mm-4379 === Subject: List of integrals 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ... Gee, me kow mat now! DUUUUUUH!!! Here's some more integrals: 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26 ... Ooo, I'll now define my own term anti-integral to be the additive inverse of each integral. Here's some anti-integrals: -1, -2, -3, -4, -5, -6, -7, -8, ... Oh my, I'll push the envelope - I'll define a non-integral as something that is neither an integral nor an anti-integral. Here's some non-integrals: 0, pi, 2/3, e, gamma, 33 1/3, i, -1.3, ... This newsgroup has been of invaluable edge-a-ma-kation! More Integrals!!! 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42 .... What a great newsgroup to have such _intelligent_ people to teach everyone about INTEGRALS!!! 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56 ... Who needs dem expensive boox on da mat? Me can read newsgroup posts definining INTEGRALS!!!! 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, ... Me SMART! Me be math-SMART! Me become an associative professor in dem skools! INTEGRALS AND MORE INTEGRALS!!!! 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89 ... Stop me if you know the next INTEGRALS in the list: 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100 .... ;-) === Subject: MTH 245 I need the solutions to Differential Equations 6th Ed. by Zill & Cullen === Subject: Complete Electronic Solution Manuals in .pdf/.doc! Get'em Quickly! I have the comprehensive solution manual, solutions manual, solutions manuals, in electronic format for the following textbooks. They include complete solutions to all the problems in the text, except where noted below in the listing. Payment is through Paypal. 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Giancoli Physics for Scientist and Engineers, 5th Edition, by Tipler, Mosca Physics: Principles with Applications, 6th Ed. by Giancoli Power System Analysis and Design, 3rd Ed., by Glover, Sarma Principles and Applications of Electrical Engineering 4th (Revised) Edition by Rizzoni Principles and Practices of Automatic Process Control, 3rd Edition by Smith, Corripio [ISBN: 0471431907] Principles of Communication: Systems, Modulation Noise, 5th Ed., Ziemer Principles of Physics, 3rd Edition, by Serway Principles of Physics, 4th Edition, by Serway Principles of Statics, 10th Ed., by Russell C. Hibbeler [ISBN: 0131866745] Probability and Statistics for Engineers and Scientists, 3rd Edition, Hayter Probability and Statistics for Engineering and the Sciences, 6th Ed., by Jay L. 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Riskin, John M. Parr Shigley's Mechanical Engineering Design, 8th Ed. by Budynas, Nisbett (No Sol. for Chapt 18 & 19) Simply C#: An Application-Driven Tutorial Approach, by Deitel, Hoey (Chapters 1-32) Soil Mechanics: Concepts and Applications, 2nd Ed., by Powrie Solid State Electronic Devices - 5th Ed by Streetman Solid State Electronic Devices - 6th Ed by Streetman Statics and Mechanics of Materials: An Integrated Approach, 2nd Ed., by Riley, Sturges, Morris Structural Analysis, 5th Edition, by Hibbeler Theory and Design for Mechanical Measurements, 4th Ed., Beasley, Figliola Thermal Physics, 2nd Edition, by Charles Kittel Thermal Physics, by Ralph Baierlein Thermodynamics: An Engineering Approach, 5th Ed., by Cengel, Boles (Missing solutions #118-149 of Chapter 7) Thermodynamics: An Engineering Approach, 6th Ed., by Cengel, Boles The Science and Engineering of Materials, 4th Ed., by Donald R. Askeland, Pradeep P. Phule Thomas' Calculus, Early Trans., Part 1, 10th Ed. by Thomas, Weir, Hass, Giordano Thomas' Calculus: Part 2, 10th Ed. (Multivariable, chs. 8-13), by Thomas, Weir, Hass, Giordano Thomas' Calculus, Early Trans., Part 1, 11th Ed. by Thomas, Weir, Hass, Giordano Thomas' Calculus: Part 2, 11th Ed. (Multivariable, chs. 11-16), by Thomas, Weir, Hass, Giordano Transport Phenomena, 1st Edition, by R. Byron Bird Transport Phenomena, 2nd Ed., by Bird University Physics 11th Edition by Young.. Vector Mechanics: Statics 7th Edition by Beer Vector Mechanics: Dynamics, 7th Ed., by Beer, Johnston, Staab, Clausen Vibrations and Stability: Advanced Theory, Analysis, and Tools, 7th Ed., by Thomsen Wireless Communications: Principles and Practice, 2nd Ed, by Rappaport === Subject: AN: GuruGram #80 now available for free download ... as http://www.tinaja.com/glib/cmindist.pdf It is on finding the minimum distance between a point and a cubic spline. Sourcecode separately available as http://www.tinaja.com/glib/cmindist.psl Additional GuruGrams at http://www.tinaja.com/gurgrm01.asp -- Don Lancaster voice phone: (928)428-4073 Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552 rss: http://www.tinaja.com/whtnu.xml email: don@tinaja.com Please visit my GURU's LAIR web site at http://www.tinaja.com === Nntp-Posting-Host: hera.cwi.nl > ... > WO = world war also known as WW. Not to the English speaking world. It is a Dutch abbreviation, not known > to anyone else. You had better not use Dutch colloquialisms when writing > in an international newsgroep. If you think otherwise, I have a doubt > for you. ... > Note to the Dutch: I did ask Tommy1729 what the O. stood for, and > never got a response. Can't get a straight answer, sometimes. Well, what does the O. stand for? And for completeness, what does the W. in W.O. stand for? Wereld Oorlog. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === WO = world war also known as WW. Not to the English speaking world. It is a Dutch abbreviation, not known > to anyone else. You had better not use Dutch colloquialisms when writing > in an international newsgroep. If you think otherwise, I have a doubt > for you. > ... > Note to the Dutch: I did ask Tommy1729 what the O. stood for, and > never got a response. Can't get a straight answer, sometimes. Well, what does the O. stand for? And for completeness, what does the W. in W.O. stand for? Wereld Oorlog. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland;http://www.cwi.nl/~dik/ Merci beaucoup, grazie, gracia, arigato, and danke (had to look that one up!) === > <15587716.1188254299380.JavaMail.jaka...@nitrogen.mathforum.org ... > WO = world war also known as WW. > Not to the English speaking world. It is a Dutch abbreviation, not known > to anyone else. You had better not use Dutch colloquialisms when writing > in an international newsgroep. If you think otherwise, I have a doubt > for you. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, > +31205924131 > home: bovenover 215, 1025 jn amsterdam, > nederland;http://www.cwi.nl/~dik/ Now this factoid is interesting. As Johnny Carson would say, I did > not know that. I apologize to any of our international friends who I may have > offended and who use W.O. for World War II. Note to the Dutch: I did ask Tommy1729 what the O. stood for, and > never got a response. Can't get a straight answer, sometimes. Well, what does the O. stand for? And for completeness, what does the W. in W.O. stand for? > Woman Orgasm => the big WO (pheasants) === <15587716.1188254299380.JavaMail.jakarta@nitrogen.mathforum.org> <46d481f5$0$97249$892e7fe2@authen.yellow.readfreenews.net > <15587716.1188254299380.JavaMail.jaka...@nitrogen.mathforum.org > ... > WO = world war also known as WW. > Not to the English speaking world. It is a Dutch abbreviation, not known > to anyone else. You had better not use Dutch colloquialisms when writing > in an international newsgroep. If you think otherwise, I have a doubt > for you. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, > +31205924131 > home: bovenover 215, 1025 jn amsterdam, > nederland;http://www.cwi.nl/~dik/ Now this factoid is interesting. As Johnny Carson would say, I did > not know that. I apologize to any of our international friends who I may have > offended and who use W.O. for World War II. Note to the Dutch: I did ask Tommy1729 what the O. stood for, and > never got a response. Can't get a straight answer, sometimes. Well, what does the O. stand for? And for completeness, what does the W. in W.O. stand for? Woman Orgasm => the big WO (pheasants)- Hide quoted text - - Show quoted text - Is that Dutch? Hey, don't they have (heh heh) legalized prostitution over there? Whoa, where's the math in that? Back to discussing James Harris' bull, folks ... === Subject: JSH: Mathematical intuition, surrogate factoring equations Having come off a tremendous research effort that has given key answers to important questions with surrogate factoring, I find myself still curious about the math world's ability to ignore a factoring idea that has many indications that it should be possible to get it to succeed! Consider that I have y^2 = x^2 mod T and k = 2x mod T to get (x+k)^2 = y^2 + 2k^2 + nT when I finally to to the explicit, and it's like, if the first and second are assumed to be true to get the last, then how can it now work to factor a target composite T? Part of the answer is that the math can get to the same place in a different way using: y^2 = x^2 mod a and k = 2x mod b to get (x+k)^2 = y^2 + 2k^2 + nT as long as 'a' and 'b' are coprime to T, but that doesn't explain why the math would CHOOSE to do so, as why would it care? What difference does it make to the mathematics if it satisfies: y^2 = x^2 mod T and k = 2x mod T or y^2 = x^2 mod a and k = 2x mod b where 'a' and 'b' are coprime to T, to get to (x+k)^2 = y^2 + 2k^2 + nT? Why should it care? There are two things really that jump out at you about my surrogate factoring idea as it's so simple: 1. Why didn't anyone think of this before, as mostly it is about completing the square? 2. Why doesn't it always work? Curiosity is not only a major human driver it is crucial to the success of any real researcher. Regardless of the failures I had with these equations I needed to understand why they didn't work all the time. In contrast, I noticed that many of you seemed satisfied with simply being told that they did not work all the time, which presents me with a puzzle, why? Why weren't any of you more curious than that? Why wouldn't you demand an explanation for why and wonder if with such relatively easy math, none were given? That is, where was your human curiosity? This post is about probing that question, which is a question of your humanity. Given the information presented here, why wouldn't this bug some of you, like a burr under your saddle until you got an answer? Where is your human curiosity? James Harris === Subject: Re: JSH: Mathematical intuition, surrogate factoring equations >when I finally to to the explicit, and it's like, if the first and >second are assumed to be true to get the last, then how can it now >work to factor a target composite T? James Harris This is just priceless writing! === Subject: Re: JSH: Mathematical intuition, surrogate factoring equations when I finally to to the explicit, and it's like, if the first and >second are assumed to be true to get the last, then how can it now >work to factor a target composite T? James Harris This is just priceless writing! First runner up, Miss South Carolina, and our new idiot of the year ......................... drumroll....................................... James Harris !!!!!!!!! (Well at least he'd be the best at something then) M === Subject: Re: JSH: Mathematical intuition, surrogate factoring equations > Having come off a tremendous research effort that has given key > answers to important questions with surrogate factoring, I find myself > still curious about the math world's ability to ignore a factoring > idea that has many indications that it should be possible to get it to > succeed! Consider that I have y^2 = x^2 mod T and k = 2x mod T to get (x+k)^2 = y^2 + 2k^2 + nT when I finally to to the explicit, and it's like, if the first and > second are assumed to be true to get the last, then how can it now > work to factor a target composite T? Part of the answer is that the math can get to the same place in a > different way using: y^2 = x^2 mod a and k = 2x mod b to get (x+k)^2 = y^2 + 2k^2 + nT as long as 'a' and 'b' are coprime to T, but that doesn't explain why > the math would CHOOSE to do so, as why would it care? What difference does it make to the mathematics if it satisfies: y^2 = x^2 mod T and k = 2x mod T or y^2 = x^2 mod a and k = 2x mod b where 'a' and 'b' are coprime to T, to get to (x+k)^2 = y^2 + 2k^2 + nT? Why should it care? There are two things really that jump out at you about my surrogate > factoring idea as it's so simple: 1. Why didn't anyone think of this before, as mostly it is about > completing the square? 2. Why doesn't it always work? Curiosity is not only a major human driver it is crucial to the > success of any real researcher. Regardless of the failures I had with these equations I needed to > understand why they didn't work all the time. In contrast, I noticed that many of you seemed satisfied with simply > being told that they did not work all the time, which presents me with > a puzzle, why? Why weren't any of you more curious than that? Why wouldn't you demand an explanation for why and wonder if with such > relatively easy math, none were given? That is, where was your human curiosity? This post is about probing that question, which is a question of your > humanity. Given the information presented here, why wouldn't this bug some of > you, like a burr under your saddle until you got an answer? Where is your human curiosity? James Harris I surprised that you haven't worked out the answer to that. The curious people have gone away to factor numbers and snoop on peoples messages and transactions. Why would they come on here to advertise their interest? Only dumb people like me who can't turn this into a working algorithm, bother to post on usenet. Are you planning to do any work on elliptic curves, or is that form of encryption safe for now? === Subject: Re: JSH: Mathematical intuition, surrogate factoring equations > Having come off a tremendous research effort that has given key > answers to important questions with surrogate factoring, I find myself > still curious about the math world's ability to ignore a factoring > idea that has many indications that it should be possible to get it to > succeed! Consider that I have y^2 = x^2 mod T and k = 2x mod T to get (x+k)^2 = y^2 + 2k^2 + nT when I finally to to the explicit, and it's like, if the first and > second are assumed to be true to get the last, then how can it now > work to factor a target composite T? Part of the answer is that the math can get to the same place in a > different way using: y^2 = x^2 mod a and k = 2x mod b to get (x+k)^2 = y^2 + 2k^2 + nT as long as 'a' and 'b' are coprime to T, but that doesn't explain why > the math would CHOOSE to do so, as why would it care? What difference does it make to the mathematics if it satisfies: y^2 = x^2 mod T and k = 2x mod T or y^2 = x^2 mod a and k = 2x mod b where 'a' and 'b' are coprime to T, to get to (x+k)^2 = y^2 + 2k^2 + nT? Why should it care? There are two things really that jump out at you about my surrogate > factoring idea as it's so simple: 1. Why didn't anyone think of this before, as mostly it is about > completing the square? 2. Why doesn't it always work? Curiosity is not only a major human driver it is crucial to the > success of any real researcher. Regardless of the failures I had with these equations I needed to > understand why they didn't work all the time. In contrast, I noticed that many of you seemed satisfied with simply > being told that they did not work all the time, which presents me with > a puzzle, why? Why weren't any of you more curious than that? Why wouldn't you demand an explanation for why and wonder if with such > relatively easy math, none were given? That is, where was your human curiosity? This post is about probing that question, which is a question of your > humanity. Given the information presented here, why wouldn't this bug some of > you, like a burr under your saddle until you got an answer? Where is your human curiosity? James Harris ----------------------------------------------------------- Here ya go: A minor bug report & some interesting observations of when surrogate factoring works. Taken from: Understanding surrogate factoring begin quote (with comment added) So k=1 is the proper choice. Looked over carefully the limits shown on my previously mentioned integer factorization page show that once you get to a large enough S, with k=1, then you are forced to have solutions without regard to n within that range. That is, there is a minimum size n, but once you get to it, _every_***This is not true*** value higher will give you non-trivial factorizations for that prime p. end quote Example: T=P1*P2=7*41=287 k=1 n = integer to be incremented Z = any integer S=2*k^2+n*T S=2*1^2+2*287 S=2+574=576 Looking for the factor 7 Need to find the factor of S equal to 7*Z+k = 8,15,22,29,36,43,50 ... = 1 mod 7 in general Starting from n=0, n=2 gives the first solution with S = 576 = F1*F2 = 72*8 X=(F1+F2-2K)/2 = 39 Y=(F2-F1)/2 = -32 GCD(X+Y,T) = 7 GCD(X-Y,T) = 1 If _every_ higher value of n will give you non-trivial factorizations for that prime p, then the next n, 3 should work. At n=3, S = 863, a prime, and it doesn't work. On the bright side, the 8, which worked at n=2 appears at n = 8*Z+2 at n=4, there are two solutions with S=1150 = 2 * 5 * 5 * 23, which has the factors that work; 50, occurrs at n = 50*Z+4 575, occurrs at n = 575*Z+4 n=5 fails with S=1437 = 3 * 479 n=6 works with S = 1724 = 2 * 2 * 431 using 2*431=862 which occurs at n = 862*Z+6 It still looks good, but 7 is a small prime. What happens when you try for larger primes? Also notice that the larger the factor of the surrogate that works, the longer the interval to its next appearance as n is incremented. Another item of interest: In this example, if you are looking for the other factor of T, 41 using the surrogate factor 42, it will never appear because 42 contains 7, a factor of T. In general, no factor of T ever appears as a factor of S. All the other primes and all their powers will appear as a factor of S for some n, somewhere, which can be calculated. A fun thing: With A an integer > 0 and B any integer If T = (4*A+1)*(2*B*(2*A+1)+1) or if T = (4*A-1)*(2*A*B+1) then k=1 and n = 2 always finds the first factor in the expressions shown. Have fun, Enrico === Subject: Re: JSH: Mathematical intuition, surrogate factoring equations Having come off a tremendous research effort that has given key > answers to important questions with surrogate factoring, I find myself > still curious about the math world's ability to ignore a factoring > idea that has many indications that it should be possible to get it to > succeed! Consider that I have y^2 = x^2 mod T and k = 2x mod T to get (x+k)^2 = y^2 + 2k^2 + nT when I finally to to the explicit, and it's like, if the first and > second are assumed to be true to get the last, then how can it now > work to factor a target composite T? Part of the answer is that the math can get to the same place in a > different way using: y^2 = x^2 mod a and k = 2x mod b to get (x+k)^2 = y^2 + 2k^2 + nT as long as 'a' and 'b' are coprime to T, but that doesn't explain why > the math would CHOOSE to do so, as why would it care? What difference does it make to the mathematics if it satisfies: y^2 = x^2 mod T and k = 2x mod T or y^2 = x^2 mod a and k = 2x mod b where 'a' and 'b' are coprime to T, to get to (x+k)^2 = y^2 + 2k^2 + nT? Why should it care? There are two things really that jump out at you about my surrogate > factoring idea as it's so simple: 1. Why didn't anyone think of this before, as mostly it is about > completing the square? 2. Why doesn't it always work? Curiosity is not only a major human driver it is crucial to the > success of any real researcher. Regardless of the failures I had with these equations I needed to > understand why they didn't work all the time. In contrast, I noticed that many of you seemed satisfied with simply > being told that they did not work all the time, which presents me with > a puzzle, why? Why weren't any of you more curious than that? Why wouldn't you demand an explanation for why and wonder if with such > relatively easy math, none were given? That is, where was your human curiosity? This post is about probing that question, which is a question of your > humanity. Given the information presented here, why wouldn't this bug some of > you, like a burr under your saddle until you got an answer? Where is your human curiosity? James Harris ----------------------------------------------------------- Here ya go: A minor bug report & some interesting > observations of when surrogate factoring works. Taken from: Understanding surrogate factoring > begin quote (with comment added) So k=1 is the proper choice. Looked over carefully the limits shown on my previously mentioned > integer factorization page show that once you > get to a large enough S, with k=1, then you are forced to have > solutions without regard to n within that range. That > is, there is a minimum size n, but once you get to it, _every_***This > is not true*** value higher will give you > non-trivial factorizations for that prime p. > end quote Example: T=P1*P2=7*41=287 > k=1 > n = integer to be incremented > Z = any integer > S=2*k^2+n*T > S=2*1^2+2*287 > S=2+574=576 Looking for the factor 7 > Need to find the factor of S equal to 7*Z+k = 8,15,22,29,36,43,50 ... > = 1 mod 7 in general Starting from n=0, n=2 gives the first solution with S = 576 = F1*F2 = > 72*8 > X=(F1+F2-2K)/2 = 39 > Y=(F2-F1)/2 = -32 GCD(X+Y,T) = 7 > GCD(X-Y,T) = 1 If _every_ higher value of n will give you non-trivial > factorizations for that prime p, then the next n, 3 > should work. At n=3, S = 863, a prime, and it doesn't work. > On the bright side, the 8, which worked at n=2 appears at n = 8*Z+2 at n=4, there are two solutions with S=1150 = 2 * 5 * 5 * 23, which > has the factors that work; > 50, occurrs at n = 50*Z+4 > 575, occurrs at n = 575*Z+4 n=5 fails with S=1437 = 3 * 479 n=6 works with S = 1724 = 2 * 2 * 431 using 2*431=862 which occurs at > n = 862*Z+6 It still looks good, but 7 is a small prime. What happens when you try > for larger primes? > Also notice that the larger the factor of the surrogate that works, > the longer the interval > to its next appearance as n is incremented. Another item of interest: In this example, if you are looking for the > other factor of T, 41 using > the surrogate factor 42, it will never appear because 42 contains 7, a > factor of T. In general, > no factor of T ever appears as a factor of S. All the other primes and > all their powers will appear as a > factor of S for some n, somewhere, which can be calculated. A fun thing: > With A an integer > 0 and B any integer > If T = (4*A+1)*(2*B*(2*A+1)+1) or > if T = (4*A-1)*(2*A*B+1) > then k=1 and n = 2 always finds the first factor in the expressions > shown. Have fun, > Enrico- Hide quoted text - - Show quoted text - Correction: n=6 fails because S = 1724 = 2 * 2 * 431; using 2*431=862 gives 862 - k = 861 = 3*287 = 3*T. GCD of the surrogate factor - k just gives you T. Enrico === Subject: Re: Mathematical intuition, surrogate factoring equations > Having come off a tremendous research effort that has given key > answers to important questions with surrogate factoring, I find > myself > still curious about the math world's ability to ignore a factoring Why is a goose? -- Clive Tooth http://www.shutterstock.com/cat.mhtml?gallery_id=61771 === Subject: Re: Mathematical intuition, surrogate factoring equations > Having come off a tremendous research effort that has given key > answers to important questions with surrogate factoring, I find > myself > still curious about the math world's ability to ignore a factoring Why is a goose? One leg is both the same. (Anyway, I thought it was a duck.) rossum === Subject: Re: Mathematical intuition, surrogate factoring equations You couldn't factor your way out of a wet paper bag. > Having come off a tremendous research effort, my weenie fell off. === Subject: Re: JSH: Mathematical intuition, surrogate factoring equations > You couldn't factor your way out of a wet paper bag. > Having come off a tremendous research effort, my weenie fell off. 1. Factoring is like division, separating a number into parts 2. Division by knife into parts is very fast. 3. Surrogate factoring is chopping somebody else's weenie up into parts === Subject: Re: Mathematical intuition, surrogate factoring equations You couldn't factor your way out of a wet paper bag. > Having come off a tremendous research effort, my weenie fell off. === Subject: Re: JSH: Mathematical intuition, surrogate factoring equations > y^2 = x^2 mod T and k = 2x mod T or y^2 = x^2 mod a and k = 2x mod b where 'a' and 'b' are coprime to T, to get to (x+k)^2 = y^2 + 2k^2 + nT? Why should it care? Help, can you send me a solution manual to equations? If you have it please let me know ASAP so we can work something out so I can get it. How much does it cost? Can you e-mail it to me? i need it as soon as posible! please friend i would really appreciate Lily === Subject: Re: Mathematical intuition, surrogate factoring equations > Having come off a tremendous research effort that has given key > answers to important questions with surrogate factoring, I find myself > still curious about the math world's ability to ignore a factoring > idea that has many indications that it should be possible to get it to > succeed! > factoring idea as it's so simple: 1. Why didn't anyone think of this before, as mostly it is about > completing the square? no workie 2. Why doesn't it always work? see, you know it not workie either > Curiosity is not only a major human driver it is crucial to the > success of any real researcher. you no real, you fake, you waka man > Regardless of the failures I had with these equations I needed to > understand why they didn't work all the time. you make it up, brain like bowl loose noodles > In contrast, I noticed that many of you seemed satisfied with simply > being told that they did not work all the time, which presents me with > a puzzle, why? why you spend time do this? silly amarican > Why weren't any of you more curious than that? Why wouldn't you demand an explanation for why and wonder if with such > relatively easy math, none were given? That is, where was your human curiosity? you missing point, you not know math. This post is about probing that question, which is a question of your > humanity. I can smell your fear. Given the information presented here, why wouldn't this bug some of > you, like a burr under your saddle until you got an answer? you silly, silly man. you need dayjob. Where is your human curiosity? > James Harris > === Subject: Re:JSH: Mathematical intuition, surrogate factoring equations > Having come off a tremendous research effort that has given key > answers to important questions with surrogate factoring, I find myself > still curious about the math world's ability to ignore a factoring > idea that has many indications that it should be possible to get it to > succeed! Confucius the chinese mathematician say: One who thinks he laid golden egg, have stinky turd instead. === Subject: I need the solutions manual for the 8th Edition of Electric Circuits I need the solutions manual for the 8th Edition of Electric Circuits by Nilsson and Riedel. How much does it cost? Can you e-mail it to me? === Subject: Re: I need the solutions manual for the 8th Edition of Electric Circuits >I need the solutions manual for the 8th Edition of Electric Circuits > by Nilsson and Riedel. How much does it cost? Can you e-mail it to me? > i sign you up to porno lists too === Subject: need solutions manual for A Modern Introduction to Diff. Equations by Henry Ricardo I am looking for the solutions manual for A Modern Introduction to Diff. Equations by Henry Ricardo. If you have it please let me know ASAP so we can work something out so I can get it. Email me @ tschloss05@aol.com === Subject: Re: need solutions manual for A Modern Introduction to Diff. Equations by Henry Ricardo >I am looking for the solutions manual for A Modern Introduction to > Diff. Equations by Henry Ricardo. If you have it please let me know > ASAP so we can work something out so I can get it. > Email me @ tschloss05@aol.com > ha, i sign you up to porno lists === Subject: Re: math is a physical process <3udlc3d29ickupj87kkddln5gj4fv7mn6t@4ax.com> <5dbb7$46cad8ce$82a1e228$32721@news1.tudelft.nl> <2limc3983vfpimvmapj1co5kknkqvgbllk@4ax.com> <32a7a$46d2a330$82a1e228$4614@news1.tudelft.nl> <87inhc7o5.fsf@hod.lan.m-e-leypold.de> <31DAi.13010$OM1.8447@trnddc07> Markus E L Math is the software running within the hardware of Reality. > I'd like to make this quote immortal! > Han de Bruijn and in to many cases > the ware of relality! > will there be no end to that stupid arrogance > of crook methematiciance ?? > and who is the crimminal teachers > that educated those pomous farters ?? > Y.P > Markus E L > And who is the criminal who tought you spelling? > - M > [hanson] >Hey Markus, lass doch den guten Mann sein! > Leave my friend Yehiel alone on this lingo-issue. > If you can converse in Yiddish, Hebrew and English > as well and good as this old Israeli, Porat, can & does, > then you may utter your critisim. Otherwise Schnauze. > [Mark] > You're looking too many WW II films. > [hanson] > So, attack him, Porat, on other grounds, like his Physik. Versteh! > [Mark] > Which physics? The physics of bad spelling, bad language > and trollish claims? Gimme a break ... > And BTW: I'm a physicist by education. So there :-). > - M > [hanson] > ... So there is your fatal flaw. If you had solid, unshakable > foundations of your ken in physics, you would NOT feel > the need to get riled up over some olde kacker's hobby So his hobby is bad spelling? Well ... whatever floats his boat. (I still don't see any physics here, though -- do you?) > notions. Give him a break. Don't ask for one... ahahaha... > Enjoy the wide world of cyber physics... NOT the world nor > physics is gonna change over opinions rendered on the Usenet. > Let'em sing!... All of'em!... It's a beautiful choir! You know that you shouldn't discontinue your medicaments w/o consulting your doctor? === Subject: Re: math is a physical process ahahahaha... AHAHAHAHA.. Leypold Markus, the physicist is not only fizzing... but he is fuming now... AHAHAHAHA...ahaha... > Markus E L I'd like to make this quote immortal! Han de Bruijn > and in to many cases the ware of relality! will there be no end to that stupid arrogance of crook methematiciance ?? and who is the crimminal teachers that educated those pomous farters ?? Y.P > Markus E L And who is the criminal who tought you spelling? - M > [hanson] Hey Markus, lass doch den guten Mann sein! Leave my friend Yehiel alone on this lingo-issue. If you can converse in Yiddish, Hebrew and English as well and good as this old Israeli, Porat, can & does, then you may utter your critisim. Otherwise Schnauze. > [Mark] You're looking too many WW II films. > [hanson] So, attack him, Porat, on other grounds, like his Physik. Versteh! > [Mark] Which physics? The physics of bad spelling, bad language and trollish claims? Gimme a break ... And BTW: I'm a physicist by education. So there :-). - M > [hanson] ... So there is your fatal flaw. If you had solid, unshakable foundations of your ken in physics, you would NOT feel the need to get riled up over some olde kacker's hobby notions. > [Mark is getting defensive and coy now... ahahaha...] So his hobby is bad spelling? Well ... whatever floats his boat. > [Mark is winding himself up... ahahaha...] (I still don't see any physics here, though -- do you?) > [hanson] Give Porat a break. Don't ask for one... ahahaha... Enjoy the wide world of cyber physics... NOT the world nor physics is gonna change over opinions rendered on the Usenet. Let'em sing!... All of'em!... It's a beautiful choir! > [Mark is fully uptight now and asks for a way out ... ahahaha...] You know that you shouldn't discontinue your medicaments w/o consulting your doctor? > [hanson] ahahaha.. Herr Landsrat M.E. Leypold, your advice here is just a dilettante and discombobulated as was your Rat to Porat. What's in it for you, junger Landser?. Hab Spass & Wiederschaun. ahahaha... ahahahanson === Subject: Re: Re: math is a physical process > ahahahaha... AHAHAHAHA.. Leypold Markus, the physicist is > not only fizzing... but he is fuming now... AHAHAHAHA...ahaha... please take this discussion out of comp.lang.functional. it does not belong here. ---- Garry Hodgson, Senior Software Geek, AT&T CSO nobody can do everything, but everybody can do something. do something. === Subject: Re: math is a physical process > But mathematics is nowhere nearly similarly exhausted. We need a new mathematics for the Quantum thing, or simple rationalization of our current maths. It works like this: our current mathematics cannot embrace the quantum theory properly. You know it. However, remember that the calculus was perfectly useful a couple hundred years before it could be rationalized. === Subject: Re: math is a physical process Math is the software running within the hardware > of >Reality. >I'd like to make this quote immortal! >Han de Bruijn You remain convinced, I see, that mathematics is > identical to computation. Tom Is Reality identical with a computer? Han de Bruijn If Nature can solve its Equations, so can We (: > D.Brian Spalding). > Of course not. How could one deduce the nature of reality by assuming it a priori? (as, BTW, you have done) Tom === Subject: Re: math is a physical process Downward Loewenheim-Skolem is a set theory result that makes it an > interesting problem to incontrovertibly prove the existence of more > than that many of anything :) What does the downward L.9awenheim-Skolem theorem have to do with > proving the existence of uncountable many objects of some kind? > -disproving- not proving. Well DLST says that any theory that has an uncountable model necessarily has a countable model, so however convinced you may be of the larger-than-countable infinities embedded in an uncountable model, it's possible to object that a Superior Being (tm) could observe you observing the theory and consistently reinterpret - and explain away - your perceptions of manifest uncountable infinity as being in fact provided by a parts of a countable model. No ? === Subject: Re: Re: math is a physical process fascinating as this is, could you all *please* remove comp.lang.functional from the distribution list for this thread, since it has nothing to do with that group? we've just gone through a months-long series of discussions that started when some idiot cross posted to comp.lang.lisp and comp.lang.functional. whether it's relevant to the others groups, i'll let them sort out. thank you for your consideration. ---- Garry Hodgson, Senior Software Geek, AT&T CSO nobody can do everything, but everybody can do something. do something. === Subject: Re: math is a physical process > fascinating as this is, could you all *please* remove comp.lang.functional > from the distribution list for this thread, since it has nothing to do with that > group? we've just gone through a months-long series of discussions > that started when some idiot cross posted to comp.lang.lisp and comp.lang.functional. > whether it's relevant to the others groups, i'll let them sort out. My apologies. I'll try to trim the newsgroups-line more carefully in the future, avoiding Harrop-like crossposting... -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: math is a physical process <3bc70$46ca97d1$82a1e228$24521@news1.tudelft.nl> <3udlc3d29ickupj87kkddln5gj4fv7mn6t@4ax.com> <5dbb7$46cad8ce$82a1e228$32721@news1.tudelft.nl> <2limc3983vfpimvmapj1co5kknkqvgbllk@4ax.com> <46d3fb00$1_6@news.bluewin.ch> <87r6ln27so.fsf@huxley.huxley.fi> <46d47745$1_2@news.bluewin.ch Well DLST says that any theory that has an uncountable model necessarily has a > countable model, so however convinced you may be of the larger-than-countable > infinities embedded in an uncountable model, it's possible to object that a > Superior Being (tm) could observe you observing the theory and consistently > reinterpret - and explain away - your perceptions of manifest uncountable > infinity as being in fact provided by a parts of a countable model. No ? No. It is possible to object in any way imaginable of course. This talk about perceptions provided by a parts of a countable model, Superior Being (tm) and so on has no apparent connection to anything in our mathematical practice and reasoning, however. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: simple question (not simple enough for me) > .... I > know this question probably was not appropriate for this forum because > there seems to be a lot of high level math here, but I appreciate the > answer.... Don't be timid about sending such a question to . It handles a wide range of topics, both elementary and more advanced. Your message was neither rude nor cranky, which raised it far above the level of many others. :-) Ken Pledger. === Subject: terminology for total number of elements in a matrix if I have a 2x3 matrix, it has 6 elements. What term is used to refer to that number 6? email cc's welcome. === Subject: Re: terminology for total number of elements in a matrix > if I have a 2x3 matrix, it has 6 elements. What term is used to refer to that number 6? cardinality. Sorry to answer my own Q, but I managed to get help elsewhere. === Subject: Re: Two results of set geometry Nntp-Posting-Host: hera.cwi.nl > Sorry to break in again, I have been away a bit too long. ... > M' does not contain a second index omega. A does > not contain an element omega. A+1 does not contain > omega + 1. But by extending all rows of M' we get a set of order omega + 1. Don't > we get at least one non-natural index, necessarily? As you do not define what *you* mean with extend (and you do apparently > not use the standard definition), it is even unclear what this means. > Normally with extend is meant adding something at the end when talking > about ordered sets. But as a set with order type omega does not have an > end, it is unclear what is meant in that case. But if you mean add an > element with index greater than all other indices, we clearly get from > an ordered set with order type omega a set with order type omega+1. I mean simply: append an element as Cantor does define it. There is no > index to be considered. You can see this if you append 1 to the set > {2,3,4,...} to get {2,3,4,...,1}. There *are* indices to be considered. All your elements are '1'! > So you don't believe in the unique bijection between digits and second > indexes? 1 1 1 1 ... and > 1,2,3,4,... But this is again thoroughly unmathematical. Can you show a bijection > is f(1)? You are talking about order preserving bijections between > ordered multi-sets. Argh. Look at the matrix. It should be obvious: 1 > 11 > 111 > 1111 > ... What is f(1)? Darn, you *say* you are talking about bijections between elements when actually you are talking about bijections between index positions (and so, yes, index positions *are* to be considered). > The elements of the sets are not second indexes, but > sequences of ones. The set of sequences of ones > in M' is not the same as the set of sequences of ones > in M. As above bijection shows, then also the set of second indexes must > differ. That is an order preserving bijection between multi-sets. You misunderstand. The bijection is between n and {1,2,3,...,n} but > the n's are written in unary representation by sets of 1's in the rows > and the initial segments are those of the first column. Now, this is again utterly confused. With this explanation I see only a single set: the set of natural numbers written out in unary. But if you consider it that way, there is only a single column, namely the list of natural numbers. > If the set of natural numbers has omega elements, then the set of rows > is in bijection with the complete first column. Yes, of course, because the set of rows is the set of natural numbers (because each element is a natural number). And it is in bijection with the first column, which is the set of rows. Quite trivial I would think. > If a bijection exists, > then appending (in the manner of Cantor) one element to each partner > of the bijection, then the bijection persists. Obvious. As each partner in the bijection is the set of natural numbers. > But this leads to omega > + 1 being in bijection with the set of natural numbers. Not at all. Why do you think so? What you actually *do* when you append a 1 to each row is changing each element of the set {1,2,3,...} to its successor, so you change it to {2,3,4,...}. You do not add something to the set the column is bijected with. > Appending a one the the all elements of M' > does not give the same result as appending a one > to all elements of M. That is because set theory is blurry. No, it is because you are blurry. Each '1' in M and M' actually > represents a pair of natural numbers (r,c), where r is the row number > and c the column number. The elements of the triangular matrix M are > defined as (r,c) as above and non-existing when c > r. The matrix M' > (of some strange format) is defined as M except when r = 1, in that > row *all* elements do exist. When defined such we can talk about > bijections. Now let us try a few actions: > (1) append an element to each row. > starting with M we get M itself with the first line omitted (each > line terminates, so there is a clear new position where we can append). > starting with M' we get a new matrix M'' (we have to define how we > append an element to the first line, but let that be with column > index 'omega+1'. > (2) append an element to each column. > starting with M we get the same problem as before with M', but if we > allow a row 'omega+1', we get an element with row number 'omega+1' in > each column. > starting with M' something similar happens (although each column has > a gap). > Now explain why (1) and (2) should provide an inconsistency, while (3) and > (4) do not. You said it already yourself: Starting with (1) M, the rows of M > remain (except the first line), while in (2) M we get omega + 1 > columns. Your points (3) an d (4) are uninteresting. Still I maintain, why is that an inconsistency? Note that here we are talking about indices! > The set of second indexes of M' (which is containing a completed > infinity) cannot be the same as the set of second indexes of M (which > is not containing a completed infinity). Note that M' violates your writing above that the rows represent natural numbers written in unary notation! So either go one way or the other, do not mix them. > Nobody says so, when you properly define what you mean with appending. Cantor has defined how an element is appended to omega. Just this is > what I am doing and what you should know - and what forces W.H. to > assume the same set of second indices in M aa in M'. But let us see. When defined as indices, your M looks like: a_ij is defined for every j in N and every i <= j and your M' looks like (I assume here replacing the first line, I disremember which it was, but it makes no difference): a_ij is defined for every j in N and whenever i = 1 or i <= j where is the difference between sets of second indices? So M and M' have the same set of second indices. Now we add elements to rows. Let us define that we add an element by inserting a new element at a second column index position that is larger than all other index positions in that same line. With M we can add all elements at finite positions and get M_app: a_ij is defined for every j in N and every i <= j - 1 for M' there is difficulty with the first line, and we get: a_ij is defined for every j in N and whenever i = 1 or i <= j - 1, and for i = 1 and j = w. I still do not see a problem. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Two results of set geometry M' does not contain a second index omega. A does > not contain an element omega. A+1 does not contain > omega + 1. > But by extending all rows of M' we get a set of order omega + 1. Don't > we get at least one non-natural index, necessarily? As you do not define what *you* mean with extend (and you do apparently > not use the standard definition), it is even unclear what this means. > Normally with extend is meant adding something at the end when talking > about ordered sets. But as a set with order type omega does not have an > end, it is unclear what is meant in that case. But if you mean add an > element with index greater than all other indices, we clearly get from > an ordered set with order type omega a set with order type omega+1. I mean simply: append an element as Cantor does define it. There is no > index to be considered. You can see this if you append 1 to the set > {2,3,4,...} to get {2,3,4,...,1}. There *are* indices to be considered. All your elements are '1'! No, the elements are sequences of 1's. > So you don't believe in the unique bijection between digits and second > indexes? > 1 1 1 1 ... and > 1,2,3,4,... But this is again thoroughly unmathematical. Can you show a bijection > is f(1)? You are talking about order preserving bijections between > ordered multi-sets. Argh. Look at the matrix. It should be obvious: 1 > 11 > 111 > 1111 > ... What is f(1)? Darn, you *say* you are talking about bijections between > elements when actually you are talking about bijections between index > positions (and so, yes, index positions *are* to be considered). See my last contribution, where I illustrated in detail the trijection which I had precisely defined earlier. > The elements of the sets are not second indexes, but > sequences of ones. The set of sequences of ones > in M' is not the same as the set of sequences of ones > in M. > As above bijection shows, then also the set of second indexes must > differ. That is an order preserving bijection between multi-sets. You misunderstand. The bijection is between n and {1,2,3,...,n} but > the n's are written in unary representation by sets of 1's in the rows > and the initial segments are those of the first column. Now, this is again utterly confused. With this explanation I see only > a single set: the set of natural numbers written out in unary. But if > you consider it that way, there is only a single column, namely the > list of natural numbers. You see only one column, but don't know what has to be understood by the expression initial segments of the first column? If the set of natural numbers has omega elements, then the set of rows > is in bijection with the complete first column. Yes, of course, because the set of rows is the set of natural numbers > (because each element is a natural number). And it is in bijection with > the first column, which is the set of rows. Quite trivial I would think. Only the fact that the first column has a lot of finite initial segments and an infinite initial segment while the rows are all finite may be raising some suspicion. If a bijection exists, > then appending (in the manner of Cantor) one element to each partner > of the bijection, then the bijection persists. Obvious. As each partner in the bijection is the set of natural numbers. That is incorrect. Each partner on the one side is a number of natural numbers and on the other side each partner is a natural number. I tried to express that by the variables {1,2,3,...,n} <--> n. But this leads to omega > + 1 being in bijection with the set of natural numbers. Not at all. Why do you think so? What you actually *do* when you append > a 1 to each row is changing each element of the set {1,2,3,...} to its > successor, so you change it to {2,3,4,...}. You do not add something to > the set the column is bijected with. In fact I do not add something to the set the column is bijected witd, but I do neither add something to the column. The initial segments on the left hand side change from 1, 11, 111, ..., 111... to 11, 111, ..., 111..., 111...1. If they rermain in bijection with the sequences in the rows, then the set {2, 3, 4, ...} has ordinal omega + 1. > Appending a one the the all elements of M' > does not give the same result as appending a one > to all elements of M. > That is because set theory is blurry. No, it is because you are blurry. Each '1' in M and M' actually > represents a pair of natural numbers (r,c), where r is the row number > and c the column number. The elements of the triangular matrix M are > defined as (r,c) as above and non-existing when c > r. The matrix M' > (of some strange format) is defined as M except when r = 1, in that > row *all* elements do exist. When defined such we can talk about > bijections. Now let us try a few actions: > (1) append an element to each row. > starting with M we get M itself with the first line omitted (each > line terminates, so there is a clear new position where we can append). > starting with M' we get a new matrix M'' (we have to define how we > append an element to the first line, but let that be with column > index 'omega+1'. > (2) append an element to each column. > starting with M we get the same problem as before with M', but if we > allow a row 'omega+1', we get an element with row number 'omega+1' in > each column. > starting with M' something similar happens (although each column has > a gap). > Now explain why (1) and (2) should provide an inconsistency, while (3) and > (4) do not. You said it already yourself: Starting with (1) M, the rows of M > remain (except the first line), while in (2) M we get omega + 1 > columns. Your points (3) an d (4) are uninteresting. Still I maintain, why is that an inconsistency? Note that here we are > talking about indices! We have a bijection between natural numbers and ordinal numbers of sets of natural numbers. > The set of second indexes of M' (which is containing a completed > infinity) cannot be the same as the set of second indexes of M (which > is not containing a completed infinity). Note that M' violates your writing above that the rows represent natural > numbers written in unary notation! So either go one way or the other, do > not mix them. M' is mere an example of a possible trijection. Nobody says so, when you properly define what you mean with appending. Cantor has defined how an element is appended to omega. Just this is > what I am doing and what you should know - and what forces W.H. to > assume the same set of second indices in M and in M'. But let us see. When defined as indices, your M looks like: > a_ij is defined for every j in N and every i <= j > and your M' looks like (I assume here replacing the first line, I disremember > which it was, but it makes no difference): > a_ij is defined for every j in N and whenever i = 1 or i <= j > where is the difference between sets of second indices? So M and M' have the same set of second indices. That means there is no difference in the set of numbers represented by the sequences of 1's in the rows of M and M'. The set of natural numbers has the same set of second indices as that set with an inbfinite number added? > Now we add elements to > rows. Let us define that we add an element by inserting a new element at > a second column index position that is larger than all other index positions > in that same line. With M we can add all elements at finite positions > and get M_app: > a_ij is defined for every j in N and every i <= j - 1 > for M' there is difficulty with the first line, and we get: > a_ij is defined for every j in N and whenever i = 1 or i <= j - 1, > and for i = 1 and j = w. > I still do not see a problem. I must say, I am not really surprised. (The problem is that a set of finite natural numbers cannot be complete, i.e., its ordinal number cannot exist as a number which is in trichotomy with the elements of the set and which is different from omega + 1.) === Subject: Re: Two results of set geometry Nntp-Posting-Host: hera.cwi.nl > Two results of set geometry Set geometry, a new branch of mathematics, is devised to investigate > relations of finite and infinite sets by means of geometrical > representations. Up to now there have been two important results. The first result is that the ordinal number of the set of natural > numbers is not unique. It has been shown that if {1, 2, 3, ...} = > omega, then {1, 2, 3, ...} = omega + 1 too. The complete set of natural numbers has omega elements. It is a valid > statement in set theory that a set with omega elements (like the set > of all positions of the first column) can be extended to a set of > order omega + 1 by adding one element. This feature can be used to > investigate the trijection between the positions of 1's of the first > column, the diagonal, and the lines in the matrix M 1000... > 11000... > 111000... > ... If there is a trijection, (You actually mean mutiple bijections between sets of indices, but I will follow your sloppy writing.) What bijection is there between the ones of the first column and of *any* line of that matrix? There is a bijection between the first column and each and every complete line of the matrix. > then it should remain a trijection after > extending the sets involved by one element each. Extending the first > column of M by one element we obtain a sequence of order type omega + > 1 (because it has the order type omega). After adding one 1 to each > sequence of 1's in the lines of M no sequence of 1's has order type > omega + 1 (because each one has a finite order type). You are doing different additions to the first column and the first line. You add an element at the end of the first column, but in the middle of the first line. So why do you expect a bijection between the first line and the first column? (Note that trijection is just obfuscation, there are simply multiple bijections.) > This can be > repeated infinitely often, such that the order type of the first > column becomes omega*2 (notation from Cantor, 1895) while the order > type of the set of lines cannot surpass omega. They *can* surpass omega, if you add elements at the end, which you do not. This is most properly seen when you use indices. At the first line you add a new element at position (1,2), shifting *all* 0's up one place. When you add a new element to the first column, you do *not* do it at position (2,1), shifting *all* 1's up one place, but you do it at a new position (w,1). > For the diagonal the > case remains undecided. But whatever the diagonal may be, there is no > trijection between the sequences of 1's in the first column, the > diagonal, and the lines - not finally and, therefore, also not > initially. Bogus. When you add an element after the first column, what is the diagonal? But we *do* know that there does exist a bijection between a set of order w and a set of order w+1. Let's get the new matrix with a new line added at position (w,1). I get the following bijections, from first column (c_i1) to first line (l_1i) and to diagonal (d_ii): if i = w: (c_w1) <-> (l_11) <-> (d_11) if i != w: (c_i1) <-> (l_[i+1]1) <-> (d_[i+1][i+1]). > The second result shows that the set of real numbers is countable. > And this is just as wrong. > Each number is given by a path > stretching over infinitely many nodes (bits). All nodes (bits) of the > tree are countable. The paths are not, according to Cantor's famous > diagonal proof. But we find that, up to line number n, there are -1 + 2^(n+1) nodes > whereas 2^n different paths arrive at and 2^(n+1) different paths > spring off from line number n. How you can state that couple of sentences with a straight face escapes me. As *all* paths are infinite in length, even at the root node your statements is wrong, because *all* paths spring off from the root node and not only two different paths. And more general, at each node arrive *all* the paths that do emanate from that node. So you should explain how it is possible that at a single node arrives a single path while two do emanate from it (all paths start at the root). In the second paragraph you are not talking about paths but about edges. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Two results of set geometry relations of finite and infinite sets by means of geometrical > representations. Up to now there have been two important results. The first result is that the ordinal number of the set of natural > numbers is not unique. It has been shown that if {1, 2, 3, ...} = > omega, then {1, 2, 3, ...} = omega + 1 too. The complete set of natural numbers has omega elements. It is a valid > statement in set theory that a set with omega elements (like the set > of all positions of the first column) can be extended to a set of > order omega + 1 by adding one element. This feature can be used to > investigate the trijection between the positions of 1's of the first > column, the diagonal, and the lines in the matrix M 1000... > 11000... > 111000... > ... If there is a trijection, (You actually mean mutiple bijections between sets of indices, but I will > follow your sloppy writing.) I gave an exact definition of trijection in the thread and explicated it in detail to MoeBlee. > What bijection is there between the ones of the first column and of *any* > line of that matrix? There is a bijection between the first column and > each and every complete line of the matrix. then it should remain a trijection after > extending the sets involved by one element each. Extending the first > column of M by one element we obtain a sequence of order type omega + > 1 (because it has the order type omega). After adding one 1 to each > sequence of 1's in the lines of M no sequence of 1's has order type > omega + 1 (because each one has a finite order type). You are doing different additions to the first column and the first line. > You add an element at the end of the first column, but in the middle of > the first line. So why do you expect a bijection between the first line > and the first column? (Note that trijection is just obfuscation, there > are simply multiple bijections.) In fact the trijection can be understood in this way. If you are unable to comprehend the principle of trijection, we can set up two bijections. There is one bijection between the initial segments (including the complete sequences) of the first column and the diagonal: 1 <--> 1 1 1 1 <--> 1 1 1 1 1 1 <--> 1 ... 1 1 1 1 1 1 . . . . . <--> . There is another bijection between the first column and the lines (if there are infinitely many finite natural numbers) 1 <--> 1 1 1 <--> 11 1 1 1 <--> 111 ... 1 1 1 . . . <--> ? Third, there is the bijection between rows and initial segments of the diagonal. You will be able to illustrate it yourself. This can be > repeated infinitely often, such that the order type of the first > column becomes omega*2 (notation from Cantor, 1895) while the order > type of the set of lines cannot surpass omega. They *can* surpass omega, if you add elements at the end, which you do > not. This is most properly seen when you use indices. At the first > line you add a new element at position (1,2), shifting *all* 0's up > one place. When you add a new element to the first column, you do *not* > do it at position (2,1), shifting *all* 1's up one place, but you do it > at a new position (w,1). Wrong description. The second bijection after appending one element reads: 1 1 <--> 11 1 1 1 <--> 111 1 1 1 1 <--> 1111 ... Of course appending one element to omega yields omega + 1. For the diagonal the > case remains undecided. But whatever the diagonal may be, there is no > trijection between the sequences of 1's in the first column, the > diagonal, and the lines - not finally and, therefore, also not > initially. Bogus. When you add an element after the first column, what is the > diagonal? But we *do* know that there does exist a bijection between > a set of order w and a set of order w+1. Let's get the new matrix with > a new line added at position (w,1). I get the following bijections, > from first column (c_i1) to first line (l_1i) and to diagonal (d_ii): > if i = w: (c_w1) <-> (l_11) <-> (d_11) > if i != w: (c_i1) <-> (l_[i+1]1) <-> (d_[i+1][i+1]). The clue however is, that in normal order N is omega + 1. The second result shows that the set of real numbers is countable. > And this is just as wrong. > Each number is given by a path > stretching over infinitely many nodes (bits). All nodes (bits) of the > tree are countable. The paths are not, according to Cantor's famous > diagonal proof. But we find that, up to line number n, there are -1 + 2^(n+1) nodes > whereas 2^n different paths arrive at and 2^(n+1) different paths > spring off from line number n. How you can state that couple of sentences with a straight face escapes > me. As *all* paths are infinite in length, even at the root node your > statements is wrong, because *all* paths spring off from the root node > and not only two different paths. And more general, at each node arrive > *all* the paths that do emanate from that node. So you should explain > how it is possible that at a single node arrives a single path while > two do emanate from it (all paths start at the root). In the second > paragraph you are not talking about paths but about edges. Only the number of separated paths is interesting, i.e., the number of different real numbers. In the binary tree there are all real numbers of the interval [0, 1] represented by paths from 0.000... to 0.111... . 0. / 0 1 / / 0 1 0 1 ... It can easily be seen that there cannot be more *separate* paths than nodes, because every path needs a node to separate itself from the other paths. Therefore there are not more different real numbers in the interval [0, 1] than nodes. But the number of nodes is countable with no doubt. And, by construction, the tree contains all separable real numbers of the interval. There is no chance for a diagonal number. === Subject: Re: Two results of set geometry > Sorry to break in again, I have been away a bit too long. ... > M' does not contain a second index omega. A does > not contain an element omega. A+1 does not contain > omega + 1. But by extending all rows of M' we get a set of order omega + 1. Don't > we get at least one non-natural index, necessarily? As you do not define what *you* mean with extend (and you do apparently > not use the standard definition), it is even unclear what this means. > Normally with extend is meant adding something at the end when talking > about ordered sets. But as a set with order type omega does not have an > end, it is unclear what is meant in that case. But if you mean add an > element with index greater than all other indices, we clearly get from > an ordered set with order type omega a set with order type omega+1. I mean simply: append an element as Cantor does define it. There is no > index to be considered. You can see this if you append 1 to the set > {2,3,4,...} to get {2,3,4,...,1}. As sets, {2,3,4,...,1} = {1,2,3,4,...}. As ordered sets, (2,3,...;1) =/= (1,2,3,...). Absent indexing, or some equivalent establishment of ordering, one has merely {2,3,4,...,1} = {1,2,3,4,...}. > Look at the matrix. It should be obvious: 1 > 11 > 111 > 1111 > ... You misunderstand. The bijection is between n and {1,2,3,...,n} but > the n's are written in unary representation by sets of 1's in the rows > and the initial segments are those of the first column. 1 <--> 1 1 > 1 <--> 11 1 > 1 > 1 <--> 111 ... If the set of natural numbers has omega elements, then the set of rows > is in bijection with the complete first column. If a bijection exists, > then appending (in the manner of Cantor) one element to each partner > of the bijection, then the bijection persists. The partners being paired are a row from the set of rows with the lead entries of that row as an entry in the first column. So appending an entry at the end of that column requires appending a new row, longer than all the present rows. Which leads to the order type of each, the extended set of column entries and the extended set of rows (including a row longer than any of the infinitely many finite rows), being of order type omega + 1. When WM tries to switch from members to sets or vice versa, as he does in his extending of order isomorphisms, his alleged proofs become invalid. === Subject: Re: Two results of set geometry Sorry to break in again, I have been away a bit too long. ... > M' does not contain a second index omega. A does > not contain an element omega. A+1 does not contain > omega + 1. But by extending all rows of M' we get a set of order omega + 1. Don't > we get at least one non-natural index, necessarily? As you do not define what *you* mean with extend (and you do apparently > not use the standard definition), it is even unclear what this means. > Normally with extend is meant adding something at the end when talking > about ordered sets. But as a set with order type omega does not have an > end, it is unclear what is meant in that case. But if you mean add an > element with index greater than all other indices, we clearly get from > an ordered set with order type omega a set with order type omega+1. I mean simply: append an element as Cantor does define it. There is no > index to be considered. You can see this if you append 1 to the set > {2,3,4,...} to get {2,3,4,...,1}. As sets, {2,3,4,...,1} = {1,2,3,4,...}. > As ordered sets, (2,3,...;1) =/= (1,2,3,...). > Absent indexing, or some equivalent establishment of ordering, > one has merely {2,3,4,...,1} = {1,2,3,4,...}. > The ordering is established by my matrix, which after appending one 1 to every partner of the bijection reads 1 1 <--> 11 1 1 1 <--> 111 ... 1 1 1 . . . <--> ? 1 1 1 . . . 1 <--> ? > Look at the matrix. It should be obvious: 1 > 11 > 111 > 1111 > ... You misunderstand. The bijection is between n and {1,2,3,...,n} but > the n's are written in unary representation by sets of 1's in the rows > and the initial segments are those of the first column. 1 <--> 1 1 > 1 <--> 11 1 > 1 > 1 <--> 111 ... If the set of natural numbers has omega elements, then the set of rows > is in bijection with the complete first column. If a bijection exists, > then appending (in the manner of Cantor) one element to each partner > of the bijection, then the bijection persists. The partners being paired are a row from the set of rows with the lead > entries of that row as an entry in the first column. So appending an entry at the end of that column requires appending a new > row, longer than all the present rows. That would be the case indeed *if the complete column existed*. However, it is not so. Which leads to the order type of each, the extended set of column > entries and the extended set of rows (including a row longer than any of > the infinitely many finite rows), being of order type omega + 1. The set of rows is extended but not by an infinite number. All numbers n + 1 are finite - for n in N. When WM tries to switch from members to sets or vice versa, as he does > in his extending of order isomorphisms, his alleged proofs become > invalid It is said that the natural numbers are the members of a set. === Subject: Re: Two results of set geometry I haven't bothered to read your proofs in detail. Its basically like being given a proof that 1=2 and told it is correct. It has been proved that N cannot be bijected with R, so unless you have proved that set theory is inconsistent your claim is obviously wrong. The chances of you doing that are so small as to make reading your proofs a waste of time. I was however struck by the following: > If the set of natural numbers has omega elements, then the set of rows > is in bijection with the complete first column. If a bijection exists, > then appending (in the manner of Cantor) one element to each partner > of the bijection, then the bijection persists. But this leads to omega > + 1 being in bijection with the set of natural numbers. > You can easily biject omega+1 with N. Here is one example: 1 <-> w+1 2 <-> 1 3 <-> 2 4 <-> 3 . . . If you think that this is paradoxical and worth creating a proof of, you are about 100 years too late. Cantor himself proved that N can be bijected with w+1, w+w, and w^127 + 93w + 17 for that matter. This makes your first result akin to proving 1+1=2 (everybody knows its true and nobody disputes it) and your second result akin to proving 1+1=3 (everybody knows its false, because it has been proven to be so). === Subject: Re: Two results of set geometry <46d49be6$0$22253$afc38c87@news.optusnet.com.au I haven't bothered to read your proofs in detail. Its basically like being > given a proof that 1=2 and told it is correct. It has been proved that N > cannot be bijected with R, so unless you have proved that set theory is > inconsistent your claim is obviously wrong. The chances of you doing that > are so small as to make reading your proofs a waste of time. That attitude may by the reason why set theory has not yet been recognized as inconsistent by many mathematicians. I was however struck by the following: If the set of natural numbers has omega elements, then the set of rows > is in bijection with the complete first column. If a bijection exists, > then appending (in the manner of Cantor) one element to each partner > of the bijection, then the bijection persists. But this leads to omega > + 1 being in bijection with the set of natural numbers. You can easily biject omega+1 with N. Here is one example: 1 <-> w+1 > 2 <-> 1 > 3 <-> 2 > 4 <-> 3 > . > . > . If you think that this is paradoxical and worth creating a proof of, you are > about 100 years too late. Cantor himself proved that N can be bijected with > w+1, w+w, and w^127 + 93w + 17 for that matter. Sorry, what you say is correct but not to the point. Of course N can be bijected in this way to ordered sets (at least it was assumed so for a long while; already Cantor gave examples like {1, 3, 5, ..., 2, 4, 6, ...}). What I found is simply that if the set of finite natural numbers in the normal order can be bijected to omega (represented by the infinite first column of the matrix 1 11 111 ... then the N *in the normal order* can also be bijected to omega + 1. This is shown by appending just one 1 to every partner of the bijection. First, according to the matrix given above, but in more detail, we have the bijection: 1 <--> 1 1 1 <--> 11 1 1 1 <--> 111 ... where also the infinite first column belongs to the bijection (although there is no infinite natural number). If this bijection exists, then we can, without destroying it, append one 1 to every partner. This leaves every natural number finite (only the first line of the bijection disappears) but it introduces as the new first column the ordinal omega + 1. Therefore the set {2, 3, 4, ...} of all but one natural numbers is in bijection with omega + 1. This makes your first result akin to proving 1+1=2 (everybody knows its true > and nobody disputes it) and your second result akin to proving 1+1=3 > (everybody knows its false, because it has been proven to be so). It has been proven so under the assumption that it is possible to attach a number omega to infinity. This is shown inconsistent. My second proof is also very simple: In the binary tree there are all real numbers of the interval [0, 1] represented by paths from 0.000... to 0.111... . 0. / 0 1 / / 0 1 0 1 ... It can easily be seen that there cannot be more separate paths than nodes, because every path needs a node to separate itself from the other paths. Therefore there are not more different real numbers in the interval [0, 1] than nodes. But the number of nodes is countable with no doubt. And, by construction, the tree contains all separable real numbers of the interval. There is no chance for a diagonal number. === Subject: Re: Two results of set geometry relations of finite and infinite sets by means of geometrical > representations. Up to now there have been two important results. The first result is that the ordinal number of the set of natural > numbers is not unique. It has been shown that if {1, 2, 3, ...} = > omega, then {1, 2, 3, ...} = omega + 1 too. > The second result shows that the set of real numbers is countable. Totally bogus. Utterings of your taste are appreciated but of no value concerning set geometry. The arguments of set geometry stand unrefuted. === Subject: Re: Two results of set geometry > Two results of set geometry Set geometry, a new branch of mathematics, is devised to investigate > relations of finite and infinite sets by means of geometrical > representations. Up to now there have been two important results. The first result is that the ordinal number of the set of natural > numbers is not unique. It has been shown that if {1, 2, 3, ...} = > omega, then {1, 2, 3, ...} = omega + 1 too. > The second result shows that the set of real numbers is countable. Totally bogus. Utterings of your taste are appreciated but of no value concerning set > geometry. The arguments of set geometry stand unrefuted. They certainly stand unestablished. At least by any standards mathematically acceptable. === Subject: Re: Two results of set geometry geometry. The arguments of set geometry stand unrefuted. They certainly stand unestablished. At least by any standards mathematically acceptable.- By the standards which assume the existence of nameless and undefinable numbers? === Subject: Re: Two results of set geometry Two results of set geometry Set geometry, a new branch of mathematics, is devised to investigate > relations of finite and infinite sets by means of geometrical > representations. Up to now there have been two important results. The first result is that the ordinal number of the set of natural > numbers is not unique. It has been shown that if {1, 2, 3, ...} = > omega, then {1, 2, 3, ...} = omega + 1 too. urce&hl=de Except that there is no proof here at all. Unless 2 = 1. The second result shows that the set of real numbers is countable. e That alleged showing has been debunked severally, both by the flaws > in its arguments and by counterproofs. In case you failed to read my answer on your false arguments, here it is again. (Just an opportunity for you to learn something. I do not intend to further discuss your uneducated assertions like infinite sets being implied by Peano etc.) ____________________________________________________________________ > If you mean that the unbounded initial segments are inclused in the > trijection along with the bounded ones, that is quite obvious, but > equally irrelevant. I beg your pardon, why should it be obvious that the unbounded initial segments, i.e., the complete sequences be included in my trichotomy? I think that need only be so if I defined it so. > (a_11, .., a_n1) <--> (a_11, .., a_nn) <--> (a_11, .., a_1n) > such that all elements belonging to an initial segment of > column, > diagonal, and line are 1's. > But there is no such trijection for n in omega including omega between > the first column, the diagonal and the n-th line > But there is no need for one either, since for all m, n IN omega, and > all m <= n, a_mn = a_nn = a_nm. It is not important though, but for the matrix given above we have 0 = a_23 =/= a_32 = 1, for example. > WM's problem is that he wants endless sequences to have two ends, when > they don't.- > The complete set of natural numbers has omega elements. It is a valid > statement in set theory that a set with omega elements (like the set > of all positions of the first column) can be extended to a set of > order omega + 1 by adding one element. > Not at all. What precisely do you deny? > It is only true that a well ordered set of omega elements > after having a new element appended to its ordering is a new set of of > order type (omega+ 1). Anything stated less precisely need not be hold. So let us consider the first column, i.e., the set of positions {a_k1 | k in N} and append one element to obtain a set of order omega + 1. - Zitierten Text ausblenden - - Zitierten Text anzeigen - > This feature can be used to > investigate the trijection between the positions of 1's of the first > column, the diagonal, and the lines in the matrix M > 1000... > 11000... > 111000... > ... > If there is a trijection, then it should remain a trijection after > extending the sets involved by one element each. Extending the first > column of M by one element we obtain a sequence of order type omega + > 1 (because it has the order type omega). After adding one 1 to each > sequence of 1's in the lines of M no sequence of 1's has order type > omega + 1 (because each one has a finite order type). This can be > repeated infinitely often, such that the order type of the first > column becomes omega*2 (notation from Cantor, 1895) while the order > type of the set of lines cannot surpass omega. > It is as easy to append a 1 to any endless line (as filled in to > endlessness by spaces or zeros) as to any endless column or diagonal. That is correct, but it does not concern my argument. If you would like to discuss another topic, it would be desirable to open a new thread in order to avoid mixing up the facts. In my argument the 1 should not be appended to the endless line but to the finite sequence of 1's in each line (because only finite sequences can represent natural numbers - not infinite sequences). In my argument it is essential that those natural numbers are to be put in trijection with the initial segments of column and diagonal. === Subject: Re: Two results of set geometry I beg your pardon, why should it be obvious that the unbounded > initial segments, i.e., the complete sequences be included in my > trichotomy? I think that need only be so if I defined it so. > (a_11, .., a_n1) <--> (a_11, .., a_nn) <--> (a_11, .., a_1n) If it does not hold for N, then there must be a first n in N for which it doe not hold. But induction proves otherwise. It is not important though, but for the matrix given above we have 0 > = a_23 =/= a_32 = 1, for example. > WM's problem is that he wants endless sequences to have two > ends, when they don't.- > The complete set of natural numbers has omega elements. It is a > valid statement in set theory that a set with omega elements > (like the set of all positions of the first column) can be > extended to a set of order omega + 1 by adding one element. > Not at all. What precisely do you deny? That prepending or inserting a new element, both forms of addition, into an endless but well-ordered set effects a change its order type. > It is only true that a well ordered set of omega elements after > having a new element appended to its ordering is a new set of of > order type (omega+ 1). Anything stated less precisely need not be > hold. > So let us consider the first column, i.e., the set of positions {a_k1 > | k in > N} and append one element to obtain a set of order omega + 1. - Zitierten Text ausblenden - - Zitierten Text anzeigen - This feature can be used to investigate the trijection between > the positions of 1's of the first column, the diagonal, and the > lines in the matrix M > 1000... 11000... 111000... ... > If there is a trijection, then it should remain a trijection > after extending the sets involved by one element each. Extending > the first column of M by one element we obtain a sequence of > order type omega + 1 (because it has the order type omega). After > adding one 1 to each sequence of 1's in the lines of M no > sequence of 1's has order type omega + 1 (because each one has a > finite order type). Apples and oranges. When one associates an element of some column with some row, adding an element to that column equates with adding an entire row, not adding an element to a row. It is as easy to append a 1 to any endless line (as filled in to > endlessness by spaces or zeros) as to any endless column or > diagonal. That is correct, but it does not concern my argument. But it scuttles your argument as you are adding apples to oranges. > If you would > like to discuss another topic That WM is too dense to realize that it is the proper topic to reveal the faults in his arguments is his own personal problem. And an irrelevance to the proof he is wrong. === Subject: Re: Two results of set geometry initial segments, i.e., the complete sequences be included in my > trichotomy? I think that need only be so if I defined it so. (a_11, .., a_n1) <--> (a_11, .., a_nn) <--> (a_11, .., a_1n) If it does not hold for N, then there must be a first n in N for which > it doe not hold. But induction proves otherwise. First: What holds for N need not hold or not hold for its elements. Second: You asserted just the opposite, namely: If you mean that the unbounded initial segments are included in the trijection along with the bounded ones, that is quite obvious, but equally irrelevant. > It is not important though, but for the matrix given above we have 0 > = a_23 =/= a_32 = 1, for example. WM's problem is that he wants endless sequences to have two > ends, when they don't.- The complete set of natural numbers has omega elements. It is a > valid statement in set theory that a set with omega elements > (like the set of all positions of the first column) can be > extended to a set of order omega + 1 by adding one element. Not at all. What precisely do you deny? That prepending or inserting a new element, both forms of addition, into > an endless but well-ordered set effects a change its order type. > As I did not consider _prepending_ at any time, why then did you say not at all? That _appending_ an element can change the order type is undisputed by mathematicians, however. > It is only true that a well ordered set of omega elements after > having a new element appended to its ordering is a new set of of > order type (omega+ 1). Anything stated less precisely need not be > hold. So let us consider the first column, i.e., the set of positions {a_k1 > | k in > N} and append one element to obtain a set of order omega + 1. - Zitierten Text ausblenden - - Zitierten Text anzeigen - This feature can be used to investigate the trijection between > the positions of 1's of the first column, the diagonal, and the > lines in the matrix M 1000... 11000... 111000... ... If there is a trijection, then it should remain a trijection > after extending the sets involved by one element each. Extending > the first column of M by one element we obtain a sequence of > order type omega + 1 (because it has the order type omega). After > adding one 1 to each sequence of 1's in the lines of M no > sequence of 1's has order type omega + 1 (because each one has a > finite order type). Apples and oranges. In set theory it does not matter whether sets consist of apples or oranges. Important is whether they can be put in bijection. > When one associates an element of some column with > some row, adding an element to that column equates with adding an entire > row, not adding an element to a row. > The bijection 1 <--> 1 1 1 <--> 11 1 1 1 <--> 111 remains a bijection after appending one 1 to every partner 1 1 <--> 11 1 1 1 <--> 111 1 1 1 1 <--> 1111 It is as easy to append a 1 to any endless line (as filled in to > endlessness by spaces or zeros) as to any endless column or > diagonal. That is correct, but it does not concern my argument. But it scuttles your argument as you are adding apples to oranges. If you would > like to discuss another topic That WM is too dense to realize that it is the proper topic to reveal > the faults in his arguments is his own personal problem. It is refreshing to hear this from a five-star-mathematician who does not know that Peano does not provide actually infinite sets and who describes the matrix 1000... 11000... 111000... ... by: for all m, n IN omega, and all m <= n, a_mn = a_nn = a_nm. Why m <= n but not n <= m? === Subject: Re: Two results of set geometry > Two results of set geometry Set geometry, a new branch of mathematics, is devised to investigate > relations of finite and infinite sets by means of geometrical > representations. Up to now there have been two important results. The first result is that the ordinal number of the set of natural > numbers is not unique. It has been shown that if {1, 2, 3, ...} = > omega, then {1, 2, 3, ...} = omega + 1 too. The second result shows that the set of real numbers is countable. > The first and foremost result of your new idiocy is this: === Subject: thread closed: A problem of set geometry sci.math.research [...] [ Moderator's note: This thread is closing. Further discussion of trijection may be done in sci.math, for example. ] [...] In other words, you have been thrown out of sci.math.research. Just in case you didn't notice, Mr. Prof. Dr. W. Mueckenheim. === Subject: Re: plane equation Given 3 points, how do you solve for the plane containing those > points? > It seems that standard plugging in yields 3 equations, but there are > 4 unknowns. Divide equation a x + b y + c z = d by d and set d/a = A, d/b = B and > d/c = C. Then x/A + y/B + z/C = 1. Does not work for planes through origin. A good way to remember this is that the plane is generated by all > vectors perpendicular to one particular vector. I'll try to get you started. Say you three points are a = (x_1 , y_1, z_1), b = (x_2 , y_2, z_2) > and c = (x_3 , y_3, z_3). Next, get vectors ab = (x_2 - x_1 , y_2 - y_1, z_2 - z_1) ac = (x_3 - x_1 , y_3 - y_1, z_3 - z_1) Now these two items are vectors in the plane. What mechanism do you > know that will give you a vector perpendicular to these two vectors? Once you have that vector use it to get the equation of your plane. I > am bound by social ethics not to divulge anymore as the end product of > such behavior is abhorrent, but everyone has his cross to bear! Brian > well if we define a matrix X a 3 x 4 matrix with row i of X being [x_i y-i z_i -1] and y be the vector [a,b,c,d]^T such the equation of the plane is ax+by+cz = d then Xy = 0 if the points are not all collinear [they determine a plane] then rank(x) = 3. and the null space of X is of dimension 1. y is in the nullspace of X so if we find a basis of nullspace(X) then the coefficients of the equation of the plane are any convienient multiple of this basis vector. This may be more stable than the direct three dimension solution using relative vectors and cross products, although computationally more complicated it does not depend on if the plane passes thru the origin or which point is chosen to translate to compute relative vectors. X = USV^T [svd decomposition] then Xy = 0 = USV^T U^TU=I V^TV = I and S = diag(s1,s2,s3,0). U = [U1 u] V = [V1 v] S1 = diag(s1,s2,s3) then Xv = [U1 u]diag(S1,0)[V1 v]^T v = [U1S1 0][V1 v]^T v = [U1S1V1^T 0] v = [U1S1V1^Tv 0] but V1^Tv = 0 (since V^TV = I) so Xv = [0 0] = 0 y = v = [a b c d]^T with a^2 + b^2 + c^2 +d^2 = 1 is one solution. y = c*v where c is a constant of choice is another. Algorithm: 1) create X. 2) find SVD of X noting rank[if its not 3 the problem is ill posed stop] save the 4-th column of V = v 3) y = v is a solution for the coefs with dot(v,v) = 1. if another normalization is desired multiplying y by any constant different from zero provides a different representation of the same equation. === Subject: quadratic with complex coefficients I have a quadratic equation with complex coefficients: x^2 + (1 - i)x + i = 0 And I've worked it out to: [-1 + i +/- sqrt([1 - i]^2 - 4i) ] / 2 Now, the square root expression can be simplified to: sqrt(-6i) which, in trigonometric form, is: sqrt(6^(1/2)*cis(45)) or sqrt(6^(1/2)*cis(105)) Which is the two roots of -6i supplied as arguments to sqrt. I'm trying to work it out to the answer my book gives, which is: (-[1 + sqrt(3)]/2 + [1 + sqrt(3)*i]/2), (-[1 - sqrt(3)]/2 - [1 - sqrt(3)*i]/2) Where have I errored? -- conrad === Subject: Re: quadratic with complex coefficients > I have a quadratic equation with > complex coefficients: > x^2 + (1 - i)x + i = 0 And I've worked it out to: > [-1 + i +/- sqrt([1 - i]^2 - 4i) ] / 2 Now, the square root expression > can be simplified to: > sqrt(-6i) which, in trigonometric > form, is: sqrt(6^(1/2)*cis(45)) > or sqrt(6^(1/2)*cis(105)) > Which is the two roots of -6i > supplied as arguments to sqrt. I'm trying to work it out > to the answer my book gives, > which is: > (-[1 + sqrt(3)]/2 + [1 + sqrt(3)*i]/2), > (-[1 - sqrt(3)]/2 - [1 - sqrt(3)*i]/2) Where have I errored? -- > conrad Independently of your problem, one advice: If you need to find the square root of a complex number and for avoiding nasty trigonometric forms, it is better to use the following method: a, b, x, y : real numbers. sqrt (a + bi)= x + iy <=> (x + iy)^2= a + bi <=> ( x^2 - y^2= a ) and ( 2xy = b ). This system provides the biquadratic equation: 4x^4 - 4ax^2 - b^2= 0 Fernando. === Subject: Re: quadratic with complex coefficients >I have a quadratic equation with >complex coefficients: >x^2 + (1 - i)x + i = 0 And I've worked it out to: >[-1 + i +/- sqrt([1 - i]^2 - 4i) ] / 2 Now, the square root expression >can be simplified to: >sqrt(-6i) which, in trigonometric >form, is: sqrt(6^(1/2)*cis(45)) >or sqrt(6^(1/2)*cis(105)) >Which is the two roots of -6i >supplied as arguments to sqrt. Error here. Since you used degrees above I'll do the same, rather than use radian measures. Ignoring the second (incorrect) square root operation, you correctly found one of the square roots of +6i, namely sqrt(6) cis(45) [the other one is sqrt(6) cis(225)]. What you want is sqrt(-6i). To get that, you can 'rotate' what you found above by 90 degrees, giving sqrt(-6i) = sqrt(6) cis(135) Putting this into a more convenient form for the problem at hand, sqrt(6) [cos(135) + i sin(135)] = sqrt(3) (-1 + i) Although not needed for this problem, the second square root of -6i is found by another 'rotation' of 180 degrees, giving sqrt(6) cis(315). Multiplying the Cartesian form by i*i = i^2 = -1 likewise gives a 180 degree rotation, so the second square root equals sqrt(3) (1 - i). Check: [ sqrt(3) (-1 + i) ]^2 = 3 (1 - 2i - 1) = -6i [ sqrt(3) ( 1 - i) ]^2 = 3 (1 - 2i - 1) = -6i I'm trying to work it out >to the answer my book gives, >which is: >(-[1 + sqrt(3)]/2 + [1 + sqrt(3)*i]/2), >(-[1 - sqrt(3)]/2 - [1 - sqrt(3)*i]/2) Where have I errored? === Subject: Re: quadratic with complex coefficients days. My association with the Department is that of an alumnus. >I have a quadratic equation with >complex coefficients: >x^2 + (1 - i)x + i = 0 And I've worked it out to: >[-1 + i +/- sqrt([1 - i]^2 - 4i) ] / 2 Now, the square root expression >can be simplified to: >sqrt(-6i) which, in trigonometric >form, is: sqrt(6^(1/2)*cis(45)) No, it is not; -6i is not equal to 6^{1/2}cis(45). What is the argument of -6i? It is not 45 degrees, and that's for sure. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: quadratic with complex coefficients >I have a quadratic equation with >complex coefficients: >x^2 + (1 - i)x + i = 0 And I've worked it out to: >[-1 + i +/- sqrt([1 - i]^2 - 4i) ] / 2 Now, the square root expression >can be simplified to: >sqrt(-6i) which, in trigonometric >form, is: sqrt(6^(1/2)*cis(45)) Take another look at the angle. You're doing sqrt(-6i), not sqrt(+6i). >or sqrt(6^(1/2)*cis(105)) >Which is the two roots of -6i >supplied as arguments to sqrt. I'm trying to work it out >to the answer my book gives, >which is: >(-[1 + sqrt(3)]/2 + [1 + sqrt(3)*i]/2), >(-[1 - sqrt(3)]/2 - [1 - sqrt(3)*i]/2) Where have I errored? === Subject: Re: quadratic with complex coefficients days. My association with the Department is that of an alumnus. >I have a quadratic equation with >complex coefficients: >x^2 + (1 - i)x + i =3D 0 >And I've worked it out to: >[-1 + i +/- sqrt([1 - i]^2 - 4i) ] / 2 >Now, the square root expression >can be simplified to: >sqrt(-6i) which, in trigonometric >form, is: sqrt(6^(1/2)*cis(45)) Take another look at the angle. You're doing sqrt(-6i), not >sqrt(+6i). Even if it were, he is writing sqrt once too many times. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Hundreds and Hundreds of Solution Manuals in PDF! Contact bwen...@hotmail.com for more! A Course in Algebraic Number Theory by Henry Cohen > A Course in Game Theory by Martin J. 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