Express the following as a single fraction: 4/3ab - 5/6bc > (m^2 + 2)/(m^2 + m) - (m - 2)/m You do it the same way you do it for fractions in arithematic. > The general formula is derived thus > a/b + r/s = as/bs + br/bs = (as + br)/bs Yep, I understand the basic priniciple, but I just don't know how to put it > in practise with these types of fractions. > Show us how you make 1/5 + 1/3 and 3/4 + 1/6 into a single fractions. Show us what you've done trying to make 4/3ab - 5/6bc (m^2 + 2)/(m^2 + m) - (m - 2)/m into a single fraction. What have you tried ==== > How do I do this? > > Express the following as a single fraction: > > 4/3ab - 5/6bc > > and > > (m^2 + 2)/(m^2 + m) - (m - 2)/m > > > > (1) find a common denominator. The least common denominator is a good one to use. (2) convert each fraction to an equivalent fraction, all having the same (common) denominator found in step 1. (3) Add (or subtract) numerators and put the result over the common denominator from step 1. (4) Reduce the result to lowest terms. Note that if you have used the least common denominator, the fraction may already be in lowest terms. ==== How do I do this? Express the following as a single fraction: 4/3ab - 5/6bc (1) find a common denominator. The least common denominator is a > good one to use. What, for example in the first one, is the lowest common denominator? Is it (3ab)(6bc)? If it is, do I then multiply that out? If not, what is it? ==== How do I do this? Express the following as a single fraction: 4/3ab - 5/6bc (1) find a common denominator. The least common denominator is a > good one to use. > > What, for example in the first one, is the lowest common denominator? Is it > (3ab)(6bc)? If it is, do I then multiply that out? If not, what is it? > > 6abc = (3ab)(2c) = (6bc)(a), so 6abc is a common multiple of 3ab and 6bc. Since (2c) and (a) have no common factor greater than 1, nothing can be left out of 6abc and still have a common multiple of 3ab and 6bc, so 6abc is the least common multiple. ==== John E escribi.97 en el mensaje > How do I do this? Express the following as a single fraction: 4/3ab - 5/6bc > (1) find a common denominator. The least common denominator is a >> good one to use. What, for example in the first one, is the lowest common denominator? > Is it (3ab)(6bc)? If it is, do I then multiply that out? If not, what > is it? Factorize all denominatars as much as possible, including numbers: 3ab = 3*a*b 6bc = 2*3*b*c The lowest common denominator is the lowest common multiple of the denominators. To get it, take all factors raised to the greatest exponent that appears. In this case, lcm(3ab, 6bc) = 2*3*a*b*c = 6abc Then multiply and divide each fraction by the quotient between the common denominator and its denominadtor: 4/(3ab) - 5/(6bc) = 4*(2c)/((3ab)*(2c) - 5*a/((6bc)*a) = (8c - 5a)/(6abc) -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== > John E escribi.97 en el mensaje > How do I do this? Express the following as a single fraction: 4/3ab - 5/6bc > (1) find a common denominator. The least common denominator is a >> good one to use. What, for example in the first one, is the lowest common denominator? > Is it (3ab)(6bc)? If it is, do I then multiply that out? If not, what > is it? Factorize all denominatars as much as possible, including numbers: 3ab = 3*a*b 6bc = 2*3*b*c The lowest common denominator is the lowest common multiple of the > denominators. To get it, take all factors raised to the greatest exponent > that appears. In this case, lcm(3ab, 6bc) = 2*3*a*b*c = 6abc Then multiply and divide each fraction by the quotient between the common > denominator and its denominadtor: 4/(3ab) - 5/(6bc) = 4*(2c)/((3ab)*(2c) - 5*a/((6bc)*a) = (8c - 5a)/(6abc) ==== Examining a table of factors and primes, I found that for any sequence of consecutive composite numbers there is always one integer that has a prime factor larger than any other prime factor of any of the other integers. Further, this prime is not raised to any power. My question is: Is this true for all consecutive composite sequences and if so, is there a proof? (Here's an example: 1500, 1501, ... 1509. The last integer has the prime factor 503.) ==== > Examining a table of factors and primes, I found that for any sequence > of consecutive composite numbers there is always one integer that has > a prime factor larger than any other prime factor of any of the others Nice conjecture! It is equivalent to saying (more neatly perhaps) that in any adjacent sequence of numbers, the largest prime factor occurs only once. It seems obviously true, though I can't see a slick proof off-hand. I'm sure there must be one. It is possible to produce a heuristic demo that can doubtless be turned into a formal proof with effort. Suppose your two largest equal prime factors are P, and we may as well put them at the ends of the interval WLOG. Then we have an interval of length P which must be filled up by factors less than P. This would mean the P interior numbers have to be removed one-by-one by ditching the evens, then the 3-multiples, then the 5-multiples, and so on up to P. But we can remove at most (1/2)*(2/3)*(4/5)*(6/7)*...*(1 - 1/P) this way; and a quick heuristic integral integral shows this to be 1/(log P). So the most numbers we can reduce to by removing those with (even at least one!) small prime factor is one log-Pth of the lot, and the lot is of size P. That's P/logP. Nowhere near zero or one. So it can't be done... not by a huge amount! Nice question though. ---------------------------------------------------------------------------- -- Bill Taylor W.Taylor@math.canterbury.ac.nz ---------------------------------------------------------------------------- -- Chebychev said it -- I'll say it again; There is always a prime between n and 2n. ---------------------------------------------------------------------------- -- ==== >Examining a table of factors and primes, I found that for any sequence >of consecutive composite numbers there is always one integer that has >a prime factor larger than any other prime factor of any of the other >integers. Further, this prime is not raised to any power. My question >is: Is this true for all consecutive composite sequences and if so, is >there a proof? >(Here's an example: 1500, 1501, ... 1509. The last integer has the >prime factor 503.) How does the Further,... apply to this example: 8, 9? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >Examining a table of factors and primes, I found that for any sequence >of consecutive composite numbers there is always one integer that has >a prime factor larger than any other prime factor of any of the other >integers. Further, this prime is not raised to any power. How does the Further,... apply to this example: 8, 9? > Perhaps he intended maximal sequence, 8,9,10 has 5 as max. However the maximal sequences 4 and 12 have 2 squared and the maximal sequence 18 has 3 squared. Thus the conjecture needs be limited to maximal sequences of composite numbers longer than 1. ==== >>Examining a table of factors and primes, I found that for any sequence >>of consecutive composite numbers there is always one integer that has >>a prime factor larger than any other prime factor of any of the other >>integers. Further, this prime is not raised to any power. >> How does the Further,... apply to this example: 8, 9? >Perhaps he intended maximal sequence, 8,9,10 has 5 as max. But the example he gave was 1500, 1501, ... 1509 which is not maximal. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >Examining a table of factors and primes, I found that for any sequence >of consecutive composite numbers there is always one integer that has >a prime factor larger than any other prime factor of any of the other >integers. Further, this prime is not raised to any power. > How does the Further,... apply to this example: 8, 9? >>Perhaps he intended maximal sequence, 8,9,10 has 5 as max. > But the example he gave was 1500, 1501, ... 1509 which is not maximal. Also, the largest prime factor of a number in that range is 751, not 503. The conjecture still holds, but not for the reason given. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== well, isn't this teensy duality the same, for any proof that one hasn't actually read, all the way through (or to the point where one can see it through, already) ?? > you're saying that the L-wing thing is just an argument > *about* an actual (R-wing) proof? --les ducs de Buffet et Schulz (R&D Chair Assoc.Intl.); vote NONE OF THE BELOW on Trickier Dick Cheney's California Recall/e-Dereg! http://larouchepub.com http://members.tripod.com/~american_almanac/ ==== it's not even wrong. it's just simple numbertheory -- skipcodes are that, and they were apparently used by (some?) Torah writers/copyists to ensure accuracy, as with the old CRC in 8-bit communications programs. I read taht they summed the letters on every 70th, or skipped to every 70th, or some thing. the computer can be set to find any message in any ring of an alphabet, and Drosnin et al know this ... or maybe they can't learn it, not because they're dumb. there was ambiguity in _The Bible Code_ taht he ignored, like with the variant translations and the fact that Old Hebrew has no vowels. repeat, _War and Peace_ or just the 26 letters in any order can be used with the infinite set of co-prime skips, with teh resulting hits being further massaged into some m by n array (or what ever). >Also, when the code was applied to the new testament, the NEW TESTAMENT >FAILED. Is this accurate? And does this say something about the New Testament >and the belief in a Christ figure? --les ducs de Buffet et Schulz (R&D Chair Assoc.Intl.); vote NONE OF THE BELOW on Trickier Dick Cheney's California Recall/e-Dereg! http://larouchepub.com http://members.tripod.com/~american_almanac/ ==== > it's not even wrong. it's just simple numbertheory -- > skipcodes are that, and they were apparently used > by (some?) Torah writers/copyists to ensure accuracy, > as with the old CRC in 8-bit communications programs. > I read taht they summed the letters on every 70th, > or skipped to every 70th, or some thing. > the computer can be set to find any message > in any ring of an alphabet, and Drosnin et al know this ... or > maybe they can't learn it, not because they're dumb. > there was ambiguity in _The Bible Code_ taht he ignored, > like with the variant translations and the fact that > Old Hebrew has no vowels. repeat, _War and Peace_ or just the 26 letters in any order > can be used with the infinite set of co-prime skips, > with teh resulting hits being further massaged > into some m by n array (or what ever). >Also, when the code was applied to the new testament, the NEW TESTAMENT >FAILED. Is this accurate? And does this say something about the New Testament >and the belief in a Christ figure? --les ducs de Buffet et Schulz (R&D Chair Assoc.Intl.); > vote NONE OF THE BELOW > on Trickier Dick Cheney's California Recall/e-Dereg! > http://larouchepub.com > http://members.tripod.com/~american_almanac/ A defective theory when applied does neither prove or disprove a hypothesis. The Code stuff has been applied successfully to Moby Dick to predict moderns events with Astounding accuracy. (NOT!) rj Pease ==== modern, but ex post facto, with a few rather repugnant exceptions (Die, Rabin, Die!) the sequel has this doofus, Drosnin, supposedly watching the WTC hit from his hotel room, and immediately running the program to post-dict it. the second book also shows (I think it was indexed, or I was just lucky in finding it, as I didn't *read* the God- am thing) that he got the original idea from one of Sharon's buddies (presumably) in hte military! yes, the NT can be used to equal effect (on everage). > repeat, _War and Peace_ or just the 26 letters in any order > can be used with the infinite set of co-prime skips, > with teh resulting hits being further massaged > into some m by n array (or what ever). > The Code stuff has been applied successfully to Moby Dick to predict > moderns events with Astounding accuracy. (NOT!) --le ducs d'Enron! http://larouchepub.com ==== R E D L 18 5 4 12 = 39 I went out to Broadway Ave. (the street with the penis poster poles) to offer my mathematical services to people passing by. Wade was the second of over 150 people to provide personal information to me during the month of July. 145 Wade 13 6 80 165/201 8517 Wade 33 Joseph 73 Redl 39 Wade was born 45% into the year, his full name adds to 145, his first name adds to 45% of his middle name. Wade was born in 80. The consonants in his given names add to 80. And the and last names add to 33 and 39, these Bible Books contain an average of 80 verses. Primes Non-Primes Numbers 2 1 1 3 4 2 5 6 3 7 8 4 11 9 5 13 10 6 17 12 7 19 14 8 23 15 9 29 16 10 31 18 11 37 20 12 41 21 13 43 <-14th-> 22 <-14th-> 14 --- --- --- 281 176 105 Wade was born with 201 days remaining in the year (Joshua 14), he has 14 letters adding to 145. He was born 36 days closer to the beginning of the year than to the end of the year, and his first and last names average 36, or 14 plus the 14th non-prime (22), the number of chapters in Bible Book 14. 73, these are the 22nd non-prime and the 21st prime, together for 43 (the 14th prime). Note that Bible Book 14 contains 822 verses, pretty as 22 is the 14th non-prime, 14 is the 8th non-prime, and 22 exceeds 8 by 14. Wade was born on day 165 (5x33), his first name adds to 33. Wade's first 7 letters add to 77, his last 7 letters add to 68 (the 7x7th to Numbers 28 (1 through 7). Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime Wade was born on the 13th (6th prime) day of the 6th month, his last name adds to 39 (13+13+13), his first and last names differ in value by 6 and adds to the 73 verses of Bible Book 29 (6+6p+6np). You will find 29 chapters in Bible Book 13 (6th prime) and 29 verses in Bible chapter 666, pretty as 29 is 6 plus the 6th prime (13) plus the 6th non-prime (10), or simply 6+6p+6np. Wade was born on day 165, or 26 plus the 26th prime (101) plus the 26th non-prime (38), or simply 26+26p+26np. The vowels in his given names add to 26. The prime valued letters in his given names add to 52 (26+26). He was born on the 13th, that's aluminum with an atomic weight of 26.981538. 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 In 1988 I pointed out that the penis poster poles on Broadway Ave. are representations of penises, and that the obelisks at the Vatican, The Whitehouse and on church roofs are Egyptian representations of penises, and I was repeatedly arrested and tortured (Protestants and Catholics were upset with my words and lobbied my abusive parents to have me arrested and treated). If you people think you have the right to arrest and torture me because I am not married, then I think I should have the right to ask the nubile sweeties to marry me, and Marcia. If Wade has a sister that ends up marrying Marcia and me, then Wade is goink to win himself a brand spankin' new Cadillac, it's now included in my suit against the Protestants, Saskatoon City Police Force, The City of Saskatoon, and the Liberal, PC and the NDP political parties. Good luck and may God bless you!!! Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! ==== what is found at your site is known as Sophistry. the fact that Fernat's Last Theorem is negative is not problem, just as proofs (of neg or pos statements) by contradiction are good. of course, three per cent is rational by definition. Negative statements about numbers are unverifiable. Take the definition of an irrational as a number that is not rational, where being rational means having periodic decimal expansion. Suppose someone wants to verify that %3 is irrational. Since the indirect proof is out he computes its decimal expansion to the trillionth digit and, satisfied that no evidence of periodicity looms on the horizon, declares that %3 is irrational. A mischievous fellow quickly forms a periodic decimal whose first period is that first trillion > http://www.users.bigpond.com/pidro/home.htm > --les ducs de Buffet et Schulz (R&D Chair Assoc.Intl.); vote NONE OF THE BELOW on Trickier Dick Cheney's California Recall/e-Dereg! http://larouchepub.com http://members.tripod.com/~american_almanac/ ==== sorry; I guess that you meant the second root of 3 by %3 -- but it's still a rather silly use of the law of large numbers. an example that may apply, since I don't recall what that is: there's a proof that li(x) goes less than pi(x), the number of prime numbers less than x, infinitely often, although there is (was) no known place in the sequence where it occurs. the second (skware) root of 3 is irrational by *definition*. > Fernat's Last Theorem is negative is not problem, just as > proofs (of neg or pos statements) by contradiction are good. > of course, three per cent is rational by definition. > > Negative statements about numbers are unverifiable. Take the --les ducs de Buffet et Schulz (R&D Chair Assoc.Intl.); vote NONE OF THE BELOW on Trickier Dick Cheney's California Recall/e-Dereg! http://larouchepub.com http://members.tripod.com/~american_almanac/ ==== D A U K 4 1 21 11 = 37 I went out busking on Broadway Ave. (the street with the penis poster poles) and met Robert, he was headed to The Yard and Flagon, probababbly to drink beer. Robert stopped long enough to inform me about his sisters, it's a high probably they are nubile sweeties, or even a yet higher probababbly. 92+ Ralph 30 10 53 303/62 +1206 Ralph 55 Dauk 37 77+ Linda 11 3 54 70/295 +1074 Linda 40 Dauk 37 166 Robert 3 12 78 337/28 7959 Robert 78 Michael 51 Dauk 37 37+ Bro 1 5 81 121/244 8839 37+ Sis 1 5 81 121/244 8839 37+ Sis 7 1 84 7/359 9820 Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 18 55 199 37 20 89 322 41 <-13th-> 21 <-13th-> 144 <-13th-> 521 --- --- 238 154 <-Lamentations Actually I went out lookink for wives for Marcia and me, and Robert came along and told me about his sisters. Dauk adds to 37 (37 chapters in the Bible contain the length of 13 verses). The parents were born in months adding to 13 and on days of the month adding to 41 (13th prime). The parents were born on days of the year adding to 373, the kids on days of the year adding to 586 (a difference of 213), prettier as 213 is the 166th non-prime while Robert's full name adds to 166. The Gospels are Books 40, 41, 42 and 43, together for 166 (the 128th or the 2 to the 7th non-prime). The kids were born on days of the year adding to 586, corresponding to Psalm 108 (the first 6 primes in prime positions). Robert was born in 78, his first name valued letters add to 84 (the first 13 primes minus the first 13 non-primes), his even valued letters add to 82 (twice the 13th prime). Robert's youngest sister was born in 84 (the first 13 primes minus the first 13 non-primes). Robert and the twins were born on days of the week adding to 13. The family was born on days 30, 11, 3, 1, 1, and 7, together these Bible Books contain 179 chapters (the 41st prime or the 13th prime in prime position). Anyway, pretty that 37 chapters in the Bible contain the length of 13 verses. for these are the 6th and 6+6th primes, and they differ in value by the 24 (6+6+6+6) chapters of Old Testament Books 6 and 10 (6th non-prime). Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 --- 108 The Dauk (37) parents were born on days of the year adding to 373, it ends in the 73 verses of Bible Book 29 (and is the Lucas numbers up to 29). The parents were born in 53 and 54, these are the 14th and 15th Books of the New Testament (together for 29), these Bible Books differ in length by 66 verses. The parents and the kids were born on days of the month adding to 41 and to 12, it's a difference of 29, prettier as 12 is 29.26% of 41. The vowels in Robert's given names add to 35 and Robert's initials add to 35, Genesis 29 contains 35 verses, Genesis 35 contains 29 verses and there are 35 chapters in the Bible that contain the length of 29 verses. In Robert's given names, his even valued letters exceed his odd valued letters by 29. All of his consonants add to 109 (29th prime). All 12 unrepresented letters add to 209. Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime Copper and Zinc are elements 29 and 30 (together for 59), and together they make Brass (59): B R A S S 2 18 1 19 19 = 59 The Dauk (37) parents were born on days of the year adding to 373. Robert Dauk (37) was born on the 337th day of the year. The Fibonacci valued letters in his given names add to 37. In Robert Dauk's (37's) given names, his unrepeated letters exceed his unrepeated letters by 37. Dad was born in 53, he gets twins on the 1st (Genesis with 1533 verses). The family was born on days of the month adding to 53. Mom gave birth when 9033, 9913, 9913 and 10894 days old, together for 39753 days. Mom was born on the 11th, she gave birth on the 121st (11x11th) day of the year. The parents were born on the 30th and on the 11th, together these Bible Books contain 31 (11th prime) chapters. Bible Books 9 and 20 (differ by 11) both contain 31 chapters (11th prime). The parents were born on days of the week adding to 11 while the family was born on days of the week adding to 31 (11th prime). The twins were born in 81, corresponding to Exodus 31 (11th prime). The kids were born in years adding to 324 (Second Kings 11), it's an average of 81, corresponding to Exodus 31 (11th prime). The parents were born in 53 and 54, together these Bible Books contain 160 verses (the first 11 primes). The twins are 101 days closer in age to Robert than to the last of the kids (Leviticus 11). Either twin and the last of the kids were born on days of the year adding to 128 (Numbers 11). Don't be happy, 11 is a number of disorder. Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 <-11th-> 18 <-11th-> 55 <-11th-> 199 --- --- --- --- 160 113 143 518 Dad was born on the 30th day of the month. Mom was born in 54, Bible Book 54 contains 113 verses (30th prime). The parents have first names adding together for the 95 verses of Bible Book 51. Robert's middle name adds to 51. Twins on the 1st day of the month and on the 11x11th day of year 81, and are separated by 1 minute. Dad's combined ages when the kids were born amounts to 40281 days, pretty as there were 2 in 81. Robert (78) was born in 78 (57th non-prime), his vowels add to 57. The parents were together born with 357 days remaining in their years (the number of verses in Daniel). The 57's are at chapters 41, 104, 220 and 1008, together for 1373, while the parents were born on days of the year adding to 373. Dad was an average age of 27.57 years old when the kids were born. 389 <-77th prime 104 <-77th non-prime 77 <-77 --- 570 The Four 57's Genesis 41 -> 41 Leviticus 14 -> 104 Judges 9 -> 220 <-I dreamt of 220 roofs blown John 11 -> 1008 off homes in the Dakotas ---- 1373 <-220th prime Chapter 57 is Exodus 7 with 25 verses Book 57 is Philemon with 25 verses -- -- 41st non-prime 16th non-prime <-together for 57-> Major Books of End-Times Prophecy (Daniel and Revelation are in part about 666 while Isaiah contains 66 chapters): Daniel - 357 verses Revelation - 404 verses <-57 plus the 57th prime plus the 57th non-prime Isaiah - 1292 verses <-an average of 19.575757... verses per chapter The second and the third (together for 5) kids are twins. These twins were born on day 121 (the 5th prime squared). The family was born on days of the month adding to the 959 verses of Bible Book 5. Pretty that Book 5 would contain 959 verses, for 9 is the 5th non-prime. Old Testament Book 9 (5th non-prime) and New Testament Book 9 (5th non-prime) together contain 959 verses, the number of verses in Bible Book 5. I am 7959 days older than Robert. Non-Primes 1 27 50 72 93 4 28 51 74 94 6 30 52 75 95 8 32 54 76 96 9 33 55 77 98 10 34 56 78 99 12 35 57 80 100 14 36 58 81 102 15 38 60 82 104 16 39 62 84 105 <-78th 18 40 63 85 106 20 42 64 86 108 21 44 65 87 110 22 45 66 88 111 24 46 68 90 112 25 48 69 91 114 26 49 70 92 115 non-primes, and to the 12th prime, together for 105 (78th non-prime). The even valued letters in his given names add to 78. 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 If you people think that you have the right to use my abusive parents as tools and arrest and torture me, then I think that I should have the right to ask women to marry me, or to marry Marcia and me (the nubile sweety was born on the 6th and has a 6 lettered first name). Isaiah is the Book with 66 chapters, pretty as it is Book 23, or the 6th prime plus the 6th non-prime (13+10=23). Isaiah 4, 12 and 20 (adds to 6x6) each contain 6 verses. Isaiah 4:1 is about Marcia (and me), and 6 other women who are capable of feeling shame rather than pride, greed or lust, or who limit their love for traditions and for people who abide by their traditions. You people have Egyptian penises on the roofs of your churches and lined city streets with representations of penises, and had me tortured for years for saying so, others just sat back in silence while they were doing this to me, and similarly you remain silent and compassionless now that the arrests and torture have ceased. Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! ==== I'm trying to prove the following theorem: Let P be a polynomial with real coefficients such that P(x) >=0 for every real x. Then, there are polynomials R and S such that P(x) = R^2(x) + S^2(x) for every complex x. It's easy to see that the degree of P must be even. If r is a real root of P, then the restriction of P to the reals has an absolute minimum at r and, from the differentiability of P, it follows there's an even number k such that the k-1 first derivatives of P vanish at r and the k_th is positive. Therefore, the k-1 derivatives admits the root r with multiplicity 1, which implies P admits r as a rooth with multiplicity k. So, we see every real root of P, when they exist, must have an even multiplicity (I think we could come to this same conclusion a bit faster, considering only the continuity of polynomial functions). Corollary - If all of the roots of P are real, then, P = Q^2 for some polynomial Q. So, for this particular case the theorem has just been proved. To prove the theorem, for the general case, I tried to use mathematical induction. It didn't work, that's why I'm asking for help. What I did is as folows: the previous paragraph, it's also enough to cover the case of even-degree polynomials with real coefficients and no real roots. It's well known that every monic trinomial T of the 2nd degree that satisfies such conditions can be written as T(x) = (x-a)^2 + C^2, where a and C<>0 are real. Therefore, for such trinomials the theorem holds trivially. Now, suppose there's a natural k such that the theorem holds for i=1,...k-1 for every 2i-degree polynomial with real coefficients and no real roots. If P is a 2k-degree polynomial with these same properties, then at some (or several) real r's the restriction of P to the reals attains an absolute minimum m>0. This implies that, for every real x, the polynomial P-m is non-negative, has one (or several) real root(s) and (i) P(x) - m = (x-r_1)^p1 *...(x-r^_n)^p_n * Q(x), where r_1, ..r_n are the real roots and the numbers p_1,...p_n are even. In addition, Q is monic, has even degree < 2k, has no real root and is strictly positive on the real line. By the induction assumption, the theorem holds for Q and, in virtue of (i), also holds for P-m. But now, to complete the induction, it remains to prove the theorem is good for P, in other words, it remains to prove that if the theorem holds for some polynomial P then it holds for P+m for every m>0. That's where I got stuck. Actually, I think I chose a very cumbersome way to prove the theorem, there certainly is a neater proof. Any suggestions are welcome. Amanda. ==== > I'm trying to prove the following theorem: > > Let P be a polynomial with real coefficients such that P(x) >=0 for > every real x. Then, there are polynomials R and S such that P(x) = > R^2(x) + S^2(x) for every complex x. Show P factors over the reals as a product of irreducible quadratics times a product of squares of linear polynomials. Show that an irreducible quadratic is a sum of 2 squares. Show that a product of two sums of two squares is a sum of two squares. -- ==== > > I'm trying to prove the following theorem: > > Let P be a polynomial with real coefficients such that P(x) >=0 for > every real x. Then, there are polynomials R and S such that P(x) = > R^2(x) + S^2(x) for every complex x. > > Show P factors over the reals as a product of irreducible quadratics > times a product of squares of linear polynomials. > > Show that an irreducible quadratic is a sum of 2 squares. > > Show that a product of two sums of two squares is a sum of two squares. Sure. I almost got there! All I had to do to complete my proof was show that a product of two sums of two squares is a sum of two squares. This is simple: (a^2+b^2)*(c^2+d^2) = (ac)^2 + (ad)^2 + (bc)^2 + (bd)^2 = (ac)^2 + 2abcd+ (bd)^2 + (ad)^2 -2abcd + (bc)^2 = (ac + bd)^2 + (ad - bc)^2. Thak you all for the help Amanda ==== I'm trying to prove the following theorem: Let P be a polynomial with real coefficients such that P(x) >=0 for >every real x. Then, there are polynomials R and S such that P(x) = >R^2(x) + S^2(x) for every complex x. It's easy to see that the degree of P must be even. If r is a real >root of P, then the restriction of P to the reals has an absolute >minimum at r and, from the differentiability of P, it follows there's >an even number k such that the k-1 first derivatives of P vanish at r >and the k_th is positive. Therefore, the k-1 derivatives admits the >root r with multiplicity 1, which implies P admits r as a rooth with >multiplicity k. So, we see every real root of P, when they exist, must >have an even multiplicity (I think we could come to this same >conclusion a bit faster, considering only the continuity of polynomial >functions). Corollary - If all of the roots of P are real, then, P = Q^2 for some >polynomial Q. So, for this particular case the theorem has just been >proved. To prove the theorem, for the general case, I tried to use >mathematical induction. It didn't work, that's why I'm asking for >help. What I did is as folows: the previous paragraph, it's also enough to cover the case of >even-degree polynomials with real coefficients and no real roots. It's >well known that every monic trinomial T of the 2nd degree that >satisfies such conditions can be written as T(x) = (x-a)^2 + C^2, >where a and C<>0 are real. Therefore, for such trinomials the theorem >holds trivially. Now, suppose there's a natural k such that the >theorem holds for i=1,...k-1 for every 2i-degree polynomial with real >coefficients and no real roots. If P is a 2k-degree polynomial with >these same properties, then at some (or several) real r's the >restriction of P to the reals attains an absolute minimum m>0. This >implies that, for every real x, the polynomial P-m is non-negative, >has one (or several) real root(s) and (i) P(x) - m = (x-r_1)^p1 >*...(x-r^_n)^p_n * Q(x), where r_1, ..r_n are the real roots and the >numbers p_1,...p_n are even. In addition, Q is monic, has even degree >< 2k, has no real root and is strictly positive on the real line. By >the induction assumption, the theorem holds for Q and, in virtue of >(i), also holds for P-m. But now, to complete the induction, it >remains to prove the theorem is good for P, in other words, it remains >to prove that if the theorem holds for some polynomial P then it holds >for P+m for every m>0. That's where I got stuck. Actually, I think I chose a very cumbersome way to prove the theorem, >there certainly is a neater proof. I believe there is. The real roots all have even order and the complex roots come in conjugate pairs - this means there exists a polynomial F (with complex coefficients) such that P is the product of F and the complex conjugate of F... >Any suggestions are welcome. >Amanda. ************************ David C. Ullrich ==== is there a non-Hausdorff space locally homeomorphic to R^n? ==== >is there a non-Hausdorff space locally homeomorphic to R^n? No. I use theorem locally closed Hausdorff S ==> S Hausdorff S locally closed Hausdorff when for all p in S, some open U nhood p for which cl U is a Hausdorff subspace Let the non-Hausdorff space be S. As I don't know definition of locally homeomorphic I'll do my best to intuit what it means. If p in S, then there's some open U nhood p homeomorphic to an a nhood V of R^n about q, the image of p As R^n is normal, we can find a closed nhood K of q contained in V. Now the homeomorphic image of K is a closed Hausdorff nhood of p. Thus I hope I've shown S is locally closed Hausdoff, hence Hausdorff. ---- ==== > >is there a non-Hausdorff space locally homeomorphic to R^n? > No. I use theorem > locally closed Hausdorff S ==> S Hausdorff So what do you do about the real line with double origin? (Quotient of R x {0,1} by the smallest equivalence ~ with (x,0) ~ (x,1) for x =/= 0 ). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== >is there a non-Hausdorff space locally homeomorphic to R^n? Sure thing. Look for a current thread which mentions doubling points in one of the posts (I think by Simeon Stefanov). Not if n=0, though. Lee Rudolph X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h973a2W15776; ==== I've been trying to find a generalized (open) formula for the integral of x^x dx.....does anyone know if it's been derived at all? Curious X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h974dnb19740; ==== Can you construct a set E in [0, 1] s.t. for every open interval I in [0,1] m(I intersect E) > 0 & m(I intersect E^c) > 0 m is lebesgue measure E^c is the complement of E This is so tricky! I was thinking something with the generalized Cantor set but everything I'm trying isn't working. Any suggestions? Ideas? ==== >Can you construct a set E in [0, 1] s.t. for every open interval I in [0,1] m(I intersect E) > 0 & m(I intersect E^c) > 0 >m is lebesgue measure >E^c is the complement of E >This is so tricky! I was thinking something with the generalized Cantor set but everything I'm trying isn't working. Any suggestions? Ideas? I have previously posted an example of such a set, which is Borel measurable. Use the Cantor set idea, but with different sized parts removed from the intervals still in, and added to the intervals already removed. It can be done in terms of the digits to any base. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University ==== > Can you construct a set E in [0, 1] s.t. for every open interval I in [0,1] m(I intersect E) > 0 & m(I intersect E^c) > 0 > > m is lebesgue measure > E^c is the complement of E A doubt. You say the Lebesgue measure. Are you implying that the constructed set must be Lebesgue measurable? Or the problem somehow gives us a notion that, even if the wording Lebesgue measure is replaced by Lebesgue outer measure,it is impossible to find a Lebesgue immeasurable set satisfying the condition as the problem indicates? > > > > This is so tricky! I was thinking something with the generalized Cantor set but everything I'm trying isn't working. Any suggestions? Ideas? ==== >> Can you construct a set E in [0, 1] s.t. for every open interval I in >[0,1] m(I intersect E) > 0 & m(I intersect E^c) > 0 >> >> m is lebesgue measure >> E^c is the complement of E >A doubt. You say the Lebesgue measure. Are you implying that the >constructed set must be Lebesgue measurable? Or the problem somehow >gives us a notion that, even if the wording Lebesgue measure is >replaced by Lebesgue outer measure,it is impossible to find a >Lebesgue immeasurable set satisfying the condition as the problem >indicates? By saying m(...) where m is Lebesgue measure, it's clearly asking for measurable sets. As far as outer measure m^* is concerned, you can do a lot better: there is a (nonmeasurable) set E such that for every open interval I, m^*(E intersect I) = m(I) and m^*(E^c intersect I) = m(I). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > Can you construct a set E in [0, 1] s.t. for every open interval I in [0,1] > m(I intersect E) > 0 & m(I intersect E^c) > 0 > > m is lebesgue measure > E^c is the complement of E take away fat disjoint Cantor sets K1 and J1. Now from B2 (K1 U J1) take away fat disjoint Cantor sets K2 and J2. Continue, and set E = ... (If this is a homework problem and you use this hint, be sure to give credit to sci.math.) X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h97D3uP19205; ==== I studied an alternative set theory which dissolves Russel's Paradox. In this theory, it is possible to get good theorems of ordinal number. More description about the set theory is available on my home page. http://boat.zero.ad.jp/~zbi74583/another02.htm I appreciate any comment about it. -------------------- ª@ 1.Axiom of Free Class. ª@ A. ma [A]A[A]B[A]x[A]y A ma x &ª@B ma y & A=B -> x=y This axiom means that any Atsumari makes only one class. A. el x[A]A[A]B ([A]a a el A <-> a el B) -> A=B This axiom means that an Atsumari is decided by its members. A.F [E]x{E]B ([A}a a el B <-> F(a)) & B ma x This is the Schema of comprehention axioms. Where A,B,C,...are variables for Atsumari, which means collection or set or pile in Japanese. a,b,c,...are variables for Class, [A] means all, [E] means exist, [E]! means exist only one, ma means makes, el means element, -> means then, <-> means equivalent, V means or, & means and, Atsu is an abbreviation of Atsumari ~ means negation, {a,b,c} means the Atsu which has members a,b,c 2.What does Free Class mean? ª@ [E]B a el B & B ma b means a el b in ZF (TR) For example if {a,c} ma b & {x,y} ma b are true, then a el {a,c} & {a,c} ma b so [E]B a el B & B ma b it means a el b in ZF. FC ZF {a,c} ma b corresponds a el b, c el b {x,y} ma b x el b, y el b 3. Russel's class [A]a a el R <-> ~([E]B a el B & B ma a) The right side of formula means ~(a el a) in ZF. This Atsu R makes Russel's Class r, so R ma r. Let a=r then, r el R <-> ~([E]B r el B & B ma r) If r el R is true, considering that R ma r is also true,then [E]B r el B & B ma r is true, then right side of formula is false. This is a contradiction. So, the following formulas are gotten. ~(r el R), [E]B r el B & B ma r. This conclusion means that the diagonal logic was dissolved. So, it is possible to think that 2^aleph(0)=aleph(0).ª@ X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 01:04 PM, zbi74583@boat.zero.ad.jp (Yasutoshi Kohmoto) said: > I studied an alternative set theory which dissolves Russel's >Paradox. There are many such. In what way is it more interesting or more useful than, e.g., GBN, ZFC? -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > I studied an alternative set theory which dissolves Russel's Paradox. > In this theory, it is possible to get good theorems of ordinal number. > More description about the set theory is available on my home page > http://boat.zero.ad.jp/~zbi74583/another02.htm I appreciate any comment about it. .81@ 1.Axiom of Free Class. > .81@ > Not using plain text in newsgroup makes reading difficult for some brousers. For example what does ^A@ mean? > A. ma [A]A[A]B[A]x[A]y A ma x &.81@B ma y & A=B -> x=y What does that mean? A makes for all A, for all B, for all x, for all y A makes x and ?? B makes y and A = B implies x = y Don't understand. What do you mean by 'A makes' ? > This axiom means that any Atsumari makes only one class. > A. el x[A]A[A]B ([A]a a el A <-> a el B) -> A=B > This axiom means that an Atsumari is decided by its members. > A.F [E]x{E]B ([A}a a el B <-> F(a)) & B ma x > This is the Schema of comprehention axioms. Where A,B,C,...are variables for Atsumari, which means collection or set or pile in Japanese. > a,b,c,...are variables for Class, > [A] means all, [E] means exist, [E]! means exist only one, ma means makes, > el means element, -> means then, <-> means equivalent, > V means or, & means and, Atsu is an abbreviation of Atsumari > ~ means negation, {a,b,c} means the Atsu which has members a,b,c > X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h97D48o19230; ==== I studied a generalized coloring problem. The ordinary coloring problem is defined as follows. ª@ ª@ª@Place different colors on two vertices which are next each other in a plane graph. ª@ª@How many colors are necessary and enough? ª@ ª@ª@And new coloring problem is the following. ª@ ª@ª@A definition ª@ª@D.1. ª@ª@Place different colors on two vertices which are near each other in a plane graph. ª@ª@How many colors are necessary and enough? ª@ ª@ª@The term near is defined as follows. ª@ª@For different two vertices a and b,ª@either condition 1. or 2. is filled. ª@ª@c.1. a is next of b ª@ª@c.2. There are three paths of length 2 between a and b. ª@ ª@ª@c.1 and c.2 is represented as (1,1) and (2,3) respectively. ª@ ª@ª@If G is a plane graph and ª@ª@if edges are added between all pairs of vertices which satisfy (2,3)ª@in G, ª@ª@then this graph is written as Near_2,3(G) ª@ ª@ª@New problem is also defined as follows. ª@ ª@ª@Another definition ª@ª@D.2. ª@ª@What is chr(Near_2,3(G))? ª@ª@ª@ª@ where chr(x) means the chromatic number of x. ª@ I proved a theorem as follows. [ Theª@ª@7 colors theorem. ] ª@ª@T.7C.ª@ª@7 colors are necessary and enough. More description about this problem is available on my HP. ª@ http://boat.zero.ad.jp/~zbi74583/another02.htm X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h97D4OF19281; ==== By a stroke of luck, I managed to find the missing piece to the puzzle of how to solve the resolvent septic. So here is the complete solution: Given: x^8-x^7+29x^2+29 = 0 Then, x1=(1+(a-b-c-d+e-f-g))/8 x2=(1-(a-b-c-d-e+f+g))/8 x3=(1-(a+b-c+d+e-f+g))/8 x4=(1+(a+b-c+d-e+f-g))/8 x5=(1-(a+b+c-d+e+f-g))/8 x6=(1+(a+b+c-d-e-f+g))/8 x7=(1-(a-b+c+d-e-f-g))/8 x8=(1+(a-b+c+d+e+f+g))/8 where the 7 variables a,b,c,d,e,f,g are the SQUARE ROOTS (positive case) of 4v+1 and the v's are the roots of the septic: 8903+47647v+39672v^2+7192v^3-522v^4-174v^5+v^7 = 0 (eq.3) with the solution: v=2(w^11+w^13+w^16+w^18)-2(w+w^12+w^17+w^28)-(w^2+w^5+w^24+w^27)+ (w^3+w^7+w^22+w^26)+(w^4+w^10+w^19+w^25)-(w^8+w^9+w^20+w^21) where w is ANY complex root of unity <>1 such that w^29=1. Note: Though there would be 28 such roots, the properties of these roots ensure that v will ONLY have 7 distinct values. I found the solution of v in Dave Rusin's website, and it's by P.Montgomery, though the solution wasn't used in the same way I used it. Eq.3 wasn't explicitly mentioned there but when I used the Integer Relations applet for a particular v, eq.3 popped out. It looked familiar and I realized i saw it before while trying to solve the resolvent (eq.2) of the prior post, namely: z^7-7z^6-2763z^5-19523z^4+1946979z^3+34928043z^2+119557031z-3247^2=0 (eq.2) and by letting z=4v+1, we get the new septic: 8903+47647v+39672v^2+7192v^3-522v^4-174v^5+v^7 = 0 (eq.3) So there it is. It's so nice to have completion. :) By the way, do SOME solvable 12-deg polys need an 11-deg resolvent? Sincerely, --Tito X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h97D4L919274; ==== > I have recently extended the Quaternions to larger sets by requiring >> some (new) group elements to commute. In doing so, I found this process >> and its results to be very asthetic. For one, the law of association is >> regained. However, the algebra involved is no longer a division algebra, >> i.e. we may not always follow x = 0 or y = 0 from xy=0 (when x and y are >> certain elements taken from a linear combination of group vectors). >... >> Has this type of thing been done before and are its conclusions of >> interest? It's obvious that there exist extensions of the quaternions H, >eg H + H (direct sum), >or algebras of matrices with quaternion elements. You'd have to say what properties your extension has >before anyone could say if it is of interest. I am neither refering to the ring of quaternion matrices nor to the >group >product... have sent you a copy of this work. My point was that you seemed surprised to find that >there were algebras extending the quaternions, >and I noted that this wasn't too surprising. So the fact that you have constructed such an algebra >could not be considered interesting in itself -- >any interest would have to lie in the special properties of the algebra. >-- >Timothy Murphy >tel: +353-86-2336090, +353-1-2842366 Absolutely no need to apologize for not having replied yet; it is good to know you recieved it, as I now asume to be the case. The group diagram picture I made on page 4 or 5 (I forgot) and its explanation in the text should tell a lot about the group's properties very quickly. By the way, the same process of extension (I call it reflection) can be used again and again to create more group elements, presumably proceeding to higher dimensions in the process- but Im not quite sure if this is somehow equivilant or to the procedure for extending Clifford Algebras (or, indeed, if my group is perhaps a Clifford algebra in disguise. Admitted, I don't know enough about Clifford algebras at the momement and am currently checking this possibility myself). Note also that many types of groups can be reflected, but this does not always give rise to a new group. For example, the trivial group {1,-1} does not change after reflection (the complex trivial group {1,-1, i, -i} does). Sincere thanks, Creighton Dement X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h97E9gZ23741; ==== >Take a chessboard (with or without infinetely many squares) let the >distance d((x_1,x_2),(y_1,y_2)) between two squares x and y of the >chessboard be defined as the minimum number of moves a knight takes >to reach y from x. >Is d a metric? Not trying to suggest that this is some new >question that hasn't been asked/answered before. Is there a general >formula for calculating d? More generally, the same question may be >asked for the other pieces (queen, king, knight, biship)? Actually, I >asked myself this question a few years ago. If I remember back to the >notes I took, I had something like (x_1-y_1, x_2-y_2)= (even number, >even number), then d(x,y) = even number. If (x_1-y_1, x_2-y_2)= (even >number, odd number), then d(x,y) = odd number. Finally, if (x_1-y_1, >x_2-y_2)= (odd number, odd number), then d(x,y) = even number. In other words, the same rules for adding natural numbers... C.Dement ==== > >Take a chessboard (with or without infinetely many squares) let the >distance d((x_1,x_2),(y_1,y_2)) between two squares x and y of the >chessboard be defined as the minimum number of moves a knight takes >to reach y from x. As others have pointed out, this obviously is a metric. As to the function (assume infinite board): 1) you undoubtedly know that the color of the square changes after each move, so you will need to separate into two cases according to the parity of x_1+x_2+y_1+y_2, and 2) the sets of squares that can be reached in exactly one or exactly two moves have a special, slightly irregular structure, BUT 3) for n>2, the set of squares that can be reached in exactly n moves has a very regular octagonal shape. DRAW IT!!!! (proof by induction) Item 1 and 3 gives you a rule to determine the distance, in the case that it is at least 3. You only need to check, whether the distance might be one or two. Jyrki Lahtonen, Turku, Finland ==== >Take a chessboard (with or without infinetely many squares) let the >distance d((x_1,x_2),(y_1,y_2)) between two squares x and y of the >chessboard be defined as the minimum number of moves a knight takes >to reach y from x. As far as I can see, this should give the distances (on an ordinary chessboard). The only exception I see is when your starting point O is at a corner, and you want to reach the next square on the diagonal. Then the distance is 4 and not 2. The paths are O -> P -> X -> A -> W -> S -> T. - - - - S W S W S W S T - - - - W A W A W S W S - - - - A W A W A W S W - - - - X A X A W A W S S X A X A X A X A W A W X A W P X P W A X A W S A W P X A X P X A W A W X A X A O A X A X A W S A X P X A X P X - - - - X A W P X P W A - - - - A X A X A X A X - - - - W A X A X A X A - - - - Karin ==== >Take a chessboard (with or without infinetely many squares) let the >distance d((x_1,x_2),(y_1,y_2)) between two squares x and y of the >chessboard be defined as the minimum number of moves a knight takes >to reach y from x. > > >Is d a metric? With this distance, the triangular equation is obviously true, and that makes it a metric. > Not trying to suggest that this is some new >question that hasn't been asked/answered before. Is there a general >formula for calculating d? More generally, the same question may be >asked for the other pieces (queen, king, knight, biship)? Actually, I >asked myself this question a few years ago. If I remember back to the >notes I took, I had something like (x_1-y_1, x_2-y_2)= (even number, >even number), then d(x,y) = even number. If (x_1-y_1, x_2-y_2)= (even >number, odd number), then d(x,y) = odd number. Finally, if (x_1-y_1, >x_2-y_2)= (odd number, odd number), then d(x,y) = even number. In other >words, the same rules for adding natural numbers... > > C.Dement > X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h97E9jb23746; ==== >on comp.lang.c: >> Why is this being discussed in comp.lang.c??? >> >Obviously because IT IS FULL OF JEWS!!! Jews are taking over >comp.lang.c because their greedy grubby need to take over everything is >finally seeping into the C language. Next thing you know they will be >trying to rewrite the standard. The entire reason for my low IQ and >inability to succeed in life can be attributed to jews. If it wasn't >for all the damn JEWS in science I wouldn't have to study! They keep >taking all the women too, being all nice and treating them with >respect and making me look like a complete ass. They took all the >jobs, now there is no point even looking for one. All I can do is sit >around all day filled with self pity and loathing for the damn JEWS who >did this to me. I hate my life and its all the fault of the Jew! > YOU are a total idiot. I thought he was being sarcastic. -- >/-- Joona Palaste (palaste@cc.helsinki.fi) ---------------------------| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| >| http://www.helsinki.fi/~palaste W++ B OP+ | >----------------------------------------- Finland rules! ------------/ >It sure is cool having money and chicks. > - Beavis and Butt-head If he was, then he is very bad at it. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h97FaaV30213; ==== > I have recently extended the Quaternions to larger sets by requiring > some (new) group elements to commute. In doing so, I found this process > and its results to be very asthetic. For one, the law of association is > regained. However, the algebra involved is no longer a division algebra, > i.e. we may not always follow x = 0 or y = 0 from xy=0 (when x and y are > certain elements taken from a linear combination of group vectors). >>... > Has this type of thing been done before and are its conclusions of > interest? >>It's obvious that there exist extensions of the quaternions H, >>eg H + H (direct sum), >>or algebras of matrices with quaternion elements. >>You'd have to say what properties your extension has >>before anyone could say if it is of interest. >>I am neither refering to the ring of quaternion matrices nor to the >group >>product... have sent you a copy of this work. >>My point was that you seemed surprised to find that >>there were algebras extending the quaternions, >>and I noted that this wasn't too surprising. >>So the fact that you have constructed such an algebra >>could not be considered interesting in itself -- >>any interest would have to lie in the special properties of the algebra. >>-- >>Timothy Murphy >>tel: +353-86-2336090, +353-1-2842366 Absolutely no need to apologize for not having replied yet; it is >good to know you recieved it, as I now asume to be the case. The >group diagram picture I made on page 4 or 5 (I forgot) and its >explanation in the text should tell a lot about the group's >properties very quickly. By the way, the same process of extension > (I call it reflection) can be used again and again to create more > group elements, presumably proceeding to higher dimensions in the >process- but Im not quite sure if this is somehow equivilant or to >the procedure for extending Clifford Algebras (or, indeed, if my >group is perhaps a Clifford algebra in disguise. Admitted, I >don't know enough about Clifford algebras at the momement and am >currently checking this possibility myself). Note also that many >types of groups can be reflected, but this does not always give rise >to a new group. For example, the trivial group {1,-1} does not change > after reflection (the complex trivial group {1,-1, i, -i} does). >Sincere thanks, >Creighton Dement X-Received: from secure.server-3.com (secure.server-3.com [64.191.20.132]) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id h97G9do32618 X-Received: from sps by secure.server-3.com with local (Exim 4.24) id 1A6uP0-0003DA-6d ====
 
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X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h97JmJl15824; ==== I know it is a bit stupid, but 1> how do you prove that a discrete topology is metrizable? 2> X is an set of all positive integers and T={{},{1,2,3,4...},{2,3,4...},{3,4...},{4...},...} Why is (X,T) not metrizable? > As a topological space spacetime is metrizable, but one does not, > in general, look at any particular metric (in the topological sense) > on it. Why Severian thinks that this Hausdorf topology on Space-Time > is adequate to its physical sence and geometrical structure? I care nothing for physical sence (sic). All I was doing was pointing > out that the term metric was being used in two different ways > (like many words in mathematics). Do you mean as two ways::::::::::::::::::::::::::::::::::::::::::::::::::::::: > ### (pseudo-)Riemann metric -- quadratic form > and > ### metric (distance) > ? Yes, they're different, but the first is applicable only to manifolds. Yes, I know that, but the original poster was confused by these two > distinct concepts having the same word. > Of course the two concepts are related this way. So? The > original poster was talking about one version of metric > in terms of the definition of the other version - that's going > to cause confusion, regardless of the fact that the two > are related, so pointing out that they are two different > things seems like a good idea. >>Yes. But who said about what NAMELY two concepts we are speaking? The original poster. In the original question. Where he said something >about metrics that applied to one version, then asked about what >he said in re an instance of the other version. >...And note how these concepts are related and how it can be applied to S.T. >>seems also a good idea... > More precisely: The standard topology of Euclidean space is induced by metric. You are failing to specify which usage of the term metric you are using > here. >>Directly, I used the metric (distance). >>But the Euclidean distance is a partial case of Riemann distance, >>so not important which I use distance or quadratic form. > The standard topology of Minkowski space isn't. Minkowski space is metrizable. > >He didn't say it _was_ metrizable by the Minkowski metric. >He said it was metrizable. It is. You said The standard topology of Euclidean space is induced >by metric. The standard topology of Minkowski space isn't. >We assumed that induced by metric meant induced by >some metric. What _did_ you mean by induced by metric??? If >induced by metric means induced by Minkowski >metric then the standard topology on euclidean >space is _not_ induced by metric... >You implicitly defined on it >>the standart (Hausdorf) topology of finite-dimension linear space >>and said that it's metrizable, as all finite-dimension linear spaces! >>Original question was NOT ABOUT SUCH TOPOLOGY, >>but IS Minkowski A METRIC on the Space-Time? . >>The answer is: >>___________________________________________________________________ >>[ Minkowski metric on the Space Time ( dx0^2 -dx1^2 -dx2^2 -dx3^2 ) >>[ not leads to any metric (distance function) in classical sence. That's correct. Several people have explained this already. Nobody has >disagreed with this. >~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ > The standard topology of Euclidean space is > the only (non-trivial) topology invariant to motions (rotation, mirror, shift). Is it? Proof? >>Sorry, in general it isn't. I mistaken. >>It's true that any non-one-point and non-discrete invariant topology >>is identical to standard IN ANY BOUNDED DOMAIN, Say _exactly_ what this means (and give a hint of the proof.) Seriously, I cant figure out what it means. Because I can't >see what it means for a topology on a BOUNDED DOMAIN >to be invariant under shifts, for one thing. Nor under rotations, >unless it happens that the domain is invariant under rotations. >but at infinity exist some different cases... >>Except standard topology, it may be some COMPACTIFIED topologies. >>One of them is like standard but accepts only bounded closed subsets >>(or, the same, only open subsets which are neighbourhoods of infinity). >>Is there another cases, I don't know yet. >>I would ignore this fact a long time... > The standard topology of Minkowski space is invariant but not unique. In Euclidean case, all smooth maps from RÕ to the Space are CONTINIOUS > and may represent a point trajectory in Newtonian mechanics. > In the Space-Time, all smooth maps RÕ->M are continious (in std. top.), > but not all are physically allowed. physically allowed! >>This mean that the 4-speed is not space-like: >>( d x / d tau )^2 >= 0 >>and that we have a correct time sign: >>( d x0 / d tau ) >= 0 > What bullshit. >>*** Anecdote: >>enter expression: cos ( pi / 2 ) >>Syntax error! >>enter expression: 1 * 0 >>Syntax error! >>enter expression: 2 + 2 >>Syntax error! >>enter expression: .8f.9b.90 .99.8c.89.8c >>.8f.9b.90 is not an argument! >>-- >>qq~~~~>/ / > /_/ / >> ____/ > ==== > I know it is a bit stupid, but > > 1> how do you prove that a discrete topology is metrizable? Write down a metric for it :-) The simplest possible metric should give the discrete topology ... > 2> X is an set of all positive integers and > T={{},{1,2,3,4...},{2,3,4...},{3,4...},{4...},...} Why is (X,T) not > metrizable? Is it Hausdorff? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h97JmW515882; ==== >>Take a chessboard (with or without infinetely many squares) let the >>distance d((x_1,x_2),(y_1,y_2)) between two squares x and y of the >>chessboard be defined as the minimum number of moves a knight takes >>to reach y from x. >>Is d a metric? Not trying to suggest that this is some new >>question that hasn't been asked/answered before. Is there a general >>formula for calculating d? More generally, the same question may be >>asked for the other pieces (queen, king, knight, biship)? Actually, I >>asked myself this question a few years ago. If I remember back to the >>notes I took, I had something like (x_1-y_1, x_2-y_2)= (even number, >>even number), then d(x,y) = even number. If (x_1-y_1, x_2-y_2)= (even >>number, odd number), then d(x,y) = odd number. Finally, if (x_1-y_1, >>x_2-y_2)= (odd number, odd number), then d(x,y) = even number. In other words, the same rules for adding natural numbers... C.Dement >Sorry for not quite completing the question above (even though, >perhaps, obvious): >Is d a metric over the product space of whole/natural numbers >(corresponding to two different cases of an infinite chess board) >or the restriction of the product space of natural numbers to 64 elements? X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h97JmTA15867; ==== A grammar would be S -> BaB B -> BB B -> aBb B -> bBa B -> lambda I doubt there is a way to prove it, or find a base case ==== >A grammar would be S -> BaB B -> BB >B -> aBb >B -> bBa >B -> lambda >I doubt there is a way to prove it, or find a base case It's easy to see that B gives any string with the same number of a's and b's, so we just need to show that if a string has one more a's than b's it can be written as BaB. Say s is such a string, and say s[n] is the initial substring of s of length n. Now s = s[length(s)] has more a's than b's, so there exists an n such that s[n] has this property. Let N be the _smallest_ n such that s[N] has more a's than b's. Then s[N-1] must have the same number of a's and b's and the N-th character must be a, QED. ************************ David C. Ullrich X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h97KV2u19406; ==== It is possible prove the Ternary Goldbach Conjecture (TGC) and the Twin Prime Conjecture (TPC) are true, if the Generalized Riemann Hypothesis (GRH) is true. See http://www.ams.org/era/1997-03-15/S1079-6762-97-00031-0/S1079-6762-97-00031- 0.pdf for the proof of GRH-->TGC. Is there a similar paper for the converse? If the TGC or TPC is true then, the GRH or (RH) is true. My question is can either of the following be proved? TGC-->GRH or TPC-->GRH John Washburn ==== > It is possible prove the Ternary Goldbach Conjecture (TGC) and the Twin Prime > Conjecture (TPC) are true, if the Generalized Riemann Hypothesis (GRH) is > true. GRH implies twin primes? News to me. > Is there a similar paper for the converse? If the TGC or TPC is true then, > the GRH or (RH) is true. Not to my knowledge. -- ==== >Suppose I want a 'large' computably enumerable collection of functions >f_i : N --> N with the following properties: >1. Each primitive recursive function is included. >2. Each f_i is total by construction. >3. Given i and n in N, there is a computable function time: N x N --N which tells me that the value of f_i(n) will take at most time(i,n) >to compute by a turing machine or equivalent. >['Take as long as you want, but PLEASE tell me when you will be >done!'] >4. The function time is computable in 'Ackermann + constant' time, >or at least it's behavior is boundedly nasty in some sense. :) There was a paper 40 years ago somewhat along these lines. If you want to be able to predict how long a computation will take, then you have severely restricted the class of functions. Here is the reference: Robert W. Ritchie Classes of predictably computable functions Transactions of the American Mathematical Society, vol. 106 (1963), pp. 139-173 It might help. --Herb Enderton ==== > I took a class in basic foundations and computability last year, but > there is a question that has been bothering me. Suppose I want a 'large' computably enumerable collection of functions > f_i : N --> N with the following properties: > 1. Each primitive recursive function is included. > 2. Each f_i is total by construction. > 3. Given i and n in N, there is a computable function time: N x N -- N which tells me that the value of f_i(n) will take at most time(i,n) > to compute by a turing machine or equivalent. > ['Take as long as you want, but PLEASE tell me when you will be > done!'] > 4. The function time is computable in 'Ackermann + constant' time, > or at least it's behavior is boundedly nasty in some sense. :) I feel rather queasy about the prospects for this, but it has been > bothering me for too long. The basic theme is this: I want to know > worst case scenario computation time. Will I not get much besides > primitive recursive functions? Rex Butler PS I've heard the term 'strongly computable.' Is this related? Its 10 lines of code (actually a 5 state Turing Machine) and there is no proof if it terminates. Only 10 lines and no value for f_i. In theory your class could be all well defined and well behaved functions, but in practice they will only be trivial. Say you are nesting loops, then the invariants at each level would have to be combined into parallel equations, you have to solve all enumerations of possible equations, and show they are satisfied for each loop to terminate. Even then the class of functions has no scope, here is a tiny 3 state TM enough but how do you categorise what function it is? ~ binary counter. Herc ==== Gauss-Jordan elimination. I know text books are constantly using one-way traffic flow analysis and such, and I really enjoyed this subject. However, I am taking a technical writing class and wanted to do a paper on using gauss-jordan elimination to solve real world problems....The problem is I can't find any resources for my research. Any help would be greatly appreciated. Keith ==== > Gauss-Jordan elimination. I know text books are constantly using > one-way traffic flow analysis and such, and I really enjoyed this > subject. I think you are asking when might one use numerical methods to solve large systems of linear equations. One application is in economic planning using Leontief input-output planning. I actually read a science fiction novel last week joked about this application. I have Gauss-Jordan elimination built into my applet for Linear Programming mentioned in my sig. LPs have a wide variety of applications. -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau ==== >The peephole you view your world through is very narrow, Longley. >There are many here who think the following matters are the crux of >building AI's: How do you recognize what you see? >How do you know how to move your arm? >How do you choose which words to say? >How do you understand what they mean? >How does commonsense reasoning work? > > Stop misquoting Lennon & McCartney! > > how do you recognize just what you've seen? > i can't tell you but i know it's mine > how do you know how to move your own arm? > i get by with a little help from my friends > how do you choose which big words to say? > lend me your ears and i'll sing you a song > how do you understand what they mean? > i try not to sing out of key > does commonsense reasoning work? > yes i'm certain that it happens all the time > > > oh building AIs with a little help from my friends > Brilliant, Lisa - I hope the originator of the quotes I gave can make use of your added insights. =============== > * * * > > > weary with foils, i fall into my bed > and sought some rest for limbs with dust attired, > but there began a trial in my head > to obsess, when body's overtired: > my thoughts then race, unlike their daily pace > when they escape and leave me little grace; > for verbal slings and arrows of bon mots > i am left gawking bereft of ripostes. > look in darkness on what might have been said > had i the wits to speak as those admired: > rejoinders, like galaxies, hang in space > and taunt me in my unrestful repose. > this sort of pesky followup is never very far > when you keep sending crosspostings to here, talk dot bizarre. > > > . > . thank you thank you i'll be here at dogberry's fencing school all weekend Why this thread was x-posted to the other forums, I don't know - but there's clearly a lot more humor on t.dot.b than on c.a.p. - bon mots, rather than non bots [... had i the wits ...]. So, I shall assume Cyrano is the patron saint of t.dot.b. ==== are the limit points of Q exactly R? well, I know this is true, and I can prove it. the reason i ask though is because i doubt myself on this. let (xn) be a sequence in the set of all rationals 0<=p<=1. Further, suppose xn is an enumeration of the rationals between 0 and 1. i think that all rationals between 0 and 1 are cluster points of xn, mainly because all are limit points. i could also work it from the definition since a rational lies between any two rationals, but i started doubting myself for some reason. can someone tell me if I am wrong about anything in the above? ==== It actually depends on the valuation you are using. If your valuation is the normal absolute value, then you are correct. But know that for each prime there exists a valuation called the p-adic valuatian, and in these cases your question can't even be asked. ==== Sure the question can be asked, it just that the larger field isn't R, it's the p-adic rationals Q_p. And indeed, Q is dense in Q_p with respect to the p-adic absolute value. Actually, if you define R to be the (unique up to isomorphism) completion of Q with respect to the usual absolute value, then it's pretty immediate that Q is dense in R, and ditto Q being dense in Q_p. More generally, if K is any field and |x| is an absolute value of K, then there is a unique (up to isomorphism) smallest field L containing K with the properties that |x| extends to an absolute value on L and L is complete wrt this absolute value. Then one can prove that K is dense in L. An interesting example if K = R(X), the field of rational functions with real coefficients, and the absolute value is |f(x)| = e^{-deg(f)}. The completion of K is often denoted L = R((X)), it's the field of formal (i.e., not necessarily convergent) Laurant series. > It actually depends on the valuation you are using. If your valuation > is the normal absolute value, then you are correct. But know that for > each prime there exists a valuation called the p-adic valuatian, and > in these cases your question can't even be asked. ==== > are the limit points of Q exactly R? > Yes. As every open interval about any r in R contains numerous rationals other than r, r is a limit point of Q. If you want to construct a sequence in Q with limit r. For each integer n > 0, let I_n = (r - 1/n, r + 1/n) and pick a rational q_n in I_n (If r is a rational, pick q_n so it isn't r.) To conclude, show lim(n->oo) q_n = r > well, I know this is true, and I can prove it. the reason i ask though > is because i doubt myself on this. let (xn) be a sequence in the set of all rationals 0<=p<=1. Further, > suppose xn is an enumeration of the rationals between 0 and 1. i think > that all rationals between 0 and 1 are cluster points of xn, mainly > because all are limit points. i could also work it from the definition > since a rational lies between any two rationals, but i started > doubting myself for some reason. > That is correct, every rational is a cluster point of an enumeration of the rationals. How is that used to show the limit points of Q are R? > can someone tell me if I am wrong about anything in the above? > ==== > are the limit points of Q exactly R? > > well, I know this is true, and I can prove it. the reason i ask though > is because i doubt myself on this. > > let (xn) be a sequence in the set of all rationals 0<=p<=1. Further, > suppose xn is an enumeration of the rationals between 0 and 1. i think > that all rationals between 0 and 1 are cluster points of xn, mainly > because all are limit points. i could also work it from the definition > since a rational lies between any two rationals, but i started > doubting myself for some reason. > > can someone tell me if I am wrong about anything in the above? Yes: you are wrong to doubt yourself. -- ==== i'm extremely confused as to how to change the system of ode's into a matrix version. let's say you have the the lorenz system. { x'[t] = - a*y[t] - a*z[t], y'[t] = r*x[t] + y[t] - x[t]*z[t] , z'[t] = x[t]*y[t]- b*z[t]} where a, r, b, are constants, how do i represent the above as a matrix differential equations system? and then find the eigenvalues and so on. any help is most appreciated. thanks john ==== > > i'm extremely confused as to how to change the system of ode's into a > matrix version. > > let's say you have the the lorenz system. > > { x'[t] = - a*y[t] - a*z[t], > y'[t] = r*x[t] + y[t] - x[t]*z[t] , > z'[t] = x[t]*y[t]- b*z[t]} > > where a, r, b, are constants, > > how do i represent the above as a matrix differential equations > system? > > and then find the eigenvalues and so on. > > You want to write [x y z]^T as a column vector: [ x ] X = [ y ] [ z ] and then express the system of DEs as follows: X' = AX where A is some 3x3 matrix. Since the ith row of the product AX is the product of the ith row of the matrix A with the single column of X, and since the individual equations are sums of (something) times (x,y,z) in various combinations, what you try to do is to populate the matrix A with suitable coefficients that make that happen. For instance, x' = -a y - a z can be written as follows: [ x ] x' = [0 -a -a] [ y ] [ z ] Note how the coefficient of x (namely, 0) got put into the x position, the coefficient of y (that is, -a), and the coefficient of z (again -a), were placed into the y and z positions, respectively, of the row vector. The equations for y' and z' are not remarkably different, except for the fact that you can't use constant coefficients, but then this isn't a DE with constant coefficients. Once you have each DE written in this form: x_i = R_i*X with a row vector R_i, then put them together to build the matrix A, mentioned earlier, together. Dale. > any help is most appreciated. > > thanks > > john ==== ... stuff deleted ... Some stuff, with a badly-spaced equation: > For instance, > > x' = -a y - a z > > can be written as follows: > > [ x ] > x' = [0 -a -a] [ y ] > [ z ] > I just have to fix the spacing here. Won't be a minute. [ x ] x' = [0 -a -a] [ y ] [ z ] Let's see whether that works. > Dale. > ... that's it, I guess ... > Dale ==== The inertia of a slowly moving object; body or mass of material substance will increase, or decrease; in proportion to an increase, or decrease in its speed, will it not? Can this be plotted as a component of its motion? Maybe called its momentum, or impetus? If so, what would this component look like for a body moving tranversely across a slope? ==== I would greatly appreciate some comments upon the correctness of the assertion about the following equation (1) under the given conditions. y = (a^m + b^m)(a^m - b^m) (1) Conditions: (y, a, b ) = 1; m is a non-integer > 0; prime p > 3. Assertion: If y is a p-th power then both a^m + b^m and a^m - b^m separately be p-th power. ==== > I would greatly appreciate some comments upon the correctness of the > assertion about the following equation (1) under the given conditions. y = (a^m + b^m)(a^m - b^m) (1) Conditions: (y, a, b ) = 1; m is a non-integer > 0; prime p > 3. Assertion: If y is a p-th power then both a^m + b^m and a^m - b^m > separately be p-th power. It all depends on what you mean by (y, a, b) = 1. Without any context to go by, I'd interpret it to be y=a=b=1, in which case the assertion is true, since whenever lhs (=1) = rhs (=0), anything at all is true. Of course, I could be misinterpreting the symbols. Why not say what you mean in words? Jon Miller Sender: phil@nonospaz.fatphil.org ==== > I would greatly appreciate some comments upon the correctness of the > assertion about the following equation (1) under the given conditions. > > y = (a^m + b^m)(a^m - b^m) (1) > > Conditions: (y, a, b ) = 1; m is a non-integer > 0; prime p > 3. > > Assertion: If y is a p-th power then both a^m + b^m and a^m - b^m > separately be p-th power. If you're happy with 1^p being a p-th power, then the first half of your assertion is unnecessary. If you want the unit to not be described as a p-th power, then your assertion is vacuously true. And what algebraic structure are you working in? I see no reason to not assume you're working in C, and therefore every number is a p-th power. Phil ==== > I would greatly appreciate some comments upon the correctness of the > assertion about the following equation (1) under the given conditions. > > y = (a^m + b^m)(a^m - b^m) (1) > > Conditions: (y, a, b ) = 1; m is a non-integer > 0; prime p > 3. > > Assertion: If y is a p-th power then both a^m + b^m and a^m - b^m > separately be p-th power. Huh? If m is a non-integer then it seems unlikely to me that a^m + b^m will be an integer. In what sense is sqrt 13 + sqrt 5 a 3rd power? -- Sender: jonam@buffalo.edu ==== Can the Riemann-Zeta Hypothesis possibly be explained to a math major who has only just begun to study Real Analysis? What is the significance of zero solutions lying on a critical line? And what substantial mathematical theorems are dependent on a hypothesis that has yet to be proven? I feel like a twelve year old leafing through Wiles' proof of Fermat's Last Theorem. Xevious ==== > Can the Riemann-Zeta Hypothesis possibly be explained to a math major > who has only just begun to study Real Analysis? What is the > significance of zero solutions lying on a critical line? And what > substantial mathematical theorems are dependent on a hypothesis that > has yet to be proven? > > I feel like a twelve year old leafing through Wiles' proof of Fermat's > Last Theorem. Interestingly, there are no fewer than four recent pop books on the Riemann Hypothesis. (Probably because of its appearance on the Clay Mathematics Institute list of problems.) One sees: Derbyshire, Prime Obsession du Sautoy, The Music of the Primes Sabbagh, The Riemann Hypothesis Edwards, Riemann's Zeta Function I've read only Derbyshire and du Sautoy, and despite the fact that Derbyshire is a journalist and columnist, he illustrates (not proves) more math than du Sautoy, who teaches math at Oxford (though he is a journalist too). Both of these have a (fairly consistent) account of the history, but Debyshire spends more time in giving a taste of complex analysis, while du Sautoy is broader in his approach to the story parts, especially more recent stuff. Dennis ==== > Interestingly, there are no fewer than four recent pop books > on the Riemann Hypothesis. (Probably because of its > appearance on the Clay Mathematics Institute list of problems.) > > One sees: > Derbyshire, Prime Obsession > du Sautoy, The Music of the Primes > Sabbagh, The Riemann Hypothesis > Edwards, Riemann's Zeta Function Whatever your concept of pop book is, it certainly does not apply to Edwards' book. Besides, it is not recent, since it was originally published in 1974. Jose Carlos Santos ==== > Xevious wonders in message > Can the Riemann-Zeta Hypothesis possibly be explained to a math major > who has only just begun to study Real Analysis? What is the > significance of zero solutions lying on a critical line? And what > substantial mathematical theorems are dependent on a hypothesis that > has yet to be proven? > > Interestingly, there are no fewer than four recent pop books > on the Riemann Hypothesis. (Probably because of its > appearance on the Clay Mathematics Institute list of problems.) > > One sees: > Derbyshire, Prime Obsession > du Sautoy, The Music of the Primes > Sabbagh, The Riemann Hypothesis > Edwards, Riemann's Zeta Function Another book that deserves mention here is Julian Havil's book, Gamma. It's about Euler's constant, gamma, but it gets around to some useful material on the zeta function. I think all of these books are reviewed on the MAA website so you can see another opinion before you jump in. Executive summary: the zeta function can be thougnt of as a generating function for the primes. Anything we learn about the zeros of zeta has immediate implications for the distribution of the primes. Just about any quantitative question about prime numbers, or things that depend on prime numbers like divisors, can be answered more precisely the more we know about the zeros of zeta. -- ==== > > Is there any algorithm to calculate the deep holes of leech lattice? > > Yes, it's not hard to find one that calculates all holes of any lattice. Could you please tell me the method? I probably don't need all the deep holes > > Any software can do that? I tried MAGMA but MAGMA gave no result > > I'm sure that on a lattice as complicated as Leech any known > algorithm would be totally intractible. > > I read SPLAG however I am not clear about it > > Pity. > > I wanna have the whole collection of leech lattice deep holes > > is the best account you're going to find. Finding > the Leech lattice holes was a major piece of research. Although > finding holes is a computable problem, just running some > general-purpose algorithm on Leech would take too much time. ==== >> >> Is there any algorithm to calculate the deep holes of leech lattice? >> >> Yes, it's not hard to find one that calculates all holes of any lattice. > Could you please tell me the method? I probably don't need all the deep Simple but impractical algorithm. Each hole is the circumcentre of some n-simplex with vertices in the lattice L (of rank n). To find all holes then find all curcumcentres of n-simplices with vertices in L. Problem: infinitely many of them! Resolution: really only want to find those in a fundamental parallelotope. If this has diameter D, then only need to consider vertices distance <= 2D from origin. Now finite problem. Simple, but hopeless in practice. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== lim n->oo ((sum n)^n - sum(n^n)) / n!^2 = 1 (sum n)^n - sum(n^n) = n! ^2 (sum n)^n = sum(n^n) + n! ^2 ((n^2+n)/2)^n = sum(n^n) + n! ^2 (n^2+n)^n = 2^n ( sum (n^n) + n! ^2 ) n^n (n+1)^n = 2^n ( sum (n^n) + n! ^2 ) The sum of the integers from {1, 2, ..., n ,...} to the n'th power, (1+2+...+n+...)^n is greater than the sum of the numbers of the n'th power {1^n, 2^n, 3^n, ..., n^n, ...}, (1^n + 2^n + ... + n^n + ... ). Their difference is the square of the factorial of n: 1^2 2^2 ... (n-1)^2 n^2 .... That is to say, lim n->oo ((sum n)^n - sum(n^n)) / n!^2 = 1. Stirling's approximation for n! is as so: lim n->oo ( n! e^n ) / ( n^n sqrt(2pi n) ) = 1 Thusly: lim n->oo ( (sum n)^n - sum(n^n) ) e^2n / ( n^2n 2 pi n) ) = 1 lim n->oo ( (n+1)^n - ( sum(n^n) / n^n ) ) e^2n / (2^n n^n 2pi n ) = 1 lim n->oo ( (n+1)^n - (sum(n^n)/n^n) ) e^2n / ( 2^(n+1) n^(n+1) pi) = 1 As described earlier in this thread, other equations describe n! in the limit. Euler lived in the 1700's. Euler did some amazing work. I wonder if there is a closed for sum(n^n). There are closed forms for sum(n^x), they are defined in terms of Bernoulli polynomials the coefficients of which are generated by recurrence relation. For small values of n, sum(n^n): n = 1, sum (n^n) = 1 n = 2, sum (n^n) = 5 n = 3, sum (n^n) = 36 n = 4, sum (n^n) = 354 n = 5, sum (n^n) = 4425 With (sum n)^n: n = 1, (sum n)^n = 1 n = 2, (sum n)^n = 9 n = 3, (sum n)^n = 216 n = 4, (sum n)^n = 10000 n = 5, (sum n)^n = 759375 With n! ^2: n = 1, n! ^2 = 1 n = 2, n! ^2 = 4 n = 3, n! ^2 = 36 n = 4, n! ^2 = 576 n = 5, n! ^2 = 14400 It appears that as n diverges that (sum n)^n - sum (n^n) ~= (sum n)^n, yet it is of course less than (sum n)^n, and it appears to be n! ^2. I guess I should start calculating ((sum n)^n - sum (n^n)) / n!^2 for small and increasing values of n and see if it holds true that it tends towards unity. I read Richmond and Merlini's paper, as mentioned earlier, about generalizations of Stirling cycle numbers [ n x ] to complex arguments where n-x is an integer, and don't understand it but it looks interesting. I'm more interested currently in finding closed form solutions for [ n+1 n-x+1 ], or rather |s(n+1, n-x+1)|, to evaluate (n+1)(n+2)...(n+n), in the strange case of the reformulated Gamma function. Ross ==== I got to thinking about sum 2^n and how it was equal to 2^(x+1) - 1. I wonder: does it work for 3? The answer is not quite, no. For example sum 3^0 = 1, 3^1-1= 2* sum 3^0, 3-1=2*1, sum 3^1 = 4, 4*2=10-2. Anyways here is a short list of values of sum(3^n): 1 = 3^0 4 = 3^0 + 3^1 13 = 3^0 + 3^1 + 3^2 40 = 3^0 + 3^1 + 3^2 + 3^3 121 = 3^0 + 3^1 + 3^2 + 3^3 + 3^4 and values of 3^n: 3 = 2*1+1 9 = 2*4+1 27 = 2*13+1 81 = 2*40+1 Thus it appears that the sum of 3^i for i from zero to n is (3^(n+1)-1) / 2 For sum 4^n: 1 5 21 85 341 4^n: 1 4 16 64 256 1024 It appears that sum ((4^n) = 4^(n+1) - 1 ) / 3. I would thus surmise that sum(x^n) = x^(n+1)-1)/(x-1), the sum of x^i for i=1 to n equal (x^(n+1)-1)/(x-1). Where should I look to find a proof of this, so I don't have to write one, and that I can read the surrounding material to which it refers to gain understanding of its concept? This doesn't help me solve sum(n^x), because that is the sum of i^x for i=1 to n, the closed form of that quantity is determined as has been noted by the Bernoulli polynomials, sum of powers. I guess after that would be sum(n^n). Anyways in the limit n! = sqrt( sum(n)^n - sum(n^n) ), another in the long list of approximations of n!. Ross ==== > I got to thinking about sum 2^n and how it was equal to 2^(x+1) - 1. > I wonder: does it work for 3? The answer is not quite, no. Look up geometric series. If you must reinvent the wheel, you need not do in so publicly. ==== > > I got to thinking about sum 2^n and how it was equal to 2^(x+1) - 1. > I wonder: does it work for 3? The answer is not quite, no. > > Look up geometric series. > > If you must reinvent the wheel, you need not do in so publicly. I guess it's kind of like how the sum for i=1 to oo of b^-i is 1/(b-1). The concept here is the repeated unit, the rep-unit, in base b. The sum for i=1 to n of b^i is the sequence 1_n 1_(n-1) ... 1_2 1_1, a sequence of 1's of length n representing an integer in base b. Then b^(n+1)-1 is a sequence of n many (b-1)'s in base b: (b-1)_n (b-1)_(n-1) ... (b-1)_2 (b-1)_1, dividing that by (b-1) has as a result the rep-unit of length n. In summing the negative integers powers of an integer, it is similar, represent the number in that base so that the sum is simply .1111.... If you multiply that by (b-1) then the result is .(b-1) (b-1) (b-1) (b-1) ... = 1. These are well-known and almost trivial, for example, the sum for i=1 to infinity of 1/25000^i is 1/24999, 1/9 in decimal is .1111..., etcetera. I assume that for real x>1 that the sum of x^-i for i from 1 to oo is 1/(x-1). I will research geometric series. I'm not here to reinvent the wheel, but it damn well better let me say that half the integers are even, as they are as the asymptotic density of the even numbers in the integers is one half. What do you think about the canonicalization of the infinite binary sequences? That is to say, for the sequence .010101(01)... with half zeroes and half ones, that you could exchange any two sequence elements any number of times and not get .001001001(001).... Is there a rule to exchange (permute) the elements of the sequence .0101(01)... to get 00010001(0001)...? Ross ==== > I will research geometric series. I'm not here to reinvent the wheel, > but it damn well better let me say that half the integers are even, as > they are as the asymptotic density of the even numbers in the integers > is one half. Doing any sort of research in mathematics with a fixed idea of what you will allow yourself to find is not really a good idea. > > What do you think about the canonicalization of the infinite binary > sequences? That is to say, for the sequence .010101(01)... with half > zeroes and half ones, that you could exchange any two sequence > elements any number of times and not get .001001001(001).... Is there > a rule to exchange (permute) the elements of the sequence .0101(01)... > to get 00010001(0001)...? It depends on what you consider an allowable permutation. The most general form of such permutation can be considered as follows. Let N be the set of positive integers, then any infinite binary sequence can be uniquely represented by a functions f:N -> {0,1} in which f(k) is the k'th digit of the sequence. Let B be the set of such functions f:N -> {0,1}. Then any bijection g:N <-> N creates a permutation G: B -> B by defining G(f) = fog, where o represents function composition, G(f) is the function whose valueat k in N is f(g(k), i.e., G(f)(k) = f(g(k)), for each k in N. Let P represent the set of all such permutations, It is clear that the number of zeros and number of ones in an expansion must remain fixed under any such permutation of digits. Then, for a given f:N -> {0,1} and h:N -> {0,1} in B, a necessary and sufficient condition for there being some permutation G in P such that G(f) = h is that card(k in N: f(k) = 0}) = card(k in N: h(k) = 0}) and card(k in N: f(k) = 1}) = card(k in N: h(k) = 1}) i.e., the infinite binary sequences for f and h have the same number, possibly infinite, of zeros and the same number, possibly infinite, of ones, and only one of these can be finite. But if the number of zeros or the number of ones differ, then there is no permutation carrying one into the other. This divides up the set B into equivalence classes of binaries permutable into each other which can be indexed by the integers, Z. Any two such binaries with infinitely many zeros and infinitely many ones in their expansions can be permuted into each other, and can be indexed by 0. If the number of ones is finite, use that number as index, and if the number of zeros is finite use the negative of that number as index. Then two binaries can be permuted into each other if and only if they have the same index. ==== > > I will research geometric series. I'm not here to reinvent the wheel, > but it damn well better let me say that half the integers are even, as > they are as the asymptotic density of the even numbers in the integers > is one half. > > Doing any sort of research in mathematics with a fixed idea of what > you will allow yourself to find is not really a good idea. > > What do you think about the canonicalization of the infinite binary > sequences? That is to say, for the sequence .010101(01)... with half > zeroes and half ones, that you could exchange any two sequence > elements any number of times and not get .001001001(001).... Is there > a rule to exchange (permute) the elements of the sequence .0101(01)... > to get 00010001(0001)...? > > It depends on what you consider an allowable permutation. > > The most general form of such permutation can be considered as > follows. > > Let N be the set of positive integers, then any infinite binary > sequence can be uniquely represented by a functions f:N -> {0,1} in > which f(k) is the k'th digit of the sequence. > > Let B be the set of such functions f:N -> {0,1}. > > Then any bijection g:N <-> N creates a permutation G: B -> B by > defining G(f) = fog, where o represents function composition, G(f) > is the function whose valueat k in N is f(g(k), > i.e., G(f)(k) = f(g(k)), for each k in N. > > Let P represent the set of all such permutations, It is clear that > the number of zeros and number of ones in an expansion must > remain fixed under any such permutation of digits. > > Then, for a given f:N -> {0,1} and h:N -> {0,1} in B, a necessary > and sufficient condition for there being some permutation G in P > such that G(f) = h is that > card(k in N: f(k) = 0}) = card(k in N: h(k) = 0}) and > card(k in N: f(k) = 1}) = card(k in N: h(k) = 1}) > i.e., the infinite binary sequences for f and h have the same > number, possibly infinite, of zeros and the same number, possibly > infinite, of ones, and only one of these can be finite. > > But if the number of zeros or the number of ones differ, then there > is no permutation carrying one into the other. > > This divides up the set B into equivalence classes of binaries > permutable into each other which can be indexed by the integers, Z. > > > Any two such binaries with infinitely many zeros and infinitely many > ones in their expansions can be permuted into each other, and can be > indexed by 0. If the number of ones is finite, use that number as > index, and if the number of zeros is finite use the negative of that > number as index. > > Then two binaries can be permuted into each other if and only if > they have the same index. B is the set of functions f(n) -> {0,1}, n E N. There are uncountably many such functions, each represents a subset of N elements or a distinct element of R[0,1) with no dual representation. I consider .00010001(0001)... as an infinite sequence and how to permute its elements to get .01010101(0101).... It seems almost simple, move the 1 in the fourth place to the second place, and move each one in the 4x'th place for x>1 to the 4(x-1)'th place. There's certainly no dearth of ones and zeros, the amount of each is infinite. Yet, the zero that was in the second place would have to go somewhere, and necessarily displace another. It's easy to describe a canonical sequence relative amounts of zeros and ones. pick a fraction p/q less than one, p and q positive, p B_(p/q +-x) < B, because g is a bijection. For example it doesn't permute the sequence of all ones to the one of all zeros. For example .0101(01)... permutes to .1001(01)... but not to .001001(001).... How does it permute to .0001(01)...? I can easily determine a bijective function from {1, 3, 5, ...} to {3, 5, ...}, and from {0, 2, 4, ...} to {0, 1, 2, 4, ...}. Changing the second element from a one to a zero means that somewhere a zero has to be changed into a one. It could be put off indefinitely but that necessarily implies that some subsequence 01 would be changed to 11. I focus upon necessary and sufficient condition. I can agree that it is a necessary condition for there to be the permutation, yet, not that it is a sufficient condition. I wonder what you mean by allowable. The sequences .(01)... and .(10)..., for half zeros and half ones, are perhaps the simplest cases with infinite numbers of ones and zeros to consider. Exchange each one and zero of the representative subsequence and it becomes the other. Consider .100..., it can't be permuted to .000.... For the same reason, (01)... can't be permuted to 00(01).... If you change the value of one symbol, you have to change the value of an occurrence of the opposite symbol, they exchange. It's simple to get an extra symbol out of the sequence to replace its opposite, the problem is putting it away, it also has to replace its opposite. 01010101.... ^ 00010101... ^^^ ^ ^ ... 10010101... 01010101... 00110101... 00011101... 0001011101... The first sequence has it's second element changed to its opposite. The second sequence is required to change one of its marked symbols to its opposite. Perhaps it's a specialization of permutation. The sequence is only changed by exchanging any of its elements any number of times, the value of a symbol is not changed without changing an occurrence of its opposite symbol. It's not possible to permute in this way 001(001)... to 011(011).... One way to see this is that there are sequence z_1=00... , z_2=00..., and o_1=11.... Take the first element from each then the second, etcetera: 001(001).... If you want to form 011 then it is a matter of taking the first element of z_1, then two elements of o_1 then an element of z_2 (to keep it even), then two more elements of o_1, etcetera. The amount taken from o_1 to form the sequence (011) from the constituent sequences of (001) rises steadily and never returns to a par of those taken from z_1 and z_2. On the other hand, forming a sequence like 111111000000000000(001) does. Given instructions: form a sequence of infinitely many ones and infinitely many zeros, there are many ways to describe a rule or method to do that. Given instructions: given a sequence with infinitely many ones and infinitely many zeros, exchange its elements to form another sequence, there are less sequences. Randomly and fairly (uniformly) select a subset of the natural numbers, there are even odds it contains zero. Half the subsets of N contain zero. Is there a rational function for x=4? Ross ==== > What do you think about the canonicalization of the infinite binary > sequences? That is to say, for the sequence .010101(01)... with half > zeroes and half ones, that you could exchange any two sequence > elements any number of times and not get .001001001(001).... Is there > a rule to exchange (permute) the elements of the sequence .0101(01)... > to get 00010001(0001)...? Yes, in the sense described below. > > It depends on what you consider an allowable permutation. > > The most general form of such permutation can be considered as > follows. > > Let N be the set of positive integers, then any infinite binary > sequence can be uniquely represented by a functions f:N -> {0,1} in > which f(k) is the k'th digit of the sequence. > > Let B be the set of such functions f:N -> {0,1}. > > Then any bijection g:N <-> N creates a permutation G: B -> B by > defining G(f) = fog, where o represents function composition, G(f) > is the function whose valueat k in N is f(g(k), > i.e., G(f)(k) = f(g(k)), for each k in N. > > Let P represent the set of all such permutations, It is clear that > the number of zeros and number of ones in an expansion must > remain fixed under any such permutation of digits. > > Then, for a given f:N -> {0,1} and h:N -> {0,1} in B, a necessary > and sufficient condition for there being some permutation G in P > such that G(f) = h is that > card(k in N: f(k) = 0}) = card(k in N: h(k) = 0}) and > card(k in N: f(k) = 1}) = card(k in N: h(k) = 1}) > i.e., the infinite binary sequences for f and h have the same > number, possibly infinite, of zeros and the same number, possibly > infinite, of ones, and only one of these can be finite. > > But if the number of zeros or the number of ones differ, then there > is no permutation carrying one into the other. > > This divides up the set B into equivalence classes of binaries > permutable into each other which can be indexed by the integers, Z. > > > Any two such binaries with infinitely many zeros and infinitely many > ones in their expansions can be permuted into each other, and can be > indexed by 0. If the number of ones is finite, use that number as > index, and if the number of zeros is finite use the negative of that > number as index. > > Then two binaries can be permuted into each other if and only if > they have the same index. > > B is the set of functions f(n) -> {0,1}, n E N. Your notation here differs from the usual f: N -> {0,1}. Do you mean something different? > There are uncountably many such functions, each represents a > subset of N elements or a distinct element of R[0,1) with no dual > representation. Not so. There are distinct functions which, as binary expansions, represent the same real in [0,1). For ach function with a positive index (a positive but finite number of 1's in its expansion) there is a function with a negative index representing the same real. > > I consider .00010001(0001)... as an infinite sequence and how to > permute its elements to get .01010101(0101).... It seems almost > simple, move the 1 in the fourth place to the second place, and move > each one in the 4x'th place for x>1 to the 4(x-1)'th place. There's > certainly no dearth of ones and zeros, the amount of each is > infinite. Yet, the zero that was in the second place would have to go > somewhere, and necessarily displace another. For each n in N, there is an nth zero in .00010001(0001)... and an nth zero in .01010101(0101).... Similarly there is an nth one in .00010001(0001)... and an nth one in .01010101(0101).... For each n in N, move the nth zero in .00010001(0001) to the position of the nth zero in .01010101(0101).... and similarly with the nth ones, and the thing is done. > > It's easy to describe a canonical sequence relative amounts of zeros > and ones. pick a fraction p/q less than one, p and q positive, p say there are that many zeros and (q-p)/q many ones, plainly the > sequence is the repeating subsequence of p zeros sequentially > (consecutively) and then q-p ones. > > Consider the canonical sequence of zero-density p/q, where finitely x > many zero elements are changed to ones. Is it possible to permute > that sequence to the canonical sequence with none of the elements > changed by exchanging its elements? No, it is not. It permutes to > its canonical sequence of the finite subsequence of ones, however many > zeros were changed into ones there were, and then the repeating > sequence representative of the ratio of the densities of zeros and > ones. > > Then there are the sequences of type b with infinitely many zeros and > ones and infinitely many more or less of of one than the other, > opposite symbol, eg sequences representing subsets of the naturals > like the primes, powers of two, etcetera. Consider the powers of > the primes. Any two sequences of all zeros and ones with the same, possibly infinite, number of zeros and the same, possibly infinite, number of ones, can be permuted into each other by the simple method of moving the nth 0 of one to the position of the nth 0 of the other and also moving the nth 1 in one to the position of the nth 1 of the other. What is so hard about that? As an alternate approach, any such sequence of zeros and ones can be represented by a subset S of N of the positions in which there is a 1, with the compliment, NS, being the set of positions of the zeros in that expansion. Given b1 and b2 as two such sequences, with S1 and S2 as their sets of locations of 1's, then there is a bijection from N to N carrying S1 to S2 and NS1 to NS2, i.e., a permutation, if and only if Card(S1) = Card(S2) and Card(NS1) = Card(NS2). > > Where you note g(n) is a bijection from N to itself, where f(n)=0 or > 1, then their composition defines a permutation G, but I think it > defines a permutation from some proper subset B_(p/q +-x) -> B_(p/q > +-x) < B, because g is a bijection. For example it doesn't permute > the sequence of all ones to the one of all zeros. > > For example .0101(01)... permutes to .1001(01)... but not to > .001001(001).... How does it permute to .0001(01)...? I can easily > determine a bijective function from {1, 3, 5, ...} to {3, 5, ...}, and > from {0, 2, 4, ...} to {0, 1, 2, 4, ...}. Changing the second element > from a one to a zero means that somewhere a zero has to be changed > into a one. It could be put off indefinitely but that necessarily > implies that some subsequence 01 would be changed to 11. > > I focus upon necessary and sufficient condition. I can agree that > it is a necessary condition for there to be the permutation, yet, not > that it is a sufficient condition. I wonder what you mean by > allowable. > > The sequences .(01)... and .(10)..., for half zeros and half ones, are > perhaps the simplest cases with infinite numbers of ones and zeros to > consider. Exchange each one and zero of the representative > subsequence and it becomes the other. > > Consider .100..., it can't be permuted to .000.... For the same > reason, (01)... can't be permuted to 00(01).... If you change the > value of one symbol, you have to change the value of an occurrence of > the opposite symbol, they exchange. > > It's simple to get an extra symbol out of the sequence to replace its > opposite, the problem is putting it away, it also has to replace its > opposite. > > 01010101.... > ^ > 00010101... > ^^^ ^ ^ ... > 10010101... > 01010101... > 00110101... > 00011101... > 0001011101... > > The first sequence has it's second element changed to its opposite. > The second sequence is required to change one of its marked symbols to > its opposite. > > Perhaps it's a specialization of permutation. The sequence is only > changed by exchanging any of its elements any number of times, the > value of a symbol is not changed without changing an occurrence of its > opposite symbol. > > It's not possible to permute in this way 001(001)... to 011(011).... > One way to see this is that there are sequence z_1=00... , z_2=00..., > and o_1=11.... Take the first element from each then the second, > etcetera: 001(001).... If you want to form 011 then it is a matter > of taking the first element of z_1, then two elements of o_1 then an > element of z_2 (to keep it even), then two more elements of o_1, > etcetera. The amount taken from o_1 to form the sequence (011) from > the constituent sequences of (001) rises steadily and never returns to > a par of those taken from z_1 and z_2. On the other hand, forming a > sequence like 111111000000000000(001) does. > > Given instructions: form a sequence of infinitely many ones and > infinitely many zeros, there are many ways to describe a rule or > method to do that. Given instructions: given a sequence with > infinitely many ones and infinitely many zeros, exchange its elements > to form another sequence, there are less sequences. > > Randomly and fairly (uniformly) select a subset of the natural > numbers, there are even odds it contains zero. Half the subsets of N > contain zero. > > Is there a rational function for x=4? > > Ross ==== I guess maybe I don't discuss a permutation, but some other combinatoric operation on the elements of a sequence. An infinite binary sequence with a beginning is presented to you, and you can modify it via this method: you can change any element from a one to a zero, or a zero to a one. If you change a one to a zero, then you must change that or some other zero to a one. If you change a zero to a one instead, you must change that or some other one to a zero. There may be allowed the interchange of identical elements as they would not change the sequence. About the dual representation, consider a crazy model where there is not dual representation and .1000... does not equal .0111.... If you must have dual representation, then the infinite binary sequence is a unique representation of a subset of the naturals but not a unique representation of a real, it is a representation but not necessarily unique, via dual representation. So anyways given a sequence (01)..., the subsequence 01 repeating infinitely, it's possible to change it to (10)..., the subsequence 10 repeating infinitely, by the operation of changing the first, second, third, etcetera elements of the sequence, as the first requires changing the value of a symbol that occurs in the second place, changing the third requires changing the value of a symbol that occurs at the fourth place, etcetera, ad infinitum. This corresponds exactly with that the density of ones and zeros in each of those sequences is one half. The sequence can be modified this way to get a result sequence of any other sequence with density 1/2 of each, yet, the sequence will never be changed to a sequence with density 1/3 ones or zeros. Using this method on, as an example, the sequence of 000..., the result will never be different from 000.... Change a zero element to a one, then, the one element has to be changed to a zero: there are no symbols with the value one in the sequence. Similarly, the sequence as input with finitely many x count of ones will always result in a sequence of output with x many ones. The class of algorithms that modify these sequences' algorithms do not change the density of ones and zeros in the sequence. That apostrophe is there because class is singular and the word algorithms is plural. So I guess it's not a permutation, I tend to trust you, and haven't found a flaw with your points there, ignoring trivialities and easily corrected dsitractions, I describe a method that has the properties that I have here described. One thing I like about this is that I can use it to say that where the density of ones in (10)... is one half, that the density of the even naturals is one half. This is where (10)... represents the subset of N {0, 2, 4, 6, 8, ...}, the even numbers. That might work better with the natural numbers not containing zero, Z+, for that the odd numbers have density 1/2 in Z+, the positive integers. Also, any sequence that I can derive as output from an input of (10)... for N has density of one half in N, e.g, {0, 2, 5, 6, 8, ...}. Given an input sequence a and some output sequence b, there is a function between the various elements they may represent, a bijection. For example if the sequence a has finitely may ones so does b, the same amount, and if sequence a has infinitely many ones so does b. For a given input sequence, generally the canonical input sequence of given density or of indeterminate density via a rule on the ones and zeros in progressive subsequences (irrational sequences), a various number of sequences are possible as output. This can be expressed as an asymptotic term of the variable n of the length of the sequence. For example, n many of all possible sequences have one on element or one off element. Anyways, this goes back to earlier discussions about canonical sequences and the canonicalization of sequences, using some form of sequence element interchange. Ross ==== This is a slight generalization of the theorem in the Prime-Derivative Puzzle thread http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&threadm=b4be2fdf .0308191725.6e5e8441%40posting.google.com&rnum=21&prev= from mid August. Let q and r be any positive integers. Let, for all x where -1 < x < 1, f(x) = (1-x)^((1-x)^(-q)) *(1+x)^((1+x)^(-r)) In ascii-art mode: f(x) = -q -r (1-x) (1+x) (1-x) *(1+x) Then: GCD(q+r ,m) always divides the (m+1)th derivative of f(x) at x = 0. (This derivative, and all derivatives, of f(x), at x =0, are integers.) Leroy Quet ==== > This is a slight generalization of the theorem in the > Prime-Derivative Puzzle thread > http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&threadm=b4be2fdf. 0308191725.6e5e8441%40posting.google.com&rnum=21&prev= > from mid August. > > > > Let q and r be any positive integers. > > > Let, for all x where -1 < x < 1, > > > f(x) = > > (1-x)^((1-x)^(-q)) *(1+x)^((1+x)^(-r)) > > > In ascii-art mode: > > f(x) = > > > -q -r > (1-x) (1+x) > (1-x) *(1+x) > > > > Then: > > GCD(q+r ,m) > > always divides > > the (m+1)th derivative of > > f(x) at x = 0. > > > (This derivative, and all derivatives, of f(x), at x =0, are > integers.) I know that I did not originally post this as a puzzle. But I believe it would be fun to try to prove. So, I am cross-posting this reply to rec.puzzles, as well as posting it to sci.math. > Leroy > Quet ==== > > I have been arguing the question, “Two coins were flipped, and > at least one is a head. What are the chances for two heads?”, in > sci.math, for some time. I argue that the correct answer is 1/2. Unambiguously!!! > > [snip] > > Our question, as written, has correct answer 1/2. Dr. Holt should > concur. > To answer 1/3, you must assume a slightly different question. When we > are willing to assume a different questions, we can get different > answers. Eldon Moritz > > Apparently there is something missing from the exposition which was not > included in the (extensive) argument I snipped. This is a standard logic > problem, which yields to the following: > > 1) There are exactly four possible outcomes for flipping two coins, HH, > HT, TH, TT. > 2) For 'fair' coins, any of the four outcomes is equally likely. > 3) There are 3 outcomes with 'at least one head'. > 4) If we choose the universe of discourse as the set of trials with 'at > least one head', the occurrence of two heads will happen 1 out of 3 times. > > In short, if you run the experiment 100 times, 75 results (should) have > 'at least one head'. 25 (should) have two heads. 25/75 = 1/3. > > Are you discussing a different problem? > There are four equally likely outcomes for the toss. Prior to the statement at least one is a head. After the statement, there are three left, they are no longer equally likely. It is more likely to get the 'heads' statement with HH, than with HT, or with TH. That is, assuming that the statement is true. As I showed earlier, assume it false and you assume a different question. Eldon > -- > There are two things you must never attempt to prove: the unprovable -- > and the obvious. > -- > Democracy: The triumph of popularity over principle. ==== Suppose we take x numbers out of y numbers in a decreasing sequence. Say, take 2 numbers from {1,2,3} and arrange as {2,1}. What is the number of possible combinations ? ==== > Suppose we take x numbers out of y numbers in a decreasing sequence. > Say, take 2 numbers from {1,2,3} and arrange as {2,1}. What is the > number of possible combinations ? y choose x (assuming the y numbers are distinct). -- ==== > > Suppose we take x numbers out of y numbers in a decreasing sequence. > Say, take 2 numbers from {1,2,3} and arrange as {2,1}. What is the > number of possible combinations ? > > y choose x (assuming the y numbers are distinct). ==== > > Suppose we take x numbers out of y numbers in a decreasing sequence. > Say, take 2 numbers from {1,2,3} and arrange as {2,1}. What is the > number of possible combinations ? > > y choose x (assuming the y numbers are distinct). > C(x + y - 1, x). Which is the number of non-negative integer solutions to X_1 + X_2 + ... + X_y = x. Using your example of 2 numbers from {1, 2, 3} you want the solutions to X_1 + X_2 + X_3 = 2 where X_1 counts the number of 1's, X_2 the 2's and X_3 the 3's. There are C(3 + 2 - 1, 2) = C(4, 2) = 6 possibilities. -- Paul Sperry Columbia, SC (USA) ==== Could you please explain these integrators differece? 1. 1/(1-z^(-1)) 2. z^(-1)/(1-z^(-1)) 3. Tz^(-1)/(1-z^(-1)) Boki. ==== > > > Could you please explain these integrators differece? > > 1. 1/(1-z^(-1)) > 2. z^(-1)/(1-z^(-1)) > 3. Tz^(-1)/(1-z^(-1)) > Write the difference equations. You can see that 1. uses the current input in the sum, 2. is one input behind and 3. is the same as 2. but scaled by T. Chuck -- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net ==== > > A propos, here are cites from Pierre Bourdieu's > _Language & Symbolic Power_ (the titles are mine). > Enjoy! > > Censorship > > There Oughta Be A Law Cool-Hand Luke > > > The Social Conditions for the Effectiveness of Ritual Discourse > > Heretical Discourse > The *Skeptron* The skeptron is passed to the orator before he begins his speech so that he may speak with authority (......). It is an attribute of the person who brings a message, a sacred personage whose mission is to transmit the message of authority. E. Benveniste, in Indo-European Language and Society If, as Austin observes, there are utterances whose role is not only to 'describe a state of affairs or state some fact', but also to 'execute an action', this is because the power of words resides in the fact that they are not pronounced on behalf of the person who is only the 'carrier' of these words: the authorized spokesperson is only able to use words to act on other agents and, through their action, on things themselves, because his speech concentrates within it the accumulated symbolic capital of the group which has delegated him and of which he is the *authorized representative*. The laws of social physics are only apparently independent of the laws of physics, and the power which certain *slogans* have to secure efforts from others without expending efforts themselves--which is the very aim of magical action--is rooted in the capital which the group has accumulated through its effort and whose effective use is subordinated to a whole set of conditions, those which define the *rituals of social magic*. Most of the conditions that have to be fulfilled in order for a performative utterance to succeed come down to the question of the appropriateness of the speaker--or better still, his social function--and of the discourse he utters. A performative utterance is destined to fail each time that it is not pronounced by a person who has the 'power' to pronounce it, or, more generally, each time that the 'particular persons and circumstances in a given case' are not 'appropriate for the invocation of the particular speaker invoked'; in short, each time that the speaker does not have the authority to emit the words that he utters. But perhaps the most important thing to remember is that the success of these operations of social magic--comprised by *acts of authority*, or, what amounts to the same, *authorized acts*--is dependent on the combination of a systematic set of interdependent conditions which constitute social rituals. It is clear that all the efforts to find, in the specifically linguistic logic of different forms of argumentation, rhetoric and style, the source of their symbolic efficacy are destined to fail as long as they do not establish the relationship between the properties of discourses, the properties of the person who pro- nounces them and the properties of the institution which authorizes him to pronounce them. The limits (and the interest) of Austin's attempt to define performative utterances lie in the fact that he does not exactly do what he thinks he is doing, and this prevents him from following it through to the end. Believing that he was contributing to the philosophy of language, he was in fact working out a theory of a particular class of symbolic expressions, of which the discourse of authority is only the paradigmatic form, and whose specific efficacy stems from the fact that they seem to possess *in themselves* the source of a power which in reality resides in the institutiional conditions of their production and reception. The specificity of the discourse of authority (e.g. a lecture, a sermon, etc.) consists in the fact that it is not enough for it to be *understood* (in certain cases it may even fail to be understood without losing its power), and that it exercises its specific effect only when it is *recognized* as such. This recognition, whether accompanied by understanding or not--is granted, in the manner of something taken for granted, only under certain conditions, namely, those which define legitimate usage; namely, it must be uttered by the person legitimately licensed to do so, the holder of the *skeptron*, known and recognized as being able and enabled to produce this particular class of discourse: a priest, a teacher, a poet, etc.; it must be uttered in a legitimate situation, that is, in front of legitimate receivers (one cannot read a piece of Dadaist poetry at a Cabinet meeting); finally, it must be enunciated according to the leg- itimate forms (syntactic, phonetic, etc.) What one might call the *liturgical* conditions, namely, the set of prescriptions which govern the *form* of the public manifestation of authority, like ceremonial etiquette, the code of gestures and officially prescribed rites, are clearly only an *element*, albeit the most visible one, in a system of conditions of which the most important and indispensable are those which produce the disposition towards recognition in the sense of misrecognition and belief, that is, the delegation of authority which confers its authority on authorized discourse. By focusing exclusively on the formal conditions for the effectiveness of ritual, one overlooks the fact that the ritual conditions that must be fulfilled in order for ritual to function and for the sacrament to be both *valid* and *effective* are never sufficient as long as the conditions which produce the recognition of this ritual are not met: the language of authority never governs without the collaboration of those that it governs, without the help of the social mechanisms capable of producing this complicity, based on misrecognition, which is the basis of all authority. In order to gauge the magnitude in Austin's and other strictly formalist analyses of symbolic systems, it suffices to show that the language of authority is only the limiting case of the legitimate language, whose authority does not reside, as the racism of social class would have it, in the set of prosodic and articulatory variations which define distinguished pronunciation, or in the complexity of the syntax or the richness of the vocabulary, in other words in the intrinsic properties of discourse itself, but rather in the social conditions of production and reproduction of the distribution between the classes of knowledge and recognition of the legitimate language. (Pierre Bourdieu, _Language & Symbolic Power_, pp. 109-113) ==== > > A propos, here are cites from Pierre Bourdieu's > _Language & Symbolic Power_ (the titles are mine). > Enjoy! > > Censorship > > There Oughta Be A Law Cool-Hand Luke > > > The Social Conditions for the Effectiveness of Ritual Discourse > > Heretical Discourse > > The *Skeptron* > Symbolic Power & the Symbolism of Power There is no symbolic power without the symbolism of power. Symbolic attributes--as is well illustrated in the paradigmatic case of the *skeptron* and the sanctions against the improper wearing of uniforms--are a public display and thereby an officialization of the contract of delegation: the ermine and the robe declare that the judge or the doctor is recognized as having just cause (in the collective recognition) for declaring himself judge or doctor, that his imposture--in the sense of the pretension expressed by his appearance--is legitimate. The competence that is specifically linguistic--the Latin once spoken by doctors or the eloquence of the spokesperson--is also one of the manifestations of competence in the sense of right to speech and to power through speech. There is a whole dimension of authorized language, its rhetoric, syntax, vocabulary and even pronunciation, which exists purely to underline the authority of its author and the trust he demands. In this respect, style is an element of the *mechanism*, in the Pascalian sense, through which language aims to produce and impose the representation of its own importance and thereby help to ensure its own credibility.[14] The symbolic efficacy of the discourse of authority always depends, in part, on the linguistic competence of the person who utters it. This is more true, of course, when the authority of the speaker is less clearly institutionalized. It follows that the exercise of symbolic power is accompanied by work on the *form* of discourse which, as is clearly demonstrated in the discourse of poets in archaic societies, has the purpose of demonstrating the orator's mastery and gaining him the recognition of the group. (This logic is also found in the popular rhetoric of insults which seeks, by flagrant overstatement and the regulated deformation of ritual formulas, to produce the expressive accomplishment which allows one to 'get those laughing on one's side'. Notes 14. The two senses of competence come together if one sees that, according to Percy Ernst Schramm, just as the crown of the medieval king designates both the thing itself and the set of rights which constitute royal dignity (as in the term 'crown property'), so too linguistic competence is a symbolic attribute of the authority which *designates* a socially recognized status as a set of rights, beginning with the right to speak and the corresponding technical capacity. (Pierre Bourdieu, _Language and Symbolic Power_ pp. 75-76) ==== > > A propos, here are cites from Pierre Bourdieu's > _Language & Symbolic Power_ (the titles are mine). > Enjoy! > > Censorship > > There Oughta Be A Law Cool-Hand Luke > > > The Social Conditions for the Effectiveness of Ritual Discourse > > Heretical Discourse > > The *Skeptron* > > Symbolic Power & the Symbolism of Power Collusion in the 'Hood For groups united by some form of collusion (such as sets of *colleagues*), it is a fundamental imperative to maintain *discretion* about, to keep *secret*, everything which concerns the intimate beliefs of the group. They fiercely condemn manifestations of cynicism displayed to the outside world, even though such manifestations are quite acceptable among *initiates* because they cannot by definition affect the fundamental belief in the value of the group--a certain free-and-easy attitude to values is often experienced as a supplementar proof of their value. . . But groups are hardly less mistrustful of those who, taking proclaimed values too seriously, refuse the compromises and shady deals which are the condition of the real existence of the group. (Pierre Bourdieu, _Language & Symbolic Power_, p. 180) ==== > > ====================================================================== > Why do you take so much trouble to expose such a reasoner as > Mr. Smith? I answer as a deceased friend of mine used to answer > on like occasions - A man's capacity is no measure of his power > to do mischief. Mr. Smith has untiring energy, which does > something; self-evident honesty of conviction, which does more; > and a long purse, which does most of all. He has made at least > ten publications, full of figures few readers can critize. A great > many people are staggered to this extend, that they imagine there > must be the indefinite something in the mysterious all this. > They are brought to the point of suspicion that the mathematicians > ought not to treat all this with such undisguised contempt, > at least. > -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan > ====================================================================== A field is a structured social space, a field of forces, a force field. It contains people who dominate and others who are dominated. Constant, permanent relationships of inequality operate inside this space, which at the same time becomes a space in which the various actors struggle for the transformation or preservation of the field. All the individuals in this universe bring to the competition all the relative power at their disposal. It is this power that defines their position in the field and, as a result, their strategies. Economic competition between networks or newspapers for viewers, readers, or for marketshare, takes place concretely in the form of a contest between journalists. This contest has its own, specific stakes - the scoop, the 'exclusive', professional reputations, and so on. This kind of competition is neither experienced nor thought of as a struggle purely for economic gain, even though it remains subject to pressures deriving from the position the news medium itself occupies within a larger set of economic and symbolic power relations. Today, invisible but objective relations connect people and parties who may never meet - nevertheless, in everything these entities do, they are led, consciously or unconsciously, to take into account the same pressures and effects, because they belong to the same world! ==== > >>Good writing their Jim Ferry, and I have to give you credit for that, >>but hey, how many years were you ridiculing me, and now you expect me >>to just go, hey, pal? >>Show me you're serious and wade in and respond to some of these >>ant-mathematicians in the current battle. >>Prove your sincerity, and um, keep posting any interesting dreams you >>have, as I found it interesting puzzling over that one. > > > Exactly! This Jim Ferry now thinks you are a genius, but has he defended > you in your current battle? No! > > I suspect that he is not *fully* committed to your view of mathematics. That > is just a feeling, and I suppose I could be wrong. > > I, however, have been trying to find out more about your Object Ring, but I > feel you are pushing me away. > > Jim sent you roses, but I sent diamonds . . . > > I shan't be drawn into a battle of pitching woo, Clive. > > It's true that I have given up trying to understand James's Object Ring. > I've come to the conclusion that it's simply beyond my comprehension. I > suppose I'm just accepting that it's true on faith. I find that surprising, but I'll take you at your word. > Oh, I can hear the derisive snorting. Faith? Mathematicians don't take > things on faith! We accept pure logic only! > > Well let me ask you this, ye snorters: Andrew Wiles proved FLT, right? > And how do you know this? Did you go through his proof? Did you go through > Ribet's proof as well? No? Interesting. Excellent points! But the problem with religion is that sometimes people who have it refuse to admit it, and from what I've seen, mathematicians refuse to admit they have a religion of often blind faith in other mathematicians, where there's this committee out there that supposedly checks everything thoroughly and declares truth. > Yes, I know the responses to this. Everyone says it's right. It's been > checked by reputable mathematicians. The infallibility of the peer review > process guarantees . . . etc. It has the herd's official brand of approval, > so any objections are moooooot. Oh, yeah, that's the committee out there those reputable mathematicians. > (Okay, that moooooot was pretty awful . . .) > > Anyway, I can't help James by going on the lecture circuit and presenting > an argument which I myself can't follow. Besides, the honor should belong to > James. What I can do, and have been doing for the last several days, is apply > myself to the problem of how one goes about convincing the world of an idea > to which they've shut their ears. Well that made me feel a lot better Jim Ferry! It's amazing how difficult of a problem it is, and my guess is that it requires the best minds on the planet to solve it. Which is probably why you're now on the team. > I have some ideas, but they're still maturing. They don't involve the > wholesale slaughter of the mathematical establishment. This would be entirely > the wrong approach. So stop thinking about it, James! No, not even a little > slaughter! I know you like direct action, but direct action hasn't convinced > anyone thus far, and besides, they'd just throw you in prison and take away > your pencil. No, I have something entirely different in mind. No frontal > assaults. No guerilla attacks. Huh? I've never been a physical threat. > Uh oh. By suggesting that war may not be the most effective way of achieving > one's ends, I think I may have violated the Patriot Act. Ungood. > > BTW, it occurs to me that you might be a Republican, and hence enjoy futile, > pointless wars. In this case all you'd be interested in is more guns and > bigger bombs, and I don't have any of those. Let me know if this is the case. > There's no need to waste my time and yours. I'm not a Republican, nor am I a Democrat as I'm an Independent. No need to talk politics Ferry, and hey, I think it's fascinating that you're trying to figure out that problem you mentioned as it's so wild that it's a difficult one!!! How do you convince the world of correct mathematics that unfortunately is so revolutionary that mathematicians either cannot or refuse to acknowledge it as correct? It's the kind of conundrum that can stress your mental abilities to their maximum. James Harris ==== > Maybe the only point is that I fear James being overwhelmed by evil. > Hmmm. > I have to ask myself, Why should I care? James may be the > reincarnation > of Gauss, but is it really any skin off my nose if he goes > unrecognized? > I've been worrying about the guy for months (ever since I realized > that > he > was not, in fact, a crank, but a genius) and defending him on this > newsgroup. > My reward? Laughter and bile from the peanut gallery. And not a > word > from > James himself. A word of advice to Prof. Connes: Don't waste your > time > on > James Harris. If he loves being the solitary genius so much, let > him > fight > his own damn battles. He's not worth losing sleep over. Well damn it, I'm losing sleep at least partly because of your scary > dream!!! Good writing their Jim Ferry, and I have to give you credit for that, > but hey, how many years were you ridiculing me, and now you expect me > to just go, hey, pal? Show me you're serious and wade in and respond to some of these > ant-mathematicians in the current battle. Prove your sincerity, and um, keep posting any interesting dreams you > have, as I found it interesting puzzling over that one. Exactly! This Jim Ferry now thinks you are a genius, but has he > defended > you in your current battle? No! I suspect that he is not *fully* committed to your view of mathematics. > That > is just a feeling, and I suppose I could be wrong. The primary focus is the odd definition error in core. Once that's accepted for what it is, and most importantly FIXED, then > it's not about committing to my view of mathematics, but about showing > your commitment to mathematics itself. There is ONE mathematics. I, however, have been trying to find out more about your Object Ring, > but I > feel you are pushing me away. I have spent some effort guiding you along at the Mega Foundation > discussion area. > > Yes, indeed. > > But, I expect you remember that I did ask you a question that I was unable > to answer. > > I was using the notation [a,b,c] to represent an ordered triple of complex > numbers. And I was wondering if the ordered triple [1,2,8] was an element of > the Object Ring. > > You replied: > ================================== > Quit being lazy!!! You have the definition, figure it out for yourself!!! > What amazes me is how often people are willing to ask someone else to do > their work for them. > If you're smart enough, answer your own question. > I'm curious to see if you can. > I've given the definition for the object ring, so no excuses. > ================================== > > Well, I am ashamed to say I still cannot figure it out. Sounds like a ploy. You have to understand that your record Clive Tooth is rather long and involves some rather...sleazy behavior...like that attack webpage, and quite a few negative postings over a period of YEARS. You don't get the benefit of the doubt from me but have to make the extra effort yourself, so quit being lazy!!! > Could you help me please? It sounds to me like you think you have some angle for even more negativity and I've given enough time explaining. > Remember mathematics is a continuing process. The object ring is fascinating in and of itself, so you can't expect > me to know all the answers just because I'm a discoverer. After all, if it were that easy, then math research would have ended > long ago with the first mathematician explaining it all. Your friend. Clive Time will tell. Yes, as I said, time will tell. James Harris ==== > message > Maybe the only point is that I fear James being overwhelmed by evil. > Hmmm. > I have to ask myself, Why should I care? James may be the > reincarnation > of Gauss, but is it really any skin off my nose if he goes > unrecognized? > I've been worrying about the guy for months (ever since I realized > that > he > was not, in fact, a crank, but a genius) and defending him on this > newsgroup. > My reward? Laughter and bile from the peanut gallery. And not a > word > from > James himself. A word of advice to Prof. Connes: Don't waste your > time > on > James Harris. If he loves being the solitary genius so much, let > him > fight > his own damn battles. He's not worth losing sleep over. > Well damn it, I'm losing sleep at least partly because of your scary > dream!!! > Good writing their Jim Ferry, and I have to give you credit for that, > but hey, how many years were you ridiculing me, and now you expect me > to just go, hey, pal? > Show me you're serious and wade in and respond to some of these > ant-mathematicians in the current battle. > Prove your sincerity, and um, keep posting any interesting dreams you > have, as I found it interesting puzzling over that one. Exactly! This Jim Ferry now thinks you are a genius, but has he > defended > you in your current battle? No! I suspect that he is not *fully* committed to your view of mathematics. > That > is just a feeling, and I suppose I could be wrong. The primary focus is the odd definition error in core. Once that's accepted for what it is, and most importantly FIXED, then > it's not about committing to my view of mathematics, but about showing > your commitment to mathematics itself. There is ONE mathematics. I, however, have been trying to find out more about your Object Ring, > but I > feel you are pushing me away. I have spent some effort guiding you along at the Mega Foundation > discussion area. Yes, indeed. But, I expect you remember that I did ask you a question that I was unable > to answer. I was using the notation [a,b,c] to represent an ordered triple of complex > numbers. And I was wondering if the ordered triple [1,2,8] was an element of > the Object Ring. You replied: > ================================== > Quit being lazy!!! You have the definition, figure it out for yourself!!! > What amazes me is how often people are willing to ask someone else to do > their work for them. > If you're smart enough, answer your own question. > I'm curious to see if you can. > I've given the definition for the object ring, so no excuses. > ================================== Well, I am ashamed to say I still cannot figure it out. Sounds like a ploy. You have to understand that your record Clive Tooth is rather long and > involves some rather...sleazy behavior...like that attack webpage, and > quite a few negative postings over a period of YEARS. That was just... just... boyish high spirits, James. And you joined in the fun by threatening to sue me for libel!! Ah... happy days... > You don't get the benefit of the doubt from me but have to make the > extra effort yourself, so quit being lazy!!! Oh... James... You have to admit that I did help you out, on the Mega board, with sqrt(i) which you didn't realize was a complex number. Can't you help me out just a little with this one? And I noticed that you said that Jim Ferry is on the team! How can I get on the team if you won't help me when I am struggling? By the way, is anybody else on the team? I think that all the team-members should have well-defined roles for the up-coming battle. > Could you help me please? It sounds to me like you think you have some angle for even more > negativity and I've given enough time explaining. > Remember mathematics is a continuing process. The object ring is fascinating in and of itself, so you can't expect > me to know all the answers just because I'm a discoverer. After all, if it were that easy, then math research would have ended > long ago with the first mathematician explaining it all. Your friend. Clive Time will tell. Yes, as I said, time will tell. Very true. -- Clive Tooth http://www.clivetooth.dk ==== > message > Maybe the only point is that I fear James being overwhelmed by > evil. > Hmmm. > I have to ask myself, Why should I care? James may be the > reincarnation > of Gauss, but is it really any skin off my nose if he goes > unrecognized? > I've been worrying about the guy for months (ever since I > realized > that > he > was not, in fact, a crank, but a genius) and defending him on > this > newsgroup. > My reward? Laughter and bile from the peanut gallery. And not > a > word > from > James himself. A word of advice to Prof. Connes: Don't waste > your > time > on > James Harris. If he loves being the solitary genius so much, > let > him > fight > his own damn battles. He's not worth losing sleep over. > Well damn it, I'm losing sleep at least partly because of your > scary > dream!!! > Good writing their Jim Ferry, and I have to give you credit for > that, > but hey, how many years were you ridiculing me, and now you expect > me > to just go, hey, pal? > Show me you're serious and wade in and respond to some of these > ant-mathematicians in the current battle. > Prove your sincerity, and um, keep posting any interesting dreams > you > have, as I found it interesting puzzling over that one. > Exactly! This Jim Ferry now thinks you are a genius, but has he > defended > you in your current battle? No! > I suspect that he is not *fully* committed to your view of > mathematics. > That > is just a feeling, and I suppose I could be wrong. The primary focus is the odd definition error in core. Once that's accepted for what it is, and most importantly FIXED, then > it's not about committing to my view of mathematics, but about showing > your commitment to mathematics itself. There is ONE mathematics. > I, however, have been trying to find out more about your Object > Ring, > but I > feel you are pushing me away. I have spent some effort guiding you along at the Mega Foundation > discussion area. Yes, indeed. But, I expect you remember that I did ask you a question that I was > unable > to answer. I was using the notation [a,b,c] to represent an ordered triple of > complex > numbers. And I was wondering if the ordered triple [1,2,8] was an > element of > the Object Ring. You replied: > ================================== > Quit being lazy!!! You have the definition, figure it out for > yourself!!! > What amazes me is how often people are willing to ask someone else to > do > their work for them. > If you're smart enough, answer your own question. > I'm curious to see if you can. > I've given the definition for the object ring, so no excuses. > ================================== Well, I am ashamed to say I still cannot figure it out. Sounds like a ploy. You have to understand that your record Clive Tooth is rather long and > involves some rather...sleazy behavior...like that attack webpage, and > quite a few negative postings over a period of YEARS. > > That was just... just... boyish high spirits, James. And you joined in the > fun by threatening to sue me for libel!! Ah... happy days... Well, it was effective writing on your part as even when you took that webpage down that other guy copied it for his own webpage, and then some other, um, person ran that robot program. And *someone* out there was at one point trying a somewhat meek denial of service attack on one of my old websites, or, weirdly enough, they were trying to convince me I had thousands of hits per day on my webpages!!! That's one of the reasons I was happy with MSN as I never even knew how many hits my pages were actually getting with them, but there was no worry about them charging me extra! > You don't get the benefit of the doubt from me but have to make the > extra effort yourself, so quit being lazy!!! > > Oh... James... You have to admit that I did help you out, on the Mega board, > with sqrt(i) which you didn't realize was a complex number. Can't you help > me out just a little with this one? That was a silly error. Like I said there though I was reaching as I really *wanted* to believe that the object ring isn't a subset of complex numbers, but didn't have the proof, so I guess I did what I've done many times before and told myself what I wanted to hear. As for your question, again, quit being lazy!!! Figure it out yourself, or even go get help from someone else besides me, as I've spent enough time with you already, and given your history, it's not sensible to spend any more. > And I noticed that you said that Jim Ferry is on the team! How can I get > on the team if you won't help me when I am struggling? By the way, is > anybody else on the team? I think that all the team-members should have > well-defined roles for the up-coming battle. So far the team as I call it are people who recognize that my work is indeed correct, and so far seem to only be very high IQ people. I guess that's not too surprising. > Could you help me please? It sounds to me like you think you have some angle for even more > negativity and I've given enough time explaining. > Remember mathematics is a continuing process. The object ring is fascinating in and of itself, so you can't expect > me to know all the answers just because I'm a discoverer. After all, if it were that easy, then math research would have ended > long ago with the first mathematician explaining it all. > Your friend. > Clive Time will tell. Yes, as I said, time will tell. > > Very true. Yup. James Harris X-Admin: news@aol.com ==== >So far the team as I call it are people who recognize that my work >is indeed correct, and so far seem to only be very high IQ people. Hey James, did I ever tell you about how the high IQ people fared when I challenged them in their annual Quiz Bowl? Undefeated in a double elimination tournament, culminating in two consectutive -- Mensanator Slayer of the Mensa ==== > As for your question, again, quit being lazy!!! > Figure it out yourself, or even go get help from someone else besides > me, as I've spent enough time with you already, and given your > history, it's not sensible to spend any more. Translation: You haven't the slightest idea how to answer the question, so you're stalling and hoping someone else will do it for you. Of course, you'd still be quite happy to take the credit for the answer. -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock ==== > That was just... just... boyish high spirits, James. And you joined in the > fun by threatening to sue me for libel!! Ah... happy days... Well, it was effective writing on your part as even when you took that > webpage down that other guy copied it for his own webpage, and then > some other, um, person ran that robot program. And *someone* out there was at one point trying a somewhat meek denial > of service attack on one of my old websites, or, weirdly enough, they > were trying to convince me I had thousands of hits per day on my > webpages!!! That's one of the reasons I was happy with MSN as I never even knew > how many hits my pages were actually getting with them, but there was > no worry about them charging me extra! You don't get the benefit of the doubt from me but have to make the > extra effort yourself, so quit being lazy!!! Oh... James... You have to admit that I did help you out, on the Mega board, > with sqrt(i) which you didn't realize was a complex number. Can't you help > me out just a little with this one? That was a silly error. Like I said there though I was reaching as I > really *wanted* to believe that the object ring isn't a subset of > complex numbers, but didn't have the proof, so I guess I did what I've > done many times before and told myself what I wanted to hear. As for your question, again, quit being lazy!!! Figure it out yourself, or even go get help from someone else besides > me, as I've spent enough time with you already, and given your > history, it's not sensible to spend any more. I am not sure who to turn to, James. I would guess that neither Arturo nor Nora would be able to help me with this. They just don't seem to have grasped the basic concepts of the Object Ring. > And I noticed that you said that Jim Ferry is on the team! How can I get > on the team if you won't help me when I am struggling? By the way, is > anybody else on the team? I think that all the team-members should have > well-defined roles for the up-coming battle. So far the team as I call it are people who recognize that my work > is indeed correct, and so far seem to only be very high IQ people. James, I know you have been in the Army. Perhaps we could organise the Team along Army lines... with ranks! You could be the General, of course... and... um... I could be a colonel, say. And that Jim Ferry could be a private. Talking about Jim Ferry, I wonder if he could help me with my question about the ordered triple of complex numbers. I don't know where to turn... -- Clive Tooth http://www.clivetooth.dk ==== > >And I noticed that you said that Jim Ferry is on the team! How can I get >on the team if you won't help me when I am struggling? By the way, is >anybody else on the team? I think that all the team-members should have >well-defined roles for the up-coming battle. Sarcasm. Sigh. >>So far the team as I call it are people who recognize that my work >>is indeed correct, and so far seem to only be very high IQ people. I must reiterate: I don't recognize that your work is indeed correct. (Maybe my IQ is not high enough...) My position is that it's not absurd to think it could be correct rather than to assume, arrogantly, that if I don't understand it, it must be wrong. I'm sickened by this carnival of dogs driven wild by raw meat, and I'm outraged that James is being treated like raw meat. James Harris is a child of God. We are all children of God. Where did it all go so wrong? I am supporting James because if he is correct (and I'm entirely unable to assign a number to the probability of that), it will have a profound impact on society. For the better, despite short-term chaos. > James, I know you have been in the Army. Perhaps we could organise the Team > along Army lines... with ranks! You could be the General, of course... > and... um... I could be a colonel, say. And that Jim Ferry could be a > private. I'd prefer that when you dream about James's privates, my name not spring to mind. James would probably prefer people not to think about his privates at all, especially people named Tooth. Or Ferry, for that matter. But enough silliness. I don't take orders from James, nor does James give them at all, as far as I know. At first I didn't realize what my role in all this was to be, but it's becoming clear to me now. James needs a plan, and for reasons not clear to me, I have been receiving a plan. It keeps me awake at night. It distracts me from my work. It is vast and beautiful, and I worry that I'll never get it typed up, or worded right. I also worry that James will simply reject it out of hand because I've been such an asshole all these years. In fact, why would his plan even come to me? Why wouldn't it just come to him? That would seem more direct. Maybe it's too much for one person to do the math and receive the plan. It is, after all, a vast plan. Already it's wearing me out, and I've only typed up some notes, sketching the broad themes. I want to just chuck it all -- I don't owe James anything, not really -- but there's a reason I can't, a reason that has nothing to do with James, or his math. I'd sort of like to say, but I can imagine the kind of reaction I'd get on *this* newsgroup. Anyway, better to save it for the final product. Here's to the revolution! > Talking about Jim Ferry, I wonder if he could help me with my question about > the ordered triple of complex numbers. I don't know where to turn... Oh poor Clive. It's really burning you up, isn't it? As James said, figure it out yourself, or just forget about it. Stop pretending to be James's buddy. Colonel Clive? Colonel of Sarcasm! Colonel of Chuckles! Colonel of, I don't know, figure something else out, wise guy. Sorry, that was uncalled for. Clive Tooth, too, is a child of God. It's just that I wish you wouldn't bother James with such nonsense at a time like this. -- | Jim Ferry | Center for Simulation | +------------------------------------+ of Advanced Rockets | | http://www.uiuc.edu/ph/www/jferry/ +------------------------+ | jferry@[delete_this]uiuc.edu | University of Illinois | ==== > > I posted something like this in sci.physics but people seem to be mainly > with their heads in the stars there.. > I have programmed, ages ago, this billiard mechanics engine, using very > basic physical laws on momentum collision. It works in the sense that it > creates : > 1. good collisions between two balls that each have a specific velocity > vector, and > 2. it never draws balls over each other, > although I'm not satisfied with the amount of calculation needed for those > two things (it involves solving a quadratic equation and then trigonometry). > But, as you may be aware, in snooker or pool one starts with the red balls > (and the pink) touching each other. > This kind of ruins this whole nice model since one can no longer use an > O(n^2) algorithm to scan the balls for possible collisions and then work > them, since..well it becomes a mess. When the white ball hits the pack of > red balls, I get a nice 4-dimensional representation of chaos, I think, > which isn't my intention. As you know, when balls touch each other then the > energy of the first ball gets transferred to the last. Most of it, anyway. > In fact I think it calls for an entirely new strategy. Does anybody have any > ideas how I could : > - devise a collision strategy that works on singular collisions as well > as on multiple simultaneous collisions > or > - treat these chained collisions separately (this would involve > sorting the touching balls with respect to the ball(s) that will hit them, > since a computer can't know in what direction the force gets transferred - meaning, this is probably impossible to accomplish. Correct. A triangle of balls is statically indeterminant, that is, they are just like a triangular pyramid of balls in static equilibrium. There are not enough constraints to specify the forces on all the balls. > > I would probably do best to treat the table of balls as a table of static > balls, with force fields on them, instead of each ball having its vx and vy, > but really, it's not that easy.. > If anybody has any ideas or experience with this, I'd be very grateful, I have toyed with this type of problem over the years, and I think the best thing is just to never let them touch. With double precision you can leave dither in the 8th place or so, and use your single collision algorithm. Lew Mammel, Jr. ==== >>I'm studying the following diophantine equation: >>a! - b! = n^2 >>Of course, a, b and n are positive integers. >>I think the only two solutions are: >>2! - 1! = 1^2 >>3! - 2! = 2^2 [snip my proof outline] >I'm not sure how easy it is to get an effective upper bound using this >line of argument. We actually don't need anything quite as strong as the >n^0.6 result; I think any prime gap smaller than, say, 0.3 n/log n will >probably do. The primes in (b/2,b] are handled quite well by the Rosser- >Schoenfeld inequalities. As luck would have it, I recently learned of Pierre Dusart's improvements to Rosser-Schoenfeld, proving that for x >= 3275 there is always a prime in the interval (x, x + x/(2 ln^2 x)]. So it looks like a full solution is easily possible (indeed, to the following more general question: when is the difference of two factorials a powerful number?). -- Erick ==== Not too long ago, someone here asked for more rigorous references for issue of The American Mathematical Monthly, there is a glowing review of the book Hubbard JH, Hubbard BB. Vector Calculus, Linear Algebra, and Differential Forms. Might be worth looking at. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu X-Received: from host.safe-hosting.com (host.safe-hosting.com [69.56.134.82]) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id h97NALo30990 X-Received: from igdnycb by host.safe-hosting.com with local (Exim 4.24) id 1A70y9-0003Ht-2r ==== X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h980fWW04789; ==== I have an odd request, here goes... I purchased an Expedition with the keypad entry - no one knows what >the combination is and I don't have time or money to take it to the >dealer... I'd like to try combinations at my leisure until I find the one that >works. What I need is to know every possible combination using the >numbers 1,3,5,7,9. Numbers may repeat, so that needs to be taken into >consideration as well... >Sincerely, >Jeanne Search the Web for sites on how to reset the code. You may not live long enough to try all the codes yourself. For example, a six-digit code may require up to 1 million attempts (half a million on average). phil X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h980fWX04793; ==== Depends how much security you are looking to achieve. You must assume, for example, that the algorithm you are using (e.g., DES) will eventually become known. Then the only security left is the key. Many other considerations enter. Do you need to protect against known plaintext attack, for example? phil X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h983NPO15303; ==== I've got an array size 100 of integer values (range: 0-5). Each plot in >the array represents a 1ms time window. I currently have these graphed >as stairs in MatLab (which looks like a bar graph). I want to make >this function smooooooth, but i'm not sure of a method to apply to it. What steps should I take to complete this? Dustin An easy to remember formula for interpolation between points on a uniform grid: Target coordinate: . . . . . . . | . . . . . . . . Binomial coefficients: 1 5 10 10 5 1 Reciprocal of offset: -1/7 -1/4 -1/1 1/2 1/5 1/8 Product times 56: -8 -70 -560 280 56 7 Alternate signs: 8 -70 560 280 -56 7 Divided by total which is 729. Thus producing coefficients to use for the interpolation. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h983NPO15307; ==== I don't get how to use the hint. Now quite sure how to attack this problem either. I dont think I understand. Any help please? Let {f_n} denote the characteristic function of the rationals in the interval [0, 1]. Show that there is no sequence of continous functions {f_n} such that lim f_n(x) = f(x) for all x in [0, 1]. HINT: If I_n is a nested sequence of closed intervals such that the length of I_n tends to 0 as n tends to infinity then (intersection n=1 to n=infinity)I_n is not empty. ==== >I don't get how to use the hint. Now quite sure how to attack this >problem either. I dont think I understand. Any help please? >Let {f_n} denote the characteristic function of the rationals in the >interval [0, 1]. Show that there is no sequence of continous functions >{f_n} such that lim f_n(x) = f(x) for all x in [0, 1]. I think you meant f is the characteristic function, not {f_n}. >HINT: If I_n is a nested sequence of closed intervals such that the >length of I_n tends to 0 as n tends to infinity then (intersection n=1 >to n=infinity)I_n is not empty. Further hints: I_n should be a closed interval on which f_n > 1/2. What do you know about any point in the intersection of these? Now how can you arrange it so that the fact that the intersection is nonempty produces a contradiction? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 X-Received: (from approve@localhost) ==== > >> z^7-7z^6-2763z^5-19523z^4+1946979z^3+34928043z^2+119557031z-3247^2=0 MAPLE tells me this has cyclic Galois group of order 7. Thus >its roots may be ordered z_0, ..., z_6 in such a way that >a^7 = b in Z[zeta] >where zeta = exp(2 pi I/7) and >a = sum_{j=0}^6 z_j zeta^j >is a Galois resolvent. I calculated the roots numerically to 50 dp and tried various orderings. >I came across one which gave >a^7 = -12392836399104 + 20856562728960 zeta + 26834412666880 zeta^2 > -13437774020608 zeta^3 + 2576491954176 zeta^4 + 25860547313664 zeta^5 > (snip) I came across this reply after I made my second post. Turns out the solution to the septic in question (a variant of it, really) was already in cyberspace, waiting to be plucked. with 11 deg resolvents? --Tito X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h985IgK22585; ==== >prove the following: lim c^1/n = 1 where c>0 >n -> inf. Do you have any pointers? ==== >I am stuck on a proof. I have gotten a series down to where I need to >prove the following: lim c^1/n = 1 where c>0 >n -> inf. Do you have any pointers? The limit of a discrete series is the same as the limit of a smooth function with those values. Why not say: lim[n->oo, c^(1/n)] = lim[x->oo,c^(1/x)] = lim[x->0,c^x] = c^0 = 1 (since c =/= 0). ==== >I am stuck on a proof. I have gotten a series down to where I need to >prove the following: lim c^1/n = 1 where c>0 >n -> inf. Do you have any pointers? The limit of a discrete series is the same as the limit of a smooth function > with those values. Why not say: lim[n->oo, c^(1/n)] = lim[x->oo,c^(1/x)] = lim[x->0,c^x] = c^0 = 1 (since c > =/= 0). Well, at least if the function has a limit, then the series has one too. Sometimes the function might not even have a limit. In this case, f(x) = c^(1/x) does have a nice limit as x->oo. ==== > >>
I am stuck on a proof.  I have gotten a series down to where I need to
>>prove the following:
>> lim     c^1/n = 1  where c>0
>>n -> inf.
>>Do you have any pointers?
>>
There is a proof of this in (among many others) What is Mathematics by Courant and Robbins. It starts with Bernoulii's inequality (1+x)^n >= 1+n*x if n is a positive integer and x >= 0. Rewrite this as (1+y)^(1/n) <= 1 + y/n and choose an appropriate y. ==== >prove the following: lim c^1/n = 1 where c>0 >n -> inf. Do you have any pointers? --I could have baked both cakes yesterday. > --I could have baked both cakes today. > --I could have baked this one yesterday > and that one today. > --I could have baked that one yesterday and > this one today. > Your example doesn't work! > How am I going to distinguish 'one cake' from 'the other cake'! Since they now are identical-with-somethings, if they shared > all the same properties they would be one cake rather than > two. THerefore, since they are two cakes, they don't share > all their properties, and I can distinguish them by means > of any property that one has and the other doesn't. > you mean argument by GPS or by just determining their global positioning > with sattellites we have a difference even if there are no other clear > attributes of different classes besides location? That is a call which no logician qua logician has any particular > competence > to make. > Are you saying it is possible for something to exist at location A and > location B at the same time? I thought it was inference by the rule of > non-contradiction. I have no idea whether this is possible or not. My suggestion that quanta may satisfy '~(x=x)' is meant to translate into logical terms the claim by some physicists that quanta 'lack individuality' (Heller suggests that quanta 'lack haecceities'). The scenario with cakes-to-be suggests that the behaviour of other entities without individuality resembles that of quanta in the relevant respect. Heller's description of that behaviour follows my sig. > > --John > JJ > > 1 I. The Problem, and the Problem with the Problem, > 2 of Identical Articles and Quantum Statistics > 3 > 4 Suppose we have a box with two qualitatively > 6 bouncing around inside. We think of the box > 7 as having a left (*l*) and a right (*r*) side. > 9 at random without interacting, so that their > 10 motions are independent; in particular we > 12 that we may neglect collisions. What are the > 13 chances for finding one or both on one side > 14 or the other? > 15 > 16 Many find the following reasoning persuasive. > 19 in *r*, and 2 in *l* and 1 in *r*. These should > 20 be equally likely, so that each has a probability > 21 of 1/4, or a probability of 1/4 for two in *l*, > 22 1/4 for two in *r*, and 1/2 for one on each side. > 23 > 24 This stylized example is a simple mock-up for a > 25 kind of situation that can occur with quantum entities > 26 and properties. For many of these situations the > 27 probabilities are in fact found to be 1/3 for each > 28 of the three cases: two in *l*, two in *r* and one > 29 on each side. Many interpreters have found this > 30 fact utterly astonishing. > 31 > 32 But on the face of it, there is a very simple > 33 resolution of the puzzle: give up supposing > 34 that there are two qualitatively identical but > 36 there are two *quanta*, as I'll put it, to which > 37 the notion of being numerically distinct does not > 38 apply. . . (p. 114) ==== > --I could have baked both cakes yesterday. > --I could have baked both cakes today. > --I could have baked this one yesterday > and that one today. > --I could have baked that one yesterday and > this one today. > Your example doesn't work! > How am I going to distinguish 'one cake' from 'the other cake'! > Since they now are identical-with-somethings, if they shared > all the same properties they would be one cake rather than > two. THerefore, since they are two cakes, they don't share > all their properties, and I can distinguish them by means > of any property that one has and the other doesn't. > > you mean argument by GPS or by just determining their global positioning > with sattellites we have a difference even if there are no other clear > attributes of different classes besides location? That is a call which no logician qua logician has any particular > competence > to make. > Are you saying it is possible for something to exist at location A and > location B at the same time? I thought it was inference by the rule of > non-contradiction. I have no idea whether this is possible or not. My suggestion that quanta may satisfy '~(x=x)' is meant to translate > into logical terms the claim by some physicists that quanta > 'lack individuality' (Heller suggests that quanta 'lack haecceities'). > The scenario with cakes-to-be suggests that the behaviour of other > entities without individuality resembles that of quanta in the > relevant respect. Heller's description of that behaviour follows > my sig. > Metaphysical Background, Thomas McTighe asserted that the quiddity of a thing is nothing other than unity itself. Hence, by virtue of its positive content, the sun differs not at all from the moon or any other particular thing. The diversity which is exhibited by the natural world is merely the product of accidental differences; no object possesses any specific form which interposes itself between a particular existing thing and the source of their being e.g. the Absolute.15 All individual entities are nothing more than differing contractions of the whole devoid of any being of their own. ...because the restricted quiddity of a thing is the thing itself. http://www.crvp.org/book/Series01/I-10/chapter_ii.htm Aristotle and Aquinas and Scotus and Bonaventura all believed that human minds can conceive and express the intelligibilities or quiddities of things and their properties, intelligibilities that are not simply mind-dependent. We can capture in thought and language the actual natures of things, spelling out their genera and specific differences. Definition brackets or delimits for us as knowers just what it is we attempt to understand and nothing else. The mind-independent thing-substance or the characteristics that we are attempting to define measure the epistemic correctness of a definition. Such real (as opposed to nominal) definition relies on the intelligible and perceptible characteristics thing-substances exhibit to perception and thought for understanding what they are and for picking out individuals of a type. In this way the epistemological realism of the definition corresponds to an ontological realism of actual formal features in mind-independent entities. http://www.sunysb.edu/philosophy/faculty/lmiller/Delinonaliud.htm After all, the three tenets that largely define Nicholas's 'metaphysic of contraction' seem altogether remote from Anselm's Scholasticism. For Anselm has no use for the triad of notions (1) that there is an infinite finite minds can never positively know what God is, given the alleged ground (3) that He is the Coincidence of opposites, i. e., is undifferentiated 'Being' itself, which, with respect to its Quiddity, can never be conceived by anyone except itself. http://www.cla.umn.edu/jhopkins/CusaAnselm.pdf --John > JJ > > 1 I. The Problem, and the Problem with the Problem, > 2 of Identical Articles and Quantum Statistics > 3 > 4 Suppose we have a box with two qualitatively > 6 bouncing around inside. We think of the box > 7 as having a left (*l*) and a right (*r*) side. > 9 at random without interacting, so that their > 10 motions are independent; in particular we > 12 that we may neglect collisions. What are the > 13 chances for finding one or both on one side > 14 or the other? > 15 > 16 Many find the following reasoning persuasive. > 19 in *r*, and 2 in *l* and 1 in *r*. These should > 20 be equally likely, so that each has a probability > 21 of 1/4, or a probability of 1/4 for two in *l*, > 22 1/4 for two in *r*, and 1/2 for one on each side. > 23 > 24 This stylized example is a simple mock-up for a > 25 kind of situation that can occur with quantum entities > 26 and properties. For many of these situations the > 27 probabilities are in fact found to be 1/3 for each > 28 of the three cases: two in *l*, two in *r* and one > 29 on each side. Many interpreters have found this > 30 fact utterly astonishing. > 31 > 32 But on the face of it, there is a very simple > 33 resolution of the puzzle: give up supposing > 34 that there are two qualitatively identical but > 36 there are two *quanta*, as I'll put it, to which > 37 the notion of being numerically distinct does not > 38 apply. . . (p. 114) ==== Are you saying it is possible for something to exist at location A and > location B at the same time? I thought it was inference by the rule of > non-contradiction. I have no idea whether this is possible or not. My suggestion that quanta may satisfy '~(x=x)' is meant to translate > into logical terms the claim by some physicists that quanta > 'lack individuality' (Heller suggests that quanta 'lack haecceities'). > The scenario with cakes-to-be suggests that the behaviour of other > entities without individuality resembles that of quanta in the > relevant respect. Heller's description of that behaviour follows > my sig. > Metaphysical Background, Thomas McTighe asserted that the quiddity of a > thing is nothing other than unity itself. Hence, by virtue of its positive > content, the sun differs not at all from the moon or any other particular > thing. The diversity which is exhibited by the natural world is merely the > product of accidental differences; no object possesses any specific form > which interposes itself between a particular existing thing and the source > of their being e.g. the Absolute.15 All individual entities are nothing more > than differing contractions of the whole devoid of any being of their own. > ...because the restricted quiddity of a thing is the thing itself. > > http://www.crvp.org/book/Series01/I-10/chapter_ii.htm I take a quiddity to be a thing's *suchness* and a haecceity to be its *thisness*. A *thisness* I take to be, as Robert Adams does, the property of self-identity, although I disagree with an assumption which might be read into Adams, that an individual can lack self-identity and yet possess the property of being some individual or other). In Primitive Thisness and Primitive Identity (_The Journal of Philosophy_, Vol. 76, No. 1. (Jan., 1979), pp. 5-26), Adams A thisness is the property of being a certain particular individual, not the property of being some individual or other, but my property of being identical with me, your property of being identical with you, etc. These properties have recently been called 'essences', but that is historically unfortunate, for essences have normally been understood to be constituted by logical properties, and we are entertaining the possibility of nonqualitative thisnesses. In defining 'thisness' as I have, I do not mean to deny that universals have analogous properties--for example, the property of being identical with the quality red. But since we are concerned here principally with the question whether the identity and distinctness of individuals is purely qualitative or not, it is useful to reserve the term 'thisness' for the identities of individuals. It may be controversial to speak of a property of being identical with me. I want the word 'property' to carry as light a metaphysical load here as possible. 'Thisness' is intended to be a synonym or translation of the traditional term 'haecceity' (in Latin, 'haecceitas'), which so far as I know was invented by Duns Scotus. Like many medieval philosophers, Scotus regarded properties as components of the things that have them. He introduced haecceities (thisnesses), accordingly, as a special sort of metaphysical component of individuals.[4] I am not proposing to revive this aspect of his conception of a haecceity, because I am not committed to regarding properties as components of individuals. To deny that thisnesses are purely qualitative is not necessarily to postulate 'bare particulars', substrata without qualities of their own, which would be what was left of the individual when all its qualitative properties were subtracted. Conversely, to hold that thisnesses are purely qualitative is not to imply that individuals are nothing but bundles of qualities, for qualities may not be components of individuals at all. (pp. 6-7) Note 4. Johannes Duns Scotus, _Quaestiones in libros metaphysicorum_, VII. xii. schol. 3; cf. _Ordinatio_, II.3.1.2, 57. I am indebted to Marilyn McCord Adams for acquainting me with these texts and views of Scotus, and for much discussion of the topics of this paragraph. > > Aristotle and Aquinas and Scotus and Bonaventura all believed that human > minds can conceive and express the intelligibilities or quiddities of things > and their properties, intelligibilities that are not simply mind-dependent. > We can capture in thought and language the actual natures of things, > spelling out their genera and specific differences. Definition brackets or > delimits for us as knowers just what it is we attempt to understand and > nothing else. The mind-independent thing-substance or the characteristics > that we are attempting to define measure the epistemic correctness of a > definition. Such real (as opposed to nominal) definition relies on the > intelligible and perceptible characteristics thing-substances exhibit to > perception and thought for understanding what they are and for picking out > individuals of a type. In this way the epistemological realism of the > definition corresponds to an ontological realism of actual formal features > in mind-independent entities. > > http://www.sunysb.edu/philosophy/faculty/lmiller/Delinonaliud.htm > > After all, the three tenets that largely define Nicholas's 'metaphysic of > contraction' seem altogether remote from Anselm's Scholasticism. For Anselm > has no use for the triad of notions (1) that there is an infinite > finite minds can never positively know what God is, given the alleged ground > (3) that He is the Coincidence of opposites, i. e., is undifferentiated > 'Being' itself, which, with respect to its Quiddity, can never be conceived > by anyone except itself. > > http://www.cla.umn.edu/jhopkins/CusaAnselm.pdf ==== > Are you saying it is possible for something to exist at location A and > location B at the same time? I thought it was inference by the rule of > non-contradiction. I have no idea whether this is possible or not. My suggestion that quanta may satisfy '~(x=x)' is meant to translate > into logical terms the claim by some physicists that quanta > 'lack individuality' (Heller suggests that quanta 'lack haecceities'). > The scenario with cakes-to-be suggests that the behaviour of other > entities without individuality resembles that of quanta in the > relevant respect. Heller's description of that behaviour follows > my sig. > Metaphysical Background, Thomas McTighe asserted that the quiddity of a > thing is nothing other than unity itself. Hence, by virtue of its positive > content, the sun differs not at all from the moon or any other particular > thing. The diversity which is exhibited by the natural world is merely the > product of accidental differences; no object possesses any specific form > which interposes itself between a particular existing thing and the source > of their being e.g. the Absolute.15 All individual entities are nothing more > than differing contractions of the whole devoid of any being of their own. > ...because the restricted quiddity of a thing is the thing itself. http://www.crvp.org/book/Series01/I-10/chapter_ii.htm I take a quiddity to be a thing's *suchness* and a haecceity to be its > *thisness*. A *thisness* I take to be, as Robert Adams does, the > property of self-identity, although I disagree with an assumption > which might be read into Adams, that an individual can lack > self-identity and yet possess the property of being some > individual or other). In Primitive Thisness and Primitive Identity (_The Journal > of Philosophy_, Vol. 76, No. 1. (Jan., 1979), pp. 5-26), Adams A thisness is the property of being a certain particular > individual, not the property of being some individual or other, > but my property of being identical with me, your property of > being identical with you, etc. These properties have recently been > called 'essences', but that is historically unfortunate, for essences > have normally been understood to be constituted by logical properties, > and we are entertaining the possibility of nonqualitative > thisnesses. In defining 'thisness' as I have, I do not mean > to deny that universals have analogous properties--for example, > the property of being identical with the quality red. But since > we are concerned here principally with the question whether > the identity and distinctness of individuals is purely > qualitative or not, it is useful to reserve the term > 'thisness' for the identities of individuals. It may be controversial to speak of a property of being > identical with me. I want the word 'property' to carry as > light a metaphysical load here as possible. 'Thisness' > is intended to be a synonym or translation of the traditional > term 'haecceity' (in Latin, 'haecceitas'), which so far as I > know was invented by Duns Scotus. Like many medieval philosophers, Scotus regarded properties as > components of the things that have them. He introduced > haecceities (thisnesses), accordingly, as a special sort > of metaphysical component of individuals.[4] I am not proposing > to revive this aspect of his conception of a haecceity, because > I am not committed to regarding properties as components of > individuals. To deny that thisnesses are purely qualitative > is not necessarily to postulate 'bare particulars', substrata > without qualities of their own, which would be what was left > of the individual when all its qualitative properties were > subtracted. Conversely, to hold that thisnesses are purely > qualitative is not to imply that individuals are nothing > but bundles of qualities, for qualities may not be components > of individuals at all. (pp. 6-7) > Acceptance of haecceities is a distinctive feature of the thought of many followers of Scotus, though there are some sixteenth-century scholastics who accept haecceities without accepting many other distinctively Scotist teachings. Having said this, some early followers of Scotus reject haecceities and the theory of the common nature altogether, and of those who accept haecceities, some found the correct understanding of the nature of the distinction between an individual's nature and its haecceity a troublesome matter. One of the earliest Scotists, Francis of Meyronnes, writing his commentary of common natures, and the claim that individuation is by haecceity. But he holds that it is inappropriate to talk of a formal distinction in this context. Formal distinction obtains only between things that have some sort of quidditative content. Haecceities have no such quidditative content, and thus cannot be formally distinct from their nature. Rather, a haecceity is modally distinct from its nature. A modal distinction, according to Meyronnes, obtains between a thing and an intrinsic mode of that thing, where an intrinsic mode is something which when added to a thing does not vary its formal definition . . . since it does not of itself imply any quiddity or formal definition. A haecceity does not affect a thing's kind; it is thus an intrinsic mode of the thing. It may look as though this is just a terminological shift, but it is not so in at least the following way: a modal distinction is a lesser kind of distinction than a formal distinction. Formal distinctions obtain between genus and specific difference; thus, the difference between species/nature and haecceity, for Meyronnes, is less than the difference between genus and difference. Scotus, contrariwise, makes no such distinction between degrees of difference in this context (for this contrast between the two thinkers, see Dumont [1987], 18). Still, without some principled way of spelling out degrees of difference, this contrast between Scotus and Meyronnes amounts to nothing of any philosophical interest. To this extent the difference between the two thinkers might as well be merely terminological, and Meyronnes needs to do more work if he is to make any significant philosophical point here. http://setis.library.usyd.edu.au/stanford/entries/medieval-haecceity/ http://plato.stanford.edu/entries/francis-marchia/ ...center, content, core, essence, gist, heart, heart and soul, inwardness, kernel, marrow, matter, meaning, means, meat, message, nitty-gritty, nub, pith, subject matter, sum, divagation, drift, drivel, element, entity, entry, excursus, extension, ferment, fluid, foamentation, food, fuel, goo, gook, grinding, guck, guidance, gunk, haecceity, hokum, humectant, humor, humour, hydrocolloid, hypostasis, idea, implication, import, import, info, information, inhibitor, inoculant, inoculum, insertion, instruction, interpolation, jelly, latent content, leaven, leavening, litter, living substance, lysin, material, meaning, meaninglessness, medium, memorial, mental object, micronutrient, mixture, muck, mush, narration, narrative, nonsense, nonsensicality, nutrient, offer, offering, ooze, opinion, packaging, parenthesis, part, petition, philosopher's stone, phlogiston, physical thing, poison, poisonous substance, portion, postulation, promotion, promotional material, propellant, propellent, proposal, protoplasm, publicity, pyrectic, pyrogen, quiddity, quintessence, reference, refrigerant, refusal, reminder, request, residue, respects, sediment, sensationalism, shocker, significance, significance, signification, slime, sludge, solid, solute, solvate, statement, story, strain, stuff, stuff, subject, submission, substrate, system, tale, tenor, theme, thought, topic, treacle, undercurrent, undertone, view, wherewithal, wit, witticism, wittiness, ylem > Note 4. Johannes Duns Scotus, _Quaestiones in libros metaphysicorum_, > VII. xii. schol. 3; cf. _Ordinatio_, II.3.1.2, 57. I am indebted > to Marilyn McCord Adams for acquainting me with these texts and > views of Scotus, and for much discussion of the topics of this > paragraph. Aristotle and Aquinas and Scotus and Bonaventura all believed that human > minds can conceive and express the intelligibilities or quiddities of things > and their properties, intelligibilities that are not simply mind-dependent. > We can capture in thought and language the actual natures of things, > spelling out their genera and specific differences. Definition brackets or > delimits for us as knowers just what it is we attempt to understand and > nothing else. The mind-independent thing-substance or the characteristics > that we are attempting to define measure the epistemic correctness of a > definition. Such real (as opposed to nominal) definition relies on the > intelligible and perceptible characteristics thing-substances exhibit to > perception and thought for understanding what they are and for picking out > individuals of a type. In this way the epistemological realism of the > definition corresponds to an ontological realism of actual formal features > in mind-independent entities. http://www.sunysb.edu/philosophy/faculty/lmiller/Delinonaliud.htm After all, the three tenets that largely define Nicholas's 'metaphysic of > contraction' seem altogether remote from Anselm's Scholasticism. For Anselm > has no use for the triad of notions (1) that there is an infinite therefore, > finite minds can never positively know what God is, given the alleged ground > (3) that He is the Coincidence of opposites, i. e., is undifferentiated > 'Being' itself, which, with respect to its Quiddity, can never be conceived > by anyone except itself. http://www.cla.umn.edu/jhopkins/CusaAnselm.pdf ==== Metaphysical Background, Thomas McTighe asserted that the quiddity > of a thing is nothing other than unity itself. Hence, by virtue > of its positive content, the sun differs not at all from the moon > or any other particular thing. The diversity which is exhibited > by the natural world is merely the product of accidental > differences; no object possesses any specific form which > interposes itself between a particular existing thing and the > source of their being e.g. the Absolute.15 All individual > entities are nothing more than differing contractions of the > whole devoid of any being of their own. ...because the > restricted quiddity of a thing is the thing itself. http://www.crvp.org/book/Series01/I-10/chapter_ii.htm I didn't understand this the first time around. This time I'll make an effort, by placing it in its context: Metaphysical Background, Thomas McTighe asserted that the quiddity of a thing is nothing other than unity itself. Hence, by virtue of its positive content, the sun differs not at all from the moon or any other particular thing.14 The diversity which is exhibited by the natural world is merely the product of accidental differences; no object possesses any specific form which interposes itself between a particular existing thing and the source of their being e.g. the Absolute.15 All individual entities are nothing more than differing contractions of the whole devoid of any being of their own. Putting it in context didn't work. I have no idea what McTighe is talking about. Do you? I take a quiddity to be a thing's *suchness* and a haecceity to be its > *thisness*. A *thisness* I take to be, as Robert Adams does, the > property of self-identity, although I disagree with an assumption > which might be read into Adams, that an individual can lack > self-identity and yet possess the property of being some > individual or other). In Primitive Thisness and Primitive Identity (_The Journal > of Philosophy_, Vol. 76, No. 1. (Jan., 1979), pp. 5-26), Adams A thisness is the property of being a certain particular > individual, not the property of being some individual or other, > but my property of being identical with me, your property of > being identical with you, etc. These properties have recently been > called 'essences', but that is historically unfortunate, for essences > have normally been understood to be constituted by logical properties, > and we are entertaining the possibility of nonqualitative > thisnesses. In defining 'thisness' as I have, I do not mean > to deny that universals have analogous properties--for example, > the property of being identical with the quality red. But since > we are concerned here principally with the question whether > the identity and distinctness of individuals is purely > qualitative or not, it is useful to reserve the term > 'thisness' for the identities of individuals. It may be controversial to speak of a property of being > identical with me. I want the word 'property' to carry as > light a metaphysical load here as possible. 'Thisness' > is intended to be a synonym or translation of the traditional > term 'haecceity' (in Latin, 'haecceitas'), which so far as I > know was invented by Duns Scotus. Like many medieval philosophers, Scotus regarded properties as > components of the things that have them. He introduced > haecceities (thisnesses), accordingly, as a special sort > of metaphysical component of individuals.[4] I am not proposing > to revive this aspect of his conception of a haecceity, because > I am not committed to regarding properties as components of > individuals. To deny that thisnesses are purely qualitative > is not necessarily to postulate 'bare particulars', substrata > without qualities of their own, which would be what was left > of the individual when all its qualitative properties were > subtracted. Conversely, to hold that thisnesses are purely > qualitative is not to imply that individuals are nothing > but bundles of qualities, for qualities may not be components > of individuals at all. (pp. 6-7) > Acceptance of haecceities is a distinctive feature of the thought of many > followers of Scotus, though there are some sixteenth-century scholastics who > accept haecceities without accepting many other distinctively Scotist > teachings. Having said this, some early followers of Scotus reject > haecceities and the theory of the common nature altogether, and of those who > accept haecceities, some found the correct understanding of the nature of > the distinction between an individual's nature and its haecceity a > troublesome matter. > > One of the earliest Scotists, Francis of Meyronnes, writing his commentary > of common natures, and the claim that individuation is by haecceity. But he > holds that it is inappropriate to talk of a formal distinction in this > context. Formal distinction obtains only between things that have some sort > of quidditative content. I'm not sure what he/you are getting at here. > > Haecceities have no such quidditative content, and thus cannot be formally > distinct from their nature. Rather, a haecceity is modally distinct from its > nature. What does Cross (or whoever) mean by Haecceities cannot be formally distinct from their nature.? > A modal distinction, according to Meyronnes, obtains between a thing > and an intrinsic mode of that thing, where an intrinsic mode is something > which when added to a thing does not vary its formal definition . . . since > it does not of itself imply any quiddity or formal definition. A haecceity > does not affect a thing's kind; it is thus an intrinsic mode of the thing. In order for this to mean something to anyone who is not sure what an intrinsic mode, or the formal definition, of something (in the sense of Meyronnes) are--and is therefore unsure what 'varying something's formal definition' might involve, illustrative examples should be provided. > > It may look as though this is just a terminological shift, but it is not so > in at least the following way: a modal distinction is a lesser kind of > distinction than a formal distinction. Formal distinctions obtain between > genus and specific difference; thus, the difference between species/nature > and haecceity, for Meyronnes, is less than the difference between genus and > difference. Scotus, contrariwise, makes no such distinction between degrees > of difference in this context (for this contrast between the two thinkers, > see Dumont [1987], 18). Still, without some principled way of spelling out > degrees of difference, this contrast between Scotus and Meyronnes amounts to > nothing of any philosophical interest. To this extent the difference between > the two thinkers might as well be merely terminological, and Meyronnes needs > to do more work if he is to make any significant philosophical point here. Without any explanation of the before-shift and after-shift terminology, how is one supposed to know which side is up? > > http://setis.library.usyd.edu.au/stanford/entries/medieval-haecceity/ Here's the concluding paragraph from that Stanford Encyclopedia Note, of course, that Scotus's account of the common nature entails something stronger than Adams is proposing: indeed, it entails precisely the sort of minimal hypostatization that Scotus proposes. And the reason for this, of course, is Scotus's view that individual substances cannot themselves be primarily diverse -- a fact that is explained by his claim that common natures have some sort of unity in their instantiations: the nature in Socrates is (non-numerically) the same as the nature in Plato. Natures, for Scotus, cannot be primarily diverse; substances must include more than natures. But individual natures in Ockham's view can indeed be primarily diverse, and this surely amounts to a form of haecceitism -- nothing other than an individual nature's own self-identity explains its distinction from all other such natures. Maintaining that individual natures are primarily diverse amounts not to having no theory of individuation, but to accepting a form of haecceitism that, like Adams's, does not involve ontological commitment to the existence of real haecceities as distinct real constituents of things. Is the author of that one any relation to Mr. McTighe? > http://plato.stanford.edu/entries/francis-marchia/ > --John ==== >> Is there a relationship known for the distance between the largest root >> of an orthogonal polynomial and the endpoint of the orthogonality >> interval? I would like to know how fast this distance shrinks as a >> function of the degree of the polynomial... > > 1) Chebyshev 1st kind, orthogonal on [-1,1]: x_n = cos((2n-1)pi/2n) > > 2) Legendre, orthogonal on [-1,1]: x_n = 1-(j_n)^2/2 * 1/n^2 + O(1/n^3). > Here j_n is the nth biggest positive zero of Bessel function J_0(x). > > All the best, > Teijo non-classical orthogonal polynomials at hand. I have the orthogonality interval and the weight function, but not a closed form of the polynomials. Is there some general theory? I've just ordered Szeg.9a's book Orthogonal Polynomials. I hope to find something in there.... Bye, gert ==== Hy all, this is my problem: I have to find *all* eigenvalue and eigenvector from very large sparse matrix (from 30000x30000 to 100000x100000 elements). Matrix has value very condensed to the diagonal, it's symmetric and it's stored in memory with skyline or CSR method (I heaven't decide yet...) Which method can I use to solve my problem? Another (sub)problem: is there a method to find eigenvalue and maintain the matrix structure (in order not to occupy much of the memory)? Attilio Gelosa. ==== > this is my problem: I have to find *all* eigenvalue and eigenvector > from very large sparse matrix (from 30000x30000 to 100000x100000 > elements). Matrix has value very condensed to the diagonal, it's > symmetric and it's stored in memory with skyline or CSR method (I > heaven't decide yet...) > > Which method can I use to solve my problem? I would suggest lanczos type methods for this. Still, it is going to be difficult just to store that many vectors. Another possible approach would be to compute all the eigenvalues with lanczos, and then repeat the process to compute blocks of eigenvectors using the precomputed expansion vectors from the tridiagonal representation. $.02 -Ron Shepard ==== > Hy all, > > this is my problem: I have to find *all* eigenvalue and eigenvector > from very large sparse matrix (from 30000x30000 to 100000x100000 > elements). This will mean your matrix of eigenvectors is 100000x100000 but full! How on earth are you going to utilize this information? Probably you need an approach where you generate and use one eigenvector after the other, to avoid storing them all? > Matrix has value very condensed to the diagonal, it's > symmetric and it's stored in memory with skyline or CSR method (I > heaven't decide yet...) > > Which method can I use to solve my problem? Another (sub)problem: is > there a method to find eigenvalue and maintain the matrix structure > (in order not to occupy much of the memory)? Probably you may use subspace iteration with (A-sI)^-1 for a set of suitable shifts s, using a factorization of A-sI. But it will be nontrivial to ensure that you get all eigenvalues in this way. Arnold Neumaier X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0DL6Sc30654; ==== A ubiquitous problem in data analysis is sorting and searching. As part of a project I am working on in time series analysis, I need to search very efficiently through a high dimensional set of vectors. I did some reading on the web and found that others described a clustering process which procedes in two phases - a processing phase and a search phase. During the processing phase, a binary tree of clusters is built that can later be searched efficiently. The processing phase needs to be done only once (unless the set of vectors being searched changes). I will outline the processing phase and afterwhich I will describe a problem I am having with the processing. All vectors belong to a root cluster. If the number of vectors in a cluster exceeds a specified threshold, divide the cluster into 2 child clusters. Repeat recursively until each leaf cluster's population of vectors is below the threshold size. For all branch and leaf clusters calculate a center of mass and a radius (the maximum distance from the vector closest to the center of mass to any other vector in the cluster). I have written a program in the C language which performs the processing phase but I have noticed that it produces binary cluster trees, that while the clusters at any particular level in the tree are disjoint, some clusters lie completely inside others. That is a leaf cluster will have a radius large enough to completely contain another leaf cluster. How can I get these pairs of clusters to combine without resorting to a brute force search for this condition after the binary tree is constructed. In case you are wondering, I have tried to implement the algorithm described in the references * C. Merkwirth, U. Parlitz, and W. Lauterborn, ãFast Exact and Approximate Nearest Neighbor Searching for Nonlinear Signal Processing, Phys. Rev. E, vol. 62, no. 2, pp. 2089ö2097, 2000. * J. McNames, ãA nearest trajectory strategy for time series prediction,ä in Proceedings of the International Workshop on Advanced Black-Box Techniques for Nonlinear Modeling. K. U. Leuven Belgium, 1998. Bob Buchanan Bob.Buchanan@millersvill.edu ==== (snip) > Think of the finite fft as an approximation to the integral fourier > transform. If you make a trapezoidal approximation to the integral, > the first and last points must have weight 0.5 relative to the > internal points. For a signal that decays to 0, the last point doesn't > matter, but the first one does. Well, there is that, and there is also, for the traditional Fourier series, Integrate from 0 to 2pi cos(nt)cos(mt) or sin(nt)sin(mt). For m==n the integral is pi if m and n are not zero, but 2pi if they are zero. -- glen ==== >I need an automated method to fit a curve of the form y = >c*x*(1+a*e^(b*x)) to test data. This is a magnetic saturation curve >for an electric machine. The exponential term is the saturation >function, and is negligible at low values of x. We have several >points at low x values. At high values of x, the exponential term >dominates. We usually have a few points out to where the exponential >term is 0.5 or more, but not always. The data at low x are often a >bit noisy. When we fit by hand, we draw a straight line approximating an >asymptote to the lower part of the curve, then we subtract that line >from the data to get approximately the exponential part of the >function (the saturation factor). By taking the logarithm of both >sides for the dataabove a certain value of x, we can get a pretty good >fit on the upper part of the curve. Combining the two functions gives >the full approximation. The problem being, it requires a bit of human judgement to pick out >the spot where the linear part ends and the exponential begins. If I >get it wrong, either the linear portion will be skewed by having part >of the exponential term mixed in, or else it won't fit the transition >if not enough data is used for the exponential part (particularly if >it is not large). The approximation only needs to fit within 1% or so of the real >values (which don't actually fit the function exactly). Accuracy near >the transition region is most important, although this form of >equation was selected because it gives reasonable results when >extrapolated to higher values than we have data for. We're doing this in LabView, so we don't have access to numerical >libraries for nonlinear curve fitting and such. Would prefer a quick >and dirty solution, though we might be able to port in some Fortran >code. I don't know how much procedural code is possible. A table of sample data follows: Voc(PU) Igf(A) >0.196 5.00 >0.304 7.85 >0.399 10.43 >0.616 16.20 >0.797 21.54 >0.906 25.89 >0.942 29.01 >1.014 32.75 >1.051 37.25 >1.105 44.80 >1.196 61.18 >1.304 105.61 This is approximately fit with: a = 0.000232 >b = 7.005 >c = 25.74 > Allen, I found another function that may be easier to fit and which gives results which seem adequate in the transition region. The model is: Ln(y)=a + bx + cx^2 + dx^3 If you fit this using an iterative descent method, without first transforming the data, you get coefficients of: a = .43769 b = 7.483 c = -8.35098 d = 3.9069 If you first take the log (Ln) of your y values and then do a standard polynomial least squares fit, you get: a = .44426 b = 7.458 c = -8.3266 d = 3.900 If I plot these on the computer, they are within a pixel or so of each other, so I think the second fitting method can be justified. Note that this model function doesn't pass through the origin, but that may be ok for you. Allen ==== <400168d2$0$1163$ba620e4c@news.skynet.be>: No. It's true that your series has _applications_ in calculus, but manipulating it as a series is algebra pure and simple -- junior- high-school algebra, in fact, when I studied it. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com My god, you can be insensitive sometimes. Sometimes? I must be slipping. -- Mrs Winterbourne (1996) ==== It seems like it is obvious yet I can't find a proof and again ask for help. The problem: Prove that the set S = [a_1,b_1]X...X[a_n,b_n] does NOT have content 0 if a_i < b_i for i=1,...,n. Note: a subset A of R^n is said to have content 0 if for each epsilon > 0 there is a finite cover of A by closed rectangles, say U_1,...,U_m, such that v(U_1) + ... + v(U_m) < epsilon, where the volume of the rectangle U_i is given by v(U_i). We are given (proven in the corresponding section of the text as theorem 3-5) that the closed interval [a_i,b_i] does not have content 0. It is clear that if we choose epsilon less than the volume of S we will prove that the set does not have content 0, but this is my difficulty, i.e. showing that v(U_1) +...+ v(U_n) >= v(S). Any assistance met with great thanks. Message-Id: ==== I got a short pbm about vectorial space: if A and B are sub vectorial spaces, proof that A union B is also a subvetorial spaces if and only if A is in B or B is in A. I think that because of a stability probleme but I don't know how to start the demo... if anyone got a hint, it would be great :) thx, johann ==== I quess the difficult implikation is: A U B subspace ==> A is in B or B is in A (U being union.) So here is a hint for that: Suppose A U B is subspace and that A is not in B. Show that B is in A. Let x in B be some element. Choose y in A but not in B. Is x in B? Is (x+y) in A? Is (x+y) in B? (Remember, that you can write x like x=(x+y)-y.) Mogens ==== > I got a short pbm about vectorial space: > if A and B are sub vectorial spaces, proof that A union B is also > a subvetorial spaces if and only if A is in B or B is in A. > I think that because of a stability probleme but I don't know how > to start the demo... You are correct (stable in French = closed in English.) If A / B is a subspace and A is not a subspace of B, let a be in A but not in B. Let b be any element of B and consider a + b which, by closure, must be in A / B. So a + b must be an element of either A or of B. It can't be an element of B or else a would be also. Consequently, a + b is an element of A and hence so is b. The if part is trivial. -- Paul Sperry Columbia, SC (USA)