration that the function and its inverse ought to be single-valued. === Subject: How could you multiply and/or divide using straightedge and compass only? posting-account=hEsTJAoAAABBB9neOo9d5YFd2CC8eBZx Gecko/20061205 Iceweasel/2.0.0.1 (Debian-2.0.0.1+dfsg-1),gzip(gfe),gzip(gfe) ~ You can certainly do this for special cases. For example you can trisect a segment, effectively dividing any type of segments by three, but I don't know of any general way of doing it for any cases ~ lbrtchx === Subject: Re: How could you multiply and/or divide using straightedge and compass only? > How could you multiply and/or divide using > straightedge and compass only? > You can certainly do this for special cases. > ... I don't know of any general way of doing > it for any cases Consider the length AB that you wish to divide into n equal segments. You'll need to draw a sketch as we go. Construct a line AC at an arbitrary angle to AB, not equal to 0 or pi. (around 45 degrees is good) Set your compass to some arbitrary length >0 (around 1/n of the length AB is good, but not essential. Anything will do.) and mark off n new equally spaced points D0=A, D1, D2, D3, ... Dn along AC. (Yes, that's n+1 points, but I'm not counting D0 as a new point because we already had it. Join Dn to B. points Ai on AB such that angle(A.Di.Ai) = angle(A.Dn,B) The points Ai divide AB into n equal segments. As stated in other posts, to divide a length by a given *length*, not a number, you need to know what number that length represents, so you need a concept of a unit length. I think, however, this is beyond what the OP was asking. === Subject: Re: How could you multiply and/or divide using straightedge and compass only? <081220080740521412%edgar@math.ohio-state.edu.invalid> posting-account=hEsTJAoAAABBB9neOo9d5YFd2CC8eBZx Gecko/20061205 Iceweasel/2.0.0.1 (Debian-2.0.0.1+dfsg-1),gzip(gfe),gzip(gfe) > What unit length did you choose in your divide by 3 construction? ~ > These methods are included in Euclid's Elements I believe. ~ Well, I was thinking of pure straightedge and compass constructions. I think they have a relatively convoluted example of how to trisect a segment without using proportions, for example, here: ~ http://en.wikipedia.org/wiki/Compass_and_straightedge ~ lbrtchx === Subject: Re: How could you multiply and/or divide using straightedge and compass only? posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1 FOH:R051,gzip(gfe),gzip(gfe) nothing could be simpler; simply, construct an arbitrary unitlength from one end of your unit-to-be-subdivided, and add two additional unitlengths to the end; then, construct parallels ... figure it out with a sketch, but it's probably in your wookiepoopeya reference. and, division is the inverse construction. > http://en.wikipedia.org/wiki/Compass and straightedge thus: there is absolutely nothing deterministic about billiards, mainly because of english. perhaps you are referring to the mathematical idealization thereof, using point masses (a.k.a. The Photon, the only truly zero-D object in Universe .-) > If you have a billiard table and there are balls bouncing around, > everything is nice and deterministic. Disregarding Plancklength and QM thus: seriously, any polynomial with a constant (or times x to the zeroth power) cannot have zero as a solution; otherwise, zero is a trivial solution, since zero times any coefficient is nothing! anyway, what's the problem, if an infinite series polynomial is not algebraic per definition? thus: Euler's identity is, what -- exp(i*pi) = 1, i*pi = ln(1) pi = -i*ln(1), but I forgot what the natural log of one is. thus: there's got to be a few Latin phrases for this kind of fallacy; this is the same thing that Are Buckafka Fullofit tried to put over, and most of the Buckywitches really believe it. anyway, the queestion is, why should line segments of polygons be so God-am existential, or of polyhedra -- because of the typical stick-drawn configuration in some edition of Euclid? >Also, from the position taken above, spheres are only possible in theory aswell. --Cheeny & Zbiggy, fo'mo' years! http://tarpley.net http://larouchepub.com http://xplodeyourmlmbiz.com/?s1=Fuller+Brush&t=LSP01&gclid=CPiar6nispcCFQu-G godYm21kA http://www.rwgrayprojects.com/synergetics/plates/plates.html === Subject: Re: How could you multiply and/or divide using straightedge and compass only? > nothing could be simpler; simply, The OP doesn't say the opposite, he just wanted to *generalize* to any ratio, for instance to multiply by 1.25, or divide by 1.25 > construct an arbitrary unitlength > from one end of your unit-to-be-subdivided, and > add two additional unitlengths to the end; then, > construct parallels ... figure it out with a sketch, but > it's probably in your wookiepoopeya reference. and, division is the inverse construction. > I think they have a relatively convoluted example of how to > trisect a segment without using proportions, for example, here: http://en.wikipedia.org/wiki/Compass_and_straightedge > But this one is **simpler** than drawing parallels. To draw parallels you need to draw several auxiliary circles and lines. To duplicate the unit length also. With the Wikipedia construction, all is in one. However this construction *does* uses proportions. Just they are hidden (everything is proportions from Euclides). Consider point C'. C'D is parallel to AB and equal. C'A = C'E/3, or EA = 2/3 EC' Hence AT = 2/3 C'D = 2/3 AB To get more intricated hidden proportions, you may also use the compass only constructions from Mascheroni : Duplicate AB to AD and BE (in cyan) : draw circle (A,AB) and (B,BA), intersecting in C draw circle (C,CA), intersects (A,AB) in C1 draw circle (C1,C1A), intersects (A,AB) in D Similarily on the other side to get E Draw circle (D,DB) and (E,ED) (green), intersecting in M and N Draw circles (M,MD) and (N,ND) (red) intersecting in T AT = AB/3, and A,B,T in line. You need just 3 radii (AB=BA=CA=C1A=...), ED and (DB=MD=ND). No straightedge at all. But one more time, this is not the original question. For instance in the 1.25 case, you have to choose a suitable unit length, so that 1.25 is a multiple of this unit length. as 1.25 = 5/4 you could consider a 1/4 unit length . +---------- 4/4 --------+ . +-----+-----+-----+-----+-----+ . +-1/4-+ . +------------ 5/4 ------------+ Then apply this to have a corresponding proportion of the 4/4 and 5/4 segments to AB and what you search. However if you want to use a ratio of pi for instance, as pi is irrational this method doesn't work. For those irrationnal which are constructible with compass and straightedge, just construct it from the base unit length = AB. For instance to construct AB*sqrt(2)... (easy, and no comment). For irrationals which are not constructible, you just can't do it with compass and straightedge. For instance you just can't construct pi*AB (this is quadrature of circle). If I GIVE you some unit length OU and if I GIVE you a length OP, saying that OP = pi*OU (constructed from some *other* tools than compass and straightedge), then it is exactly as easy to construct pi*AB from any AB and the given OU,OP, than to construct 3*AB, with the same method (parallels). -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Re: How could you multiply and/or divide using straightedge and compass only? > You can certainly do this for special cases. For example you can > trisect a segment, effectively dividing any type of segments by three, > but I don't know of any general way of doing it for any cases Hint: you also need a unit length. To multiply length x by length y, find a length z such that z:x = y:1. To divide, x:y = z:1. -- === Subject: Re: How could you multiply and/or divide using straightedge and compass only? > You can certainly do this for special cases. For example you can > trisect a segment, effectively dividing any type of segments by three, > but I don't know of any general way of doing it for any cases Hint: you also need a unit length. To multiply length x by length y, > find a length z such that z:x = y:1. To divide, x:y = z:1. Yes. As the OP said divide by 3 for example... in general ? A pure number is ... the ratio y/1 (length y over unit length) Thus we don't multiply/divide a length by a length, but a length by a ratio, in the same manner we do it when the ratio value is 3. It is equivallent to the z:x = y:1 and x:y = z:1 What unit length did you choose in your divide by 3 construction ? -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Distribution of two consecutive eigenvalues spacing in a wishard posting-account=BHQ0RQoAAACGbyHx2qDA8u1OskzVIfxx Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) i have x in C^n complex gaussian vector and X=x*x' form Hermitian matrix with rank 1. eign distribution of X can be evaluated with Wishard Distribution. But i would like to know.. the spacing distribution of two consecutive eigenvalues . you guys have any trick... === Subject: Re: Distribution of two consecutive eigenvalues spacing in a wishard > i have x in C^n complex gaussian vector and X=x*x' form Hermitian > matrix with rank 1. eign distribution of X can be evaluated with > Wishard Distribution. > But i would like to know.. the spacing distribution of two consecutive > eigenvalues . you guys have any trick... If it has rank 1, there is only one nonzero eigenvalue. -- === Subject: Re: Distribution of two consecutive eigenvalues spacing in a wishard posting-account=BHQ0RQoAAACGbyHx2qDA8u1OskzVIfxx Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) On Dec 8, 1:44pm, Robert Israel > i have x in C^n complex gaussian vector and X=x*x' form Hermitian > matrix with rank 1. eign distribution of X can be evaluated with > Wishard Distribution. > But i would like to know.. the spacing distribution of two consecutive > eigenvalues . you guys have any trick... If it has rank 1, there is only one nonzero eigenvalue. > -- > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada lets say i have two events 1 and 2 with P(1)=0.3 and P(2)=0.7. then with random generation we may get consecutive numbers) are - - - - - - - - - Y-> 1 1 1 0 0 1 1 0 0 then i would like to get P(1)=? and P(0)=? In more general, if i know a particular distribution[ P(X) ] then how could i find P(Y)? any trick? === Subject: Re: Distribution of two consecutive eigenvalues spacing in a wishard > On Dec 8, 1:44=A0pm, Robert Israel > i have x in C^n complex gaussian vector and X=3Dx*x' form Hermitian > matrix with rank 1. eign distribution of X can be evaluated with > Wishard Distribution. > But i would like to know.. the spacing distribution of two consecutive > eigenvalues . you guys have any trick... If it has rank 1, there is only one nonzero eigenvalue. > -- > Robert Israel =A0 =A0 =A0 =A0 =A0 =A0 > =A0isr...@math.MyUniversitysInitial= > s.ca > Department of Mathematics =A0 =A0 =A0 =A0http://www.math.ubc.ca/~israel > University of British Columbia =A0 =A0 =A0 =A0 =A0 =A0Vancouver, BC, > Cana= > da > lets say i have two events 1 and 2 with P(1)=3D0.3 and P(2)=3D0.7. then > with random generation we may get X-> 1 2 1 2 2 2 1 2 2 2 and their spacing(difference between two > consecutive numbers) are > - - - - - - - - - > Y-> 1 1 1 0 0 1 1 0 0 then i would like to get > P(1)=3D? and P(0)=3D? In more general, if i know a particular distribution[ P(X) ] then how > could i find P(Y)? any trick? Ah, OK, you want the distribution of the absolute value of the difference between two independent random variables with the same Wishart distribution. Well, if f_X is the probability density for X, the probabiity density for the absolute value of the difference should be f_2(s) = int_{-infty}^infty dx f_X(x) (f_X(x-s) + f_X(x+s)) I don't know if that can be evaluated in closed form. -- === Subject: Re: Analysis with Cauchy product u_n. > Hello teacher~ Let sum{n=1 to oo} a_n, sum{n=1 to oo} b_n > be convergent series of nonnegative terems, > with sums a, b, respectively, and let u_n = (a_1)(b_n) + (a_2)(b_(n-1)) + ... + (a_n)(b_1). Prove that > sum{n=1 to N} u_n <= [sum{n=1 to N} a_n][sum{n=1 to N} b_n] > <= sum{n=1 to N} u_(2n) (hence that sum{n=1 to oo} u_n converges to the sum ab.) ---------------------------------------------------------------------------- - - > ---------- > sum{n=1 to N} u_n <= [sum{n=1 to N} a_n][sum{n=1 to N} b_n] This is trivial as > N=1 ==> u_1 = a_1.b_1 <= a_1.b_1 N=2 ==> u_1 + u_2 = a_1.b_1 + a_1.b_2 + a_2.b_1 > <= (a_1 + a_2).(b_1 + b_2) N=3 ==> ... > [sum{n=1 to N} a_n][sum{n=1 to N} b_n] <= sum{n=1 to N} u_(2n) How do you show it ? N=1 ==> a_1.b_1 <= u_2 = a_1.b_2 + a_2.b_1 (?) Hint: The square [0,N] x [0, N] is contained in the triangle {x, y >= 0, x + y <= 2N} === Subject: Re: Analysis with Cauchy product u_n. Hello teacher~ Let sum{n=1 to oo} a_n, sum{n=1 to oo} b_n > be convergent series of nonnegative terems, > with sums a, b, respectively, and let u_n = (a_1)(b_n) + (a_2)(b_(n-1)) + ... + (a_n)(b_1). Prove that > sum{n=1 to N} u_n <= [sum{n=1 to N} a_n][sum{n=1 to N} b_n] > <= sum{n=1 to N} u_(2n) (hence that sum{n=1 to oo} u_n converges to the sum ab.) --------------------------------------------------------------------------- > --- > ---------- > sum{n=1 to N} u_n <= [sum{n=1 to N} a_n][sum{n=1 to N} b_n] This is trivial as > N=1 ==> u_1 = a_1.b_1 <= a_1.b_1 N=2 ==> u_1 + u_2 = a_1.b_1 + a_1.b_2 + a_2.b_1 > <= (a_1 + a_2).(b_1 + b_2) N=3 ==> ... > [sum{n=1 to N} a_n][sum{n=1 to N} b_n] <= sum{n=1 to N} u_(2n) How do you show it ? N=1 ==> a_1.b_1 <= u_2 = a_1.b_2 + a_2.b_1 (?) Hint: The square [0,N] x [0, N] is contained in the triangle {x, y >= >0, x + y <= 2N} I think you read what obviously should be there instead of the > existing typo (sum{n=1 to N} u_(2n) for sum{n=1 to 2N} u_(n).) I saw that you had already made the correction. === Subject: Re: How to show Matrix is a Tensor or not? posting-account=dV9I2goAAACWj0lUaIWiRgL71QEKja9z Gecko/20070321 Netscape/8.1.3,gzip(gfe),gzip(gfe) > If I have a 3x3 matrix A, with A_11 =1, and all other elements are > 0, how do show that it is or is not a tensor ? I tried using the transformation laws. So I did R A R' , R being a > rotation matrix, and obtained a matrix in terms of sines & cosines. I don't think your question makes sense as written. Would I be correct > in assuming that you're doing an undergraduate physics degree or > something similar? If so then by tensor you probably mean a multi- > indexed object which has certain transformation laws under the action > of rotations. In that case the question is meaningless - any matrix / > could/ be the components of a tensor, depending on how it transforms > under the action of R in SO(3). On the other hand, if you mean to ask > whether it is an /invariant/ tensor, i.e. one whose components stay > the same under the action of any R in SO(3) (like, for example, > delta_i^j or epsilon_{ijk}) then you're almost done - by showing > that R A R^{-1} is not equal to A you have shown that A is not an > invariant tensor of type (1,1). If you can find a rotation matrix R > such that R A R^T is not equal to A (where R^T is the transpose of R) > then you will also have shown that A is not an invariant tensor of > type (0,2) or (2,0). Let me know if that helps. Page 32, a rank 2 tensor can be regarded as 3x3 matrix, but not vice- versa. It's this vice-versa bit which was confusing me. So I made up that 3x3 matrix and wanted to know whether it is or isn't a tensor of rank 2. Sid. === Subject: Solutions Manuals for Sale posting-account=fQ9MdQoAAAAwsvTJs2IMvoz2FcJe0m3G 1.1.4322; .NET CLR 2.0.50727; InfoPath.2),gzip(gfe),gzip(gfe) I have the following solutions manuals for sale for these textbooks. Send inquiries to cheapste...@gmail.com: Automatic Control Systems 8th edition by Kuo Engineering Circuit Analysis /6th /William H. Hayt / McGraw-Hill Science Essential of Fluid Mechanics/ Yunus A. Cengel & John M.Cimbala / Mc GrawHill Fundametals of signals and system/ M.J Roberts/Mcgraw-Hill 2008 Heat and Mass Transfer: A Practical Approach 3rd edition by Yunus Cengel Mechanics of Fluids/3ed/Merle C.Potter David C.Wiggert/BrooksCole Design of Machinery 3rd edition by Norton Introduction to Fluid Mechanics by Fox System Dynamics by Ogata Introduction to Chemical Engineering Thermodynamics/7th/J.M. Smith/ McGRAW-HILL Mechanics of Materials 3rd edition by Beer Fundamentals of Heat and Mass Transfer Solution/6th/Frank P.Incropera,David P.DeWitt/WILEY Fundamentals of Fluids Mechanics 5th (Munson & Young) Theory and Design for Mechanical Measurements 4th edition by Figliola and Beasley Modern Control Systems by Richard Dorf === Subject: converging eigenvector expansion posting-account=E2ovhwkAAAC0OMet1FlTZzQxrztGYIiT 4.90),gzip(gfe),gzip(gfe) Hello! I am learning about Sturm Liouville theory and there is (at least) one thing that I don't understand but which I can't find in textbooks. It is about WHY the eigenvector expansion converges UNIFORMLY ABSOLUTELY. More precise: we work on the interval J=[a,b]. D={differentiable functions on J whose derivative is absolutely continuous and whose second derivative is in L^2(J)}. Let L:D-->L^2(J): h |--> h''+qh with q continuous on J, be the Sturm Liouville operator (assumed to be injective here). There is also a Green function, and a (compact) operator G:L^2(J)-->L^2(J) such that LG=1 and GL=1 on opportune domains. Let a_{i} be the countable set of non-zero eigenvalues of G and let e_ {i} be the corresponding eigenvalues. These e_{i} can bo shown to be in D (defined above). It can also be shown that for f in L^2(J) and A a number not equal to any eigenvalue a_{i}, there is a unique h in D such that Lh-Ah=f. (The QUESTION) Now, h can be expanded in eigenfunctions e_{i}, via h=sum (e_{i} | h)e_{i} =sum (A-a_{i})^{-1} (f | e_{i}) e_{i}. and the claim is: this series converges absolutely and uniformly. I don't see why this is the case, so if anyone can tell me how this wirks.. I would highly appreciate it! A reference is also OK :) cheers === Subject: differentiability of optimal value w.r.t. to constraints I have a question in convex optimization as follows: consider the following minimization problem in standard form min f(x) s.t. x in Omega, g(x) leq z, Best, YH === Subject: Re: differentiability of optimal value w.r.t. to constraints posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20081029 Firefox/2.0.0.18,gzip(gfe),gzip(gfe) I have a question in convex optimization as follows: consider the following minimization problem in standard form > min f(x) > s.t. x in Omega, g(x) leq z, > Best, > YH A Google search using the keywords differentiability of optimal value turns up many books and papers. Almost all the papers require journal subscription. Some of the books allow limited perusal via R.G. Vickson === Subject: Collatz map posting-account=WOiaSQoAAAA05vyEQ086teGIhq4g31ZC CLR 1.1.4322; .NET CLR 2.0.50727; MS-RTC LM 8),gzip(gfe),gzip(gfe) wikipedia says : The Collatz map can be viewed as the restriction to the integers of the smooth real and complex map: f(z) = 1/2 zcos^2(PI/2 z) + (3z + 1)sin^2(PI/2 z) while I see that this does provide correct results for the restriction to integers, wouldn't the following also be valid?: f(z) = 1/2 zcos(PI/2 z) + (3z + 1)sin(PI/2 z) For that matter, wouldn't other complex maps also exist that do the trick? === Subject: Re: Collatz map posting-account=Jz4DtgkAAAAZkdWvJAd__jMF7l1N5_1V .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > wikipedia says : The Collatz map can be viewed as the restriction to the integers of > the smooth real and complex map: f(z) = 1/2 zcos^2(PI/2 z) + (3z + 1)sin^2(PI/2 z) while I see that this does provide correct results for the restriction > to integers, wouldn't the following also be valid?: f(z) = 1/2 zcos(PI/2 z) + (3z + 1)sin(PI/2 z) No. What values do you get for 1, 2, 3, 4? You should get 4, 1, 10, 2. For that matter, wouldn't other complex maps also exist that do the > trick? Yes. There are many others, based on functions which are 0 for even numbers and 1 for odd numbers (and 1 and 0). === Subject: The SU(3) QCD Massive Meson Mass Spectrum, and (Really) Solving the Propagator Pole Problem Almost two years ago in the course of my work on Yang Mills, I came across what I believe is an approach by which mass spectrum of the massive mesons of QCD might be understood. I had what I still believe is the right concept, and many of the pieces, but I could not figure out the right execution of the concept in complete detail. Over the past year and a half I walked away from this to let the dust settle and to also arrive at a place where the basic principles of quantum field theory were no longer new to me but had become somewhat ingrained. The objective of my recent return to Yang Mills has been to see if this earlier line of research could be finally brought to fruition after some good breathing space. In the file linked at: http://jayryablon.files.wordpress.com/2008/12/massive-mesons.pdf In this file, I review how mass is known to be generated in SU(2), but engage in some overkill which is not strictly necessary to get the right SU(2) masses, as a template for considering SU(3) QCD where this is totally necessary and is no longer overkill. I have tried to explain as simply as possible, what I believe to be the origin of QCD meson masses, as well as to lay the foundation for theoretically predicting these. Keep in mind, finding out how the vector mesons of QCD obtain their non-zero masses, which make the QCD interaction short range despite supposedly-massless gluons, is one aspect of the so-called mass gap problem, see point 1) on page 3 of http://www.claymath.org/millennium/Yang-Mills_Theory/yangmills.pdf at http://www.claymath.org/millennium/Yang-Mills_Theory/. In the coming days, I will extend this development, in detail, to SU(3), but the above already fully summarizes how the symmetry breaking in SU(3) QCD is to take place and all the other steps to be followed, so now this is all just about turning the crank and generating mass numbers. What is already interesting about the discussion here, is that even without moving on yet to the detail of SU(3), it neatly solves the problem of propagator poles (infinities) in a manner which I believe has not heretofore been discovered. Goodbye to the +ieta prescription, off propagator. Hope you enjoy! Jay. ____________________________ Jay R. Yablon Email: jyablon@nycap.rr.com co-moderator: sci.physics.foundations Weblog: http://jayryablon.wordpress.com/ Web Site: http://home.nycap.rr.com/jry/FermionMass.htm === Subject: Re: Analysis & topology with uniformly continuous. > Hello teacher~ Let (X, d) be a metric space that is not totally bounded. Prove that there exist a sequence (x_n) in X > and a positive number c such that d(x_m, x_n) >= c > whenever m =/= n. ---------------------------------------------------------------------------- - -- > pf) > Since X is not totally bounded, > there exists an e > 0 such that > there does not exist a finite covering of X by e-balls. Let x_1 in X. Then there exists a point x_2 in X with d(x_2, x_1) >= e. > for otherwise, X subset B(x_1, e). Similarly, there exists a point x_3 in X with > d(x_3, x_1) >= e and d(x_3, x_2) >= e. > for otherwise, X subset B(x_1, e) U B(x_2, e). Continuing in this manner, > we arrive at a sequence (x_1, x_2, .... ) with the property > that d(x_m, x_n) >= c whenever m =/= n. In fact, it means that (a_n) cannot any subsequences > which converges. > so, X is not sequentially compact. ---------------------------------------------------------------------------- - --- > Let (X, d) be a metric space that is not totally bounded, > and choose (x_n) and c in above problem. For each n, > construct a uniformly continuous function > g_n : X -> [0, 1] such that > (1) g_n(x_n) = 1 and > (2) g_n(x) = 0 if d(x, x_n) >= c/3. ---------------------------------------------------------------------------- - -- > Sorry, I need your advice. Is this really the problem statement? > Consider g_n(x) = max{1 - 3 d(x,x_n), 0}. > This is uniformly continuous for *any* metric space > and does not make use of the additional structure: > If d(x,y) < eps/3 then |g_n(x) - g_n(y)| < eps. ---------------------------------------------------------------------------- - ---- > This is the original problem. Let (X, d) be a metric space that is not totally bounded, > and choose (x_n) and c in above problem. For each n, > construct a uniformly continuous function > g_n : X -> [0, 1] such that > (1) g_n(x_n) = 1 and > (2) g_n(x) = 0 if d(x, x_n) >= c/3. Given any sequence (c_n) of real numbers, > show that f = sum{n=1 to oo} (c_n).(g_n) is a well-defined > continuous function on X, > and that if (c_n) is bounded, then f is uniformly continuous on X. ---------------------------------------------------------------------------- - ------- Think... http://board-2.blueweb.co.kr/user/math565/data/math/xse.jpg g_n : X -> [0, 1] such that (1) g_n(x_n) = 1 and (2) g_n(x) = 0 if d(x, x_n) >= c/3. (3) g_n(x) = 1 - d(x, x_n) if d(x, x_n) < c/3 Let A subset X. h : X->R, h(x) = d(x,A) = inf{a in A} d(x, a) is uniformly continuous. (This is a exercise.) so, for each n, g_n(x) = 1 - d(x, x_n) if d(x, x_n) < (2/3).c is uniformly continuous. Namely, There exists s such that |x- y| < s , (all x, y in B(x_n, (2/3).c) ==> |g_n(x) - g_n(y)| < e. Let s' = min{s, c/3}. so, |x - y| < s' ,(all x, y in X) ==> |g_n(x) - g_n(y)| < e. so, g_n : X -> [0, 1] is uniformly continuous. ---------------------------------------------------------------------------- --------------- I will show that f = sum{n=1 to oo} (c_n).(g_n) is a well-defined continuous function on X. f(x) = c_1.{g_1(x)} + c_2.{g_2(x)} + c_3.{g_3(x)} + ... By jpg picture, f(x) = 0, x in [B(x_1, c/3) U B(x_2, c/3) U ...]^c f(x) = c_1.{1 - d(x, x_1)}, x in B(x_1, c/3) f(x) = c_2.{1 - d(x, x_2)}, x in B(x_2, c/3) ... Let x' in B(x_1, c/3). d(x, x') < min{c/3 - d(x', x_1), e/(c_1)} ==> x in B(x_1, c/3) ==> |f(x) - f(x')| = |c_1.{1 - d(x, x_1)} - c_1.{1 - d(x', x_1)} <= |c_1.{d(x', x_1) - d(x, x_1)}| <= |c_1|.d(x', x) < e It means that f(x) is continuous at x = x' in B(x_1, c/3). Similarly, f(x) is continuous at x = x' in B(x_2, c/3). f(x) is continuous at x = x' in B(x_3, c/3). ... Of course, f(x) is continuous at x = x' in [B(x_1, c/3) U B(x_2, c/3) U ...]^c. (for some enough small open set). so, f = sum{n=1 to oo} (c_n).(g_n) is a well-defined continuous function on X. ---------------------------------------------------------------------------- --------- I will show that if (c_n) is bounded, then f is uniformly continuous on X. Let |c_n| < M for each n. d(x, y) < min{e, c/3} |f(x) - f(y)| = |c_n.{1 - d(x, x_n)} - c_n.{1 - d(y, x_n)} = |(c_n - c_n) + {d(x, x_m) - d(y, x_n)}| <= 0 + d(x, y) < e. or (If x or y in [B(x_1, c/3) U B(x_2, c/3) U ...]^c) |f(x) - f(y)| = |c_n.{1 - d(x, x_n)} - 0| or |0 - c_n.{1 - d(x, x_n)} = |c_n|.|1 - d(x, x_n)| or |c_n|.|1 - d(x, x_n)| ??? Strange... I need your advice about (???). === Subject: Re: Another Newbie Question !!! > ... http://projecteuler.net/ ... I glanced at the first two pages there, a curious set of math puzzles. The following problem is too easy: I solved it manually, simply by drawing a graph of the first triple, then scanning down the list to find other triples that could be easily fitted into the graph by extrapolating off either end or interpolating in the middle. I ended up with a sequence of eight digits with no repeats, and going back through the list of triples, checking them *all*, I saw that indeed I had the solution already. If I had known ahead of time that no digit was repeated, then I wouldn't have needed triples at all, pairs would have been enough to establish the unique total ordering consistent with all ordered pairs. A more interesting/difficult problem, which is actually what I expected at the outset, and what my graph-extending algorithm was designed to solve, is where the password has at least one digit repeated in two non-adjacent locations, but the bank deliberately arranges to always ask for three digits not containing both of the repeats of that one digit, thus no triple shows the repeat directly. Then only some of the triples containing that repeated digit would be mutually consistent with each other as sub-strings of a single non-repeat sequence, the rest of the triples containing the repeated digit would be consistent with a different non-repeat sequence (where the single repeat digit has simply moved from one location to the other within the rest of the sequence), and all the triples not containing that repeated digit are consistent with both non-repeating sequences. === Subject: [OT] Is a straight line a circle of infinite size? Content-ID: <20081207071728.M91318@agora.rdrop.com> --------------------------------------------------------------------- Content-ID: <20081207071728.H91318@agora.rdrop.com> I don't know why you want to carry on the discussion in > the subject line. [NonBreakingSpace]But as you do In the first place, NO, I DON'T, and in the second place, > THAT IS NOT what has been happening! > with only secondary comments in the body of the message Oh, SHUT UP. > What has been going on in the body of the message is > AUGMENTATION AND CLARIFICATION, if *I* was writing it. > And if others were writing it, it was ANSWERING THE QUESTION. > Which you DID have room to do, if not in the subject, THEN IN THE > BODY. > SO SHUT THE ** UP! OR ANSWER THE QUESTION!! > Or at least say something HALFWAY RELEVANT AND INFORMATIVE! Gee whiz. What's the problem? I've got your goat? I'll tell what. Just for you, just for you George, you can have your goat back, any time you want. Honest, no strings attacked. Have a nice day, you and your goat. ---- === Subject: Re: Putnam 2008 QUESTIONS Cc: rusin@math.niu.edu posting-account=zh4IswoAAABXFrMZMjiE07tIRTtK2_hq Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) I have received a couple of comments regarding the posting of the questions and answers for this year's Putnam exam. That surprises me: I've done this many, many times in the past and have never had a negative reaction. I don't have a copy of the rule sheet handed out with this year's exam, to check for sure, but I did find the schedule and rules on line. According to these data, everyone finishes taking the exam Saturday afternoon (i.e. well before I posted anything), with the only exceptions listed being those covered in this clause: Students who for religious reasons cannot take the examination at the scheduled hours may take the examination after sundown on the scheduled Saturday, upon request by the supervisor and approval of the Director. Such students must remain under the supervision of a faculty member, rabbi, or clergyman from the official starting time for that time zone on the day of the examination. I really don't see that under these conditions anyone is going to be able to get any benefit from the fact that I posted the solutions early Sunday morning. There are also solutions posted on the web (e.g. a set has been posted by Kedlaya and Ng as of early Monday morning; I don't know when those were first posted). dave are still present at mathforum.) And the formatting screw-up was a team effort between me and Google. (My regular posting mechanism is on the fritz.) === Subject: New mathematics / physical sciences positions at http://jobs.phds.org, Dec 08, 2008 There are new job listings at http://jobs.phds.org -------------------------------------------------------------------- Title: Researchers & Developers - Derivatives Employer: Jhirad Consulting Location: New York, NY, United States The quantitative trading group of an established financial firm seeks experienced c and c++ developers and researchers to design and enhance proprietary fixed income, convertible bond, equity derivatives and FX derivatives trading systems. Some of... Full details: http://jobs.phds.org/job/1292/jhirad-consulting/researchers-developers -------------------------------------------------------------------- Title: Product Specialist - Derivatives, Fixed Income Employer: Jhirad Consulting Location: New York, NY, United States Prestigious financial firm is seeking fixed income and derivatives product managers. 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Full details: http://jobs.phds.org/job/11534/comprehensive-recruiting/quantitative-analyst -------------------------------------------------------------------- Title: Derivatives Analysis Associate Employer: Comprehensive Recruiting Location: New York, NY, United States Top tier investment bank seeks Derivatives Analysis Associate for Market Risk Management group. The Derivatives Analysis Associate will be a part of a quantitative group that focuses on derivatives valuation and risk across all product areas. The... Full details: http://jobs.phds.org/job/11533/comprehensive-recruiting/derivatives-analysis -------------------------------------------------------------------- Title: Postdoc position available in experimental nanomaterial research Department: Department of Electrical and Computer Engineering Employer: University of California-San Diego Location: La Jolla, CA, United States A Postdoctoral researcher position is available immediately at the Department of Electrical and Computer Engineering at University of California, San Diego (UCSD). Highly motivated postdoc candidates will have the opportunity to work on harnessing... 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Location: New York, NY, United States I am looking for a quantatitive analyst with 1-2 years of experience with the folowing skill sets: * Experience of credit model development in a front office or model validation environment; * A sound understanding of the principles of derivative... Full details: http://jobs.phds.org/job/11157/rimrock-associates-inc/credit-derivatives-qua n titative -------------------------------------------------------------------- Title: Doctoral or Postdoctoral Position in Numerical Mathematics Department: Numerical Mathematics, Department of Mathematics Employer: Universitíót Duisburg-Essen Location: Duisburg, Germany Your tasks: Participation in research projects with emphasis on Efficient Numerical Methods for Partial Differential Equations. Assistance in teaching and administrative tasks. An opportunity to obtain further academic quali.95å»cations (PhD,... 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Full details: http://jobs.phds.org/job/11489/large-pharmaceutical/synthetic-organic-chemis t -------------------------------------------------------------------- Title: Bioinformatics Postdoc Employer: Lawrence Berkeley National Laboratory Location: Emeryville, CA, United States This postdoctoral researcher will be part of computational research projects investigating the optimization of biofuels production from renewable lignocellulosic biomass with a focus on the enzymology of engineered enzymes as well as novel enzymes... Full details: http://jobs.phds.org/job/11488/lawrence-berkeley-national/bioinformatics-pos t doc -------------------------------------------------------------------- Title: Quant Strategist / PhD / 2+ years experience! Employer: Integrated Management Resources, Inc. Location: New York, NY, United States Top U.S. Based Investment Manager seeks Quantitative Analyst/Strategist possessing a PHD in a Quantitative field AND 2-3 years FINANCE experience. On an everyday basis, candidate will be working in the development and application of risk, PnL,... Full details: http://jobs.phds.org/job/9320/integrated-management/quant-strategist-phd -------------------------------------------------------------------- Title: Fixed Income Quantitative Analyst Employer: Integrated Management Resources, Inc. Location: Chicago, IL, United States Major Financial Firm seeks a Fixed Income Quant for a Chicago position! This is a senior level role for an experienced Quantitative Analyst with a Fixed Income background. Position entails working in front office-trading environment. Ideal candidate... Full details: http://jobs.phds.org/job/9126/integrated-management/fixed-income-quantitativ e -------------------------------------------------------------------- Title: STATISTICAL HIGH FREQUENCY TRADER Employer: Executive Search Consultants Location: Chicago, IL, United States Opportunity to join a leading Hedge Fund Developing optimal execution algorithms for High Frequency Trading. This is a challenging position within a group that is aggressively growing. Working in this high profile team will create the opportunity to... Full details: http://jobs.phds.org/job/2821/executive-search-consultants/statistical-high- f requency -------------------------------------------------------------------- Title: Quant Developer/Financial Engineer-Derivatives Employer: Integrated Management Resources, Inc. Location: New York, NY, United States Outstanding Trading Company seeks top Quant Developer for derivatives trading desk. Must have 2 to 4 years of experience in derivatives, with energy background a plus. Candidate will need to be hands-on with Matlab, solid C++ skills, and be capable... Full details: http://jobs.phds.org/job/4928/integrated-management/quant-developer-financia l -------------------------------------------------------------------- Title: Senior EFX/High Frequency Trading Desk Quant Employer: Integrated Management Resources, Inc. Location: New York, NY, United States Premier global investment bank seeks a Senior EFX/High Frequency Trading Desk Quant for New York position. Candidate will be responsible for developing new trading strategies, while taking advantage of the high frequency data to maximize trading... 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This is a great opportunity for the experienced Quantitative analyst... Full details: http://jobs.phds.org/job/11484/datacom-capital-markets/senior-quantitative-a n alyst -------------------------------------------------------------------- Title: Molecular Biology/Electrophysiology Staff Position Employer: Masonic Medical Research Laboratory Location: Utica, NY, United States The Masonic Medical Research Laboratory, a not-for-profit independent basic biomedical research institute located in the foothills of the Adirondack Mountains in Upstate New York, invites applications from outstanding investigators for a Research... Full details: http://jobs.phds.org/job/11482/masonic-medical-research/molecular-biology-el e ctrophysiology -------------------------------------------------------------------- Title: Director of Molecular Genetics Employer: Masonic Medical Research Laboratory Location: Utica, NY, United States The Masonic Medical Research Laboratory, a not-for-profit independent basic biomedical research institute located at the foothills of the Adirondack Mountains in Central New York, invites applications for an individual to head its Department of... Full details: http://jobs.phds.org/job/628/masonic-medical-research/director-of-molecular -------------------------------------------------------------------- Title: Advanced Study Program Postdoctoral Fellows Department: Advanced Study Program Employer: National Center for Atmospheric Research Location: Boulder, CO, United States The National Center for Atmospheric Research (NCAR) seeks talented individuals to conduct original research in areas broadly related to atmospheric science or other research areas at NCAR and to document that research in publications. These... Full details: http://jobs.phds.org/job/11478/national-center-for-atmospheric/advanced-stud y -program -------------------------------------------------------------------- Title: Quantitative Analyst Employer: Executive Search Consultants Location: New York, NY, United States Proprietary Trading group is looking to hire experienced motivated individuals to effect model-driven trades in a variety of strategies. Strong quantitative and analytical abilities, excellent communication, negotiation, and interpersonal skills, and... Full details: http://jobs.phds.org/job/2811/executive-search-consultants/quantitative-anal y st -------------------------------------------------------------------- Title: Junior Quantitative Analyst / PhD degree Employer: Top Tier Investment Bank Location: New York, NY, United States Top Tier Investment Bank is seeking a Quantitative Analyst with a minimum of one year experience in front office developing quantitative pricing models...any asset class. Global Team Member will be developing models within the commodity space... 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Quantitative trading opportunity where you will take part in the full lifecycle of quant strategy design, implementation,... Full details: http://jobs.phds.org/job/11476/market-making-firm/quant-strategist-with -------------------------------------------------------------------- Title: PhD Quantitative Trader Employer: European Hedge Fund > $30bn AUM Location: London, United Kingdom European fund seeks a talented entry-level systematic trader for a position within their London business. The firm runs an extremely successful mutli-asset quantitative trading platform. You will join the systematic black-box trading and statistical... 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Full details: http://jobs.phds.org/job/11471/global-hedge-fund/high-frequency-systematic -------------------------------------------------------------------- Title: Assistant or Associate Physicist - Sub-Micron Resolution X-ray Spectroscopy Beamline Department: National Synchrotron Light Source II Employer: Brookhaven National Laboratory Location: Upton, NY, United States Working with the user community and under the direction of the Submicron Resolution X-ray Spectroscopy Group Leader, will be responsible for the development of a scientific research program and its technical implementation at a state-of-the-art... 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Full details: http://jobs.phds.org/job/11059/brookhaven-national-laboratory/assistant-or-a s sociate -------------------------------------------------------------------- Title: Postdoctoral Research Associate - Beam Injection Systems Department: National Synchrotron Light Source II Employer: Brookhaven National Laboratory Location: Upton, NY, United States Requires a PhD degree in Physics or a related discipline, with a focus on experimental accelerator physics and accelerator components and equipment for both circular and linear machines. The successful candidate will work with experienced accelerator... 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The Derivatives Analysis Associate will be a part of a quantitative group that focuses on derivatives valuation and risk across all product areas. The... Full details: http://jobs.phds.org/job/11445/comprehensive-recruiting/derivatives-analysis -------------------------------------------------------------------- Title: Commodities Desk Strategist Employer: Comprehensive Recruiting Location: New York, NY, United States Top tier investment bank seeks experienced strategist for the Commodities Desk. The candidate must have a PH. D from a top university in a hard science and possess strong C++ programming skills. Commodities background a plus, but will consider... Full details: http://jobs.phds.org/job/11444/comprehensive-recruiting/commodities-desk-str a tegist -------------------------------------------------------------------- Title: Credit Risk Mgmt Assoc. Employer: Comprehensive Recruiting Location: New York, NY, United States Top tier investment bank seeks Sr. Associate for Regulatory Reporting. The candidate will be a member of the Credit Risk Management & Advisory Department and on a daily basis will lead the generation, analysis and presentation of credit exposure... Full details: http://jobs.phds.org/job/11443/comprehensive-recruiting/credit-risk-mgmt-ass o c -------------------------------------------------------------------- Title: Chair in Energy Storage Systems Employer: Heriot-Watt University Location: Edinburgh, Midlothian, United Kingdom The successful applicants will be responsible for the development of new research activities within their area and the exploitation of multi-disciplinary opportunities with other theme appointments and existing staff. Full details: http://jobs.phds.org/job/11466/heriot-watt-university/chair-in-energy-storag e -------------------------------------------------------------------- Title: Chair in Computational Biology Employer: Heriot-Watt University Location: Edinburgh, Scotland, United Kingdom The post is based in the School of Mathematical and Computer Sciences (MACS) and a major element of the role is to build multidisciplinary interactions with other groups in the university with excellent research in the Life Sciences and Interfaces. Full details: http://jobs.phds.org/job/11464/heriot-watt-university/chair-in-computational -------------------------------------------------------------------- Title: Chair / Reader in Chemical Biology and Biomedicinal Chemistry Employer: Heriot-Watt University Location: Edinburgh, Scotland, United Kingdom The Life and Physical Sciences Interface Theme at Heriot-Watt University wishes to appoint a Reader in chemical biology/biomedicinal chemistry. 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Full details: http://jobs.phds.org/job/11453/heriot-watt-university/lectureship-in-carbon -------------------------------------------------------------------- Title: Lecturer / Senior Lecturer in .89´[Thorn]Materials for Energy Employer: Heriot-Watt University Location: Edinburgh, Scotland, United Kingdom The Energy Theme at Heriot-Watt University wishes to appoint a Lecturer/Senior Lecturer in Materials for Energy based in The School of Engineering and Physical Sciences. Full details: http://jobs.phds.org/job/11452/heriot-watt-university/lecturer-senior-lectur e r -------------------------------------------------------------------- Title: Provost Postdoctoral Program at Duke University Employer: Duke University Location: Durham, NC, United States The Provost at Duke University has created a Postdoctoral Program, with competitive research appointments as a Postdoctoral Associate for two years. 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Full details: http://jobs.phds.org/job/11437/columbia-university-in/software-scientist-in -------------------------------------------------------------------- Post your job (free!): http://jobs.phds.org/job/post PhDs.org: Science, Math, and Engineering Career Resources --------------------------------------------------------- * Job Listings: http://jobs.phds.org/ - Job board with hundreds of listings for Ph.D.s - Reach tens of thousands of Ph.D.s each month * Graduate School Rankings: http://graduate-school.phds.org/ - Comprehensive, customizable rankings of graduate programs * Career Resources: http://www.phds.org/ - Pointers to the best resources on the web for: + getting into graduate school + writing your dissertation + jobs for Ph.D.s in academia and industry * Engineering Science Weblog: http://blog.phds.org/ - Building better scientists and engineers === Subject: Re: Need help with factorials > i=2,3,.... ? It's also true for i = 0,1. > Why not just (1/2)^i or (1/n) ? Did you try any tests, like n = 2? Now slow down and keep your mind in gear. Did you mean Why not just 1/i ? Did you try any tests, like n = 3? > Why does it have to be 1/2 ?? It doesn't. Anything large would suffice. For example 1. The real question is, is it the smallest possible? There by golly is a question for you. Is 1/2 the smallest real number r for which 1/n! <= r^(n-1) for all non-negative integers n = 0,1,... > Needs a proof. If so, can anyone show that > the inequality is true using induction?? What do you mean? (1/2)^i - 1 or (1/2)^(i-1) ? The first choice is obvious false. The second choice is trival to prove. 2^i <= 2i! === Subject: Set theory with p -> q. Hello teacher~ Find the 1 + 2 + 2^2 + 2^3 + ... Let x = 1 + 2 + 2^2 + 2^3 + ... ---(1) so, 2x = 2 + 2^2 + 2^3 + ... ---(2) so, (1) - (2) ==> -x = 1 so, x = -1. --------------------------------------------------------------------- Of course, false. Because, x = 1 + 2 + 2^2 + 2^3 + ... is impossible. If I write that x = 1 + 2 + 2^2 + 2^3 + ... -> x = -1, (as p -> q) is this true statement by truth table ? (Because, p is false.) === Subject: Re: Set theory with p -> q. posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > If I write that > x = 1 + 2 + 2^2 + 2^3 + ... -> x = -1, (as p -> q) is this true statement by truth table ? > (Because, p is false.) I'd rephrase it a bit more explicitly, since there are many summation methods, many interpretations of what x could be, etc. If x is a real number and x = limit(n --> infinity) of 2^0 + 2^1 + 2^2 + ... + 2^n, then x = -1. Of course, we could also, strictly on the basis of logic, use x = -3 and the statement would be true. However, the argument that results in x = -1 contains some correct steps in it (the finite geometric series sum, the idea of taking a limit when you have an infinite sum, etc.) and thus that argument is more useful mathematically than the bald conditional statement you gave. For example, we can investigate (and here I'm talking about a general situation, not the specific one you're dealing with where these questions have been dealt with in many ways) whether we can modify the proof in some way so that it becomes correct, we can investigate the utility of extracting the correct aspects and applying them to other situations, etc. Dave L. Renfro === Subject: Re: Set theory with p -> q. mina_world a .8ecrit : > Hello teacher~ Find the 1 + 2 + 2^2 + 2^3 + ... Let x = 1 + 2 + 2^2 + 2^3 + ... ---(1) so, 2x = 2 + 2^2 + 2^3 + ... ---(2) so, (1) - (2) ==> -x = 1 so, x = -1. --------------------------------------------------------------------- > Of course, false. > Because, x = 1 + 2 + 2^2 + 2^3 + ... is impossible. If I write that > x = 1 + 2 + 2^2 + 2^3 + ... -> x = -1, (as p -> q) is this true statement by truth table ? > (Because, p is false.) p isn't false, p has no sense in general... but you can give it a sense, e.g.: * working in the field of 2-adic numbers: the series of general term 2^n converges to -1 * using eulerian computing on function entire series === Subject: Geometry with a+b=c. Hello teacher~ http://board-2.blueweb.co.kr/user/math565/data/math/abc.jpg === Subject: Re: Geometry with a+b=c. > Hello teacher~ http://board-2.blueweb.co.kr/user/math565/data/math/abc.jpg A simple pure geometric proof of arctan(1) = arctan(1/2) + arctan(1/3). The one I knew is -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Re: simple question about subgame perfect nash equilibriums - game theory posting-account=G_JWxwkAAACXgEIjxxaemP14QLvS44Jj Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) *bump* === Subject: Re: simple question about subgame perfect nash equilibriums - game theory > *bump* Sci.math is not a web forum, even if you are reading and posting through a web interface. Sci.math is a Usenet forum propagated through nntp protocol. Bump and its brethren are not conventions in sci.math nor in Usenet in general. Google provides access to many forums. Some of them are web forums fabricated by Google. Some of them, such as sci.math, are independent of and predate Google. -- Michael Press === Subject: Re: simple question about subgame perfect nash equilibriums - game theory posting-account=G_JWxwkAAACXgEIjxxaemP14QLvS44Jj Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > Sci.math is not a web forum, even if you are reading > and posting through a web interface. > Sci.math is a Usenet forum propagated through nntp protocol. > Bump and its brethren are not conventions > in sci.math nor in Usenet in general. Who got out of bed on the wrong side today? Michael Press thats who! I know they aren't conventions. Just because the word didnt originate here doesn't mean I cant use it. As time goes by words come and go. > Google provides access to many forums. > Some of them are web forums fabricated by Google. > Some of them, such as sci.math, are independent of > and predate Google. that game theory question... *bump* === Subject: Re: Functions and the Doctrine of Existential Indeterminacy. > As usual I start with the following: > Axiom > Existential forms are self referential So, > That which exists exists. > That which does not exist does not exist. > That which is existentially indeterminate is existentially > indeterminate. Can you ever be sure about that? Tom Davidson > Richmond, VA Nice one. /BAH === Subject: Re: Functions and the Doctrine of Existential Indeterminacy. posting-account=tCEoyAoAAAAkltU5zxOoI8uJ4lyz5-kv .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022; Tablet PC 2.0),gzip(gfe),gzip(gfe) > As usual I start with the following: > Axiom > Existential forms are self referential > So, > That which exists exists. > That which does not exist does not exist. > That which is existentially indeterminate is existentially > indeterminate. Can you ever be sure about that? Tom Davidson > Richmond, VA Nice one. /BAH- Hide quoted text - - Show quoted text - I just got it. lol^2 === Subject: Re: Functions and the Doctrine of Existential Indeterminacy. posting-account=tCEoyAoAAAAkltU5zxOoI8uJ4lyz5-kv .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022; Tablet PC 2.0),gzip(gfe),gzip(gfe) As usual I start with the following: > Axiom > Existential forms are self referential [snip crap] According to Feynman, what does the inside of a brick look like? [just a plain old fashioned snip] There are two different kinds of voodoo. That which is definately voodoo, and that for which you cannot be sure if it is voodoo or not. To err on the side of caution is, in fact, sometimes an err. Speaking of Dick, when he gave that speech about the voodoo science - what was he talking about specifically ? I think I have some ideas, but I am not %100 clear on what he would rule in or rule out. Certainly time travel is voodoo, and remote viewing is voodoo, and StarTrek is voodoo. If anyone knows please elaborate - what was Feybman referring to in his Caltech commencement address when he spoke of ñCargo cult scienceî ? In any case, he did use probability theory, which, at it's core, is based on such things as AOC. So I dont think that anything I've said is the least bit more or less ridiculous than anything that he said. === Subject: Re: Functions and the Doctrine of Existential Indeterminacy. posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > As usual I start with the following: > Axiom > Existential forms are self referential [snip crap] According to Feynman, what does the inside of a brick look like? [just a plain old fashioned snip] There are two different kinds of voodoo. That which is definately > voodoo, and that for which you cannot be sure if it is voodoo or not. To err on the side of caution is, in fact, sometimes an err. Speaking of Dick, when he gave that speech about the voodoo science - > what was he talking about specifically ? I think I have some ideas, > but I am not %100 clear on what he would rule in or rule out. > Certainly time travel is voodoo, and remote viewing is voodoo, and > StarTrek is voodoo. If anyone knows please elaborate - what was > Feybman referring to in his Caltech commencement address when he spoke > of ñCargo cult scienceî ? In any case, he did use probability theory, which, at it's core, is > based on such things as AOC. So I dont think that anything I've said > is the least bit more or less ridiculous than anything that he said. http://en.wikipedia.org/wiki/Cargo cult science === Subject: Re: Functions and the Doctrine of Existential Indeterminacy. posting-account=tCEoyAoAAAAkltU5zxOoI8uJ4lyz5-kv .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022; Tablet PC 2.0),gzip(gfe),gzip(gfe) As usual I start with the following: > Axiom > Existential forms are self referential [snip crap] According to Feynman, what does the inside of a brick look like? [just a plain old fashioned snip] There are two different kinds of voodoo. That which is definately > voodoo, and that for which you cannot be sure if it is voodoo or not. To err on the side of caution is, in fact, sometimes an err. Speaking of Dick, when he gave that speech about the voodoo science - > what was he talking about specifically ? I think I have some ideas, > but I am not %100 clear on what he would rule in or rule out. > Certainly time travel is voodoo, and remote viewing is voodoo, and > StarTrek is voodoo. If anyone knows please elaborate - what was > Feybman referring to in his Caltech commencement address when he spoke > of ñCargo cult scienceî ? In any case, he did use probability theory, which, at it's core, is > based on such things as AOC. So I dont think that anything I've said > is the least bit more or less ridiculous than anything that he said. http://en.wikipedia.org/wiki/Cargo cult science- Hide quoted text - - Show quoted text - Feynman himself consideed prominent examples of bad science. There were alot of things happening in the 70's. ESP, timetravel, remote viewing, all kinds of things. Anyhow - from the link, we know that Feynman recommended: recommended that researchers adopt an unusually high level of honesty which is rarely encountered in everyday life The question I have is this - If you develop a system of conjectures which collectively form sub- doctrines, which are then assembled into a much larger doctrine which categorizes all of these conjectures systematically...... If each of these conjectures can be reduced to a provable statement .... Is it valid to work with conjectures and plausabilities ? Is that junk science ? I think I know the answer - that even this is indeterminate. Even if it produces correct results and predictions in physics - it is still indeterminate. Not a yes, and not a no. It is in the middle between yes and no. But of course - thats where indeterminacy lives. Right in the middle. === Subject: Re: Functions and the Doctrine of Existential Indeterminacy. posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY AppleWebKit/525.19 (KHTML, like Gecko) Chrome/0.3.154.9 Safari/525.19,gzip(gfe),gzip(gfe) > As usual I start with the following: > Axiom > Existential forms are self referential [snip crap] According to Feynman, what does the inside of a brick look like? [just a plain old fashioned snip] There are two different kinds of voodoo. That which is definately > voodoo, and that for which you cannot be sure if it is voodoo or not. To err on the side of caution is, in fact, sometimes an err. Speaking of Dick, when he gave that speech about the voodoo science - > what was he talking about specifically ? I think I have some ideas, > but I am not %100 clear on what he would rule in or rule out. > Certainly time travel is voodoo, and remote viewing is voodoo, and > StarTrek is voodoo. If anyone knows please elaborate - what was > Feybman referring to in his Caltech commencement address when he spoke > of ñCargo cult scienceî ? In any case, he did use probability theory, which, at it's core, is > based on such things as AOC. So I dont think that anything I've said > is the least bit more or less ridiculous than anything that he said. http://en.wikipedia.org/wiki/Cargo cult science-Hide quoted text - - Show quoted text - Feynman himself consideed prominent examples of bad science. There > were alot of things happening in the 70's. ESP, timetravel, remote > viewing, all kinds of things. Anyhow - from the link, we know that Feynman recommended: > recommended that researchers adopt an unusually high level of honesty > which is rarely encountered in everyday life The question I have is this - If you develop a system of conjectures which collectively form sub- > doctrines, which are then assembled into a much larger doctrine which > categorizes all of these conjectures systematically...... If each of these conjectures can be reduced to a provable > statement .... Is it valid to work with conjectures and plausabilities ? Is that junk > science ? I think I know the answer - that even this is indeterminate. Even if it produces correct results and predictions in physics - it is > still indeterminate. Not a yes, and not a no. It is in the middle > between yes and no. But of course - thats where indeterminacy lives. > Right in the middle. The answer is surprisingly stupid. You appear to ask a question because you want to know the answer. You even believe in this yourself. But you also want to hear a specific answer. This you know to be true. Now... that what you want to hear has far more weight as the actual answer. So... lets assume you are right, in fact it had been figured out a thousand times already. Do you seriously think it has any value against that what people want to hear? Personally I've been wrong lots of times when I thought I was right, I've also been right lots of times when I thought I was wrong. I've discovered the valium like state people live in goes beyond anyone's wildest imagination. I discovered how naive I was about lots of things, I've also discovered how serious I was about nonsense. The best way to hang on to the things you want to hear is by failure in connecting cause and effect. I cant stomach killing a cow, but boy do I love eating them. I know cause and effect but I'm not going to bridge the relationship. Billions of people are starving so that we can discuss trivial subjects in a newsgroup. hahahahaha I think it is pretty damn funny actually. Those poor bastards! Jet here we are, acting as-if the world is a pink cloud. Who *caused* them not to have drinking water? Was it you? me? Lets assume it is all your fault for a moment. lol Would you want to hear about it? Lets try a different example. You stated something like: Certainly time travel is voodoo, remote viewing is voodoo, StarTrek is voodoo, ESP is voodoo. First Feynman asserts voodoo is a new definition of fake. You then borrow the terminology and add examples of assertions of your own. But, by asserting something to be fake you are no longer capable of investigation. This suggests sufficient level of truth was already acquired without any investigation. Which proves that what people want to hear is of far greater value than the actual data. In the middle between yes and no the indeterminacy lives. Not just right in the middle but actual yes and actual no don't really exist. Unless you like to think they do of course :o) Lots of words but no real answer - sorry. This is funny. http://www.youtube.com/watch?v=wt3smrXkVpE === Subject: Re: Functions and the Doctrine of Existential Indeterminacy. posting-account=tCEoyAoAAAAkltU5zxOoI8uJ4lyz5-kv .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022; Tablet PC 2.0),gzip(gfe),gzip(gfe) As usual I start with the following: > Axiom > Existential forms are self referential [snip crap] According to Feynman, what does the inside of a brick look like? [just a plain old fashioned snip] There are two different kinds of voodoo. That which is definately > voodoo, and that for which you cannot be sure if it is voodoo or not. To err on the side of caution is, in fact, sometimes an err. Speaking of Dick, when he gave that speech about the voodoo science - > what was he talking about specifically ? I think I have some ideas, > but I am not %100 clear on what he would rule in or rule out. > Certainly time travel is voodoo, and remote viewing is voodoo, and > StarTrek is voodoo. If anyone knows please elaborate - what was > Feybman referring to in his Caltech commencement address when he spoke > of ñCargo cult scienceî ? In any case, he did use probability theory, which, at it's core, is > based on such things as AOC. So I dont think that anything I've said > is the least bit more or less ridiculous than anything that he said. http://en.wikipedia.org/wiki/Cargo cult science-Hidequoted text - - Show quoted text - Feynman himself consideed prominent examples of bad science. There > were alot of things happening in the 70's. ESP, timetravel, remote > viewing, all kinds of things. Anyhow - from the link, we know that Feynman recommended: > recommended that researchers adopt an unusually high level of honesty > which is rarely encountered in everyday life The question I have is this - If you develop a system of conjectures which collectively form sub- > doctrines, which are then assembled into a much larger doctrine which > categorizes all of these conjectures systematically...... If each of these conjectures can be reduced to a provable > statement .... Is it valid to work with conjectures and plausabilities ? Is that junk > science ? I think I know the answer - that even this is indeterminate. Even if it produces correct results and predictions in physics - it is > still indeterminate. Not a yes, and not a no. It is in the middle > between yes and no. But of course - thats where indeterminacy lives. > Right in the middle. The answer is surprisingly stupid. You appear to ask a question because you want to know the answer. You > even believe in this yourself. But you also want to hear a specific > answer. This you know to be true. Now... that what you want to hear > has far more weight as the actual answer. So... lets assume you are > right, in fact it had been figured out a thousand times already. Do > you seriously think it has any value against that what people want to > hear? Personally I've been wrong lots of times when I thought I was > right, I've also been right lots of times when I thought I was wrong. > I've discovered the valium like state people live in goes beyond > anyone's wildest imagination. I discovered how naive I was about lots > of things, I've also discovered how serious I was about nonsense. The best way to hang on to the things you want to hear is by failure > in connecting cause and effect. I cant stomach killing a cow, but boy do I love eating them. I know > cause and effect but I'm not going to bridge the relationship. > Billions of people are starving so that we can discuss trivial > subjects in a newsgroup. hahahahaha I think it is pretty damn funny > actually. Those poor bastards! Jet here we are, acting as-if the world > is a pink cloud. Who *caused* them not to have drinking water? Was it > you? me? Lets assume it is all your fault for a moment. lol Would you > want to hear about it? Lets try a different example. You stated something like: Certainly > time travel is voodoo, remote viewing is voodoo, StarTrek is voodoo, > ESP is voodoo. First Feynman asserts voodoo is a new definition of > fake. You then borrow the terminology and add examples of assertions > of your own. But, by asserting something to be fake you are no longer > capable of investigation. This suggests sufficient level of truth was > already acquired without any investigation. Which proves that what > people want to hear is of far greater value than the actual data. In the middle between yes and no the indeterminacy lives. Not just > right in the middle but actual yes and actual no don't really exist. > Unless you like to think they do of course :o) Lots of words but no real answer - sorry. This is funny. http://www.youtube.com/watch?v=wt3smrXkVpE- Hide quoted text - - Show quoted text - In all sincerity your reply was absolutely delicious to read, particularly the link to youtube. Best laugh I had all day, ty. So how to reply to that. Hmmmm. Well, I am constantly thinking about voodoo science because that is precisely what people would call this if it is nonsense. And, that is also what I would call it if it were nonsense. But I cannot see why it would be wrong. I know why it is not mathematics, but that does not mean that you dont have something useable. If you can construct a system by which conjecture is formalized If you have a billiard table and there are balls bouncing around, everything is nice and deterministic. Disregarding Plancklength and QM for a minute - consider an idealized perfect mathematical pool table. Perfect spheres, perfectly flat surfaces, zero friction. Of you want to have something like probable causality, the easiest way to do it is by making one or more balls probabilistically existent. There are too many examples to point to where such an approach to modelling is useful. I want to put it on a formal footing so that if someone takes that approach toward modelling, they know that the foundations are sound. I know that all of this philosophical drivel gets very boring, but it must be done in order to clarify that one is not guilty of voodoo. What I would really like to do is spend more time modelling randomness as if it were a fluid, and use that approach to model dynamic spacebending like fluid dynamics. But it makes no sense to even discuss such a thing unless ALL of the fundamentals are beaten literally to death. BTW - the secret lives of plants video - that will keep me laughing a === Subject: Re: Functions and the Doctrine of Existential Indeterminacy. I want to put it on a formal footing so that if someone takes that > approach toward modelling, they know that the foundations are sound. > I know that all of this philosophical drivel gets very boring, but it > must be done in order to clarify that one is not guilty of voodoo. > If you are *really* interested, start by taking a logics course offered by the Philosophy Dept. in your university. I don't know what they are called. Go ask the dept. head. /BAH === Subject: Re: Functions and the Doctrine of Existential Indeterminacy. posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1 FOH:R051,gzip(gfe),gzip(gfe) there is absolutely nothing deterministic about billiards, mainly because of english. perhaps you are referring to the mathematical idealization thereof, using point masses (a.k.a. The Photon, the only truly zero-D object in Universe .-) > If you have a billiard table and there are balls bouncing around, > everything is nice and deterministic. Disregarding Plancklength and QM thus: you really seem to like beating that horse; is it dead, yet? seriously, any polynomial with a constant (or times x to the zeroth power) cannot have zero as a solution; otherwise, zero is a trivial solution, since zero times any coefficient is nothing! anyway, what's the problem, if an infinite series polynomial is not algebraic per definition? > roots and all that. But if it is significant, then if transcendental > polynomials can't have zero as root then why should they have any number > as a root? thus: Euler's identity is, what -- exp(i*pi) = 1, i*pi = ln(1) pi = -i*ln(1), but I forgot what the natural log of one is. thus: there's got to be a few Latin phrases for this kind of fallacy; this is the same thing that Are Buckafka Fullofit tried to put over, and most of the Buckywitches really believe it. anyway, the queestion is, why should line segments of polygons be so God-am existential, or of polyhedra -- because of the typical stick-drawn configuration in some edition of Euclid? >Also, from the position taken above, spheres are only possible in theory aswell. --Cheeny & Zbiggy, fo'mo' years! http://tarpley.net http://larouchepub.com http://xplodeyourmlmbiz.com/?s1=Fuller+Brush&t=LSP01&gclid=CPiar6nispcCFQu-G godYm21kA http://www.rwgrayprojects.com/synergetics/plates/plates.html === Subject: Re: Functions and the Doctrine of Existential Indeterminacy. posting-account=tCEoyAoAAAAkltU5zxOoI8uJ4lyz5-kv .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022; Tablet PC 2.0),gzip(gfe),gzip(gfe) > there is absolutely nothing deterministic about billiards, > mainly because of english. perhaps you are referring > to the mathematical idealization thereof, using point masses > (a.k.a. The Photon, the only truly zero-D object > in Universe .-) If you have a billiard table and there are balls bouncing around, > everything is nice and deterministic. Disregarding Plancklength and QM thus: > you really seem to like beating that horse; > is it dead, yet? seriously, any polynomial with a constant (or times x to the zeroth > power) cannot > have zero as a solution; otherwise, zero is a trivial solution, > since zero times any coefficient is nothing! anyway, what's the problem, if an infinite series polynomial is not > algebraic > per definition? roots and all that. But if it is significant, then if transcendental > polynomials can't have zero as root then why should they have any number > as a root? thus: > Euler's identity is, what -- > exp(i*pi) = 1, > i*pi = ln(1) > pi = -i*ln(1), but > I forgot what the natural log of one is. thus: > there's got to be a few Latin phrases for this kind of fallacy; > this is the same thing that Are Buckafka Fullofit tried to put over, > and most of the Buckywitches really believe it. > anyway, the queestion is, why should line segments of polygons > be > so God-am existential, or of polyhedra -- because > of the typical stick-drawn configuration in some edition of Euclid? Also, from the position taken above, spheres are only possible in theory aswell. --Cheeny & Zbiggy, fo'mo' years!http://tarpley.nethttp://larouchepub.comhttp://xplodeyourmlmbiz.com/?s 1 =Fuller+Brush&t=LSP01&gclid=CPiar6nisp...http://www.rwgrayprojects.com/syner g etics/plates/plates.html A reading from the gospel of Wikipedia http://en.wikipedia.org/wiki/Axiom of choice For any set of non-empty sets, X, there exists a choice function f defined on X. Thus the negation of the axiom of choice states that there exists a set of nonempty sets which has no choice function. To say that one can choose to accept or not accept the AOC is tantamount to accepting or rejecting the existence of the choice function. Clearly f may or may not exist, and you get to choose which. But there is something more fundamental, which is the original Axiom that I posted above in the first message in this thread. To reiterate: Axiom Existential forms are self referential That which exists exists. That which does not exist does not exist. That which is existentially indeterminate is existentially indeterminate. If you reject this Axiom, then you have AOC If you accept this Axiom, then you still have AOC in mathematics, but you also have the couterpart to AOC which is conjecture. A conjectural AOC would look something like this: For any set of non-empty sets, X, there may exist a choice function f defined on X. And this statement, while conjectural, is convertible into a non- conjectural statement by destroying the indeterminacy of the word may. AOC must have a counterpart in this Doctrine of Existential Indeterminacy, and the Axiom given above gives rise to them both. === Subject: Re: Functions and the Doctrine of Existential Indeterminacy. posting-account=tCEoyAoAAAAkltU5zxOoI8uJ4lyz5-kv .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022; Tablet PC 2.0),gzip(gfe),gzip(gfe) ... operators which are > indeterminately either addition or multiplication. That is the logical equivalent of saying something happens, maybe, > but we can't be sure. You are trying to define something as undefined. It is a matter of perspective. Addition, viewed in a logarithmic > space, IS multiplication. Your philosophy is merely sophistry, intended to deceive yourself and > any naive others into believing you think you really understand > something. lol - I would say the same thing if it were someone else who said it. Unfortunately I seem to actually believe what I am saying. I will explain how it works in laymans terms. Mathematics is like a big thick book. The bulk of it deals with logical relationships among things which exist. Then, somewhere on the very last page there is a point sized singularity which is not there, this is nonexistence. So, you have two chapters. That which exists, and that which does not. The third chapter of this book may or may not be there, it is based totally on indeterminacy and serves as a boundary between the chapters on the existent and the nonexistent. This is the area where existential indeterminacy lives. You get this chapter of the book by ACCEPTING the Axiom which I mentioned above. Axioms are never proved. But the paradox which results it that even if you accept the Axiom, you still have the situation that this entire chapter on existential indeterminacy might not be there, and you just have to live with it. You cannot say it is not there, and you cannot say it is. You have to say : indeterminate. So, we devise a doctrine to give this chapter some structure, and it should be possible to validate components of this doctrine by imposing order - thereby destroying the uncertainty - which would yield provable statements. By reducing the conjectures contained in the doctrine to provable statements, you can validate that the doctrine is within the bounds of the plausible. The way to reduce a conjecture is by imposing order on it - with something like an anti-random variable. > Tom Davidson > Richmond, VA === Subject: Re: Functions and the Doctrine of Existential Indeterminacy. posting-account=tCEoyAoAAAAkltU5zxOoI8uJ4lyz5-kv .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022; Tablet PC 2.0),gzip(gfe),gzip(gfe) As usual I start with the following: > Axiom > Existential forms are self referential So, > That which exists exists. > That which does not exist does not exist. > That which is existentially indeterminate is existentially > indeterminate. Can you ever be sure about that? That's why I call it an Axiom. It cannot be proved, and does not need to be. I think that this is more fundamental than AOC, and I suspect it is the root cause of why you have such a thing as the AOC in the first place. > Tom Davidson > Richmond, VA === Subject: Re: why does professor david c ullrich have to put people down to Nntp-Posting-Host: hera.cwi.nl ... > Earlier in the thread you said > that I was helpful back in the 1990's but > it's obvious that then I was Changed by > the arrival of the Cranks, and I'm no > longer helpful because of the changes > their arrival made. Now the date of the > Great Change is pushed up to some time > before 2008. > > I don't know the _exact_ date, since the arrival > of the cranks was gradual. I do not know either. But AP started before David started posting. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: why does professor david c ullrich have to put people down to >... > Earlier in the thread you said > that I was helpful back in the 1990's but > it's obvious that then I was Changed by > the arrival of the Cranks, and I'm no > longer helpful because of the changes > their arrival made. Now the date of the > Great Change is pushed up to some time > before 2008. I don't know the _exact_ date, since the arrival > of the cranks was gradual. I do not know either. But AP started before David started posting. AP started even before he was AP! He used to be Ludwig Plutonium. >I preferred that name. The cranks have been around since the beginning of USENET. What has changed in recent years is the decline in the number of non-cranks posting. -- Daryl McCullough Ithaca, NY === Subject: Re: why does professor david c ullrich have to put people down to > The cranks have been around since the beginning of USENET. What has changed > in recent years is the decline in the number of non-cranks posting. Since obviously they are now in the minority, the non-cranks have become the cranks. Justice at last. Han de Bruijn === Subject: Re: why does professor david c ullrich have to put people down to posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e 2.0.50727),gzip(gfe),gzip(gfe) > The cranks have been around since the beginning of USENET. What has changed > in recent years is the decline in the number of non-cranks posting. Since obviously they are now in the minority, the non-cranks have become > the cranks. Justice at last. Han de Bruijn ***************************************************** Hi there, Han. Nonsense. Majority, minority: if someone's right about something, then he's still right about it even if ALL the rest say otherwise. And I'm talking about science, maths and stuff, NOT about subjective reasoning like religions, ethics and morals and etc. Galileo still was right when the huge bulk of the population back then didn't believe him, so 99.998% of the people then were cranks, and Galileo was not. Tonio === Subject: Re: why does professor david c ullrich have to put people down to The cranks have been around since the beginning of USENET. What has changed > in recent years is the decline in the number of non-cranks posting. > Since obviously they are now in the minority, the non-cranks have become > the cranks. Justice at last. What a depressing thought. Hi there, Han. Nonsense. Majority, minority: if someone's right about something, then > he's still right about it even if ALL the rest say otherwise. > And I'm talking about science, maths and stuff, NOT about subjective > reasoning like religions, ethics and morals and etc. (Oh dear, I really must stay out of these threads! Perhaps 50p a > post?) Are you implying that an individual can't be right about something > religiously/ethically/morally even if there isn't one single other > voice to be heard in support of what [s]he believes? Or are you just wisely staying out of the religious/ethical/moral > area altogether, and limiting your remarks to science/maths (in the > interests of simplicity and relevance)? > Why not make a top 10 list of math. heresies/heretics ? And if JSH or someone else doesn't accept some standard argument, or if the math cabal doesn't accept a JSH or other argument, at some point I'd rather just say forget it; this could last a lifetime. Admittedly, some heretics like Oliver Heaviside arrived at results that were right (I think) by heretic methods. http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Heaviside.html But in physics and cosmology, maybe some heretics are skeptics. Cf.: http://www.hiltonratcliffe.com/index.htm David Bernier === Subject: Re: why does professor david c ullrich have to put people down to posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e 2.0.50727),gzip(gfe),gzip(gfe) > The cranks have been around since the beginning of USENET. What has changed > in recent years is the decline in the number of non-cranks posting. > Since obviously they are now in the minority, the non-cranks have become > the cranks. Justice at last. What a depressing thought. Hi there, Han. Nonsense. Majority, minority: if someone's right about something, then >he's still right about it even if ALL the rest say otherwise. >And I'm talking about science, maths and stuff, NOT about subjective >reasoning like religions, ethics and morals and etc. (Oh dear, I really must stay out of these threads! Perhaps 50p a > post?) Are you implying that an individual can't be right about something > religiously/ethically/morally even if there isn't one single other > voice to be heard in support of what [s]he believes? Or are you just wisely staying out of the religious/ethical/moral > area altogether, and limiting your remarks to science/maths (in the > interests of simplicity and relevance)? > *************************************************************** I think both: I don't believe in the truth that so often is voiced in religious, political, moral, etc. issues, but I do believe in soundness and logic in mathematics in science, according to an agreed set of axioms and definition and rules of inference. Tonio > Angus Rodgers === Subject: Re: why does professor david c ullrich have to put people down to >The cranks have been around since the beginning of USENET. What has changed >in recent years is the decline in the number of non-cranks posting. Since obviously they are now in the minority, the non-cranks have become >the cranks. Justice at last. What a depressing thought. >Hi there, Han. >Nonsense. Majority, minority: if someone's right about something, then >he's still right about it even if ALL the rest say otherwise. >And I'm talking about science, maths and stuff, NOT about subjective >reasoning like religions, ethics and morals and etc. (Oh dear, I really must stay out of these threads! Perhaps 50p a >post?) Are you implying that an individual can't be right about something >religiously/ethically/morally even if there isn't one single other >voice to be heard in support of what [s]he believes? Or are you just wisely staying out of the religious/ethical/moral >area altogether, and limiting your remarks to science/maths (in the >interests of simplicity and relevance)? > *************************************************************** I think both: I don't believe in the truth that so often is voiced > in religious, political, moral, etc. issues, but I do believe in > soundness and logic in mathematics in science, according to an > agreed set of axioms and definition and rules of inference. Tonio Yes, and THAT makes your soundness incredibly boring. Yyyaaaawwwwnnn . Therefore I'll tell you once again what the _real_ world is about: Reality is that which, when you stop believing in it, doesn't go away. (Philip K. Dick) http://en.wikipedia.org/wiki/Agnosticism I only believe what I know. But you'd be surprised about what I believe. Han de Bruijn === Subject: Re: why does professor david c ullrich have to put people down to <81e4e$493e71c6$82a1e228$25265@news1.tudelft.nl> posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e 2.0.50727),gzip(gfe),gzip(gfe) >The cranks have been around since the beginning of USENET. What has changed >in recent years is the decline in the number of non-cranks posting. >Since obviously they are now in the minority, the non-cranks have become >the cranks. Justice at last. >What a depressing thought. Hi there, Han. Nonsense. Majority, minority: if someone's right about something, then >he's still right about it even if ALL the rest say otherwise. >And I'm talking about science, maths and stuff, NOT about subjective >reasoning like religions, ethics and morals and etc. >(Oh dear, I really must stay out of these threads! Perhaps 50p a >post?) >Are you implying that an individual can't be right about something >religiously/ethically/morally even if there isn't one single other >voice to be heard in support of what [s]he believes? >Or are you just wisely staying out of the religious/ethical/moral >area altogether, and limiting your remarks to science/maths (in the >interests of simplicity and relevance)? *************************************************************** I think both: I don't believe in the truth that so often is voiced > in religious, political, moral, etc. issues, but I do believe in > soundness and logic in mathematics in science, according to an > agreed set of axioms and definition and rules of inference. Tonio Yes, and THAT makes your soundness incredibly boring. Yyyaaaawwwwnnn . Therefore I'll tell you once again what the real world is about: Reality is that which, when you stop believing in it, doesn't go away. > (Philip K. Dick) http://en.wikipedia.org/wiki/Agnosticism I only believe what I know. But you'd be surprised about what I believe. Han de Bruijn- ************************************************************** Oh, after reading your rather psychodelic and picturesque participations in threads about Cantor's theorem and/or set theory, I really can't be surprised by what you believe, my good Han. Tonio === Subject: Re: why does professor david c ullrich have to put people down to The cranks have been around since the beginning of USENET. What has changed >in recent years is the decline in the number of non-cranks posting. Since obviously they are now in the minority, the non-cranks have become >the cranks. Justice at last. Han de Bruijn ***************************************************** Hi there, Han. Nonsense. Majority, minority: if someone's right about something, then > he's still right about it even if ALL the rest say otherwise. > And I'm talking about science, maths and stuff, NOT about subjective > reasoning like religions, ethics and morals and etc. Galileo still was right when the huge bulk of the population back then > didn't believe him, so 99.998% of the people then were cranks, and > Galileo was not. Tonio Nope. Galileo _was_ a crank at that time. But, sometimes, cranks win ! Han de Bruijn === Subject: Re: why does professor david c ullrich have to put people down to <36474$493e55d7$82a1e228$24936@news1.tudelft.nl> posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e 2.0.50727),gzip(gfe),gzip(gfe) >The cranks have been around since the beginning of USENET. What has changed >in recent years is the decline in the number of non-cranks posting. >Since obviously they are now in the minority, the non-cranks have become >the cranks. Justice at last. >Han de Bruijn ***************************************************** Hi there, Han. Nonsense. Majority, minority: if someone's right about something, then > he's still right about it even if ALL the rest say otherwise. > And I'm talking about science, maths and stuff, NOT about subjective > reasoning like religions, ethics and morals and etc. Galileo still was right when the huge bulk of the population back then > didn't believe him, so 99.998% of the people then were cranks, and > Galileo was not. Tonio Nope. Galileo was a crank at that time. But, sometimes, cranks win ! Han de Bruijn- ********************************************************** Ok Han, if that though comforts you then so be it...for you. Tonio === Subject: Re: why does professor david c ullrich have to put people down to Nntp-Posting-Host: hera.cwi.nl > > I know that Ullrich has posted at sci.math > > longer than anyone else > I humbly protest... > > Well, Ullrich definitely has the most posts according > to Google. For mina_world, who was looking up Google, > only checked the _number_ of posts, not the date to > see who was here the longest. So longer is actually more? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: holomorphically convex > And in particular K_n is contained in the interior of K_{n+1} in > the later book? Yes - there is apparently no subtle reason for not needing that to show that if a domain is the union of holomorphically convex nested compact sets, the holomorphic convex hull of a compact set is compact. Laura === Subject: Re: crank crank crank Nntp-Posting-Host: hera.cwi.nl > 1. My interpretation is wrong, and the standard > mathematicians are upset that I've misinterpreted > their intentions. > 2. My interpretation is right, but the standard > mathematicians don't want to admit it. > Well, I think it's time somebody breaks the news to you, and I am > happy to be that one: there is no such thing a non-standard > mathematics or non-standard mathematicians. > > There is certainly nonstandard _analysis_ (e.g. > Robinson's infinitesimals), as opposed to the > standard (classical) analysis. Aha. But non-standard analysis is standard mathematics ;-). > Mathematicians is anyone who loves and understands mathematics, and > this already implies a lot (like open mind and some kind of humbleness > when facing unknown stuff). > > I keep on using the term standard mathematicians > because I see no easy name to refer to the two > sides of this debate. Sometimes I see the word > Cantorian used for this purpose, but this word > isn't appropropiate since this particular debate > has nothing to do with Cantor. The word Cantorian is frequently used by people who do not know how to do any logical reasoning in there defense of their ill-bred theories. To wit the preface of WM's book on mathematics for teaching purposes: no axioms, no definitions, no theorems. Well, if you do not give definitions nothing can be proven in a mathematical sense. And that is the problem with the other side. They do not provide definitions, they just spout terminology. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: crank crank crank > [...] What's going on is complicated, and I think you see some part > of it that I don't (but that I want to see), but I also think > you're missing something (which I at least vaguely glimpse). Galthaea thinks it is alright for galathaea to do things >that galathaea scolds others for doing, and galathaea >You want to discuss propriety with someone such as that? Are you saying that one should never discuss anything with > anyone who has done anything of which one disapproves, and > therefore that having a conversation with anyone implies a > blanket approval of every single thing they have ever said > or done? If not, what are you saying? It's quite possible > that I'm being thick and/or insensitive (and I'm certainly > having another off-day), so I apologise in advance in case > that is so, but I'm completely mystified by your response. What is to be mystified over? Galathaea acknowledges that galathaea is justified in doing to someone the same thing that galathaea scolds that person for doing. -- Michael Press === Subject: Re: crank crank crank posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.4) Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) [...] I think that dichotomies have a place (but so do trichotomies > and all kinds of deviant logics, and God knows what else). I > don't at all understand your philosophical position, or a lot > of the mathematics that you know, but I at least vaguely suspect > that in your justifiable desire to find a place for perspectives > excluded from classical mathematics, you risk rendering invalid > some of the classical mathematics in which you yourself believe > (whether you admit it or not). I suspect there is some kind of > psychological contradiction in your system, and you rather depend > on there being other people to represent something that you have > to deny in yourself. Perhaps you secretly wish not to have to > be in this battle, and to be able to accept what you are still > denying, so that you don't have to rely on someone else accepting > it on your behalf so that you can argue with them and carry on > with your denial. Or, I may just be full of bull. What's going on is complicated, and I think you see some part > of it that I don't (but that I want to see), but I also think > you're missing something (which I at least vaguely glimpse). Galthaea thinks it is alright for galathaea to do things > that galathaea scolds others for doing, and galathaea > You want to discuss propriety with someone such as that? galathaea thinks galathaea can do things that defend herself from attacks there is a major difference between insults directed towards a person intended to lessen or belittle and insults couched on the condition if david's reasoning were correct then.. some lessons are more effective than others these are words not swords defense is much more important here -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: crank crank crank <63doj4tkjt09r8gkfuhr7e7711d72utau6@4ax.com> posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e 2.0.50727),gzip(gfe),gzip(gfe) >[...] > What's going on is complicated, and I think you see some part > of it that I don't (but that I want to see), but I also think > you're missing something (which I at least vaguely glimpse). Galthaea thinks it is alright for galathaea to do things >that galathaea scolds others for doing, and galathaea >You want to discuss propriety with someone such as that? Are you saying that one should never discuss anything with > anyone who has done anything of which one disapproves, and > therefore that having a conversation with anyone implies a > blanket approval of every single thing they have ever said > or done? If not, what are you saying? It's quite possible > that I'm being thick and/or insensitive (and I'm certainly > having another off-day), so I apologise in advance in case > that is so, but I'm completely mystified by your response. -- > Angus Rodgers ************************************************************** I think Michael merely said that galathea whinned hard about abuse and bullying while she herself abused and bullied many chatters here. I know, sometimes is easy to get lost in all that poetic-like mumbo- as IF THIS ISN'T ING WORSE THAN TOMMY BEHAVIOR THAN YOU ERS ARE BLIND!! , where we ALL at once become ers and, besides that, blind, in order to fully understand that the lady has now REALLY got pissed off. Sfter the torment the calm does NOT come over, but it is usually followed by a more or less long parade of self-righteous and show-off posts, which mostly say nothing new. **Sigh** Tonio === Subject: Matrix Equation XX^TAXX^T=B posting-account=S688EQoAAAAnFmRHjMyAK7LqLaNGIutk AppleWebKit/525.19 (KHTML, like Gecko) Chrome/0.4.154.29 Safari/525.19,gzip(gfe),gzip(gfe) which X^T is the transpose matrix of X, A and B are positive definite. Can anybody provide some clues on the solutions? Simon === Subject: Re: Matrix Equation XX^TAXX^T=B > X*X^T*A*X*X^T=B, in > which X^T is the transpose matrix of X, A and B are > positive > definite. Can anybody provide some clues on the solutions? > Simon I know that by Sylvester's law of intertia, there exists Y in GL(n,IR) such that Y^t*A*Y = B, but is it true that this Y can be chosen symmetric positive definite as indicated above ? Maybe if A and B are simultaneously diagonizable ... Best wishes Torsten. === Subject: Re: Matrix Equation XX^TAXX^T=B > X*X^T*A*X*X^T=B, in > which X^T is the transpose matrix of X, A and B > are > positive > definite. Can anybody provide some clues on the solutions? > Simon I know that by Sylvester's law of intertia, there > exists Y in GL(n,IR) such that > Y^t*A*Y = B, > but is it true that this Y can be chosen > symmetric positive definite as indicated above ? > Maybe if A and B are simultaneously diagonizable ... Best wishes > Torsten. in the sci.math.research - forum. === Subject: Re: Matrix Equation XX^TAXX^T=B posting-account=O9zR9AkAAACmp918j6u5m5plppeILcze Filter 1.2.0.72; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022; WWTClient2),gzip(gfe),gzip(gfe) > which X^T is the transpose matrix of X, A and B are positive > definite. Can anybody provide some clues on the solutions? > Simon See https://kb.osu.edu/dspace/bitstream/1811/22234/1/V074N5 273.pdf Dave === Subject: Re: Matrix Equation XX^TAXX^T=B posting-account=S688EQoAAAAnFmRHjMyAK7LqLaNGIutk Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) which X^T is the transpose matrix of X, A and B are positive > definite. 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One of the words in the subject is a give-away that it isn't. > Oh you're right. XP > Maybe I am that naive about the internet. Apparently. Such subjects are common to porno web sites. So they say. So they say? ... LOL! === Subject: TMFCS-09 call for papers posting-account=H_onOgoAAADscC8Mv_EiDpbOw2hhNnM7 .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) TMFCS-09 call for papers The 2009 International Conference on Theoretical and Mathematical Foundations of Computer Science (TMFCS-09) (website: http://www.PromoteResearch.org ) will be held during July 13-16 2009 in Orlando, FL, USA. We invite draft paper submissions. The conference will take place at the same time and venue where several other international conferences are taking place. The other conferences include: ´ International Conference on Artificial Intelligence and Pattern Recognition (AIPR-09) ´ International Conference on Automation, Robotics and Control Systems (ARCS-09) ´ International Conference on Bioinformatics, Computational Biology, Genomics and Chemoinformatics (BCBGC-09) ´ International Conference on Enterprise Information Systems and Web Technologies (EISWT-09) ´ International Conference on High Performance Computing, Networking and Communication Systems (HPCNCS-09) ´ International Conference on Information Security and Privacy (ISP-09) ´ International Conference on Recent Advances in Information Technology and Applications (RAITA-09) ´ International Conference on Software Engineering Theory and Practice (SETP-09) ´ International Conference on Theory and Applications of Computational Science (TACS-09) The website http://www.PromoteResearch.org contains more details. John Edward Publicity committee === Subject: TMFCS-09 call for papers posting-account=H_onOgoAAADscC8Mv_EiDpbOw2hhNnM7 .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) TMFCS-09 call for papers The 2009 International Conference on Theoretical and Mathematical Foundations of Computer Science (TMFCS-09) (website: http://www.PromoteResearch.org ) will be held during July 13-16 2009 in Orlando, FL, USA. We invite draft paper submissions. The conference will take place at the same time and venue where several other international conferences are taking place. The other conferences include: ´ International Conference on Artificial Intelligence and Pattern Recognition (AIPR-09) ´ International Conference on Automation, Robotics and Control Systems (ARCS-09) ´ International Conference on Bioinformatics, Computational Biology, Genomics and Chemoinformatics (BCBGC-09) ´ International Conference on Enterprise Information Systems and Web Technologies (EISWT-09) ´ International Conference on High Performance Computing, Networking and Communication Systems (HPCNCS-09) ´ International Conference on Information Security and Privacy (ISP-09) ´ International Conference on Recent Advances in Information Technology and Applications (RAITA-09) ´ International Conference on Software Engineering Theory and Practice (SETP-09) ´ International Conference on Theory and Applications of Computational Science (TACS-09) The website http://www.PromoteResearch.org contains more details. John Edward Publicity committee === Subject: Algebra with ideal R_1 x R_2. Hello teacher~ R_1 is a commutative ring with unity. R_2 is a commutative ring with unity. I is a ideal of R_1 x R_2 <==> I = I_1 x I_2 for some I_1, I_2. (I_1 is a ideal of R_1, I_2 is a ideal of R_2.) ---------------------------------------------------------------------------- pf) (<=) easy to show subring + ideal property. (=>) How do you show it ? === Subject: Re: Algebra with ideal R_1 x R_2. posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.2; .NET CLR 3.5.21022; Tablet PC 2.0),gzip(gfe),gzip(gfe) > Hello teacher~ R 1 is a commutative ring with unity. > R 2 is a commutative ring with unity. I is a ideal of R 1 x R 2 > <== I = I 1 x I 2 for some I 1, I 2. > (I 1 is a ideal of R 1, I 2 is a ideal of R 2.) ---------------------------------------------------------------------------[ CapitalEth]- > pf) > (<=) easy to show subring + ideal property. (=>) How do you show it ? **************************************************************************** ** Let J be an ideal of R 1 x R 2 . Now look at J / R 1 and J / R 2... Tonio === Subject: Re: Algebra with ideal R_1 x R_2. > R_1 is a commutative ring with unity. > R_2 is a commutative ring with unity. I is a ideal of R_1 x R_2 > <== I = I_1 x I_2 for some I_1, I_2. > (I_1 is a ideal of R_1, I_2 is a ideal of R_2.) ---------------------------------------------------------------------------- > pf) > (<=) easy to show subring + ideal property. (=>) How do you show it ? If (x,y) is in I, then (x,y) = (x,0) + (0,y). But (x,0) = (1,0).(x,y) and therefore (x,0) belongs to I. By the same argument, (0,y) belongs to I. Therefore, if you define I_1 = { x in R_1 | (x,0) in I } and I_2 = { x in R_2 | (0,x) in I } then I = I_1 x I_2. It it easy to check that I_1 and I2 are ideals. Jose Carlos Santos === Subject: Re: Algebra with ideal R_1 x R_2. > R_1 is a commutative ring with unity. > R_2 is a commutative ring with unity. I is a ideal of R_1 x R_2 > <==> I = I_1 x I_2 for some I_1, I_2. > (I_1 is a ideal of R_1, I_2 is a ideal of R_2.) ---------------------------------------------------------------------------- > pf) > (<=) easy to show subring + ideal property. (=>) How do you show it ? If (x,y) is in I, then (x,y) = (x,0) + (0,y). But (x,0) = (1,0).(x,y) and therefore (x,0) belongs to I. By the same argument, (0,y) belongs to > I. Therefore, if you define I_1 = { x in R_1 | (x,0) in I } and I_2 = { x in R_2 | (0,x) in I } then I = I_1 x I_2. It it easy to check that I_1 and I2 are ideals. Yes, good idea. (a, b) in I ==> (a, 0) in I, (0, b) in I ==> a in I_1, b in I_2 ==> (a, b) in I_1 x I_2 (a, b) in I_1 x I_2 ==> a in I_1, b in I_2 ==> (a, 0) in I, (0, b) in I ==> (a, 0) + (0, b) = (a, b) in I. if R is a field, {0}x{0}, {0}xR, Rx{0}, RxR are all ideals of RxR. === Subject: The half-baked quotient I recall seeing a book many years ago containing the author's many way-out ideas, where he assigned a number to each idea, based on the notion that a half-baked idea had quotient 0.5. Does anyone recall that book? I had a vague idea the book was by I.J. Good, but I don't see anything under his name. It strikes me this might be a good idea for sci.math ... === Subject: Re: The half-baked quotient >I recall seeing a book many years ago >containing the author's many way-out ideas, >where he assigned a number to each idea, >based on the notion that a half-baked idea had quotient 0.5. Does anyone recall that book? >I had a vague idea the book was by I.J. Good, >but I don't see anything under his name. It strikes me this might be a good idea for sci.math ... Irving John Good, > The Scientist Speculates: An Anthology of Partly-baked Ideas > I'll see if I can find that anywhere. > If I remember right, the PBIs in question weren't (or weren't > all) the author's own. I guess one would have to get into negative quotients for sci.math, or perhaps infinitesimally small ... === Subject: Re: The half-baked quotient Timothy Murphy a .8ecrit : > I recall seeing a book many years ago > containing the author's many way-out ideas, > where he assigned a number to each idea, > based on the notion that a half-baked idea had quotient 0.5. Does anyone recall that book? > I had a vague idea the book was by I.J. Good, > but I don't see anything under his name. It strikes me this might be a good idea for sci.math ... > It is The Scientist speculates (as it is an anthology *collected* by Good, you may had trouble finding it under his name indeed) === Subject: Re: The half-baked quotient Denis Feldmann a .8ecrit : > Timothy Murphy a .8ecrit : > I recall seeing a book many years ago > containing the author's many way-out ideas, > where he assigned a number to each idea, > based on the notion that a half-baked idea had quotient 0.5. Does anyone recall that book? > I had a vague idea the book was by I.J. Good, > but I don't see anything under his name. It strikes me this might be a good idea for sci.math ... > It is The Scientist speculates > (as it is an anthology *collected* by Good, you may had trouble finding > it under his name indeed) And I must add it has been quite a long time since I was first to answer seriously (and, I hope, usefully) a serious query :-) === Subject: =?windows-1252?Q?Symposium_=93Computational_Methods_in_Image_Analysis?= =?windows-1252?Q?=94_within_the_USNCCM_X_Congress_=96_Announce_=26_Call_for_ P ap?= =?windows-1252?Q?ers?= posting-account=N5wzPwoAAACfM7xTOME2HKP_WjJviHXd Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) ---------------------------------------------------------------------------- --------------------------------------------------- (Apologies for cross-posting) Symposium on ñComputational methods in image analysisî 10th US National Congress on Computational Mechanics (USNCCM X) Columbus, Ohio, USA, July 16-19, 2009 http://usnccm-10.eng.ohio-state.edu/ We would appreciate if you could distribute this information by your colleagues and co-workers. ---------------------------------------------------------------------------- --------------------------------------------------- Within the 10th US National Congress on Computational Mechanics (USNCCM X), to be held in Ohio, USA, in July 16-19, 2009, we are organizing the Symposium ñComputational methods in image analysisî. Examples of some topics that will be considered in the symposium ñComputational methods in image analysisî are: Image Analysis, Objects Modeling, Image Segmentation, Matching, Shape Reconstruction, Motion and Deformation Analysis, Objects Description and Registration, Medical imaging, Software Development for Image Analysis and Grid and High Performance Computing in Image Analysis. Due to your research activities in those fields, we would like to invite you to submit your work and participate in the Symposium ñComputational methods in image analysisî. For instructions and submission, please access to the conference website at: http://usnccm-10.eng.ohio-state.edu/abstractsub.html Please note that, when submitting your work you should select the Symposium ñ2.18.4 Computational methods in image analysisî. Important dates: - January 31, 2009: Deadline for abstract submission; - March 1, 2009: Deadline and notification of abstract acceptance; - July 16-19, 2009: Congress Events. Jo.8bo Manuel R. S. Tavares (University of Porto, Portugal, tavares@fe.up.pt) Renato Natal Jorge (University of Porto, Portugal, rnatal@fe.up.pt) Yongjie Zhang (Carnegie Mellon University, USA, jessicaz@andrew.cmu.edu) Dinggang Shen (UNC-CH School of Medicine, USA, dgshen@med.unc.edu) (Symposium organizers) === Subject: Re: Fraction as the ratio of sum of two cubes of positive integers posting-account=T18toAoAAABIMEO7Fe58Ju93zQyZ4fsR Gecko/20080528 Epiphany/2.22,gzip(gfe),gzip(gfe) > p.u.(u^2 - 3.a.u + 3.a^2) = q.v.(v^2 - 3.c.v + 3.c^2) In the special case p.u = q.v this becomes: (2a - u)^2 - (2c - v)^2 = (v^2 - u^2)/3 = w say How do you get this? I am getting something else. see below. p.u.(u^2 - 3.a.u + 3.a^2) = q.v.(v^2 - 3.c.v + 3.c^2) If pu = qv then u^2 - 3au + 3a^2 = v^2 - 3cv + 3c^2 3a^2 - 3c^2 + 3cv - 3au = v^2 - u^2 a^2 - c^2 + cv - au = (v^2 - u^2)/3 but your L.H.S. (2a - u)^2 - (2c - v)^2 is not equal to my L.H.S. i.e. a^2 - c^2 + cv - au Who is wrong here? === Subject: Optical Computing: special issue - Natural Computing, Springer posting-account=qJpM7AoAAACIRLR6aFYZipAwlqGZEJiY SV1),gzip(gfe),gzip(gfe) CALL FOR PAPERS Special issue on Optical SuperComputing Natural Computing journal, Springer-Verlag ------------------------------------------------------------------ Scope Using light, instead of electric power, for performing computations is an exciting idea whose applications can be already seen on the market. Without any doubt photons play an important role for transferring information and doing some low-level information processing. Current research concerning ultrafast-switches and detectors as well as the usage of quantum mechanical properties of photons have led to a renewed interest in optical processing. This choice is motivated by several features that light has: ´ It is very fast. Actually the fastest thing that we know, and speed is exactly what we need for our computers. ´ It can be easily manipulated (divided, transported, delayed, split, etc). ´ It is very well suited for parallelization. It is hopped that these features along with other properties can make optical computers to perform better than the electronic counterparts in the future. We are soliciting papers in the following areas (but not limited to): - Optical architectures for solving problems. - Theoretical aspects of optical devices for solving problems. - Practical implementation of existing or new optical solutions. - Analysis of the existing devices. - Logic gates implemented optically. - Hybrid devices where light plays a part but not all. - Niche applications / problems where light could perform better than the electronic devices. - Optical implementations of some well known algorithms (AI algorithms, brute-force, heuristics, etc). - Optical storage systems. - Other signal-based devices which can simulate the optical ones. Special Issue Committee Shlomi Dolev, Ben-Gurion University, Israel, http://www.cs.bgu.ac.il/~dolev/ Mihai Oltean ( CONTACT EDITOR ), Babes-Bolyai University, Romania, http://www.cs.ubbcluj.ro/~moltean/ Tobias Haist, Stuttgart Universitat, Germany, http://www.uni-stuttgart.de/ito/Institut/Institutsmitglieder/haist.htm Submission of papers: 15th of January 2009 Decision Notification: 15th of April 2009 Camera ready & revised papers: 15th of May 2009 Submission of papers The material in a paper must represent substantially new work that has not been previously published. Submit your paper through NACO submission system: http://naco.edmgr.com. You MUST choose SI: Optical Supercomputing as the category of your submission. Links: http://www.springer.com/computer/foundations/journal/11047 www.cs.ubbcluj.ro/~moltean/osc2008 Mihai Oltean www.cs.ubbcluj.ro/~moltean === Subject: Re: MatheRealism <1vocqtg8yaipk.1vqi4l7fwjm3e$.dlg@40tude.net> posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/20070309 Firefox/2.0.0.3,gzip(gfe),gzip(gfe) > Now I already know, of course, that the empty set is > the von Neumann natural 0 -- but recall that these are > Frege cardinals, not von Neumann's. The Frege 0 is the > set {{}}, and the Frege 1 is the set of all singletons. Frege-Russell cardinals ARE JUST BULL. The sooner you get this through your head, the sooner some progress can be made. The set of all singletons IS NOT a set; it is a proper class. === Subject: Re: MatheRealism posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > Nobody can ever have any more than an idea of a set, since sets (at > least mathematical ones) are only ideas. But having ideas of sets without possibly having ideas of all their > elements is nonsense [...] I have an idea of the set of all Wilhelms. But obviously I do not have an idea about each and every Wilhelm. So I am forced to conclude that there is no such thing as the set of all Wilhelms. Therefore no Wilhelm exists. Hey, I am getting pretty good at this MatheRealism stuff. === Subject: Re: MatheRealism posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > Nobody can ever have any more than an idea of a set, since sets (at > least mathematical ones) are only ideas. But having ideas of sets without possibly having ideas of all their > elements is nonsense [...] I have an idea of the set of all Wilhelms. But obviously I do not > have an idea about each and every Wilhelm. So I am forced > to conclude that there is no such thing as the set of all Wilhelms. > Therefore no Wilhelm exists. The set of all Wilhelms exists because all Wilhemls exist. That case is independent of your knowledge. But ideas that do not exist independently of your knowledge, do not exist unless you know them. Only those ideas that you can specify exist, the others do not. Is it really so hard to understand that? REegards, WM === Subject: Re: MatheRealism posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG 1.1.4322; .NET CLR 2.0.50727; MS-RTC LM 8; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > But having ideas of sets without possibly having ideas of all their > elements is nonsense [...] > I have an idea of the set of all Wilhelms. But obviously I do not > have an idea about each and every Wilhelm. So I am forced > to conclude that there is no such thing as the set of all Wilhelms. > Therefore no Wilhelm exists. The set of all Wilhelms exists because all Wilhemls exist. That case > is independent of your knowledge. But ideas that do not exist > independently of your knowledge, do not exist unless you know them. > Only those ideas that you can specify exist, the others do not. I (and you) cannot specify each an every Wilhelm, therefore I (and you) do not have an idea about each Wilhelm. Therefore, according to your requirement about specified ideas, not every Wilhelm exists. I keep getting better and better at this MatheRealism stuff. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/20070309 Firefox/2.0.0.3,gzip(gfe),gzip(gfe) > The set of all Wilhelms exists because all Wilhemls exist. > That case is independent of your knowledge. > But ideas that do not exist independently of your knowledge, > do not exist unless you know them. So? The fraction of all ideas that are dependent on any one person's knowing them is negligible; all those ideas are thoughts in that person's head. The letter a, or the idea of it, IS NOT dependent on any one person's knowing it. If all humans died, all the a's they had ever written would still be there, and could still eventually be discovered by other intelligence. We are in any case here talking NOT about ideas BUT RATHER about ABSTRACTIONS. Which ARE NOT ideas. > Only those ideas that you can specify exist, the others do not. Since we are talking about abstractions AND NOT ideas, this is entirely moot. More to the point, the claim that only what can be specified exists LEADS PROVABLY to contradictions. In other words, it is EASY TO PROVE that things that CANNOT be specified MUST exist. > Is it really so hard to understand that? It's a lot worse than so hard: IT'S PROVABLY IMPOSSIBLE. === Subject: Re: MatheRealism > Nobody can ever have any more than an idea of a set, since sets (at > least mathematical ones) are only ideas. But having ideas of sets without possibly having ideas of all their > elements is nonsense [...] I have an idea of the set of all Wilhelms. But obviously I do not > have an idea about each and every Wilhelm. So I am forced > to conclude that there is no such thing as the set of all Wilhelms. > Therefore no Wilhelm exists. The set of all Wilhelms exists because all Wilhemls exist. That case > is independent of your knowledge. But ideas that do not exist > independently of your knowledge, do not exist unless you know them. Ideas that do not exist within WM's knowledge can certainly exist, since despite his assertions, there are things he does not know. > Only those ideas that you can specify exist, the others do not. Then only the Wilhelms that you can specify exist. Is it really so hard to understand that? It is impossible, at least for those othere than WM, to accept that nothing outside of WM's own personal knowledge can exist, but that seems to be what WM is trying to assert. === Subject: Re: MatheRealism posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > Nobody can ever have any more than an idea of a set, since sets (at > least mathematical ones) are only ideas. But having ideas of sets without possibly having ideas of all their > elements is nonsense [...] I have an idea of the set of all Wilhelms. But obviously I do not > have an idea about each and every Wilhelm. So I am forced > to conclude that there is no such thing as the set of all Wilhelms. > Therefore no Wilhelm exists. The set of all Wilhelms exists because all Wilhelms exist. That case > is independent of your knowledge. But ideas that do not exist > independently of your knowledge, do not exist unless you know them. Ideas that do not exist within WM's knowledge can certainly exist, since > despite his assertions, there are things he does not know. Only those ideas that you can specify exist, the others do not. Then only the Wilhelms that you can specify exist. No, why should the others case to exist? Is it really so hard to understand that? It is impossible, at least for those othere than WM, to accept that > nothing outside of WM's own personal knowledge can exist, but that seems > to be what WM is trying to assert Did I talk of me or did I say that only those ideas that you can specify exist === Subject: Re: MatheRealism > Nobody can ever have any more than an idea of a set, since sets (at > least mathematical ones) are only ideas. But having ideas of sets without possibly having ideas of all their > elements is nonsense [...] I have an idea of the set of all Wilhelms. But obviously I do not > have an idea about each and every Wilhelm. So I am forced > to conclude that there is no such thing as the set of all Wilhelms. > Therefore no Wilhelm exists. The set of all Wilhelms exists because all Wilhelms exist. That case > is independent of your knowledge. But ideas that do not exist > independently of your knowledge, do not exist unless you know them. Ideas that do not exist within WM's knowledge can certainly exist, since > despite his assertions, there are things he does not know. Only those ideas that you can specify exist, the others do not. Then only the Wilhelms that you can specify exist. No, why should the others case to exist? In WM's world, because WM insists that things of which one has no knowledge cannot exist. Is it really so hard to understand that? It is impossible, at least for those othere than WM, to accept that > nothing outside of WM's own personal knowledge can exist, but that seems > to be what WM is trying to assert Did I talk of _me_ or did I say that only those ideas that _you_ can > specify exist Since most of the things in WM's wild wild world of MathUnRealism are outside of MY personal knowledge or ability to specify, they cannot, by WM's own criterion, exist. === Subject: Re: MatheRealism posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > A name like 10^100 does not have the first property (because there is > no set of 10^100 elements, and there is no set of such sets), [...] > I have a set just like that. > B = { x | x in N and x <= 10^100 } I keep it in my sock drawer. You may have a dream or a belief or a delusion. You may even have the > idea of a set. But you cannot have 10^100 distinct elements, unless > you are corresponding with some Gods (or believing to be corresponding > with them). I forgot to tell you about my other set: B2 = { x | x in N and x not in B } = { x | x in N and x > 10^100 } I keep it in my sock drawer, too. === Subject: Re: MatheRealism <17eaubwnye0bn.8wmst46qmx81$.dlg@40tude.net> posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > Realism in math includes awareness of the fact Oh, SHUT UP. YOU are NOT aware of any relevant facts. Math is NOT even ABOUT facts. In math, you PROVE things FROM AXIOMS. Whether the axioms or are aren't factual is not even a legitimate QUESTION. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > No one will ever be able to spell a name that needs more than 10^100 > bits - simply because there are not enough bits available. By that rationale, no one will ever be able to represent 1/3 as a decimal, because there are not enough 3s in the universe. You're an IDIOT for even TRYING this. === Subject: Re: MatheRealism <186xagtylbqm7.57hvrbumggon$.dlg@40tude.net> posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > It is WM who is into religiously accepting things on faith, like > potentially infinite sets, which are logically self-contradictory. NOthing is self-contradictory IF you define it right. It is OK for set VARIABLES to be potentially infinite, or for a class or domain of things to be potentially infinite. But most of the things that WM *calls* potentially infinite are in fact actually infinite. N is actually infinite. Every element of N is actually finite. A VARIABLE of TYPE N, which could take (after the fact) ANY element of N as a value, might reasonably be called potentially infinite. But actually, kinetically, it is always either (before) uninstantiated or undefined, or (after) finite. These are distinctions that WM not only can't be bothered to make, but in fact GLORIES in WILFULLY obscuring. === Subject: Re: MatheRealism <186xagtylbqm7.57hvrbumggon$.dlg@40tude.net> posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) Sets are abstract. You mean like a sort of virtual reality? or ... Probably he means like the Gods and Godesses in the other big branch > of theology. Oh, go yourself. VERY slowly. In your left eye-socket. I mean that symbols in general are abstract. The letter A is abstract. If you want to obsess about what realm it lies on, that's YOUR fantasy. === Subject: Re: MatheRealism <186xagtylbqm7.57hvrbumggon$.dlg@40tude.net> posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Sets are abstract. You mean like a sort of virtual reality? or ... Probably he means like the Gods and Godesses in the other big branch > of theology. They also have properties and numbers that [we] are not > allowed to question or to rationalize. In the real world, sets, properties and numbers are not allowed to ask > questions or make rationalizations. See the correction. Fact is: A set can exist as a collection of distinct material objects > or as a collection of distinct thought objects. Alas, some set > theorists who long ago have dismissed logic and rational thinking, > believe in sets the elements of which can neither be represented by > matter nor by ideas. They simply love matheology, probably as an > ersatz religion. It is WM who is into religiously accepting things on faith, like > potentially infinite sets, which are logically self-contradictory. In your logic, with no doubt. === Subject: Re: MatheRealism Sets are abstract. You mean like a sort of virtual reality? or ... Probably he means like the Gods and Godesses in the other big branch > of theology. They also have properties and numbers that [we] are not > allowed to question or to rationalize. In the real world, sets, properties and numbers are not allowed to ask > questions or make rationalizations. See the correction. > Fact is: A set can exist as a collection of distinct material objects > or as a collection of distinct thought objects. Alas, some set > theorists who long ago have dismissed logic and rational thinking, > believe in sets the elements of which can neither be represented by > matter nor by ideas. They simply love matheology, probably as an > ersatz religion. It is WM who is into religiously accepting things on faith, like > potentially infinite sets, which are logically self-contradictory. In your logic, with no doubt. In all logics that, unlike WM's, are not self-contradictory. All sets in all reasonable set theories are, in WM's terminology, completed objects, which WM's potentially infinite sets cannot be, so whatever they are, they are not sets in any reasonable set theory. === Subject: Re: MatheRealism In message >It is very easy to disprove continuity, be it as an axiom or as an >assumption or as a proven theorem. Cover all rational numbers q_n between 0 and oo by closed intervals of >length 1/2^n, such that the rational q_n resides in the center of the >interval n-th interval. (This covering has measure 1.) Every interval >has its finite length 1/2^n. So you can shrink it at both ends until is >gets an open interval with irrational ends. This procedure does not >spoil the complete covering of all rationals. All uncovered reals are now those ends of the intervals. Because: Would >there be any further uncovered reals outside of the open intervals, >then there would be also uncovered rationals (because between every >pair of reals there is a rational number) contrary to the assumption. > No. Each rational between two uncovered reals will be covered; the corresponding open interval will itself lie between the two reals. One or both of them may be endpoints, but in most cases they won't be. >Result: The countable set of uncovered interval-ends must make up >measure oo. On the other hand, it could be covered by a sequence of >aleph_0 intervals. That contradicts the assumption of existence of all >elements of the continuum. To establish the contradiction, you need to *prove* that there is a covering of this nature whose only uncovered reals are endpoints. It won't be easy! -- David Hartley === Subject: Re: MatheRealism <2OFKmuMS5CPJFwgf@212648.invalid> posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > In message >assumption or as a proven theorem. Cover all rational numbers q n between 0 and oo by closed intervals of >length 1/2^n, such that the rational q n resides in the center of the >interval n-th interval. (This covering has measure 1.) Every interval >has its finite length 1/2^n. So you can shrink it at both ends until is >gets an open interval with irrational ends. This procedure does not >spoil the complete covering of all rationals. All uncovered reals are now those ends of the intervals. Because: Would >there be any further uncovered reals outside of the open intervals, >then there would be also uncovered rationals (because between every >pair of reals there is a rational number) contrary to the assumption. No. Each rational between two uncovered reals will be covered; the > corresponding open interval will itself lie between the two reals. One > or both of them may be endpoints, but in most cases they won't be. Result: The countable set of uncovered interval-ends must make up >measure oo. On the other hand, it could be covered by a sequence of >aleph 0 intervals. That contradicts the assumption of existence of all >elements of the continuum. To establish the contradiction, you need to *prove* that there is a > covering of this nature whose only uncovered reals are endpoints. It > won't be easy! It is easy. Proof by contradiction: Assume that there is a real number r that is not covered by an interval and is not an endpoint of an interval. What does that mean? There is another real number, s, also uncovered, between r and the next interval on the right hand side of r. For instance, s could be the endpoint of an interval. Them (r, s) is uncovered. It contains a rational number that is also uncovered contrary to the assumption. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) In message >assumption or as a proven theorem. Cover all rational numbers q n between 0 and oo by closed intervals of >length 1/2^n, such that the rational q n resides in the center of the >interval n-th interval. (This covering has measure 1.) Every interval >has its finite length 1/2^n. So you can shrink it at both ends until is >gets an open interval with irrational ends. This procedure does not >spoil the complete covering of all rationals. All uncovered reals are now those ends of the intervals. Because: Would >there be any further uncovered reals outside of the open intervals, >then there would be also uncovered rationals (because between every >pair of reals there is a rational number) contrary to the assumption. No. Each rational between two uncovered reals will be covered; the > corresponding open interval will itself lie between the two reals. One > or both of them may be endpoints, but in most cases they won't be. Result: The countable set of uncovered interval-ends must make up >measure oo. On the other hand, it could be covered by a sequence of >aleph 0 intervals. That contradicts the assumption of existence of all >elements of the continuum. To establish the contradiction, you need to *prove* that there is a > covering of this nature whose only uncovered reals are endpoints. It > won't be easy! It is easy. Proof by contradiction: > Assume that there is a real number r that is not covered by an > interval and is not an endpoint of an interval. What does that mean? > There is another real number, s, also uncovered, between r and the > next interval on the right hand side of r. If you were a mountain climber, this would be the logical equivalent of FALLING OFF, TO YOUR DEATH. Just consider the series of points 1/n (the nth point is 1/n). This does NOT COVER zero! Yet there IS NO next point AFTER (to the right of ) zero! ANY distance to the right of zero covers INFINITELY many of these points, NO MATTER HOW SMALL the distance! You CAN'T HAVE a next point to the right, under this ordering! You were closer to having it right when you said that the uncovered points had to all be singular points. (It took me a while to understand what you meant by that; that is actually true, since you can prove that no interval can be uncovered, since every interval contains infinitely many covering- intervals, since there is a covering-interval for every rational, and every interval contains infinitely many rationals). But you still have not done a good job of counting those singular points. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > It is easy. Proof by contradiction: You have no proof. You just have your own personal stupidity. > Assume that there is a real number r that is not covered by an > interval OK. I choose r=1/2. > and is not an endpoint of an interval. That's right: I have an interval around every rational number except 1/2, and for none of those intervals is 1/2 an endpoint of that interval or included in that interval. Of course, these are ALL OPEN intervals to begin with, so THEY NEVER include their endpoints, but we can skip that for now; it works just as well with closed intervals. > What does that mean? Nothing YOU could grok, certainly. > There is another real number, s, also uncovered, between r and the > next interval on the right hand side of r. THERE IS NO *NEXT* interval to the right of r, DUMBASS!! If I JUST had the sequence of numbers 1/n, for all natural n, THERE IS NO *next* number to the RIGHT of ZERO!! If you go ANY distance to the right of zero WHATSOEVER, you have tripped over INFINITELY many numbers! And in MY collection of intervals, if you take ANY distance to the right of 1/2 whatsoever, you have covered INFINITELY many rational numbers and therefore INFINITELY many intervals!! THERE IS NO *NEXT* interval given that you have a DENSE set of midpoints for the intervals, !!!!!!!!!!!!!!!!!!!!!!!*DUMBASS*!!!!!!!!!!!!!!!!!!!!!!!!! > For instance, s could be > the endpoint of an interval. Them (r, s) is uncovered. No, it isn't. > It contains a > rational number that is also uncovered contrary to the assumption. Using the countable union theorem or the axiom of choice or whatever other method you may choose (including the usual parallel diagonal one), the rationals between 0 and 1 can be well-ordered. Pick an ordering that maps 1/2 to 0, thereby omitting it from {1,2,3,4,....}, onto which ALL OTHER rationals are mapped by the ordering. For positive natural i, define the i'th interval as having a midpoint of the i'th rational (call that r), and a width of 2^(-i)*((r-(1/2))^2). None of these intervals WILL EVER cover 1/2 but they WILL cover EVERY OTHER natural number. The only thing preventing you from being a ing idiot is that you are too stupid to get laid. === Subject: Re: MatheRealism <2OFKmuMS5CPJFwgf@212648.invalid> In message It is easy. Proof by contradiction: >Assume that there is a real number r that is not covered by an interval >and is not an endpoint of an interval. What does that mean? There is >another real number, s, also uncovered, between r and the next interval >on the right hand side of r. For instance, s could be the endpoint of >an interval. Them (r, s) is uncovered. It contains a rational number >that is also uncovered contrary to the assumption. Why should there be a next interval to the right? For any e > 0, there will be infinitely many intervals (from the given covering of the rationals) lying between r and r + e. -- David Hartley === Subject: Re: MatheRealism posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > In message It is easy. Proof by contradiction: >Assume that there is a real number r that is not covered by an interval >and is not an endpoint of an interval. What does that mean? There is >another real number, s, also uncovered, between r and the next interval >on the right hand side of r. For instance, s could be the endpoint of >an interval. Them (r, s) is uncovered. It contains a rational number >that is also uncovered contrary to the assumption. Why should there be a next interval to the right? Because r is not a member or an end of an interval and the real line is linear. Perhaps you allude to the fact that for r there is no next real number to the right. That is correct. However, this consideration does not apply here, because we have explicitly excluded that r belongs to an interval or is the end of an interval (in short: r is not a limit point of an interval). This implies that there is a real number between r and this interval. (Otherwise r would be a limit point of an interval.) > For any e > 0, there will be infinitely many intervals (from the given > covering of the rationals) lying between r and r + e. Then there will also be infinitely many covered reals in any surrounding of r. Then r is, by definition of limit point, a limit point and is not outside of any interval, contrary to the assumption. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/20070309 Firefox/2.0.0.3,gzip(gfe),gzip(gfe) > Why should there be a next interval to the right? Because r is not a member or an end of an interval IT WOULDN'T MATTER if r were or were not an endpoint of an interval -- THE INTERVALS ARE *OPEN*, so they NEVER include their own endpoints! > and the real line is linear. And the LINE is DENSE, DUMBASS. Given ANY point on a line, THERE IS *NEVER* a *NEXT* point on the line! THAT IS WHY there is never a next INTERVAL EITHER, DUMBASS, since the intervals ARE ALSO dense (since there is one of them for every rational). > Perhaps you allude to the fact that for r there is no next real > number to the right. That is correct. Of course. > However, this consideration does not apply here, Of course it does. > because we have explicitly excluded that r belongs to an > interval or is the end of an interval SO what?? > (in short: r is not a limit point of an interval). THERE IS NO SUCH THING as a limit point of ONE interval, DUMBASS!! A limit point is a limit OF A SEQUENCE!! r IS SO TOO a limit point of THIS WHOLE SERIES of intervals!! > This implies that That you simply don't know what you are talking about, as usual. === Subject: Re: MatheRealism <2OFKmuMS5CPJFwgf@212648.invalid> In message > (in short: r is not a limit point of an interval). THERE IS NO SUCH THING as a limit point of ONE interval, DUMBASS!! A >limit point is a limit OF A SEQUENCE!! r IS SO TOO a limit point of >THIS WHOLE SERIES of intervals!! While most of WM's argument is nonsense, here he is using the standard topological notion of the limit point of a set. -- David Hartley === Subject: Re: MatheRealism <5K3JHGKbcXPJFwQV@212648.invalid> posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > While most of WM's argument is nonsense, here he is using the standard > topological notion of the limit point of a set. HE *IS NOT*, DUMBASS! THAT notion involves a SEQUENCE or set of points that IS NOT *connected* like an INTERVAL! He said limit point of an INTERVAL, NOT of a set! That was MY whole point! WM does not have the FIRST CLUE about the standard topological notion!! And it is NOT a topological notion in any case. The topological notions are defined AFTER limit points, IN TERMS OF limit points. === Subject: Re: MatheRealism > While most of WM's argument is nonsense, here he is using the standard > topological notion of the limit point of a set. HE *IS NOT*, DUMBASS! Whether WM actually knows it or not,, he is. See http://en.wikipedia.org/wiki/Limit_point#Types_of_limit_points We have the following characterisation of limit points: x is a limit point of S if and only if it is in the closure of S {x}. WM is not consistent enough to be wrong ALL the time. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) While most of WM's argument is nonsense, here he is using the standard > topological notion of the limit point of a set. HE *IS NOT*, DUMBASS! Whether WM actually knows it or not,, he is. Seehttp://en.wikipedia.org/wiki/Limit point#Types of limit points NO. DON'T see THAT. THAT is irrelevant. See INSTEAD http://planetmath.org/encyclopedia/AccumulationPointOfTheSequence.html which will inform you (though not WM) that In a topological space X, a point x is a limit point of the sequence x 0, x 1, ... if, for every open set containing x, there are finitely many indices such that the corresponding elements of the sequence do not belong to the open set. [and] A point x is an accumulation point of the sequence x 0, x 1,... if, for every open set containing x, there are infinitely many indices such that the corresponding elements of the sequence belong to the open set. Meanwhile, back at dumbass wikipedia, we had > We have the following characterisation of limit points: x is a > limit point of S if and only if it is in the closure of S {x}. In the first place, this is completely irrelevant since 1) WM does not know what a closure is, and 2) the kind of open intervals we are dealing with have the property that ALL of their points are limit points, which makes this version of the notion IRRELEVANT. More to the point, this WAS NOT the notion that WM was using in any case! And WM is wrong enough to be confusing you and several other people into being almost as wrong as he is. A lot of you need to understand that WM is always going to respond the wrongest person. NO progress is actually going to be made until MORE PEOPLE *SHUT UP*. === Subject: Re: MatheRealism posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I For any e > 0, there will be infinitely many intervals (from the given > covering of the rationals) lying between r and r + e. Then there will also be infinitely many covered reals in any > surrounding of r. Correct. >Then r is, by definition of limit point, a limit point Correct. (Note that it is not a limit point of real numbers in the same interval) > and is not outside of any interval, Nope, to conclude this you need more than r is a limit point, you need that r is a limit point of real numbers in the same interval. - William Hughes === Subject: Re: MatheRealism posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) For any e > 0, there will be infinitely many intervals (from the given > covering of the rationals) lying between r and r + e. Then there will also be infinitely many covered reals in any > surrounding of r. Correct. Then r is, by definition of limit point, a limit point Correct. (Note that it is not a limit point of real numbers > in the same interval) Note that r cannot be a limit point of real numbers in (more than two) different intervals, because every point of the real axis has only two sides. We are not in a plane! And every interval has a finite length. and is not outside of any interval, Nope, to conclude this you need more than r is a limit > point, you need that r is a limit point of real numbers > in the same interval. You need only to know that (probably except for matheologicians) every point of the real axis has only two sides. If you go from r in positive direction, than you have to traverse a first interval of finite length that contains infinitely many real numbers. Some of them are in the immediate surrounding of r. Or would you propose that there is not more than one (or few) real per interval that is in the surrounding of r, but that infinitely many intervals are surrounding r. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > You need only to know that (probably except for matheologicians) every > point of the real axis has only two sides. WE all know this, dumbass. YOU are the one having problems with it. > If you go from r in positive direction, than you have to traverse a first interval NO, YOU DON'T. There are NO *inherent* intervals. The context of our problem involves a denumerable sequence of intervals THAT WE SPECIFIED. Since they are intervals AROUND ALL RATIONALS and ARE THEREFORE *DENSE*, there IS NEVER, at ANY point or in ANY context, a first one, except maybe in terms of the ordering by which the rationals were enumerated. In particular, because of the density, THERE IS NEVER a first interval as you go ALONG the line, left to right, NO MATTER WHERE you start. > of finite length that contains infinitely many real numbers. It is precisely because all the intervals contain infinitely many RATIONAL numbers that there are ALWAYS INFINITELY many intervals intersecting any interval and THEREFORE NEVER a first one. > Some of them are in the immediate surrounding of r. THERE IS NO SUCH THING AS the surrounding of r. I don't know whether this is a German problem or a stupidity problem, but either way, IT IS NOT ENGLISH. > Or would you propose that there is not more than one (or few) real per > interval that is in the surrounding of r, but that infinitely many > intervals are surrounding r. OBVIOUSLY there are infinitely many intervals surrounding r. In particular, the infinitely many open intervals (r-e^-n, r+e^-n), for e<1/2 and all natural n, all surround r. On the other hand, if you want to restrict yourself to the countably infinite set of intervals-around-rationals that we were talking about, then it may well be the case that NONE of THESE surround r, or even include it at all, although there will always be infinitely many of them getting arbitrarily close to it, on BOTH sides. === Subject: Re: MatheRealism For any e > 0, there will be infinitely many intervals (from the given > covering of the rationals) lying between r and r + e. Then there will also be infinitely many covered reals in any > surrounding of r. Correct. Then r is, by definition of limit point, a limit point Correct. (Note that it is not a limit point of real numbers > in the same interval) Note that r cannot be a limit point of real numbers in (more than two) > different intervals, because every point of the real axis has only two > sides. Unless those intervals have pairwise disjoint interiors, which need not be the case, WM is presuming conditions contrary to fact. > We are not in a plane! And every interval has a finite length. and is not outside of any interval, Nope, to conclude this you need more than r is a limit > point, you need that r is a limit point of real numbers > in the same interval. You need only to know that (probably except for matheologicians) every > point of the real axis has only two sides. Points do not have sides, but if WM means that point on the real line separates the remaining reals into two disconnected sets, he is correct. > If you go from r in > positive direction, than you have to traverse a first interval of > finite length that contains infinitely many real numbers. There is no such first interval of finite length as within every such interval of finite length starting at any real there is a prior (smaller) one starting at the same real. === Subject: Re: MatheRealism posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > ... If you go from r in > positive direction, than you have to traverse a first interval Nope. There is no first interval. - William Hughes === Subject: Re: MatheRealism posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) ... If you go from r in > positive direction, than you have to traverse a first interval Nope. There is no first interval. Why don't you claim that there are no real numbers? If there are infinitely many covered reals in every epsilon- surrounding of r, then infinitely many of them belong to one and the same covered interval, no? Then r is a limit point of that interval, by definition. === Subject: Re: MatheRealism ... If you go from r in > positive direction, than you have to traverse a first interval Nope. There is no first interval. Why don't you claim that there are no real numbers? Why don't you think a bit before writing, WM? There is no more a first such interval than there is a last natural number and for the same sort of reason. Given any real point and any interval of positive length having it as an endpoint, there is a smaller subinterval also having it as endpoint, so there is no first (smallest) such interval. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/20070309 Firefox/2.0.0.3,gzip(gfe),gzip(gfe) > If there are infinitely many covered reals in every epsilon- > surrounding of r, then infinitely many of them belong > to one and the same covered interval, no? Of course. Then r is a limit point of that interval, by definition. Nobody can help it if you are too damn stupid to know the definition of limit point. No individual interval CAN have any limit points. If you mean a limit point OF THE WHOLE SEQUENCE, a limit point of it THAT HAPPENS TO LIE IN THIS INTERVAL, then, whoop-de-! If the sequence is the sequence of all rationals THEN EVERY REAL, EVERY POINT ON THE LINE, is a limit point of it. Your talk about individual intervals having limit points is just breathtaking in its sheer ignorance. > === Subject: Re: MatheRealism posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > ... If you go from r in > positive direction, than you have to traverse a first interval Nope. There is no first interval. Why don't you claim that The fact that there is no first interval follows trivially from the fact that there is no first real number. > there are no real numbers? If there are infinitely many covered reals in every epsilon- > surrounding of r, then infinitely many of them belong to one and the > same covered interval, no? No. Of course not. You have to have a set K with infinitely many reals and infinitely many intervals, but you don't have to have any interval that contains more than one element of K. - William Hughes === Subject: Re: MatheRealism posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > If there are infinitely many covered reals in every epsilon- > surrounding of r, then infinitely many of them belong to one and the > same covered interval, no? No. Of course not. You have to have a set K with infinitely many > reals > and infinitely many intervals, but you don't have to have any > interval that contains more than one element of K. I don't know what you are talking about. Cauchy-sequences? But why then intervals? And even intervals with only one element? Sorry, I can't make any sense of your statement. Here is what I said: I have considered a covering of the rationals by a set of finite intervals such that the rational number q_n is covered by the closed interval I_n of length 1/2^n (which later on is shrunk somewhat, but not very much, to become the open interval I'_n). Every interval I'_n covers infinitely many real numbers. If an interval I'_n belongs to the surrounding of the real number r, then infinitely many real numbers of I'_n belong to the surrounding of r. (It is mathematically impossible to have only one element of I'_n belonging to the surrounding of r.) Then r is a limit point of I'_n (i.e., either r belongs to the interval or is one of its endpoints). If this is not the case, then r is not a limit point of I'_n and then there is an epsilon surrounding of r such that no element of I'_n belongs to this surrounding. === Subject: Re: MatheRealism posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I If there are infinitely many covered reals in every epsilon- > surrounding of r, then infinitely many of them belong to one and the > same covered interval, no? No. Of course not. You have to have a set K with infinitely many > reals > and infinitely many intervals, but you don't have to have any > interval that contains more than one element of K. I don't know what you are talking about. There are two possibilities. i: you are too stupid to tell the difference between an interval that contains one element, and an interval that contains an infinite number of elements not from K and one element from K ii: you are being a scumbag by pretending not to understand something that is obvious even to a total idiot. Answer yes or no. Is it possible to have infinitely many covered reals in every epsilon- surrounding of r with no two reals in one covered interval. - William Hughes === Subject: Re: MatheRealism posting-account=U44YcwkAAAAbGXB70Qr7gA3kornmKE4i Gecko/20080922 Ubuntu/7.10 (gutsy) Firefox/2.0.0.17,gzip(gfe),gzip(gfe) If there are infinitely many covered reals in every epsilon- > surrounding of r, then infinitely many of them belong to one and the > same covered interval, no? No. Of course not. You have to have a set K with infinitely many > reals > and infinitely many intervals, but you don't have to have any > interval that contains more than one element of K. I don't know what you are talking about. Not surprising, is it? There are two possibilities. i: you are too stupid to tell the difference between > an interval that contains one element, > and > an interval that contains an infinite number of elements > not from K and one element from K > ii: you are being a scumbag by pretending not to understand > something that is obvious even to a total idiot. > Answer yes or no. Oh, oh, let me try this one! Um, how about Yes? Brian Chandler === Subject: Re: MatheRealism If there are infinitely many covered reals in every epsilon- > surrounding of r, then infinitely many of them belong to one and the > same covered interval, no? No. Of course not. You have to have a set K with infinitely many > reals > and infinitely many intervals, but you don't have to have any > interval that contains more than one element of K. Haven't been following the thread, so I don't know what's going on, but if you have a set K of reals, then to say There is a member of K in every epsilon neigborhood of r is equivalent to saying There are an infinite number of members of K in every epsilon-neighborhood of r. -- hz > I don't know what you are talking about. Not surprising, is it? > There are two possibilities. i: you are too stupid to tell the difference between > an interval that contains one element, > and > an interval that contains an infinite number of elements > not from K and one element from K > ii: you are being a scumbag by pretending not to understand > something that is obvious even to a total idiot. > Answer yes or no. Oh, oh, let me try this one! Um, how about Yes? Brian Chandler === Subject: Re: MatheRealism > Haven't been following the thread, so I don't know what's going on, > but if you have a set K of reals, then to say There is a member of K in every epsilon neigborhood of r is equivalent to saying There are an infinite number of members of K in every > epsilon-neighborhood of r. I guess I need to add except when r itself is an element of K ? -- hz === Subject: Re: MatheRealism <493E86D2.D30DE64E@gmail.com> <493E9F0B.965542A5@gmail.com> posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > Haven't been following the thread, so I don't know what's going on, > but if you have a set K of reals, then to say There is a member of K in every epsilon neigborhood of r is equivalent to saying There are an infinite number of members of K in every > epsilon-neighborhood of r. I guess I need to add except when r itself is an element of K ? -- > hz No. If r is not in any of the covered intervals, then this makes no difference at all. Even if r is in one or more of the intervals, it may not make much of a differece, it will certainly not turn a finite number into an infinite number, (In the original problem, K consists of covered elements so r is not an element of K). Basic example, r={0}, K = { 1/2,1/3,1/4...} intervals centered on the points of K with length 1/(2*2),1/(3*3), 1/(4*4)... Note we have an infinite number of elements of K in any neighbourhood of 0, but only one element of K in each interval. - William Hughes === Subject: Re: MatheRealism Haven't been following the thread, so I don't know what's going on, > but if you have a set K of reals, then to say There is a member of K in every epsilon neigborhood of r is equivalent to saying There are an infinite number of members of K in every > epsilon-neighborhood of r. I guess I need to add except when r itself is an element of K ? -- > hz No. If r is not in any of the covered intervals, then > this makes no difference at all. Even if r is in one or more > of the intervals, it may not make much of a differece, it > will certainly not turn a finite number into an infinite number, > (In the original problem, K consists of covered elements > so r is not an element of K). Basic example, r={0}, K = { 1/2,1/3,1/4...} intervals centered on the > points of K with length 1/(2*2),1/(3*3), 1/(4*4)... > Note we have an infinite number of elements of K in any > neighbourhood of 0, Right. > but only one element of K in each interval. Right. Just so. -- hz === Subject: Re: MatheRealism <493E86D2.D30DE64E@gmail.com> posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I If there are infinitely many covered reals in every epsilon- > surrounding of r, then infinitely many of them belong to one and the > same covered interval, no? No. Of course not. You have to have a set K with infinitely many > reals > and infinitely many intervals, but you don't have to have any > interval that contains more than one element of K. Haven't been following the thread, so I don't know what's going on, > but if you have a set K of reals, then to say There is a member of K in every epsilon neigborhood of r is equivalent to saying There are an infinite number of members of K in every > epsilon-neighborhood of r. Even without following the thread you should be able to distinguish between There are an infinite number of members of K in every epsilon-neighborhood of r and Infinitely many mebers of K belong to one and the same covered interval - William Hughes === Subject: Re: MatheRealism If there are infinitely many covered reals in every epsilon- > surrounding of r, then infinitely many of them belong to one and the > same covered interval, no? No. Of course not. You have to have a set K with infinitely many > reals > and infinitely many intervals, but you don't have to have any > interval that contains more than one element of K. Haven't been following the thread, so I don't know what's going on, > but if you have a set K of reals, then to say There is a member of K in every epsilon neigborhood of r is equivalent to saying There are an infinite number of members of K in every > epsilon-neighborhood of r. Even without following the thread you should be able > to distinguish between There are an infinite number of members of K in every > epsilon-neighborhood of r and Infinitely many members of K belong to one and the > same covered interval - William Hughes Right. Those two statements are not equivalent. I already posted an emendation to my previous post. I'm wondering if I need to except the case where r e K or whether that is already ruled out by the definition of epsilon-surrounding of r. Do you happen to know? -- hz === Subject: Re: MatheRealism <2OFKmuMS5CPJFwgf@212648.invalid> In message > If there are infinitely many covered reals in every epsilon- > surrounding of r, then infinitely many of them belong to one and the > same covered interval, no? No. Of course not. You have to have a set K with infinitely many > reals > and infinitely many intervals, but you don't have to have any > interval that contains more than one element of K. Actually yes. Each of the given intervals contains infinitely many reals, all of which are obviously covered. But of course that does not mean that any interval (from the covering) in the given neighbourhood contains all the covered reals. >I don't know what you are talking about. Cauchy-sequences? But why >then intervals? And even intervals with only one element? Sorry, I >can't make any sense of your statement. Here is what I said: I have considered a covering of the rationals by a set of finite >intervals such that the rational number q_n is covered by the closed >interval I_n of length 1/2^n (which later on is shrunk somewhat, but >not very much, to become the open interval I'_n). Every interval I'_n >covers infinitely many real numbers. If an interval I'_n belongs to >the surrounding of the real number r, then infinitely many real >numbers of I'_n belong to the surrounding of r. (It is mathematically >impossible to have only one element of I'_n belonging to the >surrounding of r.) Then r is a limit point of I'_n (i.e., either r >belongs to the interval or is one of its endpoints). If this is not >the case, then r is not a limit point of I'_n and then there is an >epsilon surrounding of r such that no element of I'_n belongs to this >surrounding. Each neighbourhood of r will contain infinitely many of the intervals I'_n. But - since you specified that r is not in or an endpoint of any such interval - for any particular I'_n there will be a smaller neighbourhood of r which is disjoint from I'_n. I.e. r is a limit point of the set U{I'_n : n = 1,2, ...} but it is not a limit point of I'_n for any n. To see more easily that there is no contradiction, consider the collection of intervals {(r + 1/2^(n+1), r + 1/2^n) : n = 1,2, ...}. This covers every rational in (r, r + 1/2) (r irrational) but r is not an endpoint of any of the intervals. Of course the set of uncovered points in your example is much more complicated - having infinite measure but containing no rationals - and a lot more difficult to visualise, but you have still not proved your contradiction. -- David Hartley === Subject: Re: MatheRealism <1b+aTEM+wXPJFwQS@212648.invalid> posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Each neighbourhood of r will contain infinitely many of the intervals > I'_n. But - since you specified that r is not in or an endpoint of any > such interval - for any particular I'_n there will be a smaller > neighbourhood of r which is disjoint from I'_n. I.e. r is a limit point > of the set U{I'_n : n = 1,2, ...} but it is not a limit point of I'_n > for any n. If r is a limit point of the intervals, then r is a limit point also of the real numbers that are contained in the intervals. Please spare your matheology of intervals without contents. And r is not a limit point of a distant interval. So r cannot be a limit point of the intervals but at most the limit point of one interval. To see more easily that there is no contradiction, consider the > collection of intervals {(r + 1/2^(n+1), r + 1/2^n) : n = 1,2, ...}. > This covers every rational in (r, r + 1/2) (r irrational) but r is not > an endpoint of any of the intervals. Obviously this is a contradiction. The real axis consists of numbers, not of intervals. Of course the set of uncovered points in your example is much more > complicated - having infinite measure but containing no rationals - and > a lot more difficult to visualise, but you have still not proved your > contradiction. But you seriously think that your idea of a limit point of intervals that is not a limit point of numbers in the intervals makes any sense? === Subject: Re: MatheRealism <2OFKmuMS5CPJFwgf@212648.invalid> <1b+aTEM+wXPJFwQS@212648.invalid> In message > Each neighbourhood of r will contain infinitely many of the intervals > I'_n. But - since you specified that r is not in or an endpoint of any > such interval - for any particular I'_n there will be a smaller > neighbourhood of r which is disjoint from I'_n. I.e. r is a limit point > of the set U{I'_n : n = 1,2, ...} but it is not a limit point of I'_n > for any n. If r is a limit point of the intervals, then r is a limit point also >of the real numbers that are contained in the intervals. Please spare >your matheology of intervals without contents. And r is not a limit >point of a distant interval. So r cannot be a limit point of the >intervals but at most the limit point of one interval. You do know what U{I'_n : n = 1,2, ...} means, don't you? It *is* the set of all the real numbers contained in the intervals I'_n. That is the set that r is a limit point of. I just told George that you understood limit points; don't prove me wrong. Where did you get the idea of intervals without contents? > To see more easily that there is no contradiction, consider the > collection of intervals {(r + 1/2^(n+1), r + 1/2^n) : n = 1,2, ...}. > This covers every rational in (r, r + 1/2) (r irrational) but r is not > an endpoint of any of the intervals. Obviously this is a contradiction. The real axis consists of numbers, >not of intervals. No-one said it didn't. That collection of intervals covers part of the interval (r, r + 1/2), including all the rationals in that interval. It is quite possible that your covering should have a sub-collection of similar structure around some point r (though not around every uncovered point, because of the finite measure). The uncovered points between r and r + 1/2 are just the r + 1/2^n, for n=2,3, ... Pick one of these and call it s. If your original argument were valid, every rational between r and s would be uncovered, can you see any such? > Of course the set of uncovered points in your example is much more > complicated - having infinite measure but containing no rationals - and > a lot more difficult to visualise, but you have still not proved your > contradiction. But you seriously think that your idea of a limit point of intervals >that is not a limit point of numbers in the intervals makes any sense? I never even mentioned a limit point of intervals, only a limit point of the union of the intervals, which *is* the set of the numbers in the intervals. -- David Hartley === Subject: Re: MatheRealism posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Of course the set of uncovered points in your example is much more > complicated - having infinite measure but containing no rationals - and > a lot more difficult to visualise, but you have still not proved your > contradiction. But you seriously think that your idea of a limit point of intervals >that is not a limit point of numbers in the intervals makes any sense? I never even mentioned a limit point of intervals, only a limit point of > the union of the intervals, which *is* the set of the numbers in the > intervals. You are correct. Now you should try to prove that there are uncountably many such clusters of .92ntervals. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > You do know what U{I' n : n = 1,2, ...} means, don't you? You're LATE. NO, he DOESN'T know what this means, and more to the point, he DOESN'T EVEN CONCEDE THE EXISTENCE of actually infinite sets. In other threads he is alleging that 10^100 is not a natural number. This is a good/cop bad/cop situation. I started cussing this fool out FROM DAY 1, as a result of which he has attempted (with some success) to simply ignore everything I write. He has contented himself with talking to Virgil and Hughes, which is wildly distorting his conception of his own competence. If he will talk to you then all will be well (you are actually competent, bless you for arriving) but you are giving him WAY TOO MUCH credit. Your surmisal that he was using limit point in some sort of normal topological sense IS APPALLING. Even if he can match the technical definition, that does NOT mean he UNDERSTANDS it or understands how it relates to the problem at hand. The problem at hand TIME. You would do well to change all the parameters of the discussion TO CLOSED intervals JUST TO DISABUSE him of his infatuation with the term limit point. That term in the topological sense within open sets IS NOT RELEVANT to the issue at hand; what IS relevant is accumulation points of infinite sequences. > It *is* the > set of all the real numbers contained in the intervals I' n. That is the > set that r is a limit point of. I just told George that you understood > limit points; I've been talking to this fool for months; YOU'RE NEW. Do NOT FURTHER PRESUME to TELL *ME* . *I KNOW*. YOU are still questing. The RELEVANT point here is that if you are starting with a well- ordered infinite sequence of all the rationals, EVERY LAST REAL IN THE WHOLE INTERVAL is a limit point of THAT sequence. WM's use of limit point FOR OPEN intervals is DOUBLY inappropriate because THE WHOLE point about open sets is that THEY DON'T contain their boundaries! THEY DON'T contain their outermost limit points! If you talk about limit points of an open set THEN YOU ARE CLOSING it! You are talking about something that ALSO includes the endpoints! In order to have this argument, you need to PRESERVE the distinction between the open interval and its endpoints! You need to FOCUS on the fact that the endpoints are NOT elements of the interval. But they are also limit points, just like the things IN the interior of the interval Is BESIDE the point! THAT is NOT the point! THAT is the OPPOSITE of the point! If you really want to emphasize how the endpoints are similar THEN YOU SHOULD JUST USE *CLOSED* intervals! > I.e. r is a limit point > of the set U{I' n : n = 1,2, ...} And more to the point, r is NOT IN this union, EVEN though it is a limit point of it.. And EVEN more to the point, you are here taking a limit point OF AN INFINITE SET OF THINGS (SPARE me this collapsing it to 1 thing with U; THAT wasn't EVEN relevant). The RELEVANT use of limit point AROUND HERE *is* for INFINITE SEQUENCES of things. DO NOT keep talking about limit points of OTHER things! > To see more easily that there is no contradiction, consider the > collection of intervals {(r + 1/2^(n+1), r + 1/2^n) : n = 1,2, ...}. > This covers every rational in (r, r + 1/2) (r irrational) but r is not > an endpoint of any of the intervals. You only have a countable number of uncovered points here. You are still going to have to go a lot deeper than this to avoid his paradox about the measures not adding up. His point is that the set of uncovered points is ALWAYS going to be bijectible with the set of intervals. In THIS case that bijection is easy. You still have to prove to him that it will not usually be this way. IF, that is, we can ever get past his tendency to JUST MAKE UP about terms that already have textbook definitions (like covered or limit point). In your example, the sums of the measures of the (disjoint) intervals DO add up to the measure of the whole interval (r,r+1/2). OUR point by contrast was about a sequence of intervals-around- rationals that add up to something FAR LESS than that (something arbitrarily small, in fact). If this measure is going to be additive then the number of points left out: has to be UNcountable. But what it even MEANS for a point to be left out of (or not covered by) this series of intervals is itself highly problematic, SINCE THEY OVERLAP so much. Anyway, you need to stop giving the fool credit for parroting IRrelevant definitions and focus on the problem at hand. === Subject: Re: MatheRealism > The RELEVANT point here is that if you are starting with a well- > ordered infinite sequence of all the rationals, EVERY LAST REAL > IN THE WHOLE INTERVAL is a limit point of THAT sequence. I've not been following this thread (why would anyone follow a WM thread?), don't know what's going on, just looked at this post at random out of idle curiousity. I just wanted to mention that also there are more limit points of that sequence then there are elements of that sequence. -- But -- I'm prepared to hear that that's not true in this context -- Or -- That it's utterly irrelevant in this context. It's just a fact I find amusing in its proper context. -- hz === Subject: Re: MatheRealism <2OFKmuMS5CPJFwgf@212648.invalid> <1b+aTEM+wXPJFwQS@212648.invalid> In message >I've been talking to this fool for months; YOU'RE NEW. Do NOT >FURTHER PRESUME to TELL *ME* __. *I KNOW*. YOU are still >questing. Actually, I'm Old. I kill-filed WM years ago. Some glitch allowed him to pop out again and I foolishly mis-remembered that he occasionally responded to reason. His views on infinity are not directly relevant here. As usual he is trying to justify them by trying to prove a contradiction within modern set theory, using its own apparatus. Of course, if someone does manage to convince him that his argument is false, he will fall back to but infinite sets don't exist anyway ... I think I'll just repair my kill file. -- David Hartley === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > Actually, I'm Old. I kill-filed WM years ago. I apologize. You certainly did seem very competent. That would tend to imply that you were from the prior population. > His views on infinity are not directly relevant here. Yes, they are. His views on infinity are what this is about. > As usual he is trying to justify them by trying to prove > a contradiction within modern set theory, Well, sort of. What he is ACTUALLY doing is presenting factual situations and then claiming FALSELY that they are contradictory, expecting the fact that at least SOME people find them counter- intuitive to SUFFICE -- if the formal machinery doesn't confirm the contradiction then he wants people to dismiss the formal machinery as insufficient, on its face. > using its own apparatus. No, NEVER. You see, the problem is, the axiom of infinity IS part of that apparatus, and he's far from competent to use THE REST of it. In particular, he doesn't grok Union. > Of course, if someone does manage > to convince him that his argument is false, he will fall back to but > infinite sets don't exist anyway ... I think I'll just repair my kill > file. Well, if I had known you were going to leave, I would've been nicer. You were needed. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > I never even mentioned a limit point of intervals, only a limit point of > the union of the intervals, which *is* the set of the numbers in the It is precisely BECAUSE it is the set of numbers in the intervals that it DOESN'T MATTER what its limit points are or aren't THAT union IS JUST AN OPEN INTERVAL, so given that it is ONE thing, IT ISN'T EVEN REASONABLE TO SPEAK OF IT AS HAVING any limit points! It has points that are IN It and it has ENDpoints that it doesn't include, AND NO MENTION OF THE WORD limit IN THIS CONTEXT is justifiable! IT'S JUST PRETENSION TO COMPLEXITY! USE limit point FOR INFINITE sequences of things! MAKE IT *relevant*! MAKE IT *matter*!! Taking a union OF AN ACTUALLY INFINITE SET really does INVOLVE PASSING TO a limit! If the open set is already fully constructed then that passage has ALREADY OCCURRED, regardless of whether one then chooses to have the limit point ITSELF inside the set or out! === Subject: Re: MatheRealism Each neighbourhood of r will contain infinitely many of the intervals > I'_n. But - since you specified that r is not in or an endpoint of any > such interval - for any particular I'_n there will be a smaller > neighbourhood of r which is disjoint from I'_n. I.e. r is a limit point > of the set U{I'_n : n = 1,2, ...} but it is not a limit point of I'_n > for any n. If r is a limit point of the intervals, then r is a limit point also > of the real numbers that are contained in the intervals. Please spare > your matheology of intervals without contents. And r is not a limit > point of a distant interval. So r cannot be a limit point of the > intervals but at most the limit point of one interval. At most the limit point of one interval ??? WM demonstrated his profound ignorance of actual mathematics once again. Unless there is a non-empty open interval between any pair of his intervals, or, equivalently, their closures are pairwise disjoint, it is quite possible for two of them to have a common limit point. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > To see more easily that there is no contradiction, consider the > collection of intervals {(r + 1/2^(n+1), r + 1/2^n) : n = 1,2, ...}. > This covers every rational in (r, r + 1/2) (r irrational) but r is not > an endpoint of any of the intervals. > Obviously this is a contradiction. Obviously, the only thing preventing you from being a ing idiot is that you are too stupid to get laid. This IS NOT a contradiction: this is A FACT. You can't claim that it is impossible for something to exist WHILE YOU ARE LOOKING RIGHT AT IT. r IS NOT IN ANY of these intervals. r is TO THE LEFT of ALL of these intervals. But for ANY real s BIGGER than r, (and less than r+1/4), INFINITELY many of these intervals contain that real s. THAT'S JUST PROVABLE from this and there is NOTHING CONTRADICTORY ABOUT this. IF YOU THINK THERE IS, then all you have to do is JUST PROVE IT. The fact that you do all your arguing in a natural language that you can't even speak is a large part of your lack of credibility. ALL the arguments that MATTER around HERE are FORMAL! === Subject: Re: MatheRealism > To see more easily that there is no contradiction, consider the > collection of intervals {(r + 1/2^(n+1), r + 1/2^n) : n = 1,2, ...}. > This covers every rational in (r, r + 1/2) (r irrational) but r is not > an endpoint of any of the intervals. Obviously this is a contradiction. Obviously, the only thing preventing you from being a ing idiot > is that you are too stupid to get laid. This IS NOT a contradiction: > this is A FACT. You can't claim that it is impossible for something > to exist WHILE YOU ARE LOOKING RIGHT AT IT. > r IS NOT IN ANY of these intervals. > r is TO THE LEFT of ALL of these intervals. > But for ANY real s BIGGER than r, (and less than r+1/4), > INFINITELY many of these intervals contain that real s. > THAT'S JUST PROVABLE from this and there is > NOTHING CONTRADICTORY ABOUT this. > IF YOU THINK THERE IS, then all you have to do is JUST PROVE IT. > The fact that you do all your arguing in a natural language that you > can't even > speak is a large part of your lack of credibility. ALL the arguments > that MATTER > around HERE are FORMAL! SHOUTING at WM doesn't penetrate his invincible ignorance any better that more quietly pointing out his idiocies. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > SHOUTING at WM doesn't penetrate his invincible ignorance any better > that more quietly pointing out his idiocies. JEEzus, like anybody gives a what YOUR ignorant ass thinks. The fastest way to penetrating WM's ignorance by far would be FOR YOU TO SAY *ABSOLUTELY*NOTHING further. You're a complete embarrassment. Far worse, YOU ARE THE PRIMARY ENABLER of this BULL. Every time YOU make a mistake, WM thinks he's smart again. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > If r is a limit point of the intervals, THEN YOU ARE TALKING A COMPLETELY DIFFERENT LANGUAGE from the one you were talking before, because before, you talked about limit points OF ONE interval. If you are now going to allege that some sequence OF INTERVALS has limit points, THEN YOU STILL HAVE TO *DEFINE* what YOU mean by this! DH was trying to give you the benefit of the doubt but EVEN HE had NO idea what you were talking about! Since WE have NO idea what YOU mean by limit point of the intervals, there is going to be a lot more heat than light until you either just admit your IGNORANCE AND SHUT THE UP, OR state a definition having SOMEthing in common with SOMEthing that people who HAVE read the textbooks MIGHT recognize! > then r is a limit point also of the real numbers that are contained in the intervals. THERE IS NO SUCH THING as a limit point of the real numbers! Do you mean THE SET of real numbers that are contained in the intervals?? EVERY such real number is a limit point OF THAT set! IN ADDITION to the real numbers that are limit points because they ARE IN the set, there ARE ALSO some MORE limit points that ARE NOT IN ANY interval! But you continue to use limit point in a COMPLETELY BOGUS way withOUT clarifying ANYthing! > Please spare your matheology of intervals without contents. NOBODY HERE HAS EVER OFFERED YOU any intervals without contents!! ABSOLUTELY EVERY interval that ANYbody HAS EVER spoken to you about here is an open interval WITH content! WITH a rational midpoint and equal amounts of rational content/length ON BOTH sides of that midpoint! > And r is not a limit point of a distant interval. THERE IS NO SUCH THING as a limit point OF ONE interval! Limit points are limit points OF *infinite sequences* of things!! === Subject: Re: MatheRealism <1b+aTEM+wXPJFwQS@212648.invalid> posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > Of course the set of uncovered points in your example is much more > complicated - having infinite measure but containing no rationals Nothing around here needs to contain infinite measure. We need to confine ourselves to the interval (0,1) on the outside and to open intervals around all rationals on the inside. And you need to to tell WM what a limit point is, and more importantly, what KINDS of things CAN HAVE limit points. WM has been talking about limit points OF AN INTERVAL. That is not even grammatical. SEQUENCES OF POINTS have limit points. INTERVALS do NOT have limit points. BECAUSE sequences of points have limit points, it is also possible for sequences OF INTERVALS to have limit points. But again, those limit points are properties of A WHOLE COMPLETED ACTUALLY INFINITE SET (something WM DOES NOT EVEN CONCEDE THE EXISTENCE OF), NOT of any INDIVIDUAL interval!! === Subject: Re: MatheRealism <2OFKmuMS5CPJFwgf@212648.invalid> <1b+aTEM+wXPJFwQS@212648.invalid> In message > Of course the set of uncovered points in your example is much more > complicated - having infinite measure but containing no rationals Nothing around here needs to contain infinite measure. >We need to confine ourselves to the interval (0,1) on the outside >and to open intervals around all rationals on the inside. >And you need to to tell WM what a limit point is, and more >importantly, >what KINDS of things CAN HAVE limit points. WM has been talking >about limit points OF AN INTERVAL. That is not even grammatical. >SEQUENCES OF POINTS have limit points. INTERVALS do NOT have >limit points. BECAUSE sequences of points have limit points, it is >also >possible for sequences OF INTERVALS to have limit points. But again, >those limit points are properties of A WHOLE COMPLETED ACTUALLY >INFINITE SET >(something WM DOES NOT EVEN CONCEDE THE EXISTENCE OF), NOT of any >INDIVIDUAL interval!! > Of course we don't *need* infinite measure. WM could have constructed a similar argument within a finite interval, and still produced the same fallacy, but he chose to use the whole line. Intervals are subsets of R, subsets of topological spaces (like R) have limit points. -- David Hartley === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > Intervals are subsets of R, subsets of topological spaces (like R) have > limit points. Countable sequences ALSO have limit points, and THOSE are the limit points THAT ARE RELEVANT. EVERY point in an open subinterval of R is a limit point of that interval, so THAT is IRRELEVANT. WM in fact does NOT EVEN KNOW WHICH version of the definition of limit point: he IS EVEN TRYING to talk about.; he certainly did NOT mean for EVERY point of the interval to be a limit point of the interval. === Subject: Re: MatheRealism <2OFKmuMS5CPJFwgf@212648.invalid> <1b+aTEM+wXPJFwQS@212648.invalid> In message > Intervals are subsets of R, subsets of topological spaces (like R) have > limit points. Countable sequences ALSO have limit points, and THOSE are the limit >points THAT ARE RELEVANT. EVERY point in an open subinterval of R >is a limit point of that interval, so THAT is IRRELEVANT. >WM in fact does NOT EVEN KNOW WHICH version of the definition of >limit point: he IS EVEN TRYING to talk about.; he certainly did NOT >mean >for EVERY point of the interval to be a limit point of the interval. The following quote seems to show that WM is using the topological definition of limit point, and knows that every point of an open interval is a limit point of the interval. >Then r is a limit point of I'_n (i.e., either r >belongs to the interval or is one of its endpoints). If this is not >the case, then r is not a limit point of I'_n and then there is an >epsilon surrounding of r such that no element of I'_n belongs to this >surrounding. -- David Hartley === Subject: Re: MatheRealism posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > In message > Intervals are subsets of R, subsets of topological spaces (like R) have > limit points. Countable sequences ALSO have limit points, and THOSE are the limit >points THAT ARE RELEVANT. EVERY point in an open subinterval of R >is a limit point of that interval, so THAT is IRRELEVANT. WM in fact does NOT EVEN KNOW WHICH version of the definition of >limit point: he IS EVEN TRYING to talk about.; he certainly did NOT >mean >for EVERY point of the interval to be a limit point of the interval. The following quote seems to show that WM is using the topological > definition of limit point, and knows that every point of an open > interval is a limit point of the interval. >Then r is a limit point of I' n (i.e., either r >belongs to the interval or is one of its endpoints). If this is not >the case, then r is not a limit point of I' n and then there is an >epsilon surrounding of r such that no element of I' n belongs to this >surrounding. Of course a limit point of an interval is a point such that every epsilon surrounding of the limit point contains infinitely many points of the interval. But george is only good in cursing and grumbling. He does not know such definitions. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) >Then r is a limit point of I' n (i.e., either r >belongs to the interval or is one of its endpoints). If this is not >the case, then r is not a limit point of I' n and then there is an >epsilon surrounding of r such that no element of I' n belongs to this >surrounding. This IS NOT RELEVANT. That IS NOT THE DEFINITION of limit point that IS GOING ON here! You have got to know what to throw away and what to keep. His conitnue use of this surrounding IS NOT RELEVANT *either*!! If he really wants r to be outside the interval THEN HE CAN JUST MAKE THE INTERVALS CLOSED and say that r is NOT IN the interval AS OPPOSED to r is not a limit point of the interval! He is using limit point HERE, IF he is using it that way, as an ALIAS for JUST BEING *IN* the interval and THAT IS WORTHLESS, as a use of the term! if THAT'S all you meant then YOU COULD JUST *SAY* It's in the interval AS OPPOSED to it's a limit point of the interval!! NOBODY, ESPECIALLY not people caring about the topological sense, is going to use limit point IN THIS WAY, in this context!! GET A CLUE!!! JEEzus!! === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/20070309 Firefox/2.0.0.3,gzip(gfe),gzip(gfe) > Here is what I said: I have considered a covering of the rationals by a set of finite > intervals such that the rational number q_n is covered by the closed > interval I_n of length 1/2^n this presumably means 1/(2^n+1) ON EACH SIDE of a midpoint at q_n > (which later on is shrunk somewhat, but not very much, This is just stupid; you should just let the intervals be open to start with. > to become the open interval I'_n). Every interval I'_n > covers infinitely many real numbers. AND infinitely many RATIONAL numbers, relevantly. > If an interval I'_n belongs to the > surrounding of the real number r, This is a completely NON-ENGLISH use of belongs to and surrounding. IN ENGLISH, this would be said, If r is in the interval l'_n. > then infinitely many real numbers of I'_n belong to > the surrounding of r There IS NO SUCH THING AS the surrounding of r. Real numbers DON'T COME WITH surroundings. For any given e, (r-e,r+e) is an open interval of the real line, SO OF COURSE it contains infinitely many reals, infintiely many rationals, and infinitely many intervals q_whatever (since there is an interval around each one of the contained rationals). Your idiotic invention of the surrounding of is excruciatingly IRRELEVANT. > (It is mathematically > impossible to have only one element of I'_n belonging to the > surrounding of r.) The surrounding of r DOES NOT EXIST. Try again. === Subject: Re: MatheRealism > In message It is easy. Proof by contradiction: >Assume that there is a real number r that is not covered by an interval >and is not an endpoint of an interval. If that were the case, then anything and everything follows. === Subject: Re: MatheRealism posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) In message It is easy. Proof by contradiction: >Assume that there is a real number r that is not covered by an interval >and is not an endpoint of an interval. If that were the case, then anything and everything follows. Here you are as right as man can be! So not anything and everything follows, but only the following follows: The number of uncovered reals is at most twice the number of intervals. The numbers of intervals is countable. === Subject: Re: MatheRealism In message It is easy. Proof by contradiction: >Assume that there is a real number r that is not covered by an interval >and is not an endpoint of an interval. If that were the case, then anything and everything follows. Here you are as right as man can be! > So not anything and everything follows It does from QWM's assumptions. === Subject: Re: MatheRealism posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/20070309 Firefox/2.0.0.3,gzip(gfe),gzip(gfe) > only the following > follows: The number of uncovered reals You don't even know how to prove that ANY real is uncovered. That makes all the rest of this entirely moot. > is at most twice the number of intervals. You do not know how to prove this. > The numbers of intervals is countable. Of course it is, and the number of uncovered points can vary wildly. === Subject: 2^(n+1)+15=p (prime number) Let n>=0 If 2^n+6 =/=[(p^2-3)/2] mod.(p) then 2^(n+1)+15=p (prime number) example: n=0, p=17 n=1, p=19 n=2, p=23 n=3, p=31 n=4, p=47 n=5, p=79 n=6, 2^6+6=70; 70=59 mod.(11) n=7, p=271 n=8, p=523 n=9, p=1039 n=10, p=2063 n=11, p=4111 n=12, 2^12+6=4102; 4102=419 mod.(29) n=13, 2^13+6=8198; 8198=479 mod.(31) n=14, p=32783 n=15, p=65551 and so on Vincenzo Librandi vincenzo.librandi@tin.it === Subject: Re: 2^(n+1)+15=p (prime number) > Let n>=0 If 2^n+6 =/=[(p^2-3)/2] mod.(p) then 2^(n+1)+15=p (prime number) >2^n+6 != (p^2-3)/2 mod p >2^(n+1) + 12 != p^2 - 3 mod p >2^(n+1) + 15 != 0 mod p So, if 2^n+6 != (p^2-3)/2 mod p for all odd primes < >sqrt(2^(n+1)+15), >then 2^(n+1)+15 is indeed prime. example: n=0, p=17 > n=1, p=19 > n=2, p=23 > n=3, p=31 > n=4, p=47 > n=5, p=79 > n=6, 2^6+6=70; 70=59 mod.(11) > n=7, p=271 > n=8, p=523 >That should be 527, not 523. OOps! Sorry! for n=8 is false -- --Tim Smith Vincenzo Librandi === Subject: Re: 2^(n+1)+15=p (prime number) posting-account=DLD3MQkAAACJxulKt9xouw3DPpXKssCI SLCC1; .NET CLR 2.0.50727; .NET CLR 1.1.4322; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > <25284210.1228766665750.JavaMail.jaka...@nitrogen.mathforum.org>, Let n>=0 If 2^n+6 =/=[(p^2-3)/2] mod.(p) then 2^(n+1)+15=p (prime number) 2^n+6 != (p^2-3)/2 mod p > 2^(n+1) + 12 != p^2 - 3 mod p > 2^(n+1) + 15 != 0 mod p So, if 2^n+6 != (p^2-3)/2 mod p for all odd primes < sqrt(2^(n+1)+15), > then 2^(n+1)+15 is indeed prime. The for all odd primes is the bit that makes it computationally intractible for Mersenne prime sized numbers. So it is interesting, but not terribly useful to generate huge prime numbers. E.g. primepi(10^23) = 1,925,320,391,606,803,968,923, which is a lot of factors to divide by, even for a 23 digit number. One million digits is right out of the question. > example: n=0, p=17 > n=1, p=19 > n=2, p=23 > n=3, p=31 > n=4, p=47 > n=5, p=79 > n=6, 2^6+6=70; 70=59 mod.(11) > n=7, p=271 > n=8, p=523 That should be 527, not 523. -- > --Tim Smith === Subject: Re: 2^(n+1)+15=p (prime number) posting-account=q2ruDwoAAACW4OvF4soI61szUIc6xJHw Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > What are you saying here? That it doesn't work for > n=6? Or n=12 or n=13? The original post was somewhat cryptic. Apparently implicit in the conjecture is that if there exists any prime q such that 2^n+6 ==[(q^2-3)/2] mod(q) then the conjecture is making no assertion for 2^(n+1)+15. Therefore, the cases of n=6,12 et al are to be ignored, since for n=6, setting q=11 yields 2^n+6 ==[(q^2-3)/2] mod(q) and for n=12, setting q=29 yields 2^n+6 ==[(q^2-3)/2] mod(q). Also, the original post contained a harmless numerical error. For n=8, 2^(n+1)+15 = 527 = (17)(31), rather than 523. However, 2^8+6 = 262 == [(17^2)-3]/2 (mod 17). To demonstrate (for example) that the conjecture fails at n=16, each prime number (q) < 65542 must be examined to determine whether 65542 is congruent to [(q^2-3)/2] mod(q). Again I ask, is this conjecture true? Has anyone selected a specific candidate-counterexample (e.g. n=16) and computer-verified that no corresponding prime q exists? Alternatively, has anyone proven this conjecture? If the conjecture actually is true, it would be curious to find the pattern for computing q when 2^(n+1)+15 is composite. If q is readily calculable, then the conjecture could be used to generate large primes. === Subject: Re: 2^(n+1)+15=p (prime number) posting-account=DLD3MQkAAACJxulKt9xouw3DPpXKssCI SLCC1; .NET CLR 2.0.50727; .NET CLR 1.1.4322; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) On Dec 8, 12:03pm, Vincenzo Librandi =0 If 2^n+6 =/=[(p^2-3)/2] mod.(p) then 2^(n+1)+15=p (prime number) example: n=0, p=17 > n=1, p=19 > n=2, p=23 > n=3, p=31 > n=4, p=47 > n=5, p=79 > n=6, 2^6+6=70; 70=59 mod.(11) > n=7, p=271 > n=8, p=523 > n=9, p=1039 > n=10, p=2063 > n=11, p=4111 > n=12, 2^12+6=4102; 4102=419 mod.(29) > n=13, 2^13+6=8198; 8198=479 mod.(31) > n=14, p=32783 > n=15, p=65551 It seems this procedure may be used to find titanic primes in a simple way. Perhaps it can generate larger primes than GIMPS. === Subject: Re: 2^(n+1)+15=p (prime number) posting-account=q2ruDwoAAACW4OvF4soI61szUIc6xJHw Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > Let n>=0 > If > 2^n+6 =/=[(p^2-3)/2] mod.(p) > then > 2^(n+1)+15=p (prime number) > Wow. Is this conjecture true? Is this conjecture new? To original poster: how did you (or whoever did originate this) happen to think of it? === Subject: Re: trajectory of a bullet +M[5[U[QT7xFN%^gR=tuJw%TXXR'Fp~W;(T1(739R%m0Yyyv*gkGoPA.$b,D.w:z+<'=-lV T?6 {T?=R^:W5g|E2#EhjKCa+nt:4b}dU7GYB*HBxn&Td$@f%.kl^:7X8rQWd[NTcPu6nkisze/ Q;8 9Z{peQF,w)7UjV$c|RO/mQW/NMgWfr5*$-Z%u46/00mx-,R'fLPe.)^ > Could it be that the system just determines the direction the bullet > came from, assuming it's directed somewhere towards the microphones? > (Without slogging through it I'm not sure how sensitive that > somewhere towards would be though.) given that any three (non co-linear) points define a circle shots arriving from the same origin but aimed slightly differently will give different shockwave arrival times. arrival times of the shock wave tell a lot more about where the shot was going than about where it came from. If the microphones are improved to give directional and/or distance information as well as time information that increases the degrees of freedom in the input tuple which would help a lot as someone else has pointed out there are only two degrees of freedom for the location of the shooter (I had three) as it could lie anwhere along the previous path of the bullet, however there is another degree of freedom for the shot, it's speed, so we still need atleast 6 degrees speed and direction the bullet is going in) to hope to resolve the shot to it's origin. === Subject: Re: trajectory of a bullet Could it be that the system just determines the direction the bullet > came from, assuming it's directed somewhere towards the microphones? > (Without slogging through it I'm not sure how sensitive that > somewhere towards would be though.) given that any three (non co-linear) points define a circle > shots arriving from the same origin but aimed slightly differently > will give different shockwave arrival times. arrival times of the shock wave tell a lot more about where the shot > was going than about where it came from. If the microphones are improved to give directional and/or distance > information as well as time information that increases the degrees > of freedom in the input tuple which would help a lot as someone else has pointed out there are only two degrees of freedom > for the location of the shooter (I had three) as it could lie anwhere along the > previous path of the bullet, however there is another degree of > freedom for the shot, it's speed, so we still need atleast 6 degrees > speed and direction the bullet is going in) to hope to resolve the > shot to it's origin. At the end of the original post, it says: A 4th microphone is used to determine if the gunshot originated from above or below the plane of the first 3. Also, maybe the Mach number is given as n. Since they mention cones, I think we can forget about gravity. If a Concorde flies straight at Mach 2, I think the shock-wave looks like a cone advancing in the same direction as the plane. The Mach number would determine how pointed the cone is. Actually, maybe we can think of a Concorde flying in a straight line , not necessarily horizontal. David Bernier === Subject: circles DONT EXIST Circles can only survive in theory. In theory, circles have an infinite number of sides. However, even with computers, you cannot generate an infinite amount. Oh yes, you can do well... but even if it has 12 million trillion sides (which is quite enough most people would agree), it is NOT INFINITE. Also, from the position taken above, spheres are only possible in theory aswell. One thing about circles that often goes unnoticed, is that circles sides are STRAIGHT, though it appears to be a single curved line. === Subject: Re: circles DONT EXIST posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG 1.1.4322; .NET CLR 2.0.50727; MS-RTC LM 8; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > One thing about circles that often goes unnoticed, is that circles sides are > STRAIGHT, though it appears to be a single curved line. So the side of a circle is straight, but it only appears curved. Is it some kind of optical illusion? === Subject: Re: circles DONT EXIST posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e 2.0.50727),gzip(gfe),gzip(gfe) > Circles can only survive in theory. > In theory, circles have an infinite number of sides. > However, even with computers, you cannot generate an infinite amount. > Oh yes, you can do well... but even if it has 12 million trillion sides (which is quite enough most people would agree), it is NOT INFINITE. Also, from the position taken above, spheres are only possible in theory aswell. One thing about circles that often goes unnoticed, is that circles sides are STRAIGHT, though it appears to be a single curved line. *********************************************************** I wonder (but only for 10-15 seconds) what your bike's wheels look like... Tonio === Subject: Re: circles DONT EXIST >Also, from the position taken above, spheres are only possible in theory aswell. Soap bubbles don't exist then. This has to be Ernie's friend BURT with yet another ID to defeat kill filters. --Lynn === Subject: Re: circles DONT EXIST posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) Also, from the position taken above, spheres are only possible in theory aswell. Soap bubbles don't exist then. This has to be Ernie's friend BURT with yet another ID to defeat kill > filters. > I'm very sure this is BURT. The title circles DONT EXIST looks a lot like something BURT would come up with. === Subject: Re: circles DONT EXIST posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1 FOH:R051,gzip(gfe),gzip(gfe) there's got to be a few Latin phrases for this kiond of fallacy; this is the same thing that Are Buckafka Fullofit tried to put over, and most of the Buckywitches really believe it. anyway, teh queestion is, why should line segments of polygons be so God-am existential, or of polyhedra -- because of the typical stick-drawn configuration in some edition of Euclid? >Also, from the position taken above, spheres are only possible in theory aswell. --Cheeny & Zbiggy, fo'mo' years! http://tarpley.net http://larouchepub.com === Subject: Re: circles DONT EXIST posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1 FOH:R051,gzip(gfe),gzip(gfe) there's got to be a few Latin phrases for this kind of fallacy; this is the same thing that Are Buckafka Fullofit tried to put over, and most of the Buckywitches really believe it. anyway, the queestion is, why should line segments of polygons be so God-am existential, or of polyhedra -- because of the typical stick-drawn configuration in some edition of Euclid? >Also, from the position taken above, spheres are only possible in theory aswell. --Cheeny & Zbiggy, fo'mo' years! http://tarpley.net http://larouchepub.com http://xplodeyourmlmbiz.com/?s1=Fuller+Brush&t=LSP01&gclid=CPiar6nispcCFQu-G godYm21kA http://www.rwgrayprojects.com/synergetics/plates/plates.html === Subject: Re: circles DONT EXIST posting-account=HJnNYQoAAACtdMyfjpc6YoWX6U0epxWM Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > One thing about circles that often goes unnoticed, is that circles sides are STRAIGHT, though it appears to be a single curved line. What do you think a curve IS? === Subject: Re: circles DONT EXIST > Circles can only survive in theory. > In theory, circles have an infinite number of sides. > However, even with computers, you cannot generate an infinite amount. > Oh yes, you can do well... but even if it has 12 million trillion sides > (which is quite enough most people would agree), it is NOT INFINITE. Also, from the position taken above, spheres are only possible in theory > aswell. One thing about circles that often goes unnoticed, is that circles sides > are STRAIGHT, though it appears to be a single curved line. You're a freaken genius! === Subject: Re: circles DONT EXIST <25866502.1228771481150.JavaMail.jakarta@nitrogen.mathforum.org>, > Circles can only survive in theory. That theory is called geometry. > In theory, circles have an infinite number of sides. In MY geometry, circles only have two sides. Inside and outside. That a circle may be approximated by a regular polygon of many sides is quite a different thing. > However, even with computers, you cannot generate an infinite amount. > Oh yes, you can do well... but even if it has 12 million trillion sides > (which is quite enough most people would agree), it is NOT INFINITE. Also, from the position taken above, spheres are only possible in theory > aswell. One thing about circles that often goes unnoticed, is that circles sides are > STRAIGHT, though it appears to be a single curved line. Nonsense. === Subject: Re: ``$me$ always changes (A Poetic Force of Nature) posting-account=tNh-2AoAAACqdWo2IikcW-FF_IGUAvWf 2.0.50727),gzip(gfe),gzip(gfe) $me$ always changes is a Physics Truism. It's a given. An > Obviousism. Well, except when mass is constant. Which is lots of the time. Site an example of when and what mass is constant in Nature. Ok, er... Trafalgar Square, London. Or did you want an example cited? > If so you used the wrong poetic homonym. I accept the written word is often subject to varied interpretations. But an example of constant mass is pending. I asked to make a point. Your example of constant Sunday Mass, is, if you examine it more closely, never exactly the same. One can never go to the same river twice. Same hymn, changed incantation every time. Otherwise, repetition would be deafening. While on the influence of Languages, mostly of the poetic kind, there's this First Commandment, that mentions a $me$. Homonyms, not my invention. Nor Maths, nor words with meanings. Nor http://plato.stanford.edu/entries/spinoza/ ... there's a lot I can't take credit for. Poetic Axiom of Nature: All mass is restless. Enjo(y). -- Mahipal ``We search our Earth and her limitless skies all the while discovering in our hearts and minds is where heaven lies'' http://mysite.verizon.net/mahipalvirdy/ === Subject: Re: ``$me$ always changes (A Poetic Force of Nature) $me$ always changes is a Physics Truism. It's a given. An > Obviousism. Well, except when mass is constant. Which is lots of the time. Site an example of when and what mass is constant in Nature. Ok, er... Trafalgar Square, London. Or did you want an example cited? > If so you used the wrong poetic homonym. I accept the written word is often subject to varied interpretations. Different written words with different spelling such as site, sight, cite, slight and sleight have different meanings. But an example of constant mass is pending. I asked to make a point. Energy is relative, E = 1/2 mv^2 and v is relative. If I throw a stone away from you it will have negative energy relative to you because it cannot strike you, unlike a stone thrown toward you with constant mass m. Should you argue that the mass of the stone may be converted to energy I would not disagree, but you will not receive all that energy unless it elastically collides elsewhere and is reflected in your direction. Your example of constant Sunday Mass, is, if you examine it more closely, never exactly the same. I made no such example, you are hallucinating. === Subject: Re: ``$me$ always changes (A Poetic Force of Nature) <3wi%k.20133$wp1.2854@newsfe19.ams2> posting-account=tNh-2AoAAACqdWo2IikcW-FF_IGUAvWf 2.0.50727),gzip(gfe),gzip(gfe) $me$ always changes is a Physics Truism. It's a given. An > Obviousism. Well, except when mass is constant. Which is lots of the time. Site an example of when and what mass is constant in Nature. Ok, er... Trafalgar Square, London. Or did you want an example cited? > If so you used the wrong poetic homonym. I accept the written word is often subject to varied interpretations. > Different written words with different spelling such as site, sight, cite, > slight and sleight have different meanings. O, I wasn't thinking of mere individual words. It's rude and impolite to break down others' paragraphs, into sentences, or, even worse, into words. We already know how they breakdown into alphabits. It's just plain inconsiderate given how technology has made bits available on the order of gigabytes. > But an example of constant mass is pending. I asked to make a point. > Energy is relative, E = 1/2 mv^2 and v is relative. If I throw a stone > away from you it will have negative energy relative to you because > it cannot strike you, unlike a stone thrown toward you with > constant mass m. The kinetic energy of the stone, thrown towards or away from me, is a positive definite quantity. Only the velocity has direction. Relativity yields that the mass of the stone is variable with its changing velocity magnitude. > Should you argue that the mass of the stone may be converted to > energy I would not disagree, but you will not receive all that energy > unless it elastically collides elsewhere and is reflected in your direction. Your example of constant Sunday Mass, is, if you examine it more > closely, never exactly the same. > I made no such example, you are hallucinating. Hallucinations, built-in human mind functionality, are a nice welcomed break. admit they deserved. Enjo(y). -- Mahipal ``We search our Earth and her limitless skies all the while discovering in our hearts and minds is where heaven lies'' http://mysite.verizon.net/mahipalvirdy/ === Subject: Re: ``$me$ always changes (A Poetic Force of Nature) $me$ always changes is a Physics Truism. It's a given. An > Obviousism. Well, except when mass is constant. Which is lots of the time. Site an example of when and what mass is constant in Nature. Ok, er... Trafalgar Square, London. Or did you want an example cited? > If so you used the wrong poetic homonym. I accept the written word is often subject to varied interpretations. > Different written words with different spelling such as site, sight, cite, > slight and sleight have different meanings. O, I wasn't thinking of mere individual words. It's rude and impolite to break down others' paragraphs, into sentences, or, even worse, into words. Yeah, well, I never claimed to be otherwise. A pity if that hurts your lousy feelings. We already know how they breakdown into alphabits. It's just plain inconsiderate given how technology has made bits available on the order of gigabytes. > But an example of constant mass is pending. I asked to make a point. > Energy is relative, E = 1/2 mv^2 and v is relative. If I throw a stone > away from you it will have negative energy relative to you because > it cannot strike you, unlike a stone thrown toward you with > constant mass m. The kinetic energy of the stone, thrown towards or away from me, is a positive definite quantity. Only the velocity has direction. Relativity yields that the mass of the stone is variable with its changing velocity magnitude. > Should you argue that the mass of the stone may be converted to > energy I would not disagree, but you will not receive all that energy > unless it elastically collides elsewhere and is reflected in your > direction. Your example of constant Sunday Mass, is, if you examine it more > closely, never exactly the same. > I made no such example, you are hallucinating. Hallucinations, built-in human mind functionality, are a nice welcomed break. It's rude, impolite, dishonest and highly unethical to accuse others of making statements they did not make. Enjoy your dreaming. === Subject: Re: ``$me$ always changes (A Poetic Force of Nature) posting-account=tNh-2AoAAACqdWo2IikcW-FF_IGUAvWf 2.0.50727),gzip(gfe),gzip(gfe) > $me$ always changes is a Physics Truism. It's a given. An > Obviousism. Well, except when mass is constant. Which is lots of the time. Site an example of when and what mass is constant in Nature. Ok, er... Trafalgar Square, London. Or did you want an example cited? > If so you used the wrong poetic homonym. I accept the written word is often subject to varied interpretations. > Different written words with different spelling such as site, sight, cite, > slight and sleight have different meanings. O, I wasn't thinking of mere individual words. Ok I fumbled. You focused on Site/Cite, where I was on mass/mass. Mea cupla. I am known to make such errors of/from pronounciation. I've written mute for moot. It's in my cross-cultural upbringing, and I realize spell checkers alone are not enough. I confuse Mary for Meri; such for such. Happens. Regardless, Trafalgar Square is not a example of Show me an example... Language can lead to funny times. That's all good and expected. Enjo(y). -- Mahipal In Hindi, such means true Virdy ``We search our Earth and her limitless skies all the while discovering in our hearts and minds is where heaven lies'' http://mysite.verizon.net/mahipalvirdy/ === Subject: Re: ``$me$ always changes (A Poetic Force of Nature) > $me$ always changes is a Physics Truism. It's a given. An > Obviousism. Well, except when mass is constant. Which is lots of the time. Site an example of when and what mass is constant in Nature. Ok, er... Trafalgar Square, London. Or did you want an example cited? > If so you used the wrong poetic homonym. I accept the written word is often subject to varied interpretations. > Different written words with different spelling such as site, sight, > cite, > slight and sleight have different meanings. O, I wasn't thinking of mere individual words. Ok I fumbled. You focused on Site/Cite, where I was on mass/mass. Mea cupla. I am known to make such errors of/from pronounciation. I've written mute for moot. It's in my cross-cultural upbringing, and I realize spell checkers alone are not enough. I confuse Mary for Meri; such for such. and cupla for culpa, pronounciation for pronunciation, for which you are culpable. Happens. Regardless, Trafalgar Square is not a example of Show me an example... Correct, it is a site of when and what mass is constant in Nature. === Subject: Request Kronecker Derivation posting-account=q2ruDwoAAACW4OvF4soI61szUIc6xJHw Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) Kronecker result below. My number theory book: (1) established that if prime p congruent to 3 (mod 4), then {(p-1)/2}! congruent to +1 or -1 (mod p) (2) reported that Kronecker derived following algorithm to determine whether it is +1 or -1: L denotes {p-2^2, p-4^2, p-6^2, ...} M denotes {x : x in L, x>0, x has form r^(4i+1) * s^2 where r,i,s satisfy conditions below} 1. r,s,i are integers 2. i >=0, s>=1 3. r is prime number that does not divide s. Then value is +1 (rather than -1) <==> order of M is even. Example given is p=127, with M having exactly 1 element (namely 127 - 8^2 = 7^1 * 3^2). Thus (63)! congruent to -1 rather than +1 (mod 127). === Subject: Re: polynomial algorithme for isomorphic graphs <1745797.1228662361565.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008111217 Fedora/3.0.4-1.fc9 Firefox/3.0.4,gzip(gfe),gzip(gfe) > On 7 d.8ec, 15:05, riderofgiraffes I'm sorry, but that a translation by Google. Perhaps you should put a description of your algorithm > on the web where we can read it in the original. The > version you posted here is complete garbage. Google > is exceedingly poor at technical translations. > If I give you two graphs with 100 vertices, > every vertex of degree 4, does your algorithm > work? Yes I am ready to use my algorithm, give me > your graphs, and a bit of time (1 hour). I am unwilling to spend time generating difficult > examples without having some expectation that it > will be worth my time. I would be interested in > seeing an unmutilated description of your algorithm. ok > Document in French with a clear example is available at:www.wbabin.net/science/mimouni2f.pdf How do you sort the various n|h|m|b in the label for one of the vertices? -- m === Subject: Re: polynomial algorithme for isomorphic graphs > Perhaps you should put a description of your > algorithm on the web where we can read it in > the original. > Document in French with a clear example is > available at: www.wbabin.net/science/mimouni2f.pdf Merci - je souhaite avoir le temps pour le lire en detail dans les prochains jours. in detail in the next few days. === Subject: Re: Approximating Pi Each of the equations in which 3x^2 - 1 is embedded will > have different solutions. It has been found convenient > to call the solutions to 3x^2 + 1 = 0 (equation number 1 > above) the zeroes of the polynomial (since it is equated > there with zero). Solving for x, we have: 3x^2 - 1 = 0 Add 1 to both sides: 3x^2 = 1 Divide both sides by 3: x^2 = 1/3 Taking the square roots of both sides: x = sqrt(1/3) and x = -sqrt(1/3) we find the two real roots of the equation. These are the 'zeroes' of the the polynomial, even though, > obviously, neither 0.577350269 nor -0.577350269 is itself zero. That's all there is to it. -- > hz Yes, I thought that was what it was about. I know maths. I know about polynomials as you call them. I just call them equations. But I still don't see the significance of transcendentals not having polynomial roots and all that. But if it is significant, then if transcendental polynomials can't have zero as root then why should they have any number as a root? === Subject: Re: Approximating Pi <492B51D2.5EA50538@gmail.com> <49335E6A.17BE19D3@gmail.com> posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1 FOH:R051,gzip(gfe),gzip(gfe) you really seem to like beating that horse; is it dead, yet? seriously, any polynomial with a constant (or times x to the zeroth power) cannot have zero as a solution; otherwise, zero is a trivial solution, since zero times any coefficient is nothing! anyway, what's the problem, if an infinite series polynomial is not algebraic per definition? > don't see the significance of transcendentals not having polynomial > roots and all that. But if it is significant, then if transcendental > polynomials can't have zero as root then why should they have any number > as a root? thus: Euler's identity is, what -- exp(i*pi) = 1, i*pi = ln(1) pi = -i*ln(1), but I forgot what the natural log of one is. thus: there's got to be a few Latin phrases for this kind of fallacy; this is the same thing that Are Buckafka Fullofit tried to put over, and most of the Buckywitches really believe it. anyway, the queestion is, why should line segments of polygons be so God-am existential, or of polyhedra -- because of the typical stick-drawn configuration in some edition of Euclid? >Also, from the position taken above, spheres are only possible in theory aswell. --Cheeny & Zbiggy, fo'mo' years! http://tarpley.net http://larouchepub.com http://xplodeyourmlmbiz.com/?s1=Fuller+Brush&t=LSP01&gclid=CPiar6nispcCFQu-G godYm21kA http://www.rwgrayprojects.com/synergetics/plates/plates.html === Subject: Re: Approximating Pi Solving for x, we have: 3x^2 - 1 = 0 Add 1 to both sides: 3x^2 = 1 Divide both sides by 3: x^2 = 1/3 Taking the square roots of both sides: x = sqrt(1/3) and x = -sqrt(1/3) we find the two real roots of the equation. These are the 'zeroes' of the the polynomial, even though, > obviously, neither 0.577350269 nor -0.577350269 is itself zero. That's all there is to it. Ues, I did all that sums stuff in school. Can't see the significance of > somethinmg haveing or not having a root though. Um, John, it's just a definition. You are free to work out a > mathematical system without algebraic numbers (many others have > already done so), or you are free to accept them as numbers and > develop a mathematical model based on that assumption. Almost all > numbers are derived in one way or another from others. It would be > difficult to go directly from rational numbers to transcendental > numbers (limits of sequences, typically), but it can be done. It is as easy to go from N to R as it is to go from Q to R. R.D. Arthan, The Eudoxus Real Numbers. -- Michael Press === Subject: Re: Approximating Pi <492B51D2.5EA50538@gmail.com> <49335E6A.17BE19D3@gmail.com> posting-account=oX5FAgkAAAD2tWLfJT1nYaPn9C8kADIl Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > Solving for x, we have: 3x^2 - 1 = 0 Add 1 to both sides: 3x^2 = 1 Divide both sides by 3: x^2 = 1/3 Taking the square roots of both sides: x = sqrt(1/3) and x = -sqrt(1/3) we find the two real roots of the equation. These are the 'zeroes' of the the polynomial, even though, > obviously, neither 0.577350269 nor -0.577350269 is itself zero. That's all there is to it. Ues, I did all that sums stuff in school. Can't see the significance of > somethinmg haveing or not having a root though. Um, John, it's just a definition. You are free to work out a > mathematical system without algebraic numbers (many others have > already done so), or you are free to accept them as numbers and > develop a mathematical model based on that assumption. Almost all > numbers are derived in one way or another from others. It would be > difficult to go directly from rational numbers to transcendental > numbers (limits of sequences, typically), but it can be done. It is as easy to go from N to R as it is to go from Q to R. > R.D. Arthan, The Eudoxus Real Numbers. > -- > Michael Press What about C then? R does not contain all algebraic numbers. I suspect that many people treat 'transcendental' as equivalent to 'a member of C', but that was not the original meaning of the word. It means 'not algebraic'. === Subject: Re: Approximating Pi > Solving for x, we have: 3x^2 - 1 = 0 Add 1 to both sides: 3x^2 = 1 Divide both sides by 3: x^2 = 1/3 Taking the square roots of both sides: x = sqrt(1/3) and x = -sqrt(1/3) we find the two real roots of the equation. These are the 'zeroes' of the the polynomial, even though, > obviously, neither 0.577350269 nor -0.577350269 is itself zero. That's all there is to it. Ues, I did all that sums stuff in school. Can't see the significance of > somethinmg haveing or not having a root though. Um, John, it's just a definition. You are free to work out a > mathematical system without algebraic numbers (many others have > already done so), or you are free to accept them as numbers and > develop a mathematical model based on that assumption. Almost all > numbers are derived in one way or another from others. It would be > difficult to go directly from rational numbers to transcendental > numbers (limits of sequences, typically), but it can be done. It is as easy to go from N to R as it is to go from Q to R. > R.D. Arthan, The Eudoxus Real Numbers. > What about C then? It is not in R? :) I was only mentioning that R has transcendental numbers, and that R can be constructed directly from N. > R does not contain all algebraic numbers. No, it does not; a distinction I did not speak from. > I suspect that many people treat 'transcendental' as equivalent to 'a > member of C', but that was not the original meaning of the word. It > means 'not algebraic'. -- Michael Press === Subject: Re: Approximating Pi <492B51D2.5EA50538@gmail.com> <49335E6A.17BE19D3@gmail.com> posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1 FOH:R051,gzip(gfe),gzip(gfe) no big deal. it's just that numbers, themselves don't have roots; an *equation* in one variable has roots or solutions, that are numbers!... well, x - pi = 0 has a root, two! set a quaternion in (w,x,y,z) with coefficients; doess that have roots? > possible roots of certain types of number. Maybe there is no --USA out of Darfur Cruizade! http://larouchepub.com/pr lar/2008/lar pac/081124anti clintons.html --ROTC, your summer vacation in the Sahara Desert ( S u d a n ) ; presage the Draft for your middleschool class of '12 -- brought to you by Allstate (tm) and Oxford U.Press! http://larouchepub.com/pr/2008/080813moloch brown.html http://wlym.com --Wikipedia deletes notice of nullification of preclearance rule of Voting Rights Act in LaRouche v. Fowler, March 27, 2000; is the VRAo1965 a dead letter? http://en.wikipedia.org/wiki/Voting Rights Act#Pre-Clearance 2 === Subject: Re: a special function > ive been ' dreaming ' about this special function : f(x,y) = 1 + x + x^2 / (2!)^y + x^3 / (3!)^y + x^4 / (4!)^y + ... for complex x and real y > 0 f(x,1) is exp(x). f(x,2) can be expressed with bessel functions. ramanujan investigated f(x,3),f(x,4),f(x,5) and some others but y always > integer. ( usually for improving convergeance of series or just showing off with > his skills to produce beautiful equations ) > ever seen such f(x,y) with y non-integer in a paper ? > tommy1729 Not sure about y=R, however if y=integer, then note that F[(a1,a2,...,am),(b1,b2,...,bn),z]=sum(k=0,oo) ((a1)_k*(a2)_k....a(m)_k)/((b1)_k*(b2)_k....b(n)_k) z^k / k! where (a)_k = Gamma(a+k)/Gamma(a) is the pochammer symbol so (1)_k = k! so if y =n an integer f(x,n) = F[{},(1,1,1,...,n-1),x] === Subject: Re: insults On Dec 4, 6:36 pm, salsaonline > In a few of the popular threads, I've noticed > repeated complaints that some of the resident > experts are being overly insulting or critical when > they evaluate other people's ideas. Note that even you state that they are evaluating > IDEAS. They are not > evaluating the > person posing them. At least initially. Later on, > what mostly > happens is that > the person presenting the ideas refuses to learn > and/or listen and > simply becomes > argumentative. They are willfully ignorant. > Some times people really are being mean, and there's > no excuse for the occasional f*ck yous that >get > thrown around (I'm not accusing anyone in particular > here). > Now obviously I don't think it's right for a > teacher to insult a student. You are correct. But too many of the posters here > are not students > nor > do they have an attitude that is conducive to > learning. They just > want > a confirmation of their nonsense and remain willfully > ignorant anytime > someone tries to teach them something or suggests > that they should > study > some books. 'JSH' is one such. 'Tommy' is another. > 'Librandi' is yet > another. These people are not interested in > learning. They just want > to > prattle their 'ideas'. > more experienced people, you're going to have >to > swallow your pride every once and a while. The problem is that the cranks are NOT INTERESTED in > learning. Note also that when someone teaches a math course it > is quite > reasonable > to insist that the students have thr prerequisites. > Too many of the > people here have not even mastered high-school level > math and are > so lacking in mathematical maturity that any > discussion with them > is pointless. > Whatever you do, don't take too much offense when > you encounter the occasional insult. I think that some people go out of their way to take > offense, even > when none > is intended. And most of the cranks who get offended > simply do lack > the > prerequisites to have any intelligent discussion. I > am also sure that > they would take a simple statement such as: you lack > the > prerequisites to > discuss this subject; go read XYZ's book also as > an insult. > Let's face it, mathematicians are not known for > being great socialites. If you want to work in >this > area, you're going to have learn to let these things > slide. Remember that your job as a >student is to get > information out of people, not to get into fights > with them. > You also need an attitude of being WILLING TO LEARN. > Remember that in > most > cases the people with whom you are conversing know so > much more > mathematics > than you do that you are not even aware of how wide > the gap is. since you mention me i have to comment : so im not willing to learn ?? well recently i learned 1 = 4 from professor ullrich. and i have posted many threads were i asked for an answer : - > trying to learn new things !! also i have many open conjectures here on sci.math. so basicly its the other way around : they refuse to LEARN or think about the ideas I PRESENT. One of the best ways to get new ideas accepted is to solve old problems in new, faster ways. David Bernier > thats the truth. > the establised refuse to think about new things !! so they refuse to learn , not vice versa. ( well JSH isnt 'new' most of the time , but tetration is a good example ) > tommy1729 === Subject: Re: insults posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.4) Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > In a few of the popular threads, I've noticed repeated > complaints that some of the resident experts are being > overly insulting or critical when they evaluate other > people's ideas. Some times people really are being mean, > and there's no excuse for the occasional f*ck yous > that get thrown around (I'm not accusing anyone in > particular here). Now obviously I don't think it's right for a teacher to > insult a student. But almost any student in a doctoral > program is going to experience something like this > eventually. And, sorry to say, acting shocked and > indignant, while morally appealing, is just not going > to get you anywhere. If you want to make any progress > by learning from more experienced people, you're going > to have to swallow your pride every once and a while. The one great thing about math is that no matter how > convinced a professor is that you're wrong about > something, you can always change their opinion if you > present a coherent proof. But you do have to be > diplomatic about it. You have to say things like look, > I really want to sort this out because I really want > to understand why this works or doesn't work. When > you present a proof, you have to be crystal clear about > every step--even things you think are trivial. Trust me, > it's the only way to persuade people, who, for whatever > reason, are in doubt of your argument. Whatever you do, don't take too much offense when you > encounter the occasional insult. Let's face it, > mathematicians are not known for being great socialites. > If you want to work in this area, you're going to have > learn to let these things slide. Remember that your job > as a student is to get information out of people, not > to get into fights with them. No. That is not how human dynamics work. That is nothing like I have experienced, nor do I think that history represents it. I think it is something people wish would work, and it comes from a very good desire: decrease the hostile environment. If one person stops, isn't that better than both of them going at it? It is not always better. In my experience, it is not often better. Tolerating bullying empowers the bullying. It provides an accepted venue for unhealthy actions. It leads evaluations to change and power structures to shift, in the end benefiting the bully. Hell, that's such a universally common experience it's U. S. foreign policy. It would not be in my advantage to let Ullrich tell others that I am wrong about something I am not, or that I have illustrated I know very little about a field I have produced results in. I have seen this intellectual cowardice before in many guises. Cowardice causes those who might defend to stay quiet. Cowardice avoids fixing problems. Cowardice lets disease grow. The insult is that I had to stand up to David myself, because none of you had the balls to defend against this old careless bully. But you did have the balls to protest the defense. What type of balls is them? -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: insults Am 08.12.2008 01:52 schrieb galathaea: > What type of balls is them? > Does this question qualify someone as ballerina? === Subject: Re: insults > In a few of the popular threads, I've noticed repeated > complaints that some of the resident experts are being > overly insulting or critical when they evaluate other > people's ideas. Some times people really are being mean, > and there's no excuse for the occasional f*ck yous > that get thrown around (I'm not accusing anyone in > particular here). Now obviously I don't think it's right for a teacher to > insult a student. But almost any student in a doctoral > program is going to experience something like this > eventually. And, sorry to say, acting shocked and > indignant, while morally appealing, is just not going > to get you anywhere. If you want to make any progress > by learning from more experienced people, you're going > to have to swallow your pride every once and a while. The one great thing about math is that no matter how > convinced a professor is that you're wrong about > something, you can always change their opinion if you > present a coherent proof. But you do have to be > diplomatic about it. You have to say things like look, > I really want to sort this out because I really want > to understand why this works or doesn't work. When > you present a proof, you have to be crystal clear about > every step--even things you think are trivial. Trust me, > it's the only way to persuade people, who, for whatever > reason, are in doubt of your argument. Whatever you do, don't take too much offense when you > encounter the occasional insult. Let's face it, > mathematicians are not known for being great socialites. > If you want to work in this area, you're going to have > learn to let these things slide. Remember that your job > as a student is to get information out of people, not > to get into fights with them. No. That is not how human dynamics work. That is nothing like > I have experienced, nor do I think that history represents > it. I think it is something people wish would work, and it > comes from a very good desire: decrease the hostile > environment. If one person stops, isn't that better than both of them > going at it? It is not always better. In my experience, it is not often better. Tolerating bullying empowers the bullying. It provides > an accepted venue for unhealthy actions. It leads > evaluations to change and power structures to shift, in > the end benefiting the bully. Hell, that's such a > universally common experience it's U. S. foreign policy. It would not be in my advantage to let Ullrich tell > others that I am wrong about something I am not, or > that I have illustrated I know very little about a > field I have produced results in. I have seen this intellectual cowardice before in many > guises. Cowardice causes those who might defend to stay quiet. Cowardice avoids fixing problems. Cowardice lets disease grow. The insult is that I had to stand up to David myself, > because none of you had the balls to defend against > this old careless bully. But you did have the balls > to protest the defense. What type of balls is them? -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar Personally, and I expect I speak for many, I think that this forum will better serve it's stated purpose IF YOU DON'T POST TO IT AT ALL. Go find somewhere else to play. Maybe do a newsgroup search for some of the keywords in your signature: > -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar === Subject: Re: insults posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.4) Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > Personally, and I expect I speak for many, I think that this forum will > better serve it's stated purpose IF YOU DON'T POST TO IT AT ALL. Go > find somewhere else to play. Maybe do a newsgroup search for some of > the keywords in your signature: > -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > > galathaea: prankster, fablist, magician, liar hello sock puppet! i know it must be difficult unable to muster the bravery to post your own ideas frustrated impotent . that's okay because anyone can start the path to self improvement given only some clarity of goal and a little guidance on how to pursue them i know this can be an anxiety-ridden process people will be watching you but let me show you what it's like i think if you see how healthy mathematics may be pursued outside that hazing of the alpha framework maybe just maybe after a good cathartic cry you can find a little courage of your own... -+-+- so i've already shown how the matrices |0 1 0 ... 0| |0 0 1 ... 0| |0 0 0 ... 0| W = |. . . . .| n |. . . . .| |. . . . .| |0 0 0 ... 1| |1 0 0 ... 0| are important to the theory of multisection and how W x n e is equal to | |n-1 x |0 x |1 x |n-2 x | | |n e |n e |n e ... |n e | | | | |n-2 x |n-1 x |0 x |n-3 x | | |n e |n e |n e ... |n e | | ... | so people like me who unlike you dear sock puppet don't speak for many might in their extra time think about this some more and seeing as how the topic has been matrix exponentiation (at least among those not impotent of contribution) let me write down the next obvious step W n M for M an n x n invertible square matrix what does it mean? well W is a n-th root of I n for the multisection theorem n one of the key steps is to use W = I n to break apart the sum into n multisected pieces but writing the above in series form gives oo --- 1 / j / ---- | W ln(M) | --- (1) n / j=0 j so evaluation needs to proceed more carefully to account noncommutativity so the first step that should come to mind for the aspiring mathemagician should be to look at the commutator [W , M] = W M - M W n n n what are these two terms here? what do they do? multiplying by W on the left n cyclically shifts all the elements of M up one row we can call this operator up(M) multiplying by W on the right n cyclically shifts all elements out to the right one column we can call this operator out(M) so now that we've got some preliminary definitions let's see if we can come up with some properties of these things look at W M W n n it should be clear that we can do the operation on the inner matrix in any order so immediately we get up(out(M)) = out(up(M)) and the corresponding operators are commutative so if we have many of these applied to M we can simplify our notation by looking at the operator (j, k)_(M) which applies j ups and k outs (in any order) then look at M W M n this gives immediately M up(M) = out(M) M already we're accumulating a number of identities but they seem pretty minor maybe we should try to accelerate things a bit so there's already plenty of information to show when the commutator is 0 it's simple any matrix M with up(M) = out(M) commutes with W n | m1 m2 m3 ... mn | | mn m1 m2 ... m_(n-1) | | ... | | m2 m3 m4 ... m1 | however there are a lot of additional equivalences to explore continuing the above we can look at m M W M n then if j + k = m = j' + k' (j) (k) (j') (k') out (M) up (M) = out (M) up (M) and if j + k = n then (j) (k) 2 out (M) up (M) = M and we can continue like this for 3, 4, and more Ms however there is a nice way to express the equivalences in terms of the (j, k)_(M) twist operator mentioned above multiply two of these matrices together (j, k)_(M) (j', k')_(M) = j k j' k' W M W W M W = n n n n (j, k + p)_(M) (j' - p, k')_(M) which is something we can actually apply back on the series the terms we are looking to compute are j ((1, 0)_(M)) which we can now evaluate in many ways if we do have one of the up(M) = out(M) matrices above for the logarithmic entry in the series we get the multisection theorem n-1 --- j / |j (.) = / W | | e | (ln M) --- n |n / j=0 using my generalised trigonometric (exercise for furthur study and to ensure following: when does ln(M) have the out=up form?) but even if that isn't the case the relations might be used along with special relations of the matrix to do simplifications .. you see poor sock puppet if there comes a time in the coldness of your night you begin to realise that you never really liked math that you had fooled yourself for many years but only really ever worked against it if those nagging voices that you so valiantly subdue ever creep upon you unsuspecting revealing to you the endgame you have worked for don't worry there is redemption it starts with a blank sheet of paper and the desire to find something new out there are no guarantees of success but there are many successes to be had sure you won't ever fully be able to forget that your past tactics only dripped hatred on the machinery of math and sure you still will be reminded at times unsuspecting that it's only end was to halt perfectly legitimate mathematics even still the blank paper can be a little strength to help the coward make it through another impotent night -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: insults even still > the blank paper can be a little strength > to help the coward make it through another impotent night But how can this work? You're ostensibly protesting against alpha ball-busting behavior, but you're doing it by busting balls. That doesn't seem right. === Subject: Re: insults posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e 2.0.50727),gzip(gfe),gzip(gfe) .......................................... > you see > poor sock puppet > if there comes a time > in the coldness of your night > you begin to realise that you never really liked math > that you had fooled yourself for many years > but only > really > ever worked against it > if those nagging voices > that you so valiantly subdue > ever creep upon you unsuspecting > revealing to you the endgame you have worked for > don't worry there is redemption it starts with a blank sheet of paper > and the desire to find something new out there are no guarantees of success > but there are many successes to be had sure > you won't ever fully be able > to forget that your past tactics > only dripped hatred on the machinery of math > and sure > you still will be reminded > at times unsuspecting > that it's only end was to halt perfectly legitimate mathematics even still > the blank paper can be a little strength > to help the coward make it through another impotent night -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar *********************************************************** What a load of bull. Interesting though that you don't address the most important things here, suck poopet: that you began the insults, the attacks and the foul language. Now you cry cry ................cry *sob* *sob* *sob* .....................and whine a lot. By the way, after you explained it FINALLY, that multisection thing looks like something that might be interesting. Tonio === Subject: Re: insults posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.4) Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > .......................................... you see > poor sock puppet > if there comes a time > in the coldness of your night > you begin to realise that you never really liked math > that you had fooled yourself for many years > but only > really > ever worked against it > if those nagging voices > that you so valiantly subdue > ever creep upon you unsuspecting > revealing to you the endgame you have worked for > don't worry there is redemption it starts with a blank sheet of paper > and the desire to find something new out there are no guarantees of success > but there are many successes to be had sure > you won't ever fully be able > to forget that your past tactics > only dripped hatred on the machinery of math > and sure > you still will be reminded > at times unsuspecting > that it's only end was to halt perfectly legitimate mathematics even still > the blank paper can be a little strength > to help the coward make it through another impotent night -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar *********************************************************** What > a > load > of > bull. Interesting though that you don't address the most important things > here, suck poopet: that you began the insults, the attacks and the > foul language. Now you cry > cry > ................cry *sob* > *sob* > *sob* .....................and whine a lot. it's interesting watching you rationalise both your own and others' behavior you toni(c)o one of the more active whiners and bullies really have convinced yourself that i started the attacks you allow yourself to ignore all of david's prior attacks (and more strangely all of your prior attacks) and convince yourself so fully of this alien view that i suspect you actually feel indignant to me you don't like math toni(c)o you are regularly on the wrong side the one that discourages creative thinking which is why you don't display creative thinking on the groups just repetitive toady action > By the way, after you explained it FINALLY, that multisection thing > looks like something that might be interesting. i've been explaining it for years that finally you feel is another product of your mind's distortions it's the distant fingers of experience intruding in upon this worldview of yours but don't worry despite many having tried to excuse david's behavior phrasing it as he doesn't suffer fools lightly and a number of past posts of his explored and despite the fact that he claimed not the first time that i was wrong in my post and despite his continued use of masculist belittlement despite all that you can probably persist in deluding yourself if you try really hard turn all of your focus away from anything david's ever done and turn all of your focus on the vulgarity in my posts if you can do that and keep your eyes off the logical apparatus i inserted (those pesky ifs) you can probably maintain your indignation but you will need to stay away from these posts where i actually contribute to the theory it might cause you to realise that david's one contribution trivial as it may have been was couched in an attempt to justify a claim of his that i don't really know much about this and it might make you realise that despite the large count of words in these threads posted by yourself not a single one has contributed to the math those are the types of thoughts you need to avoid if you are going to maintain your indignation because that nagging will creep in the nagging voice that points out you don't really like math toni(c)o and stay away from sci.physics and similar groups where the end result of the bullying is on display you won't ever get rid of wrong theories no matter how hard you scream at them you won't even get rid of your own wrong theories no matter how hard you hide them from the light exploration means not being afraid to be a little foolish sometimes but you may get rid of those who do contribute because the bullying is not just sometimes wrong against valid mathematickising it is often wrong like this time (look away) and it is the bullies that killed sci.physics not the cranks -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: insults posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e 2.0.50727),gzip(gfe),gzip(gfe) .......................................... you see > poor sock puppet > if there comes a time > in the coldness of your night > you begin to realise that you never really liked math > that you had fooled yourself for many years > but only > really > ever worked against it > if those nagging voices > that you so valiantly subdue > ever creep upon you unsuspecting > revealing to you the endgame you have worked for > don't worry there is redemption it starts with a blank sheet of paper > and the desire to find something new out there are no guarantees of success > but there are many successes to be had sure > you won't ever fully be able > to forget that your past tactics > only dripped hatred on the machinery of math > and sure > you still will be reminded > at times unsuspecting > that it's only end was to halt perfectly legitimate mathematics even still > the blank paper can be a little strength > to help the coward make it through another impotent night -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar *********************************************************** What > a > load > of > bull. Interesting though that you don't address the most important things > here, suck poopet: that you began the insults, the attacks and the > foul language. Now you cry > cry > ................cry *sob* > *sob* > *sob* .....................and whine a lot. it's interesting watching you rationalise > both your own and others' behavior you > toni(c)o > one of the more active whiners and bullies > really have convinced yourself > that i started the attacks > ************************************************************ I posted in this very thread a resume of the first psots by David and you in that infamous thread of commuting matrices A,B ... and stuff. I challenge you to point, either in that post of mine or how you like, where I am wrong in my claim that it was YOU the one that started the direct attacks, first against David, then against other, until you reached the peek of ...you ers... and stuff. Common, you've challenged people about some rather weird things just a few days ago, now let's see you meeting this challenge. ********************************************************** > you allow yourself to ignore > all of david's prior attacks > (and more strangely all of your prior attacks) > and convince yourself so fully of this alien view > that i suspect you actually feel indignant to me you don't like math > toni(c)o you are regularly on the wrong side > ************************************************************ Of course. Let me guess: the right side...is YOUR side, uh? Talking of fundamentalism.... *********************************************************** > the one that discourages creative thinking which is why you don't display creative thinking on the groups just repetitive toady action By the way, after you explained it FINALLY, that multisection thing > looks like something that might be interesting. i've been explaining it for years that > finally > you feel > is another product of your mind's distortions it's the distant fingers of experience > intruding in upon this worldview of yours but don't worry despite > many having tried to excuse david's behavior > phrasing it as he doesn't suffer fools lightly > and a number of past posts of his explored > and despite the fact that he claimed > not the first time > that i was wrong in my post > and despite his continued use of masculist belittlement > ****************************************************** Oh, yes: I already noticed in the other thread this rather interesting pearl of superiority complex you suffer, perhaps due to a prior inferiority conplex, from wrt to men...because I assume that word I can't find in my dictionary nor in Merriam-Webster, masculist, is meant to be masculine...? Unless you meant emasculate, which is the same as castrate, and this then would be weird being you a woman... ************************************************** > despite all that > you can probably persist in deluding yourself if you try really hard > turn all of your focus away from anything david's ever done > and turn all of your focus on the vulgarity in my posts > if you can do that > and keep your eyes off the logical apparatus i inserted > (those pesky ifs) > you can probably maintain your indignation but you will need to stay away from these posts > where i actually contribute to the theory it might cause you to realise > that david's one contribution > trivial as it may have been > was couched in an attempt to justify a claim of his > that i don't really know much about this and it might make you realise that > despite the large count of words in these threads > posted by yourself > not a single one has contributed to the math > ******************************************************* Of course, dear: the only one who's ever contributed anything here to anything is you. We all heard that already. ******************************************************* > those are the types of thoughts > you need to avoid > if you are going to maintain your indignation > because that nagging will creep in the nagging voice that points out > you don't really like math toni(c)o and stay away from sci.physics and similar groups > where the end result of the bullying is on display you won't ever get rid of wrong theories > no matter how hard you scream at them you won't even get rid of your own wrong theories > no matter how hard you hide them from the light exploration means not being afraid > to be a little foolish sometimes but you may get rid of those who do contribute because the bullying is not just sometimes wrong > against valid mathematickising it is often wrong like this time (look away) and it is the bullies that killed sci.physics > not the cranks > **************************************************** Ok, nice piece of trashing you threw out above, but let's forget that for the time being: prove me wrong, as already stated above, that you were the one who started the personal, direct attacks agaisnt David in that thread. After that we'll be able, perhaps, to deal with all the rest. Tonio -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar- Hide quoted text - - Show quoted text - === Subject: Re: insults > .......................................... you see > poor sock puppet >if there comes a time > in the coldness of your night > you begin to realise that you never really liked math > that you had fooled yourself for many years > but only > really > ever worked against it >if those nagging voices > that you so valiantly subdue >ever creep upon you unsuspecting >revealing to you the endgame you have worked for > don't worry there is redemption it starts with a blank sheet of paper > and the desire to find something new out there are no guarantees of success > but there are many successes to be had sure > you won't ever fully be able > to forget that your past tactics > only dripped hatred on the machinery of math >and sure > you still will be reminded > at times unsuspecting > that it's only end was to halt perfectly legitimate mathematics even still > the blank paper can be a little strength >to help the coward make it through another impotent night -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- >galathaea: prankster, fablist, magician, liar > *********************************************************** What > a > load > of > bull. Interesting though that you don't address the most important things > here, suck poopet: that you began the insults, the attacks and the > foul language. Now you cry > cry > ................cry *sob* > *sob* > *sob* .....................and whine a lot. > By the way, after you explained it FINALLY, that multisection thing > looks like something that might be interesting. Tonio Hey Tonio, where is your first formula ever, published in this group ? Han de Bruijn === Subject: Re: insults posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e 2.0.50727),gzip(gfe),gzip(gfe) > .......................................... >you see > poor sock puppet >if there comes a time > in the coldness of your night > you begin to realise that you never really liked math > that you had fooled yourself for many years > but only > really > ever worked against it >if those nagging voices > that you so valiantly subdue >ever creep upon you unsuspecting >revealing to you the endgame you have worked for > don't worry >there is redemption >it starts with a blank sheet of paper > and the desire to find something new out >there are no guarantees of success > but there are many successes to be had >sure > you won't ever fully be able > to forget that your past tactics > only dripped hatred on the machinery of math >and sure > you still will be reminded > at times unsuspecting > that it's only end was to halt > perfectly legitimate mathematics >even still > the blank paper can be a little strength >to help the coward make it through another impotent night >-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- >galathaea: prankster, fablist, magician, liar *********************************************************** What > a > load > of > bull. Interesting though that you don't address the most important things > here, suck poopet: that you began the insults, the attacks and the > foul language. Now you cry > cry > ................cry *sob* > *sob* > *sob* .....................and whine a lot. By the way, after you explained it FINALLY, that multisection thing > looks like something that might be interesting. Tonio Hey Tonio, where is your first formula ever, published in this group ? Han de Bruijn- *********************************************************** Hey Han, you'll have to look for it, just as you did NOT do when I showed where EXACTLy is my thesis, remember? But don't worry, Han: it's not like you could have understood anything of it Tonio Ps And in case you missed it Han, as many other times in the past, Swoooosh! === Subject: Re: insults .......................................... you see > poor sock puppet >if there comes a time > in the coldness of your night > you begin to realise that you never really liked math > that you had fooled yourself for many years > but only > really > ever worked against it >if those nagging voices > that you so valiantly subdue >ever creep upon you unsuspecting >revealing to you the endgame you have worked for > don't worry there is redemption it starts with a blank sheet of paper > and the desire to find something new out there are no guarantees of success > but there are many successes to be had sure > you won't ever fully be able > to forget that your past tactics > only dripped hatred on the machinery of math >and sure > you still will be reminded > at times unsuspecting > that it's only end was to halt perfectly legitimate mathematics even still > the blank paper can be a little strength >to help the coward make it through another impotent night -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- >galathaea: prankster, fablist, magician, liar >*********************************************************** >What > a > load > of > bull. >Interesting though that you don't address the most important things >here, suck poopet: that you began the insults, the attacks and the >foul language. >Now you cry > cry >................cry >*sob* > *sob* > *sob* >.....................and whine a lot. >By the way, after you explained it FINALLY, that multisection thing >looks like something that might be interesting. >Tonio Hey Tonio, where is your first formula ever, published in this group ? Han de Bruijn- *********************************************************** Hey Han, you'll have to look for it, Can't find it, Tonio, can't find it .. > just as you did NOT do when I > showed where EXACTLy is my thesis, remember? > But don't worry, Han: it's not like you could have understood anything > of it Tonio Ps And in case you missed it Han, as many other times in the past, > Swoooosh! Han de Bruijn === Subject: Re: insults posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e 2.0.50727),gzip(gfe),gzip(gfe) >.......................................... >you see > poor sock puppet >if there comes a time > in the coldness of your night > you begin to realise that you never really liked math > that you had fooled yourself for many years > but only > really > ever worked against it >if those nagging voices > that you so valiantly subdue >ever creep upon you unsuspecting >revealing to you the endgame you have worked for > don't worry >there is redemption >it starts with a blank sheet of paper > and the desire to find something new out >there are no guarantees of success > but there are many successes to be had >sure > you won't ever fully be able > to forget that your past tactics > only dripped hatred on the machinery of math >and sure > you still will be reminded > at times unsuspecting > that it's only end was to halt > perfectly legitimate mathematics >even still > the blank paper can be a little strength >to help the coward make it through another impotent night >-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- >galathaea: prankster, fablist, magician, liar *********************************************************** What > a > load > of > bull. Interesting though that you don't address the most important things >here, suck poopet: that you began the insults, the attacks and the >foul language. Now you cry > cry >................cry *sob* > *sob* > *sob* .....................and whine a lot. By the way, after you explained it FINALLY, that multisection thing >looks like something that might be interesting. Tonio >Hey Tonio, where is your first formula ever, published in this group ? >Han de Bruijn- *********************************************************** Hey Han, you'll have to look for it, Can't find it, Tonio, can't find it .. > *************************************************** Of course you can't, Han...what else is new! Tonio === Subject: Re: insults posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.2; .NET CLR 3.5.21022; Tablet PC 2.0),gzip(gfe),gzip(gfe) ............................................................ > Whatever you do, don't take too much offense when you > encounter the occasional insult. Let's face it, > mathematicians are not known for being great socialites. > If you want to work in this area, you're going to have > learn to let these things slide. Remember that your job > as a student is to get information out of people, not > to get into fights with them. No. That is not how human dynamics work. That is nothing like > I have experienced, nor do I think that history represents > it. I think it is something people wish would work, and it > comes from a very good desire: decrease the hostile > environment. If one person stops, isn't that better than both of them > going at it? It is not always better. In my experience, it is not often better. Tolerating bullying empowers the bullying. It provides > an accepted venue for unhealthy actions. It leads > evaluations to change and power structures to shift, in > the end benefiting the bully. Hell, that's such a > universally common experience it's U. S. foreign policy. It would not be in my advantage to let Ullrich tell > others that I am wrong about something I am not, or > that I have illustrated I know very little about a > field I have produced results in. I have seen this intellectual cowardice before in many > guises. Cowardice causes those who might defend to stay quiet. Cowardice avoids fixing problems. Cowardice lets disease grow. The insult is that I had to stand up to David myself, > because none of you had the balls to defend against > this old careless bully. But you did have the balls > to protest the defense. What type of balls is them? -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar- **************************************************************************** ************* First of all, it's a refreshing thing to see Galathea writing as most people have agreed on: it makes her posts more easily readable. Second, and for the sake of justice and fairness, since I've also been a participant in all this, please do allow me to make a resume of the whole thing. I am referring to the thread originally called When do matrices B and C commute?, by Alainverghote, from last Nov. 19. Anyone interested in following the flow of the next stuff should open this thread in the Drexel math forum (discussions > sci.math): http://mathforum.org/kb/forum.jspa?forumID=13 Please do pay attention to the fact that I usually included in each quoted post the very first lines in the original one which belong to the post been asnwered. ** The first post by Galathea was Nov. 20, at 2:22 pm, and it's the following: > So tell me: What in the world > do you think that A^C is, if A and C are 2x2 matrices. > What is A to the power C? wouldn't that just be exp(C ln(A))? both exp and ln may be defined on square matrices using the power series (though there may be convergence to consider) Galathea's post begins at wouldn't..., and she's answering a question posted by David to Tommy's proposition to use A^C, with A,C square matrices, for whatever. So far David only asked Tommy what did he mean by A^C, and Galathea stepped in to answer David's question to Tommy. So far nothing to bizarre or extrange. ** To the above, Amy666 (One of the alter-egos of Tommy1729) replied at 5:04 pm, Nov.20: > What is A to the power C? wouldn't that just be exp(C ln(A))? yep thats it. and before objections are made , i said det =/= [0;1] both exp and ln > may be defined on square matrices > using the power series > (though there may be convergence to consider) perhaps i should try to sell a book too , like david. yeah , ill sell david ullrich a book about subjects like : continu iterations ( gamma , Barnes G , tetration , moebius transforms etc ) matrix powers ( A^C ) polysigned ( with credit to timothy ) large cardinal axioms ( it was strange nobody noticed these missing in the ' disproofs ' of TST and aleph aleph 0 is max , they assumed the large cardinals could be proven to exist just with ZF ... ) Here, tommy is trying, not very succesfully imfho, to laugh at David and his ignorance in subjects like iterations, matrix powers, polysigned [sic], etc. Anyone knowing David, and specially knowing Tommy, will probably giggle or even open laugh outloud. Nothing new, Tommy can't stand David and he's just making clear this point. ** To the above, David posts on Nov. 21, at 7:47 am: > wouldn't that just be exp(C ln(A))? yep thats it. That's funny - you need someone else to say what you meant. Why is it that and not, say, exp(ln(A) C)? And exactly what do you mean by ln(A)? I didn't bother asking galathea, since this is your amazing discovery, after all, but he was wrong in saying that ln(A) can be defined by using a power series (except under certain conditions. Can you tell us when ln(A) can be defined using a power series, and can you tell us what the definition of ln(A) is for A not satisfying that condition? If you do stumble on the definition you'll find that there are many different choices for ln(A); which one did you mean?) David makes it clear that he's realized that not Tommy but Galathea defined A^C, and even now he makes some trying to clarify stuff. Nothing really outstanding so far, imo, some little stings here and there, but now things begin to warm up and beyond. ** The next is Galathea's retort on Nov. 22, 12:17 am to David's claim that she commited a mistake. For the sake of clearness and fariness I shall copy the whole thing: [David] If you do > stumble on the definition you'll find that there > are many different choices for ln(A); which one > did you mean?) if you wanted a 1-on-1 conversation with tommy email would be more appropriate than usenet i think you have an error in your post the correct thing to say is: galathaea was right in saying that ln(A) could be defined using a power series and even pointed out that you may need to consider convergence depending on the particular matrix of course a lot of this begs some questions if you don't know about matrix exponentiation then it indicates you probably haven't read any intro texts on lie algebras (which almost all contain a good mention including campbell-baker-hausdorff and the geometric and arithmetic meanings) and if you haven't read intro lie algebra texts then your new book probably doesn't go into solutions of differential equations and the associated germs and such that's okay complex analysis often waits to explore those issues ( many don't want to have to build the whole fibration theory foundation ) but it means your coverage of monodromy is bound to be pretty superficial generating trajectories won't be given a clear differential foundation that's fine usually they call that complex differential topology or something similar i know not quite the same target audience oh and i see that you are now playing off my alleged error as a challenge to tommy which you think will hide your initial challenge in a well-i-had-a-more-subtle-point move if i was more bad hearted i might point out domination patterns and connect them to capabilities as a teacher in ways that might subvert the intellectual authority of your book but don't worry i wouldn't actually support any of those horribly offensive positions and i hope you don't also just assume i'm horribly wrong particularly when it is an area i've pursued in some detail (specifically in my generalised trigonometry where the operation generalises what happens on the 2d rotation matrix) and maybe along the way you can stop acting like a little child with tommy and start acting like a teacher i hear that's what you do (you've been corrected before about another error you seem to like to continue with) -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar This time Galathea already says stuff to David that I cannot but take as a direct and rather not-nice personal attack against him: from stop acting like a child through start acting as a teacher (and one could add and don't you dare contradict me!!! to make it sound as my sometimes frightening grandma) and all the way until domination patterns and capabilites as a teacher. I can hardly see how all this was called for here, but perhaps I don't know some issues these two had between them in the past. ** On nov 22, at 9:05 am David retorts back (only excerpts): ............................................................ the correct thing to say is: > galathaea was right in saying > that ln(A) could be defined using a power series Really? Exactly how does that go. ............................................................ >if you don't know about matrix exponentiation Why would you conclude that? Of course there's no problem defining the exponential of a matrix; I haven't said anything to the contrary. Now explain exactly how I use power series to define ln(A). ............................................................ but don't worry >i wouldn't actually support any of those horribly offensive positions > and i hope you don't also just assume i'm horribly wrong > particularly when it is an area i've pursued in some detail Fabulous. Then explaining how to use power series to define ln(A) should be no problem. ............................................................ I find it very interesting to note that David did NOT address any of the personal attacks of Galathea, and only asked rather specific questions, even if using some very, very light irony. Perhaps this is due to the fact that those alleged attacks weren't as nasty as I see them, perhaps because David just didn't want to be nasty against Galathea for whatever reason. ** Galathea retors back on Nov. 22, at 2:54 pm: Why would you conclude that? Of course there's > no problem defining the exponential of a matrix; > I haven't said anything to the contrary. well someone who wants to make fun of another for writing A^C for A and C matrices doesn't display an understanding of matrix exponentiation ........................................................ > Now explain exactly how I use power series to > define ln(A). i don't think you do ........................................................ ...but now once again if this is so hard for you to see how do you cover series definitions and analytic continuation in your book? do you do it in a way that doesn't make sense for matrices? i'm saying all this about your book because you still haven't taken back your error about my position it's a silly stand but i love to properly corrected and hate to be falsely corrected particularly by someone already on teasing kick so i'm just pointing out if you really can't see any of it it is not saying great things about your book Galathea strikes back and forth, and all is apparently because she feels she's been wrongly corrected on something she was right. It can indeed be annoying to be corrected in something you think/know you're right, but so far I can't see a single attack on Galathea. The above correspondence continues on and on, and I think it is a good example of miscomunications mixed with that rare blend of talents and characteristics Galathea has been blessed with: lots of bad faith, stubborness, hipocresy, intellectual dishonestity, narcissism and self- righteousness (she actually calls defence to her verbal attacks and gross abuse against David and many other chatters, while using very ugly and despective adjectives ), mixed with irony and some sarcastic remarks by David and by other chatters as well, who mingled in the convo. On Nov. 25, at 10:15 am Galathea uses profanity directly against individual...but then you can read the whole thread if you will. Tonio === Subject: Re: insults posting-account=tCEoyAoAAAAkltU5zxOoI8uJ4lyz5-kv .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022; Tablet PC 2.0),gzip(gfe),gzip(gfe) In a few of the popular threads, I've noticed repeated > complaints that some of the resident experts are being > overly insulting or critical when they evaluate other > people's ideas. Some times people really are being mean, > and there's no excuse for the occasional f*ck yous > that get thrown around (I'm not accusing anyone in > particular here). Now obviously I don't think it's right for a teacher to > insult a student. But almost any student in a doctoral > program is going to experience something like this > eventually. And, sorry to say, acting shocked and > indignant, while morally appealing, is just not going > to get you anywhere. If you want to make any progress > by learning from more experienced people, you're going > to have to swallow your pride every once and a while. The one great thing about math is that no matter how > convinced a professor is that you're wrong about > something, you can always change their opinion if you > present a coherent proof. But you do have to be > diplomatic about it. You have to say things like look, > I really want to sort this out because I really want > to understand why this works or doesn't work. When > you present a proof, you have to be crystal clear about > every step--even things you think are trivial. Trust me, > it's the only way to persuade people, who, for whatever > reason, are in doubt of your argument. Whatever you do, don't take too much offense when you > encounter the occasional insult. Let's face it, > mathematicians are not known for being great socialites. > If you want to work in this area, you're going to have > learn to let these things slide. Remember that your job > as a student is to get information out of people, not > to get into fights with them. A thick skin can be a very good thing. It took me a long time to develop mine, and I think that most people in math and other core sciences have such a keen sense of academic sesibility that they become unable to really understand what being thick skinned is all about. It is a terrible paradox. Because on the one hand you benefit tremendously from having a purely academic mind, and to such a mind it is unconscienable that people would be rude toward one another. On the other hand, engaging in a good fight really gets the adrenaline flowing and when two people are really slugging it out with ideas - that can be a very productive process as well. Just think about how much mathematics and technology was created as a result of people trying to kill each other. Being able to switch gears appropriately between congenialisty and conversational war - that is an art which should be required of every educated person. If both parties are willing to step into the boxing arena I say go for it. But it is probably inappropriate to suckerpunch someone who is not expecting it. That is not fair. === Subject: Re: Simply ordered set and Hausdorff space <20081206035913.F78685@agora.rdrop.com> posting-account=LdgEVwoAAADzlB0-fAT_xkiYXMrEZ-Zc Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > order topology) The open sets in X are the sets that are a union of (possibly > infinitely many) such open intervals and rays. The order topology makes X into a completely normal Hausdorff space. Not from Wikipedia, but I conclude the order topology on a finite set is > the same as the discrete topology. Simple result since linear order topologies are Hausdorff and finite > Hausdorff spaces are discrete. Exercise. Give an example of a discrete linear order topology > with arbitrarily large cardinality. ---- === Subject: any fast algorithm of calculating magnitude of a given vector? posting-account=WOMbEwoAAABSmbQhJFsLtz_F527uyH75 Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) i am trying to implement a fast algorithm of calculating the magnitude of a vector in FPGA, currently the method in hand is the alpha MAX + beta MIN way, by making the alpha and beta be reciporal power of 2s , i.e sum(1/2)^p, the hardware can be simplified to SHIFT(divide by 2) and ADD, but the accuracy is not ideal. is there any method of calculating or estimating the magnitude of a + jb ? or simply calculating the c = sqrt (a^2 + b^2); ? I want to use this module in the high speed end of A/D and digital down mux where the data is complex numbers been represented as I + jQ. So the speed and accuracy is both important. savor 20081209 === Subject: Re: any fast algorithm of calculating magnitude of a given vector? i am trying to implement a fast algorithm of calculating the magnitude > of a vector in FPGA, currently the method in hand is the alpha MAX + > beta MIN way, by making the alpha and beta be reciporal power of > 2s , i.e sum(1/2)^p, the hardware can be simplified to SHIFT(divide > by 2) and ADD, but the accuracy is not ideal. is there any method of calculating or estimating the magnitude of a + > jb ? or simply calculating the c = sqrt (a^2 + b^2); ? I want to use > this module in the high speed end of A/D and digital down mux where > the data is complex numbers been represented as I + jQ. So the speed > and accuracy is both important. If A and B are positive, A>B, you can multiply B by 2 (while B 6 (or 2B = 6) 6+6 =12 >A then A - 2B = 5, or A = 2B + 5 , 5>B, subtract another B: A - 3B = 2 < B Let C = A - 3B = 2 < B Start again with B, C B>C but 2C>B B - C = 1 < C, so B = C + 1, let D = 1 posting-account=WOMbEwoAAABSmbQhJFsLtz_F527uyH75 Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) It is a quick method used to calculate magnitude of a vector in DSP: http://en.wikipedia.org/wiki/Alpha_max_plus_beta_min_algorithm and : quote: > The approximation is expressed as: |V| = alpha,! mathbf{Max} + beta,! mathbf{Min} Where mathbf{Max} is the maximum absolute value of I and Q and mathbf {Min} is the minimum absolute value of I and Q. For the closest approximation, the optimum values for alpha,! and beta,! are alpha_0 = frac{2 cos frac{pi}{8}}{1 + cos frac {pi}{8}} = 0.96043387... and beta_0 = frac{2 sin frac{pi}{8}}{1 + cos frac{pi}{8}} = 0.39782473..., giving a maximum error of 3.96%. <<<<<<<<<<<<<<<<<< so with a=0.96 and b=0.39, the MAX error could be attained within 4%, a much appealing result but require two extra multiplexers in FPGA inorder to calculate a*MAX + b*MIN, which is critical in resource limited FPGA chips. BTW : i want to use the magnitude of fourier transform result I+jQ to estimate the peak of the signal. others method include : CORDIC and BKM algorithms which is basiclly a SHIFT-ADD method of exponential function approximation, which using pipeline to achieve high accuracy using extra clock cycles. The magnitude approximation question in sci.math(which is simply a EE problem) is to consult whether there exits a more efficient method, or point out the proper way on how to find such ? === Subject: Use cases for quiet NaN comparison? posting-account=_N1kkwoAAACM8Tv1UJHiaF7Tu29T3ZYV Gecko/2008111319 Ubuntu/8.10 (intrepid) Firefox/3.0.4,gzip(gfe),gzip(gfe) I'm approaching this from a language design perspective (my formal math knowledge is not as good as it should be). I'm trying to understand what uses cases there are (if any?) for NaN comparisons always returning false. The normal number comparison use cases I'm aware of assume a total order is present, so (for example) sorting a NaN just produces garbage, rather than a meaningful unordered. The alternative would be treating it as a dummy value, with a total order, probably sorting right by one of the infinities. I'm aware of the behaviour IEEE 754 requires (as best I can be, without having my own copy of it.) However, I'm not aware of any rationale for the behaviour. It seems to be a Parable of the Monkeys. Can anybody enlighten me? === Subject: Re: Use cases for quiet NaN comparison? Nntp-Posting-Host: hera.cwi.nl ... > I'm aware of the behaviour IEEE 754 requires (as best I can be, > without having my own copy of it.) However, I'm not aware of any > rationale for the behaviour. It seems to be a Parable of the Monkeys. > > Can anybody enlighten me? Why do you think it makes sense to compare something that is not a number with a number? Anyway, the appearance of a NaN in a calculation shows that something has gone fundamentally wrong in a calculation, and the NaN is an out of band signal to indicate this. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Use cases for quiet NaN comparison? posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG 1.1.4322; .NET CLR 2.0.50727; MS-RTC LM 8; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > I'm aware of the behaviour IEEE 754 requires (as best I can be, > without having my own copy of it.) However, I'm not aware of any > rationale for the behaviour. It seems to be a Parable of the Monkeys. Can anybody enlighten me? Why do you think it makes sense to compare something that is not a number > with a number? Anyway, the appearance of a NaN in a calculation shows > that something has gone fundamentally wrong in a calculation, and the NaN > is an out of band signal to indicate this. Exactly. Once you get a NaN, a non-number, then you no longer have a valid numeric value. So arithmetic and comparison operations make no sense, and all you can get from further operations is non-numeric values. It's not a number, so it doesn't compare to any valid number, nor to any other non-number. Or perhaps you are asking why NaN operations are silent instead of triggering exceptions? -drt === Subject: Re: Use cases for quiet NaN comparison? posting-account=_N1kkwoAAACM8Tv1UJHiaF7Tu29T3ZYV Gecko/2008111319 Ubuntu/8.10 (intrepid) Firefox/3.0.4,gzip(gfe),gzip(gfe) > I'm aware of the behaviour IEEE 754 requires (as best I can be, > without having my own copy of it.) However, I'm not aware of any > rationale for the behaviour. It seems to be a Parable of the Monkeys. > Can anybody enlighten me? Why do you think it makes sense to compare something that is not a number > with a number? Anyway, the appearance of a NaN in a calculation shows > that something has gone fundamentally wrong in a calculation, and the NaN > is an out of band signal to indicate this. Exactly. Once you get a NaN, a non-number, then you no longer > have a valid numeric value. So arithmetic and comparison operations > make no sense, and all you can get from further operations is > non-numeric values. It's not a number, so it doesn't compare to any > valid number, nor to any other non-number. Or perhaps you are asking why NaN operations are silent instead > of triggering exceptions? In general, having no sane result should produce an exception. In this case there's a clear desire for it to continue quietly, ie a quiet NaN is pretending it *is* a number. There must be some further logic in how it behaves. Btw, thank you both for replying. This has been niggling at my brain for years. === Subject: Re: The year of Cyber Tester > Ask not what computer mathematics will do for you, > but what together we can do for computer mathematics. -- Gottfried Wilhelm Leibniz Mairzy doats and dozy doats and liddle lamzy divey A kiddley divey too, wooden shoe? -- Milton Drake, Al Hoffman, Jerry Livingston, 1943 === Subject: The Right Direction posting-account=KroL8woAAAACDGpRprxyFi_gYw4Un8Xt 3.2.0; yplus 5.1.04b),gzip(gfe),gzip(gfe) A person on a long journey came to a fork in the road. And he could not decide which was the right direction. How could he solve his dilemma? Solution: He could decide by the process of elimination. If the direction on the right was not the right direction then the only direction that would be left would be the direction to his left. But if the direction on his left was not the right direction then the only direction left would be the right direction. www.intelrap.com === Subject: Re: JSH: So of course I'm right > It's as simple and direct a demonstration of a major result that you > can probably get mathematically, but the problem here is that > something is happening that's not supposed to be possible: given > various events I'm increasingly certain that leading academics around > the world are trying to hide a major result in their own field. Alert the media! > They killed a math journal. Swamp Jay, we hardly knew you. -- Michael Press === Subject: Re: JSH: So of course I'm right Let's assume that at some point (it might take years) everyone gives up > refuting your quadratic equation-based ideas. If this happens, will you > interpret silence as proof that you are right? He's doing that now, isn't he? >I've been here before: I prove with absolute mathematical rigor, Lies, you have NEVER given a proof. >posters delete out proof, talk around it, and continually say I'm >wrong. Nothing changes. you have never proven you are right, dill weed. <493c16ef$0$28669$7a628cd7@news.club-internet.fr> posting-account=3WPJYgoAAAA55VjhzK9i07RN8h8u8eEs Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) On Dec 7, 1:33pm, Denis Feldmann Possibly this is just a typo, but, 5^2+3(5)+2=42, not 56 No, youre wrong, as everybody knows that 42=6*9 Sounds like you graduated from the JSH school of math. Here are two other of its illustrious graduates: http://s151.photobucket.com/albums/s142/Rigo 75/?action=view¤t=BailOutExplained.flv ;>) M === Subject: Re: JSH: So of course I'm right <12685457.1228673640346.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=BVr-MgkAAABE4LRE1rHDnN9heo0IZZTk 2.0.50727),gzip(gfe),gzip(gfe) > On Dec 7, 1:33pm, Denis Feldmann Possibly this is just a typo, but, 5^2+3(5)+2=42, not 56 No, youre wrong, as everybody knows that 42=6*9 Sounds like you graduated from the JSH school of math. > Here are two other of its illustrious graduates: http://s151.photobucket.com/albums/s142/Rigo 75/?action=viewót=... ;>) M > Possibly this is just a typo, but, 5^2+3(5)+2=42, not 56 No, youre wrong, as everybody knows that 42=6*9 > Let me guess - Hitchhiker's Guide to the Galaxy What do you get when you multiply 6 by ............... 9 ? I always thought there was something fundamentally wrong with the universe. Enrico === Subject: Re: JSH: So of course I'm right >Let me guess - Hitchhiker's Guide to the Galaxy Correct. >What do you get when you multiply 6 by ............... 9 ? >I always thought there was something fundamentally wrong with the >universe. Try it in base 13. ;) rossum Enrico === Subject: Re: JSH: So of course I'm right > Try it in base 13. ;) I may be sick, but I don't make jokes in base 13. === Subject: Re: JSH: So of course I'm right posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081029 Firefox/2.0.0.18,gzip(gfe),gzip(gfe) [...] What math people claim is that the 7 moves around dependent on the > value of x. But if Well, let's check that. At x = 0: 7(175x^2 - 15x + 2) = 7 * 2 > 5b_1(x) + 7 = 7 > 5b_2(x)+ 2 = 2 The 7 is in 5b_1(x) + 7. How about at x = 2/7? Then: 7(175x^2 - 15x + 2) = 7 * 12 > 5b_1(x) + 7 = 12 > 5b_2(x)+ 2 = 7 [...] Hey, look at that! The 7 is now in 5b_2(x)+ 2. Oops. How about at x = 1? Then 7(175x^2 - 15x + 2) = 7 * 162 > 5b_1(x) + 7 = 22 + 5 sqrt(-26) > 5b_2(x)+ 2 = 22 - 5 sqrt(-26) Which one of those is the 7 in now, James? Similar examples can be done with 7(x^2 + 3x + 2) = (7x + 7)(x+2). How about a challenge for you? Show *any* case in the complex plane > where a number bounces around where you can verify at x=0 a > distribution and have it CHANGE with a different value for x. In context show a case where with something like 7(x^2 + 3x + 2) = (7x + 7)(x+2) the 7 moves based on the value of x ***in the complex plane***. Didn't Tim just show that, in the very example you chose, this > happens? No. He's looking at factors that shift based on value, but how you > multiply doesn't change. No, he isn't. He didn't mention factors at all in his post in his > thread. He just pointed out that, at x = 2/7, you can see that the 7 > is in the factor second factor 5b_2 + 2, since this is equal to 7. Now > you say that this is naive - but this is the *exact same argument > you used to conclude that the 7 distributed through the first factor > at x = 0*! Look, a quote from one of your own posts: so here it is where you can *see* it entire: > 7(x^2 + 3x + 2) = (7x + 7)(x+2) > notice that it is valid in the complex plane. Also note that you can > check at x=0, and *see* yet again that the 7 has multiplied in only > one way. Would you like to see some more of your own words? Look, here's a > quote from your blog: it's trivial that (5b2(x)+ 2) could not have been multiplied by 7 > as there is that 2, and if you multiply 2 times something like 7, > what do you get? 14. Let's see how this applies to Tim's example at x = 2/7, shall we? You > claim that the first factor 5b_1 + 7 was multiplied by 7 for every x. > But at x = 2/7 we have 5b_1 + 7 = 12. But that couldn't have been > multiplied by 7, as there is that 12, and if you multiply 12 times > something like 7, what do you get? 84. Why couldn't it have been? You're in the complex plane: 7(12/7) = > 12. No problem. Understand? It's not a factor argument! Given 7(x^2 + 3x + 2) = (7x + 7)(x+2) is it possible the '7' on the right hand side will move around based > on the value of x? Now if you let x=1/7, the '7' appears to go away > but does that violate a*(f(x) + b) = a*f(x) + a*b? Did the *function* move the 7? No, it was still multiplied by 7. Now move to the more complex: 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) where now you cannot just see the '7' like you could in the previous > example and in fact at this point you know nothing about how the 7 > distributed. But given that b_1(0) = b_2(0) = 0, you now know you > have the same type situation as with 7(x^2 + 3x + 2) = (7x + 7)(x+2) except now you're flying by mathematical instruments and not your eye > sight. (Why don't you know how the 7 distributed previous to knowing b_1(0) = > b_2(0) = 0? Ans. The functions can be an infinity of choices that can show the > infinite number of ways to multiply two elements by 7.) Here is the definitive question: can the 7 have distributed through > with 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) in *any* other way, if you have that at x=0, b_1(0) = b_2(0) = 0? Imagine a mathematical demon multiplying things out, he has factored > 175x^2 - 15x + 2 into two elements and multiplied one by 7, which one > did he multiply by 7? And what are the two elements? Anyone can answer. Of course the above arguments make no sense. But they're YOUR > arguments. So you have a choice: either admit those arguments are > garbage so that you have no basis for your claim, or accept those > arguments as correct so that they blow your claim out of the water. Or > continue to make a fool of yourself by posting wrong mathematics, like > you have been for years. Your call. If you factor say, x^2 + 3x + 2, and multiply it by 7, and do it this > way: 7(x^2 + 3x + 2) = (x+1)(7x + 14) is it possible that the 7 will magically *move* if you change x? Can > you not verify at x=0 how it was done, even if you couldn't see it > directly? What makes you think that with a slightly more complex example where > the 7 is only somewhat hidden as you can still see it in the > factorization that all the mathematical rules fly out the window and > it behaves like some bizarre drunken number instead of a perfectly > logical one? What do you see when you see 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) where b_1(0) = b_2(0) = 0? I'm curious. What do you see? > What I see is a guy who is stuck back in high school with factorization by inspection, and he cannot get past it because he is not willing to learn any real math. The analogy here is (n + 1)*(n + 2) in the ordinary integers. You can alway factor 2 out of it. Out of which factor? When n = 0, the 2 factors out of the second parenthesis. When n = 1, it factors out of the first parenthesis. Which of the two parentheses it factors out of is a function of n. The same thing happens in your example, but it is harder to understand. In the algebraic integers, the 7 splits up in a variable way. You cannot see it by inspection however, just as, with (n + 1)*(n + 2) in the integers, you cannot factor out 2 just by inspection. The main point in all these arguments is getting lost. I gave an explicit example. You keep turning your back on that and returning to your old erroneous arguments: nothing new. Your arguments simply contradict basic arithmetic. You keep not facing up to that. This idea, that 7 must split up in ways that are functions of x, has been around essentially since when you first submitted 'Advanced Polynomial Factorization' to question of the distributive law in the complex plane. Nothing wrong with the distributive law nor with the complex plane. The problem is, you are assuming without proof that 7 can split up in only one way regardless of the value of x. Taint true. Marcus. Marcus. > James Harris === Subject: Re: JSH: So of course I'm right > [...] > What math people claim is that the 7 moves around dependent on the > value of x. But if > Well, let's check that. At x = 0: > 7(175x^2 - 15x + 2) = 7 * 2 > 5b_1(x) + 7 = 7 > 5b_2(x)+ 2 = 2 > The 7 is in 5b_1(x) + 7. > How about at x = 2/7? Then: > 7(175x^2 - 15x + 2) = 7 * 12 > 5b_1(x) + 7 = 12 > 5b_2(x)+ 2 = 7 > [...] > Hey, look at that! The 7 is now in 5b_2(x)+ 2. > Oops. > How about at x = 1? Then > 7(175x^2 - 15x + 2) = 7 * 162 > 5b_1(x) + 7 = 22 + 5 sqrt(-26) > 5b_2(x)+ 2 = 22 - 5 sqrt(-26) > Which one of those is the 7 in now, James? > Similar examples can be done with > 7(x^2 + 3x + 2) = (7x + 7)(x+2). > How about a challenge for you? Show *any* case in the complex plane > where a number bounces around where you can verify at x=0 a > distribution and have it CHANGE with a different value for x. > In context show a case where with something like > 7(x^2 + 3x + 2) = (7x + 7)(x+2) > the 7 moves based on the value of x ***in the complex plane***. > Didn't Tim just show that, in the very example you chose, this > happens? > No. He's looking at factors that shift based on value, but how you > multiply doesn't change. > No, he isn't. He didn't mention factors at all in his post in his > thread. He just pointed out that, at x = 2/7, you can see that the 7 > is in the factor second factor 5b_2 + 2, since this is equal to 7. Now > you say that this is naive - but this is the *exact same argument > you used to conclude that the 7 distributed through the first factor > at x = 0*! Look, a quote from one of your own posts: > so here it is where you can *see* it entire: > 7(x^2 + 3x + 2) = (7x + 7)(x+2) > notice that it is valid in the complex plane. Also note that you can > check at x=0, and *see* yet again that the 7 has multiplied in only > one way. > Would you like to see some more of your own words? Look, here's a > quote from your blog: > it's trivial that (5b2(x)+ 2) could not have been multiplied by 7 > as there is that 2, and if you multiply 2 times something like 7, > what do you get? 14. > Let's see how this applies to Tim's example at x = 2/7, shall we? You > claim that the first factor 5b_1 + 7 was multiplied by 7 for every x. > But at x = 2/7 we have 5b_1 + 7 = 12. But that couldn't have been > multiplied by 7, as there is that 12, and if you multiply 12 times > something like 7, what do you get? 84. > Why couldn't it have been? You're in the complex plane: 7(12/7) = > 12. No problem. > Understand? > It's not a factor argument! > Given > 7(x^2 + 3x + 2) = (7x + 7)(x+2) > is it possible the '7' on the right hand side will move around based > on the value of x? Now if you let x=1/7, the '7' appears to go away > but does that violate a*(f(x) + b) = a*f(x) + a*b? > Did the *function* move the 7? No, it was still multiplied by 7. > Now move to the more complex: > 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) > where now you cannot just see the '7' like you could in the previous > example and in fact at this point you know nothing about how the 7 > distributed. But given that b_1(0) = b_2(0) = 0, you now know you > have the same type situation as with > 7(x^2 + 3x + 2) = (7x + 7)(x+2) > except now you're flying by mathematical instruments and not your eye > sight. > (Why don't you know how the 7 distributed previous to knowing b_1(0) = > b_2(0) = 0? > Ans. The functions can be an infinity of choices that can show the > infinite number of ways to multiply two elements by 7.) > Here is the definitive question: can the 7 have distributed through > with > 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) > in *any* other way, if you have that at x=0, b_1(0) = b_2(0) = 0? > Imagine a mathematical demon multiplying things out, he has factored > 175x^2 - 15x + 2 into two elements and multiplied one by 7, which one > did he multiply by 7? And what are the two elements? > Anyone can answer. > Of course the above arguments make no sense. But they're YOUR > arguments. So you have a choice: either admit those arguments are > garbage so that you have no basis for your claim, or accept those > arguments as correct so that they blow your claim out of the water. Or > continue to make a fool of yourself by posting wrong mathematics, like > you have been for years. Your call. > If you factor say, x^2 + 3x + 2, and multiply it by 7, and do it this > way: > 7(x^2 + 3x + 2) = (x+1)(7x + 14) > is it possible that the 7 will magically *move* if you change x? Can > you not verify at x=0 how it was done, even if you couldn't see it > directly? > What makes you think that with a slightly more complex example where > the 7 is only somewhat hidden as you can still see it in the > factorization that all the mathematical rules fly out the window and > it behaves like some bizarre drunken number instead of a perfectly > logical one? > What do you see when you see > 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) > where b_1(0) = b_2(0) = 0? > I'm curious. What do you see? > What I see is a guy who is stuck back in high school with > factorization by inspection, and he cannot get past it > because he is not willing to learn any real math. The analogy here is (n + 1)*(n + 2) in the > ordinary integers. You can alway factor 2 out of > it. Out of which factor? That's why I'm trying to hold you people to the complex plane. You > LOVE those factor arguments. Do you have a formula to count Gaussian primes inside the circle x^2 + y^2 = N, where N>0 is some integer? David Bernier === Subject: Re: JSH: So of course I'm right posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081029 Firefox/2.0.0.18,gzip(gfe),gzip(gfe) [...] What math people claim is that the 7 moves around dependent on the > value of x. But if Well, let's check that. At x = 0: 7(175x^2 - 15x + 2) = 7 * 2 > 5b_1(x) + 7 = 7 > 5b_2(x)+ 2 = 2 The 7 is in 5b_1(x) + 7. How about at x = 2/7? Then: 7(175x^2 - 15x + 2) = 7 * 12 > 5b_1(x) + 7 = 12 > 5b_2(x)+ 2 = 7 [...] Hey, look at that! The 7 is now in 5b_2(x)+ 2. Oops. How about at x = 1? Then 7(175x^2 - 15x + 2) = 7 * 162 > 5b_1(x) + 7 = 22 + 5 sqrt(-26) > 5b_2(x)+ 2 = 22 - 5 sqrt(-26) Which one of those is the 7 in now, James? Similar examples can be done with 7(x^2 + 3x + 2) = (7x + 7)(x+2). How about a challenge for you? Show *any* case in the complex plane > where a number bounces around where you can verify at x=0 a > distribution and have it CHANGE with a different value for x. In context show a case where with something like 7(x^2 + 3x + 2) = (7x + 7)(x+2) the 7 moves based on the value of x ***in the complex plane***. Didn't Tim just show that, in the very example you chose, this > happens? No. He's looking at factors that shift based on value, but how you > multiply doesn't change. No, he isn't. He didn't mention factors at all in his post in his > thread. He just pointed out that, at x = 2/7, you can see that the 7 > is in the factor second factor 5b_2 + 2, since this is equal to 7. Now > you say that this is naive - but this is the *exact same argument > you used to conclude that the 7 distributed through the first factor > at x = 0*! Look, a quote from one of your own posts: so here it is where you can *see* it entire: > 7(x^2 + 3x + 2) = (7x + 7)(x+2) > notice that it is valid in the complex plane. Also note that you can > check at x=0, and *see* yet again that the 7 has multiplied in only > one way. Would you like to see some more of your own words? Look, here's a > quote from your blog: it's trivial that (5b2(x)+ 2) could not have been multiplied by 7 > as there is that 2, and if you multiply 2 times something like 7, > what do you get? 14. Let's see how this applies to Tim's example at x = 2/7, shall we? You > claim that the first factor 5b_1 + 7 was multiplied by 7 for every x. > But at x = 2/7 we have 5b_1 + 7 = 12. But that couldn't have been > multiplied by 7, as there is that 12, and if you multiply 12 times > something like 7, what do you get? 84. Why couldn't it have been? You're in the complex plane: 7(12/7) = > 12. No problem. Understand? It's not a factor argument! Given 7(x^2 + 3x + 2) = (7x + 7)(x+2) is it possible the '7' on the right hand side will move around based > on the value of x? Now if you let x=1/7, the '7' appears to go away > but does that violate a*(f(x) + b) = a*f(x) + a*b? Did the *function* move the 7? No, it was still multiplied by 7. Now move to the more complex: 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) where now you cannot just see the '7' like you could in the previous > example and in fact at this point you know nothing about how the 7 > distributed. But given that b_1(0) = b_2(0) = 0, you now know you > have the same type situation as with 7(x^2 + 3x + 2) = (7x + 7)(x+2) except now you're flying by mathematical instruments and not your eye > sight. (Why don't you know how the 7 distributed previous to knowing b_1(0) = > b_2(0) = 0? Ans. The functions can be an infinity of choices that can show the > infinite number of ways to multiply two elements by 7.) Here is the definitive question: can the 7 have distributed through > with 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) in *any* other way, if you have that at x=0, b_1(0) = b_2(0) = 0? Imagine a mathematical demon multiplying things out, he has factored > 175x^2 - 15x + 2 into two elements and multiplied one by 7, which one > did he multiply by 7? And what are the two elements? Anyone can answer. Of course the above arguments make no sense. But they're YOUR > arguments. So you have a choice: either admit those arguments are > garbage so that you have no basis for your claim, or accept those > arguments as correct so that they blow your claim out of the water. Or > continue to make a fool of yourself by posting wrong mathematics, like > you have been for years. Your call. If you factor say, x^2 + 3x + 2, and multiply it by 7, and do it this > way: 7(x^2 + 3x + 2) = (x+1)(7x + 14) is it possible that the 7 will magically *move* if you change x? Can > you not verify at x=0 how it was done, even if you couldn't see it > directly? What makes you think that with a slightly more complex example where > the 7 is only somewhat hidden as you can still see it in the > factorization that all the mathematical rules fly out the window and > it behaves like some bizarre drunken number instead of a perfectly > logical one? What do you see when you see 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) where b_1(0) = b_2(0) = 0? I'm curious. What do you see? What I see is a guy who is stuck back in high school with > factorization by inspection, and he cannot get past it > because he is not willing to learn any real math. The analogy here is (n + 1)*(n + 2) in the > ordinary integers. You can alway factor 2 out of > it. Out of which factor? That's why I'm trying to hold you people to the complex plane. You > LOVE those factor arguments. James Harris Hey - Your desired conclusion is that 7 is a factor of one of a_1(x) and a_2(x), and has no factors in common with the other. You must LOVE those factor arguments too. And factor conclusions require factor arguments. You get nothing if all you are doing is in the complex plane. Marcus. === Subject: Re: JSH: So of course I'm right posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081029 Firefox/2.0.0.18,gzip(gfe),gzip(gfe) [...] What math people claim is that the 7 moves around dependent on the > value of x. But if Well, let's check that. At x = 0: 7(175x^2 - 15x + 2) = 7 * 2 > 5b_1(x) + 7 = 7 > 5b_2(x)+ 2 = 2 The 7 is in 5b_1(x) + 7. How about at x = 2/7? Then: 7(175x^2 - 15x + 2) = 7 * 12 > 5b_1(x) + 7 = 12 > 5b_2(x)+ 2 = 7 [...] Hey, look at that! The 7 is now in 5b_2(x)+ 2. Oops. How about at x = 1? Then 7(175x^2 - 15x + 2) = 7 * 162 > 5b_1(x) + 7 = 22 + 5 sqrt(-26) > 5b_2(x)+ 2 = 22 - 5 sqrt(-26) Which one of those is the 7 in now, James? Similar examples can be done with 7(x^2 + 3x + 2) = (7x + 7)(x+2). How about a challenge for you? Show *any* case in the complex plane > where a number bounces around where you can verify at x=0 a > distribution and have it CHANGE with a different value for x. In context show a case where with something like 7(x^2 + 3x + 2) = (7x + 7)(x+2) the 7 moves based on the value of x ***in the complex plane***. Didn't Tim just show that, in the very example you chose, this > happens? No. He's looking at factors that shift based on value, but how you > multiply doesn't change. No, he isn't. He didn't mention factors at all in his post in his > thread. He just pointed out that, at x = 2/7, you can see that the 7 > is in the factor second factor 5b_2 + 2, since this is equal to 7. Now > you say that this is naive - but this is the *exact same argument > you used to conclude that the 7 distributed through the first factor > at x = 0*! Look, a quote from one of your own posts: so here it is where you can *see* it entire: > 7(x^2 + 3x + 2) = (7x + 7)(x+2) > notice that it is valid in the complex plane. Also note that you can > check at x=0, and *see* yet again that the 7 has multiplied in only > one way. Would you like to see some more of your own words? Look, here's a > quote from your blog: it's trivial that (5b2(x)+ 2) could not have been multiplied by 7 > as there is that 2, and if you multiply 2 times something like 7, > what do you get? 14. Let's see how this applies to Tim's example at x = 2/7, shall we? You > claim that the first factor 5b_1 + 7 was multiplied by 7 for every x. > But at x = 2/7 we have 5b_1 + 7 = 12. But that couldn't have been > multiplied by 7, as there is that 12, and if you multiply 12 times > something like 7, what do you get? 84. Why couldn't it have been? You're in the complex plane: 7(12/7) = > 12. No problem. Understand? It's not a factor argument! Given 7(x^2 + 3x + 2) = (7x + 7)(x+2) is it possible the '7' on the right hand side will move around based > on the value of x? Now if you let x=1/7, the '7' appears to go away > but does that violate a*(f(x) + b) = a*f(x) + a*b? Did the *function* move the 7? No, it was still multiplied by 7. Now move to the more complex: 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) where now you cannot just see the '7' like you could in the previous > example and in fact at this point you know nothing about how the 7 > distributed. But given that b_1(0) = b_2(0) = 0, you now know you > have the same type situation as with 7(x^2 + 3x + 2) = (7x + 7)(x+2) except now you're flying by mathematical instruments and not your eye > sight. (Why don't you know how the 7 distributed previous to knowing b_1(0) = > b_2(0) = 0? Ans. The functions can be an infinity of choices that can show the > infinite number of ways to multiply two elements by 7.) Here is the definitive question: can the 7 have distributed through > with 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) in *any* other way, if you have that at x=0, b_1(0) = b_2(0) = 0? Imagine a mathematical demon multiplying things out, he has factored > 175x^2 - 15x + 2 into two elements and multiplied one by 7, which one > did he multiply by 7? And what are the two elements? Anyone can answer. Of course the above arguments make no sense. But they're YOUR > arguments. So you have a choice: either admit those arguments are > garbage so that you have no basis for your claim, or accept those > arguments as correct so that they blow your claim out of the water. Or > continue to make a fool of yourself by posting wrong mathematics, like > you have been for years. Your call. If you factor say, x^2 + 3x + 2, and multiply it by 7, and do it this > way: 7(x^2 + 3x + 2) = (x+1)(7x + 14) is it possible that the 7 will magically *move* if you change x? Can > you not verify at x=0 how it was done, even if you couldn't see it > directly? What makes you think that with a slightly more complex example where > the 7 is only somewhat hidden as you can still see it in the > factorization that all the mathematical rules fly out the window and > it behaves like some bizarre drunken number instead of a perfectly > logical one? What do you see when you see 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) where b_1(0) = b_2(0) = 0? I'm curious. What do you see? What I see is a guy who is stuck back in high school with > factorization by inspection, and he cannot get past it > because he is not willing to learn any real math. The analogy here is (n + 1)*(n + 2) in the > ordinary integers. You can alway factor 2 out of > it. Out of which factor? That's why I'm trying to hold you people to the complex plane. You > LOVE those factor arguments. James Harris Hey - Your desired conclusion is that 7 is a factor of one > of a_1(x) and a_2(x), and has no factors in common with > the other. You must LOVE those factor arguments too. Nope. That is a meaningless statement in the complex plane and false > in the ring of algebraic integers. > Oh, OK. So you are saying that you do NOT have a proof that one of a_1(x) and a_2(x) must be divisible by 7, while the other has no factors in common with 7? If that is what you are saying I am happy. That shoots down your main conclusion in Advanced Polynomial Factorization, but if you are giving up on that, fine with me. Marcus. === Subject: Re: JSH: So of course I'm right posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081029 Firefox/2.0.0.18,gzip(gfe),gzip(gfe) [...] What math people claim is that the 7 moves around dependent on the > value of x. But if Well, let's check that. At x = 0: 7(175x^2 - 15x + 2) = 7 * 2 > 5b_1(x) + 7 = 7 > 5b_2(x)+ 2 = 2 The 7 is in 5b_1(x) + 7. How about at x = 2/7? Then: 7(175x^2 - 15x + 2) = 7 * 12 > 5b_1(x) + 7 = 12 > 5b_2(x)+ 2 = 7 [...] Hey, look at that! The 7 is now in 5b_2(x)+ 2. Oops. How about at x = 1? Then 7(175x^2 - 15x + 2) = 7 * 162 > 5b_1(x) + 7 = 22 + 5 sqrt(-26) > 5b_2(x)+ 2 = 22 - 5 sqrt(-26) Which one of those is the 7 in now, James? Similar examples can be done with 7(x^2 + 3x + 2) = (7x + 7)(x+2). How about a challenge for you? Show *any* case in the complex plane > where a number bounces around where you can verify at x=0 a > distribution and have it CHANGE with a different value for x. In context show a case where with something like 7(x^2 + 3x + 2) = (7x + 7)(x+2) the 7 moves based on the value of x ***in the complex plane***. Didn't Tim just show that, in the very example you chose, this > happens? No. He's looking at factors that shift based on value, but how you > multiply doesn't change. No, he isn't. He didn't mention factors at all in his post in his > thread. He just pointed out that, at x = 2/7, you can see that the 7 > is in the factor second factor 5b_2 + 2, since this is equal to 7. Now > you say that this is naive - but this is the *exact same argument > you used to conclude that the 7 distributed through the first factor > at x = 0*! Look, a quote from one of your own posts: so here it is where you can *see* it entire: > 7(x^2 + 3x + 2) = (7x + 7)(x+2) > notice that it is valid in the complex plane. Also note that you can > check at x=0, and *see* yet again that the 7 has multiplied in only > one way. Would you like to see some more of your own words? Look, here's a > quote from your blog: it's trivial that (5b2(x)+ 2) could not have been multiplied by 7 > as there is that 2, and if you multiply 2 times something like 7, > what do you get? 14. Let's see how this applies to Tim's example at x = 2/7, shall we? You > claim that the first factor 5b_1 + 7 was multiplied by 7 for every x. > But at x = 2/7 we have 5b_1 + 7 = 12. But that couldn't have been > multiplied by 7, as there is that 12, and if you multiply 12 times > something like 7, what do you get? 84. Why couldn't it have been? You're in the complex plane: 7(12/7) = > 12. No problem. Understand? It's not a factor argument! Given 7(x^2 + 3x + 2) = (7x + 7)(x+2) is it possible the '7' on the right hand side will move around based > on the value of x? Now if you let x=1/7, the '7' appears to go away > but does that violate a*(f(x) + b) = a*f(x) + a*b? Did the *function* move the 7? No, it was still multiplied by 7. Now move to the more complex: 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) where now you cannot just see the '7' like you could in the previous > example and in fact at this point you know nothing about how the 7 > distributed. But given that b_1(0) = b_2(0) = 0, you now know you > have the same type situation as with 7(x^2 + 3x + 2) = (7x + 7)(x+2) except now you're flying by mathematical instruments and not your eye > sight. (Why don't you know how the 7 distributed previous to knowing b_1(0) = > b_2(0) = 0? Ans. The functions can be an infinity of choices that can show the > infinite number of ways to multiply two elements by 7.) Here is the definitive question: can the 7 have distributed through > with 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) in *any* other way, if you have that at x=0, b_1(0) = b_2(0) = 0? Imagine a mathematical demon multiplying things out, he has factored > 175x^2 - 15x + 2 into two elements and multiplied one by 7, which one > did he multiply by 7? And what are the two elements? Anyone can answer. Of course the above arguments make no sense. But they're YOUR > arguments. So you have a choice: either admit those arguments are > garbage so that you have no basis for your claim, or accept those > arguments as correct so that they blow your claim out of the water. Or > continue to make a fool of yourself by posting wrong mathematics, like > you have been for years. Your call. If you factor say, x^2 + 3x + 2, and multiply it by 7, and do it this > way: 7(x^2 + 3x + 2) = (x+1)(7x + 14) is it possible that the 7 will magically *move* if you change x? Can > you not verify at x=0 how it was done, even if you couldn't see it > directly? What makes you think that with a slightly more complex example where > the 7 is only somewhat hidden as you can still see it in the > factorization that all the mathematical rules fly out the window and > it behaves like some bizarre drunken number instead of a perfectly > logical one? What do you see when you see 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) where b_1(0) = b_2(0) = 0? I'm curious. What do you see? What I see is a guy who is stuck back in high school with > factorization by inspection, and he cannot get past it > because he is not willing to learn any real math. The analogy here is (n + 1)*(n + 2) in the > ordinary integers. You can alway factor 2 out of > it. Out of which factor? That's why I'm trying to hold you people to the complex plane. You > LOVE those factor arguments. James Harris Hey - Your desired conclusion is that 7 is a factor of one > of a_1(x) and a_2(x), and has no factors in common with > the other. You must LOVE those factor arguments too. Nope. That is a meaningless statement in the complex plane and false > in the ring of algebraic integers. Oh, OK. So you are saying that you do NOT have a proof > that one of a_1(x) and a_2(x) must be divisible by 7, while the > other has no factors in common with 7? If that is I said that before. It is absolutely true that NEITHER of the roots > have 7 as a factor in the ring of algebraic integers! what you are saying I am happy. That shoots down your > main conclusion in Advanced Polynomial Factorization, but > if you are giving up on that, fine with me. Nope. I am beginning to think you are REALLY impaired. Your old conclusion was that one of a_1(x) and a_2(x) is divisible by 7, and the other has no factors in common with 7. That was basically the argument in APF (except for 3 factors). Now you are saying that old conclusion was not true. So I asked you: are you now saying your old conclusion was not true? And you reply Nope. Meaning, your old conclusion WAS true, right? So you are now in the weird position of saying: 1. X is not true. 2. X is true. Same X. You are now so totally messed up on this you don't which end to wipe when you go to the bathroom. The distributive property hasn't changed, so the mathematical > proof remains. > It proves nothing of interest in the complex plane. Everything is divisible by everything except zero. > With 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) where b_1(x) = a_1(x) and b_2(x) = a_2(x) + 1, so b_1(0) = b_2(0) = 0 as it is the normalization, where the a's are > roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0 what happens at x=0 is what happens for all x in the complex plane, as a*(f(x) + b) = a*f(x) + a*b. > You have stopped listening a LONG time ago. But now you have argued yourself into the corner of saying that two contradictory things are both true. And you don't even understand that that is what has happened. Marcus. > James Harris === Subject: Re: JSH: So of course I'm right So you are now in the weird position of saying: 1. X is not true. 2. X is true. Same X. You are now so totally messed up on this you > don't which end to wipe when you go to the bathroom. You are using two valued logic. Maybe this calls for some of that three > valued logic that JSH has talked about a couple times on his blog? > Harristotelian logic! === Subject: Re: JSH: So of course I'm right > So you are now in the weird position of saying: 1. X is not true. 2. X is true. Same X. You are now so totally messed up on this you > don't which end to wipe when you go to the bathroom. You are using two valued logic. Maybe this calls for some of that three > valued logic that JSH has talked about a couple times on his blog? > > Harristotelian logic! All part of the Ring of Integers === Subject: Re: JSH: So of course I'm right posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) [...] What math people claim is that the 7 moves around dependent on the > value of x. But if Well, let's check that. At x = 0: 7(175x^2 - 15x + 2) = 7 * 2 > 5b 1(x) + 7 = 7 > 5b 2(x)+ 2 = 2 The 7 is in 5b 1(x) + 7. How about at x = 2/7? Then: 7(175x^2 - 15x + 2) = 7 * 12 > 5b 1(x) + 7 = 12 > 5b 2(x)+ 2 = 7 [...] Hey, look at that! The 7 is now in 5b 2(x)+ 2. Oops. How about at x = 1? Then 7(175x^2 - 15x + 2) = 7 * 162 > 5b 1(x) + 7 = 22 + 5 sqrt(-26) > 5b 2(x)+ 2 = 22 - 5 sqrt(-26) Which one of those is the 7 in now, James? Similar examples can be done with 7(x^2 + 3x + 2) = (7x + 7)(x+2). How about a challenge for you? Show *any* case in the complex plane > where a number bounces around where you can verify at x=0 a > distribution and have it CHANGE with a different value for x. In context show a case where with something like 7(x^2 + 3x + 2) = (7x + 7)(x+2) the 7 moves based on the value of x ***in the complex plane***. Didn't Tim just show that, in the very example you chose, this > happens? No. He's looking at factors that shift based on value, but how you > multiply doesn't change. No, he isn't. He didn't mention factors at all in his post in his > thread. He just pointed out that, at x = 2/7, you can see that the 7 > is in the factor second factor 5b 2 + 2, since this is equal to 7. Now > you say that this is naive - but this is the *exact same argument > you used to conclude that the 7 distributed through the first factor > at x = 0*! Look, a quote from one of your own posts: so here it is where you can *see* it entire: > 7(x^2 + 3x + 2) = (7x + 7)(x+2) > notice that it is valid in the complex plane. Also note that you can > check at x=0, and *see* yet again that the 7 has multiplied in only > one way. Would you like to see some more of your own words? Look, here's a > quote from your blog: it's trivial that (5b2(x)+ 2) could not have been multiplied by 7 > as there is that 2, and if you multiply 2 times something like 7, > what do you get? 14. Let's see how this applies to Tim's example at x = 2/7, shall we? You > claim that the first factor 5b 1 + 7 was multiplied by 7 for every x. > But at x = 2/7 we have 5b 1 + 7 = 12. But that couldn't have been > multiplied by 7, as there is that 12, and if you multiply 12 times > something like 7, what do you get? 84. Why couldn't it have been? You're in the complex plane: 7(12/7) = > 12. No problem. Understand? Yes! I DO understand! And you should know full well that I understand about how the 7 must have distributed at x = 0. Look, a quote from a post of yours in the Normalization issue thread, and my reply: > notice that it is valid in the complex plane. Also note that you can > check at x=0, and *see* yet again that the 7 has multiplied in only > one way. Er, James? Look: in the complex plane, x^2 + 3x + 2 = (7x + 7)(x/7 + 2/7) Huh. When one puts it like that it looks a lot like it was the second > factor in (1), not the first, which was multiplied by 7. See? You're in the complex plane: 7(2/7) = 2, so the 7 could just have well have multiplied the second factor at x = 0. No problem. I used happens at x = 0. You evidently didn't think it refuted your argument at x = 0. But now you apparently DO think it refutes Tim's argument - which is actually YOUR argument - at x = 2/7. Can't you see what's going on here? You're running yourself in circles, giving two exactly opposite arguments at different values of x. You repeatedly say that other people are claiming the distributive property changes with the value of x. But you, apparently, seem to think that *logic itself* varies with x, so that a correct argument at x = 0 is an incorrect argument at x = 2/7, and an incorrect refutation at x = 0 is a correct refutation at x = 2/7. Just sit down, take a deep breath, and try to think logically about what's going on. As it is your whole thinking on this issue is a total mess, as everyone can see from watching contradict yourself in the space of a few posts. === Subject: Re: JSH: So of course I'm right posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) [...] Why couldn't it have been? You're in the complex plane: 7(12/7) = > 12. No problem. Understand? Yes! I DO understand! And you should know full well that I understand > about how the 7 must have distributed at x = 0. Look, a quote from a > post of yours in the Normalization issue thread, and my reply: notice that it is valid in the complex plane. Also note that you can > check at x=0, and *see* yet again that the 7 has multiplied in only > one way. Er, James? Look: in the complex plane, x^2 + 3x + 2 = (7x + 7)(x/7 + 2/7) Huh. When one puts it like that it looks a lot like it was the second > factor in (1), not the first, which was multiplied by 7. See? You're in the complex plane: 7(2/7) = 2, so the 7 could just have > well have multiplied the second factor at x = 0. No problem. I used Um, but I don't see a 7 on the left hand side of the equals. Do YOU > see one? Good grief, James. Are you so mentally damaged that you can't remember your own argument? No, there's no 7 on the left hand side of the equation I give above. The 7 comes in when you multiply the LHS by 7 and then factor it, like you do in this quote, again from the Normalization issue thread: > so here it is where you can *see* it entire: 7(x^2 + 3x + 2) = (7x + 7)(x+2) notice that it is valid in the complex plane. Also note that you can > check at x=0, and *see* yet again that the 7 has multiplied in only > one way. There. You state that at x = 0 you can *see* that that the 7 has multiplied in only one way. Just like Tim did at x = 2/7. You state that the 2 is the crucial tattle-tale, and that it could not have been multiplied by 7 as there is that 2, and if you multiply 2 times something like 7, what do you get? 14. But since 7(2/7) = 7, it simply doesn't follow that the 7 can only have multiplied one way - a fact that you evidently now recognise since you use the same argument to refute Tim's example at x = 2/7. > Is it a phantom? Do you see one? Are you seeing phantom 7's on the > left hand side? I like to think that your current totally lame attempt to avoid the contradiction staring you in the face is just public bravado as it sinks in that you've been hoist with your own petard. I like to think that you're not as dumb as you're pretending to be here - please don't prove me wrong. === Subject: Re: JSH: So of course I'm right > Normalization issue thread: so here it is where you can *see* it entire: 7(x^2 + 3x + 2) = (7x + 7)(x+2) >Where there IS a 7 on the left hand side. trivial. === Subject: Re: JSH: So of course I'm right posting-account=XH7n1goAAAC_c_M3JtSDv3QFzlbwvvMO .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) [...] What math people claim is that the 7 moves around dependent on the > value of x. But if Well, let's check that. At x = 0: 7(175x^2 - 15x + 2) = 7 * 2 > 5b 1(x) + 7 = 7 > 5b 2(x)+ 2 = 2 The 7 is in 5b 1(x) + 7. How about at x = 2/7? Then: 7(175x^2 - 15x + 2) = 7 * 12 > 5b 1(x) + 7 = 12 > 5b 2(x)+ 2 = 7 [...] Hey, look at that! The 7 is now in 5b 2(x)+ 2. Oops. How about at x = 1? Then 7(175x^2 - 15x + 2) = 7 * 162 > 5b 1(x) + 7 = 22 + 5 sqrt(-26) > 5b 2(x)+ 2 = 22 - 5 sqrt(-26) Which one of those is the 7 in now, James? Similar examples can be done with 7(x^2 + 3x + 2) = (7x + 7)(x+2). How about a challenge for you? Show *any* case in the complex plane > where a number bounces around where you can verify at x=0 a > distribution and have it CHANGE with a different value for x. In context show a case where with something like 7(x^2 + 3x + 2) = (7x + 7)(x+2) the 7 moves based on the value of x ***in the complex plane***. Didn't Tim just show that, in the very example you chose, this > happens? No. He's looking at factors that shift based on value, but how you > multiply doesn't change. No, he isn't. He didn't mention factors at all in his post in his > thread. He just pointed out that, at x = 2/7, you can see that the 7 > is in the factor second factor 5b 2 + 2, since this is equal to 7. Now > you say that this is naive - but this is the *exact same argument > you used to conclude that the 7 distributed through the first factor > at x = 0*! Look, a quote from one of your own posts: so here it is where you can *see* it entire: > 7(x^2 + 3x + 2) = (7x + 7)(x+2) > notice that it is valid in the complex plane. Also note that you can > check at x=0, and *see* yet again that the 7 has multiplied in only > one way. Would you like to see some more of your own words? Look, here's a > quote from your blog: it's trivial that (5b2(x)+ 2) could not have been multiplied by 7 > as there is that 2, and if you multiply 2 times something like 7, > what do you get? 14. Let's see how this applies to Tim's example at x = 2/7, shall we? You > claim that the first factor 5b 1 + 7 was multiplied by 7 for every x. > But at x = 2/7 we have 5b 1 + 7 = 12. But that couldn't have been > multiplied by 7, as there is that 12, and if you multiply 12 times > something like 7, what do you get? 84. Why couldn't it have been? You're in the complex plane: 7(12/7) = > 12. No problem. Understand? It's not a factor argument! Given 7(x^2 + 3x + 2) = (7x + 7)(x+2) is it possible the '7' on the right hand side will move around based > on the value of x? Now if you let x=1/7, the '7' appears to go away > but does that violate a*(f(x) + b) = a*f(x) + a*b? Did the *function* move the 7? No, it was still multiplied by 7. Now move to the more complex: 7(175x^2 - 15x + 2) = (5b 1(x) + 7)(5b 2(x)+ 2) where now you cannot just see the '7' like you could in the previous > example and in fact at this point you know nothing about how the 7 > distributed. But given that b 1(0) = b 2(0) = 0, you now know you > have the same type situation as with 7(x^2 + 3x + 2) = (7x + 7)(x+2) except now you're flying by mathematical instruments and not your eye > sight. (Why don't you know how the 7 distributed previous to knowing b 1(0) = > b 2(0) = 0? Ans. The functions can be an infinity of choices that can show the > infinite number of ways to multiply two elements by 7.) Here is the definitive question: can the 7 have distributed through > with 7(175x^2 - 15x + 2) = (5b 1(x) + 7)(5b 2(x)+ 2) in *any* other way, if you have that at x=0, b 1(0) = b 2(0) = 0? Imagine a mathematical demon multiplying things out, he has factored > 175x^2 - 15x + 2 into two elements and multiplied one by 7, which one > did he multiply by 7? And what are the two elements? Anyone can answer. Of course the above arguments make no sense. But they're YOUR > arguments. So you have a choice: either admit those arguments are > garbage so that you have no basis for your claim, or accept those > arguments as correct so that they blow your claim out of the water. Or > continue to make a fool of yourself by posting wrong mathematics, like > you have been for years. Your call. If you factor say, x^2 + 3x + 2, and multiply it by 7, and do it this > way: 7(x^2 + 3x + 2) = (x+1)(7x + 14) is it possible that the 7 will magically *move* if you change x? Can > you not verify at x=0 how it was done, even if you couldn't see it > directly? What makes you think that with a slightly more complex example where > the 7 is only somewhat hidden as you can still see it in the > factorization that all the mathematical rules fly out the window and > it behaves like some bizarre drunken number instead of a perfectly > logical one? Because for a function like f(x) = x^2 + x, the number f(x) is divisible by 2 for any integer x, but f(x) factors as x(x+1), and whether 2 divides the first factor (x) or the second factor (x+1) is determined by x. So 2 divides f(x) for any integer x, but the 2 magically moves, as you put it, when you move x. This kind of phenomenon certainly does happen. What do you see when you see 7(175x^2 - 15x + 2) = (5b 1(x) + 7)(5b 2(x)+ 2) where b 1(0) = b 2(0) = 0? I'm curious. What do you see? James Harris- Hide quoted text - - Show quoted text - === Subject: Re: JSH: So of course I'm right posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081029 Firefox/2.0.0.18,gzip(gfe),gzip(gfe) > It's as simple and direct a demonstration of a major result that you > can probably get mathematically, but the problem here is that > something is happening that's not supposed to be possible: given > various events I'm increasingly certain that leading academics around > the world are trying to hide a major result in their own field. They killed a math journal. Have kept accepting public funds for > bogus research. And most tragically I think, have kept teaching new > students the flawed techniques, assigning them homework, testing them. So what is the flaw in how posters present how to look at the > argument, well I have to show the special construction again to > explain it, and remember, in the field of complex numbers, with 7(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2) where b_1(x) = a_1(x) and b_2(x) = a_2(x) + 1, which is the normalization, where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0 What math people claim is that the 7 moves around dependent on the > value of x. But if a*(f(x) + b) = a*f(x) + a*b Let's get right down to how you actually use this. You want to factor 7 out of the product (in your 'normalized' form): (5b_1(x) + 7)*(5a_2(x) + 2). There are TWO parentheses here, not just one as in the distributive law. You choose - CHOOSE - to put the 7 in front of the first parenthesis. But if the 7 were split up into two parts, say, 7 = A * B, you could consider how you might factor A out of the first parenthesis and B out of the second parenthesis. So you say: when x = 0, clearly A must be 7 and B must be 1. So if A and B are constants, you have the result in general. But that is making an assumption which you do not justify. Namely, that A and B have to be constants. You could think of 7 as split up in VARIABLE ways - ways which depend on x. That is, 7 = A(x) * B(x). If you do that, then your distributive property argument falls apart. Sure, you may be able to factor A(x) out of the first parenthesis and B(x) out of the second one, but since A(x) and B(x) are functions of x, you CANNOT conclude that A(x) = A(0) = 7. The distributive property does not give you that. The example I gave when x = 1 shows explicitly that you CAN split 7 up into two algebraic integers, 7 = A(1) * B(1), and you can factor A(1) out of the first parenthesis and B(1) out of the second parenthesis, with all the quotients being algebraic integers, and A(1) is definitely not equal to 7 and B(1) is definitely not equal to 1 and neither A(1) nor B(1) is equal to a unit. You keep denying that this can happen. But I gave the factors A(1) and B(1) explcitly. You can check the arithmetic. What you keep saying denies the arithmetic. And Dik Winter gave a similarly factorization when x = -1. Again, A(-1) and B(-1) are algebraic integers and neither one equals 7 or 1 and neither one is a unit. And again, you say it cannot be so. You contradict basic arithmetic. And it doesn't even seem to bother you. You still think you are right. Not only have we showed you are wrong, but (above) I have explained why your 'distributive property' argument does not prove what you think it proves. That is we have shown that: 1. You cannot be right: you contradict arithmetic. 2. We have identified your mistake. What more could you ask? then why does the function change what is multiplying times it? It > cannot. Trivially the equivalent can be seen with a classical factorization > (the one above is what I call a non-polynomial factorization): 7(x^2 + 3x + 2) = (7x + 7)(x+2) and if you think this math is basic, yes, you are right. It is basic > math. There is no way the 7 can bounce around It doesn't 'bounce around'. It splits up differently depending on x. > because what is being > multiplied changed. You may say, too simple! Exactly right. > Those are weird funky > roots of some quadratic functions that can be weird bizarre > mathematical beasties! > Which you still do not understand. > But the distributive property doesn't change no matter how weird the > math beastie, right? The distributive law is in fine shape. It's your assuming that the factors must be constant which is wrong. Really, it is no different from saying that if g(0) + 7 is divisible by 7, then g(x) + 7 must be divisible by 7 for all x, regardless of what the function g(x) is. a*(weird_math_beastie + b) = a*weird_math_beastie + a*b > The little 'a' and 'b' weird beasties are functions of the 'weird-math-beastie', which is a variable function of x. You fail to include that. > Notice that by shifting to factor arguments posters try to convince > you to defy the distributive property with the notion that the > *function* is changing things because the function can behave > differently based on different values, but consider 7*f(x) + 7 = x + 7 here f(x) = x/7, which is hidden to some extent, but so what? Still > doesn't change the distributive property!!! > Agreed. The distributive property holds just fine. But you are making the hidden assumption that the factors are constant. What we have shown is that that is not the case. Basic arithmetic. No Galois, no Dedekind or Hilbert, no theory. Just arithmetic which anyone here can do for themselves and see that you are wrong. > If a*(f(x) + b) = a*f(x) + a*b then I'm right. > Nope. In your application you cannot assume that 'a' and 'b' are constants. If you could, you would indeed be right. But examples prove otherwise, and if you say those examples cannot happen, you are basically saying that 2 + 2 cannot equal 4. Just write in a(x) and b(x) in your favorite equation above and see how far you get. > So why do they lie? [remainder of vicious non-math comments SNIPPED] Marcus. === Subject: How to calculate Pi in bases other than 2/10/16 posting-account=bdkY3QoAAACfjBfO8YzZ5Iz9vSDKo_mi OptusNetDSL6),gzip(gfe),gzip(gfe) Hi all, I hope this is the right group for my question - if not, would someone please point me in the right direction. I am interested - mainly as an exercise in programming - in calcuating Pi to a largish number of places (500 would be ample) in various unusual bases (eg 13, 17, 19). Can someone point me to an algorithm that could be implemented on a Windows XP or 98SE PC, and give an accurate result in a reasonable period of time (days or weeks rather than years!)? I am a reasonably competent VBasic programmer, and could probably get access to a Pascal compiler (I prefer Pascal over all the other languages I have used in more than 20 years in IT), but my math knowledge is not very strong. Looking forward to hearing from someone. johngross === Subject: Re: How to calculate Pi in bases other than 2/10/16 > Hi all, I hope this is the right group for my question - if not, would someone > please point me in the right direction. I am interested - mainly as an exercise in programming - in calcuating > Pi to a largish number of places (500 would be ample) in various > unusual bases (eg 13, 17, 19). Can someone point me to an algorithm that could be implemented on a > Windows XP or 98SE PC, and give an accurate result in a reasonable > period of time (days or weeks rather than years!)? How about milliseconds? > I am a reasonably competent VBasic programmer, and could probably get > access to a Pascal compiler (I prefer Pascal over all the other > languages I have used in more than 20 years in IT), but my math > knowledge is not very strong. Looking forward to hearing from someone. johngross In Maple: F:= (b,N) -> ListTools[Reverse](convert( round(evalf(floor(Pi*b^(N-1)), 3+ilog10(b^N))), base,b)); For example, F(13, 100); [3, 1, 10, 12, 1, 0, 4, 9, 0, 5, 2, 10, 2, 12, 7, 7, 3, 6, 9, 12, 0, 11, 11, 8, 9, 12, 12, 9, 8, 8, 3, 2, 7, 8, 2, 9, 8, 3, 5, 8, 11, 3, 7, 0, 1, 6, 0, 3, 0, 6, 1, 3, 3, 12, 10, 5, 10, 12, 11, 10, 5, 7, 6, 1, 4, 11, 6, 5, 11, 4, 1, 0, 0, 2, 0, 12, 2, 2, 11, 4, 12, 7, 1, 4, 5, 7, 10, 9, 5, 5, 10, 5, 1, 5, 5, 11, 0, 4, 10, 6] F(13, 10000) takes 0.187 seconds on my computer. -- === Subject: Re: How to calculate Pi in bases other than 2/10/16 posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1 FOH:R051,gzip(gfe),gzip(gfe) Euler's identity is, what -- exp(i*pi) = 1, i*pi = ln(1) pi = -i*ln(1), but I forgot what the natural log of one is. thus: there's got to be a few Latin phrases for this kind of fallacy; this is the same thing that Are Buckafka Fullofit tried to put over, and most of the Buckywitches really believe it. anyway, the queestion is, why should line segments of polygons be so God-am existential, or of polyhedra -- because of the typical stick-drawn configuration in some edition of Euclid? >Also, from the position taken above, spheres are only possible in theory aswell. --Cheeny & Zbiggy, fo'mo' years! http://tarpley.net http://larouchepub.com http://xplodeyourmlmbiz.com/?s1=Fuller+Brush&t=LSP01&gclid=CPiar6nispcCFQu-G godYm21kA http://www.rwgrayprojects.com/synergetics/plates/plates.html === Subject: Re: How to calculate Pi in bases other than 2/10/16 posting-account=O9zR9AkAAACmp918j6u5m5plppeILcze 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0),gzip(gfe),gzip(gfe) > Euler's identity is, what -- > exp(i*pi) = 1, No. exp(i*pi) = -1, so pi = -i*ln(-1). > i*pi = ln(1) > pi = -i*ln(1), but > I forgot what the natural log of one is. That's a silly thing to forget. Back to the books for you. Dave === Subject: Re: How to calculate Pi in bases other than 2/10/16 posting-account=DLD3MQkAAACJxulKt9xouw3DPpXKssCI SLCC1; .NET CLR 2.0.50727; .NET CLR 1.1.4322; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > Hi all, I hope this is the right group for my question - if not, would someone > please point me in the right direction. I am interested - mainly as an exercise in programming - in calcuating > Pi to a largish number of places (500 would be ample) in various > unusual bases (eg 13, 17, 19). Can someone point me to an algorithm that could be implemented on a > Windows XP or 98SE PC, and give an accurate result in a reasonable > period of time (days or weeks rather than years!)? I am a reasonably competent VBasic programmer, and could probably get > access to a Pascal compiler (I prefer Pascal over all the other > languages I have used in more than 20 years in IT), but my math > knowledge is not very strong. Just compute pi once in any base that you like (10 or 16 spring to mind) and then use a base conversion algorithm to compute pi in the new base. E.g. the algorithm is: 1. Compute pi (Or read the value in as data) 2. Set newbase to 2 3. Change base of calculated number pi to newbase 4. Increment newbase 5. Is newbase large enough to exit? Yes -> exit; No -> goto 3. === Subject: Re: How to calculate Pi in bases other than 2/10/16 posting-account=H1y7YgoAAADzGQwbcYaL9UvwttgsjOjp Gecko/20071025 Firefox/2.0.0.9,gzip(gfe),gzip(gfe) > Hi all, I hope this is the right group for my question - if not, would someone > please point me in the right direction. I am interested - mainly as an exercise in programming - in calcuating > Pi to a largish number of places (500 would be ample) in various > unusual bases (eg 13, 17, 19). Can someone point me to an algorithm that could be implemented on a > Windows XP or 98SE PC, and give an accurate result in a reasonable > period of time (days or weeks rather than years!)? You're not happy to compute it in base 10 (say) and then convert to base 13 or whatever? -- GM === Subject: Re: Abstract Algebra- Ring Theory posting-account=zF2JRwoAAACsZrL8JJc0nL5uhvkKu3iq 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Is Z[sqrt(2)] a UFD? *********************************************************************** Yes it is. In fact it is an Euclidean Domain wrt the norm N(a + b*Sqrt > (2)):= a^2 - 2*b^2, and you can find the proof in any good number > theyr or algebra book (try Dummit and Foote's Algebra, for example). Tonio I went to the library today and read a bit of Dummit and Foote. Didn't find the proof, although I really liked the layout of the book. Anyhow, I had no problem proving that Z[sqrt(2)] is an integral domain. However, I am struggling with showing proving that: delta(ab) >= delta(a) Many of the similar proofs I found showed that delta(ab)=delta(a)delta (b), yet this isnt the case with Z[sqrt(2)]. Any suggestions how to proceed? === Subject: Re: Abstract Algebra- Ring Theory posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e 2.0.50727),gzip(gfe),gzip(gfe) > Is Z[sqrt(2)] a UFD? *********************************************************************** Yes it is. In fact it is an Euclidean Domain wrt the norm N(a + b*Sqrt > (2)):= a^2 - 2*b^2, and you can find the proof in any good number > theyr or algebra book (try Dummit and Foote's Algebra, for example). Tonio I went to the library today and read a bit of Dummit and Foote. Didn't > find the proof, although I really liked the layout of the book. > Anyhow, I had no problem proving that Z[sqrt(2)] is an integral > domain. However, I am struggling with showing proving that: delta(ab) >= delta(a) Many of the similar proofs I found showed that delta(ab)=delta(a)delta > (b), yet this isnt the case with Z[sqrt(2)]. Any suggestions how to > proceed? ************************************************************** You have a proof that Z[Sqrt(2)] is an euclidean domain, and thus a UFD, in the enxt link: http://tinyurl.com/6nmpet Tonio === Subject: Re: calculus / pre-calculus faculty survey On a Nov 26th post I asked members of this group to consider participating in a survey that I'm using to collect data for my dissertation. Some of you did and I am very grateful for your participation and valuable feedback. I am posting a second time, in case some of you intended to participate and haven't yet. If so, please take a few (15 or so) minutes to complete the survey. A special plea goes to those of you that teach pre-calculus at the postsecondary level. My response rate is low for this group... Please see the links below. Best wishes, Kostas Stroumbakis. Ks2470@columbia.edu My name is Kostas Stroumbakis and I am a doctoral student at Teacher's College, Columbia University. For my dissertation, I want to compare views on what makes adequate preparation for calculus. To this end, I have developed three online surveys in which I ask participants to rate various pre-calculus topics on their importance, as prerequisites, for success in calculus. If you teach, or recently have taught, (1) Calculus I (postsecondary), (2) Pre-Calculus (postsecondary), or (3) Pre-Calculus (high school), I would really appreciate it if you took a few minutes to complete my survey. It will take about 15 minutes. Please follow the corresponding link below. Also, if you can help to snowball this or suggest other venues for reaching qualified participants, I would greatly appreciate it. Kostas Stroumbakis ks2470@columbia.edu 1) College Calculus http://www.surveymonkey.com/s.aspx?sm=US4XorzONM_2feP6m7GXGbaw_3d_3d 2) College Pre-Calculus http://www.surveymonkey.com/s.aspx?sm=Lq2hCQMwbH_2bQhSidjuo03A_3d_3d 3) High School Pre-Calculus http://www.surveymonkey.com/s.aspx?sm=K98orxnBzLkN_2fcpCRZiNFQ_3d_3d PS I have posted a similar request to other groups to which you might belong. I apologize if I am cluttering your inbox. === Subject: Information Theory posting-account=KroL8woAAAACDGpRprxyFi_gYw4Un8Xt 3.2.0; .NET CLR 2.0.50727; yplus 5.1.05b),gzip(gfe),gzip(gfe) Information is conveyed by symbols that are ñ in formationî in accordance with some orderly scheme. But Thermodynamics is a special branch of mechanics that outlaws the perpetual motion machine. Just as heat always dissipate to cooler climes the beat will seem to oscillate with the rapper's rhymes. www.intelrap.com === Subject: Algebra with kernel, ideal, field. Hello teacher~ R and R' are commutative ring with unity. f : R -> R' is a surjective ring homomorphism. Prove that there is a 1-1 correspondence from the set of all ideals in R that contain K = ker f onto the set of all ideals in R' ---------------------------------------------------------------------------- ---- pf) Let A be the set of all ideals in R that contain K = ker f. Let B be the set of all ideals in R'. Let p : A -> B, p(J) = f(J) = {f(j) | j in J}. I will show that p is bijective. and then END. For an ideal J of R that contain K, p(J) = f(J) is an ideal of R'. For an ideal N of R', f^{-1}(N) is an ideal of R. p[f^{-1}(N)] = f[f^{-1}(N)} = N. (since f is surjective.) It means that p is surjective. Let h : R/K -> R', h(r+K) = f(r). This is a isomorphism by fundamental theorem. Let f(r) = r'. so, h^{-1} : R' -> R/K such that h^{-1}(r') = h^{-1}(f(r)) = h^{-1}(h(r+K)) = r + K. p(J) = p(J') ==> f(J) = f(J') ==> h^{-1}(f(J)) = h^{-1}(f(J')) ==> J + K = J' + K ==> J = J' (since K subset J, K subset J') It means that p is 1-1. ---------------------------------------------------------------------------- ------- I will show that M : maximal ideal ==> R/M : field. (R is a commutative ring with unity.) pf) Let g : R -> R/M, g(r) = r+M. so, g is a surjective ring homomorphism. Let A be the set of all ideals in R that contain M. Let B be the set of all ideals in R/M. so, A = {M, R} (since M is maximal.) Namely, |A| = 2. By 1-1, onto, |B| = 2. It means that R/M is a simple ring. Namely, it is nontrivial and has no proper nontrivial ideals. Since Field F <==> all ideals are {0}, F. (under F is a commutative ring with unity.) (=>) nontrivial ideal N ==> 1 in N ==> F = N. (<=) nonzero a in R ==> aR is an ideal of R ==> aR = R ==> ar = 1. so, R/M is field. === Subject: Re: Algebra with kernel, ideal, field. > Hello teacher~ R and R' are commutative ring with unity. f : R -> R' is a surjective ring homomorphism. Prove that there is a 1-1 correspondence > from the set of all ideals in R that contain K = ker f > onto the set of all ideals in R' ---------------------------------------------------------------------------- - --- > pf) > Let A be the set of all ideals in R that contain K = ker f. > Let B be the set of all ideals in R'. Let p : A -> B, p(J) = f(J) = {f(j) | j in J}. > I will show that p is bijective. > and then END. For an ideal J of R that contain K, > p(J) = f(J) is an ideal of R'. For an ideal N of R', > f^{-1}(N) is an ideal of R. > p[f^{-1}(N)] = f[f^{-1}(N)} = N. (since f is surjective.) > It means that p is surjective. Let h : R/K -> R', h(r+K) = f(r). > This is a isomorphism by fundamental theorem. > Let f(r) = r'. > so, h^{-1} : R' -> R/K such that > h^{-1}(r') = h^{-1}(f(r)) = h^{-1}(h(r+K)) = r + K. p(J) = p(J') ==> f(J) = f(J') > ==> h^{-1}(f(J)) = h^{-1}(f(J')) > ==> J + K = J' + K > ==> J = J' (since K subset J, K subset J') > It means that p is 1-1. OR.. f : R -> R' is a surjective ring homomorphism. g : R -> R/K, g(r) = r+K. Of course, g is a surjective ring homomorphism. Prove that there is a 1-1 correspondence from the set of all ideals in R that contain K = ker f onto the set of all ideals in R/K. pf) For an ideal J of R that contain K, g(J) = J/K is an ideal of R/K. For an ideal N of R/K, g^{-1}(N) is an ideal of R. Of cousre, K subset g^{-1}(N). g(J) = g(J') ==> J/K = J'/K ==> J = J' (since K subset J, K subset J') It means that there is a 1-1 correspondence from the set of all ideals in R that contain K = ker f onto the set of all ideals in R/K. AND h : R/K -> R' is a isomorphism. It means that there is a 1-1 correspondence from the set of all ideals in R/K onto the set of all ideals in R'. Because, h(J) = h(J') ==> h^{-1}(h(J)) = h^{-1}(h(J')) ==> J = J' so, it means that there is a 1-1 correspondence from the set of all ideals in R that contain K = ker f onto the set of all ideals in R' === Subject: What is this angle ? posting-account=6ItQJAoAAAD4KGGM_Aq7deF5BDrwThbQ CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) http://home.megapass.net/~kt97885535/q3.jpg In above link, ABCD, CDEF, EFGH are all square. Then, What is the angle p + angle q ? I wish to know the solution not by using triangular function. === Subject: Re: What is this angle ? Herb schrieb > In above link, > ABCD, CDEF, EFGH are all square. Then, What is the angle p + angle q ? > I wish to know the solution not by using triangular function. Reflect F and the line AF in the line AH. Draw F'G. The triangle AF'G has a right angle and tow equal sides. HTH Jutta === Subject: Re: What is this angle ? > Herb schrieb > In above link, > ABCD, CDEF, EFGH are all square. > Then, What is the angle p + angle q ? I wish to know the solution not by using triangular function. Reflect F and the line AF in the line AH. > Draw F'G. > The triangle AF'G has a right angle and two equal sides. > Hallo Jutta, This corresponds to what I answered to mina_world's post (the method I knew) and can be summarized as : complete the grid. Interesting is also mina's solution as stated into her post, here in sci.math, just ... yesterday ! (hence my joke about Herb being a student of mina, or copying mina) > Hello teacher~ -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Re: What is this angle ? > Hallo Jutta, This corresponds to what I answered to mina_world's post > (the method I knew) and can be summarized as : complete the grid. Interesting is also mina's solution as stated into her post, > here in sci.math, just ... yesterday ! > (hence my joke about Herb being a student of mina, or copying mina) Yes, I know :-) But I also understand that he did not look under Geometry with a+b = c. It does not say anything about angles. Jutta === Subject: Re: What is this angle ? > Interesting is also mina's solution as stated into her post, > here in sci.math, just ... yesterday ! Yes, I know :-) But I also understand that he did not look under > Geometry with a+b = c. It does not say anything about angles. Yes. One has to read inside, and as Herb uses Google to access sci.math, this doesn't help him much in sorting the interesting posts... -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Re: What is this angle ? >http://home.megapass.net/~kt97885535/q3.jpg >In above link, >ABCD, CDEF, EFGH are all square. Then, What is the angle p + angle q ? >I wish to know the solution not by using triangular function. 45 degrees or pi/4 radians, --Lynn === Subject: Re: What is this angle ? > http://home.megapass.net/~kt97885535/q3.jpg > In above link, ABCD, CDEF, EFGH are all square. > Then, What is the angle p + angle q ? > I wish to know the solution not by using triangular function. 45 degrees or pi/4 radians, Are you (Herb) a student of mina_world ? Or a plagiar ? She gave one answer (without trig). -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Re: lets turn the tables > Well, that's not /all/ you can do; you can also dive headlong into > threads you haven't read to complain that standard mathematicians > are hostile to new concepts, while backing up your claim by > referring to facts that never actually happened. I'm sure that will > help your reputation no end. To lwalke3, there is no difference between what write. No difference. Lwalke3 is not competent in mathematics; does not discriminate between a mathematical statement and a nonsensical statement that uses terms from mathematics. It is not enough to say that a mathematical argument does not sway lwalke3; a mathematical argument sounds like gibberish to lwalke3. It is a sequence of meaningless sounds. -- Michael Press === Subject: Re: lets turn the tables To lwalke3, there is no difference between what tommy1729/amy666 >competent in mathematics; does not discriminate between a >mathematical statement and a nonsensical statement that uses terms >from mathematics. It is not enough to say that a mathematical >argument does not sway lwalke3; a mathematical argument sounds like >gibberish to lwalke3. It is a sequence of meaningless sounds. I think you're fitting him up. This might as well be a witch > hunt. If he denies that someone else is a witch - why, then, > he must be a witch! Indeed there are many salient differences between lwalke3 and tommy1729. Lwalke3's idea that the unwillingness of standard mathematicians to attempt to invest tommy's gibberish with some coherent meaning is due to some dogmatic opposition to anything but ZFC is bizarre, but his mathematical musings are not at all meaningless gibberish for most of the time. It is another matter these musings seem entirely pointless -- there is no reason to assume a random unmotivated set theory to be of any mathematical, philosophical or conceptual significance or interest, even if by some miracle it happens to agree with this or that interpretation of these or those incoherent demands from the likes of tommy. It is possible to take some delight in such pointless exercises, a bit like some people enjoy solving sudokus, of course, but such an attitude is not very becoming if combined with tedious and silly rants about the evil of standard mathematicians, about their wanting everyone to use commercial newsclients so that non-professionals are excluded, about their objecting to the very notion there could be alternative ways of conceiving mathematics, or formal theories of sets beyond ZFC etc. > If he's going to defend cranks, he's going to seem like a crank. I have in the past defended cranks in the sense of pointing out when their arguments and questions are not answered in any way by tedious formal ZFC arcana, when there is a way of reading a particular piece of their rambling that is not at all silly, when their interlocutors are themselves being very silly, and so on. I doubt this has led many people to the conclusion I myself am a crank. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: lets turn the tables <871vwhplfg.fsf@alatheia.dsl.inet.fi> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > Indeed there are many salient differences between lwalke3 and > tommy1729. Lwalke3's idea that the unwillingness of standard > mathematicians to attempt to invest tommy's gibberish with some > coherent meaning is due to some dogmatic opposition to anything but > ZFC is bizarre, but his mathematical musings are not at all > meaningless gibberish for most of the time. It is another matter these musings seem entirely pointless -- there is > no reason to assume a random unmotivated set theory to be of any > mathematical, philosophical or conceptual significance or interest, > even if by some miracle it happens to agree with this or that > interpretation of these or those incoherent demands from the likes of > tommy. It is possible to take some delight in such pointless > exercises, a bit like some people enjoy solving sudokus, of course, > but such an attitude is not very becoming if combined with tedious and > silly rants about the evil of standard mathematicians, about their > wanting everyone to use commercial newsclients so that > non-professionals are excluded, about their objecting to the very > notion there could be alternative ways of conceiving mathematics, or > formal theories of sets beyond ZFC etc. Head of nail meets hammer. MoeBlee === Subject: Re: lets turn the tables <21537183.1228433314317.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) The truth is that given two identical borderline ideas, > whether the benefit of the doubt is given depends mainly > on reputation. Period. > Yes, and that's the way it should be. History has taught us that the > borderline ideas that turn out to not be nuts almost always turn out to > come from people established in the field. > When a borderline idea is presented by someone with a track record of > good work in the field, it's a safe bet that the proponent of the idea > has made sure already that it basically works--it is consistent with > what is known, has some uses, etc. When the exact same idea comes from > someone who isn't known and can't be connected to someone known, it > usually turns out that they haven't checked it competently to make sure > it is viable, or discovered any uses for it. Of course people are going > to be much less likely to spend time checking it out, as that time is > highly likely to be wasted. > And so this means that no new theory that I, or anyone > else labeled a crank, will ever be taken seriously, > since we don't have a high enough repuation to propose > new theories. AGAIN, you show that, when it suits you, you just can't resist an opportunity to distort other's postings. MoeBlee === Subject: Re: #30 whether calculus & trigonometry are unique to Euclidean Geometry; new book 2nd edition: Math a subset of Physics, AP-adics posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1 FOH:R051,gzip(gfe),gzip(gfe) http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html and, click on pentagramma mirificun at http://wlym.com > One of the troubles of Elliptic Geometry is the triangle concept > itself, for there exists diangles or > lunes where a triangle is only two sides. So, is the trigonometry and calculus unique branches of mathematics to > that of Euclidean Geometry thus: nothing could be simpler; simply, construct an arbitrary unitlength from one end of your unit-to-be-subdivided, and add two additional unitlengths to the end; then, construct parallels ... figure it out with a sketch, but it's probably in your wookiepoopeya reference. and, division is the inverse construction. > http://en.wikipedia.org/wiki/Compass_and_straightedge thus: there is absolutely nothing deterministic about billiards, mainly because of english. perhaps you are referring to the mathematical idealization thereof, using point masses (a.k.a. The Photon, the only truly zero-D object in Universe .-) thus: seriously, any polynomial with a constant (or times x to the zeroth power) cannot have zero as a solution; otherwise, zero is a trivial solution, since zero times any coefficient is nothing! anyway, what's the problem, if an infinite series polynomial is not algebraic per definition? thus: Euler's identity is, what -- exp(i*pi) = 1, i*pi = ln(1) pi = -i*ln(1), but I forgot what the natural log of one is. thus: there's got to be a few Latin phrases for this kind of fallacy; this is the same thing that Are Buckafka Fullofit tried to put over, and most of the Buckywitches really believe it. anyway, the queestion is, why should line segments of polygons be so God-am existential, or of polyhedra -- because of the typical stick-drawn configuration in some edition of Euclid? >Also, from the position taken above, spheres are only possible in theory aswell. --Cheeny & Zbiggy, fo'mo' years! http://tarpley.net http://larouchepub.com http://xplodeyourmlmbiz.com/?s1=Fuller+Brush&t=LSP01&gclid=CPiar6nispcCFQu-G godYm21kA http://www.rwgrayprojects.com/synergetics/plates/plates.html === Subject: outline of an idea for a proof involving Riemann Integrals Let f be Riemann integrable on [a,b]. Assume that g agrees with f everywhere on [a,b] except for a finite number of points. Show that g is also Riemann integrable on [a,b] and that the integral of g(x) on the interval [a,b] g(x) equals the integral of f(x) on the interval [a,b]. The approach I was going to take was to use the definition of the Riemann integral and create a partition of [a,b] that had small fractional parts of epsilon, say, epsilon/20 as the width of the intervals in the partition containing the points in which g does not agree with f. I would then argue that the integral of g - integral of f on [a,b] is within epsilon and thus integral f =integral g. Does this make sense and sound like a good way to do this? === Subject: Re: outline of an idea for a proof involving Riemann Integrals > Let f be Riemann integrable on [a,b]. Assume that g agrees with > f everywhere on [a,b] except for a finite number of points. Show that > g is also Riemann integrable on [a,b] and that the integral of g(x) on > the interval [a,b] g(x) equals > the integral of f(x) on the interval [a,b]. The approach I was going to take was to use the definition of the > Riemann integral > and create a partition of [a,b] that had small fractional parts of > epsilon, say, epsilon/20 > as the width of the intervals in the partition containing the points in > which g does not agree with f. > I would then argue that the integral of g - integral of f on [a,b] is > within epsilon and thus > integral f =integral g. > Does this make sense and sound like a good way to do this? It makes sense, but I don't see where the value 20 comes from. Indeed, instead of using epsilon/20 you should use epsilon/something where the something depends upon the number of points at which f(x) is different from g(x) as well as some M > 0 such that |f(x)|,|g(x)| < M, for each _x_ in [a,b]. It would be simpler if you would prove it by induction on the number of points _x_ such that f(x) is different from g(x). Jose Carlos Santos === Subject: Newton Raphon's formula for one equation, three variables posting-account=a93YEwoAAAClHm9Euy--V39SJpP16NI8 Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) I'm interested in finding the nearest point on an an implicit surface F (x,y,z) = 0 , from an initial point p(x,y,z). The algorithm I'm implementing -- A marching method for the triangulation of (Implicit) surfaces by E. Hartmann, makes extensive use of Newton's method in order to find points on an implicit surface. As the algorithm glosses over details of Netwon's method in this case i.e. one (implicit surface/equation) and three variables: x, y, z, I'm having difficulty implementing the algorithm and debugging my code. I was wondering any one would be kind enough help with this formula. I've looked up a few resources e.g. Numerical recipes etc, but not much attention is given to the specific version of the method I'm interested in. - Olumide === Subject: Re: Newton Raphon's formula for one equation, three variables I'm interested in finding the nearest point on an an implicit surface F >(x,y,z) = 0 , from an initial point p(x,y,z). The algorithm I'm >implementing -- A marching method for the triangulation of (Implicit) >surfaces by E. Hartmann, makes extensive use of Newton's method in >order to find points on an implicit surface. As the algorithm glosses >over details of Netwon's method in this case i.e. one (implicit >surface/equation) and three variables: x, y, z, I'm having difficulty >implementing the algorithm and debugging my code. I was wondering any >one would be kind enough help with this formula. I've looked up a few >resources e.g. Numerical recipes etc, but not much attention is given >to the specific version of the method I'm interested in. >- Olumide besides the hint you already obtained: your problem sounds more like if you would are in need of orthogonal distance regression odrpack: (if not one, but many points are involved ) your problem , seen for one point, is the minimization of the distance 0.5*||p(x,y,z)-(x^*,y^*,z^*)||^2 under the constraint F(x^*,y^*,z^*)=0 and if you indeed want it to solve this for one point, then you could either apply constraint optimization (the big hammer) or Newtons method for the equations of the Lagrange multiplier rule. p(x,y,z)-(x^*,y^*,z^*) - grad F((x^*,y^*,z^*)) * mu =0 F((x^*,y^*,z^*))=0 four equations in the four unknowns (x^*,y^*,z^*) and mu. the computation of the Jacobian is [ -Hessian(F((x^*,y^*,z^*))*mu -I , - grad F((x^*,y^*,z^*)) ] [ - grad F((x^*,y^*,z^*))^T , 0 ] The unknowns ordered as (x^*,y^*,z^*,mu) . you could apply marching along - grad F(p(x,y,z)) until F changes sign (this assumes some regularity of your surface: I assume the on one side of the surface can be identified with the sign of F (or: ||grad(F(...)))|| nonzero everythere) in order to get a reasonable initial guess. hth peter === Subject: Re: Newton Raphon's formula for one equation, three variables posting-account=DLD3MQkAAACJxulKt9xouw3DPpXKssCI SLCC1; .NET CLR 2.0.50727; .NET CLR 1.1.4322; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) I'm interested in finding the nearest point on an an implicit surface F > (x,y,z) = 0 , from an initial point p(x,y,z). The algorithm I'm > implementing -- A marching method for the triangulation of (Implicit) > surfaces by E. Hartmann, makes extensive use of Newton's method in > order to find points on an implicit surface. As the algorithm glosses > over details of Netwon's method in this case i.e. one (implicit > surface/equation) and three variables: x, y, z, I'm having difficulty > implementing the algorithm and debugging my code. I was wondering any > one would be kind enough help with this formula. I've looked up a few > resources e.g. Numerical recipes etc, but not much attention is given > to the specific version of the method I'm interested in. The Pascal source code is on his web site: http://www.mathematik.tu-darmstadt.de/~ehartmann/ in this archive: http://www.mathematik.tu-darmstadt.de/~ehartmann/cdg0e/cdg0egv.tgz and specifically: c:dcorbit64mathcdg0egv>c:utilsgrep.exe -d -i implicit surfaces *.* File examplestriblend pov.p: writeln('*** Triangulation of implicit surfaces *** '); File examplestrisample.p: writeln(' *** Triangulation of implicit surfaces *** '); === Subject: Some simple questions about Banach's Space Good morning, I have some simple questions about Banach's Space. Below citations are from Krzysztof Maurin's book Analiza. Elementy (there is english edition Analysis. Elements but I don't have one). The translation is mine. The notation is not exactly as it is given in the book due to the natural limits of the Usenet's writings. I will be very grateful for all responses and advices. L(X, Y) is the space of linear and continous functions from X to Y. Pages 120-124 ______________________________________________________________ Theorem VII. I. Let (X, ||*||) and (Y, ||*||) to be Banach's Spaces and L: X->Y to be a linear function. Then following conditions are equivalent: 1) from sequences which convergences to zero L returns sequences which are limited what mean that for each sequence x_n->0 there exists C>0 such that: ||Lx_n||<=C, n=1, 2. 1)=>2) Let assume that L is not contiuous in the zero point. It means that there exists a sequence x_n->0 such that ~(Lx->L(0)=0). Thus we can take such subsequence x_n->0 of the sequence x_n that there exists C>0 for which ||L_x_n_k||>C is satisfied. Let b_k:=(||x_n_k||)^(-1/2). We can see that b_k->+inf. On the other hand a sequence d_k:=b_k*x_n_k->0. It is because: ||d_k||=b_k*||x_n_k||=||x_n_k||/(||x_n_k||^(1/2))->0. Thus: ||Ld_k||=||L(b_k*x_n_k)||=b_k*||Lx_n_k||>=b_k*C -> +inf. So we have contradiction with condition 1). _________________________________________________________________ I don't understand - why b_k -> +inf ? (x_n->0 but it doens't mean that ||x_n_k||->0). _____________________________________________________________________ 2)=>3) Let X to be continuous in x_o in X point. For each e>0 there exists d>0 such that ||L(x-x_0||1) If x_n->x_0 the due to the fact that function L is continuous and linear we have Lx_n->Lx_0. But we know that the sequence which is convergenced is limited. ______________________________________________________________________ OK, but in my opinion it is not important that L is linear. Am I wrong? _______________________________________________________________________ Let (X, ||*|| to be a n-dimensional Banach's Space. In X we take any base e_1, e_2, ..., e_n. We introduce a new norm: ||x_1||=sup x_i where i=1,...,n and numbers x_i are coordinates of the vector x in the base e_1, e_2, .., e_n; x=SIGMA x_i*e_i where i=1,...,n. Due to the inequality ||x||<=sup|x|*sup|e_i|*n the norm ||*||_1 is stronger than norm ||*||. Now we have to prove that the reversed implication is true. For that we consider a function F: X->R^n, X in x -> F(x)=(x_1, x_2, ..., x_n) in R^n; it's obvious that F is linear - we want to prove that it is continuous in the norm ||*||. Let assume that F is not continuous. So there exists a sequence x_n->0 such that ~(F(x_n)->0). Thus we are able to take a subsequence x_n_k such that there exists a>0 for which ||F(x_n_k)||>a is satisfied. Let y_k=(x_n_k)*(||F(x_n_k)||)^(-1). Then we have y_k->0 and ||F(y_n)||=1. But a sphere in which r=1 is compact in R^n thus we are able to take such subsequence of sequence F(y_k) that it will be convergence to z in R^n, ||z||=1. We remember that we have shown that function F^(-1) is continuous. So F^(-1)(z)=0 but on the other hand F^(-1)(z)=SIGMA z_i*e_i. Due to the fact that the zero vector has explicit representation as the linear combination of the base vectors we have z_i=0 (i=1, 2, ..., n). So we have a contradiction with equality ||z||=1. That contradiction shows that function F is continuous. So there exists such A>0 that ||F(x)||=sup|x_i|<=A||x|| where i=1,2,...,n. ____________________________________________________________________________ ____ In the fact I don't understand most of the content of the above proof. In particular I don't understand: - why inequality ||x||<=sup|x|*sup|e_i|*n is true, - why y_k->0 (x_n_k->0, but we know nothing about (||F(x_n_k)||)^(-1), am I wrong? - I completely don't understand why we can talk about the sphere, - I don't know where and how we have shown that the function F^(-1) is continuous. ____________________________________________________________________________ _______ Point 3) of the theorem VII.I implies that for L in L(X, Y) there exists such A>0 that ||Lx|<=A||x||. The smallest A>0 which satisfy that condition for all x in X we call the norm of the function L: ||L||:=inf_A {||Lx||<=A||x||; x in X} We show that ||L||=sup_(||x||<=1) ||Lx||. For that we divide both sides of our inequality ||Lx||<=||L||||x|| by ||x||; we have: ||L(x/||x||)||<=||L||. If we take supremum of both sides we have: sup_(xin X) ||L(x/||x||)<=||L||; we show now that it is impossible that < inequality is true. Let A=sup_(x in X) ||L(x/||x||)<||L||. Then: ||Lx||<=A||x||. But that is a contradiction with the definition A as the smallest constants which satisfy (1) ____________________________________________________________________________ ______ I don't understand why that is the contradiction with the definition of A. Warning: in the below proof there is that it is necessary to show only that L is completeness due to the fact that we use the above norm (that means the norm of the function L). Before that theorem there is the proof that ||L|| is the norm. ____________________________________________________________________________ ______ Theorem VII.2. Space L(X,Y) is the Banach's Space. Proof (it is necessary to show only that L is completeness). Let L in L(X, Y), n=1,2 and let L_n to be a Cauchy's sequence in L(X, Y) what means that for each e>0 there exists such N>0 that ||L_n-L_m||N. Due to the inequality ||L_n x - L_m x||<=||L_n-L_m||||x||inf) L_n x. We define a function: Lx:=lim(n->inf) L_n x. That's obvious that a function L is linear; we have to show that it is continuous. A modified triangle inequality: abs(||L_n-L_m||)<=||L_n-L_m||N shows that a sequence (||L_n|| is a Cauchy's sequence of the real numbers. Let A:=lim(n->inf) ||L_n||. We have: ||Lx||=||lim(n->inf) L_n x||=lim (n->inf) ||L_n x||<= lim(n->inf) ||L_n|*|||x||=A*||x||, but ||L_m x - L_n x||N. ____________________________________________________________________________ ______ I don't understand the condition that ||x||<=1. ____________________________________________________________________________ ______ For each e>0 and for each x in X there exists such M>0 that ||Lx-L_m x||M ____________________________________________________________________________ ______ I don't understand why above inequality is true. Page 125 ____________________________________________________________________________ ______ 3. The differentiation of the functions in the Banach's Space Definition. Let U be open set in the Banach's Space X. We say that the function T of the set U in Banach's Space Y is differentiable in point x_o in U only if there exists a linear and continuous function L_x_0 of the Banach's Space X in the Banach's Space Y such that T(x_0+h)-T(x_0)=L(x_0)(h)+r(x_0;h), where lim(h->0) (||r(x_0, h)/||h||)=0. ____________________________________________________________________________ ______ OK, but I don't understand why the condition that the set is open is essential. === Subject: Re: Some simple questions about Banach's Space posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) More answers follow... [...] > Point 3) of the theorem VII.I implies that for L in L(X, Y) there exists such > A>0 that ||Lx|<=A||x||. The smallest A>0 which satisfy that condition for all x in X > we call the norm of the function L: ||L||:=inf A {||Lx||<=A||x||; x in X} We show that ||L||=sup (||x||<=1) ||Lx||. For that we divide both sides of our > inequality ||Lx||<=||L||||x|| by ||x||; we have: ||L(x/||x||)||<=||L||. If we take supremum of both sides we have: sup (xin X) ||L(x/||x||)<=||L||; we show now that it is impossible that < inequality is true. Let A=sup (x in X) ||L(x/||x||)<||L||. Then: ||Lx||<=A||x||. But that is a contradiction with the definition A as the smallest constants which > satisfy (1) > I don't understand why that is the contradiction with the definition of A. I don't think it's a contradiction with the definition of A, I think it's a contradiction with the definition of ||L||. The author has shown that ||Lx|| <= A||x|| for all x in X. Therefore A is in the set {B | ||Lx|| <= B||x|| for all x in X}. But ||L|| is by definition a lower bound for that set, so we must have A >= ||L||, which contradicts A < ||L||. > Warning: in the below proof there is that it is necessary to show only that > L is completeness due to the fact that we use the above norm (that means the norm > of the function L). Before that theorem there is the proof that ||L|| is the norm. Theorem VII.2. Space L(X,Y) is the Banach's Space. > Proof (it is necessary to show only that L is completeness). Let L in L(X, Y), > n=1,2 and let L n to be a Cauchy's sequence in L(X, Y) what means > that for each e>0 there exists such N>0 that ||L n-L m||N. Due to the > inequality ||L n x - L m x||<=||L n-L m||||x|| sequence in Y. > I don't understand: L n x means a sequence of the values L n x for x which is > fixed or that means a sequence of the functions L n. The first one. It means the sequence L 1 x, L 2 x, L 3 x, etc. > Due to the comleteness of the Space Y there exists a point y in Y such > that y=lim(n->inf) L n x. We define a function: Lx:=lim(n->inf) L n x. That's obvious that a function L is linear; we have to show that it is continuous. > A modified triangle inequality: abs(||L n-L m||)<=||L n-L m||N shows that a sequence (||L n|| is a Cauchy's sequence of the real numbers. Let > A:=lim(n->inf) ||L n||. We have: ||Lx||=||lim(n->inf) L n x||=lim (n->inf) ||L n x||<= > lim(n->inf) ||L n|*|||x||=A*||x||, but ||L m x - L n x||N. > I don't understand the condition that ||x||<=1. ||L m x - L n x|| = ||(L m - L n)x|| <= ||L m - L n||*||x|| by definition of ||L m - L n||. The author has shown that ||L m - L n|| < e for n,m > N, so ||L m x - L m x|| < e*||x|| for all x. When ||x|| <= 1 we have ||L m x - L n x|| < e. > For each e>0 and for each x in X there exists such M>0 that ||Lx-L m x||M > I don't understand why above inequality is true. The function L is defined pointwise by Lx:=lim(n->inf) L n x. So for each x we have L n x -> L x as n -> infinity. The above inequality (or more precisely the existence of M such that the above inequality is true) is just the definition of L n x -> L x. > Page 125 > > 3. The differentiation of the functions in the Banach's Space > Definition. Let U be open set in the Banach's Space X. We say that the function T of > the set U in Banach's Space Y is differentiable in point x o in U only if there exists > a linear and continuous function L x 0 of the Banach's Space X in the Banach's Space Y > such that T(x 0+h)-T(x 0)=L(x 0)(h)+r(x 0;h), where lim(h->0) (||r(x 0, h)/||h||)=0. > OK, but I don't understand why the condition that the set is open is essential. It's just part of the definition. Presumably it will be used later in the book in the proofs of some theorems, theorems which would be false if the requirement that U be open were relaxed. But I don't know enough about this stuff to guess what those theorems might be. === Subject: Re: Some simple questions about Banach's Space posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > Good morning, > I have some simple questions about Banach's Space. Below > citations are from Krzysztof Maurin's book Analiza. Elementy > (there is english edition Analysis. Elements but I don't have > one). The translation is mine. The notation is not exactly as it > is given in the book due to the natural limits of the Usenet's > writings. I will be very grateful for all responses and advices. I'll try to answer the first two parts now, and return to the rest later if I have time. > L(X, Y) is the space of linear and continous functions > from X to Y. Pages 120-124 > ______________________________________________________________ > Theorem VII. I. Let (X, ||*||) and (Y, ||*||) to be > Banach's Spaces and L: X->Y to be a linear function. Then > following conditions are equivalent: 1) from sequences which convergences to zero L returns sequences > which are limited what mean that for each sequence x_n->0 there > exists C>0 such that: ||Lx_n||<=C, n=1, 2. The term for sequences which satisfy the above requirement is bounded. By the way, you haven't said what conditions 2), 3) and 4) are. > 1)=>2) Let assume that L is not contiuous in the zero point. It > means that there exists a sequence x_n->0 such that ~(Lx->L(0)=0). > Thus we can take such subsequence x_n->0 of the sequence x_n that > there exists C>0 for which ||L_x_n_k||>C is satisfied. Let > b_k:=(||x_n_k||)^(-1/2). We can see that b_k->+inf. On the other > hand a sequence d_k:=b_k*x_n_k->0. It is because: ||d_k||=b_k*||x_n_k||=||x_n_k||/(||x_n_k||^(1/2))->0. Thus: > ||Ld_k||=||L(b_k*x_n_k)||=b_k*||Lx_n_k||>=b_k*C -> +inf. So we have contradiction with condition 1). > _________________________________________________________________ I don't understand - why b_k -> +inf ? (x_n->0 but it doens't mean > that ||x_n_k||->0). Yes it does. x_n -> 0 means that for every e > 0 there is an N such that ||x_n - 0|| = ||x_n|| < e whenever n >= N. But this is the same as the definition of ||x_n|| -> 0. So ||x_n|| -> 0, and if x_n_k is a subsequence of x_n then n_k >= k for every k, so that if k >= N then n_k >= N. For every e > 0 we can find N such that ||x_n|| < e whenever n >= N, so also ||x_n_k|| < e whenever k >= N, so that ||x_n_k|| also converges to 0. > _____________________________________________________________________ > 2)=>3) Let X to be continuous in x_o in X point. For each e>0 there > exists d>0 such that ||L(x-x_0|| x-x_0:=u. Then we have ||Lu|| _____________________________________________________________________ I don't understand why ||(x/||x)*(d/2)|| ____________________________________________________________________ ______________________________________________________________________ > 4)=>1) If x_n->x_0 the due to the fact that function L is continuous > and linear we have Lx_n->Lx_0. But we know that the sequence which is > convergenced is limited. > ______________________________________________________________________ OK, but in my opinion it is not important that L is linear. Am I wrong? I don't think it's possible to answer this without knowing what 4) says. > _______________________________________________________________________ > Let (X, ||*|| to be a n-dimensional Banach's Space. In X we take any > base e_1, e_2, ..., e_n. We introduce a new norm: ||x_1||=sup x_i where i=1,...,n I think this should be ||x||_1=sup |x_i| where i=1,...,n. and numbers x_i are coordinates of the vector x in the base e_1, e_2, > .., e_n; x=SIGMA x_i*e_i where i=1,...,n. Due to the inequality ||x||<=sup|x|*sup|e_i|*n the norm ||*||_1 is > stronger than norm ||*||. Now we have to prove that the reversed > implication is true. For that we consider a function F: X->R^n, > X in x -> F(x)=(x_1, x_2, ..., x_n) in R^n; it's obvious that > F is linear - we want to prove that it is continuous in the norm > ||*||. Let assume that F is not continuous. So there exists a sequence > x_n->0 such that ~(F(x_n)->0). Thus we are able to take a subsequence > x_n_k such that there exists a>0 for which ||F(x_n_k)||>a is satisfied. > Let y_k=(x_n_k)*(||F(x_n_k)||)^(-1). Then we have y_k->0 and ||F(y_n)||=1. > But a sphere in which r=1 is compact in R^n thus we are able to take > such subsequence of sequence F(y_k) that it will be convergence to > z in R^n, ||z||=1. We remember that we have shown that function F^(-1) > is continuous. So F^(-1)(z)=0 but on the other hand > F^(-1)(z)=SIGMA z_i*e_i. Due to the fact that the zero vector has explicit > representation as the linear combination of the base vectors we have > z_i=0 (i=1, 2, ..., n). So we have a contradiction with equality ||z||=1. > That contradiction shows that function F is continuous. So there exists > such A>0 that ||F(x)||=sup|x_i|<=A||x|| where i=1,2,...,n. > ____________________________________________________________________________ _ ___ In the fact I don't understand most of the content of the above proof. > In particular I don't understand: > - why inequality ||x||<=sup|x|*sup|e_i|*n is true, I think this is wrongly transcribed and should read ||x|| <= sup|x_i| *sup||e_i||*n. Assuming that my correction is what you mean, here's how to prove it: ||x|| = ||x_1*e_1 + x_2*e_2 + ...|| <= |x_1|*||e_1|| + |x_2|*||e_2|| + ... <= |x_1|*sup||e_i|| + |x_2|*sup||e_i|| + ... = (|x_1| + |x_2| + ...)*sup||e_i|| <= (sup|x_i| + sup|x_i| + ...)*sup||e_i|| = n*sup|x_i|*sup||e_i|| > - why y_k->0 (x_n_k->0, but we know nothing about (||F(x_n_k)||)^(-1), am I wrong? Since x_n_k -> 0, for every e > 0 there is an N such that ||x_n_k||_1 < e whenever k >= N. Since ||F(x_n_k)|| > a we have that ||y_k||_1 = || x_n_k||_1/||F(x_n_k)|| < ||x_n_k||_1/a < e/a whenever k >= N. So given any d > 0 let e = a*d, and there exists N such that ||y_k||_1 < e/a = d whenever k >= N. > - I completely don't understand why we can talk about the sphere, The sphere means the set {x in R^n | ||x|| = 1}. It is compact by the Heine-Borel theorem. > - I don't know where and how we have shown that the function F^(-1) is continuous. Suppose that x_n is a sequence in R^n such that x_n -> 0. If x_n = (x_n_1,x_n_2,...) then sup_i |x_n_i| -> 0, since sup_i |x_i| = |x_i| for some i in {1,2,...n} = (|x_i|^2)^(1/2) <= (|x_1|^2 + |x_2|^2 + ...)^(1/2) = ||x||. Note that F^{-1}(x_n) = x_n_1*e_1 + x_n_2*e_2 + ... so that ||F^{-1} (x_n)||_1 = sup_i |x_n_i|, so we have that ||F^{-1}(x_n)||_1 -> 0. Now since ||F^{-1}(x_n)|| <= C*||F^{-1}(x_n)||_1, where C = n*sup||e_i|| is constant, it follows that ||F^{-1}(x_n)|| ->, as required to show that F^{-1} is continuous. === Subject: Re: Some simple questions about Banach's Space posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) Correction: [...] - why y_k->0 (x_n_k->0, but we know nothing about (||F(x_n_k)||)^(-1), am I wrong? Since x_n_k -> 0, for every e > 0 there is an N such that ||x_n_k||_1 > < e whenever k >= N. Since ||F(x_n_k)|| > a we have that ||y_k||_1 = || > x_n_k||_1/||F(x_n_k)|| < ||x_n_k||_1/a < e/a whenever k >= N. So given > any d > 0 let e = a*d, and there exists N such that ||y_k||_1 < e/a = > d whenever k >= N. Every occurrence of ||.||_l in the above paragraph should be replaced with ||.||. === Subject: Questions About Online Teaching Jobs I have several questions about online teaching jobs, specifically about teaching mathematics online. I am currently searching for such a job. 1. How competitive are these jobs as compared to traditional mathematics teaching jobs? I suspect the competition is nowhere as fierce, but I am wondering if I am correct. 2. How good does it look to state such a job on my vita? In other words, how much will it increase my chances, if any, of earning a permanent teaching job someday at a community college or four-year college or university? I don't plan to teach mathematics online forever, just for a while. 3. My mom thinks it's good going for an online teaching job since she suspects all colleges will be entirely online someday, maybe in the next 10 years or so. I seriously doubt that. I do believe it is very well possible that someday, maybe in the next 10 years or so, most classes will be online, but I doubt classroom teaching will ever go exinct, especially in mathematics. Classroom teaching allows students to see and get to know the professor in a way that online classes do not allow, and many students need face-to-face interaction with their professors, especially in mathematics, to have any chance to succeed. In other words, they need help from their professors in a way that is similar to visiting them in their office hours. Does anyone here agree or diagree with me? Jonathan Groves === Subject: question about algorithm for computing zeta function I've been looking at a preprint by Ghaith Ayesh Hiary, who was advised in his work by Andrew Odlyzko. The title is: Fast methods to compute the Riemann zeta function, at: http://arxiv.org/abs/0711.5005v2 The algorithm is for computing zeta(1/2 + it), t>0. In the Introduction, the Riemann-Siegel method is mentioned. I'm trying to understand the simplest of the three methods presented: the one where beta = 1/10. I understand that a pre-computed table is necessary, say 0.05 Gigabytes ( in the discussion, page 6). I know that K is an integer, and L is a lattice. I think L is a subset of [0,1]x[0,1]. The lattice L is described between equations (15) and (16), and it read: let (a~ , b~) be a point in L closest to (a,b). Maybe in the definition of L, delta and tau take on all integer values in the range? There are really three algorithms, with different values of beta. For beta = 1/10, complexity~ t^(2/5 + o(1)) , but it's relatively simple. beta = 1/6 gives a t^(1/3 + o(1)) and a bigger beta gives a method faster than Odlyzko and Schonhage. David Bernier === Subject: Re: failure functions Let phi(x) be a function of x (belonging to Z).Let the definition of a failure be a composite number. Psi(x) = x_0 + k.phi(x_0) ( k belongs to N) is a failure function since values of x generated by psi(x) will be such that when substituted in phi(x), phi(x) will generate only failures (composite numbers). ´ ´ Before furnishing a proof let me give a numerical example: ´ ´ Let phi(x) = x^2 + x + 11 . phi(1) = 13. x = psi(1) = ´ ´ 1 + k.13 = 14, 27, 40, 53.......... When we substitue these values ´ ´ of x in phi (x) we get only failures (composites). We also have the ´ ´ additional information that all of them are multiples of 13. A.K.Devaraj ( to be continued ) === Subject: Re: failure functions ? additional information that all of them are multiples >of 13. >A.K.Devaraj ( to be continued ) phi(10)=121 Vincenzo Librandi === Subject: g(x,y) variables separation and iteration... posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) I think there are some links between separability and iterates forms. Be g(x,y) a smooth function and exists f such as : f(g(x,y)) = f(x) + h(y) , f bijective . Example: g(x,y) = x/(1 +yx) gives 1/g = 1/x + y If f(u) =1/u , g(x,y) = f^ [-1] (f(x)+y) = (x/(1+x))^[y] [y] y th iterate , In case f(g(x,y)) = m(x) + h(y) , m and h non constant functions, we may write g(m^-[1](f(x) ,y) = f^[-1](f(x) + h(y)) = { f^[-1](f(x) + 1)) } ^ [h(y)] , Any comments ? Alain === Subject: Why didn't ancient Greek Mathematicians use a string instead of a compass for their constructions? posting-account=hEsTJAoAAABBB9neOo9d5YFd2CC8eBZx Gecko/20061205 Iceweasel/2.0.0.1 (Debian-2.0.0.1+dfsg-1),gzip(gfe),gzip(gfe) As you could tell from my previous question: ~ sci.math: How could you multiply and/or divide using straightedge and compass only? ~ I am interested in the details of why have we Mathematicians used certain means and ways both logical and mechanical to operate and express ourselves ~ Using a string would have let them produce not only circumferences, but also ellipses, so they wouldn't need conic sections ~ Pythagorean triplets were known and used to survey land using strings possibly way back before Pythagoras himself knew of them, so, I guess they certainly knew they could use a string. Why didn't they? ~ Is it possible that they found conics as a natural part of an extension of 2d geometry into 3d and they could produce ellipses from them and/or is it a consequence of the huge influence that regular polygons and in general Platonic philosophy had in those times? ~ lbrtchx === Subject: Re: Why didn't ancient Greek Mathematicians use a string instead of a compass for their constructions? posting-account=1wPXHwkAAACFV0NiGWX7tZb1o0HYkMjT 6.6; .NET CLR 1.1.4322) w:PACBHO60,gzip(gfe),gzip(gfe) > As you could tell from my previous question: > ~ > sci.math: How could you multiply and/or divide using straightedge and > compass only? > ~ > I am interested in the details of why have we Mathematicians used > certain means and ways both logical and mechanical to operate and > express ourselves > ~ > Using a string would have let them produce not only circumferences, > but also ellipses, so they wouldn't need conic sections > ~ > Pythagorean triplets were known and used to survey land using > strings possibly way back before Pythagoras himself knew of them, so, > I guess they certainly knew they could use a string. Why didn't they? > ~ > Is it possible that they found conics as a natural part of an > extension of 2d geometry into 3d and they could produce ellipses from > them and/or is it a consequence of the huge influence that regular > polygons and in general Platonic philosophy had in those times? > ~ > lbrtchx Geometry was the issue. A compass allows an angle to form the part. A single inference of parts then causes. A string as a part must be relatable to allow a like inference. What placment of length would allow another parts relative size. A true relative angle to part as a science of geometry defines the science. A like analogy could be constructed using a string? How? Place the string. What relation would be used? I can think of NO possible relation. A mere string as the part itself was to be caused in exact inference. A compass was examined and found a unique part effector. It is for this reason geometry IS NOT mathematical. === Subject: Re: Why didn't ancient Greek Mathematicians use a string instead of a compass for their constructions? > As you could tell from my previous question: > I am interested in the details of why have we Mathematicians used > certain means and ways both logical and mechanical to operate and > express ourselves > ~ > Using a string would have let them produce not only circumferences, > but also ellipses, so they wouldn't need conic sections I'm not sure the study of conic sections by the ancient was motivated by extending the compass and straightedge. Geometry was the issue. > ... > It is for this reason geometry IS NOT mathematical. Humm. I leave this opinion to YOU. However a string (apart from the added ellipses(*)) allows to construct a lot of things too. Of course straight lines (tightening the rope, that's the way gardners and other house builders draw straight lines, with a rope soaked with chalk). Of course a circle. So all what can be drawn with compass and straight edge can be drawn with rope alone. But as you said, it allows even more, as does just paper folding for instance (allows trisection of angles, and construction of 7 sides regular polygons). Some times ago, I played with this idea : http://mathafou.free.fr/pbm_en/pb220.html However for the history of maths, I think nobody knows *exactly* what was in the mind of ancien geometers. I've been told that the discovery of irrationnality (sqrt(2)) gave them a big shock ! Hence they tried to prevent the occurrence of such further shocks by not allowing too powerfull tools, and limiting to just compass and straightedge. The Idea is as is. In that time, geometry was quite contemplative, many proofs being just look from a well drawn figure. Hence the search of exact constructions in that time. Then much effort has been made to make the geometry more rigourous. First one being Euclides, then more recently Hilbert etc. Then the geometry today is much more algebraic, considering sets and metrics and manifolds etc. (but it IS mathematical) Considering geometry constructions as just a game, we need however to fix the rules. Changing the rules changes the results (what can and can't be constructed). As well known, the use of rope (the 12 knots rope) to construct right angles was a current practice. Why didn't they choose the 'rope only' constructions ? To mention also that even the ancient tried to overcome these limits, by using specific curves (trisectrix, quadratrix etc) instead of just circles and straight lines, also Neusis constructions like the well known angle trisection by Archimedes (humm, not sure he was the first) with a marked straightedge etc. As there are mechanical tools to draw continuously conic sections, you could use construct by conic sections instead of just compass and straightedge (straight lines and circles being degenerate cases of conic sections, we call that just conic sections). But these are quite recent discoveries. A big problem also is that a straightedge is quite hard to make. (How do you ensure it is perfectly straight ?) But a compass is much more robust. Just suffice it is enough rigid. Hence many geometers tried to discard the straightedge. I mentioned Mascheroni in another post. The study of inversion also gave a mean to *construct* a straight line from scratch (from a circle). The world of construction rules, and geometry fundaments is quite wide... (*) draw an ellipse with a rope. Proove that it is a true ellipse, considering the diameter of the pencil etc... The right method : use a LOOP of rope, going around 3 poles of same non null diameter : two fixed poles, one moving (the pencil). It is not if the 3 diameters are unequal. -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: solutions manual (To search click in keyboard Ctrl+F) posting-account=jGYvqQoAAAC8JVWtG92jlQ1UvZHQAEv9 .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) solutions manual (To search click in keyboard Ctrl+F) Solutions Manuals in Electronic (PDF)Format! Just contact with , servicepp (at) hotmail.com (my email address), these are parts of our solutions, if the solution you want is on the list, please email to me. NOTICE: if the solutions manual that in my list ,please note it in your email . 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Stevenson Jr. computer system architecture 3rd by M.Morris Mano === Subject: Geometry with paper folding, 90. Hello teacher~ http://board-2.blueweb.co.kr/user/math565/data/math/90.jpg Hello teacher~ http://board-2.blueweb.co.kr/user/math565/data/math/90.jpg How do you sure it ? Your 1st drawing looks strange : C should be on base side of course (because of the 90Á angle shown), but is OK if you consider a 3D view before the paper is fully fold. Now for the proof : This doesn't depend on the first angle being 90Á !! Consider my folding. Of course NC = ND and EC = ED (because folded one onto the other). Hence CND is isosceles, and the median NE is also the altitude. -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Re: Geometry with paper folding, 90. > Hello teacher~ http://board-2.blueweb.co.kr/user/math565/data/math/90.jpg How do you sure it ? Your 1st drawing looks strange : > C should be on base side of course (because of the 90Á angle shown), > but is OK if you consider a 3D view before the paper is fully fold. Now for the proof : > This doesn't depend on the first angle being 90Á !! > Consider my folding. Hence CND is isosceles, and the median NE is also the altitude. Yes, good point. CNE = DNE === Subject: Why Clifford Algebra ? posting-account=BHQ0RQoAAACGbyHx2qDA8u1OskzVIfxx Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) in Our Wireless Engineering Field, Clifford Algebra is famously used in generating orthogonal squared matrix at a given complex vector. Actually we dont need to understand Clifford Algebra very well, you can just use the final result to generate such matrix. But i feel that people in our filed, talk a lot Clifford algebra without understanding exactly what it is... ( cos i asked them to explain me ) .. So i doubt that Clifford algebra is just a fame or what is its real strength and why it is popular in physics and engineering... ( i know nothing abt Clifford algebra ) === Subject: Re: Why Clifford Algebra ? in Our Wireless Engineering Field, Clifford Algebra is famously used > in generating orthogonal squared matrix at a given complex vector. > Actually we dont need to understand Clifford Algebra very well, you > can just use the final result to generate such matrix. But i feel that people in our filed, talk a lot Clifford algebra > without understanding exactly what it is... ( cos i asked them to > explain me ) .. So i doubt that Clifford algebra is just a fame or > what is its real strength and why it is popular in physics and > engineering... ( i know nothing abt Clifford algebra ) > http://en.wikipedia.org/wiki/Clifford_algebra http://en.wikipedia.org/wiki/Clifford_algebra http://en.wikipedia.org/wiki/Clifford_algebra http://en.wikipedia.org/wiki/Clifford_algebra === Subject: the cardinality of the se of countable sets of reals posting-account=ShKz4goAAAB-5fxA1JyErZXtNgWVpdct Gecko/2008111318 Ubuntu/8.10 (intrepid) Firefox/3.0.4,gzip(gfe),gzip(gfe) What is the cardinality of the set A of countable sets of reals ? I know P(N) is subset of A and then aleph<|A|. What next ? === Subject: Re: the cardinality of the se of countable sets of reals > What is the cardinality of the set A of countable sets of reals ? > I know P(N) is subset of A and then aleph<|A|. What next ? You are basically asking for the cardinality of R^N, the set of all mappings from N into R. This is given by the laws of exponents as follows: |R^N| = c^aleph_0 = (2^aleph_0)^aleph_0 = 2^(aleph_0*aleph_0) = 2^aleph_0 = c. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: the cardinality of the se of countable sets of reals > What is the cardinality of the set A of countable sets of reals ? > I know P(N) is subset of A and then aleph<|A|. I am not sure what you have in mind. Using the ordinary definitions P(N) is not a subset of the set of countable sets of reals. But to see A is at least of the cardinality of the continuum, note that for every real r there is a unique countable set of reals {r}, the singleton of r, that is in A. > What next ? Consider a countable set A = {r_1, ..., r_n, ...} of reals in [0,1], expressed as infinite sequences of binary digits . Can you think of some way of encoding A in a single infinite sequence of binary digits? If this hint leaves you stumped, consider instead sets A' = {r_1, ..., r_5} of five reals in [0,1]; can you think of a way of encoding such an A' in a single real? -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Why meta diagonals are irrelevant > So, similarly if we claim that bananas are tasty we are in fact > implicitly asserting an implication, with the antecedent being the > conjunction of the axioms for bananas? *OF COURSE* we are! What are the axioms for bananas? On the face it the notion is ludicrous. > ALL words mean what they mean IN VIRTUE OF *THE DICTIONARY*, in Who stipulated these elusive axioms for bananas? Surely you're not suggesting dictionary makers simply make up arbitrary axioms for words. > How are the axioms of first-order arithmetic 1st-order PEANO arithmetic. > 1st-order arithmetic as YOU conceive it CAN'T EVEN BE axiomatized. An odd thing to say since by first-order arithmetic I meant just Peano arithmetic. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Why meta diagonals are irrelevant <871vx1uu2e.fsf@alatheia.dsl.inet.fi> <87ljupo4h5.fsf@alatheia.dsl.inet.fi> posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > What are the axioms for bananas? > On the face it the notion is ludicrous. Oh, SHUT THE ** UP, BITCH. You understand this concept PERFECTLY well. You are just being sadistic. On the face of it, you somehow turned into an asshole, which does not look pretty on a face. > ALL words mean what they mean IN VIRTUE OF *THE DICTIONARY*, in Who stipulated these elusive axioms for bananas? THE COMMUNITY, DUMBASS!! > Surely you're not > suggesting dictionary makers simply make up arbitrary axioms for > words. There is a distinction between a definition and an axiom, but yes, the dictionary DOES codify those axioms, and OF COURSE the dictionary makers did not MAKE THEM UP -- they simply codified the axioms THAT THE LINGUISTIC COMMUNITY HAD ALREADY stipulated. > An odd thing to say since by first-order arithmetic I meant just > Peano arithmetic. Liar. === Subject: Re: Why meta diagonals are irrelevant On the face of it, you somehow turned into an asshole, ... > I'll second that. Herb === Subject: Re: Why meta diagonals are irrelevant <871vx1uu2e.fsf@alatheia.dsl.inet.fi> <1u1dm6107uv9.mxmqc3zj08lu.dlg@40tude.net> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > On the face of it, you somehow turned into an asshole, ... I'll second that. You're joking, right? MoeBlee === Subject: Re: Why meta diagonals are irrelevant [what exactly the topic is ?] > Not sure if I got you right. My point is: in NO logic (logical system) > where the substitution of identicals holds, objects can change. Since we do in fact reason about objects that change, in ordinary contexts and in, say, intuitionistic analysis, this suggests that the way of modelling change in a logic you consider is not very helpful, and does not capture any aspect of the way we in fact reason about change. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Why meta diagonals are irrelevant [what exactly the topic is ?] <877i6pqvaa.fsf@alatheia.dsl.inet.fi> <1i9913uy8hc7r$.1xx3bd64rilng$.dlg@40tude.net> <87prk1o4nt.fsf@alatheia.dsl.inet.fi> Not sure if I got you right. My point is: in NO logic (logical system) > where the substitution of identicals holds, objects can change. >Since we do in fact [bla bla bla]. > Can't you read? I said: My point is: in NO logic (logical system) where the substitution of identicals holds, objects can change. This affects, for example, any system of FOPL with identity. Hence ANY theory formulated in the framework of FOPL with identity, and hence especially ALL standard systems of axiomatic set theory. (ZFC, NBG, MK, NG, NFU, ML, Ackermann's set theory, etc. etc.) And of course it also effects standard systems of _classical_ higher order logic. Herb === Subject: Re: Why meta diagonals are irrelevant [what exactly the topic is ?] Not sure if I got you right. My point is: in NO logic (logical system) > where the substitution of identicals holds, objects can change. > Since we do in fact [bla bla bla]. > Can't you read? I said: My point is: in NO logic (logical system) where the > substitution of identicals holds, objects can > change. What does this mean for ZFC ( a first order theory with identity)? David Bernier > This affects, for example, any system of FOPL with identity. Hence ANY > theory formulated in the framework of FOPL with identity, and hence > especially ALL standard systems of axiomatic set theory. (ZFC, NBG, MK, NG, > NFU, ML, Ackermann's set theory, etc. etc.) And of course it also effects > standard systems of _classical_ higher order logic. === Subject: Re: Why meta diagonals are irrelevant [what exactly the topic is ?] My point is: in NO logic (logical system) where the > substitution of identicals holds, objects can change. > What does this mean for ZFC (a first order theory with identity)? > It means, for example, that for any set A A u {a} =/= A with a !e A. (Contrary to WM's claims that sets can change, i.e. that a set can be one and the same set, even though it containts different elements... *ummm*) A simple example is the [...] set N. If we add another element, the set remains the same. (WM, sci.logic/sci.math) Herb === Subject: Re: Why meta diagonals are irrelevant [what exactly the topic is ?] <12md18qc88gr2$.101qo07xq0i16.dlg@40tude.net> My point is: in NO logic (logical system) where the > substitution of identicals holds, objects can change. > What does this mean for ZFC (a first order theory with identity)? >It means, for example, that for any set A A u {a} =/= A with a !e A. (Contrary to WM's claims that sets can change, i.e. that a set can be one > and the same set, even though it contains different elements... *ummm*) > A simple example is the [...] set N. If we add > another element, the set remains the same. (WM, sci.logic/sci.math) > With other words, from the _logical_ axiom-schema of identity theory a = b -> (Phi[a] -> Phi[b]) (substitution of identicals) we get A = B -> Ax(x e A <-> x e B). And from the latter (with contraposition) we get Ex(x e A & x !e B) v Ex(x !e A & x e B) -> A =/= B. Hence sets with different elements are not identical. With other words, a certain set cannot change (grow or shrink): if an element is added to (or removed from) some set, we get a _different_ set. Not just the same with one more (or less) element. Right, this is in contrast with our intuitive daily usage of identical when referring to macroscopic objects (of our daily life). (Actually, object persistence is an interesting philosophical problem.) Herb === Subject: Re: Why meta diagonals are irrelevant [what exactly the topic is ?] Not sure if I got you right. My point is: in NO logic (logical system) >where the substitution of identicals holds, objects can change. Since we do in fact reason about objects that change, in ordinary > contexts and in, say, intuitionistic analysis, this suggests that the > way of modelling change in a logic you consider is not very helpful, > and does not capture any aspect of the way we in fact reason about > change. Logical systems are at best a proper part of common reasoning. It's not the other way around. Han de Bruijn === Subject: Re: Why meta diagonals are irrelevant [what exactly the topic is ?] Since we do in fact reason about objects that change, in ordinary > contexts and in, say, intuitionistic analysis, this suggests that the > way of modelling change in a logic you consider is not very helpful, > and does not capture any aspect of the way we in fact reason about > change. Logical systems are at best a proper part of common reasoning. It's not > the other way around. The relevance of this observation escapes me. -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: why does professor david c ullrich have to put people down to feel good about himself? > Ullrich _was_ a good teacher in the 1990's, back > before the rise in so-called cranks. Obviously, > the rise in cranks frustrated Ullrich, so he makes > do the best he can by inserting humor/entertainment > into the situation. (For that matter, Virgil also > uses humor when dealing with cranks.) You misplaced the quatation marks. When speaking about Virgil it's humor that should be in quotes. > I guess I can't blame them. Had I been a standard > mathematician who preferred standard theories to > novel theories, I'd be just as frustrated as > Ullrich (and Virgil) are. I don't know whether I > would (be able to) find entertainment in cranks. These speculations of yours are bizarre. Is there some reason you wish to attribute entirely arbitrary motives to people without any apparent justification? -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: why does professor david c ullrich have to put people down to feel good about himself? <87hc5do4c4.fsf@alatheia.dsl.inet.fi> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) > These speculations of yours [lwal's] are bizarre. Is there some reason you wish > to attribute entirely arbitrary motives to people without any apparent > justification? Another one right on the head. MoeBlee === Subject: 2^p-1 =Mp Let p, p_1 = prime number let n>0 2^p-1 = Mp is prime if and only if isn't verified the congruence: 2^p = (p_1)^2 + 1 (mod 2p_1) with p_1 searching between the prime numbers proceeds, like suggested from Fermat, the shape 2pn+1, to varying of n. Vincenzo Librandi vincenzo.librandi@tin.it === Subject: Re: Is a straight line a circle of infinite size? > I.e., of infinite radius or infinite diameter or infinite > circumference? > I mean, you must admit that, as the radius gets bigger and bigger, > the curvature gets less and less, i.e., the circumference gets > locally > flatter and flatter, more nearly linear. Is there some way of having an infinitely big circle in (e.g.) > projective geometry, > in some place where there could actually be a point at infinity? The topology isn't the same. A circle of infinite curvature isn't a point, and a circle of infinite diameter isn't a straight line. === Subject: Re: Is a straight line a circle of infinite size? posting-account=S6jUlgkAAAAS0KYO9CfNqTx523v1YxGt CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > The topology isn't the same. Of COURSE, but when the topology IS the same, THEN what?? THEN HOW can you have a circle of infinite radius? Remember, YOU are the one who said we COULD draw an infinite circle in the Cantorian matrix!! > A circle of infinite curvature isn't a point, > and a circle of infinite diameter isn't a straight line. In the USUAL AFFINE Euclidean geometry, IT IS SO TOO. Of course, in the usual Euclidean geometry, NEITHER OF THESE THINGS IS DEFINED AT ALL, but the logical conclusion (for a circle of radius 0) is that it IS a point -- it is DEFINED as the set of all points at a distance r from the center, so if r is 0, the center is the only point you get. As for the other case, AGAIN, YOU have to define infinite circle. If you leave it to the usual definitions then NO SUCH THING IS POSSIBLE, so YOU look stupid, since YOU said there could be one. === Subject: Re: Is a straight line a circle of infinite size? > The topology isn't the same. Of COURSE, but when the topology IS the same, THEN what?? > THEN HOW can you have a circle of infinite radius? Remember, > YOU are the one who said we COULD draw an infinite circle in the > Cantorian matrix!! A circle of infinite curvature isn't a point, > and a circle of infinite diameter isn't a straight line. In the USUAL AFFINE Euclidean geometry, IT IS SO TOO. > Of course, in the usual Euclidean geometry, NEITHER OF THESE THINGS > IS DEFINED AT ALL, but the logical conclusion (for a circle of radius > 0) is that > it IS a point -- it is DEFINED as the set of all points at a distance > r from the center, > so if r is 0, the center is the only point you get. As for the other case, AGAIN, YOU have to define infinite circle. > If you leave it to the usual definitions then NO SUCH THING IS > POSSIBLE, > so YOU look stupid, since YOU said there could be one. Shutup. The topology is not the same. You can't convert a teacup or polomint into a sphere, a circle into a point, etc. === Subject: turing machine posting-account=qc3zgAoAAAA1qc5KUfDcFvlOTvCRiTRq AppleWebKit/525.19 (KHTML, like Gecko) Chrome/0.4.154.29 Safari/525.19,gzip(gfe),gzip(gfe) Can anybody help me to design turing machine of subtraction of two binary numbers? f(x)=x-y. === Subject: A problem in Geometry posting-account=diAU6QoAAAB3YxIYj_oXXMwehKojL-C9 Gecko/2008111318 Ubuntu/8.04 (hardy) Firefox/3.0.4,gzip(gfe),gzip(gfe) Given a positive number A find the curve with minimal perimeter and area A. Anil === Subject: Re: A problem in Geometry >Given a positive number A find the curve with minimal perimeter and >area A. Anil === Subject: Re: A problem in Geometry posting-account=diAU6QoAAAB3YxIYj_oXXMwehKojL-C9 Gecko/2008111318 Ubuntu/8.04 (hardy) Firefox/3.0.4,gzip(gfe),gzip(gfe) Given a positive number A find the curve with minimal perimeter and >area A. Anil === Subject: Geometry with middle school. Hello teacher~ http://board-2.blueweb.co.kr/user/math565/data/math/sand.jpg === Subject: Re: Geometry with middle school. posting-account=G_G-iQoAAAB08LNQidt_LsMkopmIb4ZS Gecko/20060111 Firefox/1.5.0.1 Mnenhy/0.7.3.0,gzip(gfe),gzip(gfe) > Hello teacher~ http://board-2.blueweb.co.kr/user/math565/data/math/sand.jpg ABD and ABC are similar triangles. Then AB/AD = AC/AE. OR AD/AB = AE/AC. AD = BD-AB; AE = CE-AC. Then (BD-AB)/AB = (CE-AC)/AC BD/AB -1 = CE/AC - 1 BD/AB = CE/AC === Subject: Oh, Christ As if things couldn't get any worse, I got into a really scary car accident last night, and I've been depressed ever since. I should consider myself lucky, because I was hit directly on my driver's side door by a van several times larger than my Camry, but there has been so much happening that I really don't need any more hassles. Fortunately I walked away with just a light soreness in my side, but now I'll have to spend maybe about a thousand dollars on body work, which is money I needed to use for other things. Between the health problems I already had, problems finding a job in the Detroit area, and problems with my house causing my health problems, this is just one more headache to deal with that I don't need right now. Everything's going wrong at once. ;( === Subject: Re: Oh, Christ As if things couldn't get any worse, I got into a really scary car accident >last night, and I've been depressed ever since. I should consider myself >lucky, because I was hit directly on my driver's side door by a van several >times larger than my Camry, but there has been so much happening that I >really don't need any more hassles. Fortunately I walked away with just a >light soreness in my side, but now I'll have to spend maybe about a >thousand dollars on body work, which is money I needed to use for other >things. Between the health problems I already had, problems finding a job >in the Detroit area, and problems with my house causing my health problems, >this is just one more headache to deal with that I don't need right now. >Everything's going wrong at once. ;( You have a $1000 deductible on your insurance? -- One of the advantages of being disorderly is that one is constantly making exciting discoveries. - AA Milne === Subject: Re: Oh, Christ > As if things couldn't get any worse, I got into a really > scary car accident last night, and I've been depressed > ever since. I should consider myself lucky, because I > was hit directly on my driver's side door by a van > several times larger than my Camry, but there has been > so much happening that I really don't need any more > hassles. Fortunately I walked away with just a light > soreness in my side, but now I'll have to spend maybe > about a thousand dollars on body work, which is money I > needed to use for other things. Between the health > problems I already had, problems finding a job in the > Detroit area, and problems with my house causing my > health problems, this is just one more headache to deal > with that I don't need right now. Everything's going > wrong at once. ;( You have a $1000 deductible on your insurance? Lot of people don't have liability. === Subject: Re: Oh, Christ As if things couldn't get any worse, I got into a really scary car accident >last night, and I've been depressed ever since. I should consider myself >lucky, because I was hit directly on my driver's side door by a van several >times larger than my Camry, but there has been so much happening that I >really don't need any more hassles. Fortunately I walked away with just a >light soreness in my side, but now I'll have to spend maybe about a >thousand dollars on body work, which is money I needed to use for other >things. Between the health problems I already had, problems finding a job >in the Detroit area, and problems with my house causing my health problems, >this is just one more headache to deal with that I don't need right now. >Everything's going wrong at once. ;( Uh, if you got hit and it was the other drivers fault...why did you pay for the body work? -- Onideus Mad Hatter mhm Í x Í http://www.backwater-productions.net http://www.backwater-productions.net/hatter-blog Hatter Quotes ------------- === Subject: Best heuristics (smallest number of tests) I have a problem - I want to maximize a function y = f(x1,x2,x3 ... xn) . n ( number of input variables) will be about 20 , and these will be real values with some constraints (min and max values for each input value). The function y is black box function - I have all input values specified and then I calculate y. I think, that this function is unimodal - when I find a local maximum, it will be a global maximum. The problem is, each calculation of 'y' is very time-consuming (about 1 minute), so I want to have a very good algorithm of finding input values, that will find a maximum with small number of tests. Which heuristics functions are good for such problems? Julek === Subject: Re: JSH: Then consider this example > It seems that I haven't convinced with my talk of the distributive > property so here is an example which should test your understanding of > Galois Theory to the limit. In the ring of algebraic integers, let x=1+sqrt(-6), with 7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0. I picked x so that it is a factor of 7, as (1+sqrt(-6))(1-sqrt(-6)) = > 7. Also notice that despite x having a factor in common with 7: 7*(175x^2 - 15x + 2) still only has 7 as a factor. Further note that a_1(x)*a_2(x) has 7(1+sqrt(-6)) as a factor, which > is the point, as if the a's do not share factors in common with 7 > carefully, then you contradict with only 7 being a factor of (5a_1(x) + 7)(5a_2(x)+ 7). Give the a's and solve for the factors in common with the a's for this > value of x. Have fun! Preserve your deluded view of Galois Theory--if you can. The challenge is in front of you, are any of you good enough? Smart > enough? > Smart enough, yes. But the computations here are formidable. First, you need to find a polynomial with integer coefficients which the roots of a^2 - (7x - 1)a + (49x^2 - 14x) = 0 satisfy. Just substituting in x = 1 + sqrt(-6) gives you a 2nd degree polynomial in a, but it does not have integer coefficients. However after some manipulation you can obtain a^4 - 12a^3 - 518a^2 - 3108a + 116767 = 0, which is 4th degree with integer coefficients and is irreducible. Note that 116767 = 7^2 * 2383. Galois theory and, independently, algebraic number theory tell you that all the roots of this 4th degree polynomial have nonunit factors in common with 7 and none are divisible by 7. Actually finding what those factors are requires a LOT of computation, and I don't have software that does it. Perhaps someone else here does. It is easy to write down problems which in theory can be solved numerically, but for which the computations are very difficult. That is what you have done. The fact that I do not produce the algebraic integer factors of the roots in this case does NOT prove that it cannot be done. It can be. But in the more tractible case where x = 1, I HAVE shown the factors of the roots explicitly. Neither root is divisible by 7 in the algebraic integers and neither is coprime to 7. You have seen this. It is a matter of arithmetic to check that it is correct. You claim that you have a proof that it cannot be correct. Your claim is contradicted by arithmetic. It boils down to your saying that 2 + 2 cannot equal 4. So clearly your claim is wrong. Further, I have pointed out where your argument based on the distributive property is flawed. Dik Winter showed a similar factorization when x = 1. This example also contradicts your claim. By no means does your current example test anyone's understanding of Galois theory to the limit. This is a straightforward application of Galois theory. It can also be approached via class field theory, which provides an algorithm for constructing the greatest common divisor of the roots of the 4th degree polynomial and 7. However that algorithm can involve an enormous amount of calculation. Again, the fact that I have not written down the factors of the roots is not evidence that it cannot be done. To show me wrong, you need to PROVE that it cannot be done. You have not done that. Unless you do that this example leads nowhere. One counterexample can shoot down a thousand pages of proof. The simple examples that we have given here are sufficient to prove that your claims are wrong. Your claims ccntradict basic arithmetic. You need to deal with simpler things before you go challenging people with computations that are out of reach. Marcus. > James Harris === Subject: Re: JSH: Then consider this example posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081029 Firefox/2.0.0.18,gzip(gfe),gzip(gfe) > It seems that I haven't convinced with my talk of the distributive > property so here is an example which should test your understanding of > Galois Theory to the limit. In the ring of algebraic integers, let x=1+sqrt(-6), with 7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0. I picked x so that it is a factor of 7, as (1+sqrt(-6))(1-sqrt(-6)) = > 7. Also notice that despite x having a factor in common with 7: 7*(175x^2 - 15x + 2) still only has 7 as a factor. Further note that a_1(x)*a_2(x) has 7(1+sqrt(-6)) as a factor, which > is the point, as if the a's do not share factors in common with 7 > carefully, then you contradict with only 7 being a factor of (5a_1(x) + 7)(5a_2(x)+ 7). Give the a's and solve for the factors in common with the a's for this > value of x. Have fun! Preserve your deluded view of Galois Theory--if you can. The challenge is in front of you, are any of you good enough? Smart > enough? Smart enough, yes. But the computations here are formidable. First, you need to find a polynomial with integer coefficients > which the roots of a^2 - (7x - 1)a + (49x^2 - 14x) = 0 satisfy. Just substituting in x = 1 + sqrt(-6) gives you a 2nd > degree polynomial in a, but it does not have integer coefficients. > However after some manipulation you can obtain a^4 - 12a^3 - 518a^2 - 3108a + 116767 = 0, which is 4th degree with integer coefficients and is > irreducible. Note that 116767 = 7^2 * 2383. But there are only two a's: a_1(x) and a_2(x). > Right. There are four roots. Two of them are roots of your quadratic. > If you can't do it, fine. Wait for someone who can. > If you cannot do the arithmetic for the x = 1 case, why not admit it? > Galois theory and, independently, algebraic number theory tell > you that all the roots of this 4th degree polynomial have > nonunit factors in common with 7 and none are divisible by 7. > Actually finding what those factors are requires a LOT of > computation, and I don't have software that does it. Perhaps > someone else here does. It is easy to write down problems which in theory can be > solved numerically, but for which the computations are very > difficult. That is what you have done. The fact that I do not > produce the algebraic integer factors of the roots in this > case does NOT prove that it cannot be done. It can be. But in the more tractible case where x = 1, I HAVE shown > the factors of the roots explicitly. Neither root is divisible Well yeah, the easier example WAS easy. x=0 is even easier but you > claim what it shows doesn't matter. > The point being, ONLY ONE example is sufficient to prove that your claims are wrong. That example has been given. If I gave factors for the roots for your present example you could just create another one where the computations are even more difficult. And so on ad nauseam. The fact that you cannot refute the x = 1 and x = -1 examples is more than enough to demonstrate unequivocally that your proof has errors. One example is worth a thousand proofs. Here you have two simple easily computed examples and you have not refuted either of them. If you did, you would be saying 2 + 2 = 5. > You want easy. Real life doesn't always give you easy. > The shoe is on the other foot. You have not succeeded in dealing with the easier examples that were given by me and Dik Winter. Those are enough to prove you wrong. Why don't you deal with those ? > My point here is that what you believe does not work, but it takes an > example a step beyond easy to show it. > Your example shows absolutely nothing except that I don't happen to have software to do the calculations. My example, on the other hand, shows, via arithmetic that any high school student could carry out, that your claims are wrong. That is, what YOU believe does not work. And it is easily within your powers of proving it. > If modern number theorists can't handle this example, fine. I'm not a number theorist. > Admit > it. I admitted I don't have the software. I know that it can be done. The fact that I have not done it proves nothing. You do not have a proof that it cannot be done. Now how about you dealing with the x = 1 or x = -1 cases? Are you going to admit that the arithmetic there proves you wrong? Go ahead. Admit it. > I may find another way to show your view of Galois Theory doesn't > work. > Go for it. > So I'm giving you an out: admit failure here, and then we all move > on. I'll look for another way. > The calculation is beyond what I can do right now. So what? It's your turn: 1. Admit that my and Dik Winter's examples when x = 1 or x = 1- show by simple arithmetic that your claims are wrong. No elaborate calculations needed, just a little pencil and paper. 2. Admit that you cannot prove even with your own example, that one of the roots must be divisible by 7 in the ring of algebraic integers. > Physicists know that real life doesn't always give you easy. > Which proves nothing. Marcus. > James Harris === Subject: Re: JSH: Then consider this example posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > It seems that I haven't convinced with my talk of the distributive > property so here is an example which should test your understanding of > Galois Theory to the limit. In the ring of algebraic integers, let x=1+sqrt(-6), with 7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0. I picked x so that it is a factor of 7, as (1+sqrt(-6))(1-sqrt(-6)) = > 7. Also notice that despite x having a factor in common with 7: 7*(175x^2 - 15x + 2) still only has 7 as a factor. Absolute piffle. E.g. 7*(175x^2 - 15x + 2) has x as a factor. - William Hughes === Subject: Re: JSH: Then consider this example posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > Also notice that despite x having a factor in common with 7: 7*(175x^2 - 15x + 2) still only has 7 as a factor. Absolute piffle. E.g. 7*(175x^2 - 15x + 2) has x as a factor. - William Hughes > - William Hughes === Subject: Re: JSH: Then consider this example posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > Also notice that despite x having a factor in common with 7: 7*(175x^2 - 15x + 2) still only has 7 as a factor. Absolute piffle. E.g. 7*(175x^2 - 15x + 2) has x as a factor. - William Hughes - William Hughes What evasion? You noticed something trivial: 7*(175x^2 - 15x + 2) > has 7 as a factor, x is a factor of 7, so from what is in my original > post, anyone would know that it has x as a factor. > Apparently anyone does not include JSH JSH: Also notice that despite x having a factor in common with 7: JSH: 7*(175x^2 - 15x + 2) JSH: still only has 7 as a factor. - William Hughes === Subject: Re: JSH: Then consider this example posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > Also notice that despite x having a factor in common with 7: 7*(175x^2 - 15x + 2) still only has 7 as a factor. Absolute piffle. E.g. 7*(175x^2 - 15x + 2) has x as a factor. - William Hughes - William Hughes What evasion? You noticed something trivial: 7*(175x^2 - 15x + 2) > has 7 as a factor, x is a factor of 7, so from what is in my original > post, anyone would know that it has x as a factor. Apparently anyone does not include JSH JSH: Also notice that despite x having a factor in common with 7: JSH: 7*(175x^2 - 15x + 2) JSH: still only has 7 as a factor. - William Hughes You moron. I mean that once you divide off 7 it is still coprime to > 7. Well this is 1: not what you said, 2: not quite right (how can it be still coprime if it was not coprime before). 3: not at all obvious. How do you know that (175x^2 - 15x + 2) is coprime to 7? (Yes it is easy to see that it is coprime to x, but this is not the same thing as coprime to 7.) - William Hughes === Subject: Re: JSH: Then consider this example Also notice that despite x having a factor in common with 7: > 7*(175x^2 - 15x + 2) > still only has 7 as a factor. > Absolute piffle. E.g. 7*(175x^2 - 15x + 2) has x as a factor. > - William Hughes > What evasion? You noticed something trivial: 7*(175x^2 - 15x + 2) > has 7 as a factor, x is a factor of 7, so from what is in my original > post, anyone would know that it has x as a factor. > Apparently anyone does not include JSH JSH: Also notice that despite x having a factor in common with 7: JSH: 7*(175x^2 - 15x + 2) JSH: still only has 7 as a factor. - William Hughes anyone meant anyone with a basic knowledge of mathematics. === Subject: Re: JSH: Then consider this example > It seems that I haven't convinced with my talk of the distributive > property so here is an example which should test your understanding of > Galois Theory to the limit. In the ring of algebraic integers, let x=1+sqrt(-6), with 7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0. I picked x so that it is a factor of 7, as (1+sqrt(-6))(1-sqrt(-6)) = > 7. Also notice that despite x having a factor in common with 7: 7*(175x^2 - 15x + 2) still only has 7 as a factor. Further note that a_1(x)*a_2(x) has 7(1+sqrt(-6)) as a factor, which > is the point, as if the a's do not share factors in common with 7 > carefully, then you contradict with only 7 being a factor of (5a_1(x) + 7)(5a_2(x)+ 7). Give the a's and solve for the factors in common with the a's for this > value of x. Have fun! Preserve your deluded view of Galois Theory--if you can. How come you're the one person in the world who sees a problem with ... whatever about albebraic integers? > The challenge is in front of you, are any of you good enough? Smart > enough? > James Harris === Subject: IICAI-09 Call for papers IICAI-09 Call for papers The 4th Indian International Conference on Artificial Intelligence (IICAI-09) will be held in Tumkur (near Bangalore), India during December 16-18 2009. The conference consists of paper presentations, special workshops, sessions, invited talks and local tours, etc. We invite draft paper submissions. Please see the website: http://www.iiconference.org for more details of the conference. Edward Publicity Committee === Subject: boolean algebra posting-account=o0hF9QoAAAAbaA3UBy-XaadGRKk5ij7E 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) HELP: in this solution how does the author distribute out the b term in line 7? ab + abÍ + aÍb a(b + b') + a'b a(1) + a'b a + a'b a + a'b + 0 a + a'b + aa' a + b(a + aÍ) a + b´1 a + b === Subject: just want to double check Hi all, I'm pretty sure that I understood the whole solution space of the two equations beneath, but would like to double check my ideas. If I understood the problem correctly, there should be only the trivial solution for c1=c2=0, and otherwise there should be four solutions for any non-zero c1,c2. Two of the solutions I would expect to be real, two complex. c1 and c2 can be any real number. equ1= 2*x1*x2-c1; equ2= (x2^2-x1^2)-c2; Peter === Subject: Opa asdtyuj News posting-account=2RM9pAoAAABHmAlb9u_vAwKAKfA8IrZC iOpus-Web-Automation; MRSPUTNIK 2, 0, 1, 26 SW; MRA 5.2 (build 02400); .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) The decision comes after the review of a six-month pilot program that extended the TIS National to pharmacies. http://bmool.blogsome.com/ http://asdmk.blogsome.com/ The pilot program, conducted by the department and the Pharmacy Guild of Australia, involved 331 pharmacies having access to the service 24 hours a day. http://steppo.blogsome.com/ Participating pharmacies will be able to use the Translating and Interpreting Service (TIS National), operated by the Department of Immigration, 24 hours a day, seven days a week. Pharmacy Guild of Australia national president Kos Sclavos implored pharmacists to use the service to prevent patients from misinterpreting important usage instructions. http://mygotar.blogsome.com/ The interpreter service is not a full solution. The best solution is to always have a staff member of that language. Pharmacies have been very good at going out of their way to employ perhaps one staff member fluent in another language. But because the trading hours of pharmacies, it's very hard to have a staff member on at all times with that language skills base. So this service fills in or augments current practices. Mr Sclavos said there were inherent problems with other multilingual initiatives, such as printing instruction labels in other languages. The tendency to revert to a first language as experienced by some older Australians, who have acquired English as a second language, is one of the greatest issues facing our ageing population. This was demonstrated by the heavy usage of the TIS services by post-war aged European communities throughout the duration of the pilot. Pharmacies who participated in the pilot used interpreting services more frequently and reported improved outcomes in terms of client understanding of medications. http://rollz.blogsome.com/ You need to be getting that verbal and non-verbal feedback from the patient that they have indeed understood the instructions and that's why the interpreter service is so important because the interpreter on the other end of the phone will ask 'now repeat what I've just explained' or 'have you understood what I just said', Mr Sclavos said. http://aermk.blogsome.com/ As a pharmacist, if I was checking the prescription later and unless it's dual-labelled, I couldn't be sure because I couldn't read that language or couldn't understand the translation. The difficulty we have with the dual-labelling approach is that we would be getting complaints that the packaging is getting smaller and smaller and there is no room to put a sticker on the packet. That's why the GuildÍs policy is not to support automatic translations because it increases the risk of errors. http://twoaac.blogsome.com/ Parliamentary Secretary for Multicultural Affairs Laurie Ferguson said health care needs of recent migrants and elderly former migrants necessitated the extension of the service to pharmacies. http://viewpharmacy.blogsome.com/ Interpreting services are crucial for the proper distribution and usage of prescription medicines by non-English speaking Australians, Mr Ferguson said.