mm-4569 === Subject: Re: JSH: Frustrating nonsense > here to end humanity by stopping its progress mathematically and > scientifically but now I think I'm just dealing with people who are > doing this for minor, meager selfish gain. >Please don't take offense, but have you ever seriously considered the >possibility that you might be mentally ill? >I'm asking because people who seriously think that alien conspiracies >exist usually are. > James has indeed considered that possibility. He has diagnosed > himself with Narcissistic Personality Disorder > (NPD):http://en.wikipedia.org/wiki/Narcissistic_personality_disorder > rossum I'm not a doctor, but this looks to me more like a psychotic disorder. > Anyway, if he acknowledges that he is ill, why doesn't he seek medical > help instead of making those posts? Rule #1) One with NPD will never understand he has NPD (i.e. can't unlearn NPD) it is a view of life that NPDer has that he/she is the center of the known universe - and the rest of us are subservient ghosts. JSH is a troll with some NPD, not that bad, yet. Usually severe cases end up living in small trailers in remote locations where they interact with few people and remain king/queen of their kingdom (or Queendom?) reality. I am not making that one up folks. === Subject: Re: JSH: Frustrating nonsense posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > So that the only thing that will work is to collapse the security > system of the Internet world by factoring an RSA number or a big > enough number that it's clear that it is reachable, as otherwise you > people will just keep going on like nothing is happening? Point of order: factoring an RSA number /will not/ collapse the security > system. Several RSA numbers have already been factored. Stuff that needs > to be secure has moved on to ECC or better already (and stuff that > doesn't as well). Besides, we already use a mind-boggling number of > cracked methods. Most downloads use MD5 for verification, despite > being now ridiculously easy to collide*. The basis for wireless > encryption? RC5, already cracked. A certificate I'm looking at uses > 1024-bit security, and that's starting to become a low end of > industrial-grade encryption. Your algorithm would have to be extremely > fast in order to make such security worthless. Just you wait. JSH has already gotten to 45 bits. * Okay, I admit: finding a second message to hash to the same value is > still difficult to construct. But the definition of cracking a method is > collision, and collision has been proven, so MD5 is nonetheless cracked. === Subject: Re: JSH: Frustrating nonsense > So that the only thing that will work is to collapse the security > system of the Internet world by factoring an RSA number or a big > enough number that it's clear that it is reachable, as otherwise you > people will just keep going on like nothing is happening? > Point of order: factoring an RSA number /will not/ collapse the security > system. Several RSA numbers have already been factored. Stuff that needs > to be secure has moved on to ECC or better already (and stuff that > doesn't as well). Besides, we already use a mind-boggling number of > cracked methods. Most downloads use MD5 for verification, despite > being now ridiculously easy to collide*. The basis for wireless > encryption? RC5, already cracked. A certificate I'm looking at uses > 1024-bit security, and that's starting to become a low end of > industrial-grade encryption. Your algorithm would have to be extremely > fast in order to make such security worthless. >Just you wait. JSH has already gotten to 45 bits. When was that? Bet that was just one 45 bit number, and not any 45 bit number. I bet for any 45 bit number, I can convert to decimal, look it up on the internet, FASTER than JSH could factor it. === Subject: Re: JSH: Frustrating nonsense <47cddcfe$2@news.x-privat.org> posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > So that the only thing that will work is to collapse the security > system of the Internet world by factoring an RSA number or a big > enough number that it's clear that it is reachable, as otherwise you > people will just keep going on like nothing is happening? > Point of order: factoring an RSA number /will not/ collapse the security > system. Several RSA numbers have already been factored. Stuff that needs > to be secure has moved on to ECC or better already (and stuff that > doesn't as well). Besides, we already use a mind-boggling number of > cracked methods. Most downloads use MD5 for verification, despite > being now ridiculously easy to collide*. The basis for wireless > encryption? RC5, already cracked. A certificate I'm looking at uses > 1024-bit security, and that's starting to become a low end of > industrial-grade encryption. Your algorithm would have to be extremely > fast in order to make such security worthless. >Just you wait. JSH has already gotten to 45 bits. When was that? æ I thought I saw mention of that in his Test Factorization thread. > Bet that was just one 45 bit number, and not any 45 bit > number. Sure. Just like his his previous attemp that couldn't factor EVERY six digit number with two 3-digit factors (where there's a comment in the source code to the effect of I can't factor it, so it must be prime. I bet for any 45 bit number, I can convert to decimal, look it up on the > internet, FASTER than JSH could factor it. === Subject: Re: z constraint and factoring posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > I realized later that the z constraint is crucial in showing with a > fairly easy result that my line of research pushes mathematical > knowledge at the heart of factoring beyond what was previously known. But I need to make some corrections to my original post. Below... > One tidbit result that has come out of the research into the concept I > call surrogate factoring has been a constraint on z, where z^2 = y^2 + nT which I consider something of a weak constraint though it is an > absolute one in a key way I'll explain is z = (1 + 2¬¢^2)k/(2¬¢) where k is just some non-zero integer, so it just boils down to z must > have 1 + 2¬¢^2 as a factor for some non-zero ¬¢ that is an integer, > which I consider kind of interesting. The constraint applies if k and ¬¢ exist for an odd prime p such that Turns out that z having 3 as a factor is a necessity for the constraint to be generally correct, as in always correct. Otherwise it is conditionally correct and only applies if an integer x exists such that x^2 = y^2 mod p and z = x + ¬¢k. Oddly enough, otherwise ¬¢ and k do not exist as rationals, but you CAN still factor as they can still give you z and y, which is a neat thing I noticed while puzzling over this area. Then you are in my object ring, and can factor T using non-rationals. Those doubting might try T = 23(29), with n=1, and p=17, and you'll find that alpha=1, while k = 6, and z = 9 mod 17, or explicitly 26. And no x and k explicitly can be found that are integers that will fulfill all the conditions, while you can factor with f 1 = ¬¢k mod 17 = 6 mod 17 = 23. > k^2 = (.a6'^2+1)^{-1}(nT) mod p and positive factors f 1 and f 2 of nT, where f 1*f 2 = nT and f 1 is > the smaller factor are greater than p or p - f 1 is greater than f 1. Remember that z = (f 1 + f 2)/2. So the limitation is on small values for nT, as, for instance, the > constraint will not work with nT = 15, because there are no odd primes > available. > That is correct. And since z can be forced to be divisible by 3, the secondary issue with the use of non-rationals is just pure math. But it is some fun math which may someday for real have practical application as it is a linking of non-rationals and rationals for a rational factorization. May connect to some truly super advanced science or something. Sigh, but you'll all probably be long dead before anyone figures that out, just like Newton and Archimedes and all the others before including Gauss died long before lasers, computers, and yes, nuclear bombs. The math comes way, way, way before the best that can result from it-- and the worst--but without the math you don't get the rest. James Harris === Subject: Re: z constraint and factoring posting-account=axlf9QgAAADH9qSnMXr4KE9F8O1Dk5OI CLR 1.1.4322),gzip(gfe),gzip(gfe) I didn't look at what Enrico had to say about it; empirically, I already can tell from a brief scansion of the ur-text, that it is paralinguistic math speak, what ever the viability of the following manipulations of some equations in ecstasy. of course, doug didn't bother with any math, either but, then, neither did the august journal, that you destroyed with your bare two-fingered typing! so, anyway, your typing prowess is such, you should be able to do one-fingered push-ups, after a short trial of spotting from your student(s). you & us are all victims of General Bourbaki, who will be shot at high noon for treason to la Republique quel-que-nombre de France: he was just doing le math -- mais, non! > James Harris- Hide quoted text when we learn, who you *really* are, it may not be so amuzing. === Subject: Re: List of solutions manual (thousands) 1.0.3705; Media Center PC 4.0; .NET CLR 2.0.50727; Seekmo 10.0.341.0),gzip(gfe),gzip(gfe) > hey i was looking for the solution manual for Fluid Mechanics fundemental and application Solutions 1st Ed by Cengel > and Cimbala + ebook i will pay for it > my email is saif uddin(at)hotmail.com > hi > i want the solution for æRadiative Heat Transfer (2nd Ed., Michael > Modest) > how much is it? > question: > what is the format for the solution, well printed, handwrite or e- > file? > does the solutions included all the problem ( even and odd )? > thank you > My List of Solutions Manual > > contact me to : æ newbergh...@yahoo.com > If your wanted solutions manual ins't on this list, also can ask me if > is available . These are some only. > This list (not links) is available from : >http://rapidshare.com/files/59002351/List of solutions manual.txt > > - Mechanics, Mechanical Engineering & Aerospace Engineering: > > Classical mechanics (2nd Ed., Goldstein) > Classical Mechanics (Douglas Gregory) + original Ebook > Advanced Dynamics (Greenwood) + original Ebook > Advanced Engineering Dynamics (2nd Ed., Jerry Ginsberg) + Ebook > Classical Dynamics (Jorge V. Jos?) + Ebook > Impact Mechanics (W.J. Stronge) > Introduction to Mechanical Engineering (Rizza) > Mechanical Engineering Principles (Bird & Ross) + original Ebook > Mechanics of Fluids (8th Ed., Massey) + original Ebook > Fluid Mechanics (5th Ed., White) + Ebook > Fluid Mechanics (6th Ed., White) > Viscous Fluid Flow (3rd Ed., White) + Ebook > Fundamentals of Thermal-Fluid Sciences (1st Ed., Cengel) + original > Ebook > Fundamentals of Thermal-Fluid Sciences (2nd Ed., Cengel) + original > Ebook > Fundamentals of Thermal-Fluid Sciences with Student Resource CD (3rd > Ed., Cengel & Turner) > Thermodynamics: An Engineering Approach (5th Ed., Cengel) + original > Ebook > Thermodynamics: An Engineering Approach (6th Ed., Cengel) + original > Ebook > Essentials of Fluid Mechanics: Fundamentals and Applications (1st Ed., > Cengel) + original Ebook > Fluid Mechanics (1st Ed., Cengel) + original Ebook > Heat Tranfer (2nd Ed., Cengel) + original Ebook > Heat and Mass Transfer: A Practical Approach (3rd. Ed., Cengel) + > original Ebook > Design and Simulation of Thermal Systems (Suryanarayana & Arici) > Introduction to Fluid Mechanics (6th Ed., Robert Fox, Alan McDonald & > Philip Pritchard) > Fluid Mechanics (5th Ed., Douglas) > Fluid Mechanics (3rd Ed., Kundu) > Fluid Mechanics with Engineering Applications (Finnemore) > Fundamentals of Fluid Mechanics, 4th Ed (Bruce R. Munson, Donald F. > Young, Theodore H. Okiishi) + original ebook > Fundamentals of Fluid Mechanics, 5th Ed (Bruce R. Munson, Donald F. > Young, Theodore H. Okiishi) > A Brief Introduction to Fluid Mechanics, 3rd Ed (Donald F. Young, > Bruce R. Munson, Theodore H. Okiishi) > A Brief Introduction to Fluid Mechanics, 4th Ed (Donald F. Young, > Bruce R. Munson, Theodore H. Okiishi, Wade W.) > Engineering Fluid Mechanics, 7th Ed (Clayton T. Crowe, Donald F. > Elger, John A. Roberson) > Engineering Fluid Mechanics, 8th Ed (Clayton T. Crowe, Donald F. > Elger, John A. Roberson) > Mechanics of Fluids (3rd Ed., Potter) > Mechanics of Fluids (4th Ed., Shames) > Extended Irreversible Thermodynamics (3rd Ed., D. Jou, J. Casas- > Vazquez & G. Lebon) > Thermodynamics: An Integrated Learning System (Schmidt, Ezekoye, > Howell & Baker) > Introduction to Thermal and Fluids Engineering (Kaminski & Jensen) > Heating, Ventilating and Air Conditioning Analysis and Design (6th > Ed., McQuiston) > An Introduction to Fluid Dynamics: Principles of Analysis and Design > (Middleman) > Introduction to Mass and Heat Transfer: Principles of Analysis and > Design (Middleman) > Heat Transfer (2nd Ed., Mills) > Convective Heat and Mass Transfer (4th Ed., Kays & Crawford) > Advanced Engineering Thermodynamics (3rd Ed., Bejan) > Convection Heat Transfer (2nd Ed., Bejan) > Convection Heat Transfer (3rd Ed., Bejan) > Thermal Design and Optimization (Bejan) > Shape and Structure, from Engineering to Nature (Bejan) > An Introduction to Combustion: Concepts and Applications (2nd Ed., > Turns) > Thermodynamics: Concepts and Applications (Stephen Turns) > Thermal-Fluid Sciences: An Integrated Approach (Stephen Turns) > Principles of Heat Transfer (Kaviany) > Heat Convection (Latif M. Jiji) + original Ebook > Heat Transfer (9th Ed., Holman) > Fundamentals of Momentum, Heat and Mass Transfer (4th Ed., Welty) > Momentum, Heat, and Mass Transfer Fundamentals (Kessler) + original > Ebook > Analytical Methods for Heat Transfer and Fluid Flow Problems (Bernhard > Weigand) > Heat Tranfer (Rao) > Heat Conduction (kakac) > Heat Exchanges (Kakac) > Convective Heat Transfer (kakac) > Heat Exchangers: Selection, Rating and Thermal Design (2nd Ed. Sadik > Kakac & Hongtan Liu) > Fundamentals of Engineering Thermodynamics, 5th Ed (Michael J. Moran, > Howard N. Shapiro) + original Ebook > Fundamentals of Engineering Thermodynamics, 6th Ed (Michael J. Moran, > Howard N. Shapiro) > Fundamentals of Heat and Mass Transfer (5th Ed., Incropera, DeWitt) > Fundamentals of Heat and Mass Transfer (6th Ed., Incropera, DeWitt) > Introduction to Heat Transfer (4th Ed., Incropera, DeWitt) > Introduction to Heat Transfer (5th Ed., Incropera, DeWitt) > Radiation Detection and Measurement (3rd Ed., Glenn Knoll) > Radiative Heat Transfer (2nd Ed., Michael Modest) > Engineering Heat Transfer (2nd Ed., Janna) > Engineering Thermodynamics: Work and Heat Transfer (4th Ed., G.F.C. > Rogers & Y.R. Mayhew) > Elements of Heat Transfer (Yildiz Bayazitoglu and M. Necati Ozisik) > Inverse Heat Transfer: Fundamentals and Applications (M.N. Ozisik & > Helcio R.B. Orlande) > Thermal Radiation Heat Transfer (4th Ed.,Robert Siegel & John R. > Howell) > Computational Heat Transfer (2nd Ed., Jaluria) > Principles of Combustion (2nd Ed., Kenneth Kuan-yun Kuo) > Incompressible Flow (3rd Ed., Panton) > Modern Compressible Flow: With Historical Perspective (3rd Ed., John > D. Anderson) > Non-Newtonian Flow : Fundamentals and Engineering Applications (R P > Chhabra & J F Richardson) + original Ebook > Computational Techniques for Fluid Dynamics (Srinivas, K., Fletcher, > C.A.J.) > Ebook > Theory of Applied Robotics: Kinematics, Dynamics and Control (Reza N. > Jazar) > Kinematic Chains and Machine Components Design (Dan B. Marghitu) + > original Ebook > Kinematics and Dynamics of Machinery (3rd Ed., Wilson & Sadler) > Kinematics, Dynamics, and Design of Machinery (2nd Ed., Waldron & > Kinzel) > Mechanism Design: Analysis and Synthesis-Volume 1 (4th Ed., Erdman & > Sandor) > Machines and Mechanisms: Applied Kinematic Analysis (3rd Ed., > Myszka) > Mechanical Design: A Components Approach (Peter Childs) > Mechanical Design of Machine Elements and Machines: A Failure > Prevention Perspective (Collins) > Fundamentals of Machine Component Design (3rd Ed., Juvinall) > Fundamentals of Machine Component Design (4th Ed., Juvinall) > Design of Machine Elements (8th Ed., Spotts) > Machine Design (Wentzell) > Solutions Manual to the text : Problems on the Design of Machine > Elements (Faires) > Machine Elements in Mechanical Design (4th Ed., Mott) > Mechanical Design: An Integrated Approach (1st Ed., Ugural) > Design of Machinery (3rd Ed., Norton) > Design of Machinery (4th Ed., Norton) > Machine Design (2nd Ed., Norton) > Machine Design : An Integrated Approach (3rd Ed., Norton) > Mechanical Engineering Design (6th Ed., Shigley) > Mechanical Engineering Design (7th Ed., Shigley) > Shigley's Mechanical Engineering Design (8th Ed., Budynas) > Fundamentals of Machine Elements (1st Ed., Hamrock) > Fundamentals of Machine Elements (2nd Ed., Hamrock) > Mechanics of Materials: A Modern Integration of Mechanics and > Materials in Structural Design (Christopher Jenkins & Sanjeev Khanna) > Mechanics of Materials (3th Ed., Beer) > Mechanics of Materials (5th Ed., Gere) > Mechanics of Materials (6th Ed., Gere) > Mechanics of Materials (Ugural) > Simplified Mechanics and Strength of Materials (6th Ed., James > Ambrose) > Engineering Mechanics, Statics, 2nd Ed (William F. Riley, Leroy D. > Sturges) > Engineering Mechanics, Dynamics, 2nd Ed (William F. Riley, Leroy D. > Sturges) > Engineering Mechanics - Statics, 5th Ed (J. L. Meriam, L. G. Kraige) + > Ebook > Engineering Mechanics - Statics, 6th Ed (J. L. Meriam, L. G. Kraige) > Engineering Mechanics - Dynamics, 5th Ed (J. L. Meriam, L. G. Kraige) > Engineering Mechanics - Dynamics, 6th Ed (J. L. Meriam, L. G. Kraige) > Vector Mechanics for Engineers: Statics (7th Ed., Ferdinand P. Beer) > Vector Mechanics for Engineers: Statics (8th Ed., Ferdinand P. Beer) > Vector Mechanics for Engineers: Dynamics (7th Ed., Ferdinand P. Beer) > Vector Mechanics for Engineers: Dynamics (8th Ed., Ferdinand P. Beer) > Statics: Analysis and Design of Systems in > ... > read more Hi there > I am Extremely interested in the solution manual for > Statics and Strengths of Materials Sixth Edition by H.W Morrow and > Robert P.Kokernak > Is there any possibility I might be able to receive that from you? > Francisco if you have solution manual for Kinematics, Dynamics, and Design of Machinery (2nd Ed., Waldron & Kinzel) please i want o buuy respond me back as soon as you got this massage by e mail ebbisa23@gmail.com with price of it === Subject: Help! posting-account=-70AUwoAAABCySSUPRLY5HdbR59agHhG AppleWebKit/523.15.1 (KHTML, like Gecko) Version/3.0.4 Safari/523.15,gzip(gfe),gzip(gfe) Would it be possible to get the solutions manuel for applied statics and strength of materials 4th Ed. by leonard spiegal & george limbrunner. my email is redant737@mac.com === Subject: The z constraint, redux posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) While writing a long post about the current state of the art on surrogate factoring, I found myself talking about an equation that constrains a variable I call z, where z^2 = y^2 + nT and T is a target to be factored, while n is a free integer variable. Seemed like a minor thing to me at the time but soon I found myself consumed with properly defining what I called the z constraint, and just figured out the full theory today, as the motivation has been to have a complete surrogate factoring theory. That constraint is that for any factorizations of T into two positive integers f 1 and f 2, where there exists an odd prime p less than both, or if f 1 is the smaller factor p - f 1 is less than f 1, where also exists an .87 such that k^2 = (1 + .87^2)^{-1}(nT) mod p then z = (1 + 2.87^2)k/(2.87) in general when nT mod 3 = 2, but conditionally if not, as then z is coprime to 3, and the z constraint applies only if also x exists as an integer where x^2 = y^2 mod p and z = x + .87k. The requirement that p be less than the smaller factor f 1, or that p - f 1 be less than f 1 comes from solutions for the factors modulo p, as if you have positive integer factors f 1 and f 2 where f 1*f 2 = nT: f 1 = .87k mod p and f 2 = .87^{-1}(1 + .87^2)k mod p. (Oh yeah, that's cool in and of itself. If you play with these equations at all you'll find that with z divisible by 3, you can always find an odd prime p for which those hold.) As an example of the conditional case when z is not divisible by 3, consider, n=1, T = 23(29)=667, p=17, so I can directly calculate that z = (23 + 29)/2 = 26. Notice that k^2 = (1 + .87^2)^{-1}(667) mod 17 = (1 + .87^2)^{-1}(4) mod 17 so .87 = 1 works to give k^2 = 2 mod 17, so k = 6 mod 17 is a solution. And notice then that f 1 = .87k mod p = 6 mod 17, corresponding to the solution 23. And f 2 = .87^{-1}(1 + .87^2)k mod p = 12 mod 17, giving the residue modulo 17 of 29. But intriguingly enough z = (1 + 2.87^2)k/(2.87) is not valid for any integers k and .87. And if you dig deeply into the surrogate factoring equations you find that x cannot exist as an integer where x^2 = y^2 mod 17 and z = x + .87k. Note that y is given by (29-23)/2, so y=3, for this example. As x is part of the full derivation it follows that x is not rational, and further as no integers k and .87 can give z, it follows that both of them are non-rational as well. because k^2 = (1 + .87^2)^{-1}(nT) mod p so .87 = 1 mod 17, is what is actually happening, but it's not rational. So you factored with the help of non-rationals, as well as the helper prime 17. For more mundane examples and to see the z constraint in action, either use T = 2 mod 3, or if T = 1 mod 3, let n=5 or n=11 to force nT mod 3 = 2, or any other n that you choose though I do stay away from n=2, as I stay away from even nT. James Harris === Subject: Re: JSH: The z constraint, repukachoo >While writing a long post about the current state of the art on >surrogate factoring, I found myself talking about an equation that >constrains a variable I call z, where >z^2 = y^2 + nT Let T = RSA-100 = 1522605027922533360535618378132637429718068114961380688657908494580122963258 952897654000350692006139 >and T is a target to be factored, while n is a free integer variable. So, how you Picka n ? And what does that have to do with y ? Do you just Picka y too ? If you picka n and picka y, then you gotta z for sure. But how you gonna factor T? even since you picka n, picka y and you gotta z ?? What next? >Seemed like a minor thing to me at the time but soon I found myself >consumed with properly defining what I called the z constraint, and >just figured out the full theory today, as the motivation has been to >have a complete surrogate factoring theory. sounda likea bsa to me. >That constraint is that for any factorizations of T into two positive >integers f_1 and f_2, where there exists an odd prime p less than >both, or if f_1 is the smaller factor p - f_1 is less than f_1, where >also exists an .87 such that >k^2 = (1 + .87^2)^{-1}(nT) mod p Picka? Lotsa Picka! Picka k Picka .87 Picka p >then >z = (1 + 2.87^2)k/(2.87) >in general when nT mod 3 = 2, but conditionally if not, as then z is >coprime to 3, and the z constraint applies only if also x exists as an >integer where >x^2 = y^2 mod p Picka? Picka! Picka x >and z = x + .87k. Nope you can't do that, cause then z = x + .87k = (1 + 2.87^2)k/(2.87) = +-sqrt ( y^2 + nT ) and now you gotta real mess. You multiply an RSA with some integer, then take the + or - square root of that + an additional term (y^2) you just clusterfcked your RSA number. How big is x ? and k? and .87 ? At least 25 to 50 digits EACH. How you gonna Picka that ? Puka-CHOO!! Your problem is you are looking at really very very tiny small itty bitty numbers (and it dosen't work there either). Hint: http://en.wikipedia.org/wiki/RSA-100 >James Harris === Subject: Re: Engineering Mechanics: Statics, 6th edition. Meriam, Kraige posting-account=EeR93woAAAA2ohvSpF_LCH8ZFE2KjfY- > I am in need as well. If anyone has it please send a copy my way. I got my hands on a copy. If anyone still needs it e-mail me @ gumert@epsilongraphics.com === Subject: Re: JSH: Beyond the cutting edge > My research takes you beyond what was previously known now in > factoring as it did in other mathematical areas. There simply is no way known outside of what I call surrogate > factoring to get a general constraint on z, when z^2 = y^2 mod N Picka? Picka! Picka y > to use the more conventional usage, where N is the target to be > factored, where you also get the factors mod p, where p is some picked > odd prime. Picka? Picka! Picka p some odd prime (WHICH ONE ?) > The knowledge simply transcends what mathematicians previously thought > possible. When did you aquire the thoughts and knowledge of all the mathematicians that ever existed ? This afternoon ? > Now then, you can all pretend to not see and go about your lives, pay > your bills and act as if the end is not in sight to this silly little > game of deception that many of you have played on the world, but then > you cannot believe in anything nor anybody. no pretention here, *We challange you to factor the RSA-100* Else take your little piddly usless equation and stuff it. > And I think many of you do not. But you do not know what any of us are really thinking now do you ? > But then, what do you do when they come for you, and you're sitting in > a little box trying desperately to figure a way out, and you do not > believe in anything nor anybody? You work in a cube too? How big ? Bet it is a lot smaller than mine, like micro smaller. You walked into this trap. The mathematics laid it for you. Sorry to burst your ego bubble like this, JSH. (Try not to put it on your resume that you failed again.) And now She will finish you. Wrong sex, it is Mr Math to you. > James Harris - The Flying Priest of Monkey Mathematics, keeping the serfs > working and parasites outta here for 12 years === Subject: JSH: White collar welfare posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Religion makes stuff up all the time and will even send people to die for it. People making up mathematics to keep warm and full stomachs is not beyond imagining. The surprise though has been with you. I can blow modern mathematicians out of the water across multiple subject lines, from prime numbers to logic, and now to factoring but you dutifully keep following along and I think it has to be because you wish to live as they do--getting money for nothing. I call it white collar welfare. And my job is ending it. Some of you may work hard just in time to have no degree at the end of it. James Harris === Subject: Re: JSH: White collar welfare > Religion makes stuff up all the time and will even send people to die > for it. Religion is Faith, you have Faith in your Math, therefore your Religion is Your Math. Stupid isn't it ? > People making up mathematics to keep warm and full stomachs is not > beyond imagining. It is what you strive for, a fool's goal. > The surprise though has been with you. Wrong again, how can you know what suprises us ? You are mind reading, which is a thinking distortion, which causes severe communication problems. This can be unlearned, just ask direct questions. > I can blow modern mathematicians out of the water across multiple > subject lines, from prime numbers to logic, and now to factoring but > you dutifully keep following along and I think it has to be because > you wish to live as they do--getting money for nothing. Well, you can't because *you* do not know Math. And you will never get any money, even though *you have nothing*. > I call it white collar welfare. Is that what you want to be on ? Join the Democrappic party. > And my job is ending it. Your work is self-blocking, your job is self-ending. Nobody else has to do anything at all. > Some of you may work hard just in time to have no degree at the end of > it. Still angry you flunked out of Algebra ? let it go man. It has been more than 12 years, and you are not interested in it, as you would have completed a course by now. > James Harris === Subject: Re: JSH: White collar welfare > Religion makes stuff up all the time and will even send people to die > for it. Religion is Faith, you have Faith in your Math, therefore your Religion > is Your Math. > Stupid isn't it ? > People making up mathematics to keep warm and full stomachs is not > beyond imagining. It is what you strive for, a fool's goal. > The surprise though has been with you. Wrong again, how can you know what suprises us ? You are mind reading, > which is a thinking distortion, which causes severe communication > problems. This can be unlearned, just ask direct questions. > I can blow modern mathematicians out of the water across multiple > subject lines, from prime numbers to logic, and now to factoring but > you dutifully keep following along and I think it has to be because > you wish to live as they do--getting money for nothing. Well, you can't because *you* do not know Math. > And you will never get any money, even though *you have nothing*. > I call it white collar welfare. Is that what you want to be on ? Join the Democrappic party. > And my job is ending it. Your work is self-blocking, your job is self-ending. Nobody else has to > do anything at all. > Some of you may work hard just in time to have no degree at the end of > it. Still angry you flunked out of Algebra ? let it go man. It has been more > than 12 years, and you are not interested in it, as you would have > completed a course by now. > James Harris > The parasites will assemble and finish JSH off, with fava beans and a nice chianti. http://www.youtube.com/watch?v=iVlkZVAw8Gc === Subject: Re: JSH: White collar welfare Religion makes stuff up all the time and will even send people to die > for it. People making up mathematics to keep warm and full stomachs is not > beyond imagining. Not beyond imagining, that's true because people can imagine all sorts of strange things. But note that religious believers are gullible--why else would they believe in religion?--mathematicians are smarter. > The surprise though has been with you. I can blow modern mathematicians out of the water across multiple > subject lines, from prime numbers to logic, and now to factoring but > you dutifully keep following along and I think it has to be because > you wish to live as they do--getting money for nothing. I call it white collar welfare. And my job is ending it. Get on and do it then. > Some of you may work hard just in time to have no degree at the end of > it. James Harris -- === Subject: Re: JSH: White collar welfare Just curious: Do you know of anyone who doesn't laugh at these posts about how you're going to get us all fired? >Religion makes stuff up all the time and will even send people to die >for it. People making up mathematics to keep warm and full stomachs is not >beyond imagining. The surprise though has been with you. I can blow modern mathematicians out of the water across multiple >subject lines, from prime numbers to logic, and now to factoring but >you dutifully keep following along and I think it has to be because >you wish to live as they do--getting money for nothing. I call it white collar welfare. And my job is ending it. Some of you may work hard just in time to have no degree at the end of >it. >James Harris David C. Ullrich === Subject: Statics Solutions posting-account=dQLiRgoAAACVB9fB8D1if1z1n0jNhCmJ Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Hey could you send me the solution to Engineering Mechanics Statics === Subject: Re: Fluid Mechanics by Cengel 1st Ed. solution manual + Ebook posting-account=j9qGKgoAAADKxJauf-vqaHqZDFJTUozc Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > I have the solution manual + ebook for Fluid Mechanics by Cengel 1st > Ed. let me know if anybody wants it. my email is binbin...@hotmail.com ------------------ Hi I really want the solution manual + ebook of Fluid Mechanics by Cengel my email is khosravi.s.m@gmail.com plz email me. === Subject: Re: compound interest; taking the limit [...] Can you give me some examples of where the number e pops > up? http://en.wikipedia.org/wiki/E_%28mathematical_constant%29 http://mathworld.wolfram.com/e.html and any text on analysis (I like Burkill's First Course which has the advantage of being less than 200 pages long). -- === Subject: Re: JSH: Thinking about equals > Now I finally figured out that modern math society is a society of > fiction writers partly by thinking about equals as in the = sign > as it is supposed to mean that what is on the left is equal to what is > on the right. So you can practically define any full mathematical statement as one > where if you put in numbers and simplify you will end up with an > identity: > It's been painful to watch the deterioration of your mind over the past > few years; but especially so recently. I look forward to the time when James's mind has deteriorated so much that he can no longer use a computer. -- === Subject: Re: JSH: Thinking about equals posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > This post must be saved in the Google archives. It may well be the most > astounding, over the top post JSH has ever written, and after a decade > of these harangues, that's saying a lot. > I needed to explain lying on the scale that I've seen where not only did publication in a peer reviewed mathematical journal not matter, along with argument explaining in detail the exact argument itself, and my ability to explain exactly why it's correct even now, but the damn journal died, and that didn't matter either. And Cameron University deleted all reference to the journal from its websites. And that didn't matter either. And I realized I had a big problem and religion is similar in that you get a lot of wacky behavior around religion including people willing to blow themselves up for prizes in heaven. And I connected the dots--modern science took away much of the power of religion by letting people do things like actually fly versus imagining it, but priests had a nice cushy job where people gave them food and money just to tell them stuff. Think about it, did priests actually WORK for a living? So would they just quit when science outdid religion in most people's eyes? I suggest to you no, they just branched out and discovered something odd---they could actually get into science and into math in certain areas. In science they went into cosmology which was kind of a variant on mythological tales anyway where the great thing was no one could go out there and just measure things like how far quasars are from us. Look-up Halton Arp to see what those people did to a notable in their own field who questioned them. Priests. In mathematics they stayed mostly away from applied math and pushed pure math until they had critical mass and could just make things up. So like, Andrew Wiles's supposed proof of Fermat's Last Theorem is so damn dumb I can collapse it by noting it fails by a logical error: Cum hoc, ergo propter hoc Which is kind of ironic. I've had arguments explaining the error on newsgroups and posters ultimately appeal to the experts and never answer the direct refutation of his approach. Priests did not want to give up on the free food. That's all. So they came up with pure math and modern cosmology, oh, and they've been working at physics too with string theory. Some people---always looking for a handout. James Harris === Subject: Re: JSH: Thinking about equals > This post must be saved in the Google archives. It may well be the most > astounding, over the top post JSH has ever written, and after a decade > of these harangues, that's saying a lot. I needed to explain lying on the scale that I've seen where not only > did publication in a peer reviewed mathematical journal not matter, > along with argument explaining in detail the exact argument itself, > and my ability to explain exactly why it's correct even now, but the > damn journal died, and that didn't matter either. And Cameron University deleted all reference to the journal from its > websites. > And that didn't matter either. And I realized I had a big problem and religion is similar in that you > get a lot of wacky behavior around religion including people willing > to blow themselves up for prizes in heaven. And I connected the dots--modern science took away much of the power > of religion by letting people do things like actually fly versus > imagining it, but priests had a nice cushy job where people gave them > food and money just to tell them stuff. Think about it, did priests actually WORK for a living? So would they just quit when science outdid religion in most people's > eyes? I suggest to you no, they just branched out and discovered something > odd---they could actually get into science and into math in certain > areas. In science they went into cosmology which was kind of a variant on > mythological tales anyway where the great thing was no one could go > out there and just measure things like how far quasars are from us. Look-up Halton Arp to see what those people did to a notable in their > own field who questioned them. Priests. In mathematics they stayed mostly away from applied math and pushed > pure math until they had critical mass and could just make things > up. So like, Andrew Wiles's supposed proof of Fermat's Last Theorem is so > damn dumb I can collapse it by noting it fails by a logical error: Cum hoc, ergo propter hoc Which is kind of ironic. I've had arguments explaining the error on > newsgroups and posters ultimately appeal to the experts and never > answer the direct refutation of his approach. Priests did not want to give up on the free food. That's all. So they came up with pure math and modern cosmology, oh, and they've > been working at physics too with string theory. Some people---always looking for a handout. > James Harris So, Priests caused you to fail Algebra? Because they are hungry ? === Subject: Re: JSH: Thinking about equals posting-account=mgs1FwoAAABD3j5T_RLZ06yrgt2dghDu Gecko/20050915,gzip(gfe),gzip(gfe) > You put in a number for x, so if you have x=0, then it's valid, as you > get 0 < 1 but if you try x=2, then it's false. Notice then that with any valid mathematical statement if you plug > numbers in then you reduce to a truth, like 0<1. Actually though I wouldn't categorize x<1 as a mathematical statement > under a full system, but as a logical statement about mathematical > objects. I would prefer to define mathematical statements around the equals, > and consider other varieties to be logical statements about > mathematical objects. A bit of rigor in the definitions that might escape you, but you can > start by doing a Google search on my definition of mathematical proof, > by doing a Google search on definition of mathematical proof to get > a sense of how I connect truth to mathematics. James, could you please put numbers into the following mathematical statement and reduce it to an identity? There is a problem with the ring of algebraic integers. -- Academics have truly made me a very angry person -- James Harris === Subject: Re: guidance..Show using Riemann sums d/dr[A]=2pi*r (A=area circle) It was given verbally. he said something like this. It is interesting that when you take the derivative of a the area of a circle with respect to the radius you get the circumference and similarly you get the same thing when you take the derivative of the volume of a sphere you get the surface area of a sphere. > he said show me using Riemann sums why this is true. Consider the circle case first. Write down the derivative. Get from that a claim about an integral. You need to prove, using Riemann sums, that that claim is true. Remember that Riemann sums refer to definite integrals, so it's a definite integral that you need. Do _not_ change to Cartesian co-ordinates. You will be integrating w.r.t. a variable called r, that it isn't called x is irrelevant. -- === Subject: Re: Complete Solutions Manuals For College and Graduate Textbooks posting-account=ZvklQwoAAACQlg6-WQyUGYEJOrpbQsUB .NET CLR 2.0.50727; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Updated list per December 18th, 2007 > I have the current edition of these comprehensive solutions manual for > the following textbooks in electronic format (PDF/Word). The solutions > manual are comprehensive with answers to both even & odd problems in > the text. The price is US$35 for each. > The methods of payment is through PAYPAL (It is easy, safe, and you > can use debit or credit card to pay even if you don't have an account) > Email me at sbooks4sale[at]hotmail[dot]com if you are interested. If > you could not find the book you are looking for, please let me know, I > P.S. I halso have some Test Banks available. > -------------------------- > International Financial Management, Abridged Edition - Jeff Madura > (8th ed) (ISBN 0324365632) > International Management: Managing Across Borders and Cultures - Helen > Deresky (6th ed) (ISBN 0136143261) > International Money and Finance - Michael Melvin (7th ed) (ISBN > 0201770288) > Introduction to Abstract Algebra - Olympia Nicodemi (1st ed) (ISBN > 0131019635) > Introduction to Business Law - Jeffrey F. Beatty (2nd ed) (ISBN > 0324311427) > Introduction to Business Statistics - Ronald M. Weiers (6th ed) (ISBN > 0324381433) > Introduction to Chemical Principles - Stephen Stoker (9th ed) (ISBN > 0132379945) > Introduction to Computing Systems - Sanjay J. Patel, Yale Patt (2nd > ed) (ISBN 0072467509) > Introduction to Corporate Finance - William L. Megginson (1st ed) > (ISBN 0324379862) > Introduction to Cryptography with Coding Theory - Wade Trappe (2nd ed) > (ISBN 0131862391) > Introduction to Derivatives and Risk Management - Don M. Chance (7th > ed) (ISBN 0324321392) > Introduction to Econometrics - James H. Stock (2nd ed) (ISBN > 0321278879) > Introduction to Econometrics Brief Edition - James H. Stock (1st ed) > (ISBN 0321432517) > Introduction to Economic Reasoning - William D. Rohlf (7th ed) (ISBN > 0321416112) > Introduction to Environmental Engineering and Science - Gilbert M. > Masters (3rd ed) (ISBN 0131481932) > Introduction to Financial Accounting - Charles Horngren (9th ed) (ISBN > 0131479725) > Introduction to Fourier Optics - Joseph Goodman (3rd ed) (ISBN > 0974707724) > Introduction to Graph Theory - Douglas West (2nd ed) (ISBN 0130144002) > Introduction to Law - Joanne Hames (3rd ed) (ISBN 0131183818) > Introduction to Linear Algebra - Lee Johnson, Dean Riess, Jimmy Arnold > (5th ed) (ISBN 0201658593) > Introduction to Linear Programming - Leonid Vaserstein (1st ed) (ISBN > 0130359173) > Introduction to Management Accounting - Charles T. Horngren (14th ed) > (ISBN 0136129218) > Introduction to Management Accounting, Chap. 1-17: International > Edition - Charles Horngren (13th ed) (ISBN 0131273078) > Introduction to Management Science and Student - Bernard Taylor (8th > ed) (ISBN 0131050524) > Introduction to Management Science and Student - Bernard Taylor (9th > ed) (ISBN 0131888099) > Introduction to Mathematical Statistics and Its Applications - Richard > J. Larsen (4th ed) (ISBN 0131867938) > Introduction to Optics - Frank Pedrotti et al. (3rd ed) (ISBN > 0131499335) > Introduction to Quantum Mechanics - David Griffiths (2nd ed) (ISBN > 0131118927) > Introduction to Risk Management and Insurance - Mark Dorfman (8th ed) > (ISBN 0131449583) > Introduction to Technical Mathematics - Allyn J. Washington, Mario > Triola (5th ed) (ISBN 0321374177) > Introduction to Telecommunications - Martha Rosengrant (2nd ed) (ISBN > 0131126156) > Introduction to the Design and Analysis of Algorithms - Anany Levitin > (1st ed) (ISBN 0201743957) > Introduction to the Design and Analysis of Algorithms - Anany Levitin > (2nd ed) (ISBN 0321358287) > Introduction to Vacuum Technology - David M. Hata (1st ed) (ISBN > 0130450189) > Introductory Algebra - Marvin L. Bittinger (10th ed) (ISBN 0321269470) > Introductory Chemistry - Steve Russo, Michael Silver, Mike Silver (2nd > ed) (ISBN 032104634X) > Introductory Circuit Analysis - Robert Boylestad (11th ed) (ISBN > 0131730444) > Introductory Linear Algebra: An Applied First Course - Bernard Kolman > (8th ed) (ISBN 0131437402) > Introductory Mathematical Analysis - Ernest F Haeussler (12th ed) > (ISBN 0132404222) > Introductory Statistics - Neil A. Weiss (8th ed) (ISBN 0321393619) > Investments - Frank K. Reilly (7th ed) (ISBN 0324288999) > Java Software Solutions: Foundations of Program Design - John Lewis > (5th ed) (ISBN 0321409493) > John E. Freund's Mathematical Statistics with Applications - Irwin > Miller (7th ed) (ISBN 0131427067) > Labor Relations - Arthur A Sloane (12th ed) (ISBN 013196223X) > Labor Relations and Collective Bargaining: Cases, Practice, and Law - > Michael R. Carrell (8th ed) (ISBN 0131868721) > Law for Business - John D. Ashcroft (16th ed) (ISBN 0324381573) > Linear Algebra and Its Applications - David C. Lay (3rd ed) (ISBN > 0321287134) > Linear Algebra for Engineers and Scientists Using Matlab - Kenneth > Hardy (1st ed) (ISBN 0139067280) > Linear Algebra with Applications - Otto Bretscher (3rd ed) (ISBN > 0131453343) > Linear Algebra with Applications - Steven Leon (7th ed) (ISBN > 0131857851) > Logic and Computer Design Fundamentals - M. Morris Mano (4th ed) (ISBN > 013198926X) > Machine Design: An Integrated Approach - Robert L. Norton (3rd ed) > (ISBN 0131481908) > Machines and Mechanisms: Applied Kinematic Analysis - David H. Myszka > (3rd ed) (ISBN 0131837761) > Macroeconomics - Michael Parkin (7th ed) (ISBN 032124608X) > Macroeconomics - Michael Parkin (8th ed) (ISBN 0321416570) > Macroeconomics - Richard Froyen (8th ed) (ISBN 0131435825) > Macroeconomics - Robert Gordon (10th ed) (ISBN 0321278801) > Macroeconomics: Principles and Policy - William J. Baumol (10th ed) > (ISBN 0324537034) > Macroeconomics: Principles and Tools - Arthur O'Sullivan, Steven > Sheffrin (4th ed) (ISBN 0131536184) > Macroeconomics: Principles, Applications, and Tools - æArthur > O'Sullivan (5th ed) (ISBN 013232928X) > Management - Stephen P Robbins (9th ed) (ISBN 0132257734) > Managerial Accounting - Carl Warren (9th ed) (ISBN 0324381913) > Managerial Accounting - Linda Bamber (1st ed) (ISBN 0132284634) > Managerial Accounting: An Introduction to Concepts, Methods and Uses - > Michael W. Maher (10th ed) (ISBN 0324639767) > Managerial Economics: A Problem Solving Approach - Luke M. Froeb (1st > ed) (ISBN 0324359810) > Managerial Economics: Applications, Strategies, and Tactics - James R. > McGuigan (11th ed) (ISBN 0324421605) > Managerial Statistics A Case-Based Approach - Peter Klibanoff (1st ed) > (ISBN 0324226454) > Managing Human Resources - Luis Gomez-Mejia (5th ed) (ISBN 013187067X) > Managing in a Global Economy: Demystifying International > Macroeconomics (1st ed) (ISBN 0324395507) > Market Regulation - Roger Sherman (1st ed) (ISBN 0321322320) > Materials Science and Engineering: An Introduction - William Callister > (6th ed) (ISBN 0471135763) > Mathematical Ideas - Charles D. Miller (11th ed) (ISBN 0321361466) > Mathematical Methods for Economics - Michael Klein (2nd ed) (ISBN > 0201726262) > Mathematics for Business - Stanley A. Salzman (8th ed) (ISBN > 0321357434) > Mathematics for Physicists - Susan Lea (1st ed) (ISBN 0534379974) > Mathematics of Interest Rates and Finance - Gary Guthrie (1st ed) > (ISBN 0130461822) > Medical Imaging Signals and Systems - Jerry L. Prince (1st ed) (ISBN > 0130653535) > Microbiology: An Introduction - Gerard J. Tortora (9th ed) (ISBN > 0805347909) > Microeconomics - Jeffrey Perloff (4th ed) (ISBN 0321414527) > Microeconomics - Michael Parkin (7th ed) (ISBN 0321454944) > Microeconomics - Michael Parkin (8th ed) (ISBN 0321416600) > Microeconomics - Robert Pindyck, Daniel Rubinfeld (6th ed) (ISBN > 0130084611) > Microeconomics: Principles and Tools - Arthur O'Sullivan, Steven > Sheffrin (4th ed) (ISBN 0131536060) > Microeconomics: Principles, Applications, and Tools - Arthur > O'Sullivan (5th ed) (ISBN 0131572830) > Microeconomics: Theory and Applications with Calculus - Jeffrey M. > Perloff (1st ed) (ISBN 0321277945) > Microwave Engineering - David Pozar (3rd ed) (ISBN 0471448788) > Modern Control Systems - Richard C Dorf (11th ed) (ISBN 0132270285) > Modern Electronic Communication - Jeff Beasley (9th ed) (ISBN > 0132251132) > Modern Industrial Organization - Dennis Carlton, Jeffrey Perloff æ(4th > ed) (ISBN 0321180232) > Modern Labor Economics: Theory and Public Policy - Ronald Ehrenberg, > Robert Smith (9th ed) (ISBN 0321305035) > Modern Management - Samuel C. Certo (10th ed) (ISBN 0131494708) > Modern Physics - Raymond Serway (3rd ed) (ISBN 0534493394) > Modern Wireless Communications - Simon Haykin (1st ed) (ISBN > 0130224723) > Money, Banking and Financial Markets - Roger LeRoy Miller (3rd ed) > Money, the Financial System, and the Economy - R. Glenn Hubbard (6th > ed) (ISBN 0321426703) > Multinational Business Finance - David K. Eiteman (11th ed) (ISBN > 0321357965) > Multinational Management - John B. Cullen (4th ed) (ISBN 032442177X) > Numerical Analysis - Timothy Sauer (1st ed) (ISBN 0321268989) > Numerical Methods for Engineers - Bilal Ayyub, Richard McCuen (1st ed) > (ISBN 0133373614) > Numerical Methods Using Matlab - John Mathews (4th ed) (ISBN > 0130652482) > Operating Systems Principles - æLubomir F. Bic (1st ed) (ISBN > 0130266116) > Operating Systems: Internals and Design Principles - William Stallings > (5th ed) (ISBN 0131479547) > Operations Management - Jay Heizer (8th ed) (ISBN 0131554441) > Operations Management - Nigel Slack (5th ed) (ISBN 0273708473) > Operations Management: Process and Value Chains - Lee J. Krajewski > (8th ed) (ISBN 0131697390) > Organic Chemistry - Paula Bruice (Test Bank only) (5th ed) (ISBN > 0131963163) hey there i was wondering if you have the Complete Solutions Manuals > For College book ( LAW FOR BUSINESS 9th ed) I need Operations Management: Process and Value Chains - Lee J. > Krajewski> (8th ed) (ISBN 0131697390). How and when can I get it? I am interested in getting the Intro to Management Science 8th edition by Bernard Taylor solutions manual and also test bank. Do you still have it and how can I pay for it. Do you have a website? === Subject: Re: Complete Solutions Manuals For College and Graduate Textbooks > > I need Operations Management: Process and Value Chains - Lee J. > Krajewski> (8th ed) (ISBN 0131697390). How and when can I get it? > I am interested in getting the Intro to Management Science 8th edition > by Bernard Taylor solutions manual and also test bank. Do you still > have it and how can I pay for it. Do you have a website? How good of all three of you to ignore the op's Email me at sbooks4sale[at]hotmail[dot]com if you are interested. And how very good of all three of you not to snip the post you're replying to. -- === Subject: Re: mathematical statistics with applications solutions Hi i'm looking for solutions from mathematical statistics with > applications (6th ed) > Pls can someone help me? The following is quoted from 'Notes on Exercises' in Joseph H Silvermans _The Arithmetic of Elliptic Curves_: Except for an occasional computational problem, we have not included solutions (nor even hints). Indeed, since it is hoped that this book will lead the student on into the realm of active mathematics, the benefits of working without aid clearly outweigh any advantage that might be gained by having solutions readily available. -- === Subject: Re: mathematical statistics with applications solutions posting-account=usqYmgoAAAD13pQjMhCPQWv5tdmCU11q .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) On 5 mrt, 15:13, Frederick Williams applications (6th ed) > Pls can someone help me? The following is quoted from 'Notes on Exercises' in Joseph H Silvermans > The Arithmetic of Elliptic Curves : Except for an occasional computational problem, we have not included > solutions (nor even hints). æIndeed, since it is hoped that this book > will lead the student on into the realm of active mathematics, the > benefits of working without aid clearly outweigh any advantage that > might be gained by having solutions readily available. -- Strange. There are solution books available. But the shops don't sell the book any more. Only US-shops have the book but there aren't shops who deliver also to European country's. === Subject: What is Euclidean 3-space? In Barrett O'Neil's _Elementary Differential Geometry_, 1966, we read on page 3: Euclidean 3-space E^3 is the set of all ordered triples of real numbers. Why is such a set called 'Euclidean'? Surely it only deserves the name Euclidean when the distance between points x = (x_1, x_2, x_3) and y = (y_1, y_2, y_3) is introduced and given by d(x, y) = sqrt((x_1 - y_1)^2 + (x_2 - y_2)^2 + (x_3 - y_3)^2)? Only on page 43 does O'Neil define d in a way that is equivalent to the above identity. It seems to me that Euclidean 3-space should be defined as (R^3, d) and that E^3 should be used as a shorter name for that if one wishes. -- === Subject: manual solution for cost accounting managerial emphasis for foster posting-account=V64brQoAAAD9TZtTYXZeXBwdngcSFnM4 2.0.50727),gzip(gfe),gzip(gfe) pls send me the manual solution for cost accounting managerial emphasis for foster === Subject: Re: JSH: Making funding reductions smart posting-account=mgs1FwoAAABD3j5T_RLZ06yrgt2dghDu Gecko/20050915,gzip(gfe),gzip(gfe) > Consider in an integral ring c*P(x)=(f_1(x) + c)*(f_2(x) + c) where P(x) is a polynomial with integer coefficients, and c is a non- > zero integer, where also P(0) is nonzero and coprime to c in Z. That > is, c is coprime to P(0) in the ring of integers. Then it cannot be true that both f_1(0) = 0 and f_2(0)=0. Proof: Assume both DO equal 0 when x=0, then you have c*P(0) = c^2 which contradicts with P(0) being coprime to c in Z. Since they both cannot equal 0 when x=0, let's normalize with f_1(x) = g_1(x) + b_1 and f_2(x) = g_2(x) + b_2 where g_1(0) = 0 and g_2(0) = 0, where b_1 and b_2 are not functions > of x and both cannot equal 0 as proven above. Then making the substitutions I have c*P(x)=(g_1(x) + b_1 + c)*(g_2(x) + b_2 + c) and notice that at x=0 I have c*P(0)=(b_1 + c)*(b_2 + c) so b_1 and b_2 must be products that include factors of c, and let > those factors be d_1 and d_2, where Let: c=3 P(x)=10 f_1(x)=2 f_2(x)=3 This meets all your conditions. It follows that: g_1(x)=0 g_2(x)=0 b_1=2 b_2=3 Substituting into c*P(0)=(b_1 + c)*(b_2 + c): 3*10 = (2+3)*(3+3) 30 = 5*6 Yet, b_1 is not a product that includes factors of c. > c = d_1*d_2. Then it follows from the distributive property that g_1(x) and g_2(x) > must be products as well so let g_1(x) = d_1*h_1(x) and g_2(x) = d_2*h_2(x) and making those substitutions gives d_1*d_2*P(x)=(d_1*h_1(x) + b_1 + d_1*d_2)*(d_2*h_2(x) + b_2 + d_1*d_2) and now for reference return to our original expression c*P(x)=(f_1(x) + c)*(f_2(x) + c) and notice that we now have the result that d_1 has distributed > through f_1(x) + c and d_2 has distributed through f_2(x) + c. Proof complete. That simple proof destroys the foundations of much of modern number > theory. So you people lie about it. James Harris === Subject: Re: JSH: Making funding reductions smart posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Consider in an integral ring c*P(x)=(f_1(x) + c)*(f_2(x) + c) where P(x) is a polynomial with integer coefficients, and c is a non- > zero integer, where also P(0) is nonzero and coprime to c in Z. That > is, c is coprime to P(0) in the ring of integers. Then it cannot be true that both f_1(0) = 0 and f_2(0)=0. Proof: Assume both DO equal 0 when x=0, then you have c*P(0) = c^2 which contradicts with P(0) being coprime to c in Z. Since they both cannot equal 0 when x=0, let's normalize with f_1(x) = g_1(x) + b_1 and f_2(x) = g_2(x) + b_2 where g_1(0) = 0 and g_2(0) = 0, where b_1 and b_2 are not functions > of x and both cannot equal 0 as proven above. Then making the substitutions I have c*P(x)=(g_1(x) + b_1 + c)*(g_2(x) + b_2 + c) and notice that at x=0 I have c*P(0)=(b_1 + c)*(b_2 + c) so b_1 and b_2 must be products that include factors of c, and let > those factors be d_1 and d_2, where Let: > c=3 > P(x)=10 > f_1(x)=2 > f_2(x)=3 This meets all your conditions. It follows that: g_1(x)=0 > g_2(x)=0 > b_1=2 > b_2=3 Substituting into c*P(0)=(b_1 + c)*(b_2 + c): 3*10 = (2+3)*(3+3) > 30 = 5*6 Yet, b_1 is not a product that includes factors of c. I think you have fallen prey to JSHterminology. When he says that a number b includes factors of c I believe he actually means that b and c share a common factor (in this case 1). Quite why he insists on sticking to this terminology after years of people telling him it causes confusion is a mystery, though my best guess is that he intentionally (though perhaps subconsciously) makes his technical counterexamples tend to get posted within minutes. === Subject: Re: JSH: Making funding reductions smart posting-account=mgs1FwoAAABD3j5T_RLZ06yrgt2dghDu Gecko/20050915,gzip(gfe),gzip(gfe) > I think you have fallen prey to JSHterminology. When he says that a > number b includes factors of c I believe he actually means that b and > c share a common factor (in this case 1). Quite why he insists on > sticking to this terminology after years of people telling him it > causes confusion is a mystery, though my best guess is that he > intentionally (though perhaps subconsciously) makes his technical > counterexamples tend to get posted within minutes. So, when James says, b_1 and b_2 must be products that include factors of c, what that really means is b_1, b_2, and c are numbers. That oughta destroy the foundations of much of modern number theory any day now! -- I like to be very thorough in checking the mathematics of my ideas -- James Harris === Subject: Re: Mechanics of Materials solutions manual posting-account=1UfMtgoAAAB_gW0XeRXMgMklSRzwhAhM Firefox/1.5.0 Opera 9.26,gzip(gfe),gzip(gfe) > I am looking for this as well. me three. === Subject: Re: JSH: Whooda thunk it >As I have done previously, I tested James' latest factoring method, as >per his March 5th blog entry, on 500 random composite odd numbers that >are multiples of two different primes, each in the range 500 to 1000. >The results are compared to Fermat's method, trial factorisation (both >forward and reverse) and random picking. Fermat average = 6.96 probes. > JSH average = 1.11 probes. > Probe ratio = 1 : 0.160 > Trial average = 120.15 probes. > Reverse average = 11.43 probes. > Random average = 757.46 probes. > 500 trials, 0 misfactors found. Average alphas tried per factorisation: 5.184 This is much better than usual James, your current method filters out >unlikely values before calculating the GCD (counted as probes), it >is good at focusing in on the likely possibilities. Three points. Firstly I think there is a typo in your March 5th blog >entry, at one point you talk about f_1 - p must be less than f_1, >but later you say p - f_1 is smaller than it [= f_1]. I have gone >with the second version since the first version just says that p is >positive. Secondly, since I do not know what f_1 is I just started p at the cube >root of nT and worked downwards. This is probably OK for RSA style >numbers, but will not work in general. For a more general method you >would probably have to start p low and work upwards just to be sure >that p < f_1 as required. Thirdly, you say that k is incremented from k_0 by steps of abs(2ap). >When I tried this about 300 out of 500 trials resulted in a misfactor. >I changed the loop to step through all possible values of k between >k_0 and the maximum you state, k_0 + abs(2ap) * k_0 / 2p. With this >change there were no misfactors. You may want to revisit this part of >your work. I will now do some work on absolute, rather than relative, timings. Good work James, much better than your usual results. rossum > I have now done some work on absolute timings for James latest method. I set up an array of 200 semiprimes of a given size and timed James' method factoring them in a loop. I took sizes from 20 bits to 30 bits in 2 bit steps. Average timings over 200 numbers: 20 bits: 2.030 mSec average per number. 22 bits: 5.235 mSec average per number. 24 bits: 32.110 mSec average per number. 26 bits: 47.185 mSec average per number. 28 bits: 104.920 mSec average per number. 30 bits: 2038.830 mSec average per number. method, and hence too slow to be any danger to RSA numbers (500 bits or more). Looking at the two extreme numbers, there is a factor of 1000 difference and the range covers 10 bits: 2^10 = 1024 which is pretty much spot-on exponential growth with size in bits. So James, a much better piece of work than before, but it is still an exponential method. You need a sub-exponential method if you are to factor an RSA number in a reasonable time. To see why try extrapolating the expected timing for a 500 bit number. rossum === Subject: Re: JSH: Whooda thunk it posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) As I have done previously, I tested James' latest factoring method, as >per his March 5th blog entry, on 500 random composite odd numbers that >are multiples of two different primes, each in the range 500 to 1000. >The results are compared to Fermat's method, trial factorisation (both >forward and reverse) and random picking. Fermat average = 6.96 probes. > JSH average = 1.11 probes. > Probe ratio = 1 : 0.160 > Trial average = 120.15 probes. > Reverse average = 11.43 probes. > Random average = 757.46 probes. > 500 trials, 0 misfactors found. Average alphas tried per factorisation: 5.184 This is much better than usual James, your current method filters out >unlikely values before calculating the GCD (counted as probes), it >is good at focusing in on the likely possibilities. Three points. Firstly I think there is a typo in your March 5th blog >entry, at one point you talk about f_1 - p must be less than f_1, >but later you say p - f_1 is smaller than it [= f_1]. I have gone >with the second version since the first version just says that p is >positive. Secondly, since I do not know what f_1 is I just started p at the cube >root of nT and worked downwards. This is probably OK for RSA style >numbers, but will not work in general. For a more general method you >would probably have to start p low and work upwards just to be sure >that p < f_1 as required. Thirdly, you say that k is incremented from k_0 by steps of abs(2ap). >When I tried this about 300 out of 500 trials resulted in a misfactor. >I changed the loop to step through all possible values of k between >k_0 and the maximum you state, k_0 + abs(2ap) * k_0 / 2p. With this >change there were no misfactors. You may want to revisit this part of >your work. I will now do some work on absolute, rather than relative, timings. Good work James, much better than your usual results. rossum I have now done some work on absolute timings for James latest method. > I set up an array of 200 semiprimes of a given size and timed James' > method factoring them in a loop. I took sizes from 20 bits to 30 bits > in 2 bit steps. Average timings over 200 numbers: > 20 bits: 2.030 mSec average per number. > 22 bits: 5.235 mSec average per number. > 24 bits: 32.110 mSec average per number. > 26 bits: 47.185 mSec average per number. > 28 bits: 104.920 mSec average per number. > 30 bits: 2038.830 mSec average per number. method, and hence too slow to be any danger to RSA numbers (500 bits > or more). Looking at the two extreme numbers, there is a factor of > 1000 difference and the range covers 10 bits: 2^10 = 1024 which is > pretty much spot-on exponential growth with size in bits. So James, a much better piece of work than before, but it is still an > exponential method. You need a sub-exponential method if you are to > factor an RSA number in a reasonable time. To see why try > extrapolating the expected timing for a 500 bit number. rossum newsgroups about my research. I'm out of theory mode again and back to experiment so things can tone down, thankfully. Though maybe there is still some theory to work through if incrementing by 2*alpha*p gives problems. I'm tired though. Probably will just do experimental work and take a long break from theory. Theory is more exciting but it's very tiring, and, um, I kind of go a little bonkers when I'm doing theory. Helps me figure things out. Or as I like to say, I dare to walk that line between sanity and insanity and cross it more than once, hoping that I'll always be able to get back across... Nuff said. Off to go play with some math. James Harris === Subject: Re: JSH: Whooda thunk it posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20071009 SeaMonkey/1.1.5,gzip(gfe),gzip(gfe) > As I have done previously, I tested James' latest factoring method, as > per his March 5th blog entry, on 500 random composite odd numbers that > are multiples of two different primes, each in the range 500 to 1000. > The results are compared to Fermat's method, trial factorisation (both > forward and reverse) and random picking. Fermat average = 6.96 probes. > JSH average = 1.11 probes. > Probe ratio = 1 : 0.160 > Trial average = 120.15 probes. > Reverse average = 11.43 probes. > Random average = 757.46 probes. > 500 trials, 0 misfactors found. Average alphas tried per factorisation: 5.184 This is much better than usual James, your current method filters out > unlikely values before calculating the GCD (counted as probes), it > is good at focusing in on the likely possibilities. Three points. Firstly I think there is a typo in your March 5th blog > entry, at one point you talk about f_1 - p must be less than f_1, > but later you say p - f_1 is smaller than it [= f_1]. I have gone > with the second version since the first version just says that p is > positive. Secondly, since I do not know what f_1 is I just started p at the cube > root of nT and worked downwards. This is probably OK for RSA style > numbers, but will not work in general. For a more general method you > would probably have to start p low and work upwards just to be sure > that p < f_1 as required. Thirdly, you say that k is incremented from k_0 by steps of abs(2ap). > When I tried this about 300 out of 500 trials resulted in a misfactor. > I changed the loop to step through all possible values of k between > k_0 and the maximum you state, k_0 + abs(2ap) * k_0 / 2p. With this > change there were no misfactors. You may want to revisit this part of > your work. I will now do some work on absolute, rather than relative, timings. Good work James, much better than your usual results. rossum This is interesting and encouraging - I was starting to suspect this also. Since Harris refuses to spell out his algorithm clearly and all in one place, I wonder if you might be willing to do so? It is going to be somewhat harder I would guess to make comparisons with the quadratic sieve, unless there is already a program out there somewhere that does it. It seems unlikely that Harris's method by itself will outperform the QS, but it's possible that his method could speed up the QS in a useful way. Marcus. === Subject: Re: JSH: Whooda thunk it > As I have done previously, I tested James' latest factoring method, as > per his March 5th blog entry, on 500 random composite odd numbers that > are multiples of two different primes, each in the range 500 to 1000. > The results are compared to Fermat's method, trial factorisation (both > forward and reverse) and random picking. > Fermat average = 6.96 probes. > JSH average = 1.11 probes. > Probe ratio = 1 : 0.160 > Trial average = 120.15 probes. > Reverse average = 11.43 probes. > Random average = 757.46 probes. > 500 trials, 0 misfactors found. > Average alphas tried per factorisation: 5.184 > This is much better than usual James, your current method filters out > unlikely values before calculating the GCD (counted as probes), it > is good at focusing in on the likely possibilities. > Three points. Firstly I think there is a typo in your March 5th blog > entry, at one point you talk about f_1 - p must be less than f_1, > but later you say p - f_1 is smaller than it [= f_1]. I have gone > with the second version since the first version just says that p is > positive. > Secondly, since I do not know what f_1 is I just started p at the cube > root of nT and worked downwards. This is probably OK for RSA style > numbers, but will not work in general. For a more general method you > would probably have to start p low and work upwards just to be sure > that p < f_1 as required. > Thirdly, you say that k is incremented from k_0 by steps of abs(2ap). > When I tried this about 300 out of 500 trials resulted in a misfactor. > I changed the loop to step through all possible values of k between > k_0 and the maximum you state, k_0 + abs(2ap) * k_0 / 2p. With this > change there were no misfactors. You may want to revisit this part of > your work. > I will now do some work on absolute, rather than relative, timings. > Good work James, much better than your usual results. > rossum This is interesting and encouraging - I was starting >to suspect this also. Since Harris refuses to spell >out his algorithm clearly and all in one place, I >wonder if you might be willing to do so? It is going to be somewhat harder I would guess to >make comparisons with the quadratic sieve, unless there >is already a program out there somewhere that does it. >It seems unlikely that Harris's method by itself will >outperform the QS, but it's possible that his method >could speed up the QS in a useful way. Marcus. This post give the version of James' method I used in my tests. I took it from the post in James' blog (http://mymath.blogspot.com/) dated 5 March - Putting it all together. Algorithm --------- To factor an odd semiprime T: 1 Check T mod 3. If zero, then set f <- 3 and go to step 21, if 1 then set n <- 5, if 2 then set n <- 1. 2 Set pMax <- ceil(cubeRoot(n * T)). (See Notes 1, 2) 3 Set alpha <- 0. 4 Increment alpha by one. 5 Set p <- pMax. 6 Set p <- previousPrime(p). If p <= 2 then go to step 4. 7 Set kSquared <- ((1 + alpha^2)^(-1) * nT) mod p. 8 If kSquared is not a quadratic residue mod p then go to step 6. 9 Set kMin <- floor(sqrt((nT) / (1.0 + alpha^2))). (See Note 3) 10 Set kStep <- 2 * alpha * p. 11 Set kMax <- kMin + kStep * ceil(kMin / (2.0 * p)). 12 Set k <- kMin. 13 If k > kMax then go to step 6. 14 Set z <- floor(((1 + 2 * alpha^2) * k) / (2 * alpha)). 15 Set ySquared <- abs(z^2 - nT). 16 If ySquared is not a square number then increment k by 1 and go to step 13. (See Note 4) 17 Set y <- sqrt(ySquared). 18 If 1 < GCD(y + z, T) < T then set f <- GCD(y + z, T) and go to step 21. 19 If 1 < GCD(abs(y - z), T) < T then set f <- GCD(abs(y - z), T) and go to step 21. 20 Increment k by 1 and go to step 13. (See Note 4) 21 Return f and T/f as factors and exit. Notes ----- 1 We have to set p less than f_1 where f_1 is the smallest factor of nT. I have ignored n, since with n = 5 we could only use primes less than 5. With T an RSA style number it is reasonable to assume that f_1 is less than the cube root of T. This assumption will not be true for a general composite T. 2 James has two conditions here, p < f_1 and p - f_1 < f_1. The second equates to p < 2f_1, which is less restrictive than the first. I have used the first so the second is automatically followed also. Just double pMax to follow the second restriction only. 3 I use kMin where James has k_0 for the minimum value of k. 4 James said to step k by kStep. When I tried this misfactors resulted so I changed it to step k by 1 up to the same limit, kMax. Further Work ------------ For further study I want to look at which values of alpha, p and k actually give factors to see if there are any simple heuristics. For example does k have to be odd or even? rossum === Subject: Re: JSH: Whooda thunk it posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20071009 SeaMonkey/1.1.5,gzip(gfe),gzip(gfe) > As I have done previously, I tested James' latest factoring method, as > per his March 5th blog entry, on 500 random composite odd numbers that > are multiples of two different primes, each in the range 500 to 1000. > The results are compared to Fermat's method, trial factorisation (both > forward and reverse) and random picking. > Fermat average = 6.96 probes. > JSH average = 1.11 probes. > Probe ratio = 1 : 0.160 > Trial average = 120.15 probes. > Reverse average = 11.43 probes. > Random average = 757.46 probes. > 500 trials, 0 misfactors found. > Average alphas tried per factorisation: 5.184 > This is much better than usual James, your current method filters out > unlikely values before calculating the GCD (counted as probes), it > is good at focusing in on the likely possibilities. > Three points. Firstly I think there is a typo in your March 5th blog > entry, at one point you talk about f_1 - p must be less than f_1, > but later you say p - f_1 is smaller than it [= f_1]. I have gone > with the second version since the first version just says that p is > positive. > Secondly, since I do not know what f_1 is I just started p at the cube > root of nT and worked downwards. This is probably OK for RSA style > numbers, but will not work in general. For a more general method you > would probably have to start p low and work upwards just to be sure > that p < f_1 as required. > Thirdly, you say that k is incremented from k_0 by steps of abs(2ap). > When I tried this about 300 out of 500 trials resulted in a misfactor. > I changed the loop to step through all possible values of k between > k_0 and the maximum you state, k_0 + abs(2ap) * k_0 / 2p. With this > change there were no misfactors. You may want to revisit this part of > your work. > I will now do some work on absolute, rather than relative, timings. > Good work James, much better than your usual results. > rossum This is interesting and encouraging - I was starting >to suspect this also. Since Harris refuses to spell >out his algorithm clearly and all in one place, I >wonder if you might be willing to do so? It is going to be somewhat harder I would guess to >make comparisons with the quadratic sieve, unless there >is already a program out there somewhere that does it. >It seems unlikely that Harris's method by itself will >outperform the QS, but it's possible that his method >could speed up the QS in a useful way. Marcus. This post give the version of James' method I used in my tests. I > took it from the post in James' blog (http://mymath.blogspot.com/) > dated 5 March - Putting it all together. Algorithm > --------- To factor an odd semiprime T: 1 Check T mod 3. If zero, then set f <- 3 and go to step 21, if 1 > then set n <- 5, if 2 then set n <- 1. 2 Set pMax <- ceil(cubeRoot(n * T)). (See Notes 1, 2) 3 Set alpha <- 0. 4 Increment alpha by one. 5 Set p <- pMax. 6 Set p <- previousPrime(p). If p <= 2 then go to step 4. 7 Set kSquared <- ((1 + alpha^2)^(-1) * nT) mod p. 8 If kSquared is not a quadratic residue mod p then go to step 6. > See note below - > 9 Set kMin <- floor(sqrt((nT) / (1.0 + alpha^2))). (See Note 3) 10 Set kStep <- 2 * alpha * p. 11 Set kMax <- kMin + kStep * ceil(kMin / (2.0 * p)). 12 Set k <- kMin. 13 If k > kMax then go to step 6. > This section of the algorithm looks very strange to me. In step 8, you check whether ksquared is a quadratic residue. But then it appears you don't use the resulting k - instead you start again with kMin and increment it etc., and you don't use, as far as I can see, that the expression in step 7 really has any role. Why not just skip steps 7 and 8 and go directly to step 9? Marcus. > 14 Set z <- floor(((1 + 2 * alpha^2) * k) / (2 * alpha)). 15 Set ySquared <- abs(z^2 - nT). 16 If ySquared is not a square number then increment k by 1 and go to > step 13. (See Note 4) 17 Set y <- sqrt(ySquared). 18 If 1 < GCD(y + z, T) < T then set f <- GCD(y + z, T) and go to > step 21. 19 If 1 < GCD(abs(y - z), T) < T then set f <- GCD(abs(y - z), T) and > go to step 21. 20 Increment k by 1 and go to step 13. (See Note 4) 21 Return f and T/f as factors and exit. Notes > ----- 1 We have to set p less than f_1 where f_1 is the smallest factor of > nT. I have ignored n, since with n = 5 we could only use primes less > than 5. With T an RSA style number it is reasonable to assume that > f_1 is less than the cube root of T. This assumption will not be true > for a general composite T. 2 James has two conditions here, p < f_1 and p - f_1 < f_1. The > second equates to p < 2f_1, which is less restrictive than the first. > I have used the first so the second is automatically followed also. > Just double pMax to follow the second restriction only. 3 I use kMin where James has k_0 for the minimum value of k. 4 James said to step k by kStep. When I tried this misfactors > resulted so I changed it to step k by 1 up to the same limit, kMax. Further Work > ------------ For further study I want to look at which values of alpha, p and k > actually give factors to see if there are any simple heuristics. For > example does k have to be odd or even? rossum === Subject: Re: JSH: Whooda thunk it > As I have done previously, I tested James' latest factoring method, as > per his March 5th blog entry, on 500 random composite odd numbers that > are multiples of two different primes, each in the range 500 to 1000. > The results are compared to Fermat's method, trial factorisation (both > forward and reverse) and random picking. > Fermat average = 6.96 probes. > JSH average = 1.11 probes. > Probe ratio = 1 : 0.160 > Trial average = 120.15 probes. > Reverse average = 11.43 probes. > Random average = 757.46 probes. > 500 trials, 0 misfactors found. > Average alphas tried per factorisation: 5.184 > This is much better than usual James, your current method filters out > unlikely values before calculating the GCD (counted as probes), it > is good at focusing in on the likely possibilities. > Three points. Firstly I think there is a typo in your March 5th blog > entry, at one point you talk about f_1 - p must be less than f_1, > but later you say p - f_1 is smaller than it [= f_1]. I have gone > with the second version since the first version just says that p is > positive. > Secondly, since I do not know what f_1 is I just started p at the cube > root of nT and worked downwards. This is probably OK for RSA style > numbers, but will not work in general. For a more general method you > would probably have to start p low and work upwards just to be sure > that p < f_1 as required. > Thirdly, you say that k is incremented from k_0 by steps of abs(2ap). > When I tried this about 300 out of 500 trials resulted in a misfactor. > I changed the loop to step through all possible values of k between > k_0 and the maximum you state, k_0 + abs(2ap) * k_0 / 2p. With this > change there were no misfactors. You may want to revisit this part of > your work. > I will now do some work on absolute, rather than relative, timings. > Good work James, much better than your usual results. > rossum > This is interesting and encouraging - I was starting >to suspect this also. Since Harris refuses to spell >out his algorithm clearly and all in one place, I >wonder if you might be willing to do so? > It is going to be somewhat harder I would guess to >make comparisons with the quadratic sieve, unless there >is already a program out there somewhere that does it. >It seems unlikely that Harris's method by itself will >outperform the QS, but it's possible that his method >could speed up the QS in a useful way. > Marcus. > This post give the version of James' method I used in my tests. I > took it from the post in James' blog (http://mymath.blogspot.com/) > dated 5 March - Putting it all together. > Algorithm > --------- > To factor an odd semiprime T: > 1 Check T mod 3. If zero, then set f <- 3 and go to step 21, if 1 > then set n <- 5, if 2 then set n <- 1. > 2 Set pMax <- ceil(cubeRoot(n * T)). (See Notes 1, 2) > 3 Set alpha <- 0. > 4 Increment alpha by one. > 5 Set p <- pMax. > 6 Set p <- previousPrime(p). If p <= 2 then go to step 4. > 7 Set kSquared <- ((1 + alpha^2)^(-1) * nT) mod p. > 8 If kSquared is not a quadratic residue mod p then go to step 6. See note below - > 9 Set kMin <- floor(sqrt((nT) / (1.0 + alpha^2))). (See Note 3) > 10 Set kStep <- 2 * alpha * p. > 11 Set kMax <- kMin + kStep * ceil(kMin / (2.0 * p)). > 12 Set k <- kMin. > 13 If k > kMax then go to step 6. This section of the algorithm looks very strange >to me. In step 8, you check whether ksquared is >a quadratic residue. But then it appears you don't >use the resulting k - instead you start again with >kMin and increment it etc., and you don't use, as >far as I can see, that the expression in step 7 >really has any role. Why not just skip steps 7 and >8 and go directly to step 9? Marcus. As you have earlier noticed yourself, James gives two ways to calculate k, the first by finding my kSquared and then evaluating the modular square root, the second by minimising the expression abs(nT - (1 + alpha^2) * kMin^2) to get a value for kMin. What James calls k I have split into my kSquared and kMin. Since I was sticking closely to his March 5th blog I followed that as closely as I could. He calculates kSquared and requires that it be a quadratic residue mod p, if not then you need to cycle back to the next lower p, to quote You pick some odd prime p, so that k exists for a given alpha. Having found a suitable p James then moves on to the other way to calculate kMin by minimising the given expression. I know that it is confusing, but I am following James pretty closely. It would be clearer to use two different variable names for James' two different versions of what he calls k. The importance of calculating kSquared is at step 8 where me may short-circuit the rest of the p-loop (up to step 20) and go straight to the next value of p if kSquared is not a quadratic residue mod p. rossum > 14 Set z <- floor(((1 + 2 * alpha^2) * k) / (2 * alpha)). > 15 Set ySquared <- abs(z^2 - nT). > 16 If ySquared is not a square number then increment k by 1 and go to > step 13. (See Note 4) > 17 Set y <- sqrt(ySquared). > 18 If 1 < GCD(y + z, T) < T then set f <- GCD(y + z, T) and go to > step 21. > 19 If 1 < GCD(abs(y - z), T) < T then set f <- GCD(abs(y - z), T) and > go to step 21. > 20 Increment k by 1 and go to step 13. (See Note 4) > 21 Return f and T/f as factors and exit. > Notes > ----- > 1 We have to set p less than f_1 where f_1 is the smallest factor of > nT. I have ignored n, since with n = 5 we could only use primes less > than 5. With T an RSA style number it is reasonable to assume that > f_1 is less than the cube root of T. This assumption will not be true > for a general composite T. > 2 James has two conditions here, p < f_1 and p - f_1 < f_1. The > second equates to p < 2f_1, which is less restrictive than the first. > I have used the first so the second is automatically followed also. > Just double pMax to follow the second restriction only. > 3 I use kMin where James has k_0 for the minimum value of k. > 4 James said to step k by kStep. When I tried this misfactors > resulted so I changed it to step k by 1 up to the same limit, kMax. > Further Work > ------------ > For further study I want to look at which values of alpha, p and k > actually give factors to see if there are any simple heuristics. For > example does k have to be odd or even? > rossum === Subject: Re: JSH: Whooda thunk it posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > As I have done previously, I tested James' latest factoring method, as > per his March 5th blog entry, on 500 random composite odd numbers that > are multiples of two different primes, each in the range 500 to 1000. > The results are compared to Fermat's method, trial factorisation (both > forward and reverse) and random picking. > Fermat average = 6.96 probes. > JSH average = 1.11 probes. > Probe ratio = 1 : 0.160 > Trial average = 120.15 probes. > Reverse average = 11.43 probes. > Random average = 757.46 probes. > 500 trials, 0 misfactors found. > Average alphas tried per factorisation: 5.184 > This is much better than usual James, your current method filters out > unlikely values before calculating the GCD (counted as probes), it > is good at focusing in on the likely possibilities. > Three points. Firstly I think there is a typo in your March 5th blog > entry, at one point you talk about f_1 - p must be less than f_1, > but later you say p - f_1 is smaller than it [= f_1]. I have gone > with the second version since the first version just says that p is > positive. > Secondly, since I do not know what f_1 is I just started p at the cube > root of nT and worked downwards. This is probably OK for RSA style > numbers, but will not work in general. For a more general method you > would probably have to start p low and work upwards just to be sure > that p < f_1 as required. > Thirdly, you say that k is incremented from k_0 by steps of abs(2ap). > When I tried this about 300 out of 500 trials resulted in a misfactor. > I changed the loop to step through all possible values of k between > k_0 and the maximum you state, k_0 + abs(2ap) * k_0 / 2p. With this > change there were no misfactors. You may want to revisit this part of > your work. > I will now do some work on absolute, rather than relative, timings. > Good work James, much better than your usual results. > rossum This is interesting and encouraging - I was starting >to suspect this also. Since Harris refuses to spell >out his algorithm clearly and all in one place, I >wonder if you might be willing to do so? It is going to be somewhat harder I would guess to >make comparisons with the quadratic sieve, unless there >is already a program out there somewhere that does it. >It seems unlikely that Harris's method by itself will >outperform the QS, but it's possible that his method >could speed up the QS in a useful way. Marcus. This post give the version of James' method I used in my tests. I > took it from the post in James' blog (http://mymath.blogspot.com/) > dated 5 March - Putting it all together. Algorithm > --------- To factor an odd semiprime T: 1 Check T mod 3. If zero, then set f <- 3 and go to step 21, if 1 > then set n <- 5, if 2 then set n <- 1. 2 Set pMax <- ceil(cubeRoot(n * T)). (See Notes 1, 2) 3 Set alpha <- 0. 4 Increment alpha by one. 5 Set p <- pMax. 6 Set p <- previousPrime(p). If p <= 2 then go to step 4. 7 Set kSquared <- ((1 + alpha^2)^(-1) * nT) mod p. 8 If kSquared is not a quadratic residue mod p then go to step 6. 9 Set kMin <- floor(sqrt((nT) / (1.0 + alpha^2))). (See Note 3) 10 Set kStep <- 2 * alpha * p. 11 Set kMax <- kMin + kStep * ceil(kMin / (2.0 * p)). 12 Set k <- kMin. 13 If k > kMax then go to step 6. 14 Set z <- floor(((1 + 2 * alpha^2) * k) / (2 * alpha)). 15 Set ySquared <- abs(z^2 - nT). 16 If ySquared is not a square number then increment k by 1 and go to > step 13. (See Note 4) 17 Set y <- sqrt(ySquared). 18 If 1 < GCD(y + z, T) < T then set f <- GCD(y + z, T) and go to > step 21. 19 If 1 < GCD(abs(y - z), T) < T then set f <- GCD(abs(y - z), T) and > go to step 21. 20 Increment k by 1 and go to step 13. (See Note 4) 21 Return f and T/f as factors and exit. Notes > ----- 1 We have to set p less than f_1 where f_1 is the smallest factor of > nT. I have ignored n, since with n = 5 we could only use primes less > than 5. With T an RSA style number it is reasonable to assume that > f_1 is less than the cube root of T. This assumption will not be true > for a general composite T. 2 James has two conditions here, p < f_1 and p - f_1 < f_1. The > second equates to p < 2f_1, which is less restrictive than the first. > I have used the first so the second is automatically followed also. > Just double pMax to follow the second restriction only. I focus on solutions in all integers while you can factor with some of the numbers being non-rational, which can be confusing as you consider experimental results. When I say all integers, I mean that you have all integers with the following x^2 = y^2 mod p, z^2 = y^2 + nT, and 2ax = k, and z = x+ak. Otherwise there are different rules where my favorite example is T = 101(103), which gives a k_0 = 72, but will factor with k = 68, which is less than k_0, but if you solve out for all the variables you find that x is non-rational and that x^2 = y^2 mod p does not exist with all integers, while z and y are rational--and they're the only ones that are. Here 'a' can surprise because you have its residue, so actually with a=1, you have a = 1 mod p, so 'a' can be non-rational despite have 1 as its residue modulo p. I'm not sure how big of a deal the non-rational factorizations are. > 3 I use kMin where James has k_0 for the minimum value of k. When all the variables are integers k_0 is the minimum k. For some nT is may not be possible to have all integers. > 4 James said to step k by kStep. When I tried this misfactors > resulted so I changed it to step k by 1 up to the same limit, kMax. Those were non-rational factorizations. > Further Work > ------------ For further study I want to look at which values of alpha, p and k > actually give factors to see if there are any simple heuristics. For > example does k have to be odd or even? rossum For all integer solutions k must be even. However as I've noted you can still factor using non-rationals which skeptical readers can determine is true by solving for all the variables with x^2 = y^2 mod p, z^2 = y^2 + nT, and 2ax = k, and z = x+ak. I focus on integer solutions where k increments by 2ap with a minimum of k_0, and the first k given by the nearest k to that value that has the calculated residue modulo p and has 2a as a factor. James Harris === Subject: Re: JSH: Whooda thunk it > As I have done previously, I tested James' latest factoring method, as > per his March 5th blog entry, on 500 random composite odd numbers that > are multiples of two different primes, each in the range 500 to 1000. > The results are compared to Fermat's method, trial factorisation (both > forward and reverse) and random picking. > Fermat average = 6.96 probes. > JSH average = 1.11 probes. > Probe ratio = 1 : 0.160 > Trial average = 120.15 probes. > Reverse average = 11.43 probes. > Random average = 757.46 probes. > 500 trials, 0 misfactors found. > Average alphas tried per factorisation: 5.184 > This is much better than usual James, your current method filters out > unlikely values before calculating the GCD (counted as probes), it > is good at focusing in on the likely possibilities. > Three points. Firstly I think there is a typo in your March 5th blog > entry, at one point you talk about f_1 - p must be less than f_1, > but later you say p - f_1 is smaller than it [= f_1]. I have gone > with the second version since the first version just says that p is > positive. > Secondly, since I do not know what f_1 is I just started p at the cube > root of nT and worked downwards. This is probably OK for RSA style > numbers, but will not work in general. For a more general method you > would probably have to start p low and work upwards just to be sure > that p < f_1 as required. > Thirdly, you say that k is incremented from k_0 by steps of abs(2ap). > When I tried this about 300 out of 500 trials resulted in a misfactor. > I changed the loop to step through all possible values of k between > k_0 and the maximum you state, k_0 + abs(2ap) * k_0 / 2p. With this > change there were no misfactors. You may want to revisit this part of > your work. > I will now do some work on absolute, rather than relative, timings. > Good work James, much better than your usual results. > rossum > This is interesting and encouraging - I was starting >to suspect this also. Since Harris refuses to spell >out his algorithm clearly and all in one place, I >wonder if you might be willing to do so? > It is going to be somewhat harder I would guess to >make comparisons with the quadratic sieve, unless there >is already a program out there somewhere that does it. >It seems unlikely that Harris's method by itself will >outperform the QS, but it's possible that his method >could speed up the QS in a useful way. > Marcus. > This post give the version of James' method I used in my tests. I > took it from the post in James' blog (http://mymath.blogspot.com/) > dated 5 March - Putting it all together. > Algorithm > --------- > To factor an odd semiprime T: > 1 Check T mod 3. If zero, then set f <- 3 and go to step 21, if 1 > then set n <- 5, if 2 then set n <- 1. > 2 Set pMax <- ceil(cubeRoot(n * T)). (See Notes 1, 2) > 3 Set alpha <- 0. > 4 Increment alpha by one. > 5 Set p <- pMax. > 6 Set p <- previousPrime(p). If p <= 2 then go to step 4. > 7 Set kSquared <- ((1 + alpha^2)^(-1) * nT) mod p. > 8 If kSquared is not a quadratic residue mod p then go to step 6. > 9 Set kMin <- floor(sqrt((nT) / (1.0 + alpha^2))). (See Note 3) > 10 Set kStep <- 2 * alpha * p. > 11 Set kMax <- kMin + kStep * ceil(kMin / (2.0 * p)). > 12 Set k <- kMin. > 13 If k > kMax then go to step 6. > 14 Set z <- floor(((1 + 2 * alpha^2) * k) / (2 * alpha)). > 15 Set ySquared <- abs(z^2 - nT). > 16 If ySquared is not a square number then increment k by 1 and go to > step 13. (See Note 4) > 17 Set y <- sqrt(ySquared). > 18 If 1 < GCD(y + z, T) < T then set f <- GCD(y + z, T) and go to > step 21. > 19 If 1 < GCD(abs(y - z), T) < T then set f <- GCD(abs(y - z), T) and > go to step 21. > 20 Increment k by 1 and go to step 13. (See Note 4) > 21 Return f and T/f as factors and exit. > Notes > ----- > 1 We have to set p less than f_1 where f_1 is the smallest factor of > nT. I have ignored n, since with n = 5 we could only use primes less > than 5. With T an RSA style number it is reasonable to assume that > f_1 is less than the cube root of T. This assumption will not be true > for a general composite T. > 2 James has two conditions here, p < f_1 and p - f_1 < f_1. The > second equates to p < 2f_1, which is less restrictive than the first. > I have used the first so the second is automatically followed also. > Just double pMax to follow the second restriction only. I focus on solutions in all integers while you can factor with some of >the numbers being non-rational, which can be confusing as you consider >experimental results. When I say all integers, I mean that you have all integers with the >following x^2 = y^2 mod p, z^2 = y^2 + nT, and 2ax = k, and z = x+ak. Otherwise there are different rules where my favorite example is T = >101(103), which gives a k_0 = 72, but will factor with k = 68, which >is less than k_0, but if you solve out for all the variables you find >that x is non-rational and that x^2 = y^2 mod p does not exist with >all integers, while z and y are rational--and they're the only ones >that are. Here 'a' can surprise because you have its residue, so actually with >a=1, you have a = 1 mod p, so 'a' can be non-rational despite have 1 >as its residue modulo p. I'm not sure how big of a deal the non-rational factorizations are. > 3 I use kMin where James has k_0 for the minimum value of k. When all the variables are integers k_0 is the minimum k. For some nT is may not be possible to have all integers. > 4 James said to step k by kStep. When I tried this misfactors > resulted so I changed it to step k by 1 up to the same limit, kMax. Those were non-rational factorizations. > Further Work > ------------ > For further study I want to look at which values of alpha, p and k > actually give factors to see if there are any simple heuristics. For > example does k have to be odd or even? > rossum For all integer solutions k must be even. That is not what I am getting. When I look at the statistics for the k's which give solutions I get: k mod 2: k mod 2 = 0: 385 k mod 2 = 1: 115 Most, but not all, solutions use even k. If you restrict yourself to even k then you will miss some solutions. rossum >However as I've noted you >can still factor using non-rationals which skeptical readers can >determine is true by solving for all the variables with x^2 = y^2 mod p, z^2 = y^2 + nT, and 2ax = k, and z = x+ak. I focus on integer solutions where k increments by 2ap with a minimum >of k_0, and the first k given by the nearest k to that value that has >the calculated residue modulo p and has 2a as a factor. >James Harris === Subject: Re: JSH: Whooda thunk it posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig AppleWebKit/523.12.2 (KHTML, like Gecko) Version/3.0.4 Safari/523.12.2,gzip(gfe),gzip(gfe) > As I have done previously, I tested James' latest factoring method, as > per his March 5th blog entry, on 500 random composite odd numbers that > are multiples of two different primes, each in the range 500 to 1000. > The results are compared to Fermat's method, trial factorisation (both > forward and reverse) and random picking. æ Fermat average = æ æ æ6.96 probes. > æ JSH average = æ æ æ æ 1.11 probes. > æ Probe ratio = æ æ æ æ 1 : 0.160 > æ Trial average = æ æ æ 120.15 probes. > æ Reverse average = æ æ 11.43 probes. > æ Random average = æ æ æ757.46 probes. > æ 500 trials, 0 misfactors found. æ Average alphas tried per factorisation: 5.184 This is much better than usual James, your current method filters out > unlikely values before calculating the GCD (counted as probes), it > is good at focusing in on the likely possibilities. Three points. æFirstly I think there is a typo in your March 5th blog > entry, at one point you talk about f 1 - p must be less than f 1, > but later you say p - f 1 is smaller than it [= f 1]. æI have gone > with the second version since the first version just says that p is > positive. Secondly, since I do not know what f 1 is I just started p at the cube > root of nT and worked downwards. æThis is probably OK for RSA style > numbers, but will not work in general. æFor a more general method you > would probably have to start p low and work upwards just to be sure > that p < f 1 as required. Thirdly, you say that k is incremented from k 0 by steps of abs(2ap). > When I tried this about 300 out of 500 trials resulted in a misfactor. > I changed the loop to step through all possible values of k between > k 0 and the maximum you state, k 0 + abs(2ap) * k 0 / 2p. æWith this > change there were no misfactors. æYou may want to revisit this part of > your work. I will now do some work on absolute, rather than relative, timings. Good work James, much better than your usual results. OK, this is interesting but doesn't address the scaling issue. James himself has never offered anything but two-digit examples as far as I can see. You've tested the 3-digit cases. Now how about some larger numbers, up to say 6-9 digits? It sounds like you have decent software. - Randy === Subject: Nilsson, Riedel's electric circuit posting-account=Rvx4LgoAAACg9rlO74XDshoakbrkIdG9 SV1),gzip(gfe),gzip(gfe) I'm also looking for the 8th solutions manual, did you find it? If you did can you email it to me as well. My e-mail address is kakihyu@gmail.com === Subject: Wanted! Soluiton Manual Electric Circuits 8th Edition - Nilsson! posting-account=Rvx4LgoAAACg9rlO74XDshoakbrkIdG9 SV1),gzip(gfe),gzip(gfe) I'm also looking for the solutions manual of Electric Circuits 8th Edition by Nilsson, Riedel Did you get it? If did you get, can you email it to me. thank you!! My e-mail address is kakihyu@gmail.com === Subject: JSH: You have to argue posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) It is so tiring to chase these mathematical results. It is so easy to just stop and try to rest with something that is wrong. Arguing with people who just want you to be wrong no matter what helps. It helps you keep going. And it helps you push to a solution. It is so tiring. So, so very tiring. But mathematics is nothing if it's not right. But human nature is such that you may convince yourself and convince others because it feels good to be right. So toss all of that out the window by having everyone that is critiquing you, hate you. And argue. I don't give a damn who I convince if I'm wrong. But if I'm right, I mostly wish to convince myself. James Harris === Subject: Re: You have to argue > It is so tiring to chase these mathematical results. It is so easy to > just stop and try to rest with something that is wrong. Arguing with > people who just want you to be wrong no matter what helps. It helps > you keep going. And it helps you push to a solution. It is so tiring. So, so very tiring. But mathematics is nothing if it's not right. But human nature is such that you may convince yourself and convince > others because it feels good to be right. So toss all of that out the window by having everyone that is > critiquing you, hate you. And argue. I don't give a damn who I convince if I'm wrong. But if I'm right, I mostly wish to convince myself. > James Harris rossum is doing the real work of testing for you. drop the remote, get off your butt and get the theory done, now! it is needed for more testing this time going up an order of magnitude. === Subject: Re: JSH: Measuring influence > I did a couple. I get my blog posting at #1 on the UK and of course > the US one, but on the Germany one it comes up only #5. > So yeah, wherever you are, you're getting different results. I assume > you're not in the US. > I've done previous testing to confirm that it comes up number one in > the US on both coasts. > Where are you in the world? > Heh. You're such an expert on the internet that you can't find that > information yourself? I'm not an expert on the Internet. Huh? At the top of the thread you said One group, versus one man who understand the Internet, I think, far better than any of you can possibly come close to comprehending, with, your, I'm afraid, um, limited intellects. >The point of this thread is influence. I can put out ideas for a Superman movie or ideas about math or ideas >about, oh, whatever suits me, and for some odd reason Google searches >seem to often say those are important to some people out there in the >big, wide, real world. Now then, do you think more people care about Superman movies or math? The point here is that ideas I like, like cutting down funding for >mathematical research worldwide can have an impact. Regarding a question you asked in that survey: No, please don't stop posting. This stuff is better than anything on TV. You really think that you have an impact on funding for mathematical research? Wow. >In the US more and more attention is being paid to large endowments. You people are affecting the academic world across the board, not just >in mathematics, so English professors or professors of History can >thank mathematicians for giving me the reason to care about how much >money you ALL get, or for me to want to end tenure. They can thank their mathematical colleagues and people like you on >Usenet for making it worth my time. It's your society's choice: play the odds by what you think you know >and have fewer funds, or learn. >James Harris David C. Ullrich === Subject: Re: JSH: Measuring influence > Well I just did that search and the first page consists mainly of > pages that have nothing to do with you, but the second hit is this > thread in Google Groups. Pretty impressive. > Hmmm...may be country dependent then, which is something I've been > curious about, Curious? I thought you understood the internet far better than any of >us can come close to comprehending? This is not hard to explain. In fact he understands the internet better than the internet does. You and I think of the internet as this bunch of stuff on servers all over the world - with his superior intelligence he realizes that the _real_ internet, the one that really matters, is the one that exists just in his head. I mean really. This would not be so obvious with most people. But given the nature of his superior understanding of _mathematics_ it's really a no-brainer. > oh, you did NOT put it in quotes, did you? Nope. > The search should be on: Superman plot idea. > No quotes, just those three words and it has to be Google and not > another search engine. > Try again. In the United States readers will get a plot idea I had > for a Superman movie. > Outside this country results I guess may vary. not appear on the front page of any of the searches, though perhaps it >would be different if I were elsewhere. See what you get from these >links (I'm genuinely curious as to whether the search results vary >with the searcher's location as well as the Google domain): >By the way, what ever happened to DMESE? David C. Ullrich === Subject: Re: Holman Heat transfer send it to me to( holman heat transfer manual) plz to : hammster@email.si tnx === Subject: writing a book posting-account=hHBMMgoAAAAdnToQ6KL7YyAqhOuu-mnp Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) hey . im writing a book on Nanomaterials...and am looking for a co-author. Those interested may contact at r.ravish.k@gmail.com === Subject: Maths Info and Quotations posting-account=UKPBwAoAAAACmI0Tado5DEX6bGSZsK7M InfoPath.2; Alexa Toolbar),gzip(gfe),gzip(gfe) Hi ! I am a new member .So I thought I'd let you know that I've a small website www.tipsinmaths.ind.in with certain maths info. Preview copy one ebook A Few Easy Steps in Multiplications And Squares is also available thru the site. Though a few chapters of this ebook would require registration and a small charge, even the other chapters available for free viewing could be of interest and of help to a few. I think the first story of the village boy in the ebook and info about Aaryabhata in Is Maths difficult ? page of the website could make a useful reading for many. Website also has a page for a collection of Good Quotations. For preview copy of the ebook simply click following link : : http://tipsinmaths.ind.in/AFEWEASYSTEPS.EXE I hope many would enjoy it. Mrudula === Subject: Solutions of mechanical vibrations e/4 by Singiresu s rao posting-account=pZWmhgoAAAAZd0_fbbP02tXJjxxQt7ms CLR 1.1.4322),gzip(gfe),gzip(gfe) > hello frndz...plz plz plz if anybody find the solutions of Mechanical > Vibrations Edition 4 by Singiresu s rao plz do mail me....i m vry > much in need....plz if u cn help me... === Subject: Liquid ammonia in space posting-account=p0JNqwkAAAChY16-5zbk2O2xWfBB6K-z Gecko/20070508 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) Liquid ammonia in space Liquids on planetary surfaces are interesting in themselves, but also as possible media for life - possibly the only place where complex life could arise. Only a few liquids are reasonable candidates to exist on a planetary surface. One, we know, is water - it exists on Earth, and is likely on any (sufficiently oxidised; see below) planet with approximately the same temperature as Earth. The other that exists in our solar system is liquid methane, on Titan. This is also likely very common in the galaxy; and at still lower temperatures liquid nitrogen is possible (which will contain substantial amounts of methane and/or CO). [1] But perhaps more interesting are those possible liquids that could exist at 'high' temperatures, let's say > -100C/-150F, which are more likely to possibly host life - though, if extraterrestrial life is at all common in the galaxy, I wouldn't be surprised to find methane-based life somewhere. Liquid ammonia is the most commonly proposed alternative and I here examine its possibilities for existence. What sort of planets might have an ammonia ocean? The first problem is that water is far more common than ammonia in space, less volatile, and much more stable to photodecomposition, so one would expect any planet that had surface ammonia to have it only as a solution in water. This would be normally 10-15% ammonia; it could be concentrated to the eutectic (33-35%) by freezing but this would be locked under ice. To have ammonia on the surface, then, it is necessary for ammonia to be able to exist on the planet, but for water to not exist. The main condition affecting the chemical composition of terrestrial planets is the C/O ratio, though perhaps we should better refer to the effective C/O ratio as planetary accretion is not homogeneous. The stable form of C in a nebula at T>700 K (at which the non-volatile portion of terrestrial planets condenses) is CO, and then graphite. If C/O>1 (actually >~0.95 due to refractory metals taking some O), then graphite will exist and condense due to its low volatility. The stable form of C at low temperatures is always CH4, of course, but the reaction between solid graphite and H2 is far too slow for conversion. The solar nebula had a mean C/O~0.5, and for C/O<~0.85 chemistry is basically the same - there is enough excess oxygen to fully oxidise all metals more electropositive than H, and thus form considerable H2O also. At intermediate C/O ratios, roughly 0.850.85. They would be easily distinguished from true gas giants by their radius, and the very low levels of H2 and He in their atmospheres. However, we know from studying the history of the Earth that its initial atmosphere was lost to space, and that its current atmosphere is almost entirely from outgassing subsequently. Hence, the large amounts of methane discussed above could be lost during this initial loss, and some ammonia would be saved by being dissolved in the silicate mantle (the solubility of NH3 in silicates should be > nonpolar CH4, due to the formation of silicamines). This is of course how Earth retained its water; and the mechanism works if the loss of primary atmosphere happens after accretion is largely complete, which is thought to be the case. I have not considered the possible chemistry of life in liquid ammonia, to avoid making this post longer than it already is. Nevertheless, in pure liquid ammonia, it seems likely that rough analogies to the important processes of terrestrial biochemistry can exist, making ammonia-based life not so different. The situation, I think, is actually worse for life in ammonia water, as the mixture is quite corrosive and many of the molecules stable either in pure water or in pure ammonia rapidly hydrolyse. It is interesting to contemplate, if there were chemists on an ammonia-based world, the developement of chemistry. For example, before electrolysis, how would our hypothetical chemists obtain acids? There are no convenient acidic minerals; I decided that to make any acid stronger than guanidine, the practical solution would be ammonium chloride from volcanoes. In an ammonia world, they would emit much higher levels of ammonia, due to the high concentrations in the mantle, and about the same level of HCl as on Earth. Of course, if this civilisation wanted to produce it on an industrial scale, electrolysis would be used I think - NaCl solution and a suitable membrane - as on Earth - would produce chlorine and sodium solutions, which give strong acid and strong base respectively. Of course there are other possibilities than ammonia and water for high-temperature liquids; however, I don't think I have enough knowledge to predict their occurrence. The likely candidates seem to be liquid CO2 and liquid sulphur. The former would require special conditions to avoid its reacting with silicates, but seems to be a possibility for life, though considerably different due to its non- polarity. [3] Sulphur, though, does not seem to be suitable for life due to its reactivity with organic substances. [4] Its presence would require some level of oxidation, as primordial sulphur exists exclusively as sulphide, but this can be provided by the familiar method - loss of water to photolysis. This must be what happened on Io, which has lost its water entirely. Are any further liquids a possibility? I mentioned guanidine above - how stable is that? I can almost imagine a dying ammonia world, now tectonically dead, having finally lost its ammonia to photolysis, with guanidine remaining behind ... [1] Almost every cold planet or moon that can hold a sufficient atmosphere should have liquid N2/CO/CH4 if the temperature is right. [2] In fact, the only elements which would be concentrated in the O- rich phase are Be, B, Mg, Al, Si, Ti, Zr, Hf. B and Ti are even questionable as they may occur most frequently as the nitrides. All other elements would be partitioned out. [3] Membranes in liquid CO2 I believe could be made from amphiphiles, but turned inside out - these would not have sufficient solubility in the cold liquids (N2, CH4). [4] Any alternatives to carbon-based chemistry would require lower temperatures, not higher, due to lower stabilities. Sulphur melts at 113C/235F. Andrew Usher === Subject: Re: Liquid ammonia in space > Liquid ammonia in space Liquids on planetary surfaces are interesting in themselves, but also > as possible media for life - possibly the only place where complex > life could arise. Only a few liquids are reasonable candidates to > exist on a planetary surface. One, we know, is water - it exists on > Earth, and is likely on any (sufficiently oxidised; see below) planet > with approximately the same temperature as Earth. The other that > exists in our solar system is liquid methane, on Titan. This is also > likely very common in the galaxy; and at still lower temperatures > liquid nitrogen is possible (which will contain substantial amounts of > methane and/or CO). [1] But perhaps more interesting are those > possible liquids that could exist at 'high' temperatures, let's say -100C/-150F, which are more likely to possibly host life - though, if > extraterrestrial life is at all common in the galaxy, I wouldn't be > surprised to find methane-based life somewhere. Liquid ammonia is the most commonly proposed alternative and I here > examine its possibilities for existence. What sort of planets might > have an ammonia ocean? The first problem is that water is far more > common than ammonia in space, less volatile, and much more stable to > photodecomposition, so one would expect any planet that had surface > ammonia to have it only as a solution in water. This would be normally > 10-15% ammonia; it could be concentrated to the eutectic (33-35%) by > freezing but this would be locked under ice. To have ammonia on the > surface, then, it is necessary for ammonia to be able to exist on the > planet, but for water to not exist. The main condition affecting the > chemical composition of terrestrial planets is the C/O ratio, though > perhaps we should better refer to the effective C/O ratio as planetary > accretion is not homogeneous. The stable form of C in a nebula at T>700 K (at which the non-volatile > portion of terrestrial planets condenses) is CO, and then graphite. If > C/O>1 (actually >~0.95 due to refractory metals taking some O), then > graphite will exist and condense due to its low volatility. The stable > form of C at low temperatures is always CH4, of course, but the > reaction between solid graphite and H2 is far too slow for conversion. > The solar nebula had a mean C/O~0.5, and for C/O<~0.85 chemistry is > basically the same - there is enough excess oxygen to fully oxidise > all metals more electropositive than H, and thus form considerable H2O > also. At intermediate C/O ratios, roughly 0.85 of graphite is (at equilibrium) prevented but reduced minerals are > still present, and water does not form at high temperatures. Water and > ammonia form at lower temperatures at _any_ C/O ratio due to the > reactions between CO/N2 and hydrogen, which is always in excess as > long as the nebula exists. Thus we will always have water and ammonia > and methane in the solar system, capable of being delivered to an > accreting planet. The mineral assemblages formed under intermediate C/ > O ratio do have an analogue in our solar system, this being the > enstatite chondrites. They are not wholly in equilibrium but they give > an idea of the solid phases that will form. Now this chemistry will affect whether this water and ammonia will > survive on the planet. Under Earth-like chemistry, both water and > ammonia are stable; though ammonia can of course be lost to > photodecomposition as it has on the terrestrial planets. If reduced > minerals are present, water is not stable due to reaction with them. > If graphite is present (and it will be on the surface due to its > density), ammonia is not stable as it is dehydrogenated by graphite; > this will be rapid under mantle conditions. Thus we see that only at > intermediate C/O ratio can we have liquid ammonia, for then ammonia is > stable but water is not. The geochemistry of such a planet would be > somewhat familiar as it, like Earth, would be differentiated into a > silicate mantle and an iron core. Because of the highly reducing > conditions, nearly all Fe,Co,Ni and chalcophiles would partition into > the core; any of these elements in the crust and mantle would likely > be part of the 'late veneer' which may account for ~0.1% of a > terrestrial planet. Thus iron would be about a thousand times rarer > than on Earth; however, it would occur as the free metal (or rather an > alloy with Co and Ni). Sulphide minerals would be common but they > would be almost entirely of the active metals, not chalcophiles. > Indeed nearly all the alkalis, Ca, Sr, Ba, rare earths would be in > sulphides, along with enough Mg to use up the rest of the sulphur. > Nitrides of Ti and Si are also stable. The silicate phase of the > planet would be in the main similar to Earth's but rather limited > chemically, due to the depletion of active metals to sulphides (and > halides) and of transition metals to the core. [2] The second problem with an ammonia ocean is photodecomposition. > Ammonia decomposes more readily than water, and the reverse reaction > is much slower due to the stability of N2. The thermal loss of > hydrogen to space makes this difference significant; on the gas > giants, where no H escapes, ammonia is in equilibrium. On the present > Earth the loss of water by this mechanism is held to an extremely low > value by a 'cold trap' mechanism where little water rises to the upper > atmosphere where it can be photolysed and thus lost. However, this > 'cold trap' is relatively ineffective for ammonia as it more easily > decomposes in the lower atmosphere and does not rapidly re-form. Thus, > for ammonia to be retained on geological time-scales it is likely > necessary for the formation of ammonia from H2 to compete with its > escape, and since light of sufficient energy to cleave N2 is found > only at the top of the atmosphere, escape rates of H there must be > quite low. This could be achieved by a sufficiently massive planet > combined with low exospheric temperature, which is not unlikely in > reduced atmospheres around M-type suns. This formation of ammonia is > also necessary for another reason: the source of the NH3 is likely to > be volcanic emission (see below) and at the conditions in a volcano of > high temperature and low pressure ammonia is mainly decomposed and > thus must form in the atmosphere anyway. The third problem with liquid ammonia, which turns out only to be a > problem under certain conditions, is methane accretion. Methane is > more stable than ammonia under all conditions; thus any ammonia > accreted is accompanied by a much larger (7-10x) amount of methane. If > a sufficient amount of ammonia were accreted to form oceans, hundreds > or thousands of bars of methane would accompany it. This methane could > not be lost without losing the ammonia also, and further, photolytic > loss of the methane would produce compounds that dehydrogenate > ammonia. However, if the methane were retained, the planet would > become in fact a miniature gas giant. These 'failed gas giants' may > actually exist in systems with C/O>0.85. They would be easily > distinguished from true gas giants by their radius, and the very low > levels of H2 and He in their atmospheres. However, we know from > studying the history of the Earth that its initial atmosphere was lost > to space, and that its current atmosphere is almost entirely from > outgassing subsequently. Hence, the large amounts of methane discussed > above could be lost during this initial loss, and some ammonia would > be saved by being dissolved in the silicate mantle (the solubility of > NH3 in silicates should be > nonpolar CH4, due to the formation of > silicamines). This is of course how Earth retained its water; and the > mechanism works if the loss of primary atmosphere happens after > accretion is largely complete, which is thought to be the case. I have not considered the possible chemistry of life in liquid > ammonia, to avoid making this post longer than it already is. > Nevertheless, in pure liquid ammonia, it seems likely that rough > analogies to the important processes of terrestrial biochemistry can > exist, making ammonia-based life not so different. The situation, I > think, is actually worse for life in ammonia water, as the mixture is > quite corrosive and many of the molecules stable either in pure water > or in pure ammonia rapidly hydrolyse. It is interesting to > contemplate, if there were chemists on an ammonia-based world, the > developement of chemistry. For example, before electrolysis, how would > our hypothetical chemists obtain acids? There are no convenient acidic > minerals; I decided that to make any acid stronger than guanidine, the > practical solution would be ammonium chloride from volcanoes. In an > ammonia world, they would emit much higher levels of ammonia, due to > the high concentrations in the mantle, and about the same level of HCl > as on Earth. Of course, if this civilisation wanted to produce it on > an industrial scale, electrolysis would be used I think - NaCl > solution and a suitable membrane - as on Earth - would produce > chlorine and sodium solutions, which give strong acid and strong base > respectively. Of course there are other possibilities than ammonia and water for > high-temperature liquids; however, I don't think I have enough > knowledge to predict their occurrence. The likely candidates seem to > be liquid CO2 and liquid sulphur. The former would require special > conditions to avoid its reacting with silicates, but seems to be a > possibility for life, though considerably different due to its non- > polarity. [3] Sulphur, though, does not seem to be suitable for life > due to its reactivity with organic substances. [4] Its presence would > require some level of oxidation, as primordial sulphur exists > exclusively as sulphide, but this can be provided by the familiar > method - loss of water to photolysis. This must be what happened on > Io, which has lost its water entirely. Are any further liquids a possibility? I mentioned guanidine above - > how stable is that? I can almost imagine a dying ammonia world, now > tectonically dead, having finally lost its ammonia to photolysis, with > guanidine remaining behind ... [1] Almost every cold planet or moon that can hold a sufficient > atmosphere should have liquid N2/CO/CH4 if the temperature is right. [2] In fact, the only elements which would be concentrated in the O- > rich phase are Be, B, Mg, Al, Si, Ti, Zr, Hf. B and Ti are even > questionable as they may occur most frequently as the nitrides. All > other elements would be partitioned out. [3] Membranes in liquid CO2 I believe could be made from amphiphiles, > but turned inside out - these would not have sufficient solubility in > the cold liquids (N2, CH4). [4] Any alternatives to carbon-based chemistry would require lower > temperatures, not higher, due to lower stabilities. Sulphur melts at > 113C/235F. Andrew Usher This post made The Best of Usenet's sci.astro.amateur, located in Moderated sci.astro.amateur (www.moderatedsciastroamateur.org) and can also be responded to at that location. Guests may reply to all topics at Moderated sci.astro.amateur. -- Martin R. Howell Moderated sci.astro.amateur www.moderatedsciastroamateur.org -- === Subject: Re: Combinatorical question I want to calculate the probability that two people, say Bob and Jane, in a bus with 14 people will sit next to each other if the people are being > seated randomly. I'm assuming the bus has 14 seats, not just 14 people, arranged in pairs so each seat is next to exactly one other seat (presumably the driver's seat doesn't count, and the driver is not a person for the purposes of this puzzle). > My idea is the calculate how many ways Bob and Jane can sit next to each > other divided by the total number of ways people can be placed in the bus. > The total number of ways 2 people can be placed next to each other among 14 people is 14! / (14 - 2)! = 182 ways. In how many of these do Bob and Jane > sit next to eachother? The solution says that the probability is 1/13, but why? This implies that > Bob and Jane can sit next to each other in 14 of the possibilities, but > why? The simplest way is to think of Bob as getting on the bus first, and choosing any seat. Then Jane gets on: she sees 13 possible seats, of which one is next to Bob, so if she chooses at random the probability that she sits next to Bob is 1/13. Then the other 12 people choose their seats, and we don't care where they sit. -- === Subject: Re: Combinatorical question > I want to calculate the probability that two people, say Bob and Jane, in > a > bus with 14 people will sit next to each other if the people are being > seated randomly. I'm assuming the bus has 14 seats, not just 14 people, arranged in pairs > so > each seat is next to exactly one other seat (presumably the driver's seat > doesn't count, and the driver is not a person for the purposes of this > puzzle). > My idea is the calculate how many ways Bob and Jane can sit next to each > other divided by the total number of ways people can be placed in the > bus. > The total number of ways 2 people can be placed next to each other among > 14 > people is 14! / (14 - 2)! = 182 ways. In how many of these do Bob and > Jane > sit next to eachother? > The solution says that the probability is 1/13, but why? This implies > that > Bob and Jane can sit next to each other in 14 of the possibilities, but > why? The simplest way is to think of Bob as getting on the bus first, and > choosing > any seat. Then Jane gets on: she sees 13 possible seats, of which one is > next > to Bob, so if she chooses at random the probability that she sits next to > Bob is 1/13. Then the other 12 people choose their seats, and we don't > care > where they sit. That sounds very reasonable. But is it also possible to derive the result the way I imagined? === Subject: Re: Combinatorical question > That sounds very reasonable. But is it also possible to derive the result > the way I imagined? There are 14! different seating orders. There are choose(14,1) = 14 different places where Bob can sit. There is choose(1,1) = 1 different places where Jane can sit next to Bob. There are 12! different places the remaining people can sit. Thus, the number of ways that Jane can sit next to Bob is 14 * 1 * 12!. So there is a 14 * 1 * 12! / 14! = 1/13 chance that Bob and Jane sit next to each other if everyone chooses a seat randomly. Dave === Subject: Re: Combinatorical question > I want to calculate the probability that two people, say Bob and Jane, in > a > bus with 14 people will sit next to each other if the people are being > seated randomly. > My idea is the calculate how many ways Bob and Jane can sit next to each > other divided by the total number of ways people can be placed in the > bus. > The total number of ways 2 people can be placed next to each other among > 14 > people is 14! / (14 - 2)! = 182 ways. In how many of these do Bob and > Jane > sit next to eachother? > The solution says that the probability is 1/13, but why? This implies > that > Bob and Jane can sit next to each other in 14 of the possibilities, but > why? The answer very much depends on how the seats are placed > in the bus, what `sit next to each other' means, and > how many seats there are... There are 14 seats in the bus. I'm sorry for not mentioning that in my first post. The problem does not say how the seats are placed, but I assume that people sit together two and two. === Subject: Re: Combinatorical question > I want to calculate the probability that two people, say Bob and Jane, in > a > bus with 14 people will sit next to each other if the people are being > seated randomly. > My idea is the calculate how many ways Bob and Jane can sit next to each > other divided by the total number of ways people can be placed in the > bus. > The total number of ways 2 people can be placed next to each other among > 14 > people is 14! / (14 - 2)! = 182 ways. In how many of these do Bob and > Jane > sit next to eachother? > The solution says that the probability is 1/13, but why? This implies > that > Bob and Jane can sit next to each other in 14 of the possibilities, but > why? >Suppose the bus is emppty when Bob steps aboard. How many >different seats can he choose from? The number of seats in the bus = 14 >Suppose Jane boards >second. How many seats can she choose from such that she >ends up next to Bob? She can choose from every seat except the one that Bob is occupying. That is 13 seats but only one of them is next to Bob? So is this why the probability is 1/13? === Subject: Re: Combinatorical question I want to calculate the probability that two people, say Bob and Jane, in > a > bus with 14 people will sit next to each other if the people are being > seated randomly. > My idea is the calculate how many ways Bob and Jane can sit next to each > other divided by the total number of ways people can be placed in the > bus. > The total number of ways 2 people can be placed next to each other among > 14 > people is 14! / (14 - 2)! = 182 ways. In how many of these do Bob and > Jane > sit next to eachother? > The solution says that the probability is 1/13, but why? This implies > that > Bob and Jane can sit next to each other in 14 of the possibilities, but > why? >Suppose the bus is emppty when Bob steps aboard. How many >different seats can he choose from? The number of seats in the bus = 14 >Suppose Jane boards >second. How many seats can she choose from such that she >ends up next to Bob? She can choose from every seat except the one that Bob is occupying. That is >13 seats but only one of them is next to Bob? So is this why the probability >is 1/13? Yes. quasi === Subject: Re: What is the name of gcd-like quantity? Suppose f(x),g(x) in Z[x] are co-prime. > Is there a name for the smallest d > 0 such that u(x) f(x) + v(x) g(x) = d for some u(x),v(x) in Z[x] ? It's fairly easy to see that d | R(f,g) (the resolvent of f and g). > I assume they are not always equal? > Is there any simple condition that ensures equality? I think you mean 'resultant', not 'resolvent'. It's simply the ideal contraction (f,g)Z[x] / Z which can be computed e.g. by Grobner bases. See also my prior post [1]. --Bill Dubuque === Subject: Re: What is the name of gcd-like quantity? > Suppose f(x),g(x) in Z[x] are co-prime. > Is there a name for the smallest d > 0 such that > > u(x) f(x) + v(x) g(x) = d > > for some u(x),v(x) in Z[x] ? > > It's fairly easy to see that d | R(f,g) (the resolvent of f and g). > I assume they are not always equal? > Is there any simple condition that ensures equality? I think you mean 'resultant', not 'resolvent'. I see that the Wikipedia agrees with you: , and I see no support for my term. However, I am reasonably sure that the term resolvent was used in the college courses I attended in my youth (a long time ago). > It's simply the ideal contraction (f,g)Z[x] / Z > which can be computed e.g. by Grobner bases. I'm not sure that definition is any clearer than what I said. In any case my question was if the concept had a name. I would have thought the Euclidean Algorithm was the easiest way to determine the result? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: What is the name of gcd-like quantity? days. My association with the Department is that of an alumnus. Suppose f(x),g(x) in Z[x] are co-prime. > Is there a name for the smallest d > 0 such that u(x) f(x) + v(x) g(x) = d for some u(x),v(x) in Z[x] ? It's fairly easy to see that d | R(f,g) (the resolvent of f and g). > I assume they are not always equal? > Is there any simple condition that ensures equality? [...] >I would have thought the Euclidean Algorithm was the easiest way >to determine the result? The Euclidean algorithm does not work in Z[x], at least not with respect to Euclidean norm. -- 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: Desperately need test bank for College Physics, Volume 2: 7th Edition, by Serway, Faughn posting-account=DN59fwoAAADDdibo8idhQLyUT6tZsChj CLR 2.0.50727),gzip(gfe),gzip(gfe) > I desperately needtestbankforCollegePhysics,Volume2:7th > Edition, bySerway,Faughn. I have looked everywhere and have only > found the solutions manual forvolume2but not thetestbank. Please > if anyone knows where I can get one or buy one I would greatly > appreciate it. Please email ASAP I have theTestBankforCollegePhysicsVolume2. I bought it for > $100 but I am willing to sell it for $80. If you want it please reply > back to me to cbehr...@satx.rr.com. If you are San Antonio, Texas we > can arrange a place, if not, I can post it on Amazona and you can > purchase it there. -Christina Book is sold.- Hide quoted text - - Show quoted text - Oh no! Ok thank you though. Do you have ANY idea where I can get another one? I am willing to pay a lot for it. === Subject: solution manuals posting-account=2D06tQoAAACGdzKiQ7VNedpD7xT38Wmc Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) can someone send me the solution manual for Engineering Economy and the Decision Making Process 2007? Please send to klai@virginia.edu === Subject: home tutoring posting-account=-jiXkQgAAAAZhbLPR1Fhvi1sNwKqv9WA FunWebProducts),gzip(gfe),gzip(gfe) I hope I will be forgiven for asking this mundane question. I am interested in doing home tutoring in math - algebra through integral calc. What is a reasonable rate to charge in MA/NH? TIA, Connie === Subject: Re: home tutoring posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > I hope I will be forgiven for asking this mundane > question. I am interested in doing home tutoring > in math - algebra through integral calc. What is > a reasonable rate to charge in MA/NH? TIA, That's going to depend greatly on your local market and your background: If there are graduate math programs in your area, you may find the supply side to be high, and hence the going rate relatively low. This is not just due graduate students on the market. Any department large enough to offer a graduate program is definitely going to have a fair number of undergraduate majors also, and many of these will be tutoring people. However, if you expect to be tutoring high school students, then past high school teaching experience will put you ahead of most of the undergraduate and graduate student competition, at least with parents, who you'll most likely be dealing with initially in the case of high school tutoring. [Notable exception: Parent(s) with relatively strong math backgrounds, especially those in a math related field but not math itself (such as electrical engineering or industrial chemistry), are more likely to look at your math background and not whether you taught high school.] I'll take a rough stab and say that, for your area, $20 per hour is probably as lowest you want to go (if you charge too little, people will think you're not any good), while it's hard for me to imagine you getting any takers for more than $80 per hour, unless you have something really unique going for you. The thing to do is try to find out what others are charging. Call teachers at local high schools -- they'll probably have a good idea. I'd guess somewhere around $40 an hour is a good start if you don't have anything particularly impressive in your background, and then go up if you don't have trouble getting people. Unless, that is, you find that the people are coming to you because you're charging $40 an hour and everyone else is charging much more. In this case, increasing your rate is more problematic. Here's an interesting collection of web pages by someone in a city I used to live in. http://www.charlottemathtutor.com/ Incidentally, I'm amazed that he's able to charge as much as he does. I was charging $5 an hour in 1981-82 and $8 an hour in 1984-86, and I was the top referred tutor at the university (meaning when a parent called the math department for a tutor, I was the most highly recommended). On the other hand, his web pages definitely show a lot of dedication to selling his services, so I suppose he falls into the category I mentioned earlier (have something really unique going for you). Dave L. Renfro === Subject: Re: Question about gambling To the extent that it is true, two factors seem most relevant ... (1) The house edge. (2) The gambler's ruin phenomenon. My point was that without a house edge (and assuming, for simplicity, >games of pure chance -- no skill required), the house eventually goes >broke against the set of _all_ players, assuming the total betting >capital dedicated to gambling against the casino by the set of all >players exceeds the capital of the casino. In other words, the gambler's ruin phenomenon is not what makes >Casinos profitable. It's the house edge (paired with poor play by the >players). >Interesting point but very suspect. >Now the question becomes, can a large number of players ruin the house >by the cumulative effect of their betting? Assuming (1) no house edge (2) games of no skill (hence optimal play is forced) (3) the total capital of the set of players exceeds that of the house then yes, it's likely (i.e. -- probability > 1/2) that house will go >bankrupt before the set of players loses all its capital. Effectively, the set of players can be regarded as a single player, >hence the house is now in a losing random walk. Let m,n be positive integers, m > n. Consider the following scenario ... (1) The house has capital of n dollars (2) There m players lined up outside the casino, each waiting for >their chance to play, each having only 1 dollar. (3) Only one player is allowed in the casino at a time. The player >must keep on playing until either he/she has no money left or the >house is bankrupt. (4) Similarly the house must give each of the m players a chance to >play. As soon as one player taps out, the next player is invited in. (5) For simplicity, all bets are exactly 1 dollar, paying even money >-- thus, no house edge. Now consider the wealth w(t) of the house at time t (after t totals >bets, regardless of which player). Just an ordinary random walk with >absorbing barriers at w = -n and w = m. Thus, the house goes bankrupt with probability > 1/2. Bottom line ... Barring suboptimal play by the players, and without a house edge, the >house is more than 50-50 to go bankrupt if the set of players has more >capital than the house. > Now if A places a single bet using his entire stake a at odds that are chosen such that R = b, the payout will equal B's capital, breaking the bank. For this case, a little algebra leads to Pa = a/(a + b/beta). Which implies, using your scenario, that if the house has an edge beta, the house would go bankrupt with a 50-50 probability if the set of players has more capital that the house times 1/beta. So where is the advantage of the house edge. It seems no different from the house's advantage of capital. John === Subject: Re: Question about gambling >To the extent that it is true, two factors seem most relevant ... >(1) The house edge. >(2) The gambler's ruin phenomenon. >My point was that without a house edge (and assuming, for simplicity, >games of pure chance -- no skill required), the house eventually goes >broke against the set of _all_ players, assuming the total betting >capital dedicated to gambling against the casino by the set of all >players exceeds the capital of the casino. >In other words, the gambler's ruin phenomenon is not what makes >Casinos profitable. It's the house edge (paired with poor play by the >players). Interesting point but very suspect. >Now the question becomes, can a large number of players ruin the house >by the cumulative effect of their betting? >Assuming >(1) no house edge >(2) games of no skill (hence optimal play is forced) >(3) the total capital of the set of players exceeds that of the house >then yes, it's likely (i.e. -- probability > 1/2) that house will go >bankrupt before the set of players loses all its capital. >Effectively, the set of players can be regarded as a single player, >hence the house is now in a losing random walk. >Let m,n be positive integers, m > n. >Consider the following scenario ... >(1) The house has capital of n dollars >(2) There m players lined up outside the casino, each waiting for >their chance to play, each having only 1 dollar. >(3) Only one player is allowed in the casino at a time. The player >must keep on playing until either he/she has no money left or the >house is bankrupt. >(4) Similarly the house must give each of the m players a chance to >play. As soon as one player taps out, the next player is invited in. >(5) For simplicity, all bets are exactly 1 dollar, paying even money >-- thus, no house edge. >Now consider the wealth w(t) of the house at time t (after t totals >bets, regardless of which player). Just an ordinary random walk with >absorbing barriers at w = -n and w = m. >Thus, the house goes bankrupt with probability > 1/2. >Bottom line ... >Barring suboptimal play by the players, and without a house edge, the >house is more than 50-50 to go bankrupt if the set of players has more >capital than the house. Now if A places a single bet using his entire stake a at odds that >are chosen such that R = b, the payout will equal B's capital, >breaking the bank. For this case, a little algebra leads to Pa = a/(a >+ b/beta). Which implies, using your scenario, that if the house has an edge >beta, the house would go bankrupt with a 50-50 probability if the set >of players has more capital that the house times 1/beta. So where is >the advantage of the house edge. It seems no different from the >house's advantage of capital. The betting limits (and the maximum number of customers who can play at a given time) is what protects the house edge from gambler's ruin, allowing the house edge to dominate. The house edge _trumps_ the superior capital of the set of players, provided the house can set a sufficiently low value for the maximum bet -- low relative to the size of their capital. It also depends on the strength of the house edge -- the higher their edge, the higher they can set the maximum bet relative to their total capital. Thus, without an edge, the house is doomed (against the set of all players, assuming the players have more total capital than the house). With an edge, they only need to protect themselves from the gambler's ruin phenomenon, which they can do by setting appropriate betting limits. quasi === Subject: Re: Question about gambling >Which implies, using your scenario, that if the house has an edge >beta, the house would go bankrupt with a 50-50 probability if the set >of players has more capital that the house times 1/beta. So where is >the advantage of the house edge. It seems no different from the >house's advantage of capital. The betting limits (and the maximum number of customers who can play >at a given time) is what protects the house edge from gambler's ruin, >allowing the house edge to dominate. The house edge _trumps_ the >superior capital of the set of players, provided the house can set a >sufficiently low value for the maximum bet -- low relative to the size >of their capital. It also depends on the strength of the house edge -- >the higher their edge, the higher they can set the maximum bet >relative to their total capital. Thus, without an edge, the house is doomed (against the set of all >players, assuming the players have more total capital than the house). With an edge, they only need to protect themselves from the gambler's >ruin phenomenon, which they can do by setting appropriate betting >limits. P(A ruined) = 1-(1-((1-p)/p)^a)/(1-((1-p)/p)^(a+b) (where p is the probability of winning, a is the house capital and b is the player's capital. Adding assumptions: total capital of gamblers greater than the capital of the house, betting limits, etc doesn't prove the case. The house is doomed if their capital is less than the normalized capital of the collective population of gamblers. Gambler's Ruin provides the basis for calculating the result. === Subject: Re: Question about gambling > >Which implies, using your scenario, that if the house has an edge >beta, the house would go bankrupt with a 50-50 probability if the set >of players has more capital that the house times 1/beta. So where is >the advantage of the house edge. It seems no different from the >house's advantage of capital. The betting limits (and the maximum number of customers who can play >at a given time) is what protects the house edge from gambler's ruin, >allowing the house edge to dominate. The house edge _trumps_ the >superior capital of the set of players, provided the house can set a >sufficiently low value for the maximum bet -- low relative to the size >of their capital. It also depends on the strength of the house edge -- >the higher their edge, the higher they can set the maximum bet >relative to their total capital. Thus, without an edge, the house is doomed (against the set of all >players, assuming the players have more total capital than the house). With an edge, they only need to protect themselves from the gambler's >ruin phenomenon, which they can do by setting appropriate betting >limits. P(A ruined) = 1-(1-((1-p)/p)^a)/(1-((1-p)/p)^(a+b) (where p is the > probability of winning, a is the house capital and b is the player's > capital. i.e. p is the probability of the house winning, A is the house. > Adding assumptions: total capital of gamblers greater than the capital > of the house, betting limits, etc doesn't prove the case. The house > is doomed if their capital is less than the normalized capital of the > collective population of gamblers. Gambler's Ruin provides the basis > for calculating the result. Let w = (1-p)/p, so we're assuming w < 1 and P(A ruined) = 1 - (1-w^a)/(1-w^(a+b)). This formula is assuming each bet is 1 in the units used for a and b. If the house changes the amount bet to epsilon, the effect is the same as changing a and b to a/epsilon and b/epsilon. Now as epsilon -> 0, w^(a/epsilon) -> 0 and w^((a+b)/epsilon) -> 0 so P(A ruined) -> 1 - 1/1 = 0, as quasi said. -- === Subject: Re: Question about gambling Which implies, using your scenario, that if the house has an edge >beta, the house would go bankrupt with a 50-50 probability if the set >of players has more capital that the house times 1/beta. So where is >the advantage of the house edge. It seems no different from the >house's advantage of capital. >The betting limits (and the maximum number of customers who can play >at a given time) is what protects the house edge from gambler's ruin, >allowing the house edge to dominate. The house edge _trumps_ the >superior capital of the set of players, provided the house can set a >sufficiently low value for the maximum bet -- low relative to the size >of their capital. It also depends on the strength of the house edge -- >the higher their edge, the higher they can set the maximum bet >relative to their total capital. >Thus, without an edge, the house is doomed (against the set of all >players, assuming the players have more total capital than the house). >With an edge, they only need to protect themselves from the gambler's >ruin phenomenon, which they can do by setting appropriate betting >limits. P(A ruined) = 1-(1-((1-p)/p)^a)/(1-((1-p)/p)^(a+b) (where p is the >probability of winning, a is the house capital and b is the player's >capital. Adding assumptions: total capital of gamblers greater than the capital >of the house, betting limits, etc doesn't prove the case. The house >is doomed if their capital is less than the normalized capital of the >collective population of gamblers. No, that's wrong. The formula above is based on a fixed step size of 1. Betting limits can be set so that the step size is less than one. Alternatively, the values of a and b can be scaled up proportionally, leaving the step size as 1. With the following assumptions, the house is favored _not_ to go broke before the player does. (1) The house starts with a dollars, the player starts with b dollars, where b > a, and the house knows the value of b (and of course, the house also knows the value of a). (2) p > 1/2 (i.e. -- the house has an edge), and the value of p is known to the house. (3) The house can set the bet size in advance to any positive real number (i.e. -- possibly less than 1). The bet size is announced by the house before the play begins, and is not allowed to be changed during the game as the wealth changes. (4) In each game, the player bets one betting unit as specified by the house, as indicated in assumption (3). Let me reemphasize -- one betting unit can be less than one dollar (this is key). The payoff is also one betting unit. Of course, each game favors the house based on the value of p. (5) Play continues until either the player or the house is bankrupt. Here's the claim: Given the assumptions as specified above, the house can set the betting unit sufficiently low to achieve _any_ probability greater than 1/2 and less than 1 (i.e. -- as close to 1 as desired) of winning the complete contest. To recap ... To insure the house _never_ goes bankrupt -- well, almost never (probability arbitrarily close to 0), all the house needs is (1) p > 1/2 (2) known values of a,b (3) ability to set the betting limits to an arbitrarily small positive real number (to be chosen based on p,a,b and the desired probability of winning the complete contest). Bottom line: An edge together with control of betting limits trumps gambler's ruin. quasi === Subject: Re: poor mathemathicians posting-account=W7RQ6gkAAACLDC5JWapQU1UV8ot8KkAS 1.0.3705; Dealio Toolbar 3.1.1),gzip(gfe),gzip(gfe) > hi , > what is the best way for a poor mathematician that has not enough time > to work at > mathematics: > 1. use drug to decrease my mental activity. > 2. continue with the sense of death often. It's not that difficult. Just come to the realizarion that what the vast plurality of mathematicians study, isn't mathematics, but rather some very off-the-wall theories of evolution. === Subject: Re: Question regarding squared norm posting-account=kZPjSwoAAAAjy1fDQ4KhxZbo4WaXTVGd InfoPath.1; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) hi, i would definitely eloborate more this is basically an optimization problem min || SX||^2 where S is a matrix consisting of vectors s1,s2,s3 so S=[s1,s2,s3] where s1,s2,s3 are vectors s1=[a11,a21,a31]^T s2=[a12,a22,a32]^T s3=[a13,a23,a33]^T where as X=[x1,x2,x3] is a row vector or matrix of 1*3 stacking the s1,s2,s3 vectors in to S matrix i get a11 a12 a13 a21 a22 a23 a31 a32 a33 and SX equals a11 a12 a13 x1 a21 a22 a23 x2 a31 a32 a33 x3 SX= a11*x1+a12*x2+a13*x3 a21*x1+a22*x2+a23*x3 a31*x1+a32*x2+a33*x3 i solves up to here but what is squared norm of the above matrix i-e || SX||^2 hope now you will be able to understand what i meant to say === Subject: Re: Question regarding squared norm posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > hi, i would definitely eloborate more > this is basically an optimization problem min || SX||^2 where S is a matrix consisting of vectors s1,s2,s3 so S=[s1,s2,s3] where s1,s2,s3 are vectors s1=[a11,a21,a31]^T > s2=[a12,a22,a32]^T > s3=[a13,a23,a33]^T where as X=[x1,x2,x3] is a row vector or matrix of 1*3 OK, but it had better be a column vector, since if not the product SX is not defined. stacking the s1,s2,s3 vectors in to S matrix i get No, stacking them would indicate they are row vectors; you want them to be the columns of S, so you concatenate them (put them side-by- side). a11 a12 a13 > a21 a22 a23 > a31 a32 a33 and SX equals a11 a12 a13 x1 > a21 a22 a23 x2 > a31 a32 a33 x3 SX= a11*x1+a12*x2+a13*x3 > a21*x1+a22*x2+a23*x3 > a31*x1+a32*x2+a33*x3 i solves up to here but what is squared norm of the above matrix i-e || > SX||^2 > hope now you will be able to understand what i meant to say Actually, I was pretty sure all along what you meant, but I was absolutely not going to help you until you took the time to present your problem properly. OK, so ||SX||^2 = (a11*x1 + a12*x2 + a13*x3)^2 + (a21*x1 + a22*x2 + a23*x3)^2 + (a31*x1 + a32*x2 + a33*x3)^2. I presume there are some other, unstated, constraints on X, because if not the minimum occurs at x1 = x2 = x3 = 0. (If the rank of the matrix S is less than three, there may also be some other minima in which not all the xj are zero; you just need (SX)[1] = a11*x1 + a12*x2 + a13*x3 = 0, etc.) If there are linear constraints on the xj, you have a convex quadratic programming problem, for which there exist many algorithms. Google 'quadratic program' to turn up lots of links. For example, see http://en.wikipedia.org/wiki/Quadratic_programming (especially the link NEOS Guide to QP). A simplex type method for quadratic programs is developed and illustrated by example in http://www.me.utexas.edu/~jensen/ORMM/supplements/methods/nlpmethod/S2_quadr atic.pdf If you have /nonlinear/ constraints the problem may be much harder. If you have access to the EXCEL spreadsheet, you can use the Solver tool to formulate and solve such problems with reasonable reliability (reliability, at least, for the case of linear constraints). R.G. Vickson === Subject: Re: Question regarding squared norm posting-account=SQDT2QoAAAARANqaHM5kmoBtkqydZTV9 InfoPath.1; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) hi, i would definitely eloborate more > æthis is basically an optimization problem min || SX||^2 where S is a matrix consisting of vectors s1,s2,s3 so S=[s1,s2,s3] where s1,s2,s3 are vectors s1=[a11,a21,a31]^T > s2=[a12,a22,a32]^T > s3=[a13,a23,a33]^T where as X=[x1,x2,x3] is a row vector or matrix of 1*3 OK, but it had better be a column vector, since if not the product SX > is not defined. stacking the s1,s2,s3 vectors in to S matrix i get No, stacking them would indicate they are row vectors; you want them > to be the columns of S, so you concatenate them (put them side-by- > side). a11 a12 a13 > a21 a22 a23 > a31 a32 a33 and SX equals a11 a12 a13 æ æx1 > a21 a22 a23 æ æx2 > a31 a32 a33 æ æx3 SX= æ æ æ æ æa11*x1+a12*x2+a13*x3 > æ æ æ æ æ æ æa21*x1+a22*x2+a23*x3 > æ æ æ æ æ æ æa31*x1+a32*x2+a33*x3 i solves up to here but what is squared norm of the above matrix i-e || > SX||^2 > hope now you will be able to understand what i meant to say Actually, I was pretty sure all along what you meant, but I was > absolutely not going to help you until you took the time to present > your problem properly. OK, so ||SX||^2 = (a11*x1 + a12*x2 + a13*x3)^2 + (a21*x1 + a22*x2 + > a23*x3)^2 + (a31*x1 + a32*x2 + a33*x3)^2. I presume there are some other, unstated, constraints on X, because if > not the minimum occurs at x1 = x2 = x3 = 0. (If the rank of the matrix > S is less than three, there may also be some other minima in which not > all the xj are zero; you just need (SX)[1] = a11*x1 + a12*x2 + a13*x3 > = 0, etc.) If there are linear constraints on the xj, you have a > convex quadratic programming problem, for which there exist many > algorithms. Google 'quadratic program' to turn up lots of links. For > example, seehttp://en.wikipedia.org/wiki/Quadratic programming > (especially the link NEOS Guide to QP). A simplex type method for > quadratic programs is developed and illustrated by example inhttp://www.me.utexas.edu/~jensen/ORMM/supplements/methods/nlpmethod/S... > If you have /nonlinear/ constraints the problem may be much harder. If you have access to the EXCEL spreadsheet, you can use the Solver > tool to formulate and solve such problems with reasonable reliability > (reliability, at least, for the case of linear constraints). R.G. Vickson- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - thankyou sir very much yes you were very right ....i wasnt actually presenting my problem properly .......... i have two constraints on the above stated problem 1) 0.52<=X<= 1.47 /* where X represents allthe elements of X i- e x1,x2,x3 2) x1^2+x2^2+x3^2=3 so my problem becomes x1 >= 0.52, <= 1.47; x2 >= 0.52, <= 1.47; x3 >= 0.52, <= 1.47; x4 >= 0.52, <= 1.47; minimize : ( ((-0.22*x1)+(0.13*x2)+(-0.09*x3)+(0.07*x4))^2 + ((-0.13*x1)+(0.09*x2)+(-0.07*x3)+(0.058*x4))^2 +((-0.09*x1)+(0.07*x2)+ (-0.058*x3)+(0.049*x4))^2); subject to: -4+x1^2+x2^2+x3^2+x4^2=0; which tool should i try .......( ineed your suggestion because i tried neosolver but i didnt get the right answers) === Subject: Re: Question regarding squared norm posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > hi, > i would definitely eloborate more > this is basically an optimization problem > min || SX||^2 > where S is a matrix consisting of vectors s1,s2,s3 > so S=[s1,s2,s3] > where s1,s2,s3 are vectors > s1=[a11,a21,a31]^T > s2=[a12,a22,a32]^T > s3=[a13,a23,a33]^T > where as X=[x1,x2,x3] is a row vector or matrix of 1*3 OK, but it had better be a column vector, since if not the product SX > is not defined. > stacking the s1,s2,s3 vectors in to S matrix i get No, stacking them would indicate they are row vectors; you want them > to be the columns of S, so you concatenate them (put them side-by- > side). > a11 a12 a13 > a21 a22 a23 > a31 a32 a33 > and SX equals > a11 a12 a13 x1 > a21 a22 a23 x2 > a31 a32 a33 x3 > SX= a11*x1+a12*x2+a13*x3 > a21*x1+a22*x2+a23*x3 > a31*x1+a32*x2+a33*x3 > i solves up to here but what is squared norm of the above matrix i-e || > SX||^2 > hope now you will be able to understand what i meant to say Actually, I was pretty sure all along what you meant, but I was > absolutely not going to help you until you took the time to present > your problem properly. OK, so ||SX||^2 = (a11*x1 + a12*x2 + a13*x3)^2 + (a21*x1 + a22*x2 + > a23*x3)^2 + (a31*x1 + a32*x2 + a33*x3)^2. I presume there are some other, unstated, constraints on X, because if > not the minimum occurs at x1 = x2 = x3 = 0. (If the rank of the matrix > S is less than three, there may also be some other minima in which not > all the xj are zero; you just need (SX)[1] = a11*x1 + a12*x2 + a13*x3 > = 0, etc.) If there are linear constraints on the xj, you have a > convex quadratic programming problem, for which there exist many > algorithms. Google 'quadratic program' to turn up lots of links. For > example, seehttp://en.wikipedia.org/wiki/Quadratic_programming > (especially the link NEOS Guide to QP). A simplex type method for > quadratic programs is developed and illustrated by example inhttp://www.me.utexas.edu/~jensen/ORMM/supplements/methods/nlpmethod/S... > If you have /nonlinear/ constraints the problem may be much harder. If you have access to the EXCEL spreadsheet, you can use the Solver > tool to formulate and solve such problems with reasonable reliability > (reliability, at least, for the case of linear constraints). R.G. Vickson- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - thankyou sir very much > yes you were very right ....i wasnt actually presenting my problem > properly .......... i have two constraints on the above stated problem 1) 0.52<=X<= 1.47 /* where X represents allthe elements of X i- > e x1,x2,x3 > 2) x1^2+x2^2+x3^2=3 so my problem becomes x1 >= 0.52, <= 1.47; > x2 >= 0.52, <= 1.47; > x3 >= 0.52, <= 1.47; > x4 >= 0.52, <= 1.47; minimize : ( ((-0.22*x1)+(0.13*x2)+(-0.09*x3)+(0.07*x4))^2 + > ((-0.13*x1)+(0.09*x2)+(-0.07*x3)+(0.058*x4))^2 +((-0.09*x1)+(0.07*x2)+ > (-0.058*x3)+(0.049*x4))^2); > subject to: > -4+x1^2+x2^2+x3^2+x4^2=0; which tool should i try .......( ineed your suggestion because i tried > neosolver but i didnt get the right answers) > I have already suggested that you could try the EXCEL Solver (but it might not be reliable); solving several times from several randomly- generated starting points might be a way to increase the reliability. Alternatively, you can try putting the nonlinear constraint up into the objective, using a Lagrange multipllier and/or a penalty function; the best method would be to use BOTH a penalty and a multiplier. To minimize f(x) subject to a <= x <= b and g(x) = 0 (where g is a nonlinear function), the so-called Augmented Lagrangian would be to minimize F(x) = f(x) + u*g(x) + M*[g(x)]^2, subject to a <= x <= b. Here, u is essentially the Lagrange multiplier and M > 0 is the penalty parameter. You need to read some material on this method---it is too lengthy to be dealt with well in a newsgroup. See, eg., http://www.cs.ubc.ca/~ascher/542/chap10.pdf or http://www-fp.mcs.anl.gov/otc/Guide/OptWeb/continuous/constrained/nonlinearc on/auglag.html or http://www.csee.umbc.edu/~liyiou1/YLiGRC04.pdf or http://scholar.lib.vt.edu/theses/available/etd-05022001-131450/unrestricted/ etd.pdf Basically, the method is iterative, with values of u and/or M being adjusted from one step to the next. In each step we have a problem with a nonlinear objective and linear constraints (upper and lower bounds in your case), and there are effective methods for dealing with that type of problem. R.G. Vickson === Subject: Re: finding radius of inscribed triangle posting-account=5lDnxwoAAAB1g4JJgAmaTok_iTr9_MSe .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > How does a triangle come to have a radius? Sloppy wording on my part. === Subject: Re: finding radius of inscribed triangle posting-account=5lDnxwoAAAB1g4JJgAmaTok_iTr9_MSe .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) >I'm trying to find the radius from >the following image,http://i28.tinypic.com/t98az9.jpg What's the catch with this problem? >I'm assuming there is some way >to relate the size of the blocks >with the arc length to find the radius? Here's a solution using Maple ... æ æ>a:=Pi-arcsin((-r+1)/r); > æ æ>b:=2*Pi+arcsin((-r+2)/r); > æ æ>eq:=r*(b-a)=6; > æ æ>r=fsolve(eq,r); result: r = 2.768080732 > How did you derive arcsin((-r+1)/r) and arcsin((-r+2)/r) ? I wish to understand how that information is figured out based on the image. === Subject: Re: finding radius of inscribed triangle >I'm trying to find the radius from >the following image,http://i28.tinypic.com/t98az9.jpg >What's the catch with this problem? >I'm assuming there is some way >to relate the size of the blocks >with the arc length to find the radius? > Here's a solution using Maple ... > æ æ>a:=Pi-arcsin((-r+1)/r); > æ æ>b:=2*Pi+arcsin((-r+2)/r); > æ æ>eq:=r*(b-a)=6; > æ æ>r=fsolve(eq,r); > result: r = 2.768080732 How did you derive arcsin((-r+1)/r) >and arcsin((-r+2)/r) ? First, let me say -- the solution given by Philippe 92 is better -- more geometric. The way I did it was to use polar coordinates. Placing the origin at the center of the circle, let the points of contact of the pipe with the blocks of height 1 and 2 be (r*cos(a),r*sin(a)) and (r*cos(b),r*sin(b)), respectively, where a,b satisfy the constraints Pi <= a < (3/2)*Pi and (3/2)*Pi < b <= 2*Pi Focusing just on the y-cordinates, r*sin(a) = - (r - 1) r*sin(b) = - (r - 2) Thus sin(a) = (-r + 1)/r sin(b) = (-r + 2)/r Since a is in quadrant 3, a = Pi - arcsin((-r + 1)/r) Since b is in quadrant 4, b = 2*Pi + arcsin((-r + 2)/r) Given the arclength of 6, as shown in the diagram, we get the equation r*(b-a) = 6 Replacing a,b by the expressions above, and then solving (numerically) for r, yields the approximate value r = 2.768080732 quasi === Subject: Re: finding radius of inscribed triangle posting-account=5lDnxwoAAAB1g4JJgAmaTok_iTr9_MSe .NET CLR 2.0.50727; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) >I'm trying to find the radius from >the following image,http://i28.tinypic.com/t98az9.jpg >What's the catch with this problem? >I'm assuming there is some way >to relate the size of the blocks >with the arc length to find the radius? > Here's a solution using Maple ... > æ æ>a:=Pi-arcsin((-r+1)/r); > æ æ>b:=2*Pi+arcsin((-r+2)/r); > æ æ>eq:=r*(b-a)=6; > æ æ>r=fsolve(eq,r); > result: r = 2.768080732 How did you derive arcsin((-r+1)/r) >and arcsin((-r+2)/r) ? First, let me say -- the solution given by Philippe 92 is better -- > more geometric. The way I did it was to use polar coordinates. Placing the origin at the center of the circle, let the points of > contact of the pipe with the blocks of height 1 and 2 be > (r*cos(a),r*sin(a)) and (r*cos(b),r*sin(b)), respectively, where a,b > satisfy the constraints Pi <= a < (3/2)*Pi and (3/2)*Pi < b <= 2*Pi Focusing just on the y-cordinates, æ ær*sin(a) = - (r - 1) æ ær*sin(b) = - (r - 2) Thus æ æsin(a) = (-r + 1)/r æ æsin(b) = (-r + 2)/r Since a is in quadrant 3, a = Pi - arcsin((-r + 1)/r) Since b is in quadrant 4, b = 2*Pi + arcsin((-r + 2)/r) Given the arclength of 6, as shown in the diagram, we get the equation æ ær*(b-a) = 6 Replacing a,b by the expressions above, and then solving (numerically) > for r, yields the approximate value æ ær = 2.768080732 quasi- Hide quoted text - > Your method is clever. While your method was oriented around polar coordinates, what was Philippe's method based on? In particular, I can't make sense of his argument to arccos, why the 1 - r/1 and 1-r/2? Why divide radius by 1 and 2? etc. === Subject: Re: finding radius of inscribed triangle >I'm trying to find the radius from >the following image,http://i28.tinypic.com/t98az9.jpg >What's the catch with this problem? >I'm assuming there is some way >to relate the size of the blocks >with the arc length to find the radius? > Here's a solution using Maple ... > æ æ>a:=Pi-arcsin((-r+1)/r); > æ æ>b:=2*Pi+arcsin((-r+2)/r); > æ æ>eq:=r*(b-a)=6; > æ æ>r=fsolve(eq,r); > result: r = 2.768080732 >How did you derive arcsin((-r+1)/r) >and arcsin((-r+2)/r) ? > First, let me say -- the solution given by Philippe 92 is better -- > more geometric. > The way I did it was to use polar coordinates. > Placing the origin at the center of the circle, let the points of > contact of the pipe with the blocks of height 1 and 2 be > (r*cos(a),r*sin(a)) and (r*cos(b),r*sin(b)), respectively, where a,b > satisfy the constraints Pi <= a < (3/2)*Pi and (3/2)*Pi < b <= 2*Pi > Focusing just on the y-cordinates, > æ ær*sin(a) = - (r - 1) > æ ær*sin(b) = - (r - 2) > Thus > æ æsin(a) = (-r + 1)/r > æ æsin(b) = (-r + 2)/r > Since a is in quadrant 3, a = Pi - arcsin((-r + 1)/r) > Since b is in quadrant 4, b = 2*Pi + arcsin((-r + 2)/r) > Given the arclength of 6, as shown in the diagram, we get the equation > æ ær*(b-a) = 6 > Replacing a,b by the expressions above, and then solving (numerically) > for r, yields the approximate value > æ ær = 2.768080732 Your method is clever. to find angles given enough information. >While your method was oriented around polar coordinates, what was >Philippe's method based on? Well, I'll guess his steps ... Let a,b be the central angles of the 2 little sectors. Extend a horizontal line segment from the left contact point to yield a right triangle with angle a at the top, hypotenuse of length r, and vertical leg of length r - 1. Thus, cos(a) = (r - 1)/r, so a = arccos((r - 1)/r) Do the same for the second sector, yielding b = arrcos((r - 2)/r). Finally, r*(a+b) = 6 is used to solve for r. Thus, the solution involves just lengths and angles, no coordinates. Of course, the above is how Philippe's solution _appears_ to work, but let's see what Philippe says. quasi === Subject: Re: finding radius of inscribed triangle > I'm trying to find the radius from > the following image,http://i28.tinypic.com/t98az9.jpg > What's the catch with this problem? > I'm assuming there is some way > to relate the size of the blocks > with the arc length to find the radius? > While your method was oriented around polar coordinates, what was > Philippe's method based on? Well, I'll guess his steps ... Let a,b be the central angles of the 2 little sectors. Extend a horizontal line segment from the left contact point to yield > a right triangle with angle a at the top, hypotenuse of length r, and > vertical leg of length r - 1. Thus, cos(a) = (r - 1)/r, so a = arccos((r - 1)/r) Do the same for the second sector, yielding b = arccos((r - 2)/r). Finally, r*(a+b) = 6 is used to solve for r. Thus, the solution involves just lengths and angles, no coordinates. Of course, the above is how Philippe's solution _appears_ to work, > but let's see what Philippe says. > You got it. Then (r-1)/r is written as 1 - 1/r in my formulas. -- Philippe C., mail : chephip+news@free.fr site : http://chephip.free.fr/ (recreational mathematics) === === === === === Subject: Is checking necessary when the principle of powers is used with an odd power n? Why or why not? posting-account=qmp-BQoAAADCEZibMUeeXL2YDY54cONR 4.2.1.0; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) This question was posed to our class and so far, no one, including me, can give a direct answer and explanation as to why or why not... === Subject: Re: Is checking necessary when the principle of powers is used with an odd power n? Why or why not? >This question was posed to our class and so far, no one, including me, >can give a direct answer and explanation as to why or why not... What principle? Can you give some examples to illustrate the issue? quasi === Subject: Re: Is checking necessary when the principle of powers is used with an odd power n? Why or why not? >This question was posed to our class and so far, no one, including me, >can give a direct answer and explanation as to why or why not... > What principle? > Can you give some examples to illustrate the issue? He might mean: if for real x,y and integer n > 0 x^n = y^n implies x=y if n is even, you need to check if x and y have opposite signs if n is odd, you do not need the check === Subject: Re: Is checking necessary when the principle of powers is used with an odd power n? Why or why not? >This question was posed to our class and so far, no one, including me, >can give a direct answer and explanation as to why or why not... > What principle? > Can you give some examples to illustrate the issue? He might mean: > if for real x,y and integer n > 0 x^n = y^n implies x=y if n is even, you need to check if x and y have opposite signs if n is odd, you do not need the check Yes, I realized the OP might mean something like that. But I think it's reasonable to expect the OP (presumably a teacher) to state the question more precisely, and hopefully give an example or two to make the issue clear. Thus, in this case, rather than try to guess the actual question (I often try to do that, but not this time), I chose to wait for a clarification. quasi === Subject: =?ISO-8859-1?Q?analyse_critique_des_m=E9thodologies_inn=E9istes_des_?= =?ISO-8859-1?Q?successeurs_de_Piaget?= <47bc8649$0$864$ba4acef3@news.orange.fr> posting-account=1qbAGAkAAADcUtlizzXUEb5jUjfAdE2y Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > Avez vous lu ceci ?http://www.lire.fr/critique.asp/idC=51596/idR=213/idG=8 Je reviens .88 l'instant de la page Dehaene (Stanislas) http://www.college-de-france.fr/media/psy cog/UPL29739 Cours2008 1SD.pdf Conclusions * Au moins trois syst.8fmes distincts contribuent au << sens du nombre >: 1. La subitisation pour les ensembles de 1, 2 ou 3 objets 2. L'estimation de la num.8erosit.8e, bien au-del.88 de 3 objets 3. Le comptage, fond.8e sur la correspondance terme-.88-terme, pour parvenir .88 la cardinalit.8e exacte * La subitisation et l'estimation sont pr.8esents tr.8fs pr.8ecocement chez l'enfant, et existent .8egalement chez de nombreuses esp.8fces animales * Ces processus num.8eriques conf.8frent .88 l'enfant, tr.8fs pr.8ecocement, le << sens du nombre > et une capacit.8e de calcul approximatif * A l'.89ge adulte, nous continuons .88 acc.8eder rapidement .88 la repr.8esentation analogique des nombres, m.90me lorsque ceux-ci nous sont pr.8esent.8es sous forme de symboles Dans ce fouillis on ne voit pas clairement la diff.8erenciation des diverses strat.8egies possibles - ordinales ou cardinale - et l'.8evaluation de leur efficacit.8e. On en est au m.90me point qu'avec Piaget : conclusions h.89tives d'exp.8eriences b.89cl.8ees. Quand il faut mettre en oeuve des STRATEGIES d'appuis p.8edagogiques pour transmettre ou faire acqu.8erir les savoirs, les comp.8etences. CES gens l.88 en sont encore ... .88 relever les indices d'une esp.8fce d'inn.8eisme dans la construction du nombre La courbe de Dehane sur les performances de reconnaissance de 4 ou 5 .8el.8ements (chute consid.8erable de la performance) a bien fait rire celle qui utilise la m.8ethode des doigts avec les .8el.8fves : Les siens reussissent .88 montrer 5 doigts mieux plus vite et mieux que 3 doigts Un imb.8ecile dirait (je l'entend au fond ) - oui ? - mais !!! les cinq doigts , cela fait la main enti.8fre c'est plus facile l !!! Et alors, esp.8fce de cr.8etin !!! (on est revenu 50 ans en arri.8fre ! les m.90mes idioties que Piaget En supposant que tous les enfants assimilent 5 plus vite que 3 parce cela correspond .88 la main enti.8fre (et QU'UN PEDAGOGUE INTELLIGENT LEUR CONSEILLE DE L'UTILISER !!!) Quand bien m.90me cela se prolongerait et.... que les enfants toute leur vie verraient plus facilement un groupe de 5 qu'un groupe dee 3 ... quelle cons.8equence cela aurait il ??? Prouver la fausset.8e des sources de Dehaene ? on s'en fout La solution passe par la construction d'outils p.8edagogiques et l'.8evaluation de masse de r.8esultats qui en sont obtenus (sans pr.8ejuger du fait que des r.8esultats importants puissent ne pas .90tre per.8dus) Yanick Toutain Je n'envisage que difficilement le fait que quelqu'un qui gagne peut- .90tre 10 fois 537 euros puisse avoir l'humilit.8e d'aller chercher des CONSEILS chez ceux qui s'y connaissent (m.90me connaisseur d.8ebutante pourvue d'une solide connaissance de la transformation quantit.8e qualit.8e !) -------------------------------- PS POUR LES CRETINS : voil.88 j'AI PUBLIE ! Ca s'appelle analyse critique des m.8ethodologies inn.8eistes des successeurs de Piaget abstract: Les partisans de Piaget croient qu'il existe une construction du nombre a-historique et a-p.8edagogique, ind.8ependament de la qualit.8e, de la productivit.8e des m.8ethodes et outils mis .88 la disposition des apprenants par les chercheurs- constructeurs d'outils p.8edagogiques ! === Subject: Re: analyse critique des m.8ethodologies inn.8eistes des successeurs de Piaget > La courbe de Dehane sur les performances de reconnaissance de 4 ou 5 > .8el.8ements (chute consid.8erable de la performance) a bien fait rire celle > qui utilise la m.8ethode des doigts avec les .8el.8fves : Les siens > reussissent .88 montrer 5 doigts mieux plus vite et mieux que 3 doigts C'est pourtant une constante .8etudi.8ee par les neurologues sur le cerveau de diff.8erents animaux. Jusqu'.88 3-4, on .8evalue sans compter. Apr.8fs il faut compter. === Subject: =?ISO-8859-1?Q?Re:_analyse_critique_des_m=E9thodologies_inn=E9istes_?= =?ISO-8859-1?Q?des_successeurs_de_Piaget?= posting-account=1qbAGAkAAADcUtlizzXUEb5jUjfAdE2y Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) en r.8eponse .88 <47bc8649$0$864$ba4acef3@news.orange.fr> <495373af-424f-470b-8c6c- qui me r.8epondait en coupant La courbe de Dehane sur les performances de reconnaissance de 4 ou 5 > .8el.8ements (chute consid.8erable de la performance) a bien fait rire celle > qui utilise la m.8ethode des doigts avec les .8el.8fves : Les siens > reussissent .88 montrer 5 doigts mieux plus vite et mieux que 3 doigts C'est pourtant une constante .8etudi.8ee par les neurologues sur le cerveau de > diff.8erents animaux. Jusqu'.88 3-4, on .8evalue sans compter. Apr.8fs il faut > compter. Vous n'avez pas lu le texte : L'exp.8erience montre que les .8el.8fves de Julie Amadis imitent plus rapidement 5 que 3. Votre r.8eflexion n'est que l'ILLUSTRATION parfaite du titre de ma publication. Vous comparez les enfants aux animaux ! Vous cherchez un inn.8eisme dans la construction du nombre. Mais on s'en fout. Je veux DISSUADER les parents de DEFORMER le potentiel de leurs enfants en leur ENUMERANT la suite des CHIFFRES un, deux trois et ....... Mon exp.8erience me prouve que si l'on INTERDIT .88 l'entourage (comme je l'ai fait en 1980 pour mon fils) de lui faire ANONNER cette suite de chiffre..... .... le r.8esultat sera que l'enfant (avec son ARGENT DE POCHE) R.83INVENTERA TOUT SEUL LA MULTIPLICATION PAR DEUX EN EN MEMORISANT (SANS QUE L ON S'EN SOIT APERCU) LE DEBUT DE LA TABLE (TABLE DE DEUX) ... YT ---------------------------------------------------------------------------- ------- fr.soc.politique, fr.sci.maths, sci.math, fr.sci.physique, fr.education.divers === Subject: Re: analyse critique des m.8ethodologies inn.8eistes des successeurs de Piaget Yanick Toutain .8ecrivait > Je veux DISSUADER les parents de DEFORMER le potentiel de leurs > enfants en leur ENUMERANT la suite des CHIFFRES Vous .90tes un nouveau Gallil.8ee... === Subject: =?ISO-8859-1?Q?Re:_analyse_critique_des_m=E9thodologies_inn=E9istes_?= =?ISO-8859-1?Q?des_successeurs_de_Piaget?= posting-account=1qbAGAkAAADcUtlizzXUEb5jUjfAdE2y Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) here, I published! That is called Analysis criticizes methodologies inneists of successors of Piaget abstract: The partisans of Piaget believe that there exists one construction of the not-history number and not-teaching, independently of quality, the productivity of the methods and tools placed at the disposal of those which learn by the researchers manufacturers from educational tools! voil.88, J'ai publi.8e ! Cela s'appelle Analyse critique des m.8ethodologies inn.8eistes des successeurs de Piaget abstract: Les partisans de Piaget croient qu'il existe une construction du nombre non-historique et non-p.8edagogique, ind.8ependament de la qualit.8e, de la productivit.8e des m.8ethodes et outils mis .88 la disposition de ceux qui apprennent par les chercheurs- constructeurs d'outils p.8edagogiques ! === Subject: =?ISO-8859-1?Q?Re:_analyse_critique_des_m=E9thodologies_inn=E9istes_?= =?ISO-8859-1?Q?des_successeurs_de_Piaget?= posting-account=1qbAGAkAAADcUtlizzXUEb5jUjfAdE2y Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) http://svsurl.systransoft.com/?t=outputframeset&link=trans&task=11d1def534ea 1be0--80c1747-11833e7c344--99 > read this? http://www.lire.fr/critique.asp/idC=51596/idR=213/idG=8 I return at the moment of the page Dehaene (Stanislas) http://www.college-de-france.fr/media/psy cog/UPL29739 Cours2008 1SD.pdf Conclusions * At least three distinct systems contribute to the << direction of the number >: 1. The subitisation for the whole of 1,2 or 3 objects 2. The estimate of the numerosity, well beyond 3 objects 3. Counting, founded on the correspondence term-with-term, to arrive with the exact cardinality * The subitisation and the estimate are present very precociously at the child, and also exist at many animal species * These numerical processes confer on the child, very precociously, it << direction of the number > and a capacity of approximate calculation * At the adulthood, we continue to reach quickly analog representation of the numbers, even when those are to us presented in the form of symbols In this tumble one does not see the differentiation clearly of various possible strategies - ordinal or cardinal - and the evaluation of their effectiveness. One is at the same point as with Piaget: hasty conclusions b.89cl.8ees experiments. When it is necessary to put in oeuve teaching STRATEGIES of supports to transmit or make acquire the knowledge, competences. THESE people there of it are still... to raise the indices of a species of inneism in the construction of the number The curve of Dehane on the performances of recognition of 4 or 5 elements (falls considerable of the performance) made well laugh that who uses the method of the fingers with the pupils: His succeed in showing 5 fingers better more quickly and better than 3 fingers An imbecile would say (I hear it at the bottom) - yes? - but!!! the five fingers, that makes the whole hand it is more easy L!!! And then, species of cretin!!! (one returned 50 years behind! same idiocies as Piaget By supposing that all the children assimilate 5 more quickly than 3 parce that corresponds to the whole hand (and THAT an INTELLIGENT PEDAGOGUE THEM ADVISE TO USE IT!!!) When well even that would be prolonged and.... that children all their life would more easily see a group of 5 qu ' a dee group 3... which consequence that would have it??? To prove the falseness of the sources of Dehaene? one of fout The solution passes by the construction of educational tools and the evaluation of mass of results which are obtained from it (without prejudging owing to the fact that important results can not be perceived) Yanick Toutain I consider only with difficulty the fact that somebody who gains can to be 10 times 537 euros can have humility to go to seek COUNCILS at those which know each other (even expert beginner there equipped with a solid knowledge of the transformation quantity quality!) -------------------------------- PS FOR THE CRETINS: here is I AI PUBLISH! Ca is called critical analysis of methodologies inneists of successors of Piaget abstract: The partisans of Piaget believe that there exists one construction of the have-history number and have-teaching, independently quality, productivity of the methods and tools put at provision of learning by the researchers manufacturers from tools teaching! automatic translator : systran === Subject: Re: analyse critique des m.8ethodologies inn.8eistes des successeurs de Piaget Yanick Toutain .8ecrivait > http://svsurl.systransoft.com/?t=outputframeset&link=trans&task=11d1def > 534ea1be0--80c1747-11833e7c344--99 > > read this? http://www.lire.fr/critique.asp/idC=51596/idR=213/idG=8 I return at the moment of the page Dehaene (Stanislas) > http://www.college-de-france.fr/media/psy_cog/UPL29739_Cours2008_1SD.pd > f > * > At least three distinct systems contribute to the << direction of the > number >: > 1. > The subitisation for the whole of 1,2 or 3 objects > 2. > The estimate of the numerosity, well beyond 3 objects > 3. > Counting, founded on the correspondence term-with-term, to arrive > with the exact cardinality > * > The subitisation and the estimate are present very precociously at > the child, and also exist at many animal species > * > These numerical processes confer on the child, very precociously, it > << > direction of the number > and a capacity of approximate calculation > * > At the adulthood, we continue to reach quickly > analog representation of the numbers, even when those are to us > presented in the form of symbols In this tumble one does not see the differentiation clearly of > various possible strategies - ordinal or cardinal - and > the evaluation of their effectiveness. > One is at the same point as with Piaget: hasty conclusions > b.89cl.8ees experiments. When it is necessary to put in oeuve teaching STRATEGIES of supports > to transmit or make acquire the knowledge, competences. > THESE people there of it are still... to raise the indices of a > species of inneism in the construction of the number The curve of Dehane on the performances of recognition of 4 or 5 > elements (falls considerable of the performance) made well laugh that > who uses the method of the fingers with the pupils: His > succeed in showing 5 fingers better more quickly and better than 3 > fingers An imbecile would say (I hear it at the bottom) > - yes? > - but!!! the five fingers, that makes the whole hand it is more > easy L!!! And then, species of cretin!!! > (one returned 50 years behind! same idiocies as Piaget By supposing that all the children assimilate 5 more quickly than 3 > parce > that corresponds to the whole hand (and THAT an INTELLIGENT PEDAGOGUE > THEM > ADVISE TO USE IT!!!) When well even that would be prolonged and.... that children all > their life would more easily see a group of 5 qu ' a dee group > 3... > which consequence that would have it??? To prove the falseness of the sources of Dehaene? one of fout The solution passes by the construction of educational tools and > the evaluation of mass of results which are obtained from it (without > prejudging > owing to the fact that important results can not be perceived) Yanick Toutain > I consider only with difficulty the fact that somebody who gains can > to be 10 times 537 euros can have humility to go to seek > COUNCILS at those which know each other (even expert beginner there > equipped with a solid knowledge of the transformation quantity > quality!) -------------------------------- PS FOR THE CRETINS: here is I AI PUBLISH! Ca is called critical analysis of methodologies inneists of > successors of Piaget abstract: The partisans of Piaget believe that there exists one > construction of the have-history number and have-teaching, > independently > quality, productivity of the methods and tools put at > provision of learning by the researchers manufacturers from tools > teaching! _________________________________ > automatic translator : systran > Veuillez poster votre foullis ailleurs. === Subject: electrical properties of materials solution manual 7e posting-account=ROMivgkAAACp4prn3daB6k2syAQbQjmV InfoPath.2),gzip(gfe),gzip(gfe) I am looking for electrical properties of materials solution manual 7e === === Subject: Re: Restriction of a map > On 20 Feb., 18:10, Mariano Su.87rez-Alvarez > If C and D are categories and Fun(C, D) is the category > of functors C --> D with morphisms the natural > transformations of functors, then a subfunctor > of a functor F : C --> D is simply a subobject > G : C --> D of F in Fun(C, D). If you pick > specific categories C and D, you can usually work > out exactly what this means in that instance. Ok but this works only, if C and D are small categories, right? Fun(C, D) is locally small iff C is equivalent to a small category. Otherwise you have to be careful since proper classes appear. I don't understand what you want but if you want a notion of subobjects in a more general setting, you should consider the case where D is a topos. === Subject: JSH: Finding k Surprising answer with surrogate factoring that focuses on finding k, and leverages a rather intriguingly simple little result to factor. As consider 2x = k + 3r when z^2 = y^2 + T where T is the target to be factored, is odd and coprime to 3, and T mod 3 = 2, as then z must have 3 as a factor, so z = x+k gives x^2 + 2xk + k^2 = y^2 + T which is x^2 = y^2 + T - 2xk - k^2, and I can substitute out 2x, to get x^2 = y^2 + T - 2k^2 - 3kr and that's where a nifty thing pops in, as, you want r=0, but in general, r will be NEGATIVE if you start with the optimal k when r=0 and move about modulo 6, as that k will be even. That's because you'd have x^2 = y^2 + T - 2(k+6j)^2 - 3(k+6j)r and if you have positive k (no reason to use negative) and try to move with positive j, then r will be negative, and even if you move with negative j, the 36j^2 term will tend to dominate, forcing r to be negative to compensate. So r = 0 should be near the value at which abs(T-2k^2) is a minimum. Trouble is, if you continue the analysis you find that k at that point is the minimum that might work--remember k is positive--but the actual k MUST be within k/6 steps from that value. But it gives you a sense of what is possible, just with p=3. I generalized to p odd prime though, but found that it's still important to have z with 3 as a factor, but then you have as a crucial requirement: k^2 = (nT)(2)^{-1} mod p so you need a prime for which k exists, and then you find a maximal k modulo that prime, as the same argument above works, except now you'd have 2x = k + pr and you have a solution within k/(2p) steps from the maximal k. If T mod 3 = 2, then you'd use n=1, else you'd use n=5 or maybe 2, I'm still not sure, to force nT mod 3 = 2, as there is a crucial requirement that z = 3x so z has to have 3 as a factor. Oddly enough then for me, after over four years of trying to find an alternate factoring method with a concept I call surrogate factoring it all depended on this little thing that with x^2 = y^2 + T - 2(k+6j)^2 - 3(k+6j)r the j^2 will tend to dominate as you move around the correct value, which forces r to be negative to compensate. That is just so amazing to me. How such a little thing is so important. Without that, there'd be no clue about how to get close to k, and best you could loop through searching modulo p, which is what I worked at before, but the other crucial thing was realizing that k^2 = (nT)(2)^{-1} mod p was the REALLY important way to go as generalizing I used a variable I call alpha, but it turns out that with the generalization z = (1+2alpha^2)x and it's just easier to force z divisible by 3 than it is some of the other values that can be, like 19, or 73. All easy math, and easy to play with thankfully. Something of a subtle result though that requires you just do this odd thing of looking at 2x = k + pr along with z = x+k, and z^2 = y^2 + nT, and then it's easy, as long as you do all those things and can argue it out for months until it all comes together. Wow. Cool. What a result. James Harris === Subject: Re: JSH: Finding k posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) > Surprising answer with surrogate factoring that focuses on finding k, > and leverages a rather intriguingly simple little result to factor. As consider 2x = k + 3r when z^2 = y^2 + T where T is the target to be factored, is odd and coprime to 3, and T > mod 3 = 2, as then z must have 3 as a factor, so z = x+k gives x^2 + 2xk + k^2 = y^2 + T which is x^2 = y^2 + T - 2xk - k^2, and I can substitute out 2x, to > get x^2 = y^2 + T - 2k^2 - 3kr and that's where a nifty thing pops in, as, you want r=0, but in > general, r will be NEGATIVE if you start with the optimal k when r=0 > and move about modulo 6, as that k will be even. That's because you'd > have x^2 = y^2 + T - 2(k+6j)^2 - 3(k+6j)r and if you have positive k (no reason to use negative) and try to move > with positive j, then r will be negative, and even if you move with > negative j, the 36j^2 term will tend to dominate, forcing r to be > negative to compensate. So r = 0 should be near the value at which abs(T-2k^2) is a minimum. > The value of k is going to be close to sqrt(T/2). So now suppose T has 200 digits or so. Then k is going to have, say, 100 digits; k ~ 10^100. So you are thinking you just need to search through a range of values which are, you say, k/2p steps away from k. Of course, for p = 3, k/2p = k/6 is also going to be a large number: maybe 99 decimal digits. How long do you think that kind of search is going to take? Marcus. > Trouble is, if you continue the analysis you find that k at that point > is the minimum that might work--remember k is positive--but the actual > k MUST be within k/6 steps from that value. But it gives you a sense of what is possible, just with p=3. I generalized to p odd prime though, but found that it's still > important to have z with 3 as a factor, but then you have as a crucial > requirement: k^2 = (nT)(2)^{-1} mod p so you need a prime for which k exists, and then you find a maximal k > modulo that prime, as the same argument above works, except now you'd > have 2x = k + pr and you have a solution within k/(2p) steps from the maximal k. If T mod 3 = 2, then you'd use n=1, else you'd use n=5 or maybe 2, I'm > still not sure, to force nT mod 3 = 2, as there is a crucial > requirement that z = 3x so z has to have 3 as a factor. Oddly enough then for me, after over four years of trying to find an > alternate factoring method with a concept I call surrogate factoring > it all depended on this little thing that with x^2 = y^2 + T - 2(k+6j)^2 - 3(k+6j)r the j^2 will tend to dominate as you move around the correct value, > which forces r to be negative to compensate. That is just so amazing to me. How such a little thing is so > important. Without that, there'd be no clue about how to get close to k, and best > you could loop through searching modulo p, which is what I worked at > before, but the other crucial thing was realizing that k^2 = (nT)(2)^{-1} mod p was the REALLY important way to go as generalizing I used a variable I > call alpha, but it turns out that with the generalization z = (1+2alpha^2)x and it's just easier to force z divisible by 3 than it is some of the > other values that can be, like 19, or 73. All easy math, and easy to play with thankfully. Something of a > subtle result though that requires you just do this odd thing of > looking at 2x = k + pr along with z = x+k, and z^2 = y^2 + nT, and then it's easy, as long as > you do all those things and can argue it out for months until it all > comes together. Wow. Cool. What a result. James Harris === Subject: Re: JSH: Finding k > Surprising answer with surrogate factoring that focuses on finding k, > and leverages a rather intriguingly simple little result to factor. > As consider > 2x = k + 3r > when > z^2 = y^2 + T > where T is the target to be factored, is odd and coprime to 3, and T > mod 3 = 2, as then z must have 3 as a factor, so > z = x+k > gives > x^2 + 2xk + k^2 = y^2 + T > which is x^2 = y^2 + T - 2xk - k^2, and I can substitute out 2x, to > get > x^2 = y^2 + T - 2k^2 - 3kr > and that's where a nifty thing pops in, as, you want r=0, but in > general, r will be NEGATIVE if you start with the optimal k when r=0 > and move about modulo 6, as that k will be even. That's because you'd > have > x^2 = y^2 + T - 2(k+6j)^2 - 3(k+6j)r > and if you have positive k (no reason to use negative) and try to move > with positive j, then r will be negative, and even if you move with > negative j, the 36j^2 term will tend to dominate, forcing r to be > negative to compensate. > So r = 0 should be near the value at which abs(T-2k^2) is a minimum. The value of k is going to be close to sqrt(T/2). So now suppose T has 200 digits or so. Then k is > going to have, say, 100 digits; k ~ 10^100. So you are thinking you just need to search through > a range of values which are, you say, k/2p steps away > from k. Of course, for p = 3, k/2p = k/6 is also going to be > a large number: maybe 99 decimal digits. How long do you think that kind of search is going > to take? Marcus. > therein lies the problem. JSH is thinking of number 119 or less to factor, but RSA is that many digits long. I think JSH has made fundimental mistake, he is only trying to factor very small numbers, and you can look those upon the internet. === Subject: Re: JSH: Finding k >JSH is thinking of number 119 or less to factor, but RSA is that many digits >long. I think JSH has made fundimental mistake, >he is only trying to factor very small numbers, >and you can look those upon the internet. Since James is leery of factoring a real RSA sized number (>500 bits), how about something smaller: 45 bits: 17008656535007 46 bits: 38535186426049 47 bits: 66224816531197 48 bits: 115976333497043 49 bits: 308526162022031 50 bits: 429386409732223 51 bits: 1340026057765453 52 bits: 3695249821766513 53 bits: 8115064944675869 54 bits: 11169459453931717 55 bits: 19302767134139309 All or these are square free and have two odd prime factors. They are big enough to show up issues that factoring 119 (7 bits) might miss. rossum === Subject: Re: JSH: Finding k >JSH is thinking of number 119 or less to factor, but RSA is that many >digits >long. >I think JSH has made fundimental mistake, >he is only trying to factor very small numbers, >and you can look those upon the internet. > Since James is leery of factoring a real RSA sized number (>500 bits), > how about something smaller: 45 bits: 17008656535007 > 46 bits: 38535186426049 > 47 bits: 66224816531197 > 48 bits: 115976333497043 > 49 bits: 308526162022031 > 50 bits: 429386409732223 > 51 bits: 1340026057765453 > 52 bits: 3695249821766513 > 53 bits: 8115064944675869 > 54 bits: 11169459453931717 > 55 bits: 19302767134139309 All or these are square free and have two odd prime factors. They are > big enough to show up issues that factoring 119 (7 bits) might miss. rossum > I think the best he could do now is a 3 or 4 digit number. But he is way off track === Subject: Re: JSH: Finding k > Surprising answer with surrogate factoring that focuses on finding k, > and leverages a rather intriguingly simple little result to factor. As consider 2x = k + 3r when z^2 = y^2 + T where T is the target to be factored, is odd and coprime to 3, and T > mod 3 = 2, as then z must have 3 as a factor, so z = x+k gives x^2 + 2xk + k^2 = y^2 + T which is x^2 = y^2 + T - 2xk - k^2, and I can substitute out 2x, to > get x^2 = y^2 + T - 2k^2 - 3kr and that's where a nifty thing pops in, as, you want r=0, but in > general, r will be NEGATIVE if you start with the optimal k when r=0 > and move about modulo 6, as that k will be even. That's because you'd > have x^2 = y^2 + T - 2(k+6j)^2 - 3(k+6j)r and if you have positive k (no reason to use negative) and try to move > with positive j, then r will be negative, and even if you move with > negative j, the 36j^2 term will tend to dominate, forcing r to be > negative to compensate. So r = 0 should be near the value at which abs(T-2k^2) is a minimum. Trouble is, if you continue the analysis you find that k at that point > is the minimum that might work--remember k is positive--but the actual > k MUST be within k/6 steps from that value. But it gives you a sense of what is possible, just with p=3. I generalized to p odd prime though, but found that it's still > important to have z with 3 as a factor, but then you have as a crucial > requirement: k^2 = (nT)(2)^{-1} mod p so you need a prime for which k exists, and then you find a maximal k > modulo that prime, as the same argument above works, except now you'd > have 2x = k + pr and you have a solution within k/(2p) steps from the maximal k. If T mod 3 = 2, then you'd use n=1, else you'd use n=5 or maybe 2, I'm > still not sure, to force nT mod 3 = 2, as there is a crucial > requirement that z = 3x so z has to have 3 as a factor. Oddly enough then for me, after over four years of trying to find an > alternate factoring method with a concept I call surrogate factoring > it all depended on this little thing that with x^2 = y^2 + T - 2(k+6j)^2 - 3(k+6j)r the j^2 will tend to dominate as you move around the correct value, > which forces r to be negative to compensate. That is just so amazing to me. How such a little thing is so > important. Without that, there'd be no clue about how to get close to k, and best > you could loop through searching modulo p, which is what I worked at > before, but the other crucial thing was realizing that k^2 = (nT)(2)^{-1} mod p was the REALLY important way to go as generalizing I used a variable I > call alpha, but it turns out that with the generalization z = (1+2alpha^2)x and it's just easier to force z divisible by 3 than it is some of the > other values that can be, like 19, or 73. All easy math, and easy to play with thankfully. Something of a > subtle result though that requires you just do this odd thing of > looking at 2x = k + pr along with z = x+k, and z^2 = y^2 + nT, and then it's easy, as long as > you do all those things and can argue it out for months until it all > comes together. Wow. Cool. What a result. > James Harris The amazing answer with Austauschfactorisation that sourly k increases entrance halls to find and a small simple result rather plotter at the factor. Like 2x = k + 3r considers, if z2 = y2 is + T, where T is the goal which can be charged for, ungerad and more copremier at 3, and mod have 3 = must 2 of T as then z 3 as factor, then z = x+k gives x2 + 2xk + k2 = y2 + T, which is x2 = y2 + T - 2xk - k2, and I can replace T - 2k2 - outside of the 2x, around to receive x2 = y2 + 3kr and it is, where a skillful thing inside, As, you want reversal r=0, but generally r becomes NEGATIVE its whether you begin with it are, because you would have x2 = y2 + T - 2(k+6j)2 - 3(k+6j)r and if you have positive k (no reason to use itself the negative), and with positive j shifting attempt then r are negative, and even if you travel with negative j, the border to it 36j2 to be inclined prevail, the r forced, in order to be negative, in order to compensate. Like that r = would have to be 0 in the proximity of the value, to abs (T-2k2), a minimum is. The problem is, if you continue the analysis the fact that you find this k to this point are the minimum, that could to function remind of the fact that k is positive, but the real k MUST be in the stages k/6 of this value. What k? -- === Subject: Re: JSH: Finding k > Surprising answer with surrogate factoring that focuses on finding k, > and leverages a rather intriguingly simple little result to factor. > As consider > 2x = k + 3r > when > z^2 = y^2 + T > where T is the target to be factored, is odd and coprime to 3, and T > mod 3 = 2, as then z must have 3 as a factor, so > z = x+k > gives > x^2 + 2xk + k^2 = y^2 + T > which is x^2 = y^2 + T - 2xk - k^2, and I can substitute out 2x, to > get > x^2 = y^2 + T - 2k^2 - 3kr > and that's where a nifty thing pops in, as, you want r=0, but in > general, r will be NEGATIVE if you start with the optimal k when r=0 > and move about modulo 6, as that k will be even. That's because you'd > have > x^2 = y^2 + T - 2(k+6j)^2 - 3(k+6j)r > and if you have positive k (no reason to use negative) and try to move > with positive j, then r will be negative, and even if you move with > negative j, the 36j^2 term will tend to dominate, forcing r to be > negative to compensate. > So r = 0 should be near the value at which abs(T-2k^2) is a minimum. > Trouble is, if you continue the analysis you find that k at that point > is the minimum that might work--remember k is positive--but the actual > k MUST be within k/6 steps from that value. > But it gives you a sense of what is possible, just with p=3. > I generalized to p odd prime though, but found that it's still > important to have z with 3 as a factor, but then you have as a crucial > requirement: > k^2 = (nT)(2)^{-1} mod p > so you need a prime for which k exists, and then you find a maximal k > modulo that prime, as the same argument above works, except now you'd > have > 2x = k + pr > and you have a solution within k/(2p) steps from the maximal k. > If T mod 3 = 2, then you'd use n=1, else you'd use n=5 or maybe 2, I'm > still not sure, to force nT mod 3 = 2, as there is a crucial > requirement that > z = 3x > so z has to have 3 as a factor. > Oddly enough then for me, after over four years of trying to find an > alternate factoring method with a concept I call surrogate factoring > it all depended on this little thing that with > x^2 = y^2 + T - 2(k+6j)^2 - 3(k+6j)r > the j^2 will tend to dominate as you move around the correct value, > which forces r to be negative to compensate. > That is just so amazing to me. How such a little thing is so > important. > Without that, there'd be no clue about how to get close to k, and best > you could loop through searching modulo p, which is what I worked at > before, but the other crucial thing was realizing that > k^2 = (nT)(2)^{-1} mod p > was the REALLY important way to go as generalizing I used a variable I > call alpha, but it turns out that with the generalization > z = (1+2alpha^2)x > and it's just easier to force z divisible by 3 than it is some of the > other values that can be, like 19, or 73. > All easy math, and easy to play with thankfully. Something of a > subtle result though that requires you just do this odd thing of > looking at > 2x = k + pr > along with z = x+k, and z^2 = y^2 + nT, and then it's easy, as long as > you do all those things and can argue it out for months until it all > comes together. > Wow. Cool. What a result. > James Harris The amazing answer with Austauschfactorisation that sourly k increases > entrance halls to find and a small simple result rather plotter at the > factor. Like 2x = k + 3r considers, if z2 = y2 is + T, where T is the goal > which can be charged for, ungerad and more copremier at 3, and mod have 3 > = must 2 of T as then z 3 as factor, then z = x+k gives x2 + 2xk + k2 = y2 > + T, which is x2 = y2 + T - 2xk - k2, and I can replace T - 2k2 - outside > of the 2x, around to receive x2 = y2 + 3kr and it is, where a skillful > thing inside, As, you want reversal r=0, but generally r becomes NEGATIVE > its whether you begin with it are, because you would have x2 = y2 + T - > 2(k+6j)2 - 3(k+6j)r and if you have positive k (no reason to use itself > the negative), and with positive j shifting attempt then r are negative, > and even if you travel with negative j, the border to it 36j2 to be > inclined prevail, the r forced, in order to be negative, in order to > compensate. Like that r = would have to be 0 in the proximity of the > value, to abs (T-2k2), a minimum is. The problem is, if you continue the > analysis the fact that you find this k to this point are the minimum, that > could to function remind of the fact that k is positive, but the real k > MUST be in the stages k/6 of this value. What k? > from the first equation, k = 2x-3r Harristonian logic is always self-recursive. > -- > how did you get a FREE account on teranews ? === Subject: Re: Finding k (**note Picka = Guess**) > Surprising answer with surrogate factoring that focuses on finding k, > and leverages a rather intriguingly simple little result to factor. > picka ? > As consider 2x = k + 3r picka! Picka k Picka r you getta x ! when z^2 = y^2 + T picka? Picka! Picka T Picka y Picka equation outta air! you getta two z's !! where T is the target to be factored, is odd and coprime to 3, and T > mod 3 = 2, as then z must have 3 as a factor, so picka? Picka!! Picka T odd and coprime to 3 z = x+k No picka, we alreadya gota z and x and k you already Picka them. Trya some others. gives x^2 + 2xk + k^2 = y^2 + T Picka? Substutuiona to ah many ah variables nowa to get anywhere, moving like slugs with salt on. > which is x^2 = y^2 + T - 2xk - k^2, and I can substitute out 2x, to > get No you already Picka them, no fair. x^2 = y^2 + T - 2k^2 - 3kr Picka? Picka r !! Likea magic outta air! > and that's where a nifty thing pops in, as, you want r=0, Picka r = 0 >but in > general, r will be NEGATIVE Picka r negitive > if you start with the optimal k when r=0 > and move about modulo 6, as that k will be even. That's because you'd > have Picka? Picka!! Picka k even! Picka r = 0 x^2 = y^2 + T - 2(k+6j)^2 - 3(k+6j)r Picka? Picka!! Picka j ! Welcome to the club j, where the f*ck did you come from ?? > and if you have positive k (no reason to use negative) Picka k postive UnPicka k negative NoPicka k =0, a shameful discrimination of Zeros here. > and try to move > with positive j, Picka j positive > then r will be negative, Picka r negitive >and even if you move with > negative j, Wait! Quick Picka j negitive! > the 36j^2 term will tend to dominate, forcing r to be > negative to compensate. now r, you bad, you all negitive, being *forced* by nasty negitive j too. Bastards! > So r = 0 should be near the value at which abs(T-2k^2) is a minimum. Picka? Picka! Picka r = 0 Picka should instead of is Picka value at which abs(T-2k^2) is a minimum > Trouble is, if you continue the analysis you find that k at that point > is the minimum that might work--remember k is positive--but the actual > k MUST be within k/6 steps from that value. Picka? Picka! Picka k/6 But it gives you a sense of what is possible, just with p=3. WTF ? where did p come from? outta air ? Picka p=3 > I generalized to p odd prime though, Picka p odd prime (could use your prime solving techniques to finda p) > but found that it's still > important to have z with 3 Picka? Picka! Picka z = 3 >as a factor, but then you have as a crucial > requirement: k^2 = (nT)(2)^{-1} mod p Picka? Picka! Picka k^2 = (nT)(2)^{-1} mod p so you need a prime for which k exists, and then you find a maximal k > modulo that prime, as the same argument above works, except now you'd > have 2x = k + pr Picka? Picka! Picka 2x = k + pr > and you have a solution within k/(2p) steps from the maximal k. Picka? Picka! Picka k/(2p) If T mod 3 = 2, then you'd use n=1, else you'd use n=5 or maybe 2, I'm > still not sure, to force nT mod 3 = 2, as there is a crucial > requirement that Picka? Picka! Picka T mod 3 = 2 Picka n=1 Picka n=5 Picka n=5 or maybe 2 > z = 3x so z has to have 3 as a factor. > Picka z has to have 3 as a factor > Oddly enough then for me, after over four years of trying to find an > alternate factoring method with a concept I call surrogate factoring > it all depended on this little thing that with x^2 = y^2 + T - 2(k+6j)^2 - 3(k+6j)r Picka? Picka! Picka x^2 = y^2 + T - 2(k+6j)^2 - 3(k+6j)r > the j^2 will tend to dominate as you move around the correct value, > which forces r to be negative to compensate. Picka? Picka! Picka r negitive > That is just so amazing to me. How such a little thing is so > important. Without that, there'd be no clue about how to get close to k, and best > you could loop through searching modulo p, which is what I worked at > before, but the other crucial thing was realizing that k^2 = (nT)(2)^{-1} mod p was the REALLY important way to go as generalizing I used a variable I > call alpha, but it turns out that with the generalization z = (1+2alpha^2)x Picka? Picka! Picka z = (1+2alpha^2)x Picka alpha (alpha just flew in outta the air) > and it's just easier to force z divisible by 3 than it is some of the > other values that can be, like 19, or 73. Picka? Picka! Picka z = 3 Picka z = 19 Picka z = 73 > All easy math, and easy to play with thankfully. Something of a > subtle result though that requires you just do this odd thing of > looking at 2x = k + pr Picka? Picka! Picka 2x = k + pr > along with z = x+k, and z^2 = y^2 + nT, and then it's easy, as long as > you do all those things and can argue it out for months until it all > comes together. Picka? Picka! Picka z = x+k Picka z^2 = y^2 + nT > Wow. Cool. What a result. > James Harris Puka? Puka! PukaCHOO on James Harris === Subject: Re: Finding k James Harris Has this guy lost his marbles? === Subject: Re: Finding k James Harris Has this guy lost his marbles? Yes. Killfile him, and killfile all threads with 'JSH' in the subject line. Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: Finding k <47bd408b$0$12542$afc38c87@news.optusnet.com.au> posting-account=sFP0HgkAAADJMwhdrXAaC5VX7Tc3BtzY Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > James Harris Has this guy lost his marbles? Yes. --- J K Haugland http://home.no.net/zamunda === Subject: How to find out the upper / lower bound for recurrences posting-account=QptImAoAAACiaI361JEfCGkacXp-7v35 CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727; FDM),gzip(gfe),gzip(gfe) How can I know the asymptotic upper and lower bound for recurrences?? 1) T(n) = 2T(n/3) + n lg n 2) T(n) = 7T(n/2) + n^3 === Subject: Solutions Manuals & Test Banks AVAILABLE - Accounting, Finance, Taxation posting-account=Oef_qQoAAAC_nhbXm5pHPNx-3IJ2uJp1 I have the COMPLETE SOLUTIONS MANUALS & TEST BANKS in PDF and WORD formats for the following textbooks: 1. Intermediate Accounting, 12th edition by Kieso, Weygandt, Warfield; ISBN-10: 0471749559; ISBN-13: 978-0471749554 For this one the TEST BANK is available too. 2. Auditing and Assurance Services, 12th Edition Arens, Elder, Beasley ISBN-10: 0135132126 ISBN-13: 9780135132128 ISBN-10: 0136128270 ISBN-13: 9780136128274 For this one the TEST BANK is available too. 3. Advanced Accounting, 9th edition by Fischer, Taylor, Cheng; ISBN-10: 0324304013; ISBN-13: 978-0324304015 For this one the TEST BANK is available too. 4. Contemporary Financial Management, 10th edition by Moyer McGuigan, Kretlow ISBN: 0-324-28908-1 5. West Federal Taxation 2007, 30th edition by Willis, Hoffman, Maloney, Raabe ISBN-10: 0324313489 ISBN-13: 978-0324313482 Send an email to prodigy615(at)gmail(dot)com to order. === Subject: Re: Curvature of a function at all in reading. Google groups changes quoted text. > it _hides_ from you the illicit changes. > Google groups is lying to you. > In addition, Google inserts short lines of words that are it's editing format tokens. Don't Use Google. === Subject: Re: Curvature of a function > > at all in reading. Google groups changes quoted text. > it _hides_ from you the illicit changes. > Google groups is lying to you. In addition, Google inserts short lines of words that > are it's editing format tokens. Don't Use Google. is on Usenet. For that reason, and that it's the injection vector for the gibberings of the vast majority of the idiots filed, with a few cherry-picked gg posters reinstated. Phil -- -- Microsoft voice recognition live demonstration === Subject: [] Curvature of a function <87zltu4t0j.fsf@nonospaz.fatphil.org > at all in reading. > Google groups changes quoted text. > it _hides_ from you the illicit changes. > Google groups is lying to you. > In addition, Google inserts short lines of words that > are it's editing format tokens. Don't Use Google. is on Usenet. For that reason, and that it's the injection > vector for the gibberings of the vast majority of the idiots > filed, with a few cherry-picked gg posters reinstated. > When I hear that Google was going to issue stock, I knew that would be the end of Google's excellent quality. Immediately it went down hill will unneeded modifications that failed to work correctly, and has been going down hill ever sense into greasy greedy crapitalism. Down with Google!! What web searchers have you tried and which do you use? === Subject: Re: [] Curvature of a function [...] > Don't Use Google. is on Usenet. [...] When I hear that Google was going to issue stock, I knew that would be the > end of Google's excellent quality. Immediately it went down hill will unneeded > modifications that failed to work correctly, and has been going down hill > ever sense into greasy greedy crapitalism. Down with Google!! What web > searchers have you tried and which do you use? there is, it's one of the things that they have done almost right. (Google maps too.) However, the fact that when I to be stateful, I want it to perform the same search for me as it would for any other web user) it redirects me to its equivalent page in Finnish (I don't speak Finnish, I simply use an IP address owned by a Finnish ISP) pisses me off - it's trying to be 'clever', but is in fact being completely stupid. (The same thing goes for all sites that change their language depending on who your IP address is registered to, determining a user's prefered language is not what IP addresses were designed for at all.) Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: [] Curvature of a function > I want it to perform the same search for me as it would for any > other web user) it redirects me to its equivalent page in Finnish (I > don't speak Finnish, I simply use an IP address owned by a Finnish > ISP) pisses me off - it's trying to be 'clever', but is in fact > being completely stupid. I too find that quite annoying, even though I do speak Finnish. > (The same thing goes for all sites that change their language > depending on who your IP address is registered to, determining a > user's prefered language is not what IP addresses were designed for > at all.) Indeed. That should be done on the protocol level, and HTTP does in fact allow for it. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: [] Curvature of a function > I want it to perform the same search for me as it would for any > other web user) it redirects me to its equivalent page in Finnish (I > don't speak Finnish, I simply use an IP address owned by a Finnish > ISP) pisses me off - it's trying to be 'clever', but is in fact > being completely stupid. I too find that quite annoying, even though I do speak Finnish. I'm pretty sure that it skews the search results in favour of Finnish things too. Searching for fairly universal things, I'm sure I've seen Finnish results unnaturally high in the results list. > (The same thing goes for all sites that change their language > depending on who your IP address is registered to, determining a > user's prefered language is not what IP addresses were designed for > at all.) Indeed. That should be done on the protocol level, and HTTP does in > fact allow for it. Absolutely. I've not tried tweaking my headers to send a English or US English _ONLY_ and see if it changes behaviour. I suspect that it will make no difference. Phil -- -- Microsoft voice recognition live demonstration === Subject: -- One input, two bugs: The VM machine discovers a shocking quadrature failure in Mathematica 6 posting-account=ubyIWAkAAABW-OTbVB1QiN1oZlu0qUgw CLR 2.0.50727),gzip(gfe),gzip(gfe) This is yet another part of the Cyber Tester's demo. The VM machine discovers, Mathematica 6 numeric integrator is afflicted with a new regression bug. (* 2 Sin[z] Cos[z] = Sin[2 z] *) NIntegrate[Sin[z^2] Cos[z^2] Cos[z^4], {z, 0, 7}] (* = NIntegrate[(2 Sin[z^2] Cos[z^2])/2 Cos[z^4], {z, 0, 7}] = *) NIntegrate[Sin[2 z^2]/2 Cos[z^4], {z, 0, 7}] 0.291311 <- No warning message about potentially false result 0.226657 So the 1st bug is that the results are not identical (I can foresee heated comments on this statement). The 2nd bug is that no warning message is produced to alert the customer. ----------------------------------------------------------------- VERSION OUTPUT RESOLUTION ----------------------------------------------------------------- Mathematica 6.0 1) Dictinct results <---------------- BUG #1 2) No warning message <---------------- BUG #2 Mathematica 5.2 Identical results OK Mathematica 4.2 Identical results OK Mathematica 3.0 Identical results OK ----------------------------------------------------------------- Now, in Mathematica 6, even this fails NIntegrate[Sin[z^2] Cos[z^2] Cos[z^4], {z, 0, 3}] NIntegrate[Sin[2 z^2]/2 Cos[z^4], {z, 0, 3}] 0.266632 <------------------------------------------------ BUG 0.229319 OK Compare with Mathematica 5.2/4.2/3.0 perfect results NIntegrate[Sin[z^2] Cos[z^2] Cos[z^4], {z, 0, 3}] NIntegrate[Sin[2 z^2]/2 Cos[z^4], {z, 0, 3}] 0.229319 OK 0.229319 OK ----------------------------------------------------------------- Best wishes, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing ----------------------------------------------------------------- We must understand that technologies like these are the way of the future. ----------------------------------------------------------------- === Subject: Re: Help on integrals posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) I want to compute the following integrals int_{-1}^1 sin (z x) T_{2n+1}(x) d x = ? and int_{-1}^1 cos (z x) T_{2n}(x) d x = ? where T_{2n+1}(x) and T_{2n}(x) are Chebyshev polynomials of first > kind. I have checked these integrals in Gradshteyn and Ryzhik's book (http://www.mathtable.com/gr/), see the item 7.354 in it. However, > unfortunately, I find their formulas are wrong. In their formulas, the > integrands should contain the weight 1/sqrt{1-x*x}. Do some guys get familiar with these integrals? > === Subject: Riemann-Stieltjes integral help posting-account=gc2kDQoAAADMsLO9kJjQL9hCJkI0D8qJ CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Let f,g be two functions from [a,b] to the real numbers, let P be a partition of [a,b] and let Q be a partition of [a,b] obtained from the fixed points of each subinterval in P by adding the points a and b. I need to simplify the riemann-stieltjes sums: S_f ( P ; g) + S_g (Q; f) , i.e the sum of g with respect to f considering the partition P and the sum of f with respect to g considering the partition Q. So, I seem to get at the end f(b)g(b) - f(a)g(a) as the result of the sum. Is this correct? And a last question, how can I show that if f is Riemann-Stieltjes integrable with respect to g in [a,b] then g is riemann stieltjes integrable with respect to f in [a,b] ?? === Subject: Re: Riemann-Stieltjes integral help >Let f,g be two functions from [a,b] to the real numbers, let P be a >partition of [a,b] and let Q be a partition of >[a,b] obtained from the fixed points of each subinterval in P by >adding the points a and b. Huh? What is Q again? (If the fixed points of a subinterval are the endpoints then I don't see the difference between P and Q. If the fixed points of a subinterval are not the endpoints then what the heck do you mean by fixed points of each subinterval?) >I need to simplify the riemann-stieltjes sums: S_f ( P ; g) + S_g (Q; f) , i.e the sum of g with respect to f >considering the partition P and the sum of f with respect to g >considering the partition Q. So, I seem to get at the end f(b)g(b) - f(a)g(a) as the result of the >sum. Is this correct? And a last question, how can I show that if f is Riemann-Stieltjes >integrable with respect to g in [a,b] then g is riemann stieltjes >integrable with respect to f in [a,b] ?? > David C. Ullrich === === Subject: An exact 1-D integration challenge - 52 - Proud Earthling, raise and defeat all the silly CASs! Hello dear computer algebra buff the Earthling, Recently, I've sent an alarm message to the mankind! As you (ah, hopefully) have already realized, those sinister (striped!) Computers are coming!... Are you a patriot of our old good carbon-based life? If yes, then run now and fight for our humankind liberties and show your unbeatable superiority over all the moronic chips... and some professors! Get prepared for the Final Combat, - Train hard, fight easy! Look, none of the single damned chip cracks this (no wonder, they lack small grey cells badly!). Mathematica/Maple/Derive/MuPAD/AXIOM returns this unevaluated. Is there a Valiant Integrator the Warrior who can design the computer algebra steps to get to the exact value of int(1/sqrt(1+z^4+z^6+z^10), z= 0..infinity); ? Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing ----------------------------------------------------------------- We must understand that technologies like these are the way of the future. ----------------------------------------------------------------- === === === === === Subject: Re: USA full of huge numbers of ILLEGAL immigrants from India posting-account=q4scdQoAAAA-QqP7eskxIeHCXNX8Ytan SV1),gzip(gfe),gzip(gfe) > Yesterday's USA Today had a summary of a study about a major problem > with illegal immigrants in the USA. æThe illegal immigrants are from > INDIA! The Department of Homeland Security estimates there are over 290,000 > people from India in the USA ILLEGALLY. æAnother organization believes > there are more: over 400,000 illegals from India. on H-1B visas, then when the visa expires, the people from India stay > in the USA instead of the proper action of leaving. I believe the best way to prevent future illegal immigration from > India is to ban all H-1B visas. Employers can help by firing visa holders from India, as that should > make them have to go back to the USA. Unfortunately, fewer than 1000 of these illegals are deported each > year. I would like to see ALL of them removed, with their liquid assets > confiscated to cover the fines for their illegal activity and the > costs of finding and removing them. The study shows that people from India living in the USA tend to make > 60% more income than American citizens. æUnlike dirt-poor immigrants > from Mexico and other poor countries, people from India have no > financial need to be in the USA. æMost are upper caste Hindus or > middle class Hindus who can live like royalty in India. æSend them > back to India. Send these folks back to India where the worst they can do is make annoying telemarketing calls to the States. ted === Subject: Extensions of DVR posting-account=-qlJtAkAAADgsMLtrN9n4qex_kHnFTZO 1.1.4322),gzip(gfe),gzip(gfe) new to this theory, so I'm actually concentrated only on first lemma (which I have proved with full details) and on first theorem. He speaks about a complete DVR V (with field of fraction K, char K =0, residue field k, char k =p). He consider a family of extensions V subset V_1 subset V_2 subset... (each ring is the normalization of V in some finite extension of K and so it is a complete DVR). If W is another extension (finite too) he consider the tensor product V_n otimes_V W and speaks about its normalization. Here is the problem. Normalization in which ring? I think that V_n otimes_V W is a product of DVR (or at least a product of Dedekind domain) and the normalization is in V_n otimes_V Fr(W), that should be a product of field (so the normalization will be component by component), but I cannot prove it. Any suggestion? Any idea will be appreciated. Ricky === Subject: Re: Extensions of DVR > Hodge Theory. I'm > new to this theory, so I'm actually concentrated only > on first lemma > (which I have proved with full details) and on first > theorem. > He speaks about a complete DVR V (with field of > fraction K, char K =0, > residue field k, char k =p). He consider a family of > extensions V > subset V_1 subset V_2 subset... (each ring is the > normalization of > V in some finite extension of K and so it is a > complete DVR). If W is > another extension (finite too) he consider the tensor > product V_n > otimes_V W and speaks about its normalization. Here > is the problem. > Normalization in which ring? One can always normalize in the total ring of fractions, that is in the localization of the ring with respect to all non-zerodivisors. > I think that V_n otimes_V W is a product of DVR (or > at least a > product of Dedekind domain) and the normalization is > in V_n otimes_V > Fr(W), that should be a product of field (so the > normalization will be > component by component), but I cannot prove it. Any > suggestion? Any > idea will be appreciated. > Ricky Let W,W' be finite extensions of the valuation domain V. Then W|V and W'|V are faithfully flat. Therefore the natural maps W --> W otimes W' =: T and W' --> T are injective and the images (w otimes 1) and (1 otimes w') are non-zerodivisors. Consequently the total ring of fractions Frac(T) of T contains S := L otimes_K L', where K,L,L' are the fields of fractions of V,W,W' respectively. In your situation L|K is separable, thus S is a product of finitely many fields (extending K). In particular every non-zerodivisor is invertible. Since the natural map T --> S is injective it follows that Frac(T) = S. The normalization of V in S is the product of the normalizations in the factors of S. In your situation the normalization of V in S thus is a product of finitely many DVRs. However the reasoning above works without asuming V to be discrete or complete. If V is a general valuation ring, then the normalizations in S is a product of finitely many Pruefer domains. In particular if V is discrete these Pruefer domains are Dedekind domains. H === Subject: Re: Extensions of DVR <33460382.1203604640175.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=-qlJtAkAAADgsMLtrN9n4qex_kHnFTZO Gecko/20080207 Ubuntu/7.10 (gutsy) Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Hodge Theory. I'm > new to this theory, so I'm actually concentrated only > on first lemma > (which I have proved with full details) and on first > theorem. > He speaks about a complete DVR V (with field of > fraction K, char K =0, > residue field k, char k =p). He consider a family of > extensions V > subset V_1 subset V_2 subset... (each ring is the > normalization of > V in some finite extension of K and so it is a > complete DVR). If W is > another extension (finite too) he consider the tensor > product V_n > otimes_V W and speaks about its normalization. Here > is the problem. > Normalization in which ring? One can always normalize in the total ring of fractions, > that is in the localization of the ring with respect to all > non-zerodivisors. I think that V_n otimes_V W is a product of DVR (or > at least a > product of Dedekind domain) and the normalization is > in V_n otimes_V > Fr(W), that should be a product of field (so the > normalization will be > component by component), but I cannot prove it. Any > suggestion? Any > idea will be appreciated. > Ricky Let W,W' be finite extensions of the valuation domain V. > Then W|V and W'|V are faithfully flat. Therefore the natural > maps W --> W otimes W' =: T and W' --> T are injective and > the images (w otimes 1) and (1 otimes w') are non-zerodivisors. > Consequently the total ring of fractions Frac(T) of T contains > S := L otimes_K L', where K,L,L' are the fields of fractions of > V,W,W' respectively. In your situation L|K is separable, thus S is a product > of finitely many fields (extending K). In particular every > non-zerodivisor is invertible. Since the natural map T --> S > is injective it follows that Frac(T) = S. The normalization of V in S is the product of the normalizations > in the factors of S. In your situation the normalization of V in S thus is a product > of finitely many DVRs. However the reasoning above works > without asuming V to be discrete or complete. If V is a general valuation ring, then the normalizations in > S is a product of finitely many Pruefer domains. In particular > if V is discrete these Pruefer domains are Dedekind domains. H I don't understant what you mean with > The normalization of V in S is the product of the normalizations > in the factors of S., I'm studying the normalization of T, not of V... are they the same? T is not a DVR, so I think I have to decompose it as a product of DVR, using the decomposition of S, is this correct? If S=prod K_i I think that T=prod T_i, where T_i=T cap K_i, but I'm not sure. Also if this is true, how to prove that each T_i is a DVR? Ricky === Subject: Re: Extensions of DVR posting-account=-qlJtAkAAADgsMLtrN9n4qex_kHnFTZO 1.1.4322),gzip(gfe),gzip(gfe) Note that W is without torsion, so it is a free V-module, and V_n otimes_V W is, as V-module, a product of a number of V_n, but this tells nothing about the structure of V-algebra. === Subject: Re: -- infinite product >hello, The problem is to show lim P = lim Product (q=-n..m) (1+z/(pi*q)) = k^(z/pi) >* sin(z)/z where q=0 is omitted from the product and it is assumed lim >(m/n) = k. lim here means m & n ->oo ??? That's not very well put, but looking below it seems like I understand what it means. If you take k = 1 you can take m = n, and then (by the manipulations below) you see the limit is sin(z)/z, which is _not_ k^(z/pi). >I can get partway to the answer, but my reasoning/understanding needs >improvement: let m=nk, then P= Product ( q=n+1..n*k)(1+z/(pi*q)) * Product(q=1..n) ( 1- z^2/(q^2*pi^2) ) At this point because its finite there is no problems in rearranging the >terms to get the above, and we see that in the limit the last product = >sin(z)/z as n->oo. The problem i have is in justifying the following: I know that Product ( q=1..oo)(1+z/(pi*q)) is not convergent, and to make it convergent >(absolutely convergent) I can do the following, >lim n->oo exp[sum(m=1..nk-n) z/(pi*(n+m)] (1+z/(pi*(n+1))) >exp(-z/(pi*(n+1))(1+z/(pi*(n+2))) exp(-z/(pi*(n+2))....(1+z/(pi*(nk))) >exp(-z/(pi*(nk)), because (1+z/(pi*(n+1))) exp(-z/(pi*(n+1)) yeilds terms that make the >product absolutely convergent. Each term in that product then ->1 as n->oo >and we also have sum(m=1..nk-n) z/(pi*(n+m) = x/pi*(H[kn] - H[n]) H[n] = harmonic number. As n->oo this goes to z*ln(k)/pi and so we get the >k^(z/pi). I'm not entirely sure of the correct arguments that are going on >here. For example, why could i have not argued the following As n becomes large each term in the product Product ( >q=n+1..n*k)(1+z/(pi*q)) behaves like (1+z/(pi*n)) and we have nk of them >so we get n->oo of (1+z/(pi*n))^(nk) = exp(kz/pi) . I mean i know its >obviously wrong since it doesn't yield the correct answer to the problem, >but I'd like to better understand why. My guess is that Product ( >q=n+1..n*k)(1+z/(pi*q)) as n->oo simply is not convergent, and so we need >to make it convergent by adding in the exponentials. Is there any other way >of making it convergent, besides exponentials? cheers >moth > David C. Ullrich === Subject: Re: Question regarding eigenvalue > In the above line, you surely meant has nontrivial > solutions rather > than has no trivial solutions. > >A-kI has no maximum range <= > Same wording error. I would reword the above line as the range of A - kI is a proper subspace of C^n. > >det(A-kI)=0. quasi Right. Explanation of my mistakes: 1.- I translated incorrectly to English the correct Spanish sentence el sistema tiene soluciones no triviales 2.- The Spanish word rango means rank and range and so I translated it literally and incorrectly. 3.- C^n<> C^(nxn) (if n>=2) unless you use copy and paste. Fernando. === Subject: Re: Factorization of a sum of complex numbers with complex exponents in Q[I]? <30230458.1203529890177.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=PSzFRAoAAAARszS8zeFmxtqyivK9-1_f CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 3.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) On Feb 20, 1:41æpm, tommy1729 On Feb 19, 11:32æpm, tommy1729 > On Feb 19, 6:27 am, Gerry > On Feb 18, 3:24 pm, Pubkeybreaker > Srange that op's 4,5 cannot give a more > specific answer. > I guess they are only looking for > mutlivalued > solutions not in Q[I]. > Your original expression is NOT an > element of > Q(i). æ For example, > (4/10)^i æ = exp(log(4/10) * i) æ æ= > cos(4/10) + > i sin(4/10), ætaking > the principal > branch. æBut this last complex number is > NOT > an > element of Q(i). > Until you give a full, well defined > explanation > of the domain in which > you are > working, the very concept of factoring > is > meaningless. > Hi > (4/10)^i æ = exp(log(4/10) * i) æ æ= > cos(4/10) + > i sin(4/10), ætaking > ?? i thought that exp(log(4/10)*I)=cos(log(4/10))+I*sin(log(4/10)) > Until you give a full, well defined > explanation > of the domain in which > you are > working, the very concept of factoring > is > meaningless. > Yes Q[I]^Q[I} like Tommy defined. > I don't think factoring this way is > meaningless. > and like i said it can be extended further > Q[I]^Q[I]^Q[I]^.... > *What* can be extended? > What exactly do you mean by factoring `this > way'? > You > have not *yet* told us! What is it that you > want > to > do > `in Q[i]^Q[i]'? > -- m > what gerry meant is this > (a + bi)^(c + di) + (e + fi)^(g + hi) + ... > where i is sqrt(-1) and the variables are > rationals. > the sums can be taken as long or short as we > want. > and factoring into numbers of the same type. > tommy1729 > topology = folding x dimension tommy1729- > Hide > quoted text - > - Show quoted text - > Hi all, > let me explain how i came to this problem first. > I hope you don't mind that i'm using Pari/GP > representations. > I found a product identity for the sum > x^n+(x+c)^n > and called it the Tangent Product Identity (TPI) > because: > (1) x^n+(x+c)^n=2*prod(i=1,n,x+c/2*(1-I*tan((2*i-1+n%2)*Pi > /(2*n)))) > which also can be written as: > (2) x^n+(x+c)^n=2*prod(i=1,n,x+c/(exp((2*i-1+n%2)*Pi*I/n)+ > 1)) > So for example for c=1 and n=3 we get the > reducible > æpolynomial : > 2*x^3 + 3*x^2 + 3*x + 1 > which factors into:[2*x + 1, 1; x^2 + x + 1, 1] > and for c=1 and n=8 we get the irreducible > æpolynomial : > 2*x^8 + 8*x^7 + 28*x^6 + 56*x^5 + 70*x^4 + 56*x^3 > + > 28*x^2 + 8*x + 1 > which doesn't factor in Q > Reducible or irreducible the ROOTS xi ,i=1..n for > any > n in N and c in > C ARE: > æxi=-c/(exp((2*i-1+n%2)*Pi*I/n)+1) > So regardless of the fact reducible or > irreducible > the factorization as shown in the righthand > component > of (1) and (2) is UNIQUE !;-) > Now lets assign c=(1 + 2*I) > We get for SUM(x,n,c)=x^n+(x+c)^n for n=3 > SUM(x,3,1 + 2*I)=2*x^3 + (3 + 6*I)*x^2 + (-9 + > 12*I)*x + (-11 - 2*I) > a product which factors into:[x + (1/2 + I), 1; > x^2 + > (1 + 2*I)*x + > (-3 + 4*I), 1] > So for ANY x we have the sum product > relationship. > Also for x=(a > +b*I)^(c+d*I) > And this is an answer to my question. > The problem is in case we choose > c=(e+f*I)^(g+h*I) > for example let c=(1 + 2*I)^(3 + 4*I) > We get for SUM(x,n,c)=x^n+(x+c)^n for n=3 > SUM(x,3,0.12901 + 0.033924*I)=2*x^3 + (0.38703 + > 0.10177*I)*x^2 + > (0.046478 + 0.026259*I)*x + (0.0017018 + > 0.0016548*I) > a product which factors into:[x + (0.064505 + > 0.016962*I), 1; x + > (0.093884 - 0.094764*I), 1; x + (0.035126 + > 0.12869*I), 1] > Pari cannot write the coefficients in Q[I]^Q[I] > and > just outputs the > roots. > I am interested to know how and when it is > possible > to write the > coefficients in Q[I]^Q[I]. > By using a more general identity then (1) i > already > found a way to > deal with this case but > still i would like to have your valuable > opinion.:-) > Gerry > 2^x + 3^x does not factor into elements n^x > 2^x + 4^x does factor into elements n^x : > 2^x(1 + 2^x) > the same applies to complex rings. > the exponents have to be Q(i) powers of eachother. > a^x + b^x => factors if => b = a ^ Q(i) or a = b ^ > Q(i). > tommy1729- Hide quoted text - > - Show quoted text - Hi Tommy, Of course the example you gave assumes both terms > have a common > factor. > This case i'm not considering. It's more like: æ4^3+(4+1)^3=(2*4+1)*(4^2+4+1) i see. but in your example you did give the same base. something alike (1+2i)^(5+3i) + 2 * (1+2i)^(8+3i) so 4^3+(4+1)^3=(2*4+1)*(4^2+4+1) holds > but (4+i)^(3+i)+(4+1+i)^(3+i) probably does not factor. more general for rational x (x+i)^(3+i)+(x+1+i)^(3+i) probably does not factor. ..hmmm wait -> if (x+i)^z = x+1+i and z is a + bi (E Q(i)) then we do have the same exponent. perhaps your idea is intresting afterall. ill come back to it later , im hungry. --- hmm z cant be E Q(i) i think. since (x+i)^i not E Q(i). [ (x+i)^i = exp(log(x+i)*i) ] your turn :) > Gerry tommy1729- Hide quoted text - - Show quoted text - Hi Tommy, Yes i did use the same base in the first example which i should not have done. Let me use the 1st example i gave in #18 for n=3 and c=(1+2*I) and lets set x=(3+4*I)^(5+6*I) for this example For the left component we have the sum: (3+4*I)^(3*(5+6*I)) + ((3+4*I)^(5+6*I)+(1+2*I))^3 which is equal to the product based on the factorization to: 2* ((3+4*I)^(5+6*I)+1/2+I) * ((3+4*I)^(2*(5+6*I))+ (1+2*I)*(3+4*I)^(5+6*I) + (-3+4*I)) The factorization is done by Pari for any c=(a+b*I) Which means that the roots as defined by my TPI identity somehow generate the coefficients of the factors in Q[I] The problem is when i set c=(a+b*I)^(c+d*I) Like i pointed out in the 2nd example of #18 for c=(1+2*I)^(3+4*I) Pari doesn't factor in Q[I]^Q[I]. and just outputs the roots numerically. But somehow the roots as defined by the TPI identity must relate to factors with coeficients in Q[i]^Q[I] Can you please show me how?(Your turn;-) Gerry === Subject: Re: About Zero. > 1*1 = 1 > 1 x 1 = 0.5 x 2 = 0.25 x 4 = 0.125 x 8 = ... = zero x infinity AB: Yes, the delta function, so useful in signal processing, by > definition. > As such zero times infinity is undefined. > 1/0 = infinity > 0/0 = undefined > 0 over 0 is not undefined but finite random AB: Undefined is better for understanding. Randomness relates to > probability or chances, which while very popular for obvious reasons has > no > scope here. Nobody is tossing a coin. > infinity/infinity = undefined > that too is finite random except zero> 0 * infinity = undefined that too is finite random (could be 1 or 2 or 3 or so) > finite * infinity = infinity > finite / infinity = 0- Hide quoted text - > - Show quoted text - > I would only comment that 0 does have some connection to > indeterminacy. > AB: Only if we select it as a part of a dimension, signifying origin. > To explain: > If you have zero bananas, and you also have zero coconuts, there is no > way to distinguish between the banana and the coconut. > AB: True. I have nothing. I have 0 bananas and 0 coconuts. So when I > want bananas, I have none. When I want coconuts, I have none. Thus when > I want to explain to my infant son the difference between bananas and > coconuts, what I get as reaction is also zero meaning nothing. > In general, to have zero of something, you cannot be sure of exactly > what object you have because zero of anything is identical to zero of > anything else. > AB: I am not sure of this. What is zero or loose change for Mr Bill > Gates is a million dollars for the rest of us. > So you do have indeterminacy, something like a singularity. Very > fascinating really. Actually, the fascinating aspect of a singularity is its unpredictabliltiy. To give some examples: - zero defects means great quality and low prices, and who knows what else - zero crime means bad news for law enforcement agencies, and who knows what else - zero drug usage means bad news for both drug runners and law enforcement agencies... - zero immorality means bad news for fallen women... - zero money means either Star Trek society or hunter-gatherer society... Such are the practical results from the concept of zero. In my case, it is the state what the poet in Urdu calls baykhubree or unawareness of information. It is the essence of creativity, and also unpopular with the wife. > AB: To understand zero, I think in terms of beneath measurement, just as > to understand infinity I think in terms of beyond measurement. > The police stopped a man in Alalbama. He said that he had no pies. But > the police said that zero pies is indistinguishable from zero > marijuana cigarettes. He is still in jail. > AB: Churi hoyay gechay raajkoshay! Chor chai, jay korei hok, Hok na > shay jay kono lok, noilay moder jaabay maan That was a line from a > policeman in a Bengali play, meaning, the treasury has been robbed, we > must find the thief, may it be anyone (guilty or innocent) otherwise we > shall lose prestige. Interesting, and while it makes a point and gives > some illumination of how a mathematician might view > the world, it has little application to real life > where we separate quantity and quality or in > this case quantity and identity of something > real. If the cop ascertains that the perp has zero > bananas at the moment with the assumption that he > has had more than zero bananas at some point in > time extends this argument to say that because the > perp has zero bananas, he also has zero joints, and > because he has zero joints, he has had as some point > more than zero joints, I wouldn't need a lawyer to > get him out of jail. Making your arguments about zero by using examples > from everyday life does not work. === Subject: Re: About Zero. > Zero is an exact quantiy meaning precisely NOTHING. In practice it > means > smaller than anything measurable, in both the theoretical and > practical > aspect. In theory, it is such a number that any measurable quantity > multiplied by it is also zero, or something which cannot be measured. > And > If zero is nothing then why do we incorporate it geometrically as an > origin? > Do we really do that? When we say (0,0,0) represents the origin of our > co-ordinate system, that basically means that we arbitrarily or for > practical purpose fix upon a particular point in space (and there could > be > infinite such points) as the basis for estimating distances between > points, > and related functionalities. Selecting a proper origin is very important > for simplicity of the maths. For instance, it is convenient to select > the > centre of the circle as the origin, to describe the circle in more simple > mathematics. Just as the choice of rectangular, cylindrical or polar > co-ordinates makes the maths a lot simpler, for the type of system we > want > represented analytically. > This notion of zero distance as being local is somewhat > counterintuitive or at least there is room for reinterpretation. If > things very distant have little interaction with things local then why > should they be considered large? > That things very distant have little or no interaction with things local > is > an assumption, not warranted by the physics relating to gravity. No > matter > how distant objects are, if the distance is finite, there will be a > finite > force acting between them which can be calculated and represented > mathematically. > Shouldn't those far away things be > zero? > Now, this is the traditional European Aristotlean-Einsteinian mind > speaking! > This underlines what I have said earlier, the concept of infinity is > unclear > to the dyed-in-the-wool occidental. No, far away things are not zero, > they > are near-infinite distance away if they are beyond the scope of > measurement. > Like, before Hubble in seen, there were galaxies not seen - their > apparent > non-existence owing to our lack of powerful telescopes did not render > their > presence zero, only unseen. Like, America was not really zero in the > times > before Colombus; it was merely unknown to the Europeans. That they > thought > it did not exist did not alter the reality of its existence. > Then when we consider our locality unity seems appropriate. Also > in terms of the product you will see that we still maintain symmetry > since > (1)(1) = 1, > (0)(0) = 0, > and (inf)(inf) = inf Oh no, this is completely wrong. > What is wrong above here? > Beneath here you are using division within a definitional context. In > that division is the inverse of product then these relations are not > fundamental. Especially because of the exception within the division > clause of the field behaviors I think that division is not as general > as product and this then exposes one of your lines > 1 / 0 = infinity > as in conflict with another line > 0 * infinity = undefined > and within the field behaviors the first expression is denied. > The arithmetic product is somewhat mysterious since we can take two > oranges and multiply them by three oranges and have the theoretical > six oranges yet there are still only five oranges. Superposition is > natural and the Cartesian geometry is a superpositional space. Still, > from physics we do see some product relationships yet they come out in > the second derivative of this superpositional space. I still grapple > with this and hope that this is a signal toward a new semiclassical > theory. > The actual mathematics as I learnt at school (in India) are by > definition: > 1*1 = 1 > 1/0 = infinity > 0/0 = undefined > infinity/infinity = undefined > 0 * infinity = undefined > finite * infinity = infinity > finite / infinity = 0 > As metaphysics, these definitions are of the utmost philosophical > relevance. > All existence is a play between zero and infinity, as that operation > relates > to various and unlimited sorts of definition. For those with a religious > or > mythological bent, Zero relates to Shiva, the Greatest God, in His search > to > investigate the core of everything to the very minutest of detail, that > is > Aum and Infinity or sum of all existence relates to Shakti, the Great > Goddess. All existence results from their union, and our consciousness > to > perceive same is Their great gift to us. It is comforting that the eastern religions are concerned with > exposing reality. I believe they are superior to the exclusive > Abrahamic faiths whose factions are warring with each other to this > day. I do not for a moment claim to speak for eastern religions. I have no business to do that. I am speaking of my personal perspective, derived from Grace and experience and knowledge of some scripture. > Let's see if I can try this reinterpretation of zero from a different > angle. We all have been schooled in the real number as a > generalization of the discrete integers. Really? I have not been so schooled. How do you think a computer actually treats numbers, as reals or integers? Reals are appropriately positioned or formatted binaries (that are integers) from a computer scientist's perspective, and that should make them a special case of integers. Given a chance, I *always* program with integers. Our work on cryptography has nothing to do with reals. So, when we talk of .5 of one integer which is set to zero, when we talk of .25 we talk of two integers, both set to zero. When we talk of .75 we have one integer set to one, the other to zero. === Subject: Re: About Zero. posting-account=n26igQkAAACeF9xA2Ms8cKIdBH40qzwr Gecko/20070505 Iceape/1.0.9 (Debian-1.0.11~pre071022-0etch1),gzip(gfe),gzip(gfe) > Zero is an exact quantiy meaning precisely NOTHING. In practice it > means > smaller than anything measurable, in both the theoretical and > practical > aspect. In theory, it is such a number that any measurable quantity > multiplied by it is also zero, or something which cannot be measured. > And > If zero is nothing then why do we incorporate it geometrically as an > origin? OK Arindam. I think the eastern religions are more philosphically open than are the Abrahamics. Since you enjoy computers then you might enjoy a structured approach to the treatment of sign: http://bandtechnology.com/PolySigned The consequences are pretty interesting. But this does not address zero except that in terms of dimension there is a zero dimensional space that corresponds with time. Still, this level of geometry is a bit different than people are going at on this thread w.r.t zero. While the polysign system works well with ordinary magnitude there is a signal within the polysign numbers that allow for emergent spacetime and this relies upon an arithmetic product. It is this product which has led me to consider the Y = 1 / ( X + 1 ) transform of ordinary distance X to the unity style origin in Y with zero far away. Again I will just reinforce the simple point that our notion of one meter versus one banana are unique concepts, and especially the consideration of zero meters against zero bananas. In terms of physics the classical force equations blow up at this position of adjacency. The transformed system does not. On other threads I've tried to consider concepts of the infinitesimal distance and adjacency and it seems that the real valued number does not inherently allow for a concept of adjacency. Toying in continuous distance is very different from toying in natural numbers and these ideas suggest that alternative metrics could be useful and so the interpretation of zero does seem appropriate to me in that way. Still, superpositionally the standard real valued metric works well if only it had a natural unit. This natural unit could then be an inherent form of adjacency and here again considering the origin as unity may be meaningful. A continuum concept can still be valid, it's just that the unitary distance cannot be exceeded. The polysign construction taken in this adjacent context suggests that the number of adjacent points is related to the dimension of the space. for instance it is clear that on P2 (the real line) that there will be two adjacent positions. The polysign format suggests that three adjacent positions insinuates a 2D space, four positions a 3D space, etc. whereas the traditional Cartesian form extended would yield 2 for 1D, 4 for 2D, 6 for 3D, etc. This also overlooks the one-signed nubers which alot for one adjacent position. Some of this is exposed in the lattice analysis http://bandtechnology.com/PolySigned/Lattice/Lattice.html but still I try to get a continuum with this discrete adjacency going into physics so I do study threads like this one. - Tim > Do we really do that? When we say (0,0,0) represents the origin of our > co-ordinate system, that basically means that we arbitrarily or for > practical purpose fix upon a particular point in space (and there could > be > infinite such points) as the basis for estimating distances between > points, > and related functionalities. Selecting a proper origin is very important > for simplicity of the maths. For instance, it is convenient to select > the > centre of the circle as the origin, to describe the circle in more simple > mathematics. Just as the choice of rectangular, cylindrical or polar > co-ordinates makes the maths a lot simpler, for the type of system we > want > represented analytically. > This notion of zero distance as being local is somewhat > counterintuitive or at least there is room for reinterpretation. If > things very distant have little interaction with things local then why > should they be considered large? > That things very distant have little or no interaction with things local > is > an assumption, not warranted by the physics relating to gravity. No > matter > how distant objects are, if the distance is finite, there will be a > finite > force acting between them which can be calculated and represented > mathematically. > Shouldn't those far away things be > zero? > Now, this is the traditional European Aristotlean-Einsteinian mind > speaking! > This underlines what I have said earlier, the concept of infinity is > unclear > to the dyed-in-the-wool occidental. No, far away things are not zero, > they > are near-infinite distance away if they are beyond the scope of > measurement. > Like, before Hubble in seen, there were galaxies not seen - their > apparent > non-existence owing to our lack of powerful telescopes did not render > their > presence zero, only unseen. Like, America was not really zero in the > times > before Colombus; it was merely unknown to the Europeans. That they > thought > it did not exist did not alter the reality of its existence. > Then when we consider our locality unity seems appropriate. Also > in terms of the product you will see that we still maintain symmetry > since > (1)(1) = 1, > (0)(0) = 0, > and (inf)(inf) = inf > Oh no, this is completely wrong. > What is wrong above here? > Beneath here you are using division within a definitional context. In > that division is the inverse of product then these relations are not > fundamental. Especially because of the exception within the division > clause of the field behaviors I think that division is not as general > as product and this then exposes one of your lines > 1 / 0 = infinity > as in conflict with another line > 0 * infinity = undefined > and within the field behaviors the first expression is denied. > The arithmetic product is somewhat mysterious since we can take two > oranges and multiply them by three oranges and have the theoretical > six oranges yet there are still only five oranges. Superposition is > natural and the Cartesian geometry is a superpositional space. Still, > from physics we do see some product relationships yet they come out in > the second derivative of this superpositional space. I still grapple > with this and hope that this is a signal toward a new semiclassical > theory. > The actual mathematics as I learnt at school (in India) are by > definition: > 1*1 = 1 > 1/0 = infinity > 0/0 = undefined > infinity/infinity = undefined > 0 * infinity = undefined > finite * infinity = infinity > finite / infinity = 0 > As metaphysics, these definitions are of the utmost philosophical > relevance. > All existence is a play between zero and infinity, as that operation > relates > to various and unlimited sorts of definition. For those with a religious > or > mythological bent, Zero relates to Shiva, the Greatest God, in His search > to > investigate the core of everything to the very minutest of detail, that > is > Aum and Infinity or sum of all existence relates to Shakti, the Great > Goddess. All existence results from their union, and our consciousness > to > perceive same is Their great gift to us. It is comforting that the eastern religions are concerned with > exposing reality. I believe they are superior to the exclusive > Abrahamic faiths whose factions are warring with each other to this > day. I do not for a moment claim to speak for eastern religions. I have no > business to do that. I am speaking of my personal perspective, derived from > Grace and experience and knowledge of some scripture. Let's see if I can try this reinterpretation of zero from a different > angle. We all have been schooled in the real number as a > generalization of the discrete integers. Really? I have not been so schooled. How do you think a computer actually > treats numbers, as reals or integers? Reals are appropriately positioned or > formatted binaries (that are integers) from a computer scientist's > perspective, and that should make them a special case of integers. Given a > chance, I *always* program with integers. Our work on cryptography has > nothing to do with reals. So, when we talk of .5 of one integer which is set to zero, when we talk of > .25 we talk of two integers, both set to zero. When we talk of .75 we have > one integer set to one, the other to zero. === Subject: computational experiments with sine products For Quadrant I, we can construct the disjoint subsets: M_1 = {theta, pi/4=1. U_k = {theta, 0 log(|sin(theta)|) is differentiable in a neighborhood of theta_0, and that dg/dtheta (theta) = cot(theta). If x in M_1, then dg/dtheta = cot(theta) is in (0, 1). If x in M_2, then dg/dtheta = cot(theta) is in (1, 2.415). If x is in M_10, then dg/dtheta = cot(theta) is in (325, 652) and so on. Using function g: theta |-> log(|sin(theta)|) may help in studying products of sines. For the January 2007 problem of L. Wapner concerning the sequence of products: a(n)=(2 sin 1)(2 sin 2)(2 sin 3)...(2 sin n) it was discussed in the thread: < > Also of interest here are the statements contained in formulas (1), (2) and (3) of a paper by Erdos and Szekeres from 1959: < http://www.renyi.hu/~p_erdos/1959-17.pdf > The proofs of (1) and (3) are given in outline. ---- For an irrational number c in (0, 1), we can define the sequence B_c (n) = 2^n product_{k=1 ... n} ( | sin(c*n*pi) | ). If we put c = 1/pi, then we get the problem posed above by Wapner. We can study the case c = (sqrt(5)-1)/2. Then c+1 is the Golden ratio. One convergent to (sqrt(5)-1)/2 is 610/987. Here we get prod(X=1,986,2*sin(X*Pi*610/987)) = -987 (proved by Robert Israel in the Jan. 2007 thread; was known before that). and prod(X=1,986,2*sin(c*X*Pi)) ~= -845.35 . The question we can ask is how often c*n*pi and 610/987 *n*pi are in the same set U_k , for n = 1 to 986. Out of 986, they are in different U_k six times. The other 980 times, they are in the same U_k with counts as follows: k Count 1 493 2 246 3 123 4 62 5 30 6 15 7 7 8 3 9 1 ------ 980 For all 986 values of n, sin(c*n*pi) and sin(610/987 *n*pi) have the same sign. So I believe this implies that the sine function has no zero between min(c*n*pi, 610/987 *n*pi) and max(c*n*pi, 610/987 *n*pi). For any n between 1 and 986, | (c-610/987)*n*pi | < 0.0015 . (***) We can now get upper bounds on log( | sin(c*n*pi) |) and log( |sin(610/987 *n*pi) | ) by combining the upper bound (***) with upper bounds on |dg/dtheta| = | cot(theta)| . For x in U_1, |cot(x)|<= cot(pi/4) = 1. For x in U_2, |cot(x)|<=cot(pi/8)<= 2.5 For x in U_3, |cot(x)| <= cot(pi/16) <= 5.1 For x in U_4, <= 10.2 For x in U_5, <= 20.4 For x in U_6, <= 40.8 For x in U_7, <= 81.5 For x in U_8, <= 163 For x in U_9, <= 326 For x in U_10, <= 652 For n=233, we have: 233: one number in U_9, one in U_8 ==> use U_9 |cot(.)| bound. So now we can combine the six exceptions with the Count by U_k above: k Adjusted Count Upper bound on |cot(.)| Product: 1 493 1 493 2 247 2.5 617.5 3 124 5.1 632.4 4 62 10.2 632.4 5 30 20.4 612 6 15 40.8 612 7 8 81.5 652 8 4 163 652 9 2 326 652 10 1 652 652 986 6207.3 Now we multiply 6207.3 by 0.0015 (from (***) above) to conclude that the absolute value of difference between the sum of the logs of |sin(.)| in one series and in the other is less than 6207.3 * 0.0015 <= 9.32 . On the other hand, log(987) = 6.8946 ... which is about 2.4 smaller than the upper bound. So c = (sqrt(5)-1)/2 was chosen because it doesn't have good rational approximations. Maybe something can be done practically for irrational c with better rational approximations (such as e). David Bernier === Subject: Re: computational experiments with sine products > For Quadrant I, we can construct the disjoint subsets: > M_1 = {theta, pi/4 M_2 = {theta, pi/8 and in general, M_k = {theta, pi/(2^(k+1)) < theta < pi/(2^k) } , for integers k>=1. U_k = {theta, 0 sin(x) }. > It can help to sketch a few regions M_k and U_k. It is easy to show that, if sin(theta_0) is non-zero, then > g: theta |-> log(|sin(theta)|) is differentiable in a neighborhood > of theta_0, and that dg/dtheta (theta) = cot(theta). If x in M_1, then dg/dtheta = cot(theta) is in (0, 1). > If x in M_2, then dg/dtheta = cot(theta) is in (1, 2.415). > If x is in M_10, then dg/dtheta = cot(theta) is in (325, 652) > and so on. Using function g: theta |-> log(|sin(theta)|) may help in studying > products of sines. For the January 2007 problem of L. Wapner concerning > the sequence of products: > a(n)=(2 sin 1)(2 sin 2)(2 sin 3)...(2 sin n) > it was discussed in the thread: > < > > Also of interest here are the statements contained in formulas > (1), (2) and (3) of a paper by Erdos and Szekeres from 1959: > < http://www.renyi.hu/~p_erdos/1959-17.pdf > The proofs of (1) and (3) are given in outline. ---- For an irrational number c in (0, 1), we can define the sequence > B_c (n) = 2^n product_{k=1 ... n} ( | sin(c*n*pi) | ). If we put c = 1/pi, then we get the problem posed above by > Wapner. We can study the case c = (sqrt(5)-1)/2. Then c+1 is the Golden ratio. One convergent to (sqrt(5)-1)/2 is 610/987. Here we get > prod(X=1,986,2*sin(X*Pi*610/987)) = -987 > (proved by Robert Israel in the Jan. 2007 thread; was known before > that). and > prod(X=1,986,2*sin(c*X*Pi)) ~= -845.35 . [...] > Now we multiply 6207.3 by 0.0015 (from (***) above) to conclude > that the absolute value of difference between the sum of > the logs of |sin(.)| in one series and in the other is less than > 6207.3 * 0.0015 <= 9.32 . On the other hand, > log(987) = 6.8946 ... which is about 2.4 smaller than > the upper bound. In other words, if A = sum_{n=1 ... 986} log(|sin(610/987*pi*n)|) and B = sum_{n=1 ... 986} log(|sin(c*pi*n)|) , then |A - B| <= 9.32 and we know that A = log(987)-986*log(2) , so Now B + 986*log(2) = log(prod_{n=1 ... 986} 2*|sin(c*n*pi)| ) and B + 986*log(2) >= log(987) - 9.32, so prod_{n=1 ... 986} 2*|sin(c*n*pi)| >= exp (log(987) - 9.32) >= 0.088 . In the case of arbitrary c, I'm interested in values of c for which one has or might have limsup_{k->oo} prod_{n=1 ... k} 2*|sin(c*n*pi)| = oo. David Bernier > So c = (sqrt(5)-1)/2 was chosen because it doesn't have good > rational approximations. Maybe something can be done practically > for irrational c with better rational approximations (such as e). === Subject: Re: Solution Manual West Federal Taxation 2008 Comprehensive posting-account=6iQmRAoAAACKCahjOmkffHizF9u7d3Ge CLR 1.1.4322),gzip(gfe),gzip(gfe) > I have the solutions manual to West Federal Taxation 2008 > Comprehensive Appendix E. æThis is seperate then the solution manual > for the book as this is 81 pages in itself. æIf you are interested > drop me an email I'm interested. What do u want for it & how quickly can u send it. Request acknowledged email sent Do you have the solutions to e-15 for CD, Inc. This is the only one I === Subject: Re: jerk > This is probably completely trivial, but somehow I > can't seem to > puzzle it out. I'm wheeling my kid's stroller in a circle at a > constant speed (let's > say the position vector is r(t)= (cos t, sin t) ). > Then the > acceleration vector is always pointing into the > circle, and has > constant length, so my kid feels no change in > acceleration or > jerk (which is why I wheel the stroller this way > when I want to put > him to sleep. But no matter.) But the change in > acceleration is of > course (sin t, -cos t), backward along the path of > travel, which would > imply (since it's nonzero) that my kid *should* feel > it. What's going > on? > Consider a projectile problem model. An airplane at uniform velocity drops an unpowered bomb that remains directly below the plane during its entire accelerated descent to Earth. There is no horizontal accleration of the airplane, only the vertical acceleration of gravity on the dropped projectile. I think that, as in the projectile problem, acceleration toward the center of the stroller's curvilinear path is the only acceleration component, given the constant speed of the stroller. That is, unless you speed up the stroller's movement, I would think that the stroller moves as if in a flat plane at uniform velocity. The situation would be different if you stood in the center of the circle with a line tied to the stroller and whipped it around the perimeter (please don't do this! :-) ) Tom > Michael Hamm > http://www.math.wustl.edu/~msh210/ === Subject: Re: jerk > I'm wheeling my kid's stroller in a circle at a constant speed (let's > say the position vector is r(t)= (cos t, sin t) ). Then the > acceleration vector is always pointing into the circle, and has > constant length, so my kid feels no change in acceleration or > jerk (which is why I wheel the stroller this way when I want to put > him to sleep. But no matter.) But the change in acceleration is of > course (sin t, -cos t), backward along the path of travel, which would > imply (since it's nonzero) that my kid *should* feel it. > Because it's small. You're talking about feeling the > rotation around your own axis as you move in a > circle and thus change orientations. For modest > speeds, this is not a large number. Not enough > to make you feel dizzy. I'm not following. I grant that the rotation of the passenger around > his axis is small (at low speeds); and whether it's enough to make him > feel dizzy is a question for a medical newsgroup. But what does any > of this have to do with the change in acceleration? But the change in acceleration is of course (sin t, -cos t), backward along the path of travel The magnitude of the acceleration is constant, w^2R, where w is the angular frequency and R is the radius. The direction is changing (rotating), but so is the baby - he experiences a constant pull to one side. This is small and constant, therefore easily ignored. === Subject: Re: tensor product of group algebras > The tensor product of the complex group algebra CG for > a group G with the group algebra CH for the group H is > simply the group algebra for the group G x H. To get an > isomorphism, simply map the obvious basis elements in C(G > x H) to the obvious elements in the tensor product of CG > by CH. > the group algebra of CG are the product too. Interesting. For group algebras there are three standard tensor products. The tensor product of CG-modules CG (x)_CG CG is just CG. The tensor product of C-algebras CG (x)_C CG is C[G x G]. The tensor product of G-modules is a G-module whose underlying vector space is C[G x G] with G action extended linearly from g*(h,k) = (gh,gk). In the first case, the CG irreducibles are exactly the same. In the second case, it may not make sense to talk about the CG irreducibles, but if you did, then I think it would be more accurate to say it contained |G| times as many copies of each irreducible, since C[G x G] as a G x 1 module is C[G] tensored with |G| copies of the trivial (central) module. In the third case, a direct sum decomposition of G-modules can be given where the summands are all pairwise tensor products of irreducible G-modules, repeated a certain number of times according to the dimensions of the factors. The tensor product of irreducible representations is rarely irreducible. In the second case, the irreducible G x G-modules appearing are just the (second style of) tensor products of G x 1-modules with 1 x G-modules. === Subject: lebesgue f is a nonnegative and integrable over a measurable set E. Then epsilon greater than 0 and delta greater than 0 s.t. over a measurable set A subset of E with lamda(A) f is a nonnegative and integrable over a measurable set E. Then > epsilon greater than 0 and delta greater than 0 s.t. over a measurable > set A subset of E with > lamda(A) then, > lambdaA=0 and f is measurable integral_R f X_A dlambda=0 we know that if A subset of E > integral_A f dlambda= integral_R X_A f dlambda<=integral_R X_E > dlambda= > integral_E f dlambda from this > 0<= integral_R X_E dlambda=integral_E f dlambda What should I do now? That depends on what you're trying to do, which is not at all clear. What exactly is the problem you're trying to solve? Please be careful to include the quantifiers (for all) and (there exists), and in the correct order: they are very important. -- === Subject: Re: lebesgue > f is a nonnegative and integrable over a measurable set E. Then > epsilon greater than 0 and delta greater than 0 s.t. over a measurable > set A subset of E with > lamda(A) then, > lambdaA=0 and f is measurable integral R f X A dlambda=0 we know that if A subset of E > integral A f dlambda= integral R X A f dlambda<=integral R X E > dlambda= > integral E f dlambda from this > 0<= integral R X E dlambda=integral E f dlambda What should I do now? That depends on what you're trying to do, which is not at all clear. > What exactly is the problem you're trying to solve? æPlease be careful > to include the quantifiers (for all) and (there exists), and in the correct > order: they are very important. > -- === Subject: Re: lebesgue > f is a nonnegative and integrable over a measurable set E. Then > epsilon greater than 0 and delta greater than 0 s.t. over a measurable > set A subset of E with > lamda(A) it means A has measure zero.If A has measure zero we can say lambdaA=0 > then, > lambdaA=0 and f is measurable integral_R f X_A dlambda=0 > we know that if A subset of E > integral_A f dlambda= integral_R X_A f dlambda<=integral_R X_E > dlambda= > integral_E f dlambda > from this > 0<= integral_R X_E dlambda=integral_E f dlambda > What should I do now? That depends on what you're trying to do, which is not at all clear. > What exactly is the problem you're trying to solve? .8cæPlease be careful > to include the quantifiers (for all) and (there exists), and in the correct > order: they are very important. > -- > Robert Israel .8cæ .8cæ .8cæ .8cæ .8cæ .8cæ .8cæisr...@math.MyUniversitysInitials.ca > Department of Mathematics .8cæ .8cæ .8cæ .8cæhttp://www.math.ubc.ca/~israel > University of British Columbia .8cæ .8cæ .8cæ .8cæ .8cæ .8cæVancouver, BC, Canada- Alıntıyı > gizle - - Alıntıyı gí.a6ster - I'm trying to show that .89Ì.82f dlambda A State the problem precisely. === Subject: Re: lebesgue > On 21 ?ubat, 13:06, Robert Israel > f is a nonnegative and integrable over a measurable set E. Then > epsilon greater than 0 and delta greater than 0 s.t. over a measurable > set A subset of E with > lamda(A) it means A has measure zero.If A has measure zero we can say lambdaA=0 > then, > lambdaA=0 and f is measurable integral R f X A dlambda=0 > we know that if A subset of E > integral A f dlambda= integral R X A f dlambda<=integral R X E > dlambda= > integral E f dlambda > from this > 0<= integral R X E dlambda=integral E f dlambda > What should I do now? > That depends on what you're trying to do, which is not at all clear. > What exactly is the problem you're trying to solve? æPlease be careful > to include the quantifiers (for all) and (there exists), and in the correct > order: they are very important. > -- > Robert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > Department of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > University of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Al.9bnt.9by.9b > gizle - > - Al.9bnt.9by.9b g.9aster - I'm trying to show that .bcf dlambda A State the problem precisely.- Al.9bnt.9by.9b gizle - - Al.9bnt.9by.9b g.9aster - f is a nonnegative and integrable over a measurable set E. Then .b9>0 and .83å>0 s.t. over a measurable set A subset of E with .83.83(A)<.83å, show that .81.8dfd.83.83<.b9 A === Subject: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) Flaming Thunder, at www.flamingthunder.com, is the world's most powerful computer programming language. Flaming Thunder is designed for everyone from elementary school students to scientists, mathematicians and engineers. Elementary school students can start programming in seconds using Flaming Thunder. The program to write Hello world! is: Write Hello world!. Scientists, mathematicians and engineers will appreciate that Flaming Thunder is programmed entirely in assembly language, takes less than 120K of disk space, supports interval arithmetic, supports number theoretic transforms for O(N log(N)) multiplication of high precision numbers, supports 64-bit platforms, and cross-compiles for FreeBSD, Linux, Mac OS X (Intel-based) and Windows. Educators will appreciate the horizontal and vertical curriculum integration provided by Flaming Thunder. Horizontally, Flaming Thunder is appropriate not only for math and computer classes - but also for English classes. Look back at the Hello world! example: it is a simple English sentence ending with a period. Flaming Thunder is the only major programming language to combine literacy with numeracy. Vertically, students can use Flaming Thunder from the very beginning of their educational careers and on into their professional lives. Using the same tool from year to year compounds knowledge and pays big dividends. More info, and free versions for Windows, Mac (Intel-based), Linux, and FreeBSD at: http://www.flamingthunder.com/index.html === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX 5.0),gzip(gfe),gzip(gfe) > Flaming Thunder In it's current state, I would say it really is flaming thunder! === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) Flaming Thunder In it's current state, I would say it really is flaming thunder! Did you ever see the flick Taras Bulba? A critic panned it as Thud and Blunder. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) Flaming Thunder In it's current state, I would say it really is flaming thunder! :) Yep, there are lots of features we'll be adding in the coming weeks and months, plus integrating the graphics from DPGraph. That's why we're offering 4 extra months on a one-year subscription right now, and only 10 cents (or less) per student for a site license. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX 5.0),gzip(gfe),gzip(gfe) > Flaming Thunder In it's current state, I would say it really is flaming thunder! :) Yep, there are lots of features we'll be adding in the coming > weeks and months, plus integrating the graphics from DPGraph. > That's why we're offering 4 extra months on a one-year > subscription right now, and only 10 cents (or less) per > student for a site license. The writeup compares it to C/C++ and Java which is ridiculous. This looks like a fallback to COBOL - and lousy at that. Lots of stuff missing. No networking, no bit manipulation (that I can find), no modular compilation, no address/hardware access, I didn't see support for any data structures or objects, access to OS system calls, and so on... Why would you expect anyone to pay for that when there are free alternatives? Why not use a Computer-Algebra-System which can run circles around this 'flaming' idea? Do I believe programming languages need to be simplified - well yes - no argument there. Is this idea going to resolve that - not in this lifetime. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language > no bit manipulation (that I can find), Actually, we're intentionally giving that low priority in the early versions. The reason is that from a user's viewpoint, who cares what base the computer hardware uses? Base 2, base 7, base 10, their programs should run the same no matter what the base is. That's the sort of irrelevance that we want to insulate our biggest base of users from (K-12 teachers, mathematicians, scientists, engineers, etc). Computer scientists are not are largest target market. Not yet, anyway. Flaming Thunder originally started as an assembler -- but we all know how big the market for assemblers is. ;) Eventually, even assembly language will appear. We've already ensured that the syntax is compatible with assembly language. It's not a far syntactic jump from: Write Please enter your name: . Read UserName. Write Your name is , username. to: Push 0. Call imp ExitProcess@4. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Why would you expect anyone to pay for that when there are free > alternatives? Really? Name another language with built-in support for interval arithmetic, that's easy enough for elementary school teachers to use, and that can cross-compile directly to executables on both 32 and 64 bit versions of FreeBSD, Linux, Mac OS X and Windows -- with no extra effort on the part of the user. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > Flaming Thunder, atwww.flamingthunder.com, is the world's most > powerful computer programming language. Flaming Thunder is designed > for everyone from elementary school students to scientists, > mathematicians and engineers. Elementary school students can start programming in seconds using > Flaming Thunder. The program to write Hello world! is: æ æWrite Hello world!. Scientists, mathematicians and engineers will appreciate that Flaming > Thunder is programmed entirely in assembly language, takes less than > 120K of disk space, supports interval arithmetic, supports number > theoretic transforms for O(N log(N)) multiplication of high precision > numbers, supports 64-bit platforms, and cross-compiles for FreeBSD, > Linux, Mac OS X (Intel-based) and Windows. Educators will appreciate the horizontal and vertical curriculum > integration provided by Flaming Thunder. Horizontally, Flaming Thunder > is appropriate not only for math and computer classes - but also for > English classes. Look back at the Hello world! example: it is a > simple English sentence ending with a period. Flaming Thunder is the > only major programming language to combine literacy with numeracy. Vertically, students can use Flaming Thunder from the very beginning > of their educational careers and on into their professional lives. > Using the same tool from year to year compounds knowledge and pays big > dividends. More info, and free versions for Windows, Mac (Intel-based), Linux, > and FreeBSD at:http://www.flamingthunder.com/index.html How come you have a bunch of stupid stuff that no one needs, like hyperbolic trig, yet you omit important number theoretical stuff like GreatestCommonDenominator, LeastCommonMultiple, ModularInverse, IsPrime, IsSquare, etc.? A language that's worthless, i.e., of no value to me, certainly can't claim to be the most powerful. Come back when you support all the features of the GMP library. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) On Feb 21, 12:59æpm, mensana...@aol.compost like hyperbolic trig, yet you omit important number theoretical > stuff like GreatestCommonDenominator, LeastCommonMultiple, > ModularInverse, IsPrime, IsSquare, etc.? Those will be coming soon. The reason for the trig is that those are used in junior high (lots of customers) and adding the hyperbolic trig was easy enough to do along with the trig. > A language that's worthless, i.e., of no value to me, certainly > can't claim to be the most powerful. Certainly it can. Power is integrated over all users, not determined by a single user. By using simple English syntax, Flaming Thunder empowers more users, from elementary school on up. > Come back when you support > all the features of the GMP library. Will do. By the way, does the GMP library support interval arithmetic yet, like Flaming Thunder? === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language > How come you have a bunch of stupid stuff that no one needs, > like hyperbolic trig, yet you omit important number theoretical > stuff like GreatestCommonDenominator, LeastCommonMultiple, > ModularInverse, IsPrime, IsSquare, etc.? Those will be coming soon. æ Real Soon Now, eh? > The reason for the trig is that those > are used in junior high (lots of customers) and adding the > hyperbolic trig was easy enough to do along with the trig. A language that's worthless, i.e., of no value to me, certainly > can't claim to be the most powerful. Certainly it can. æPower is integrated over all users, not > determined by a single user. æ And people don't eat in the long run, they eat every day. Power integrated across all users is worthless to the individual. > By using simple English syntax, > Flaming Thunder empowers more users, from elementary school > on up. Come back when you support > all the features of the GMP library. Will do. By the way, does the GMP library support interval arithmetic yet, > like Flaming Thunder? Who asked for that? I didn't say GMP could do everything. For example, it doesn't do factorization. But if you expect anyone to change from using say, Python, then you have to replace it with someting equally powerful, preferably something more powerful to justify converting my existing algorithms that use Python/GMP and C/GMP. When you reach that point, then maybe you can claim most powerful, maybe. And don't forget to add stuff like lists, list slicing, dictionaries, sets, database interface, etc. Can't write programs without those things either. Study what can be done in Python and make sure your language can do the equivalent. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) On Feb 21, 3:06æpm, mensana...@aol.compost anyone to change from using say, Python, then you have to > replace it with someting equally powerful, preferably something > more powerful to justify converting my existing algorithms > that use Python/GMP and C/GMP. When you reach that point, > then maybe you can claim most powerful, maybe. Right now, more elementary school students are using Flaming Thunder than Python. They're the future. Matlab, Mathematica, etc, hook people on their software by offering student versions at low prices, then charging them a lot when they go into business. In the long run, whoever captures the youngest students wins. Matlab, Mathematica, C, Java, Python, are unusable by elementary school students, or more particularly, unusable by elementary school teachers. > Study what can be done > in Python and make sure your language can do the equivalent. Already have. We're adding new features as quickly as possible -- in the order that we think that will help us catch the broadest market as quickly as possible. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language > Flaming Thunder, at www.flamingthunder.com, is the world's most > powerful computer programming language. Can it do CGI easily, so that anyone who has written a neat program can easily put it up on the Web to show off for others what the person has done? Do *you* have any CGI demos programmed in Flaming Thunder so that *you* can show off what Flaming Thunder can be used for? > Flaming Thunder is programmed entirely in assembly language ... Why isn't it written in itself? Is Flaming Thunder not powerful enough to be used to write a compiler for itself?? > supports interval arithmetic That's nice. Do you have a CGI demo of that? If not, why not? Can't Flaming Thunder be used for CGI demos? === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) On Feb 21, 12:23æpm, rem6...@yahoo.com (Robert Maas, see http://tinyurl.com/uh3t) > Can it do CGI easily, so that anyone who has written a neat program > can easily put it up on the Web to show off for others what the > person has done? Do *you* have any CGI demos programmed in Flaming > Thunder so that *you* can show off what Flaming Thunder can be used > for? Yes. In fact, I am working on some examples today. It's not ready for public release yet (I'm writing it as we speak), but we're among friends here, right? ;) Here is a link to a non-public page of what I have right at this momemnt at: http://www.flamingthunder.com/examples.html The CGI scripts are at the bottom. You should be able to click on the links to execute them. Let me know if you run into any problems with them. > Why isn't it written in itself? Is Flaming Thunder not powerful > enough to be used to write a compiler for itself?? That's one of the features we're adding. For our first real release last weekend, we aimed for a general audience first. > That's nice. Do you have a CGI demo of that? If not, why not? > Can't Flaming Thunder be used for CGI demos? Please see the above CGI script demos. The free version of FTNotepad (available from the main page at http://www.flamingthunder.com/) already has a demo of interval arithmetic, and later today I'll add a demo of interval arithmetic to the CGI scripts above. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language > ready for public release yet (I'm writing it as we speak), but > we're among friends here, right? ;) æHere is a link to a non-public > page of what I have right at this momemnt at: http://www.flamingthunder.com/examples.html Oops! Looks like they're not working right now, at least not from my browser. Since I'm in the middle of editing and testing, I guess I didn't sync things up. But, if you are really want to try right now, here's a link to another site I use for my builds, which might have better verions (or have the permissions set correctly): http://www.archilingua.com/examples.html Later on today everything will be synced up with the right permissions; I'll post a note here (probably this evening) when it goes public. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Oops! æLooks like they're not working right now, at least > not from my browser. Oops again, I was wrong. The CGI scripts were working just fine, but the server was overloaded by too many people trying to access them at the same time, so it was throwing an error 500. If the CGI scripts didn't work for you earlier, you might want to go back and try them again now that the rush is probably over. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language > Flaming Thunder [--] is the world's most powerful computer programming > language. No it isn't. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) On Feb 21, 8:04æam, Aatu Koskensilta Yes it is. No it isn't. Flaming Thunder appears to be basically a very limited BASIC variant. It's thoroughly silly to describe it as the world's most powerful computer programming language. > The power of a computer language lies in it's ability to translate > human desires into actions by the machine. How does Flaming Thunder do at translating my desire for a juicy hamburger into actions by the machine? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language <44jvj.300141$0G5.258148@reader1.news.saunalahti.fi> posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) On Feb 21, 11:05æam, Aatu Koskensilta BASIC variant. Flaming Thunder is also an English variant, a Pascal variant, a C variant, etc. Everything in the world borrows something of the old to create the new. There are seveal things that make Flaming Thunder different. For example, like English, sentences end with periods. The keywords are English (as opposed to DIM etc). Flaming Thunder frees you from 32 or 64 bitness: all integers, reals, etc are arbitrary precision. > It's thoroughly silly to describe it as the world's > most powerful computer programming language. No it's not. It's the world's most powerful language because anyone who already knows English -- in particular, elementary school children and their teachers -- can use it. Power can be measured by those it empowers, making Flaming Thunder by far the world's most powerful programming language. > How does Flaming Thunder do at translating my desire for a juicy > hamburger into actions by the machine? I've added it to our list of features to add. ;) === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language There are seveal things that make Flaming Thunder different. > For example, like English, sentences end with periods. The > keywords are English (as opposed to DIM etc). Different from what? It sounds like you're describing COBOL. Is Flaming Thunder a giant step back to the 1950s? > It's the world's most powerful language because > anyone who already knows English -- in particular, elementary > school children and their teachers -- can use it. This is total nonsense. If someone knows English and wants to use Flaming Thunder, they still needs to learn the language. Otherwise they're just going to waste time writing dozens of English sentences which don't do anything in Flaming Thunder. I think the industry's experience with COBOL has demonstrated that English-like syntax doesn't make a language better (and certainly not more powerful). --Mark === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) On Feb 21, 12:03æpm, Mark Nudelman æIs > Flaming Thunder a giant step back to the 1950s? No, but it does carry forward some of the best lessons from the 1950's, and later. After all, we still speak English today. However, if you look at C, Java, etc, you'll see that they still carry forward the proud Lisp tradition of lots of parentheses, and have added curly and square brackets. > If someone knows English and wants to use > Flaming Thunder, they still needs to learn the language. True. Flaming Thunder builds on their English skills, but isn't identical. > Otherwise > they're just going to waste time writing dozens of English sentences > which don't do anything in Flaming Thunder. I think not. Most people program by example, so I think most elementary school students will read a Flaming Thunder example or two, and then write something very similar. >æI think the industry's > experience with COBOL has demonstrated that English-like syntax doesn't > make a language better (and certainly not more powerful). English-like syntax on it's own doesn't make a language more powerful, but it helps. Flaming Thunder adds arbitrary precision math, interval arithmetic, cross-compilation, etc. And an English-like syntax does give you a much wider market. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language > Flaming Thunder is also an English variant, a Pascal variant, > a C variant, etc. Everything in the world borrows something of > the old to create the new. By no stretch of imagination is Flaming Thunder an English variant. > No it's not. It's the world's most powerful language because > anyone who already knows English -- in particular, elementary > school children and their teachers -- can use it. You seem to take an extremely dim view on the intelligence of elementary school children and their teachers. But surely if they're utterly incapable of comprehending why, say, the following Python statement does not end in a period or why it uses print instead of write print Hello world! they're equally mystified by the failure of Flaming Thunder to obey their simple order Write out the multiplication table. Further, their poor minds will be thoroughly addled by the bizarre omission of spaces between words in e.g. WriteLine or GetEnvironmentVariableCount, and will fall into catatonic stupor when pondering the enigma of the quire ungrammatical #-signs cluttering the example programs. > I've added it to our list of features to add. ;) Since Flaming Thunder is powerless on face of my desire for a juicy cheeseburger, let's instead consider how it does at translating, to actions by machine, my desire for a Lisp interpreter. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) On Feb 21, 11:49æam, Aatu Koskensilta You seem to take an extremely dim view on the intelligence of > elementary school children and their teachers. Not at all. But they already know English, so there's no point wasting their time teaching them: main() { printf(Hello world!); } when they already know: Write Hello world!. > But surely if they're > utterly incapable of comprehending why, say, the following Python > statement does not end in a period or why it uses print instead of > write æprint Hello world! Okay, why doesn't it have a period at the end? And why doesn't it say write instead of print? > they're equally mystified by the failure of Flaming Thunder to obey > their simple order æWrite out the multiplication table. One of our internal research goals is to incorporate more and more English syntax into Flaming Thunder. I've added your example to the list. > Further, their poor minds will be thoroughly addled by the bizarre > omission of spaces between words in e.g. WriteLine or > GetEnvironmentVariableCount, True. To partially overcome that, Flaming Thunder ignores underscores and case, so that WriteLine, WRITE LINE, etc, are all equivalent. However, we are already testing multi-word procedure/ function names. It will be appearing as a feature at some point. > and will fall into catatonic stupor when > pondering the enigma of the quire ungrammatical #-signs cluttering the > example programs. That's why we chose them. They're clearly ungrammatical, and the comments aren't part of the grammar of the program. > Since Flaming Thunder is powerless on face of my desire for a juicy > cheeseburger, let's instead consider how it does at translating, to > actions by machine, my desire for a Lisp interpreter. Or would that be dethire? ;) === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language > On Feb 21, 8:04æam, Aatu Koskensilta human desires into actions by the machine. In Flaming Thunder, the program to write Hello world! is: Write Hello world!. That makes Flaming Thunder powerful enough for even elementary > school children. Which is far more powerful than C, C++, Java, > etc, since none of those can translate the human desires of > elementary school children into actions by the machine. What you seem to be claiming is that Flaming Thunder is easy to use or easy to learn. It is very nonstandard to call this a measure of how powerful the language is. What makes Write Hello world!. better than main() { printf(Hello world!); } ? Is it the smaller number of characters? In Perl, the same program is print hello world! Does the lack of punctuation after the statement make Perl that much more powerful than Flaming Thunder? Isn't Perl easier to use because it doesn't need to be compiled? Do you really believe that elementary school children are incapable of prgramming in C but can easily program in Flaming Thunder? Writing a non-trivial program requires skills that are mostly independent of the language. Do you believe that children can start programming in Flaming Thunder without learning the language first? Do you have any empirical evidence that Flaming Thunder is easier to learn that C or Perl? --Mark === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) On Feb 21, 10:31æam, Mark Nudelman easy to learn. Yes. It builds off of the lessons that elementary school students are already learning in their English and math classes. > It is very nonstandard to call this a measure of how > powerful the language is. Perhaps. But since the machines exist to serve us, and we don't exist to serve the machines (well ... at least not yet ;) ), then the appropriate measure of the power of a language is how directly can it translate a human's desires into action by the machine. > What makes > æ æ æ æ Write Hello world!. > better than > æ æ æ æ main() { printf(Hello world!); } > ? Write Hello world! compounds the interest already invested in teaching someone English. C discards that investment. > Is it the smaller number of characters? No, it's the fact that everyone, even children, already learn that the most logical way to tell someone or something to write Hello world! is to say: write hello world!. > In Perl, the same program is > æ æ æ æ print hello world! And when you walk over to your printer, is that what comes out on a sheet of paper? Children all learn that sentences end with periods. Where is the period at the end of the Perl sentence? Instead of compounding the knowledge investments already made in children, Perl (like C) discards them. > Does the lack of punctuation after the statement make Perl that much > more powerful than Flaming Thunder? No, it makes it less powerful. It means you have to waste extra brain cells to learn Perl's special rules, instead of using the rules you already know from English class. >æIsn't Perl easier to use because > it doesn't need to be compiled? No. Interpreted languages are less powerful. Any compiled language can have its compilation phase hidden to make it look like an interpreted language. For example, Flaming Thunder has an interpreted version called FTNotepad that you can run directly from the internet (it's right on the home page at www.flamingthunder.com). What it does when you press the Run button is compile in the background, then run the compiled code, in a way that the user never sees. > Do you really believe that elementary school children are incapable of > prgramming in C but can easily program in Flaming Thunder? Yes. For example, elementary school kids like to play word games called mad-libs where they are asked for random words and then plug them into stories. An elementary school student can easily write a mad-lib in Flaming Thunder: Write Please enter your name: . Read yourname. Write Please enter an animal: . Read animal. Write The wild , animal, ate , yourname. Notice how no mathematical symbols were used in that non-mathematical program? Math symbols tend to scare away English teachers, so Flaming Thunder doesn't use math symbols unless you're actually doing math. >æWriting a > non-trivial program requires skills that are mostly independent of the > language. I disagree. A powerful language can make non-trivial tasks easy. For example, writing a C program to do general interval arithmetic is non-trivial. Doing it in Flaming Thunder is trivial even for elementary school students: Set x to interval( 1.101, 1.102 ). Set y to interval( 2.233, 2.234 ). Write x^2 + y^2. >æDo you believe that children can start programming in > Flaming Thunder without learning the language first? Yes. They already know English: Write Hello world!. > Do you have any > empirical evidence that Flaming Thunder is easier to learn that C or > Perl? Yes. English is already taught in all the schools in the US. Therefore, they already know most of what they need to use Flaming Thunder. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language > On Feb 21, 10:31æam, Mark Nudelman æ æ æ æ Write Hello world!. > better than > æ æ æ æ main() { printf(Hello world!); } > ? Write Hello world! compounds the interest already invested in > teaching someone English. C discards that investment. Is it the smaller number of characters? No, it's the fact that everyone, even children, already learn that > the most logical way to tell someone or something to write Hello > world! > is to say: write hello world!. In Perl, the same program is > æ æ æ æ print hello world! And when you walk over to your printer, is that what comes out on a > sheet of paper? Oh, come now. If I tell someone Print your name, do you expect them to go looking for a computer? Print and write both have several meanings. What a computer does when it runs a hello world program seems more like printing than writing to me, unless you're using a cursive font. > Children all learn that sentences end with periods. > Where is the period at the end of the Perl sentence? Instead of compounding the knowledge investments already made in > children, > Perl (like C) discards them. Children learn that you join nouns with and when you want to do something to both of them. So does this work in Flaming Thunder? Write hello and goodbye. Children learn that the then is optional in many if statements. So does this work in Flaming Thunder? If x<5 Write ok. Children learn that the order of phrases can be inverted in an if statment. So does this work in Flaming Thunder? Write ok if x<5. Obviously I could go on for pages inventing examples like this. English syntax is vastly vastly larger than Flaming Thunder syntax, and without training a person is most likely to produce constructs that are valid in English but not in FT. English syntax *may* make it easier to remember the language details (but you haven't provided any evidence of it), but it goes way too far to claim that Flaming Thunder's syntax is somehow obvious to an English speaker, in the sense that this is how someone would naturally tell a computer to do something. > No. Interpreted languages are less powerful. Any compiled language > can > have its compilation phase hidden to make it look like an interpreted > language. And an interpreted language can look like a compiled one, by providing a dummy compiler program that just sleeps for a few seconds. > I disagree. A powerful language can make non-trivial tasks easy. For example, writing a C program to do general interval arithmetic is > non-trivial. > Doing it in Flaming Thunder is trivial even for elementary school > students: Set x to interval( 1.101, 1.102 ). > Set y to interval( 2.233, 2.234 ). > Write x^2 + y^2. I grant that the built-in support for interval arithmetic and arbitrary precision math is nice, even powerful. But what if I want to do something other than the built-in stuff? If I want to add support for complex numbers, can I end up with syntax like this? Set x to complex(1,2). The way the language stands today, without functions (not to mention classes), I doubt there's any easy way to deal with complex numbers at all. BTW, what is the largest program that's been written in FT? It's hard for me to imagine a program doing anything really useful given the language limitations. > Do you have any > empirical evidence that Flaming Thunder is easier to learn that C or > Perl? Yes. English is already taught in all the schools in the US. > Therefore, > they already know most of what they need to use Flaming Thunder. I'm sure you realize that this is a totally evasive answer. You have no evidence that English knowledge has any bearing on the ease of learning Flaming Thunder. You're just assuming that. When I ask for empirical evidence, I mean I want to see an experiment where you take equivalent groups of students and teach one group Flaming Thunder and teach another group another language, and compare the outcomes. --Mark === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language something to both of them. æSo does this work in Flaming Thunder? > æ æ æ æ Write hello and goodbye. Yes. I typed it into the free Flaming Thunder Notepad version, hello and goodbye ;) Flaming Thunder can do symbolic math, so when it can't calculate an explicit answer for an operation (such as hello anded with goodbye), it returns the expression evaluated up the parts that it couldn't do. > English syntax is vastly vastly larger than Flaming Thunder syntax, and > without training a person is most likely to produce constructs that are > valid in English but not in FT. True. Flaming Thunder tries as hard as possible to be a subset of English, not encompass all possible variations. As opposed to C, Java, etc, which barely intersect with English -- wasting all the linguistic knowledge that people have already acquired by elementary school. > English syntax *may* make it easier to remember the language details It sure does. For example, even FORTRAN, C, Java, etc, with their minimal intersection with English are easier for English speakers to comprehend. > (but you haven't provided any evidence of it) Try translating all the Java keywords and package names into Pashtun, then see how well an English speaker can comprehend the program. > but it goes way too far > to claim that Flaming Thunder's syntax is somehow obvious to an English > speaker, in the sense that this is how someone would naturally tell a > computer to do something. English teacher to student: Write Hello world!. Student to Flaming Thunder: Write Hello world!. It seems obvious and natural to me that if you want a computer program to write Hello world! that you would tell it to Write Hello world!. Since Flaming Thunder is trying as hard as possible to be a subset of English, that seemed like a good part of English to be a subset of. Another way to tell the computer to write hello world! might be to tell it: Yo computer, lay some Hello world! on me. ... but write Hello world! seemed at least a little bit more obvious and natural a way to tell the computer to write hello world!. > And an interpreted language can look like a compiled one, by providing > a dummy compiler program that just sleeps for a few seconds. No, because you're not left with an executable that you can distribute. You have to distribute the compiler to your customers. Flaming Thunder compiles to executables that you can distribute; your customers don't have to license Flaming Thunder, too. > I grant that the built-in support for interval arithmetic and arbitrary > precision math is nice, even powerful. æBut what if I want to do > something other than the built-in stuff? æIf I want to add support for > complex numbers, can I end up with syntax like this? > æ æ æ æ Set x to complex(1,2). Yes. In fact, as soon as we get the complex support in, that's exactly how it will look. And ... > The way the language stands today, without functions (not to mention > classes), I doubt there's any easy way to deal with complex numbers at > all. ... you are correct. As it stands today, you can't define your own functions. We can do it internally (in assembly), but we haven't finished the front end on the parser that allows users to create their own. It will be coming soon. That's one of the reasons why were adding 4 extra free months to a one-year subscription, and giving everyone a free copy of DPGraph, and site licensing it for only 10 cents per student. > BTW, what is the largest program that's been written in FT? Well, if by large you mean number of lines -- not very large at all. But if by large you mean lots of output, it's easy to do big factorials. ;) > It's hard > for me to imagine a program doing anything really useful given the > language limitations. Yes, the lack of user-defined functions, and as an earlier poster pointed out, the relative lack of built-in functions, is limiting right now. But those, obviously, are things high on our list to add. But to someone who's trying to calculate the precision of their results given the imprecision of their data, the current interval arithmetic alone is probably reason enough for buying it. Also, for elementary/junior high students starting to learn the basics of programming, it's hard to beat Flaming Thunder for ease-of-use. Even a computer/math phobic elementary school teacher could probably master clicking one/two/or three times (depending on their browser) from the Flaming Thunder homepage to bring up the Flaming Thunder Notepad version. > You have > no evidence that English knowledge has any bearing on the ease of > learning Flaming Thunder. Sure I do. I've shown Flaming Thunder to lots of people who have never programmed, and they've been able to write simple programs. That's a lot of how the language evolved to its current form. > When I ask for > empirical evidence, I mean I want to see an experiment where you take > equivalent groups of students and teach one group Flaming Thunder and > teach another group another language, and compare the outcomes. Do you mean a published study? Perhaps in the future; the 1st major release of Flaming Thunder was just last weekend. But from my own experience, it's easier to sell elementary school teachers Flaming Thunder than C -- because to most elementary school teachers (who are, after all, the ones who would be doing the teaching) C (and Java, et aL) looks like gobbledy-geek. Write Hello world! looks like English. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language > ... everyone, even children, already learn that the most logical > way to tell someone or something to write Hello world! is to say: > write hello world!. If I ask somebody to write something for me, I expect them to pick up a pen or pencil and physically write it on a piece of paper. For example, if I meet somebody and ask them to write their name and e-mail address, I expect those two items of information to be written on a piece of paper that I can carry home in my pocket or backpack. I don't expect their face to light up with a visual display of their name and e-mail address. > Flaming Thunder has an interpreted version called FTNotepad > that you can run directly from the internet (it's right on the home > page at www.flamingthunder.com). I searched that Web page. There's no mention of FTNotepad anywhere on that Web page, in fact the string FTN appears nowhere there. > What it does when you press the Run button is compile in the > background, then run the compiled code, in a way that the user > never sees. When I tell somebody to do something, I expect it to be done right away, not wait until I press the RUN button on them somewhere. Did you go to a school where each student had a RUN button on them, and none of the students did anything the teather had said to do until after the teacher pressed the particular student's RUN button? > Write The wild , animal, ate , yourname. I don't recall ever seeing that particular syntax in elementary school. It seems your wonderful new programming language isn't well modeled after what students already learned after all. > For example, writing a C program to do general interval > arithmetic is non-trivial. C is not a good example of a high-level programming language. You are attacking a so-called straw man. > Yes. They already know English: Write Hello world!. We're back to your mistaken example, where the English says to pick up a pencil or pen and write that text on a piece of paper, but your stupid programming language does no such thing, so the newbie must learn the *new* meaning of the old words, i.e. must un-learn what was learned in school as to what the words normally mean in English. Psychological studies show it's more difficult to un-learn something to replace with new meaning than to just learn something new that doesn't mask something else previously known. > English is already taught in all the schools in the US. > Therefore, they already know most of what they need to use > Flaming Thunder. Two problems: - As I mentionned, they must un-learn the old meaning before they can learn the new meaning of exactly the same syntax. - What about the rest of the world where English isn't commonly known? ---------- +--+ +---+ / | | | | /| +--+ +---+ / | | | | | / | +--+ +---+ | / | | / _| - Why should Chinese people need to learn the nitpicking stupid details of English syntax, such as capital letter at start and period at end of sentence, before they can write software? (Wouldn't it be so much easier for *everyone* to just learn that there are matching parentheses around each expression?) (Even Chinese use parentheses sometimes nowadays!!) (And the first word in each expression is the verb, everything else are nouns, or noun-sub-expressions, that feed into that verb.) (+ (interval 4.6 4.7) (interval 3.14 3.15)) (Looking up individual verbs in a English-Chinese dictionary is a lot easier than trying to learn how to write foreign language syntax!) === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) On Feb 21, 1:14æpm, rem6...@yahoo.com (Robert Maas, see http://tinyurl.com/uh3t) > If I ask somebody to write something for me, I expect them to pick > up a pen or pencil and physically write it on a piece of paper. When my kids write their emails, the hardest part is cramming the paper through the RJ45 jack. > I searched that Web page. There's no mention of FTNotepad > anywhere on that Web page, in fact the string FTN appears nowhere > there. If you are running Windows, you can access it by clicking on the link that says Windows users, click here Windows users, click here to start programming in seconds using the Free Edition of Flaming Thunder Notepad. > When I tell somebody to do something, I expect it to be done right > away ... So does my wife. ;) > C is not a good example of a high-level programming language. > You are attacking a so-called straw man. Okay, you provide the man. Here is a complete program to calculate x^2 + y^3 using interval arithmetic in Flaming Thunder: Set x to ( 1.001, 1.002 ). Set y to ( 4.202, 4.257 ). Write x^2 + y^3. Please provide your example of any high level level language that performs the same calculation in a way that's simpler and easier to understand. > We're back to your mistaken example, where the English says to pick > up a pencil or pen and write that text on a piece of paper ... So when kids write emails on computers, they're actually drawing graffiti on the case? > - What about the rest of the world where English isn't commonly known? We're already contemplating Spanish, French, etc, versions. > - Why should Chinese people need to learn the nitpicking stupid > æ ædetails of English syntax, such as capital letter at start and > æ æperiod at end of sentence, before they can write software? They don't need to learn that for Flaming Thunder. Flaming Thunder ignores the case of letters. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Set x to ( 1.001, 1.002 ). > Set y to ( 4.202, 4.257 ). > Write x^2 + y^3. Please provide your example of any high level level language > that performs the same calculation in a way that's simpler and > easier to understand. Oops, that should be: Set x to interval( 1.001, 1.002 ). Set y to interval( 4.202, 4.257 ). Write x^2 + y^3. The challenge still stands. If you know of any high level language where the program to perform that interval calculation is simpler and easier to understand, I'd like to see it. Actually, in Flaming Thunder one could also do it in a single line: Write interval(1.001,1.002)^2 + interval(4.202,4.257)^3. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language > Writing a non-trivial program requires skills that are mostly > independent of the language. Indeed, and such newfangled inventions as the ability to define functions and procedure of your own and to otherwise structure your code, apparently missing in Flaming Thunder, are considered to be of some use by some. Probably elitists, the whole lot of them, unable to appreciate the beauty and elegance of Set result to 1. For year from 2000 to 2008 do set result to result*year. Write result. or encoding the logic of your program in a bunch of imaginatively placed GOTOs, while, for and until loops and other such bundles of joy. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) On Feb 21, 11:13æam, Aatu Koskensilta functions and procedure of your own and to otherwise structure your > code, apparently missing in Flaming Thunder, are considered to be of > some use by some. Correct. They are actually working internally (we use them to write some of the procedures such as Write, etc), but we're still adjusting a few things before releasing them for general use (since Flaming Thunder runs on FreeBSD, Linux, Macs and Windows, and can cross-compile for all of them, testing all the combinations can be a little time-consuming). > ... encoding the logic of your program in a bunch of imaginatively > placed GOTOs, while, for and until loops and other such bundles of > joy. In Flaming Thunder, like English, that would be go to. Goto was used by early language designers because it made life easier for the compiler writer, instead of making life easier for the users, who then had to learn a new word. Kind of like how C's case-sensitivity makes life easier for the compiler writer, but harder for the users. === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language > Writing a non-trivial program requires skills that are mostly > independent of the language. Indeed, and such newfangled inventions as the ability to define > functions and procedure of your own and to otherwise structure your > code, apparently missing in Flaming Thunder, are considered to be of > some use by some. Probably elitists, the whole lot of them, unable to > appreciate the beauty and elegance of Set result to 1. > For year from 2000 to 2008 do > set result to result*year. > Write result. or encoding the logic of your program in a bunch of imaginatively > placed GOTOs, while, for and until loops and other such bundles of > joy. This seems like an instance of the COBOL school of language design, the philosophy being that the closer a program looks to ordinary English, the better it is. --Mark === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language > Writing a non-trivial program requires skills that are mostly > independent of the language. > > Indeed, and such newfangled inventions as the ability to define > functions and procedure of your own and to otherwise structure your > code, apparently missing in Flaming Thunder, are considered to be of > some use by some. Probably elitists, the whole lot of them, unable to > appreciate the beauty and elegance of > > Set result to 1. > For year from 2000 to 2008 do > set result to result*year. > Write result. > > or encoding the logic of your program in a bunch of imaginatively > placed GOTOs, while, for and until loops and other such bundles of > joy. > This seems like an instance of the COBOL school of language design, the > philosophy being that the closer a program looks to ordinary English, > the better it is. It's a candygrammar. -- Dave Seaman Oral Arguments in Mumia Abu-Jamal Case heard May 17 U.S. Court of Appeals, Third Circuit === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language > This seems like an instance of the COBOL school of language > design, the philosophy being that the closer a program looks to > ordinary English, the better it is. COBOL actually had at least one good idea: PICture clauses. In COBOL's formative years, 80-column Hollerith cards, with all data laid out in fixed columns, were the standard for input and output. Even printed reports used fixed format. When a data layout was specified once, it could be used both to parse incoming data and to generate outgoing data in the same format. The modern equivalent for variable-length fields with nested structure would be BNF. (The more popular regular expressions, aren't powerful enough, IMO.) I haven't seen any programming language that directly used BNF for its standard way to specify input/output parsing/formatting, which I see as a deficiency in *all* programming languages to date. So, here's a task for the English-like folks: Try to find an English-like syntax for expressing BNF, then incorporate it into a modern programming language. Here's a strawman starting idea: A WFF can be a 'p' or 'q' or 'r' or 's', or an 'N' followed by a WFF, or (a 'C' or A or 'K' or 'E') followed by two WFFs. (That idea is not original with me. Shake-a-WFF was a fun lively game!) === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language posting-account=qIx73woAAADp5gKOJOA4ARqKqxrzLb1s .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) On Feb 21, 1:41æpm, rem6...@yahoo.com (Robert Maas, see http://tinyurl.com/uh3t) > I haven't seen any > programming language that directly used BNF for its standard way to > specify input/output parsing/formatting, which I see as a > deficiency in *all* programming languages to date. In the first version of the documenation for Flaming Thunder, I used BNF form. Also, I used regular expression diagrams. But some people who read it over said the simplest for them to understand was just a line giving the basic syntax, a little description, and a couple of examples. > (That idea is not original with me. Shake-a-WFF was a fun lively game!) My wife and kids and I have played it a couple of times, but for some reason they prefer the XBox or Wii. ;) === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language This seems like an instance of the COBOL school of language design, the > philosophy being that the closer a program looks to ordinary English, > the better it is. Yes, in part. Beyond that, Flaming Thunder adds symbolic computation, arbitrary precision math, interval arithmetic, cross-compilation, etc. The main reason for lots of the weirdness in C, C++, Java, etc, is to make life easier for the compiler writer instead of the language users. Flaming Thunder is focused on making life easier for the users. === === === Subject: Circle center from point on circle Hello! I'm a programmer and need some math help! In a cartesian coordinate system a point is specified (x1,y1). This point is on the circle (with the radius r) and I also now the tangent of the point (t). I need to calculate the center of the circle as coordinates in the same coordinate system (x2,y2) The point may be anywhere in the coordinate system in all four quadrants. I.e. x and y may be positive and negative or zero. E.g. (-1,0), (0,23), (25,-11), ... Please Help!! // Anders -- English is not my first, or second, language so anything strange, or insulting, is due to the translation. Please correct me so I may improve my English! === Subject: Re: Circle center from point on circle > Hello! I'm a programmer and need some math help! In a cartesian coordinate system a point is specified (x1,y1). This point > is on the circle (with the radius r) and I also now the tangent of the > point (t). You know the tangent at the specified point, and so you should also know the slope m of that tangent. > I need to calculate the center of the circle as coordinates in the same > coordinate system (x2,y2) The point may be anywhere in the coordinate system in all four quadrants. > I.e. x and y may be positive and negative or zero. E.g. (-1,0), (0,23), > (25,-11), ... You have not given enough information to determine the center of the circle uniquely; there are possible centers on either side of the tangent line at distance r from point (x1, y1). The possible centers are (x1 -/+ r/sqrt(1 + 1/m^2) , y1 +/- r/sqrt(m^2 + 1) ) Note: Opposite signs are to be used for the terms involving the square roots. David === Subject: Re: Circle center from point on circle > Hello! I'm a programmer and need some math help! In a cartesian coordinate system a point is specified (x1,y1). This > point is on the circle (with the radius r) and I also now the tangent > of the point (t). You know the tangent at the specified point, and so you should also know > the slope m of that tangent. I need to calculate the center of the circle as coordinates in the same > coordinate system (x2,y2) The point may be anywhere in the coordinate system in all four > quadrants. I.e. x and y may be positive and negative or zero. E.g. (-1, > 0), (0,23), (25,-11), ... You have not given enough information to determine the center of the > circle uniquely; there are possible centers on either side of the tangent > line at distance r from point (x1, y1). The possible centers are (x1 -/+ r/sqrt(1 + 1/m^2) , y1 +/- r/sqrt(m^2 + 1) ) Note: Opposite signs are to be used for the terms involving the square > roots. Oops. I made a mistake. Sorry! What I gave above is incorrect if m < 0. Correctly, the possible centers are (x1 -/+ sign(m) r/sqrt(1 + 1/m^2) , y1 +/- r/sqrt(m^2 + 1) ). For completeness, note that, if the tangent line is vertical, the possible centers are (x1 +/- r, y1). David === Subject: Re: Circle center from point on circle > Hello! > I'm a programmer and need some math help! > In a cartesian coordinate system a point is specified (x1,y1). This > point is on the circle (with the radius r) and I also now the tangent > of the point (t). > You know the tangent at the specified point, and so you should also know > the slope m of that tangent. > I need to calculate the center of the circle as coordinates in the same > coordinate system (x2,y2) > The point may be anywhere in the coordinate system in all four > quadrants. I.e. x and y may be positive and negative or zero. E.g. (-1, > 0), (0,23), (25,-11), ... > You have not given enough information to determine the center of the > circle uniquely; there are possible centers on either side of the tangent > line at distance r from point (x1, y1). > The possible centers are > (x1 -/+ r/sqrt(1 + 1/m^2) , y1 +/- r/sqrt(m^2 + 1) ) > Note: Opposite signs are to be used for the terms involving the square > roots. Oops. I made a mistake. Sorry! What I gave above is incorrect if m < 0. >Correctly, the possible centers are (x1 -/+ sign(m) r/sqrt(1 + 1/m^2) , y1 +/- r/sqrt(m^2 + 1) ). For completeness, note that, if the tangent line is vertical, the possible >centers are (x1 +/- r, y1). We can get rid of the sign() and unify the denominators to make things look a little prettier with ( x1 -/+ m r/sqrt(m^2 + 1) , y1 +/- r/sqrt(m^2 + 1) ) Rob Johnson take out the trash before replying === Subject: Re: Circle center from point on circle > Hello! > I'm a programmer and need some math help! > In a cartesian coordinate system a point is specified (x1,y1). This > point is on the circle (with the radius r) and I also now the > tangent of the point (t). > You know the tangent at the specified point, and so you should also > know the slope m of that tangent. > I need to calculate the center of the circle as coordinates in the > same coordinate system (x2,y2) > The point may be anywhere in the coordinate system in all four > quadrants. I.e. x and y may be positive and negative or zero. E.g. > (-1, 0), (0,23), (25,-11), ... > You have not given enough information to determine the center of the > circle uniquely; there are possible centers on either side of the > tangent line at distance r from point (x1, y1). > The possible centers are > (x1 -/+ r/sqrt(1 + 1/m^2) , y1 +/- r/sqrt(m^2 + 1) ) > Note: Opposite signs are to be used for the terms involving the square > roots. Oops. I made a mistake. Sorry! What I gave above is incorrect if m < 0. >Correctly, the possible centers are (x1 -/+ sign(m) r/sqrt(1 + 1/m^2) , y1 +/- r/sqrt(m^2 + 1) ). For completeness, note that, if the tangent line is vertical, the >possible centers are (x1 +/- r, y1). We can get rid of the sign() and unify the denominators to make > things look a little prettier with ( x1 -/+ m r/sqrt(m^2 + 1) , y1 +/- r/sqrt(m^2 + 1) ) There is a particular reason I chose not to give that form: It doesn't work when the tangent happens to be vertical. You might object, saying that the form I gave doesn't work when the tangent is vertical either. But it actually does work. The slope of a vertical line is oo, positive-or-negative infinity. For the center's y-coordinate, y1 +/- r/sqrt(oo^2 + 1) = y1 +/- r/oo = y1 +/- 0 = y1; for the center's x-coordinate, x1 -/+ sign(oo) r/sqrt(1 + 1/oo^2) = x1 -/+ (+/-1) r/sqrt(1 + 0) = x1 +/- r having taken sign(oo) to be bivalued. BTW, the reason I had said For completeness, note... is merely that I thought Anders might not want to use extended arithmetic. But my preference is to present a single result which works in _all_ cases. On further reflection, I regret having mentioned slope at all. Here's a nicer method. Let (x2, y2) be any point on the tangent line, distinct from (x1, y1). Then the possible centers of the circle are (x1 -/+ (y2 - y1) r/sqrt((x2 - x1)^2 + (y2 - y1)^2), y1 +/- (x2 - x1) r/sqrt((x2 - x1)^2 + (y2 - y1)^2)). David === Subject: Re: Circle center from point on circle > Hello! I'm a programmer and need some math help! In a cartesian coordinate system a point is specified (x1,y1). This > point is on the circle (with the radius r) and I also now the > tangent of the point (t). You know the tangent at the specified point, and so you should also > know the slope m of that tangent. I need to calculate the center of the circle as coordinates in the > same coordinate system (x2,y2) The point may be anywhere in the coordinate system in all four > quadrants. I.e. x and y may be positive and negative or zero. E.g. > (-1, 0), (0,23), (25,-11), ... You have not given enough information to determine the center of the > circle uniquely; there are possible centers on either side of the > tangent line at distance r from point (x1, y1). The possible centers are (x1 -/+ r/sqrt(1 + 1/m^2) , y1 +/- r/sqrt(m^2 + 1) ) Note: Opposite signs are to be used for the terms involving the square > roots. >Oops. I made a mistake. Sorry! What I gave above is incorrect if m < 0. >Correctly, the possible centers are >(x1 -/+ sign(m) r/sqrt(1 + 1/m^2) , y1 +/- r/sqrt(m^2 + 1) ). >For completeness, note that, if the tangent line is vertical, the >possible centers are (x1 +/- r, y1). > We can get rid of the sign() and unify the denominators to make > things look a little prettier with > ( x1 -/+ m r/sqrt(m^2 + 1) , y1 +/- r/sqrt(m^2 + 1) ) There is a particular reason I chose not to give that form: It doesn't work >when the tangent happens to be vertical. You might object, saying that the >form I gave doesn't work when the tangent is vertical either. But it >actually does work. The slope of a vertical line is oo, >positive-or-negative infinity. For the center's y-coordinate, y1 +/- r/sqrt(oo^2 + 1) = y1 +/- r/oo = y1 +/- 0 = y1; for the center's x-coordinate, x1 -/+ sign(oo) r/sqrt(1 + 1/oo^2) = x1 -/+ (+/-1) r/sqrt(1 + 0) = x1 +/- r having taken sign(oo) to be bivalued. BTW, the reason I had said For completeness, note... is merely that I >thought Anders might not want to use extended arithmetic. But my preference >is to present a single result which works in _all_ cases. On further reflection, I regret having mentioned slope at all. Here's a >nicer method. Let (x2, y2) be any point on the tangent line, distinct from >(x1, y1). Then the possible centers of the circle are (x1 -/+ (y2 - y1) r/sqrt((x2 - x1)^2 + (y2 - y1)^2), > y1 +/- (x2 - x1) r/sqrt((x2 - x1)^2 + (y2 - y1)^2)). Indeed, your formula works in all cases, if one makes a few, widely accepted, functional extensions when using the extended reals. For x = oo, extend the following functions: 1/x^2 = 0 1/sqrt(x^2 + 1) = 0 sign(x) = +1 or -1 (the reader chooses) My formula works in all cases as well if one makes similar functional extensions. For x = oo, extend the following functions: 1/sqrt(x^2 + 1) = 0 x/sqrt(x^2 + 1) = +1 or -1 (the reader chooses) ( x1 -/+ r m/sqrt(m^2 + 1) , y1 +/- r/sqrt(m^2 + 1) ) so that m/sqrt(m^2 + 1) was together, to be interpreted as a single function; but then I swapped the r and m to make it look closer to your formula. I had assumed that you added the completeness note to handle the case m = oo. Since you had not mentioned extended reals, I did not realize that you had intended your formula to apply to the extended reals. Both formulas work in all cases, if the proper extensions are made so that the functions are defined on the extended reals. Rob Johnson take out the trash before replying === Subject: Re: Circle center from point on circle > Hello! I'm a programmer and need some math help! In a cartesian coordinate system a point is specified (x1,y1). > This point is on the circle (with the radius r) and I also now > the tangent of the point (t). You know the tangent at the specified point, and so you should also > know the slope m of that tangent. I need to calculate the center of the circle as coordinates in > the same coordinate system (x2,y2) The point may be anywhere in the coordinate system in all four > quadrants. I.e. x and y may be positive and negative or zero. > E.g. (-1, 0), (0,23), (25,-11), ... You have not given enough information to determine the center of > the circle uniquely; there are possible centers on either side of > the tangent line at distance r from point (x1, y1). The possible centers are (x1 -/+ r/sqrt(1 + 1/m^2) , y1 +/- r/sqrt(m^2 + 1) ) Note: Opposite signs are to be used for the terms involving the > square roots. >Oops. I made a mistake. Sorry! What I gave above is incorrect >if m < 0. Correctly, the possible centers are >(x1 -/+ sign(m) r/sqrt(1 + 1/m^2) , y1 +/- r/sqrt(m^2 + 1) ). >For completeness, note that, if the tangent line is vertical, the >possible centers are (x1 +/- r, y1). > We can get rid of the sign() and unify the denominators to make > things look a little prettier with > ( x1 -/+ m r/sqrt(m^2 + 1) , y1 +/- r/sqrt(m^2 + 1) ) There is a particular reason I chose not to give that form: It doesn't >work when the tangent happens to be vertical. You might object, saying >that the form I gave doesn't work when the tangent is vertical either. >But it actually does work. The slope of a vertical line is oo, >positive-or-negative infinity. For the center's y-coordinate, y1 +/- r/sqrt(oo^2 + 1) = y1 +/- r/oo = y1 +/- 0 = y1; for the center's x-coordinate, x1 -/+ sign(oo) r/sqrt(1 + 1/oo^2) = x1 -/+ (+/-1) r/sqrt(1 + 0) = x1 +/- r having taken sign(oo) to be bivalued. BTW, the reason I had said For completeness, note... is merely that I >thought Anders might not want to use extended arithmetic. But my >preference is to present a single result which works in _all_ cases. On further reflection, I regret having mentioned slope at all. Here's a >nicer method. Let (x2, y2) be any point on the tangent line, distinct >from (x1, y1). Then the possible centers of the circle are (x1 -/+ (y2 - y1) r/sqrt((x2 - x1)^2 + (y2 - y1)^2), > y1 +/- (x2 - x1) r/sqrt((x2 - x1)^2 + (y2 - y1)^2)). Indeed, your formula works in all cases, if one makes a few, widely > accepted, functional extensions when using the extended reals. Actually, I was talking about just extended _arithmetic_, as used by a computer algebra system such as Derive. (Furthermore, if all we're wanting to do is to get the possible coordinates for the center correctly, then one could use _positive_ infinity for the slope of a vertical line. Although that is generally a bad idea, it would allow one to do the programming in standard floating-point arithmetic.) > For x = oo, extend the following functions: 1/x^2 = 0 1/sqrt(x^2 + 1) = 0 In both Derive and in floating-point arithmetic, we get the above results when x is 1/0. > sign(x) = +1 or -1 (the reader chooses) In Derive, sign(1/0) = +/- 1. (The program does not choose.) In floating-point arithmetic, sign(1/0) = 1. In any event, the formula works. > My formula works in all cases as well if one makes similar functional > extensions. Functional extensions, yes. But that goes beyond extended arithmetic. For Anders' computer work, I think it's better to deal with just arithmetic. > For x = oo, extend the following functions: 1/sqrt(x^2 + 1) = 0 x/sqrt(x^2 + 1) = +1 or -1 (the reader chooses) For x/sqrt(x^2 + 1), when x = 1/0, Derive yields ? and floating-point arithmetic yields NaN. In neither case can we continue with a useful calculation. ( x1 -/+ r m/sqrt(m^2 + 1) , y1 +/- r/sqrt(m^2 + 1) ) so that m/sqrt(m^2 + 1) was together, to be interpreted as a single > function; but then I swapped the r and m to make it look closer to > your formula. I had assumed that you added the completeness note to > handle the case m = oo. Since you had not mentioned extended reals, > I did not realize that you had intended your formula to apply to the > extended reals. Both formulas work in all cases, if the proper extensions are made so > that the functions are defined on the extended reals. Right. But if we are to have m/sqrt(m^2 + 1) extended as your formula requires, then we must use something beyond extended arithmetic, using, say, something like if(m = oo, 1, m/sqrt(m^2 + 1)). OTOH, 1/sqrt(1 + 1/m^2) evaluates as it should using merely standard extended arithmetic. David === Subject: Re: Circle center from point on circle posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20070530 Fedora/1.5.0.12-1.fc5 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) > Hello! I'm a programmer and need some math help! In a cartesian coordinate system a point is specified (x1,y1). This point > is on the circle (with the radius r) and I also now the tangent of the > point (t). I need to calculate the center of the circle as coordinates in the same > coordinate system (x2,y2) What do you mean by I know the tanget of the point? If youmean that you know the direction of the tangent line to the circle at the point (x1, y1), a few drawings should convince of the fact that information does not completely determine the center of the circle. -- m === Subject: Re: -- more factoring <8o9pr394am1l4a5ip4pt1iauc727d5r3r3@4ax.com> posting-account=PSzFRAoAAAARszS8zeFmxtqyivK9-1_f InfoPath.2),gzip(gfe),gzip(gfe) All polynomials discussed are of deg >= 2 and univariate. ... Let gcd(a,b) = 1 Let F be an irreducible integer polynomial that only has primefactors 1 mod a. Let G be an irreducible integer polynomial that only has primefactors 1 mod b. Let n and m be positive integers and for all n and m the image of F and G are positive integers > 1. Now it is clear that: gcd(F(n),G(m)) = 1 for all n and m. No, it's not clear. A prime can be congruent to both 1 (mod a) and 1 (mod b). Thus, your condition on the prime factors of F,G does not imply æ ægcd(F(n),G(m)) = 1 for all m,n. In fact, I'll make the following claim ... Conjecture: There do not exist nonconstant univariate integer polynomials f,g such > that gcd(f(m),g(n)) = 1 for all m,n in Z. quasi Hi i have a question related to this topic : What are the conditions for common factors(f(m),g(n))=A033203 A033203 : integer sequence of Primes congruent to {1, 2, 3} mod 8; or primes of form x^2+2*y^2; or primes p such that x^2 = -2 has a solution mod p. Example: f(x)=41x+41 g(x)=11449x^4-15622x^3+2333x^2+2044x+228 Gerry === Subject: Milenko Kindl owns Rays' 'sailboat' stadium posting-account=DYXx1goAAABIvegUL8AUDJpQUSIjdqNv 5.1),gzip(gfe),gzip(gfe) Don't adjust your monitors, people. Pictured above is a rendering of the Rays' proposed $450 million waterfront stadium in St. Petersburg. The photos of this plan have been available for some time, but the team just released this image yesterday, which features the park with the roof in the closed position. As you can tell from the photo, translucency is king. And after seeing those Lindsay Lohan photos in New York magazine, I have to ask: Who doesn't love translucency? The roof is supposed to resemble a sail and I suppose it does. But it also kinda looks like a kids' pillow fort. Not to mention that the long post in the outfield originally reminded me of Montreal's Olympic Stadium. Still, I really like this plan, which could be completed as early as 2012. It'll be unique to the Bay area and it's obvious that anything has to be better than Tropicana Field. The sad truth is that until the Rays and Marlins get new stadiums, I'll never attempt to hit every ballpark in the bigs. This design would make the Florida swing worth it, though. Now if only they could figure out how to equip it with A/C. Milenko Kindl Banja Lula Banjaluka === Subject: Milenko Kindl hit satellite posting-account=DYXx1goAAABIvegUL8AUDJpQUSIjdqNv 5.1),gzip(gfe),gzip(gfe) WASHINGTON - Debris from an obliterated U.S. spy satellite is being tracked over the Pacific and Atlantic oceans but appears to be too small to cause damage on Earth, a senior military officer said Thursday, just hours after a Navy missile scored a direct hit on the failing spacecraft. ADVERTISEMENT Marine Gen. James Cartwright, vice chairman of the Joint Chiefs of Staff and an expert on military space technologies, told a Pentagon news conference that officials have a high degree of confidence that the missile launched from a Navy cruiser Wednesday night hit exactly where intended. It was an unprecedented mission for the Navy, so extraordinary that the final go-ahead to launch the missile Wednesday was reserved for Defense Secretary Robert Gates rather than a military commander. Cartwright estimated there was an 80 percent to 90 percent chance that the missile struck the most important target on the satellite -- its fuel tank, containing 1,000 pounds of hydrazine, which Pentagon officials say could have posed a health hazard to humans if it had landed in a populated area. Alluding to a video clip of the missile smashing into the satellite, which he showed at the news conference, Cartwright said, We have a fireball, and given that there's no fuel (on the tip of the missile), that would indicate that that's a hydrazine fire. The video showed the three-stage SM-3 missile launching from the USS Lake Erie at 10:26 p.m. EST, northwest of Hawaii, and of the missile's small kill vehicle -- a non-explosive device at the tip -- maneuvering into the path of the satellite and colliding spectacularly. He said the satellite and the kill vehicle collided at a combined speed of 22,000 mph about 130 miles above Earth's surface, and that the collision was confirmed at a space operations center at 10:50 p.m. EST. Asked about the satisfaction felt among those in the military who had organized the shootdown on short notice by modifying missile software and other components, Cartwright smiled widely. Yes, this was uncharted territory. The technical degree of difficulty was significant here, Cartwright said. You can imagine that at the point of intercept there were a few cheers that went up. He cautioned, however, that more technical analysis was required to determine for certain what debris was created and where it might go. The satellite was described as the size of a school bus and weighed about 5,000 pounds. Unlike most spacecraft that fall out of orbit and re-enter the atmosphere, this satellite had an almost full fuel tank because it lost power and became uncontrollable shortly after it reached its initial orbit in December 2006. Cartwright said the hydrazine alone was justification for undertaking the unprecedented effort to use a Navy missile interceptor to attempt to destroy the satellite in orbit. Cartwright said experts were still watching the debris fields and he could not yet rule out that hazardous material would fall to Earth. But he said that as of Thursday morning, debris had only been seen in the atmosphere -- and none had been detected surviving re-entry. He indicated that debris appeared unlikely to pose a problem. Thus far we've seen nothing larger than a football, he said, referring to debris in the atmosphere spotted by radars and other sensors. The military concluded that the missile had successfully shattered the satellite because trackers detected a fireball. Cartwright said it was unlikely that the fireball could have been caused by anything other than the hydrazine in the tank. And Cartwright cited two other sources of information that indicate the fuel tank was hit: the appearance of a vapor cloud and the results of spectral analysis, or the study of light emissions, from devices aboard two aircraft that operate from the Pacific test range associated with the Pentagon's missile defense testing. Debris from the satellite has started re-entry and will continue through Thursday and into Friday, Cartwright said. Milenko Kindl Banja Luka Banjaluka === Subject: Milenko Kindl claims fall by 9,000 posting-account=DYXx1goAAABIvegUL8AUDJpQUSIjdqNv 5.1),gzip(gfe),gzip(gfe) WASHINGTON - The number of newly laid off workers filing claims for unemployment benefits fell last week, but the larger-than-expected drop was seen as only a temporary improvement. ADVERTISEMENT The Labor Department reported Thursday that the number of jobless claims dropped by 9,000 last week to a total of 349,000. While that was bigger than the decline that had been expected, analysts noted that claims offices in California, the largest state, were closed for one day last week for a state holiday, giving laid off workers one less day to file claims.. The four-week average for claims, which gives a better picture of labor market trends, rose to 360,500, which was the highest level since claims spiked in October 2005 in the aftermath of Hurricane Katrina. Analysts said the rise in the four-week average was depicting a labor market that is coming under increasing strains because of the slowing economy. The Federal Reserve released a revised economic forecast on Wednesday that slashed growth prospects for this year but still maintained that the country could avoid a recession. However, many private economist believe the country has already entered a downturn that began this quarter and will last through the spring. They are forecasting that the overall economy, which skidded to growth at a barely discernible annual rate of 0.6 percent in the final three months of last year, will turn negative in the first and second quarter this year. The classic definition of a recession is two consecutive quarters of negative economic growth. President Bush last week signed a $168 billion economic stimulus bill which is designed to provide rebate checks of $600 for individual and $1,200 for couples with the checks scheduled to begin arriving in mailboxes this spring as a way to give the economy a boost. But even with the rebate checks, which Bush called a booster shot for the economy, analysts are forecasting a period of slower growth, reflecting the blows that have been dealt by a prolonged downturn in housing and a severe credit squeeze, which has made it harder for consumers and businesses to get loans. The economy shed 17,000 jobs in January, the first monthly job loss in more than four years. Analysts believe that the unemployment rate, which currently stands at 4.9 percent, will rise to 6 percent before the current slowdown has run its course. The performance of jobless claims for the week ending Feb. 9 was revised to show 10,000 more benefit applications during that week than previously reported, reflecting a sharp revision from data supplied by California. For that week, 37 states and territories had increases in claims while 16 had declines. California had the largest increase, a jump of 7,857 that was attributed to higher layoffs in trade, service and manufacturing industries. The biggest decline in claims occurred in Ohio, which saw a drop of 2,752. Milenko Kindl Banja Luka Banjaluka === Subject: =?ISO-8859-1?Q?Re:_Les_Nombres_R=E9els_et_leur_Position_Centrale.?= posting-account=1qbAGAkAAADcUtlizzXUEb5jUjfAdE2y Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > Le plus grand r.8eel n.8egatif est il alg.8ebrique ou transcendant ? > Mohwali Awamar. Point de vue (n.8eo) pythagoriste : seuls existent les cardinaux de 1 jusqu'au nombre d'atomos de l'univers. O n'est pas un nombre. 1 .8el.8ephant 1 chien 1 fleuve z.8ero quoi ? 1 .8el.8ephant = 1 .8el.8ephant z.8ero quoi =n'importe quel z.8ero ----------------------------------------- VOTRE SOUCIS Votre .8eglise devait r.8epondre moins aleph 1 -------------------------------------- PIEGE Combien y a t il des transcendants ? Si l'on inclut ..... pi / 4 si l'on inclut pi puissance pi ---------------------------------------- DEMONTAGE DE L 'IMPOSTURE R n'existe que si Cantor a raison si et seulement si Card (N) = Card des pairs = Card des impairs et donc si et seulement si il existe une bijection en l'ensemble des entiers et l'ensemble des nombre pairs -------------------------------------- VULGARISATION D'apr.8fs Cantor il associe 1 avec 2 2 4 3 6 et il ose pr.8etendre que c'est une bijection ----------------------------------- Mais ces farceurs (qui ont PRIS LE POUVOIR EN DEUG de maths et FORCENT nos enfants .88 REPETER DES CONNERIES pour avoir leur module) ------------------------------------------------------ IMPOSTURE le 2 de la deuxi.8fme colonne, le 2 de l'ensemble des nombres pairs est le m.90me nombre que le 2 de l'ensemble des nombres entiers Il n'y pas deux 2 diff.8erents ILS TRICHENT ---------------------------------------------------------------- Ils font une partition et s'empressent de l'oublier ----------------------------------------------------------- TOUTE LA NOTION DE CONTINUITE S'EFFONDRE DANS CETTE PAGE ------------------------------------------------------------- SEULE L'ARITHMETIQUE EST LOGIQUE IL N EXISTE QUE DES ESCALIERS DES INTEGRALES PAPIER PEINT (SOUS L'ESCALIER) LES EQUATIONS DIFFERENTIELS SONT LES RELATIONS ENTRE LA HAUTEUR DE L' ESCALIER ET LA LONGUEUR DE CHAQUE MARCHE (MARCHES è HAUTEUR CONSTANTE) Cela fait 8 ans que je r.8ep.8fte cela ici (sous les insultes et les quolibets des cr.8etins ignares et d.8epourvus d'argument) Il faut retrouver dans les archives mon manifeste Bon anniversaire monsieur Planck Il y a une premi.8fre publication faite par Jacques Barot (+ depuis) sur internet. On voit dans ce forum le premier cr.8etin abruti s'empresser de d.8enigrer sans r.8efl.8echir Voila le texte de l'empress.8e Alain Frisch La post.8erit.8e retiendra son nom (mais certainement pas pour ses travaux !!!) Pour les anonymes cr.8etins consultables sur les liens, leur affichage sera fait par les sites sp.8ecialis.8es dans l'.8epist.8emologie plus tard. Yanick Toutain Rappeler les exploits d'Alain Frisch a aussi pour fonction (discr.8fte) de montrer aux .8eventuels amateurs, que leur prose continuiste aura le m.90me sort en 2016. Qu'ils apprennent la politesse et .88 argumenter ! === Subject: Re: Les Nombres =?ISO-8859-1?Q?R=E9els_et_leur_Position_?= =?ISO-8859-1?Q?Centrale=2E?= > Le plus grand r.8eel n.8egatif est il alg.8ebrique ou transcendant ? > Mohwali Awamar. Point de vue (n.8eo) pythagoriste : seuls existent les cardinaux de 1 jusqu'au nombre d'atomos de > l'univers. > O n'est pas un nombre. 1 .8el.8ephant 1 chien 1 fleuve > z.8ero quoi ? Mais TOUTAIN !!! Toutain = z.8ero !, Toutain, nul, nullard, Khmer Rouge et Garde Rouge .88 la Mao ASSASSIN de la culture (comme les autres, comme ceux de 1917... les historiens ont mis en .8evidence suffisamment d'.8el.8ements ! comme ceux de toutes les r.8evolutions, ou presque toutes) ! Allez, Toutain, vive le grand camarade Staline, vive la science prol.8etarienne de Lyssenko, .88 bas la science bourgeoise ou juive, la science sous la dictature de l'id.8eologie, la science subordonn.8ee aux d.8elires des pseodo-philosophes scientifiques Marx-Engels et autres comme cet avocat v.8ereux de L.8enine, etc. ! RJ === Subject: Re: Les Nombres =?ISO-8859-1?Q?R=E9els_et_leur_Position_?= =?ISO-8859-1?Q?Centrale=2E?= > Le plus grand r.8eel n.8egatif est il alg.8ebrique ou transcendant ? > Mohwali Awamar. Point de vue (n.8eo) pythagoriste : seuls existent les cardinaux de 1 jusqu'au nombre d'atomos de > l'univers. > O n'est pas un nombre. 1 .8el.8ephant 1 chien 1 fleuve > z.8ero quoi ? 1 .8el.8ephant = 1 .8el.8ephant > z.8ero quoi =n'importe quel z.8ero ----------------------------------------- > VOTRE SOUCIS > Votre .8eglise devait r.8epondre moins aleph 1 > -------------------------------------- > PIEGE > Combien y a t il des transcendants ? Si l'on inclut ..... pi / 4 > si l'on inclut pi puissance pi ---------------------------------------- > DEMONTAGE DE L 'IMPOSTURE R n'existe que si Cantor a raison > si et seulement si > Card (N) = Card des pairs = Card des impairs et donc si et seulement si il existe une bijection en l'ensemble des > entiers et l'ensemble des nombre pairs > -------------------------------------- > VULGARISATION > D'apr.8fs Cantor > il associe 1 avec 2 > 2 4 > 3 6 et il ose pr.8etendre que c'est une bijection > ----------------------------------- > Mais ces farceurs (qui ont PRIS LE POUVOIR EN DEUG de maths et FORCENT > nos enfants .88 REPETER DES CONNERIES pour avoir leur module) > ------------------------------------------------------ > IMPOSTURE le 2 de la deuxi.8fme colonne, le 2 de l'ensemble des nombres pairs > est > le > m.90me > nombre > que le > 2 de l'ensemble des nombres entiers Il n'y pas deux 2 diff.8erents > __________________________________________ > ILS TRICHENT > ---------------------------------------------------------------- > Ils font une partition et s'empressent de l'oublier > ----------------------------------------------------------- TOUTE LA NOTION DE CONTINUITE S'EFFONDRE DANS CETTE PAGE > ------------------------------------------------------------- SEULE L'ARITHMETIQUE EST LOGIQUE > IL N EXISTE QUE DES ESCALIERS > DES INTEGRALES PAPIER PEINT (SOUS L'ESCALIER) LES EQUATIONS DIFFERENTIELS SONT LES RELATIONS ENTRE LA HAUTEUR DE L' > ESCALIER ET LA LONGUEUR DE CHAQUE MARCHE (MARCHES è HAUTEUR > CONSTANTE) ___________________ Cela fait 8 ans que je r.8ep.8fte cela ici (sous les insultes et les > quolibets des cr.8etins ignares et d.8epourvus d'argument) > Il faut retrouver dans les archives mon manifeste > Bon anniversaire monsieur Planck > Il y a une premi.8fre publication faite par Jacques Barot (+ depuis) sur > internet. > On voit dans ce forum le premier cr.8etin abruti s'empresser de d.8enigrer > sans r.8efl.8echir Voila le texte de l'empress.8e Alain Frisch La post.8erit.8e retiendra son nom (mais certainement pas pour ses > travaux !!!) > Pour les anonymes cr.8etins consultables sur les liens, leur affichage > sera fait par les sites sp.8ecialis.8es dans l'.8epist.8emologie plus tard. Yanick Toutain Rappeler les exploits d'Alain Frisch a aussi pour fonction > (discr.8fte) de montrer aux .8eventuels amateurs, que leur prose > continuiste aura le m.90me sort en 2016. > Qu'ils apprennent la politesse et .88 argumenter ! Vive l'homo sapiens connardus ! Exemple .8eminent : le camarade Yanick ToutCr.8etin ! RJ === Subject: help with a limit posting-account=r5Mu_AoAAAA4Or1qD_TiZuru2F3PYCEL Gecko/20080201 Firefox/2.0.0.12 eMusic DLM/4.0_1.0.0.1,gzip(gfe),gzip(gfe) Hi everybody, I have a question regarding a limit. f is defined as f (x1, x2) = (x1 * x2)/(x1^2+ x2^2) when What is lim f (x1, 0) when x1 goes to 0? I am getting lim (0)/(x1^2+ 0) when x1 goes to 0 And I have to use Lhospital's rule to get lim (0)/(2*x1) when x1 goes to 0 and Lhospital's rule again to get lim (0)/(2) =0 I am not sure it is legal to use Lhospital's rule twice. But I can not think of something else. === Subject: Re: help with a limit > Hi everybody, I have a question regarding a limit. f is defined as > f (x1, x2) = (x1 * x2)/(x1^2+ x2^2) when What is lim f (x1, 0) when x1 goes to 0? I am getting > lim (0)/(x1^2+ 0) when x1 goes to 0 And I have to use Lhospital's rule to get > lim (0)/(2*x1) when x1 goes to 0 Why in the world would you use LHR on the funtion that is identically 0? > and Lhospital's rule again to get > lim (0)/(2) =0 I am not sure it is legal to use Lhospital's rule twice. But I can > not think of something else. > === Subject: Re: help with a limit posting-account=CjtEGAoAAABpabeCpB2FB6Na4w5v-20u 5.1),gzip(gfe),gzip(gfe) Hi. Here f(x1, 0) = 0, a constant function. So Lim x1-->0 f(x1, 0)= 0. === Subject: Re: help with a limit posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig Gecko/20070515 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > Hi everybody, I have a question regarding a limit. f is defined as > f (x1, x2) = (x1 * x2)/(x1^2+ x2^2) when What is lim f (x1, 0) when x1 goes to 0? I am getting > lim (0)/(x1^2+ 0) when x1 goes to 0 And I have to use Lhospital's rule to get > lim (0)/(2*x1) when x1 goes to 0 and Lhospital's rule again to get > lim (0)/(2) =0 I am not sure it is legal to use Lhospital's rule twice. But I can > not think of something else. > Yes, that's OK. Here's another way to see it. f(x1,0) = 0/x1^2 = 0 for all nonzero x1. Thus consider the sequence { f(x_k) } for any sequence of values x_k -> 0. Since each term in { f(x_k) } is 0, the limit is 0. - Randy === Subject: NEO NEWTONISTE THEORIE posting-account=1qbAGAkAAADcUtlizzXUEb5jUjfAdE2y Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > The Observational Collapse of Einsteinian Physics Observations made in the field of Quantum Physics have already > established that the concept that space and time are different aspects of > the same entity - space-time is flawed. This concept was based on the > recognition that the Lorentz Transformations resulting from velocity and our > inability to transmit information faster than the velocity of light made it > impossible to establish our absolute velocity through space observationally. > Physicists who were more enamored with mathematical elegance than with > establishing a better understanding of the way the Universe works extended > that impossibility to generate the belief that an absolute velocity did not > exist! This leads to logical contradictions, as Dr. Einstein recognized > instinctively, causing him to resist the idea of space-time for about 25 > years after the publication of Special Relativity and asserting that the > space-time concept required the existence of an infinite number of > Aethers! Lately, Quantum physicist have made observations which demonstrate > that, of the two current explanations of quantum theory, the one which > requires that the velocity of propagation of quantum effects be infinite > velocity rather than being limited to the velocity of light, is consistent > with the results of observation. Experiments have shown that the velocity of > the coupling of the polarization angle (the quantum number of the photon) of > paired photons occurs at a velocity of at least four times the velocity of > light and perhaps at an infinite velocity. Such a coupling velocity prevents > physicists from hiding behind the idea that there is no absolute velocity > reference frame and forcibly re-establishes the idea of Aether and > demolishes the idea of space-time as other than as a conventional > mathematical abstraction. The idea that information can travel at a velocity greater than the > velocity of light is not inconsistent with the requirements of Special > Relativity. If one examines all of the conventional means of transmission of > information one finds that they involve the transmission of energy (e.g.- > the Lorentz Transformation for energy (F-L-T system of units), one finds > that the units of energy are the product of the Lorentz Transformations > Force and for Length. Since the Lorentz Transformation for Force (on axis) > is unity, the Lorentz Transformation for Energy is equal to the > Transformation for Length and is 1/(1-V^2/C^2)^0.5). This transformation > becomes infinite at the velocity of light and becomes imaginary at > velocities above the velocity of light. Clearly, no observable process which > involves energy can proceed faster than the velocity of light! The limitation imposed on the velocity of energy propagation does not > apply to the velocity of propagation of the polarization angle of paired > photons. Since changing the polarization angle of a photon does not change > its energy content, the limitation that Special Relativity imposes on the > transmission of energy does not apply. To see how Special Relativity does > impose a velocity limitation on the propagations of the polarization angle > of paired photons one must apply the Lorentz Transformation for Angle to > the problem. Angle is measured as the length along an arc divided by the > radius of that arc. As a result, the Lorentz Transformation for Angle is > (1-V^2/C^2)/(1-V^2/C^2), and is equal to unity at all velocities between > +/-C! A simple calculus procedure shows that the Transformation for Angle > remains unity even at the velocity of light, where the Transformation for > Energy becomes Infinite!. Unlike the Transformation for Energy, at > velocities above that of light, the Transformation for Angle does not become > seems reasonable to conclude that the > velocity limit for the propagation of polarization angle of such photons is > infinity. It is the writer's belief that this results applies to all quantum > numbers and that the current interpretation of Quantum Theory which asserts > that quantum effects must pervade through all of space at an infinite > velocity is the correct one. (The alternative idea, that alternate universes > co-existing in the same space would seem to be a concept that is so absurd > that it seems reasonable to wonder how much LSD its adherents absorbed > during their college years.) Quite significant is the fact that the original big bang fireball > from which our Universe is believed to have begun has been observed. We know > its location and we know our velocity with respect to it. As a result we > have actually observed our absolute velocity through space. Since this is > impossible under the space-time concept, we seem to be forced back to the > Lorentz Transformation - Aether Theory! When one recognizes that information CAN propagate at a velocity > significantly greater that the velocity of light, the idea of a ABSOLUTE > TIME cannot be dismissed, just as the idea cannot be dismissed by those who > consider the virtually instantaneously travel between locations using > wormholes. A successful use of a wormhole in this manner would also > effectively establish ABSOLUTE SIMULTANIETY. Unfortunately for the > intellectual status quo, ABSOLUTE TIME requires that the special case > solution of Special Relativity represented by the Lorentz-Transformation > Aether Theory represent our reality. The mathematical abstraction of > space-time is just that, > an abstraction which makes the computations of velocity effects easier, it > has no physical significance. Nature doesn't actually give a damn about the > abstractions we use to make computations simpler, it does care, however, > about the useful function that the classical Aether performs. (Dr. Einstein > is reported to have held on to the belief in absolute time, identical to a > belief in the Aether theory, for 25 years after he published Special > Relativity and to have warned remember gentlemen, we have not proven that > the Aether does not exist, we have merely proven that we do not need it for > computations.) We have reached a point where the existence of the Aether has > been observationally proven though the paired photon experiments. It is > time for the mathematical idiot-savants to step aside and let intelligent > men attack the problem objectively without fear of repression. Recently, astronomers in Australia announced that they have observed > that the velocity of light in the early stages of the Universe was greater > than it is now. At present, their observations are considered to be errors > since the velocity of light is known to be constant. WRONG! The velocity > of light is a constant only when measured with local units of measurement. > It cannot be both constant and a constant unless the Lorentz > Transformations for length and time are either equal or are unity. THEY > AREN'T, they are reciprocal! In addition, cosmologists are struggling to > make sense of the apparent speeding up of the expansion of the Universe at > extreme distances from the original fireball. It must be pointed out that > both these > observations would result if the correct gravitational transformations were > employed, i.e- Force = 1 > Length = 1/G*M1*M2/(R*C)^2 > Time = G*M1*M2/(R*C)^2 instead of the erroneous ones which result from the naive error made in the > original derivation of General Relativity which was plastered over with the > fakery of curved space. (There is no transformation for length in General > Relativity. It is puzzling as to why there is a search for a mysterious additional > repulsive force to explain the observed rate of expansion of the Universe. > The reason that this is puzzling is the fact that nowhere in the discussions > has the writer found that the radiation pressure of the 3K background, and a > possibly much large pressure caused by neutrinos, has been taken into > account. Since it is currently estimated that the energy represented by that > radiation is at least 10 times the energy present in the matter contained in > the Universe, it is easy to believe that the pressure produced by the > radiation can easily produce the required espansive force. Is there proof > that this is the case, probably not, but this source of expansive force has > not been included in the calculations. With this viewpoint, the Universe > would be analogous to a gas (radiation) filled balloon containing a > Scientific American describes the Universe as just such a dist filled > gas-bag.) If one examines the big bang observations one finds that they are > consistent with the Universe being formed within a gravitationally collapsed > object when the correct gravitational transformations are employed. Such an > object, as observed internally without accounting for the relativistically > changed size of the units of measurement, would collapse and then, when that > collapse had proceeded to a radius equal to four times the horizon radius, > would be observed to expand explosively at the beginning and then settle > down to a more moderate rate of expansion. In directions away from the > observed source of the expansion, the rate of that expansion should also > seem to be increasing due to the fact that we are also looking into the > past. These effects are to be expected from the application of the corrected > gravitational transformations to the problem. The effects described are > rigorously derived inhttp://einsteinhoax.com/gravity.htm. The source material for this posting may be found inhttp://einsteinhoax.com/hoax.htm(1997);http://einsteinhoax.com/gravity.htm > (1987); andhttp://einsteinhoax.com/relcor.htm(1997). EVERYTHING WHICH WE > ACCEPT AS TRUE MUST BE CONSISTENT WITH EVERYTHING ELSE WE HAVE ACCEPTED AS > TRUE, IT MUST BE CONSISTENT WITH ALL OBSERVATIONS, AND IT MUST BE > MATHEMATICALLY VIABLE. PRESENT TEACHINGS DO NOT ALWAYS MEET THIS > REQUIREMENT. THE WORLD IS ENTITLED TO A HIGHER STANDARD OF WORKMANSHIP FROM > THOSE IT HAS GRANTED WORLD CLASS STATUS. All of the Newsposts made by this site may be viewed at thehttp://einsteinhoax.com/postinglog.htm. Please make any response via E-mail as Newsgroups are not monitored on > a regular basis. Objective responses will be treated with the same courtesy > as they are presented. To prevent the wastage of time on both of our parts, > please do not raise objections that are not related to material that you > have read at the Website. This posting is merely a summary. E-mail:- einsteinh...@isp.com. If you wish a reply, be sure that your > mail reception is not blocked. The material at the Website has been posted continuously for over 8 > years. In that time THERE HAVE BEEN NO OBJECTIVE REBUTTALS OF ANY OF THE > MATERIAL PRESENTED. There have only been hand waving arguments by > individuals who have mindlessly accepted the prevailing wisdom without > questioning it. If anyone provides a significant rebuttal that cannot be > objectively answered, the material at the Website will be withdrawn. > Challenges to date have revealed only the responder's inadequacy with one > exception for which a correction was provided. J'ai essay.8e de lire votre texte .88 l'aide d'un traducteur automatique http://svsurl.systransoft.com/?t=outputframeset&link=trans&task=11d1def534ea 1be0--80c1747-11833e7c344-34c3 Je vous fais une premi.8fre r.8eponse en fran.8dais Apr.8fs publication, je mettrai en dessous un lien qui vous permettra de me lire en anglais (ou dans une autre langue pour ceux quoi voudraient utiliser la page detraductionsystran {je suis publiphobe et utilise gratuitement leur page : je travaille pour des fruitware} ---------------------------------------------------------- Vous semblez r.8ecuser Einstein en voulant supprimer le plafond 299 792 458 m/s C'est inutile. Si vous souhaitez examiner la grille th.8eorique n.8eo-newtonienne, allez sur mon site. monsyte Cela consiste .88 reprendre TEMPS ABSOLU (TEMPS OBJECTIF) LIEUX ABSOLUS (LIEUX OBJECTIFS) DEPLACEMENTS ABSOLUS (DEPLACEMENTS OBJECTIFS) VITESSES ABSOLUES (VITESESS OBJECTIVES) ------------------------------------------------------------------------- les .8equations vectorielles de Newton du type Vx2 = Vx1 - G M / ( c * t1) power 3 * X1 * d.8elai Vx1 vitesse absolue (sur l'axe des X) G constante M masse (calcul provisoire cfci-dessous) t1 d.8elai du signal X1 coordonn.8ee en X d.8elai : d.8elai entre deux mesures -------------------------------- DANS LE VIDE DE DEMOCRITOS NEWTON En y ajoutant le DELAI DU TRAJET DU SIGNAL En consid.8erant les photons (les atomos .88 l'int.8erieur) comme les causes de la gravitation --------------------------------------------------------- extraits de la page : http://site.voila.fr/monsyte/de/SCIENCES/scphys/Recherche/La veritable equation de la gravitation helio planetaire.htm CONSTANTE DE D.83VIATION H.83LIOPLAN.83TAIRE La constante de d.8eviation gravitationnelle h.8elioplan.8etaire est .8egale .88 Kdevia = 2,17004 10-6 m 3 s -1 J -1 Sa signification : Toute plan.8fte recevant une .8energie locale, Eloc, en provenance du Soleil, aura sa trajectoire d.8evi.8ee d'une quantit.8e .8egale .88 : Eloc * Kdevia Eloc en J s -1 m -2 Eloc calcul.8ee sur un plan perpendiculaire .88 la direction des photons depuis le Soleil jusqu'.88 la plan.8fte Cette quantit.8e a les dimensions d'une acc.8el.8eration (m.8ftres par secondesî) mais correspond simplement .88 la quantit.8e de d.8eviation (en direction du Soleil) de la trajectoire de la plan.8fte. m s -2 Son usage est strictement identique .88 celui d'Isaac Newton concernant l'acc.8el.8eration gravitationnelle. Le vecteur vitesse est diminu.8e d'une quantit.8e .8egale .88 Eloc*Kdevia*unit.8edetemps Pour une valeur Eloc = 1367 Il suffit de multiplier 1367 par Kdevia Soit 1367 * 2,17004 10-6 = Et l'on trouve 0.002966448 Ce r.8esultat indique que la trajectoire de la Terre est d.8evi.8ee de 2.97 millim.8ftres par seconde J s -1 m -2 ---------------------------------------------------------------------------- - Cette grille rend erron.8ee l'ellipse (les d.8elais de r.8eception ne sont pas .8egaux aux d.8elais d'.8emission des signaux gravitationnels) ---------------------------------------------------------------------- Je vous renvoie .88 l'.8equation qui va prendre la place des .89neries positivistes de Lorentz Elles sont Newton-compatibles Merci Yanick Toutain Les cr.8etins relativistes mal-polis sont dispens.8es de r.8epondre (idem pour les partisans du fasciste Heisenberg et de la m.8ecanique quantique) http://site.voila.fr/monsyte/de/SCIENCES/scphys/COURS/objectif/Cours de science objective Les equations.htm === Subject: Re: NEO NEWTONISTE THEORIE posting-account=1qbAGAkAAADcUtlizzXUEb5jUjfAdE2y Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) You can read the English automatic translation on the following link http://svsurl.systransoft.com/?t=outputframeset&link=trans&task=11d1def534ea 1be0--80c1747-11833e7c344-382a I make you a first French answer After publication, I will put in lower part a bond which will allow you to read me in English (or in another language for those what would like to use the page detraductionsystran {I am publiphobe and uses free their page: I work for fruitware} ---------------------------------------------------------- You seem to challenge Einstein while wanting to remove ceiling 299.792 458 m/s It is useless. If you wish to examine the n.8eo-Newtonian theoretical grid, go on my site. monsyte That consists in beginning again ABSOLUTE TIME (OBJECTIVE TIME) ABSOLUTE PLACES (OBJECTIVE PLACES) ABSOLUTE DISPLACEMENTS (OBJECTIVE DISPLACEMENTS) ABSOLUTE VELOCITIES (OBJECTIVE VITESESS) ------------------------------------------------------------------------- vectorial equations of Newton type Vx2 = Vx1 - G M/(C * T1) power 3 * X1 * time Vx1 absolute velocity (on the x axis) G constant M masses (provisional calculation cfci-dessous) T1 time of the signal X1 coordinated in X time: deadline between two measurements -------------------------------- IN THE VACUUM OF DEMOCRITOS NEWTON By adding to it the TIME OF the WAY OF the SIGNAL By considering the photons (atomos inside) as causes gravitation --------------------------------------------------------- extracts of the page: http://site.voila.fr/monsyte/de/SCIENCES/scphys/Recherche/La veritable equation de la gravitation helio planetaire.htm CONSTANT OF DEVIATION H.83LIOPLAN.83TAIRE The constant of gravitational deviation h.8elioplan.8etaire is equal to Kdevia = 2,17004 10-6 m3 S -1 J -1 Its significance: Any planet receiving a local energy, Eloc, coming from Eloc * Kdevia Eloc in J S -1 m -2 Eloc calculated on a level perpendicular to the direction of the photons since the Sun to planet This quantity has dimensions of an acceleration (meters by seconds î) but corresponds simply to the quantity of deviation (in direction of the Sun) of the trajectory of planet. m S -2 Its use is strictly identical to that of Isaac Newton concerning gravitational acceleration. The Flight Path Vector is decreased by a quantity equalizes with Eloc*Kdevia*unit.8edetemps For a Eloc value = 1367 It is enough to multiply 1367 by Kdevia That is to say 1367 * 2,17004 10-6 = And 0.002966448 are found This result indicates that the trajectory of the Earth is deviated of 2.97 millimetres a second J S -1 m -2 ---------------------------------------------------------------------------- - This grid makes erroneous the ellipse (the times of reception are not not equal to the times of emission of the gravitational signals) ---------------------------------------------------------------------- I return you to the equation which will take the place of the stupid things positivists of Lorentz They are Newton-compatible Yanick Toutain The relativistic cretins impolite fellows are exempted to answer (idem for the partisans of the Heisenberg fascist and mechanics quantum) http://site.voila.fr/monsyte/de/SCIENCES/scphys/COURS/objectif/Cours de science objective Les equations.htm === Subject: Re: Existence of the function >Sorry, I meant this: Does there exist a function f: R -> R with the property: >FOR ANY a in R and for any decreasing sequence {x_n} ( x_1 >= x_2 >= x_3 >= ... ) >with x_n --> a we have |f(x_n) - f(a)| --> oo ? There's a simple cardinality argument for this. Suppose f is a function such as you describe. You can use f to construct an injective map t:R -> N x Q. It goes like this. For any real x, let t(x) = (n,q) where n = ceil(|f(x)|), and where q is a rational such that x < q, and such that |f(z)| > n whenever x < z < q. (We know that such a q must exist, else we could find a descending sequence x_i -> x such that |f(x)-f(x_i)| is bounded.) To see that t is injective suppose x < y. If ceil(|f(x)|) <> ceil(|f(y)|) then t(x) <> t(y) so we're done. So suppose t(x) = (n,q) and t(y) = (n,r). We know that |f(z)|>n for all z satisfying x < z < q, so y cannot be between x and q. It follows that x < q <= y, and since y < r, we have that q <> r, and thus in this case also t(x) <> t(y). RS === Subject: Re: Existence of the function >Sorry, I meant this: >Does there exist a function f: R -> R with the property: >FOR ANY a in R and for any decreasing sequence {x_n} ( x_1 >= x_2 >= x_3 >= ... ) >with x_n --> a we have |f(x_n) - f(a)| --> oo ? There's a simple cardinality argument for this. Suppose f is a function >such as you describe. You can use f to construct an injective map >t:R -> N x Q. It goes like this. For any real x, let t(x) = (n,q) where >n = ceil(|f(x)|), and where q is a rational such that x < q, >and such that |f(z)| > n whenever x < z < q. (We know that >such a q must exist, else we could find a descending >sequence x_i -> x such that |f(x)-f(x_i)| is bounded.) To see that t is injective suppose x < y. If ceil(|f(x)|) <> ceil(|f(y)|) >then t(x) <> t(y) so we're done. So suppose t(x) = (n,q) and t(y) = (n,r). >We know that |f(z)|>n for all z satisfying x < z < q, so y cannot be >between x and q. It follows that x < q <= y, and since y < r, we have >that q <> r, and thus in this case also t(x) <> t(y). Very nice. quasi === Subject: nesheph (twilight) In the eighty-sixth emblem of Minerva Britanna, Henry Peacham dedicates the emblem to a woman whom he refers to as E:L. http://f01.middlebury.edu/FS010A/STUDENTS/n086.htm In my numbering system E = 8 and L =37 *A =26,47 B=56 C=1 D=9 E=8 F=30 G=6 H=4 I=10 J=5 *K=29 L=37 M=18 N=3 O=0.125 P=3.14..... Q=14.5 R=11 *S=19 *T=7 *U=23 V=13.7 W=80 X=20 Y=43 Z=91 45 'Abiy-`albown ab-ee-al-bone' from 1 and and an unused root of uncertain. derivation; probably, father of strength (i.e. valiant); Abialbon, an Israelite:--Abialbon. 45. agkura ang'-koo-rah from the same as 43; an anchor (as crooked):--anchor If the spelling of her name is correct, Fawza Falih (the woman due to be beheaded any day now in Saudi Arabia) has a numerical representation below.. Fawza (253) Falih (107) ///////////// Coindicence? === Subject: Grothendieck topologies posting-account=R7AgUAoAAADVFAtIe36IBmgohoHjZsKW Gecko/20071204 Ubuntu/7.10 (gutsy) Firefox/2.0.0.11,gzip(gfe),gzip(gfe) is every Grothendieck topology generated by a Grothendieck pre suppose that this is not the case. If not, what would be the definition of a sheaf on a Grothendieck site? The usual definition with the equalizer sequence works only for a Grothendieck pre topology. Otherwise one could say that 'the natural map Hom(Hom(.81|, X), F) .81¬ Hom(S, F) induced by the inclusion of S into Hom(.81|, X) is a bijection' but I don't know what that means if the categories are not small: Hom(Hom(.81|, X), F) considers the functor category! Can anybody help me with that? S. === Subject: Need solutions manual Probability and statistics walpole 8th edition posting-account=6aFgtQoAAAB_uj0qs3eJZd9g6ZxTQPzf .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) thatsmecox@yahoo.com === Subject: Re: Axiom of choice > What is necessary is countable choice for countable sets. > I am quite wiling to accept that, but the wiki reference > cited to establish that did not say that. > At the risk of saying the obvious as a result of > misunderstanding your comment, I think Rubin is > far more of an authority on this issue than a > wiki page. >I have already accepted that countable choice is required. >justifying the claim that countable choice was required. >to have said. >Since then, several references which actually did say countable choice >was required have been cited by various people. Countable choice is sufficient, but it is stronger than countable choice for countable sets, which is necessary, One can have choice for sets of certain sizes, and these are not all equivalent. For the various equivalences and non-equivalences, see _Consequences of the Axiom of Choice_, by Paul Howard and Jean Rubin. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Axiom of choice > At the risk of saying the obvious as a result of > misunderstanding your comment, I think Rubin is > far more of an authority on this issue than a > wiki page. > I will state the obvious: Anyone can edit Wikipedia >My comment had more to do with Rubin's background >than with any faults of Wikipedia. There are very, >very few people in the world who have more background >than Rubin does on the topic of his comment. (He and >consequences.) >Dave L. Renfro equivalents, not consequences. However, I have done some work on consequences and their equivalents, but I was involved to some extent in the production of that magnum opus. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Axiom of choice > > At the risk of saying the obvious as a result of > misunderstanding your comment, [...] > I do not recall who it was, but someone cited the > choice was required. I only pointed out that the wiki Well, it's a good thing I put in the part as a result > of misunderstanding your comment, since this is obviously > what I did, and through no fault of yours, as looking > at your original post shows that you said exactly this > and not what I thought I read. Dave L. Renfro I admit I was wrong about the necessity of countable choice, but I am having been caught in the one error, I was trying to be extra careful === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=-XRI1AgAAABBR-yHg7m_4HCGi05mtX1k Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > I'm wondering about the Church-Kleene least nonrecursive ordinal > omega_1^CK. > Rather than changing the subject, try to concentrate > on one thing for a while. > - William Hughes > I've read a few proofs of the existence of a least uncountable > ordinal. For example, let lambda be the initial ordinal of Aleph_1, > the least uncountable cardinal, then as lambda is an ordinal, and > union of all lesser ordinals, take lambda union the set containing > lambda (i.e. its order type) disjoint all the countable ordinals, > where the disjoint thus contains only uncountable ordinals, which are > naturally well-ordered so that subset has a least element, which is > the least uncountable ordinal, which would be lambda, called omega_1 > the least uncountable ordinal, where Aleph_1 is the least uncountable > cardinal. > So, assume there is a least uncountable ordinal, w_1, omega_1. Then, > there are only countable ordinals that comprise its membership, as a > set, else minimality would be violated. To be uncountable, there are > uncountably many countable ordinals less than w_1, because otherwise > it would be a countable union of countable sets, thus countable. So, > there are uncountably many ordinals less than w_1. Of those, not all > of them can be countable, because each contains all lesser, for there > to be uncountably many, at least one of them is uncountable, > Absolute nonsense. > A countable oridinal can contain only finite ordinals, > e.g {0,1,2,3,...}. Each is finite, but there are a (countably) > infinite number of them > Similarly, an uncountable ordinal can contain only countable > ordinals. Each is countable, but there are an uncountable infinite > number of them. > The least uncountable ordinal contains only countable ordinals. > This is not a contradiction. > - William Hughes I agree, Then stop arguing that some ordinal smaller than > the least uncountable ordinal must be uncountable. - William Hughes About a definition of subtraction for ordinals, I should have been more clear about the definition of subtration over _limit_ ordinals. That is to say, it's simple and clear to define subtraction on ordinals, particularly polynomials in omega then maybe some other elements of the Veblen hierarchy, but as zero, for example, is a limit ordinal, with no predecessor, as pure ordinals: 0 - 1 = 0. A simple definition of subtraction for limit ordinals would have, with ordinal alpha < omega the order type of finite ordinals, that for limit ordinal lambda, lambda - alpha = lambda. For polynomials on omega with finite ordinal coefficients, componentwise subtraction is simple to define, just not across limit ordinals. Is there a definition in Enderton of w - 1 as anything other than w, particularly anything less? What I intended to say, and was hopefully clear in context, was that subtraction _across limit ordinals_ was undefined, because the limit ordinal is the limit of the sequence of lesser ordinals, which in arithmetic successor progression have no last nor greatest element, i.e., lambda-1 nor lambda. Still I get to thinking some more about partitioning the countable ordinals. My first idea was to partition them by putting zero at the back of the list (actually well-ordering), so an ordinal with one less element than omega_1 would be uncountable yet an ordinal different than omega_1. Then there was the notion of zero and all the other finite ordinals, to make the difference the least countable infinite limit ordinal, towards that even if there was no last element, that there would be a last limit ordinal. That led into notions of this countable cumulative limit hierarchy, basically the limit ordinals of the countable cumulative hierarchy, in simply reordering them to define a single boundary in the ordinals thus that the boundary ordinal was less than omega_1, yet had uncountably many elements less than it, towards illustrating there not being a least uncountable ordinal. That led as well to consideration in general of the notion and use of limit ordinals, vis-a-vis foundations for number systems designed in terms of symmetry and balanced, i.e., counting upwards from zero and downwards from infinity. Then, there was a notion to simply interleave the even and odd ordinals, basically those whose component representing accumulation of finite ordinals represented an even or odd integer respectively. So, now I have been thinking about instead partitioning the countable ordinals not at one particular boundary, because it's not simple to define any particular boundary having uncountably many ordinals less than it, besides omega_1 defined as the initial ordinal of the least uncountable cardinal, and instead partitioning the countable ordinals by countably infinitely many boundaries. This would basically have that the ordinals in order map to the limit ordinals in order, by having a well-ordering of the (countable) limit ordinals in order, then their successors then theirs etcetera ad infinitum. Then the ordinals would be mapped to the limit ordinals. That has there then being as many limit ordinals as ordinals, in the countable ordinals. Then, as is obviously the case, there would be limit ordinals of those and etcetera, basically forever. Basically what that does is, up to one of the larger countable ordinals, partition the countable ordinals into that many partitions, each of which is uncountable if there are uncountably many countable ordinals. It would do that, if a partition can be defined for particular large and probably nonrecursive and impredicative ordinals, up to omega_1 itself, which variously would turn all the partitions from uncountably infinite to zero or perhaps one in extent, compared to any countable ordinal. There isn't an ordinal omega_x between omega and omega_1 which partitions omega_1 into uncountably many countably infinite partitions, in terms of partitioning at those multiples of omega_x, as it were. Here the notion of partitioning in ordinals is much different than partitioning in sets (except in ZFC, where for these large sets, they have particular ordinal quantities as sets, in a general sense.) Without establishing an order on the elements of a set, partitioning can be enumerated in various sets of subsets of the set, but a except for all the possible combinations that may be encoded and the products of all combinations and permutations and other combinatoric results that derive from combinatorial enumeration of initial ordinals of cardinals (in ZFC) and mapping of the elements to ordinals. The ordinals are well-ordered, by definition, and there are obvious ways to construct them up to nonrecursive ordinals, given omega as a constant in terms of, for example, polynomials for relatively small countable ordinals. Here's another tack: In ZF it's undecideable if 2^Aleph_0 is the least uncountable cardinal. It's unknown, in ZF, unknowable, how many cardinals there are between Aleph_0 and 2^Aleph_0. So, if omega_Y is the initial ordinal of 2^Aleph_0 in ZFC, it's undecideable whether omega_Y is omega_1. That there is a least uncountable ordinal would have it be the initial ordinal of the least uncountable cardinal. If 2^Aleph_0 >= Aleph_w, which is neither proven nor disproven in ZFC, using model theory with large (and maximal) ordinals modeling the universe of ZFC along the lines of Cohen's development of the independence of the Continuum Hypothesis, then omega_1 is still the initial ordinal of Aleph_1. Ross -- Finlayson Consulting === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > I'm wondering about the Church-Kleene least nonrecursive ordinal > omega_1^CK. > Rather than changing the subject, try to concentrate > on one thing for a while. > - William Hughes > I've read a few proofs of the existence of a least uncountable > ordinal. For example, let lambda be the initial ordinal of Aleph_1, > the least uncountable cardinal, then as lambda is an ordinal, and > union of all lesser ordinals, take lambda union the set containing > lambda (i.e. its order type) disjoint all the countable ordinals, > where the disjoint thus contains only uncountable ordinals, which are > naturally well-ordered so that subset has a least element, which is > the least uncountable ordinal, which would be lambda, called omega_1 > the least uncountable ordinal, where Aleph_1 is the least uncountable > cardinal. > So, assume there is a least uncountable ordinal, w_1, omega_1. Then, > there are only countable ordinals that comprise its membership, as a > set, else minimality would be violated. To be uncountable, there are > uncountably many countable ordinals less than w_1, because otherwise > it would be a countable union of countable sets, thus countable. So, > there are uncountably many ordinals less than w_1. Of those, not all > of them can be countable, because each contains all lesser, for there > to be uncountably many, at least one of them is uncountable, > Absolute nonsense. > A countable oridinal can contain only finite ordinals, > e.g {0,1,2,3,...}. Each is finite, but there are a (countably) > infinite number of them > Similarly, an uncountable ordinal can contain only countable > ordinals. Each is countable, but there are an uncountable infinite > number of them. > The least uncountable ordinal contains only countable ordinals. > This is not a contradiction. > - William Hughes > I agree, Then stop arguing that some ordinal smaller than > the least uncountable ordinal must be uncountable. - William Hughes About a definition of subtraction for ordinals, Complete non sequitor. -William Hughes === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) >æWhat I intended to say, and was hopefully > clear in context, was that subtraction across limit ordinals was > undefined. Subtraction is defined on ordinals. I told you one place to find out about it. When it's been pointed out to you that you don't know what you're talking about, and a reference is given for you to inform yourself, why don't you just shut up while you actually do inform yourself rather than to continue as usual, blathering nonsense and misinformation? You go on and on and on posting piles and piles of a completely uninformed arbitrary rot. I just don't see what you get out of continually acting the buffoon. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > You go on and on and on posting piles and piles of a completely > uninformed arbitrary rot. It's not uninformed arbitrary rot, it's abstract poetry. It is a grave error to try to read it as mathematics, just as you would be disappointed if you expected to find set theoretic arguments in _Finnegans wake_. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) On 21 Feb., 19:28, Aatu Koskensilta You go on and on and on posting piles and piles of a completely > uninformed arbitrary rot. It's not uninformed arbitrary rot, it's abstract poetry. It is a grave > error to try to read it as mathematics, just as you would be > disappointed if you expected to find set theoretic arguments in > _Finnegans wake_. Have you recognized the true character of set theory? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > On 21 Feb., 19:28, Aatu Koskensilta You go on and on and on posting piles and piles of a completely > uninformed arbitrary rot. It's not uninformed arbitrary rot, it's abstract poetry. It is a grave > error to try to read it as mathematics, just as you would be > disappointed if you expected to find set theoretic arguments in > _Finnegans wake_. Have you recognized the true character of set theory? WM clearly hasn't. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > On 21 Feb., 19:28, Aatu Koskensilta You go on and on and on posting piles and piles of a completely > uninformed arbitrary rot. It's not uninformed arbitrary rot, it's abstract poetry. It is a grave > error to try to read it as mathematics, just as you would be > disappointed if you expected to find set theoretic arguments in > Finnegans wake . Have you recognized the true character of set theory? No, that won't be revealed until the Evil Cantorian Army of Devils confronts the Archangels of the Goodness of Reality on the plains of Armeggedon. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) On Feb 21, 10:28æam, Aatu Koskensilta You go on and on and on posting piles and piles of a completely > uninformed arbitrary rot. It's not uninformed arbitrary rot, it's abstract poetry. It is a grave > error to try to read it as mathematics, just as you would be > disappointed if you expected to find set theoretic arguments in > Finnegans wake . Whatever it is, indeed it is a Howl. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > If the tree can contain more paths than nodes, then æthe sequence > 1,1,1,... may not be restricted to members with magnitude 1. In total > there may be more 2's than 1's, although, of course, at ever finite > position, there is a 1. That may be counter intuitive but it's not a > contradiction. > Hahaha. Good one, Wolfgang. Yeah, it's funny, but in a tragic kind of way. If the sequence has a 1 at every finite position, then at what > positions > can those 2's be found? If every finite binary tree has more nodes than paths, and if the infinite tree is nothing but the union of all finite trees, then at what positions does the overtaking of nodenumber by pathnumber happen? And what mechanism causes the miraculous increase in pathnumber? > It is a contradiction to say that the > sequence > is composed of anything /except/ numbers at finite positions. æ But it's not a contradiction to say that the sequence (a n) = 1/2, 1/2, 1/2, ... has the limit oo? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > If the tree can contain more paths than nodes, then æthe sequence > 1,1,1,... may not be restricted to members with magnitude 1. In total > there may be more 2's than 1's, although, of course, at ever finite > position, there is a 1. That may be counter intuitive but it's not a > contradiction. > Hahaha. Good one, Wolfgang. Yeah, it's funny, but in a tragic kind of way. If the sequence has a 1 at every finite position, then at what > positions > can those 2's be found? If every finite binary tree has more nodes than paths, and if the > infinite tree is nothing but the union of all finite trees, then at > what positions does the overtaking of nodenumber by pathnumber happen? There are exactly as many paths in a finite tree as there are leaf (terminal) nodes, so by WM's logic, the number of paths in the infinite tree cannot be larger than the number of terminal nodes in such a tree, which is zero. > And what mechanism causes the miraculous increase in pathnumber? The sudden vanishing of every terminal node might be the mechanism, since as long as every path has one, the number of paths remains finite. It is a contradiction to say that the > sequence > is composed of anything /except/ numbers at finite positions. æ But it's not a contradiction to say that the sequence (a_n) = 1/2, > 1/2, 1/2, ... has the limit oo? If WM's notion of limit produces that result, he should stick with it, but ours does not. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <47bc8e0b@news2.lightlink.com> posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > I also wonder where all those balls went. Maybe there is a hole in the > vase.... ;) You have solved the problem, Tony! === Subject: Re: A consideration concerning the diagonal argument of G. Cantor I also wonder where all those balls went. Maybe there is a hole in the > vase.... ;) > You have solved the problem, Tony! > If balls couldn't get in to start with, there wouldn't be any problem. Vases all have holes to let things, balls in this case, in. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=iBgNeAoAAADRhzuSC4Ai7MUeMmxtwlM7 Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > fed this fish to the penguins: > 1. Potential > 2. Actual >3. A deep fried banana. > Fried banana I can agree, but is it *potentially* fried or *actually* > fried. Big difference you know. And is it to be fried in trans-finite fat? Not very healthful, according to current beliefs. 1. Potential 2. Actual http://pespmc1.vub.ac.be/INFINITY.html === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=iBgNeAoAAADRhzuSC4Ai7MUeMmxtwlM7 Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) 1. Potential 2. Actual === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > 1. Potential 2. Actual 3. A deep fried banana. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A consideration concerning the diagonal argument of G. Cantor fed this fish to the penguins: > 1. Potential > 2. Actual 3. A deep fried banana. Fried banana I can agree, but is it *potentially* fried or *actually* fried. Big difference you know. G. Rodrigues === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <3Qfvj.300045$Je3.68039@reader1.news.saunalahti.fi> <507rr39olv1svuslcoe7ma6136tk4ic2uc@4ax.com> posting-account=oTDIagkAAACTxHurtPutBWvNQS8ZCNO9 Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > fed this fish to the penguins: > 1. Potential > 2. Actual 3. A deep fried banana. Fried banana I can agree, but is it *potentially* fried or *actually* > fried. Big difference you know. > And is it to be fried in trans-finite fat? Not very healthful, according to current beliefs. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <47bb0316@news2.lightlink.com> <47bc736b@news2.lightlink.com> <873arnphjz.fsf@phiwumbda.org> posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > That wasn't my argument. Pay attention. There are only finite levels > in the infinite binary tree. At each finite level, there are twice as > many nodes in the tree as paths. Hence, there is no point in the tree, > even when fully complete, where paths exceed nodes. I have no idea what this means, since I don't no what you mean by a > point in the tree [...] where paths exceed nodes. æ So let's ask a simpler question. æThe full binary tree has paths which > are infinitely long, yes? æ Is the following true or false? æ The set of infinite paths in the full binary tree has greater > æ cardinality than the set of nodes. True In other words: No. There are other means to test cardinality. æ There is no injection from the set of infinite paths in the full > æ binary tree to the set of nodes. æThere is an injection from the set > æ of nodes to the set of infinite paths. For instance: 1) There can be no separation of distinct paths without a node per path. 2) The countable unuion of countable sets is countable. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <47bb0316@news2.lightlink.com> <47bc736b@news2.lightlink.com> <873arnphjz.fsf@phiwumbda.org> posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) CORRECTION > Is the following true or false? æ The set of infinite paths in the full binary tree has greater > æ cardinality than the set of nodes. > FALSE. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > That wasn't my argument. Pay attention. There are only finite levels > in the infinite binary tree. At each finite level, there are twice as > many nodes in the tree as paths. Hence, there is no point in the tree, > even when fully complete, where paths exceed nodes. I have no idea what this means, since I don't no what you mean by a > point in the tree [...] where paths exceed nodes. æ So let's ask a simpler question. æThe full binary tree has paths which > are infinitely long, yes? æ Is the following true or false? æ The set of infinite paths in the full binary tree has greater > æ cardinality than the set of nodes. True In other words: No. There are other means to test cardinality. WM says true then immediately implies false. æ There is no injection from the set of infinite paths in the full > æ binary tree to the set of nodes. æThere is an injection from the set > æ of nodes to the set of infinite paths. For instance: > 1) There can be no separation of distinct paths without a node per > path. There can be no separation of infinite paths without each path having infinitely many nodes different from the other. > 2) The countable unuion of countable sets is countable. True but irrelevant. The only question relevant to their relative cardinality is in which direction(s) can injections/surjections exist between the set of nodes and the set of paths. One can easily inject nodes to paths, with lots of paths left over, but paths to nodes is impossible. If WM thinks paths to nodes IS possible, let him try it. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Is the following true or false? > æ The set of infinite paths in the full binary tree has greater > æ cardinality than the set of nodes. True > In other words: No. There are other means to test cardinality. I didn't *give* any alternative means. I spelled out the *meaning* of the above statement. You're confused. > æ There is no injection from the set of infinite paths in the full > æ binary tree to the set of nodes. æThere is an injection from the set > æ of nodes to the set of infinite paths. For instance: > 1) There can be no separation of distinct paths without a node per > path. > 2) The countable unuion of countable sets is countable. These statements are evidently irrelevant (well, the first one seems to be meaningless). -- Jesse F. Hughes You're ketchup, so I'll put you on meatloaf! -- Quincy P. Hughes, age five, tries his hand at insults === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <47bb0316@news2.lightlink.com> <47bc736b@news2.lightlink.com> <873arnphjz.fsf@phiwumbda.org> <87y79e8c2x.fsf@phiwumbda.org> posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Is the following true or false? > æ The set of infinite paths in the full binary tree has greater > æ cardinality than the set of nodes. True > In other words: No. There are other means to test cardinality. I didn't *give* any alternative means. æI spelled out the *meaning* of > the above statement. Sorry if you are confused. *I* gave the alternative means (see below). For instance: > 1) There can be no separation of distinct paths without a node per > path. > 2) The countable unuion of countable sets is countable. These statements are evidently irrelevant (well, the first one seems > to be meaningless). Sorry if you cannot comprehend the meaning. Here is an easier means: As long as trees and nodes stretch, there are less paths than nodes. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Is the following true or false? > æ The set of infinite paths in the full binary tree has greater > æ cardinality than the set of nodes. > True > In other words: > No. There are other means to test cardinality. I didn't *give* any alternative means. æI spelled out the *meaning* of > the above statement. Sorry if you are confused. *I* gave the alternative means (see below). > For instance: > 1) There can be no separation of distinct paths without a node per > path. > 2) The countable unuion of countable sets is countable. These statements are evidently irrelevant (well, the first one seems > to be meaningless). Sorry if you cannot comprehend the meaning. Here is an easier means: As long as trees and nodes stretch, there are less paths than nodes. In order to be stretchable, paths must have terminal nodes from which they can be stretched further. So the number of paths in WM's infinite tree must not exceed the number of terminal nodes in that tree. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) On 20 Feb., 13:55, Aatu Koskensilta Here in sci.logic we have a fine example of this, in the poster Andrew > abo Boucher, a true skeptic, going to the baffling extremes of > doubting that every natural has a successor. What is baffling about that? It is fact. And if you had the courage of your convictions, you would now say > that the set of natural numbers is not potentially infinite. I said it already many times. Compare the quote on mhtml:http://www.hs-augsburg.de/~mueckenh/MR.mht or the last pages of http://arxiv.org/ftp/math/papers/0505/0505649.pdf But the restriction of its cardinality to less than 10^100 does not imply the existence of a greates number. Further, even N was potentially infinite, set theory was inconsistent. Therefore I sometimes argue without regard to physical restrictions. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > On 20 Feb., 13:55, Aatu Koskensilta Here in sci.logic we have a fine example of this, in the poster Andrew > abo Boucher, a true skeptic, going to the baffling extremes of > doubting that every natural has a successor. > What is baffling about that? It is fact. And if you had the courage of your convictions, you would now say > that the set of natural numbers is not potentially infinite. I said it already many times. Compare the quote on > mhtml:http://www.hs-augsburg.de/~mueckenh/MR.mht or the last pages of > http://arxiv.org/ftp/math/papers/0505/0505649.pdf > But the restriction of its cardinality to less than 10^100 does not > imply the existence of a greates number. It does imply the existence of one, but not its accessibility. > Further, even N was potentially infinite, set theory was inconsistent. But when N is actually infinite, WM cannot show any /internal/ inconsistencies in, say ZFC, or NBG. All WM's alleged inconsistencies are actually incompatibilities between WM's own, though often unstated, axioms and those of systems he doesn't like. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > On 20 Feb., 13:55, Aatu Koskensilta Here in sci.logic we have a fine example of this, in the poster Andrew > abo Boucher, a true skeptic, going to the baffling extremes of > doubting that every natural has a successor. > What is baffling about that? It is fact. And if you had the courage of your convictions, you would now say > that the set of natural numbers is not potentially infinite. I said it already many times. In Wolkenmeukenheim the set of natural numbers is not potentially infinite but the set of even natural numbers is. - William Hughes === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > >Look, even though Quantum Mechanics and Relativity will give >inevitably give rise to confusion, I've managed to get over it. Not >so with mathematics, lasting for more than 35 years, especially >mathematical logic and set theory, infinities in particular. It is apparent infinitary set theory does not agree with your > constitution. Such things happen. Why you should turn your personal > dislike of set theory into obsessive search for elusive contradictions > is less understandable. After all, no one is forcing you to take part > in any infinitary tomfoolery. Yeah, suppose you're right ... I've decided to let it rest for a while. Maybe come back when I have an argument that's substantially different. Han de Bruijn === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Yeah, suppose you're right ... I've decided to let it rest for a while. > Maybe come back when I have an argument that's substantially different. > You might use the time to LEARN some mathematics (in general) and (especially) set theory. Read a good introduction, etc. F. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig AppleWebKit/523.12.2 (KHTML, like Gecko) Version/3.0.4 Safari/523.12.2,gzip(gfe),gzip(gfe) > No? Of what does the tree consist, besides finite levels with finitely > numbered nodes and edges? How do you define includes besides contains > all parts? There is no difference between the tree and all of its > finitely numbered nodes and edges. Nothing included in that totality > causes paths to outnumber nodes. > Agree or disagree? > The infinite tree = the n-th finite subtree for > some n > If we have an infinite sum of natural numbers, each less than 10. Can > this sum be uncountable? You would have to define that infinite sum. Probably > (even though you don't say it) you really mean the > cardinality of some certain set, which is a union > of disjoint sets of cardinality < 10. No, I said and meant numbers between 1 and 9, inclusive. Yes, but you didn't say what you meant by infinite sum. Normally the context in which such a thing might arise is as a representation of the cardinality of a set. A sum of infinitely many natural numbers is not a thing defined in the natural numbers. So presumably you have something else in mind. > If you have > infinitely many, then the sum is infinite, The sum is undefined. We have definitions for the sum of two natural numbers, which leads to a definition of any finite number of natural numbers. But we don't have a definition for what you might mean by an infinite sum of natural numbers. So you're going to have to provide a context for where this sum arises and what you mean by it. Is it the cardinality of some set? Or what? > namely it is at least > infinite, 1*oo, and at most 9*oo. But it is not uncountable. Neither 1*oo or 9*oo is defined in the natural numbers. What axiomatic system are you working in? What is represented by oo? Is it the cardinality of something? > Your turn. Agree or disagree? The infinite tree = > the n-th finite subtree for some n It is the union of all finite binary trees with n levels. Is it the n-th finite subtree for some n? That question has two possible answers. One of them is yes and the other is no. You didn't provide either one. - Randy === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > No? Of what does the tree consist, besides finite levels with finitely > numbered nodes and edges? How do you define includes besides contains > all parts? There is no difference between the tree and all of its > finitely numbered nodes and edges. Nothing included in that totality > causes paths to outnumber nodes. > Agree or disagree? > The infinite tree = the n-th finite subtree for > some n > If we have an infinite sum of natural numbers, each less than 10. Can > this sum be uncountable? > You would have to define that infinite sum. Probably > (even though you don't say it) you really mean the > cardinality of some certain set, which is a union > of disjoint sets of cardinality < 10. No, I said and meant numbers between 1 and 9, inclusive. Yes, but you didn't say what you meant by infinite sum. The series a_1 + a_2 +..., where said numbers appear. Normally the context in which such a thing might arise > is as a representation of the cardinality of a set. A sum of infinitely many natural numbers is not a thing > defined in the natural numbers. So presumably you have > something else in mind. You should have heard of convergent series. The value of a series is also called its sum. If you have > infinitely many, then the sum is infinite, The sum is undefined. We have definitions for the > sum of two natural numbers, which leads to a definition > of any finite number of natural numbers. Simply apply this definition infinitely often. But we don't have a definition for what you might mean > by an infinite sum of natural numbers. Then you should look it up. There are even theorems about such series. Springer online encyclopedia is a good source. > namely it is at least > infinite, 1*oo, and at most 9*oo. But it is not uncountable. Neither 1*oo or 9*oo is defined in the natural > numbers. What axiomatic system are you working in? What > is represented by oo? Is it the cardinality of something? No. Cardinalities have nothing to do with oo, since Cantor recognized that this would raise contradictions. > Your turn. Agree or disagree? The infinite tree = > the n-th finite subtree for some n It is the union of all finite binary trees with n levels. Is it the n-th finite subtree for some n? For the answer, please read the line above your question. > That question > has two possible answers. One of them is yes and the > other is no. You didn't provide either one. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > No? Of what does the tree consist, besides finite levels with > finitely > numbered nodes and edges? How do you define includes besides > contains > all parts? There is no difference between the tree and all of > its > finitely numbered nodes and edges. Nothing included in that > totality > causes paths to outnumber nodes. > Agree or disagree? > The infinite tree = the n-th finite subtree for > some n > If we have an infinite sum of natural numbers, each less than 10. Can > this sum be uncountable? > You would have to define that infinite sum. Probably > (even though you don't say it) you really mean the > cardinality of some certain set, which is a union > of disjoint sets of cardinality < 10. > No, I said and meant numbers between 1 and 9, inclusive. Yes, but you didn't say what you meant by infinite sum. The series a_1 + a_2 +..., where said numbers appear. While there are, for numbers a1 and a2, definitions of a1 + a2, I am not aware of any definition of a_1 + a_2 +... If WM has one, he should provide it. Note that there is sometimes a definition for lim_[n --> oo] (a_1 + ...+ a_n), but none for a_1 + a_2 +.... Normally the context in which such a thing might arise > is as a representation of the cardinality of a set. A sum of infinitely many natural numbers is not a thing > defined in the natural numbers. So presumably you have > something else in mind. You should have heard of convergent series. I know of a variety of definitions for having a series converge, but absent a series satisfying at least one of those definitoins, it has no determinable value. By what definition of convergence does WM allege a series of positive naturals converges? > The value of a series is > also called its sum. No convergence => no value <=> no sum. But we don't have a definition for what you might mean > by an infinite sum of natural numbers. Then you should look it up. There are even theorems about such series. > Springer online encyclopedia is a good source. Unless you can refer us to a definition, other than your own, of an infinite sum of positive naturals, we decline accepting that any such thing exists. > Your turn. Agree or disagree? The infinite tree = > the n-th finite subtree for some n > It is the union of all finite binary trees with n levels. Is it the n-th finite subtree for some n? For the answer, please read the line above your question. Non-responsive. That question > has two possible answers. One of them is yes and the > other is no. You didn't provide either one. > Still no answer. Wm avoids the question as he knows any honest answer will show the holes in his logic. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Note that there is sometimes a definition for > lim_[n --> oo] (a_1 + ...+ a_n), but none for a_1 + a_2 +.... Definition: lim_[n --> oo] (a_1 + ...+ a_n) := a_1 + a_2 + .... === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) Note that there is sometimes a definition for > lim [n --> oo] (a 1 + ...+ a n), but none for a 1 + a 2 +.... Definition: lim [n --> oo] (a 1 + ...+ a n) := a 1 + a 2 + .... Hilarious!!! MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig Gecko/20070515 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > Note that there is sometimes a definition for > lim_[n --> oo] (a_1 + ...+ a_n), but none for a_1 + a_2 +.... Definition: lim_[n --> oo] (a_1 + ...+ a_n) := a_1 + a_2 + .... Hilarious!!! I'm perfectly willing to accept this as a statement that the notation a1 + a2 + ... means limit of the sequence of partial sums, if it exists. - Randy MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Note that there is sometimes a definition for > lim_[n --> oo] (a_1 + ...+ a_n), but none for a_1 + a_2 +.... Definition: lim_[n --> oo] (a_1 + ...+ a_n) := a_1 + a_2 + .... When lim_[n --> oo] (a_1 + ...+ a_n) is not defined, and it is not when the a_i are positive naturals, then a_1 + a_2 +... is not either. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig Gecko/20070515 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) Note that there is sometimes a definition for > lim_[n --> oo] (a_1 + ...+ a_n), but none for a_1 + a_2 +.... Definition: lim_[n --> oo] (a_1 + ...+ a_n) := a_1 + a_2 + .... That's OK, but that means that the thing on the right, like the thing on the left, does not always exist. In particular, it doesn't exist unless a_n ->0. And while it is written with three dots ..., like the thing on the right its existence is defined in terms of the sequence of partial sums and the Cauchy criterion. - Randy === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Note that there is sometimes a definition for > lim [n --> oo] (a 1 + ...+ a n), but none for a 1 + a 2 +.... Definition: lim [n --> oo] (a 1 + ...+ a n) := a 1 + a 2 + .... That's OK, but that means that the thing on the right, > like the thing on the left, does not always exist. In > particular, it doesn't exist unless a n ->0. Let a n = 1 and a n+1 = -1. Tseries 1-1+1-1+1-+... is summable by the Ces.88ro summation method and its sum is equal to 1/2. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Note that there is sometimes a definition for > lim_[n --> oo] (a_1 + ...+ a_n), but none for a_1 + a_2 +.... Definition: lim_[n --> oo] (a_1 + ...+ a_n) := a_1 + a_2 + .... That's OK, but that means that the thing on the right, > like the thing on the left, does not always exist. In > particular, it doesn't exist unless a_n ->0. And while > it is written with three dots ..., like the thing on > the right its existence is defined in terms of the > sequence of partial sums and the Cauchy criterion. For some sums like 1/2 + 1/4 + 1/8 + ... we need no Cauchy criterion to prove that it is less than 100. And for the sum of the paths which can be distinguished from each other by nodes, we need no further criterion either, to decide that their number is less than the number of nodes used for this purpose. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig Gecko/20070515 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > No? Of what does the tree consist, besides finite levels with finitely > numbered nodes and edges? How do you define includes besides contains > all parts? There is no difference between the tree and all of its > finitely numbered nodes and edges. Nothing included in that totality > causes paths to outnumber nodes. > Agree or disagree? > The infinite tree = the n-th finite subtree for > some n > If we have an infinite sum of natural numbers, each less than 10. Can > this sum be uncountable? > You would have to define that infinite sum. Probably > (even though you don't say it) you really mean the > cardinality of some certain set, which is a union > of disjoint sets of cardinality < 10. > No, I said and meant numbers between 1 and 9, inclusive. Yes, but you didn't say what you meant by infinite sum. The series a_1 + a_2 +..., where said numbers appear. Normally the context in which such a thing might arise > is as a representation of the cardinality of a set. A sum of infinitely many natural numbers is not a thing > defined in the natural numbers. So presumably you have > something else in mind. You should have heard of convergent series. The value of a series is > also called its sum. You claim familiarity with convergent series. Then surely you're familiar with the fact that sum(n=1,oo) a_n means lim {S_1, S_2, S_3, ...} where S_k is the k-th partial sum, right? And surely you're aware that defining the limit, if it exists, is in terms of the value of S_k at finite values of k? So surely you're aware that the value of an infinite sum doesn't actually mean adding anything up infinitely many times, right? Yet while lecturing me that I should be familiar with what sum(n=1,oo) a_n means, you make statements like this... > Simply apply this definition infinitely often. So you don't actually understand the concept of limit of the sequence of partial sums, do you? - Randy === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig Gecko/20070515 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > No? Of what does the tree consist, besides finite levels with finitely > numbered nodes and edges? How do you define includes besides contains > all parts? There is no difference between the tree and all of its > finitely numbered nodes and edges. Nothing included in that totality > causes paths to outnumber nodes. > Agree or disagree? > The infinite tree = the n-th finite subtree for > some n > If we have an infinite sum of natural numbers, each less than 10. Can > this sum be uncountable? > You would have to define that infinite sum. Probably > (even though you don't say it) you really mean the > cardinality of some certain set, which is a union > of disjoint sets of cardinality < 10. > No, I said and meant numbers between 1 and 9, inclusive. Yes, but you didn't say what you meant by infinite sum. The series a_1 + a_2 +..., where said numbers appear. Not a defined notation, so it has no particular value. > Normally the context in which such a thing might arise > is as a representation of the cardinality of a set. A sum of infinitely many natural numbers is not a thing > defined in the natural numbers. So presumably you have > something else in mind. You should have heard of convergent series. The value of a series is > also called its sum. Yes, that is the meaning of an infinite sum when you have a convergent series. It is defined in terms of limits. When the series converges, the limit is defined to be the value of the infinite sum. The series you specified is not convergent, so this does not apply. > If you have > infinitely many, then the sum is infinite, The sum is undefined. We have definitions for the > sum of two natural numbers, which leads to a definition > of any finite number of natural numbers. Simply apply this definition infinitely often. No matter how many times you add one more number, you will not get infinitely many terms. No matter how many times you add 1 to a natural number, you will not get an infinite quantity. There is no point at which adding one more term changes from finite to infinite. - Randy === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > If we have an infinite sum of natural numbers, each less than 10. Can > this sum be uncountable? > You would have to define that infinite sum. Probably > (even though you don't say it) you really mean the > cardinality of some certain set, which is a union > of disjoint sets of cardinality < 10. > No, I said and meant numbers between 1 and 9, inclusive. > Yes, but you didn't say what you meant by infinite sum. The series a_1 + a_2 +..., where said numbers appear. Not a defined notation, so it has no particular > value. > Normally the context in which such a thing might arise > is as a representation of the cardinality of a set. > A sum of infinitely many natural numbers is not a thing > defined in the natural numbers. So presumably you have > something else in mind. You should have heard of convergent series. The value of a series is > also called its sum. Yes, that is the meaning of an infinite sum when > you have a convergent series. It is defined in terms > of limits. When the series converges, the limit is defined > to be the value of the infinite sum. The series you specified is not convergent, so this > does not apply. There are definitions and rules for diverging series. Consult a text book or Springer's online encyclopedia. > If you have > infinitely many, then the sum is infinite, > The sum is undefined. We have definitions for the > sum of two natural numbers, which leads to a definition > of any finite number of natural numbers. Simply apply this definition infinitely often. No matter how many times you add one more > number, you will not get infinitely many terms. Correct. But you will get more than any given number. And that's the meaning of infinity. No matter how many times you add 1 to a natural > number, you will not get an infinite quantity. > Correct. But you will get more than any given number. And that's the meaning of infinity. > There is no point at which adding one more > term changes from finite to infinite. Correct. But you will get more than any given number. And there is no other infinity. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > If we have an infinite sum of natural numbers, each less than 10. > Can > this sum be uncountable? > You would have to define that infinite sum. Probably > (even though you don't say it) you really mean the > cardinality of some certain set, which is a union > of disjoint sets of cardinality < 10. > No, I said and meant numbers between 1 and 9, inclusive. > Yes, but you didn't say what you meant by infinite sum. > The series a_1 + a_2 +..., where said numbers appear. Not a defined notation, so it has no particular > value. > Normally the context in which such a thing might arise > is as a representation of the cardinality of a set. > A sum of infinitely many natural numbers is not a thing > defined in the natural numbers. So presumably you have > something else in mind. > You should have heard of convergent series. The value of a series is > also called its sum. Yes, that is the meaning of an infinite sum when > you have a convergent series. It is defined in terms > of limits. When the series converges, the limit is defined > to be the value of the infinite sum. The series you specified is not convergent, so this > does not apply. There are definitions and rules for diverging series. Consult a text > book or Springer's online encyclopedia. Those rules do not make them convergent, and absent convergence, they do not have a value. > If you have > infinitely many, then the sum is infinite, > The sum is undefined. We have definitions for the > sum of two natural numbers, which leads to a definition > of any finite number of natural numbers. > Simply apply this definition infinitely often. No matter how many times you add one more > number, you will not get infinitely many terms. Correct. But you will get more than any given number. And that's the > meaning of infinity. Perhaps your meaning, but not necessarily ours. We have several meanings for infinity depending on context. And for series, infinity means no value at all. No matter how many times you add 1 to a natural > number, you will not get an infinite quantity. Correct. But you will get more than any given number. And that's the > meaning of infinity. Perhaps your meaning, but not necessarily ours. We have several meanings for infinity depending on context. And for series, infinity means having no value at all. There is no point at which adding one more > term changes from finite to infinite. Correct. But you will get more than any given number. And there is no > other infinity. We have several meanings for infinity depending on context. And for series, infinity means having no value at all. > === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig Gecko/20070515 Firefox/2.0.0.4,gzip(gfe),gzip(gfe) > If we have an infinite sum of natural numbers, each less than 10. Can > this sum be uncountable? > You would have to define that infinite sum. Probably > (even though you don't say it) you really mean the > cardinality of some certain set, which is a union > of disjoint sets of cardinality < 10. > No, I said and meant numbers between 1 and 9, inclusive. > Yes, but you didn't say what you meant by infinite sum. > The series a_1 + a_2 +..., where said numbers appear. Not a defined notation, so it has no particular > value. > Normally the context in which such a thing might arise > is as a representation of the cardinality of a set. > A sum of infinitely many natural numbers is not a thing > defined in the natural numbers. So presumably you have > something else in mind. > You should have heard of convergent series. The value of a series is > also called its sum. Yes, that is the meaning of an infinite sum when > you have a convergent series. It is defined in terms > of limits. When the series converges, the limit is defined > to be the value of the infinite sum. The series you specified is not convergent, so this > does not apply. There are definitions and rules for diverging series. Yes there are. But since you seem unaware of the definition of a convergent series, I have serious doubts you could make a meaningful comment about diverging series. > Consult a text > book or Springer's online encyclopedia. > If you have > infinitely many, then the sum is infinite, > The sum is undefined. We have definitions for the > sum of two natural numbers, which leads to a definition > of any finite number of natural numbers. > Simply apply this definition infinitely often. No matter how many times you add one more > number, you will not get infinitely many terms. Correct. But you will get more than any given number. And that's the > meaning of infinity. You will not get anything since you don't actually do the sum infinitely many times. You will get more than any given number at a FINITE PARTIAL SUM. That's what it means for a series to diverge. More precisely, for any given number, all the FINITE partial sums exceed that number for large enough n. No statement that you actually add infinitely many terms, no statement that the value is infinity, and no meaningful answer to your question is the sum a countable or an uncountable infinity. Remember, that was your original question: is the sum of infinitely many numbers [1-9] a countable or uncountable infinity? All we can say about that question is that the partial sums exceed any finite value. We're not equating a total (there is none) to aleph_0 or aleph_1. Aleph_0 and aleph_1 are set cardinalities, yet you claimed your question had nothing to do with set cardinalities. Again showing you have no idea what half the terms you throw around mean. - Randy === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > The series you specified is not convergent, so this > does not apply. There are definitions and rules for diverging series. Yes there are. But you don't know them? Although I don't consider it my task to teach you higher math, try the following (and don't forget to read till the end). E. Borel, Le.8dons sur les series divergentes , Gauthier-Villars (1928) G.H. Hardy, Divergent series , Clarendon Press (1949) R.G. Cooke, Infinite matrices and sequence spaces , Macmillan (1950) A. Peyerimhoff, Lectures on summability , Springer (1969) K. Knopp, Theorie und Anwendung der unendlichen Reihen , Springer (1964) ((Incomplete English translation: Blackie, 1928)) K. Zeller, W. Beekmann, Theorie der Limitierungsverfahren , Springer (1970) === Subject: Re: A consideration concerning the diagonal argument of G. Cantor The series you specified is not convergent, so this > does not apply. > There are definitions and rules for diverging series. Yes there are. But you don't know them? As they are irrelevant here, it is of no consequence whether one knows them or not. Divergent series do not have sums. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > The series you specified is not convergent, so this > does not apply. > There are definitions and rules for diverging series. > Yes there are. But you don't know them? As they are irrelevant here, it is of no consequence whether one knows > them or not. Divergent series do not have sums. There exist divergent series that are summable. For example, the series 1-1+1-1+1-+... is summable by the Ces.88ro summation method and its sum is equal to 1/2. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > Not if there is any colinearity between the points at which it has not > happened yet > the points in [0,1] are colinear > Not if this interval is taken on a circle. Even then, as long as they do not cover the whole circle, they are > linearly ordered, so may be regarded a colinear in themselves, even if > not in the circle. :) Virgil, I almost said something similar, but decided it was irrelevant banter. However, we are in agreement on that point. Have a nice day. Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > Not if there is any colinearity between the points at which it has not > happened yet > the points in [0,1] are colinear > Not if this interval is taken on a circle. Even then, as long as they do not cover the whole circle, they are > linearly ordered, so may be regarded a colinear in themselves, even if > not in the circle. :) Virgil, I almost said something similar, but decided it was irrelevant > banter. However, we are in agreement on that point. Have a nice day. Amost everyone, WM being excepted of course, is in agreement that WM is wrong about most things. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > That wasn't my argument. Pay attention. There are only finite levels > in the infinite binary tree. At each finite level, there are twice as > many nodes in the tree as paths. Hence, there is no point in the tree, > even when fully complete, where paths exceed nodes. I have no idea what this means, since I don't no what you mean by a > point in the tree [...] where paths exceed nodes. Does that mean you yes it? Interesting. You're coming around. ;) I meant a level of the tree. So let's ask a simpler question. The full binary tree has paths which > are infinitely long, yes? As countably infinite as the number of levels in the tree, so yes. Is the following true or false? The set of infinite paths in the full binary tree has greater > cardinality than the set of nodes. In other words: There is no injection from the set of infinite paths in the full > binary tree to the set of nodes. There is an injection from the set > of nodes to the set of infinite paths. There is a bijection between the set of rationals of the form 1-m*(2^-n) for m and n in N and the set of all finite paths within the tree. Since no node in the tree is ever infinitely distant from the root, no irrational is ever specified fully within the tree, and therefore doesn't exist within that tree. For every node in the tree, there is a rational specified, and an infinite number of irrationals, not to mention rationals not of the form 1-m*(2^-n) for m and n in N, yet to be specified. There are no other nodes in the tree, and so many reals are missed. The tree includes a number of levels equal to omega, a number of nodes equal to 2^omega-1, and a number of paths equal to 2^(omega-1), whatever value this phantom omega may represent. You probably feel I didn't answer your question, but read again, and see if you don't change your mind. Then rinse and repeat. :) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > That wasn't my argument. Pay attention. There are only finite levels > in the infinite binary tree. At each finite level, there are twice as > many nodes in the tree as paths. Hence, there is no point in the tree, > even when fully complete, where paths exceed nodes. I have no idea what this means, since I don't no what you mean by a > point in the tree [...] where paths exceed nodes. Does that mean you yes it? Interesting. You're coming around. ;) I meant a level of the tree. So are you saying that there is no level in the tree where the number of paths ending at that level exceed the number of nodes up to that level? While quite true, it is totally irrelevant, as the number of infinite paths which include (contain as a subpath) any single such finite path is uncountable, just like the in whole tree. > So let's ask a simpler question. The full binary tree has paths which > are infinitely long, yes? As countably infinite as the number of levels in the tree, so yes. > Is the following true or false? The set of infinite paths in the full binary tree has greater > cardinality than the set of nodes. In other words: There is no injection from the set of infinite paths in the full > binary tree to the set of nodes. There is an injection from the set > of nodes to the set of infinite paths. There is a bijection between the set of rationals of the form 1-m*(2^-n) > for m and n in N and the set of all finite paths within the tree. There is a bijection between ANY countably infinite set and the set of all finite sub-paths (starting at the root and ending at a node). All of which has nothing to do with the issue. Since > no node in the tree is ever infinitely distant from the root, no > irrational is ever specified fully within the tree, and therefore > doesn't exist within that tree. For every node in the tree, there is a > rational specified, and an infinite number of irrationals, not to > mention rationals not of the form 1-m*(2^-n) for m and n in N, yet to be > specified. There are no other nodes in the tree, and so many reals are > missed. All the irrationals, and some of the rationals, are represented by paths having both infinitely many left branchings and infinitely many right branchings. > The tree includes a number of levels equal to omega, a number of > nodes equal to 2^omega-1, and a number of paths equal to 2^(omega-1), > whatever value this phantom omega may represent. Wrong The 'number' of nodes is easily provably to be no greater than the 'number' of naturals. To wit: For any given node start a binary numeral with 1 and thereafter append 1 for every right branch or 0 for each left branch along the finite path path to the given node. This is easily seen to establish a surjection from the binary naturals to the set of all nodes. Q.E.D. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > That wasn't my argument. Pay attention. There are only finite levels > in the infinite binary tree. At each finite level, there are twice as > many nodes in the tree as paths. Hence, there is no point in the tree, > even when fully complete, where paths exceed nodes. > I have no idea what this means, since I don't no what you mean by a > point in the tree [...] where paths exceed nodes. Does that mean you yes it? Interesting. You're coming around. ;) No. It means I do not know what it means. How could you possibly take that as agreement? > So let's ask a simpler question. The full binary tree has paths > which > are infinitely long, yes? As countably infinite as the number of levels in the tree, so yes. > Is the following true or false? > The set of infinite paths in the full binary tree has greater > cardinality than the set of nodes. > In other words: > There is no injection from the set of infinite paths in the full > binary tree to the set of nodes. There is an injection from the set > of nodes to the set of infinite paths. There is a bijection between the set of rationals of the form > 1-m*(2^-n) for m and n in N and the set of all finite paths within the > tree. Since no node in the tree is ever infinitely distant from the > root, no irrational is ever specified fully within the tree, and > therefore doesn't exist within that tree. For every node in the tree, > there is a rational specified, and an infinite number of irrationals, > not to mention rationals not of the form 1-m*(2^-n) for m and n in N, > yet to be specified. There are no other nodes in the tree, and so many > reals are missed. The tree includes a number of levels equal to omega, > a number of nodes equal to 2^omega-1, and a number of paths equal to > 2^(omega-1), whatever value this phantom omega may represent. You probably feel I didn't answer your question, but read again, and > see if you don't change your mind. Then rinse and repeat. They were true/false questions. You didn't even try to answer them. Instead, you rambled a bit about nonsensical values such as omega - 1. I'll ask again. Are the above statements true or false? -- And that's what's wrong with Usenet. You people [...] can reply to a post [...] and then convince yourselves that you're great. Because you can open your mouths you think what you have is worth saying that you have proven your value. -- James S. Harris masters self-diagnosis === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > So, in other words, you don't want to consider the identity function > to apply to anything other than real numbers, so you're just not going > to do it. I am not allowed to say w=w or elephant=elephant, because > you don't want to talk about anything but reals. Gotcha. How does saying w=w require an identity function with no domain or > range? How does it require any identity function at all? f(x)=x. f(w)=w? w=w <-> f(w)=w? While it is not true that w=w -> f(w)=w (though it might be under a given rule that just occurred to me, intuitively), the converse is true. Does w=w REQUIRE the identity function? No. Does the identity function relate element count and value, thus shedding light on this question? Yes. Can a function have domain V? This one does. You're mightily confused. About some things, but nothing so fundamental. And Moe *never* said that the only legal identity function is the one > with domain and codomain R. For each set X, there is a corresponding > identity function id_X: X -> X. There is *not* a function > id:Set -> Set in ZFC (that is, a single function id such that id(x) is > defined for *every* set x). Your response is utterly irrelevant. > I'm obviously not restricting myself to ZFC. This thread isn't even about ZFC per se, but about a more fundamental argument, the diagonal argument. Like WM, I am appealing to more fundamental concepts than ZFC for my argument, which is the same, under the hood, as his. Of course he and I diverge in our choice of how to deal with the theory, but it's pretty hard to accept any objection to the identity function based on its domain. :) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <47b7beec@news2.lightlink.com> <47b88d62@news2.lightlink.com> <47bc70ef@news2.lightlink.com> <877igzphu8.fsf@phiwumbda.org> <47bce9a8@news2.lightlink.com> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > So, in other words, you don't want to consider the identity function > to apply to anything other than real numbers, so you're just not going > to do it. I am not allowed to say w=w or elephant=elephant, because > you don't want to talk about anything but reals. Gotcha. How does saying w=w require an identity function with no domain or > range? æHow does it require any identity function at all? f(x)=x. > f(w)=w? > w=w <-> f(w)=w? What is the point of you writing those formulas?. While it is the case that for every set there is the identity element on that set, it is not required to belabor all of that just to prove Ax x=x, which is theorem of identity theory itself. 'function' and 'identity function' come LATER. > While it is not true that w=w -> f(w)=w (though it might be under a > given rule that just occurred to me, intuitively), the converse is true. Of course the converse is true, since the consequent of the converse is true. So what? > Does w=w REQUIRE the identity function? No. Does the identity function > relate element count and value, thus shedding light on this question? Yes. Can a function have domain V? This one does. What is 'V'? Do you mean the universal set? In Z set theory there is no object that answers to that description. And you've given no theory in which to provide meaning to such a universal set. > You're mightily confused. About some things, but nothing so fundamental. That you are so thoroughly confused at such a fundamental level would be remarkable except that it's explained by your continued refusal to even study this fundamental material though you spout on about it at will. > And Moe *never* said that the only legal identity function is the one > with domain and codomain R. æFor each set X, there is a corresponding > identity function id X: X -> X. æThere is *not* a function > id:Set -> Set in ZFC (that is, a single function id such that id(x) is > defined for *every* set x). æ Your response is utterly irrelevant. I'm obviously not restricting myself to ZFC. This thread isn't even > about ZFC per se, but about a more fundamental argument, the diagonal > argument. It is in Z set theories (or similar or related theories) that we finally formalize the diagonal argument. And if your remarks are in the context of some other theory, then please state what theory. Meanwhile, a bunch of symbol gibberish posted by you is just cause for more embarrasement felt for you. > Like WM, I am appealing to more fundamental concepts than ZFC Your fundamental concepts are a confused mix of your undefined terminology and symbol gibberish. > for my argument, which is the same, under the hood, as his. Of course he > and I diverge in our choice of how to deal with the theory, but it's > pretty hard to accept any objection to the identity function based on > its domain. No one is objecting to the claim that for every set there exists the identity function whose domain is that set. We knew that, and (trivially) proved it long before you. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Can a function have domain V? Not unless V is a set. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > So, in other words, you don't want to consider the identity function > to apply to anything other than real numbers, so you're just not going > to do it. I am not allowed to say w=w or elephant=elephant, because > you don't want to talk about anything but reals. Gotcha. > How does saying w=w require an identity function with no domain or > range? How does it require any identity function at all? f(x)=x. > f(w)=w? > w=w <-> f(w)=w? No. In any first order language with equality, we have the axiom (A x)(x=x). That doesn't depend on the existence of an identity function at all. > While it is not true that w=w -> f(w)=w (though it might be under a > given rule that just occurred to me, intuitively), the converse is > true. Does w=w REQUIRE the identity function? No. Does the identity > function relate element count and value, thus shedding light on this > question? Yes. Shedding light on this trivial claim that w=w? No. > Can a function have domain V? This one does. There is no function with domain V in ZFC. It's perfectly acceptable to speak about an operation Set -> Set (or a functor), but that's not a function in the set theoretic sense. > You're mightily confused. About some things, but nothing so fundamental. > And Moe *never* said that the only legal identity function is the one > with domain and codomain R. For each set X, there is a corresponding > identity function id_X: X -> X. There is *not* a function id:Set -> Set in ZFC (that is, a single function id such that id(x) is > defined for *every* set x). Your response is utterly irrelevant. I'm obviously not restricting myself to ZFC. This thread isn't even > about ZFC per se, but about a more fundamental argument, the diagonal > argument. Like WM, I am appealing to more fundamental concepts than > ZFC for my argument, which is the same, under the hood, as his. Of > course he and I diverge in our choice of how to deal with the theory, > but it's pretty hard to accept any objection to the identity function > based on its domain. The objection has to do with whether the identity operator Set -> Set is a set-theoretical function. It isn't. This is just a simple terminological fact, not deep or controversial in the least. But your response was a lot stranger than you let on. First, you claimed Moe disallowed any identity function other than id_R:R -> R. Then you implied that equations like w=w depend on the existence of a single identity *function* with domain V. This response was plainly nutty, a lot nuttier than what you've said in your current post. -- Jesse F. Hughes If you hadn't noticed, basically every result I have destroys some precious belief of mathematicians and they have from what I've gathered basically gone collectively bonkers. -- James S. Harris === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > In my assertions of infinite case induction, I made clear rules as to > what kind of properties could be derived You haven't made a clear rule since the fouth grade when you said that > anything hit over Mr. Wilson's fence but not past his garden is a > ground rule double. MoeBlee Good one, but do you realize that that's become the standard rule for backyard wiffleball across the country since then? Not that I make any money off of it, but at least I made a contribution to the axioms of summer recreation. (pats self on back, injures shoulder) It's certainly better than your policy of kicking the football whenever Lucy van Cantor promised not to trick you and pull the ball away. But seriously, I stated that it applied to formulaic equalities, and formulaic inequalities that were not based on a difference not with a limit of 0 as n->oo. It doesn't apply to is finite, has a last element, likes chocolate eclairs, or has an attitude. So, you know, I kinda don't appreciate it being characterized as, infinite sets are the same as finite ones. That disingenuous. Not to get pissy or anything. BTW, my son Tom says hi. :) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > In my assertions of infinite case induction, I made clear rules as to > what kind of properties could be derived You haven't made a clear rule since the fouth grade when you said that > anything hit over Mr. Wilson's fence but not past his garden is a > ground rule double. > Good one (pats self on back, injures shoulder) But seriously, I stated that it applied to formulaic equalities, and > formulaic inequalities that were not based on a difference not with a > limit of 0 as n->oo. Can you unscramble that particular line of double talk? > It doesn't apply to is finite, has a last > element, likes chocolate eclairs, or has an attitude. So, you know, > I kinda don't appreciate it being characterized as, infinite sets are > the same as finite ones. That disingenuous. Not to get pissy or anything. You are saying precisely that there are contexts in which they are alike, but fail to be specific about which contexts. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > In any case, the function WM addresses, f(x)=x, does not have any such > undefinedness at any defined x. If omega exists, f(omega)=omega. I don't > see how this identity function, a direct unit linear relationship, can > ever be considered discontinuous at any point that exists on that line. > Do you? > The function f has a certain DOMAIN. In Z set theories there is NO > function that has EVERYTHING in its domain. If w is not in the domain > of f, then, in ordinary mathematics, 'f(w)' is taken as undefined > (though, in more formal context, with the Fregean method, we can avoid > having anything undefined like that). And continuity applies to f > evaluated at points in the domain of f - such a domain as the set of > real numbers in which continuity can make sense as defined in terms of > some metric or whatever. Continuity doesn't apply to EVERYTHING you > want to shove in as an argument even though it is not in the domain of > the function. > MoeBlee > So, in other words, you don't want to consider the identity function to > apply to anything other than real numbers, No, I've not suggested anything so ridiculous, and I don't appreciate > you suggesting that I have. :0 Uhhhhh.... What is your objection, then? Didn't it have something to do with the domain of the identity function? Perhaps you were confused. Maybe it's me. Let me reiterate: f(x)=x -> f(w)=w The nth natural is n, that is, N_n=n, thus index is equal to magnitude. A set of size n has an nth element, that is, N_|N| e N. I leave the rest of the exercise to the reader. Hint: what in the universe is not equal to itself? > > so you're just not going to > do it. I am not allowed to say w=w or elephant=elephant, because you > don't want to talk about anything but reals. Gotcha. Utter rubbish. My remarks entail no such thing. Please just go to the > blazes with this garbage, would you? MoeBlee > To the blazes? Not sure I believe in any such thing, either. Guess both questions will be answered one of these days... Besides, I'm a seasoned garbage picker. People throw out the darnedest stuff! ;) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <47b7beec@news2.lightlink.com> <47b88d62@news2.lightlink.com> <47bce2af@news2.lightlink.com> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > In any case, the function WM addresses, f(x)=x, does not have any such > undefinedness at any defined x. If omega exists, f(omega)=omega. I don't > see how this identity function, a direct unit linear relationship, can > ever be considered discontinuous at any point that exists on that line. > Do you? > The function f has a certain DOMAIN. In Z set theories there is NO > function that has EVERYTHING in its domain. If w is not in the domain > of f, then, in ordinary mathematics, æ'f(w)' is taken as undefined > (though, in more formal context, with the Fregean method, we can avoid > having anything undefined like that). And continuity applies to f > evaluated at points in the domain of f - such a domain as the set of > real numbers in which continuity can make sense as defined in terms of > some metric or whatever. Continuity doesn't apply to EVERYTHING you > want to shove in as an argument even though it is not in the domain of > the function. > MoeBlee > So, in other words, you don't want to consider the identity function to > apply to anything other than real numbers, No, I've not suggested anything so ridiculous, and I don't appreciate > you suggesting that I have. :0 Uhhhhh.... What is your objection, then? It's been explained to you by me in my original post, and then by at least two other posters after me. A function has a domain; it's nonsense to argue about the values of a function for arguments not in the domain. > Didn't it have something to do with the > domain of the identity function? In Z set theories, there is no the identity function. There is an identity function for every set. The set is the domain of the identity function on that set. And there is no function that has EVERY set in the domain of the function. > Perhaps you were confused. Maybe it's me. Not maybe. > Let me reiterate: f(x)=x -> f(w)=w WHAT IS THE DOMAIN OF f? > The nth natural is n, that is, N n=n, thus index is equal to magnitude. What is 'N' in 'N n'? I suppose in that context you mean for N to be the identity function on w (but then you take |N| as an argument, so it's not clear what you intend as the domain of N). As to 'magnitude', if you DEFINE it, then we can discuss whether your statement about it holds or not. > A set of size n has an nth element, Yes, where n is a natural number. > that is, N |N| e N. That is NONSENSE. You're using 'N' in TWO DIFFERENT senses in just one short formula and then pile even more nonsense on to it. First, you use 'N' as some kind of function. That's what 'N n' indicates. N n is the value of the function N at the argument n. Then you take the cardinality of 'N', which has to be the cardinality of that function. Then you claim that cardinality is an ARGUMENT to the function (but then what is the DOMAIN) of that function? Then you claim that the function at the argument that is the cardinality of that function is a MEMBER of that function, which is NONSENSE. It is PATHETIC that you're becomming even MORE mathematically infantile. > I leave the rest of the exercise to the reader. WHAT exercise? > Hint: what in the universe is not equal to itself? In ordinary mathematical theories, in any given universe, there is no object not equal to itself. What is your point? MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <47b7beec@news2.lightlink.com> <47b88d62@news2.lightlink.com> posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > It's been explained to you by me in my original post, and then by at > least two other posters after me. A function has a domain; it's > nonsense to argue about the values of a function for arguments not in > the domain. Oh, I see you have learned since our corresponding discussion. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor It's been explained to you by me in my original post, and then by at > least two other posters after me. A function has a domain; it's > nonsense to argue about the values of a function for arguments not in > the domain. Oh, I see you have learned since our corresponding discussion. But we see that WM has not learned anything in all these discussions. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <47b7beec@news2.lightlink.com> <47b88d62@news2.lightlink.com> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) It's been explained to you by me in my original post, and then by at > least two other posters after me. A function has a domain; it's > nonsense to argue about the values of a function for arguments not in > the domain. Oh, I see you have learned since our corresponding discussion. What are you talking about? What I just posted is nothing I haven't known long before I came across you. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > In any case, the function WM addresses, f(x)=x, does not have any such > undefinedness at any defined x. If omega exists, f(omega)=omega. I don't > see how this identity function, a direct unit linear relationship, can > ever be considered discontinuous at any point that exists on that line. > Do you? > The function f has a certain DOMAIN. In Z set theories there is NO > function that has EVERYTHING in its domain. If w is not in the domain > of f, then, in ordinary mathematics, 'f(w)' is taken as undefined > (though, in more formal context, with the Fregean method, we can avoid > having anything undefined like that). And continuity applies to f > evaluated at points in the domain of f - such a domain as the set of > real numbers in which continuity can make sense as defined in terms of > some metric or whatever. Continuity doesn't apply to EVERYTHING you > want to shove in as an argument even though it is not in the domain of > the function. > MoeBlee > So, in other words, you don't want to consider the identity function to > apply to anything other than real numbers, No, I've not suggested anything so ridiculous, and I don't appreciate > you suggesting that I have. :0 Uhhhhh.... What is your objection, then? Didn't it have something to do with the > domain of the identity function? Perhaps you were confused. Maybe it's me. Let me reiterate: f(x)=x -> f(w)=w The nth natural is n, that is, N_n=n, thus index is equal to magnitude. A much better scheme is to use the von Neumann naturals, as otherwise the empty set does not have a natural number cardinality. And with the von Neumann naturals, {} = 0 is a natural first natural. A set of size n has an nth element, that is, N_|N| e N. Then the set of the first n naturals does not contain n. With the one-origin naturals, you have |{1,2,...,n}| = n, so that each n has to be a member of the initial set having it as cardinal, which is more than a bit circular. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > No Largest Finite!!! GONG! Huyah huyah huyah Ommmmmega!!! > nice mantra Would you please stop that? It's puerile, stupid and annoying. MoeBlee I did stop for a long time, and I know it's annoying, and that's why. But, when someone responds to a point I'm making that has nothing to do with any largest element by saying, A finite set isn't the same as an infinite set, I feel compelled to point out that knee-jerk reaction for what it is. That doesn't justify inventing transitions that occur within some linear order, but not at any identifiable point therein, just ... because. Sorry to be so puerile, but I do have a personal motto I'll share in this context: Old since birth, young till death. Kind of an operating principle. Keeps me young, even though I never felt young. Besides, you gotta admit the mantra has a nice ring to it, and you know it's sometimes satisfying, if unproductive, to be annoying. Or, should I just have said, I'm not doing anything!!! ? ;) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <47bce027$1@news2.lightlink.com> posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I Gecko/20060418 Firefox/1.0.8 (Ubuntu package 1.0.8),gzip(gfe),gzip(gfe) > No Largest Finite!!! GONG! Huyah huyah huyah Ommmmmega!!! > nice mantra Would you please stop that? It's puerile, stupid and annoying. MoeBlee > I did stop for a long time, and I know it's annoying, and that's why. > But, when someone responds to a point I'm making that has nothing to do > with any largest element by saying, A finite set isn't the same as an > infinite set, I feel compelled to point out that knee-jerk reaction for > what it is. The problem is that just about every one of your aguments goes It holds for every finite set that makes up this infinite set, so it must hold for the infinite set. The fact that you get the same answer is not due to a knee-jerk reaction on the part of your critics, but because your armgument never changes, (Sometimes you try new words). It is obvious that if P holds for every finite set that makes up this infinite set, therefore P must hold for this infinite set were true, then there would be a largest integer. So whenever you present P holds for every finite set that makes up this infinite set, therefore P must hold for this infinite set, someone points out that this implies the existence of a largest natural. Despite your claim, what you say does have something to do with the existence of a largest integer. - William Hughes === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > The problem is that just about every one of your aguments > goes æIt holds for every finite set that makes > up this infinite set, so it must hold for the infinite set. > The fact that you get the same answer is not due to > a knee-jerk reaction on the part of your critics, > but because your armgument never changes, (Sometimes > you try new words). It is obvious that if P holds for every finite > set that makes up this infinite set, therefore > P must hold for this infinite set were true, > then there would be a largest integer. Of course, for every finite set there is a largest integer. The union, i.e., the infinite set has not a largest integer. This is completely different from the silly assertion: In every finite set there are more A's than B's, but in the union, i.e., in the infinite set, there are more B's than A's. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor The problem is that just about every one of your aguments > goes æIt holds for every finite set that makes > up this infinite set, so it must hold for the infinite set. > The fact that you get the same answer is not due to > a knee-jerk reaction on the part of your critics, > but because your armgument never changes, (Sometimes > you try new words). It is obvious that if P holds for every finite > set that makes up this infinite set, therefore > P must hold for this infinite set were true, > then there would be a largest integer. Of course, for every finite set there is a largest integer. The union, > i.e., the infinite set has not a largest integer. > This is completely different from the silly assertion: > In every finite set there are more A's than B's, but in the union, > i.e., in the infinite set, there are more B's than A's. In every finite tree, there are exactly as many finite paths (including on-maximal ones, so includes one path ending on each node) as nodes, and this carries over to infinite trees. In any finite tree there are absolutely no infinite paths, but this does not carry over to infinite trees. So that conflating finite paths with infinite paths, as you do, is a mugs game. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > In every finite tree, there are exactly as many finite paths (including > on-maximal ones, so includes one path ending on each node) as nodes, and > this carries over to infinite trees. In any finite tree there are absolutely no infinite paths, but this does > not carry over to infinite trees. In a tree every path ends at a last node. If there are no last nodes in infinite trees, this can be taken as an indication that there are no paths in infinite trees and, probably, no infinite trees. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor In every finite tree, there are exactly as many finite paths (including > on-maximal ones, so includes one path ending on each node) as nodes, and > this carries over to infinite trees. In any finite tree there are absolutely no infinite paths, but this does > not carry over to infinite trees. In a tree every path ends at a last node. In WM's trees, perhaps every path ends in a node, but we have seen that his mytheology has little, if anything ,to do with actual mathmatics, so that he cannot allow himself to consider anything his myths do not cover. > If there are no last nodes > in infinite trees, this can be taken as an indication that there are > no paths in infinite trees If there is no last natural, then by WM's argument he can claim there can be no naturals at all. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > Of course, for every finite set there is a largest integer. The union, > i.e., the infinite set has not a largest integer. > This is completely different from the silly assertion: > In every finite set there are more A's than B's, but in the union, > i.e., in the infinite set, there are more B's than A's. What we have is In every finite tree there are more nodes than finite paths. In the infinite tree there are more infinite paths than nodes. So we have In every finite set there are more A's than B's In the union there are more C's than A's. An infinite path is the union of finite paths. It is not a finite path. Alternately we can say In every finite tree there are more unions of finite paths than nodes. In the infinite tree there are more unions of finite paths than nodes. - William Hughes === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Of course, for every finite set there is a largest integer. The union, > i.e., the infinite set has not a largest integer. > This is completely different from the silly assertion: > In every finite set there are more A's than B's, but in the union, > i.e., in the infinite set, there are more B's than A's. What we have is æ æIn every finite tree there are more nodes than finite paths. æ æIn the infinite tree there are more infinite paths than nodes. So we have æ æ In every finite set there are more A's than B's > æ æ In the union there are more C's than A's. Sophisticated nonsense. You could also state that in every tree with four levels we have only paths over four levels. A path by definition is passing the whole tree. So what we have is: In every finite tree there are more nodes than paths which pass the whole tree. In the infinite tree there are more paths which pass the whole tree than nodes. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Of course, for every finite set there is a largest integer. The union, > i.e., the infinite set has not a largest integer. > This is completely different from the silly assertion: > In every finite set there are more A's than B's, but in the union, > i.e., in the infinite set, there are more B's than A's. What we have is æ æIn every finite tree there are more nodes than finite paths. æ æIn the infinite tree there are more infinite paths than nodes. So we have æ æ In every finite set there are more A's than B's > æ æ In the union there are more C's than A's. > Sophisticated nonsense. Perhaps sophisticated, but hardly nonsense. The A's are nodes, the B's are finite paths, the C's are infinite paths. In every finite tree there is a bijection between the set of terminal nodes and the set of paths. In our infinite tree there re zero terminal nodes, so that WM's logic requires that there be zero paths. . You could also state that in every tree with four levels we have only > paths over four levels. A path by definition is passing the whole tree. Is that anything like passing gas? > So what we have is: In every finite tree there are more nodes than paths which pass the > whole tree. In every finite tree, of more that none node, there are more nodes than terminal nodes. > In the infinite tree there are more paths which pass the whole tree > than nodes. In the infinite tree there are more paths than terminal nodes. > === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > Of course, for every finite set there is a largest integer. The union, > i.e., the infinite set has not a largest integer. > This is completely different from the silly assertion: > In every finite set there are more A's than B's, but in the union, > i.e., in the infinite set, there are more B's than A's. What we have is In every finite tree there are more nodes than finite paths. In the infinite tree there are more infinite paths than nodes. So we have In every finite set there are more A's than B's > In the union there are more C's than A's. Sophisticated nonsense. You could also state that in every tree with four levels we have only > paths over four levels. A path by definition is passing the whole tree. So what we have is: In every finite tree there are more nodes than paths which pass the > whole tree. > In the infinite tree there are more paths which pass the whole tree > than nodes. > When you have a path that passes the whole finite tree you have a path with an end. When you have a path that passes the whole infinite tree you have a path without an end. A path with and end and a path without an end are two different things. You use the same word, path to refer to two different things. - William Hughes === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > When you have a path that passes the whole finite tree you > have a path with an end. When you have a path that passes the whole > infinite tree you have a path without an end. A path with > and end and a path without an end are two different things. > You use the same word, path to refer to two different things. When you have a path in a tree with less than 10 levels, then you have a path with less than 10 nodes. When you have a path in a tree with more than 10 levels, then you have a path with more than 10 nodes. These are two different things. Nevertheless for both sorts of paths it is true that there are less paths than nodes in the respective trees. Why do we know that? Because we can calculate it. We can calculate even more. When we have a path which reaches not farther than the nodes reach, then it is true that there are less paths than nodes in the tree. Thus we have in fact two different things but we can calculate directly by a simple formula that paths which stretch not farther than the tree and its nodes, cannot gain a greater cardinality than the nodes. So the question is not whether the paths are finite or infinite but whether they consist only of nodes which have a finite distance from the root or not. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor When you have a path that passes the whole finite tree you > have a path with an end. When you have a path that passes the whole > infinite tree you have a path without an end. A path with > and end and a path without an end are two different things. > You use the same word, path to refer to two different things. When you have a path in a tree with less than 10 levels, then you have > a path with less than 10 nodes. When you have a path in a tree with > more than 10 levels, then you have a path with more than 10 nodes. > These are two different things. Nevertheless for both sorts of paths > it is true that there are less paths than nodes in the respective > trees. Why do we know that? Because we can calculate it. Every finite path ends in a terminal node, so that the number of finite paths in any tree equals the number of terminal nodes. It follows by WM's sort of illogic that then number of paths in an infinite tree is the number of terminal nodes in that tree. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I When you have a path that passes the whole finite tree you > have a path with an end. When you have a path that passes the whole > infinite tree you have a path without an end. A path with > and end and a path without an end are two different things. > You use the same word, path to refer to two different things. When you have a path in a tree with less than 10 levels, then you have > a path with less than 10 nodes. When you have a path in a tree with > more than 10 levels, then you have a path with more than 10 nodes. > These are two different things. Nevertheless for both sorts of paths > it is true that there are less paths than nodes in the respective > trees. Why do we know that? Because we can calculate it. > We can calculate even more. When we have a path which reaches not > farther than the nodes reach, then it is true that there are less > paths than nodes in the tree. Not true. When you have a tree where all paths end it is true that there are less paths than nodes in the tree. You have a direct argument for this case. However if you have a tree where no path ends you do not have a direct argument. All you say is that if it is true for the paths that end it must be true for the paths that don't end (after all paths that don't end are made up of paths that end). But paths that end and paths that don't end are very different things. - William Hughes === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > When you have a path that passes the whole finite tree you > have a path with an end. æWhen you have a path that passes the whole > infinite tree you have a path without an end. æA path with > and end and a path without an end are two different things. > You use the same word, path to refer to two different things. When you have a path in a tree with less than 10 levels, then you have > a path with less than 10 nodes. When you have a path in a tree with > more than 10 levels, then you have a path with more than 10 nodes. > These are two different things. Nevertheless for both sorts of paths > it is true that there are less paths than nodes in the respective > trees. Why do we know that? Because we can calculate it. > We can calculate even more. When we have a path which reaches not > farther than the nodes reach, then it is true that there are less > paths than nodes in the tree. Not true. æWhen you have a tree where all paths end it is true that > there > are less paths than nodes in the tree. You have a direct argument for > this case. Not the ending is decive but only the fact that all path are formed by nodes. > However if you have a tree where no path ends æyou do not have > a direct argument. æAll you say is that if it is true for the paths > that end it must be true for the paths that don't end (after all > paths that don't end are made up of paths that end). æBut paths > that end and paths that don't end are very different things. If you have a sequence of terms a n = 1/2^n, then you can conclude that even the sum of their terms cannot surpass 1 although you have only a formula for finite partial sums. There are no steps in infinity. You had recognized that some posts ago. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > When you have a path that passes the whole finite tree you > have a path with an end. æWhen you have a path that passes the whole > infinite tree you have a path without an end. æA path with > and end and a path without an end are two different things. > You use the same word, path to refer to two different things. > When you have a path in a tree with less than 10 levels, then you have > a path with less than 10 nodes. When you have a path in a tree with > more than 10 levels, then you have a path with more than 10 nodes. > These are two different things. Nevertheless for both sorts of paths > it is true that there are less paths than nodes in the respective > trees. Why do we know that? Because we can calculate it. > We can calculate even more. When we have a path which reaches not > farther than the nodes reach, then it is true that there are less > paths than nodes in the tree. Not true. æWhen you have a tree where all paths end it is true that > there > are less paths than nodes in the tree. You have a direct argument for > this case. Not the ending is decive but only the fact that all path are formed by > nodes. A path without edges is not a path. So it is equally valid to claim that a path is formed by its edges. In any case, a finite path is one with a terminal node/edge, and an infinite path is one without any terminal node/edge. A finite tree has only the former, and a completed infinite tree only the latter. And counting the former is irrelevant to counting the latter. However if you have a tree where no path ends æyou do not have > a direct argument. æAll you say is that if it is true for the paths > that end it must be true for the paths that don't end (after all > paths that don't end are made up of paths that end). æBut paths > that end and paths that don't end are very different things. If you have a sequence of terms a_n = 1/2^n Irrelevant to the numbering of paths. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > When you have a path that passes the whole finite tree you > have a path with an end. When you have a path that passes the whole > infinite tree you have a path without an end. A path with > and end and a path without an end are two different things. > You use the same word, path to refer to two different things. > When you have a path in a tree with less than 10 levels, then you have > a path with less than 10 nodes. When you have a path in a tree with > more than 10 levels, then you have a path with more than 10 nodes. > These are two different things. Nevertheless for both sorts of paths > it is true that there are less paths than nodes in the respective > trees. Why do we know that? Because we can calculate it. > We can calculate even more. When we have a path which reaches not > farther than the nodes reach, then it is true that there are less > paths than nodes in the tree. Not true. When you have a tree where all paths end it is true that > there > are less paths than nodes in the tree. You have a direct argument for > this case. Not the ending is decive but only the fact that all path are formed by > nodes. However if you have a tree where no path ends you do not have > a direct argument. All you say is that if it is true for the paths > that end it must be true for the paths that don't end (after all > paths that don't end are made up of paths that end). But paths > that end and paths that don't end are very different things. If you have a sequence of terms a_n = 1/2^n, Please do not change the subject. - William Hughes === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > When you have a path that passes the whole finite tree you > have a path with an end. æWhen you have a path that passes the whole > infinite tree you have a path without an end. æA path with > and end and a path without an end are two different things. > You use the same word, path to refer to two different things. > When you have a path in a tree with less than 10 levels, then you have > a path with less than 10 nodes. When you have a path in a tree with > more than 10 levels, then you have a path with more than 10 nodes. > These are two different things. Nevertheless for both sorts of paths > it is true that there are less paths than nodes in the respective > trees. Why do we know that? Because we can calculate it. > We can calculate even more. When we have a path which reaches not > farther than the nodes reach, then it is true that there are less > paths than nodes in the tree. > Not true. æWhen you have a tree where all paths end it is true that > there > are less paths than nodes in the tree. You have a direct argument for > this case. Not the ending is decisive but only the fact that all path are formed by > nodes. > However if you have a tree where no path ends æyou do not have > a direct argument. æAll you say is that if it is true for the paths > that end it must be true for the paths that don't end (after all > paths that don't end are made up of paths that end). æBut paths > that end and paths that don't end are very different things. If you have a sequence of terms a n = 1/2^n, Please do not change the subject. It is the same as before. It shows that there are no steps in the infinite. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > No? Of what does the tree consist, besides finite levels with > finitely > numbered nodes and edges? How do you define includes besides > contains > all parts? There is no difference between the tree and all of its > finitely numbered nodes and edges. Nothing included in that totality > causes paths to outnumber nodes. > Agree or disagree? > The infinite tree = the n-th finite subtree for > some n > If we have an infinite sum of natural numbers, each less than 10. Can > this sum be uncountable? You would have to define that infinite sum. Probably > (even though you don't say it) you really mean the > cardinality of some certain set, which is a union > of disjoint sets of cardinality < 10. No, I said and meant numbers between 1 and 9, inclusive. If you have > infinitely many, then the sum is infinite, namely it is at least > infinite, 1*oo, and at most 9*oo. But it is not uncountable. It does not exist at all. Any infinite sequence of natural numbers between 1 and 9 inclusive fails to converge to anything, thus any such alleged sum can not exist at all. Sums of finite sets of naturals can exist by iteration of summing of two such naturals, but such iteration is not defined otherwise. WM's mythematics is getting more mythtical daily. . The same is true for the infinite binary tree which is nothing but the > countable union of all finite binary trees. When WM can prove the existence of a surjection from N to the set of all binary sequences, preferably by an explicit surjection from N to the set of such sequences, only the can he claim to count the paths in a binary tree. Absent such proof, he is only within his mythematics, not in mathematics. > If you have a union of uncountably many such sets, > the union is uncountable. In the binary tree we have a countable union of countable sets of > paths. Which leaves out most of them. Your turn. Agree or disagree? The infinite tree = > the n-th finite subtree for some n It is the union of all finite binary trees with n levels. In such a tree, each node is the terminal node of a finite path at some level, from the root to node itself (the root itself being a zero-edged path) so that WM is claiming that set of finite paths should surject to the set of infinite paths. Since Cantor has disproved this, at least to the satisfaction of everyone who counts, WM's claim can only be established by explicit construction of such a surjection of nodes(or finite paths) to infinite paths. Well, WM? === Subject: Re: SOLUTION MANUALS CHEAP!!! posting-account=qyS_IQoAAACeGEcm1MebDCnbi85IEqwT 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) I am selling solution manuals for $5.00. I only accept payments > through PAYPAL. Please do not post back on this site, send me an > email > if you are interested and I will reply with instructions. You will > have your solutions manual within 24 hours, USUALLY WITHIN THE HOUR!! dc9...@comcast.net I have the following solutions manuals: Single Variable Calculus Early Trancendentals (5th Edition) // > Anderson, Cole, Drucker Principles and Applications of Electrical Engineering (5th Edition) // > Giorgio Rizzoni Principles and Applications of Electrical Engineering 4th Edition) // > Giorgio Rizzoni Mechanics of Materials (3rd Edition... same as 4th Edition, just > different numbers) // Beer, Johnston, Dewolf Fundamentals of Modern Manufacturing (3rd Edition) // Groover Thermodynamics: an engineering approach (5th Edition) // Cengel, > Boles Thermodynamics: an engineering approach (6th Edition) // Cengel, > Boles Design of Machinery (3rd Edition) // Norton Physics for Scientists and Engineers with Modern Physics (3rd Edition > Volume 11) // Giancoli Introduction to Fluid Mechanics (6th Edition) // Fox Introduction to Fluid Mechanics (5th Edition) // Fox Shigley's Mechanical Engineering Design (8th edition) // Budynas Fundamentals of Heat and Mass Transfer (6th Edition) // F.P.Incropera > and D.P.DeWitt Probability, Random Variables and Stochastic Processes (4th > Edition) // Papoulis, Unnikrishna Pillai Analytical Mechanics (7th Edition) // Fowles Fundamentals of Engineering Thermodynamics // Moran, Shapiro Electric Machinery Fundamentals (4th Edition) // Chapman Digital Image Processing (2nd Edition) æ// Gonzalez Classical Electrodynamics (2nd Edition) // Jackson, Kasper, Wijk Communication System Engineering (2002) // Proakis Advanced Engineering Mathematics (8th Edition) // Kreyszig Calculus (5th Edition) // Stewart Introduction to Linear Algebra (3rd Edition) // Strang Physics (5th Edition)// Halliday, Resnick, Krane Applied Statistics and Probability for Engineers (3rd Edition) // > Montgomery, Runger Communication Systems (4th Edition) // Haykin Digital Signal Processing: A computer Based Approach (1st Edition) // > Mitra Econometric Analysis (5th Edition) // Greene Electric Machinery (6th Edition) // Fitzgerald, Kingsley, Uman Elementary Mechanics & Thermodynamics (2000) // Norbury Engineering Fluid Mechanics (7th Edition) // Crowe, Elger, Roberson Fundamentals Of Fluid Mechanics (3rd Edition) Fundamentals of Fluid Mechanics (4th Edition) Signals and Systems (2nd Edition) // Oppenheim and Wilsky Signals and Systems (2nd Edition) // Haykin > Goldstein Classical Mechanics (2nd Edition) // Reid Chemical and Engineering Thermodynamics (3rd Edition) // Wiley Engineering Fluid Mechanics (7th Edition) // Elger, Crowe Probability Random Variables and Stochastic Processes (4th Edition) // > Papoulis, Pillai Communication System Engineering (2nd Edition) // Proakis, Salehi Physics for Scientists and Engineers (6th Edition) // Serway, Jewett Microwave Engineering (3rd Edition) // Pozar Engineering Electromagnetics (6th Edition) // Hayt, Buck Elementary Differential Equations and Boundary Value Problems (7th > Edition) // Diprima, Boyce Solved and Unsolved Problems in Number Theory (2nd Edition) // Shanks Fundamentals of Physics 1,2,3,4 (4th Edition) CALCULUS Early Transcendentals æ(7th Edition) // Davis Applied Statistics and Probability for Engineers (3rd Edition) // > Montgomery, Runger Calculus (10th Edition) Volume 1,2// Thomas Control Systems Engineering // Nise Engineering Electromagnetics (6th Edition) // Buck, Hayt I have others that may not be listed, please email with other > requests. Again, my email is dc9...@comcast.net Email me what solutions you want and i will reply back with > instructions for PAYPAL payment I am interested in the solutions manual for Principles and > Applications of Electrical Engineering 5th edition by Giorgio Rizzoni My name is Dan. I am interested in the solutions for Principles and Applications of Electrical Engineering (5th Edition) // > Giorgio Rizzoni If you could send me some information about the manual and how the transaction would take place. === Subject: $45us per HID kit posting-account=Z75xJAoAAACANiFqJxw7bwdYch7PjnrO TencentTraveler ),gzip(gfe),gzip(gfe) The detailed information about our HID kits are following: 1) before you get a HID Kit, you just paypal us $33us as UPS charges. 2) after you get the item, you can check its quality, if you feel satisfied, please paypal us $45us, if you think it is not good, please resend to our partner who is in US, you donot need to pay for that resent postage, it will be freight collect, not prepaid. 3) the quota for each person should be one HID kit. 4) this is just for the first deal, for the further order, you need to pay total payment before the shipping. our promotion always think about you, most of foreign buyer may worried about the reliability when they make the deal with Chinese sellers. for us, we try our best to let you feel comfortable to make the deal with us. On the other hand, we use paypal account to receive your payment, that should give you confidence to buy from us. Colors we have are blue, pink, green, white purple and gold. All of our different kinds of HID kits are posted on www.housinghouse.com, please go there for picking up your favorite. If you have any question question, do not hesitate to contact me. Here is my personal MSN at vicky_vi@hotmail.com Looking forwards to hearing you soon. Vicky HOUSINGHOUSE.COM === === === === === === === Subject: Permutation sum posting-account=Ic7o8QkAAABRTcr6iMEJQ_yKDBSQpNPG Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Is there a closed form for the following sum? Sum{i=0 to n} [(n!/(n-i)!)*x^i] This is almost the form of the binomial theorem, with the coefficients being the number of permutations rather than combinations. === Subject: Re: Permutation sum > Is there a closed form for the following sum? Sum{i=0 to n} [(n!/(n-i)!)*x^i] This is almost the form of the binomial theorem, with the coefficients > being the number of permutations rather than combinations. According to Mathematica, there is a closed form using the incomplete Gamma function, as described at : Your sum is Exp[1/x]*x^n*Gamma[1 + n, 1/x] David === Subject: Riemann - Stieltjes problem posting-account=gc2kDQoAAADMsLO9kJjQL9hCJkI0D8qJ CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Let f, g be two functions from [a,b] to the real numbers, let P be a partition {x_0, x_1, ..., x_n } of [a,b] with a=x_0 <= x_1 <= ... <= x_n = b and let t_i be such that t_i is in the interval [x_i , x_ (i+1) ] where i runs from 0 to n-1. Define Q as the partition obtained from the points t_i of P by adding the endpoints a and b. I need to simplify the riemann-stieltjes sums: S_f ( P ; g) + S_g (Q; f) , i.e the sum of g with respect to f considering the partition P and the sum of f with respect to g considering the partition Q. So, I seem to get at the end f(b)g(b) - f(a)g(a) as the result of the sum. Is this correct? And a last question, how can I show that if f is Riemann-Stieltjes integrable with respect to g in [a,b] then g is riemann stieltjes integrable with respect to f in [a,b] ?? === Subject: -- Lucky with Erdos-Woods numbers? According to Neil Sloanes comment in http://www.research.att.com/~njas/sequences/A059756 he is searching for some K. Lakki, author of a book about number theory, dated 1984. I suspect a spelling problem. If we are lucky, we might get a helpful hint. TIA Rainer Rosenthal r.rosenthal@web.de === Subject: Re: Barack Obama greeting card posting-account=CQYTlwkAAABjykn0H7FFPT4EUFxvJmg- .NET CLR 2.0.50727; InfoPath.2; IEMB3),gzip(gfe),gzip(gfe) > www.geocities.com/gmbajszar/barack_obama2.JPG I changed the picture a little, so if you go back hit the page 'refresh' button. === Subject: God posting-account=BHeHJAkAAACt3ULyVGLO6mXc32ulONA9 CLR 1.1.4322),gzip(gfe),gzip(gfe) Girls and boys who are good need to look at my video many times to increase number. It could be true if a lot of people look at my video the truth might be exposed. Start video. Go eat food. Watch tv. replay is bad do not use replay it is probably true replay will not increase number increase number - leave youtube and go back to youtube http://www.youtube.com/watch?v=BV_RaVVGrt8 if a person does not do their part fighting bad people they are guilty Kurt Stocklmeir === Subject: Re: A little Cantor puzzle posting-account=6xUtvgkAAAD_jypmLa2oo2HnrV0e8X9q My opinion is that the limit isn't well defined in math in every respect. E.g. the number 0.1111... is defined as the limit of the series 0.1+0.01+0.001+... but it is an infinite sum 0.1+0.01+0.001+... which is running against the limit 0.11111... = 1.0 too. In usual use there is no problem in math. But considering e.g. the infinite sequence A 0.011111111... 0.101111111... 0.110111111... 0.111011111... ... and its antidiagonal AD = 0.11111... I claim that there is in doubt that the antidiagonal is different from _any_ number of the infinite sequence. If two numbers are different, you must be able to give a value which is the difference between the two. In the case of the sequence A and the antidiagonal AD one should be able to give the value which is the difference between _any_ number of the sequence and the antidiagonal. But there is no such difference since the sequence converge against 0.11111... as the antidiagonal converge against 0.11111... . I know well that the number 0.11111... isn't element of the infinite sequence A. And I know also that the antidiagonal AD usually is considered as 0.11111... = 1.0 . But I claim that the antidiagonal AD is never 0.11111... = 1.0 as the sequence A doesn't contain a number 0.11111... = 1.0 although it runs infinitely against this number because the value of AD 0.1+0.01+0.001+... = 0.11111... also runs infinitely against this number. Assume there is no real actual infinity in the meaning of G. Cantor. Then the string 0.11111... can only symbolize the number 1.0, but it can't be a number which is exact identic with the number 1.0 . Because of a little infinitesimal difference the diagonal argument of G. Cantor fails. Albrecht S. Storz === Subject: Re: A little Cantor puzzle posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > My opinion is that the limit isn't well defined in math in every > respect. E.g. the number 0.1111... is defined as the limit of the series > 0.1+0.01+0.001+... but it is an infinite sum 0.1+0.01+0.001+... which > is running against the limit 0.11111... = 1.0 too. > In usual use there is no problem in math. > But considering e.g. the infinite sequence A 0.011111111... > 0.101111111... > 0.110111111... > 0.111011111... > ... and its antidiagonal AD = 0.11111... I claim that there is in doubt that the antidiagonal is different from > _any_ number of the infinite sequence. If two numbers are different, > you must be able to give a value which is the difference between the > two. > In the case of the sequence A and the antidiagonal AD one should be > able to give the value which is the difference between _any_ number of > the sequence and the antidiagonal. But there is no such difference > since the sequence converge against 0.11111... as the antidiagonal > converge against 0.11111... . > I know well that the number 0.11111... isn't element of the infinite > sequence A. Correct. No matter what infininte sequence you take the anti-diagonal will not be an element of the sequence. Thus there is no sequence which includes all endless sequences of 1's and 0's. Thus the enless sequences of 1's and 0's are uncountable. - William Hughes === Subject: SHA1/MD5 of S-Record File I'd like to define a message digest of an S-record file to be something that is independent of the specific formatting of the file. For my purposes, two S-record files are identical if they specify precisely the same memory locations with the same contents. For those not familiar with an S-record file: http://www.amelek.gda.pl/avr/uisp/srecord.htm http://www.seattlerobotics.org/encoder/jun99/dougl.html The issue with S-record files is that it is possible to specify an identical memory load in more than one way. For example, just editing the S-record file and swapping two lines will result in a file that is logically identical. Or, one might have records of varying lengths but the end result of what they specify might be the same. My idea is that I should: a)Parse the S-record file to form a set of (address, data) 2-tuples, i.e. one 2-tuple for each specified byte. b)Sort the 2-tuples by address. c)Concatenate the 2-tuples in the sorted order and take the SHA1 or MD5 of that concatenation. The SHA1 or MD5 should be the same for S-record files that specify the same memory load but are perhaps formatted differently. Questions: a)Is there any advantage or disadvantage to using the raw binary data in the 2-tuples? b)Anything I'm not thinking about? -- David T. Ashley (dta@e3ft.com) http://www.e3ft.com (Consulting Home Page) http://www.dtashley.com (Personal Home Page) http://gpl.e3ft.com (GPL Publications and Projects) === Subject: Re: SHA1/MD5 of S-Record File Originator: daw@taverner.cs.berkeley.edu (David Wagner) >I'd like to define a message digest of an S-record file to be something that >is independent of the specific formatting of the file. The generic principle is to canonicalize (so that any two files that you want to consider equivalent, will have the same canonicalized form), then hash. I don't know enough about S-records to know what is the right or best way to canonicalize, but that's the idea. I wouldn't use MD5, if this is for security. === Subject: Re: 2000 availables Solutions manual posting-account=aKsC7gkAAABYfxToNGrOsC0aoVssTiPR Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) I need solutions manual to Dynamics: Analysis and Design of Systems in Motion (Sheppard & Tongue) Please send to h4nsm0l3m4n@hotmail.com === Subject: Re: math grad admissions > I have no real research experience. This may be due to ignorance > on my part, but I don't see how mathematical research is even > possible at the MS level. IMO that's a petty excuse. I did math research as an undergraduate. I demonstrated (by construction) the existance of singly-generated differential ideas of all possible finite exponents. If I could do research at the undergraduate level, then I see no validity to your denial of math research possible at MS level. Maybe you went to the wrong school, where undergrad/MS research isn't encouraged by any of the professors? > I was not disadvantaged when taking the GRE, except by my own > overconfidence and lack of preparation. That'll do it every time. :-) > In any case, what is the best way to proceed? If you can find a professor at your school that encourages students to do research: Switch to that professor. otherwise: Transfer to a school that encourages students to do research. Then in either case: Do some research. Then get back to your main plan. === Subject: Re: Surrogate factoring and helper primes > And finding k such that the absolute value of 119 - 2k^2 is a > minimum > gives k=8. >And that was with k=1 mod 3. If that didn't work, then there is >k=-1 >mod 3 to check with as well. > James, if k = 8, then k = 2 mod 3, or k = -1 mod 3, either way it is > not 1 mod 3. Could you please explain more clearly here. > rossum > A misstatement. Notice at the start of the post I say k = -1 mod 3. > Sorry that was another post. > It was a misstatement though. > And the same method COULD factor an RSA public key, which is so damn > weird. If the key is 2 mod 3, and if the sum of its prime factors has > 3 as a factor, then it will factor with this approach, and the proof, > as I noted in my post, is trivial. > The proof is trivial but also subtle. then, where is it ? show how you factored an RSA with it or some very short number > There is a detail in the part where I talk about why you need the > value of k^2 such that abs(T-2k^2) is a minimum, which is kind of big. > That allows k to vary from the minimum by up to 12 if k=-1 mod 3, or 1 > if k=1 mod 3. > I discovered this after trying the method on 11(103), and finding that > k equals 38 when 26 gives the maximal abs(T-2k^2). Freaked out a bit > and then figured it out. Worked out more detail. Turns out you have a range near that maximal > k, where the maximum value for k such that abs(T-2k^2) is a minimum is itself a minimum value for k, and you can find the > correct value within k/6 with p=3. In general it is within k/2p, so > with a large target p=3 isn't enough. But importantly, using 3 is so crucial that if T mod 3 = 1, then you > need to force things by using nT, with n=5, or maybe n=2, but I'm so > used to staying away from evens that I think 5. then what about 7, 11, 13, 17, 19, 23, ........... seems like your scheme is not handling primes well at all. So if T mod 3 = 1, then you'd use 5T to force a composite which has a > residue of 2 modulo 3. Then you'd know that z has 3 as a factor for > your target non-trivial factorization, and then you'd know that k and > x exist. Then you can just pick a prime such that (nT/2) mod p is a quadratic > residue (yes, use the modular inverse for 2), as then k^2 = (nT/2) mod p so you get its residue modulo p, and you find the maximal k with that > residue and go from there as explained above, where the maximum number > of steps will be k/2p. And I think that covers everything. Rather neat. Somewhat convoluted. And definitely not exactly how I > thought it would all work. The primes are helpers though. They just step in and then step out. Leaving the answer. leaving an answer, the wrong answer in 99% of the cases. > James Harris === Subject: Re: Surrogate factoring and helper primes posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) The real story is that I was finishing out the theory on surrogate > factoring. z^2 = y^2 mod T I wondered a few years ago if there weren't another way where you > could connect factorizations, and after years of research I've found > that with z^2 = y^2 + nT, where n is some non-zero integer I can add a few more variables using z = x+.87k, so (x+.87k)^2 = y^2 + nT and if you have k and .87, and if k = 2.87x then you find x^2 = y^2 + nT - (1 + .87^2)k^2. You may recall I talked a lot recently about factors mod p, where p is > an odd prime, as now is where helper primes come in allowing you to > solve for k modulo p: k^2 = (.87^2+1)^{-1}(nT) mod p. So how did I introduce prime p? Actually with 2.87x = k + pr 2 so the full equation is actually x^2 = y^2 + nT - (1 + .87^2)k^2 + kpr 2 but if you find k such that the absolute value of nT - (1 + .87^2)k^2 is > a minimum, then you minimize the absolute value of r 2 as well, and > make it 0. That maximal k for the minimum must also have 2.87 as a factor, since > then 2.87x = k so k will always be even. So the primes help you find the factorization and then vanish without > a trace, which is why I call them helper primes. To see one in operation, let T=119 and n=1, and p = 3. Then k^2 = (.87^2+1)^{-1}(nT) mod p = (.87^2+1)^{-1}(119) mod 3. And .87 = 1 works giving k^2 = 1 mod 3, so I can use any integer k > coprime to 3. And finding k such that the absolute value of 119 - 2k^2 is a minimum > gives k=8. And that was with k=1 mod 3. If that didn't work, then there is k=-1 > mod 3 to check with as well. The main correction here is emphasizing that x must be an integer. Then I have x = 4, just like that, and substituting 16 = y^2 + 119 - 128 = y^2 - 9 so y = 5, and z = x + .87k = 4 + 8 = 12, and z-y = 12-5 = 7, and z+y = 12+5 = 17. Notice that p=3 just helped and then vanished without a trace. And yes, it IS possible than an RSA public key could be factored by > p=3. That may be surprising but the math now is elementary since you can > just look back with the surrogate factoring answer. Consider R and RSA public key, such that R mod 3 = 2. Then you have from the surrogate factoring theory that k = 1 mod 3, or > k = -1 mod 3 and .87 = 1, and k is found such that R - 2k^2 is a minimum and k is even, as that is required to make x an integer. Now here's the scarily simple proof that it will work. Let z^2 = y^2 + R. Then if z = x + .87k, and 2.87x = k, it is true then that x = z/3 so if z is divisible by 3, then x exists as an integer. Now substituting x^2 + 2.87kx + (.87k)^2 = y^2 + R, so x^2 = y^2 + R - 2.87kx - (.87k)^2 and I can get rid of x, just on the right side to have x^2 = y^2 + R - (1 + .87^2)k^2 Now if k is coprime to 3 and .87 = 1, R - 2k^2 = 0 mod 3, so x^2 = y^2 + R - 2k^2 leaving just the result that k is the maximal values such that the > absolute value of R - 2k^2 is a minimum. Remember k exists as an integer and x exists as an integer if z has 3 > as a factor, and R mod 3 = 2. Since R is a public key it is odd with > only two prime factors. To finish the proof use 2.87x = k + 3j, instead above, and get to x^2 = y^2 + R - 2k^2 - 3kj and note when j is 0. The approach I used was to note that if you > assume you have an absolute minimum and increase or decrease k, which > has to move mod 3, then j must increase, regardless of which direction > you move in, so the minimum absolute value gives the correct k, as > long as that k is even, and you have two possibles: k = 1 mod 3 and k > = -1 mod 3, so BOTH must be checked. I don't know if that's subtle or not but I'll leave it for now. Note then that ANY RSA public key that has a residue of 2 mod 3 can be > attacked with this approach to factor it with p=3, where if z is > divisible by 3, it WILL fall. But regardless, other primes can be used if those conditions aren't > met. The main point was to emphasize the power of this technique. It CAN factor an RSA public key with p = 3, and very trivial > equations. James Harris Numbers for which this factoring scheme works when p = 3 are of the form (3k - y) * (3k + y). For example: R = 11 * 31 = 341. The value of k which minimizes abs(R - 2k^2) and is even and not divisible by 3 is: k = 14. This yields x = 7 and z = 21 and x^2 = y^2 + R - 2k^2 reduces to 49 = y^2 + 341 - 2*196, or y^2 = 49 + 51 = 100, so y = 10, and the factors of R are (z - 10)*(z + 10) = (21 - 10)*(21 + 10) = 11 * 31. So in this case again, the method works. But: does the algorithm ALWAYS work for T of the form (3*x - y)*(3*x + y)? Let R = 11 * 61 = 671. This is of the form (3*x - y)*(3*x + y) for x = 12 and y = 25. However, the value of k which is not divisible by 3 and which minimizes abs(R - 2k^2) is k = 20. Thus Harris's method would give x = 10 and z = 30, and 100 = y^2 + 671 - 2*400, or y^2 = 229, which is not a perfect square. So no integer factorization in this case. Note that k = 20 = 2 mod 3. Choosing the optimal value of k which is even and congruent to 1 mod 3 yields k = 16. Thus x = 8, z = 24, and 64 = y^2 + 671 - 2*256 or y^2 = 64 - 159 = -95 which is negative, and abs(y^2) is not a perfect square anyway. So this does not give an integer factorization either. What is not clear here is why Harris wants to choose k such that k is not equal to 0 mod 3 and abs(R - 2k^2) is a minimum. As just shown, in general this does not provide an integer factorization. Chances are that y^2 as defined above is not a perfect square of an integer. Thus in general even if you know that R is of the form (3x - y)*(3x + y), you will have to search for y. If R is very large, this search is may need to cover a wide range. If R is NOT of the form (3x - y)*(3x + y), but rather, say, of the form R = (17x - y)*(17x + y), then the search based on p = 3 will not succeed at all. Here is another example: R = 2437 * 3407 = 8302859 This product is of the form (3x - y)*(3x + y). If you use the Harris algorithm to try to find x and y, here is how it works: The value of k (even integer, coprime to 3) which minimizes abs(R - 2*k^2) is k = 2038. This yields x = k/2 = 1019 and z = 3x = 3057. This in turn gives 1019^2 = y^2 + R - 2*k^2, which produces y^2 = 1019^2 - R + 2*k^2 or y^2 = 1038361 - 8302859 + 8306888 = 1043290, and y = 1020.975024, that is, y is not an integer. It is true that (3x - y)*(3x + y) = R, but this is not an integer factorization. The CORRECT values of x and y are not 3057 and 1020.975024 respectively, but rather x = 2922, y = 485. You might start with a 'y' in the neighborhood of 1021 and search above and below this value until you found y = 485, but this would take a while. For T very large (e.g. T ~ 10^100), a lot of such searching would likely be needed. Plus in an actual problem, you would not have the advantage of knowing in advance that T factored in the form (3x - y)*(3x + y). Conclusion: If you are lucky and R is of the form above, the Harris approach might give the right answer, but very likely will not do so if R is large, and you will need to search. If R is not of the expected form, you have more searching to do over primes > 3. In general this method has some resemblance to Fermat's method, but is likely less efficient. Marcus. === Subject: Re: Surrogate factoring and helper primes posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > The real story is that I was finishing out the theory on surrogate > factoring. > z^2 = y^2 mod T > I wondered a few years ago if there weren't another way where you > could connect factorizations, and after years of research I've found > that with > z^2 = y^2 + nT, > where n is some non-zero integer I can add a few more variables using > z = x+.87k, so > (x+.87k)^2 = y^2 + nT > and if you have k and .87, and if k = 2.87x then you find > x^2 = y^2 + nT - (1 + .87^2)k^2. > You may recall I talked a lot recently about factors mod p, where p is > an odd prime, as now is where helper primes come in allowing you to > solve for k modulo p: > k^2 = (.87^2+1)^{-1}(nT) mod p. > So how did I introduce prime p? Actually with > 2.87x = k + pr 2 > so the full equation is actually > x^2 = y^2 + nT - (1 + .87^2)k^2 + kpr 2 > but if you find k such that the absolute value of nT - (1 + .87^2)k^2 is > a minimum, then you minimize the absolute value of r 2 as well, and > make it 0. That maximal k for the minimum must also have 2.87 as a factor, since > then 2.87x = k so k will always be even. > So the primes help you find the factorization and then vanish without > a trace, which is why I call them helper primes. > To see one in operation, let T=119 and n=1, and p = 3. > Then > k^2 = (.87^2+1)^{-1}(nT) mod p = (.87^2+1)^{-1}(119) mod 3. > And .87 = 1 works giving k^2 = 1 mod 3, so I can use any integer k > coprime to 3. > And finding k such that the absolute value of 119 - 2k^2 is a minimum > gives k=8. And that was with k=1 mod 3. If that didn't work, then there is k=-1 > mod 3 to check with as well. The main correction here is emphasizing that x must be an integer. > Then I have x = 4, just like that, and substituting > 16 = y^2 + 119 - 128 = y^2 - 9 > so y = 5, and > z = x + .87k = 4 + 8 = 12, and z-y = 12-5 = 7, and z+y = 12+5 = 17. > Notice that p=3 just helped and then vanished without a trace. > And yes, it IS possible than an RSA public key could be factored by > p=3. That may be surprising but the math now is elementary since you can > just look back with the surrogate factoring answer. Consider R and RSA public key, such that R mod 3 = 2. Then you have from the surrogate factoring theory that k = 1 mod 3, or > k = -1 mod 3 and .87 = 1, and k is found such that R - 2k^2 is a minimum and k is even, as that is required to make x an integer. Now here's the scarily simple proof that it will work. Let z^2 = y^2 + R. Then if z = x + .87k, and 2.87x = k, it is true then that x = z/3 so if z is divisible by 3, then x exists as an integer. Now substituting x^2 + 2.87kx + (.87k)^2 = y^2 + R, so x^2 = y^2 + R - 2.87kx - (.87k)^2 and I can get rid of x, just on the right side to have x^2 = y^2 + R - (1 + .87^2)k^2 Now if k is coprime to 3 and .87 = 1, R - 2k^2 = 0 mod 3, so x^2 = y^2 + R - 2k^2 leaving just the result that k is the maximal values such that the > absolute value of R - 2k^2 is a minimum. Remember k exists as an integer and x exists as an integer if z has 3 > as a factor, and R mod 3 = 2. Since R is a public key it is odd with > only two prime factors. To finish the proof use 2.87x = k + 3j, instead above, and get to x^2 = y^2 + R - 2k^2 - 3kj and note when j is 0. The approach I used was to note that if you > assume you have an absolute minimum and increase or decrease k, which > has to move mod 3, then j must increase, regardless of which direction > you move in, so the minimum absolute value gives the correct k, as > long as that k is even, and you have two possibles: k = 1 mod 3 and k > = -1 mod 3, so BOTH must be checked. I don't know if that's subtle or not but I'll leave it for now. Note then that ANY RSA public key that has a residue of 2 mod 3 can be > attacked with this approach to factor it with p=3, where if z is > divisible by 3, it WILL fall. But regardless, other primes can be used if those conditions aren't > met. The main point was to emphasize the power of this technique. It CAN factor an RSA public key with p = 3, and very trivial > equations. James Harris Numbers for which this factoring scheme works > when p = 3 are of the form (3k - y) * (3k + y). For example: R = 11 * 31 = 341. The value of k > which minimizes abs(R - 2k^2) and is even and not > divisible by 3 is: k = 14. This yields x = 7 and > z = 21 and x^2 = y^2 + R - 2k^2 reduces to 49 = y^2 + 341 - 2*196, or y^2 = 49 + 51 = 100, so y = 10, and the factors of R are (z - 10)*(z + 10) = (21 - 10)*(21 + 10) = 11 * 31. So in this case again, the method works. But: does the algorithm ALWAYS work for T of the > form (3*x - y)*(3*x + y)? Yup, but I made a mistake in saying that the optimal k is always the one that makes abs(T-2k^2) a minimum as it can be that value but must be very close. The reason it must be close is that it can be shown (trivially) that x^2 = y^2 + T - 2k^2 - pr when z=x+k, and 2x = k + pr, where the goal is r=0. Now if you assume you have the correct k and r=0, then perturbing that system will happen modulo 6, so you'd have x^2 = y^2 + T -2(k+6j)^2 - pr and r will tend to be negative, as the k^2 term will dominate. It can be shown that the maximum k such that abs(T-2k^2) is a minimum is the minimum value for k when k is positive (as why use a negative k) and that k/6 is the maximum number of steps to a solution as you move modulo 6. Let R = 11 * 61 = 671. This is of the form (3*x - y)*(3*x + y) for x = 12 and y = 25. However, the value of k which > is not divisible by 3 and which minimizes abs(R - 2k^2) is k = 20. Thus Harris's method would give x = 10 and > z = 30, and 100 = y^2 + 671 - 2*400, or y^2 = 229, which is not a perfect square. So no integer > factorization in this case. Note that > k = 20 = 2 mod 3. Choosing the optimal value of k > which is even and congruent to 1 mod 3 yields k = 16. > Thus x = 8, z = 24, and 64 = y^2 + 671 - 2*256 or y^2 = 64 - 159 = -95 which is negative, and abs(y^2) is not a > perfect square anyway. So this does not give an > integer factorization either. What is not clear here is why Harris wants > to choose k such that k is not equal to 0 mod 3 > and abs(R - 2k^2) is a minimum. As just shown, in general this does > not provide an integer factorization. Chances are that > y^2 as defined above is not a perfect square of an integer. > Thus in general even if you know that R is of the form > (3x - y)*(3x + y), you will have to search for y. > If R is very large, this search is may need to cover > a wide range. If R is NOT of the form > (3x - y)*(3x + y), but rather, say, of the form R = (17x - y)*(17x + y), then the search based on p = 3 will not succeed at all. Here is another example: R = 2437 * 3407 = 8302859 This product is of the form (3x - y)*(3x + y). If you use the Harris algorithm to try to find x > and y, here is how it works: The value of k (even integer, coprime to 3) which > minimizes abs(R - 2*k^2) is k = 2038. This yields x = k/2 = 1019 and > z = 3x = 3057. This in turn gives 1019^2 = y^2 + R - 2*k^2, which produces y^2 = 1019^2 - R + 2*k^2 or y^2 = 1038361 - 8302859 + 8306888 = 1043290, and y = 1020.975024, that is, y is not an integer. It is true that (3x - y)*(3x + y) = R, but this is not an integer factorization. The > CORRECT values of x and y are not 3057 and > 1020.975024 respectively, but rather x = 2922, y = 485. You might start with a 'y' in the neighborhood of > 1021 and search above and below this value until you > found y = 485, but this would take a while. For T > very large (e.g. T ~ 10^100), a lot of such searching > would likely be needed. Plus in an actual problem, you > would not have the advantage of knowing in advance > that T factored in the form (3x - y)*(3x + y). Conclusion: If you are lucky and R is of the form > above, the Harris approach might give the right > answer, but very likely will not do so if R is large, and > you will need to search. If R is not of the expected form, > you have more searching to do over primes > 3. In general > this method has some resemblance to Fermat's method, > but is likely less efficient. Marcus. Well, in general the minimum value for k will be given by the maximum k such that abs(T-(a^2+1)k^2) is a minimum where 'a' is given as I've talked about before, and you can use other primes p than 3. I was using 3 as an example as I found it intriguing that it might factor an RSA public key. In general, the maximum number of steps from your k found using the minimum is given by k/2p in the positive direction. Fascinating math as I figure it all out. Still making some missteps but it's getting a lot clearer as time goes on. That k/2p number of steps can be HUGE so I'm having to back down a bit as I contemplate more. James Harris === Subject: Re: Surrogate factoring and helper primes posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 SV1),gzip(gfe),gzip(gfe) > The real story is that I was finishing out the theory on surrogate > factoring. > z^2 = y^2 mod T > I wondered a few years ago if there weren't another way where you > could connect factorizations, and after years of research I've found > that with > z^2 = y^2 + nT, > where n is some non-zero integer I can add a few more variables using > z = x+.87k, so > (x+.87k)^2 = y^2 + nT > and if you have k and .87, and if k = 2.87x then you find > x^2 = y^2 + nT - (1 + .87^2)k^2. > You may recall I talked a lot recently about factors mod p, where p is > an odd prime, as now is where helper primes come in allowing you to > solve for k modulo p: > k^2 = (.87^2+1)^{-1}(nT) mod p. > So how did I introduce prime p? æActually with > 2.87x = k + pr 2 > so the full equation is actually > x^2 = y^2 + nT - (1 + .87^2)k^2 + kpr 2 > but if you find k such that the absolute value of nT - (1 + .87^2)k^2 is > a minimum, then you minimize the absolute value of r 2 as well, and > make it 0. > That maximal k for the minimum must also have 2.87 as a factor, since > then > 2.87x = k > so k will always be even. > So the primes help you find the factorization and then vanish without > a trace, which is why I call them helper primes. > To see one in operation, let T=119 and n=1, and p = 3. > Then > k^2 = (.87^2+1)^{-1}(nT) mod p = (.87^2+1)^{-1}(119) mod 3. > And .87 = 1 works giving k^2 = 1 mod 3, so I can use any integer k > coprime to 3. > And finding k such that the absolute value of 119 - 2k^2 is a minimum > gives k=8. > And that was with k=1 mod 3. æIf that didn't work, then there is k=-1 > mod 3 to check with as well. > The main correction here is emphasizing that x must be an integer. > Then I have x = 4, just like that, and substituting > 16 = y^2 + 119 - 128 = y^2 - 9 > so y = 5, and > z = x + .87k = 4 + 8 = 12, and z-y = 12-5 = 7, and z+y = 12+5 = 17. > Notice that p=3 just helped and then vanished without a trace. > And yes, it IS possible than an RSA public key could be factored by > p=3. > That may be surprising but the math now is elementary since you can > just look back with the surrogate factoring answer. > Consider R and RSA public key, such that R mod 3 = 2. > Then you have from the surrogate factoring theory that k = 1 mod 3, or > k = -1 mod 3 and .87 = 1, and k is found such that > R - 2k^2 > is a minimum and k is even, as that is required to make x an integer. > Now here's the scarily simple proof that it will work. > Let z^2 = y^2 + R. > Then if z = x + .87k, and 2.87x = k, it is true then that > x = z/3 > so if z is divisible by 3, then x exists as an integer. > Now substituting > x^2 + 2.87kx + (.87k)^2 = y^2 + R, so > x^2 = y^2 + R - 2.87kx - (.87k)^2 > and I can get rid of x, just on the right side to have > x^2 = y^2 + R - (1 + .87^2)k^2 > Now if k is coprime to 3 and .87 = 1, > R - 2k^2 = 0 mod 3, so > x^2 = y^2 + R - 2k^2 > leaving just the result that k is the maximal values such that the > absolute value of R - 2k^2 is a minimum. > Remember k exists as an integer and x exists as an integer if z has 3 > as a factor, and R mod 3 = 2. æSince R is a public key it is odd with > only two prime factors. > To finish the proof use æ2.87x = k + 3j, instead above, and get to > x^2 = y^2 + R - 2k^2 - 3kj > and note when j is 0. æThe approach I used was to note that if you > assume you have an absolute minimum and increase or decrease k, which > has to move mod 3, then j must increase, regardless of which direction > you move in, so the minimum absolute value gives the correct k, as > long as that k is even, and you have two possibles: k = 1 mod 3 and k > = -1 mod 3, so BOTH must be checked. > I don't know if that's subtle or not but I'll leave it for now. > Note then that ANY RSA public key that has a residue of 2 mod 3 can be > attacked with this approach to factor it with p=3, where if z is > divisible by 3, it WILL fall. > But regardless, other primes can be used if those conditions aren't > met. > The main point was to emphasize the power of this technique. > It CAN factor an RSA public key with p = 3, and very trivial > equations. > James Harris æ Numbers for which this factoring scheme works > when p = 3 are of the form æ æ æ æ(3k - y) * (3k + y). æ For example: R = 11 * 31 = 341. æThe value of k > which minimizes abs(R - 2k^2) and is even and not > divisible by 3 is: k = 14. This yields x = 7 and > z = 21 and æ æ æ x^2 = y^2 + R - 2k^2 æ æreduces to æ æ æ 49 = y^2 + 341 - 2*196, or æ æ æ y^2 = 49 + 51 = 100, so y = 10, and the factors of R are æ æ æ(z - 10)*(z + 10) = (21 - 10)*(21 + 10) = 11 * 31. æ So in this case again, the method works. æ But: does the algorithm ALWAYS work for T of the > form (3*x - y)*(3*x + y)? Yup, but I made a mistake in saying that the optimal k is always the > one that makes abs(T-2k^2) a minimum as it can be that value but must > be very close. The reason it must be close is that it can be shown (trivially) that x^2 = y^2 + T - 2k^2 - pr when z=x+k, and 2x = k + pr, where the goal is r=0. Now if you assume you have the correct k and r=0, then perturbing that > system will happen modulo 6, so you'd have x^2 = y^2 + T -2(k+6j)^2 - pr and r will tend to be negative, as the k^2 term will dominate. It can be shown that the maximum k such that abs(T-2k^2) is a minimum > is the minimum value for k when k is positive (as why use a negative > k) and that k/6 is the maximum number of steps to a solution as you > move modulo 6. > This is nonsense. It is trivial to find examples where the value of k which works can be vastly different from what you call optimal. This means that you are not going to readily find the correct factorization without an extensive search. Marcus. > æ Let R = 11 * 61 = 671. æThis is of the form æ æ æ æ(3*x - y)*(3*x + y) for x = 12 and y = 25. æ However, the value of k which > is not divisible by 3 and which minimizes æ æ æ æ abs(R - 2k^2) is k = 20. æThus Harris's method would give x = 10 and > z = 30, and æ æ100 = y^2 + 671 - 2*400, or æ æ æy^2 = 229, which is not a perfect square. æSo no integer > factorization in this case. æNote that > k = 20 = 2 mod 3. æChoosing the optimal value of k > which is even and congruent to 1 mod 3 yields k = 16. > Thus x = 8, z = 24, and æ æ æ64 = y^2 + 671 - 2*256 æ or æ æ æ æy^2 = 64 - 159 = -95 which is negative, and abs(y^2) is not a > perfect square anyway. æSo this does not give an > integer factorization either. æ What is not clear here is why Harris wants > to choose k such that k is not equal to 0 mod 3 > and æ æ æabs(R - 2k^2) is a minimum. æAs just shown, in general this does > not provide an integer factorization. æChances are that > y^2 as defined above is not a perfect square of an integer. > Thus in general even if you know that R is of the form > (3x - y)*(3x + y), you will have to search for y. > If R is very large, this search is may need to cover > a wide range. æIf R is NOT of the form > (3x - y)*(3x + y), but rather, say, of the form æ æR = (17x - y)*(17x + y), then the search based on p = 3 will not succeed at all. æ Here is another example: æ æ æ R = 2437 * 3407 = 8302859 æ This product is of the form æ æ æ (3x - y)*(3x + y). æ If you use the Harris algorithm to try to find x > and y, here is how it works: æ The value of k (even integer, coprime to 3) which > minimizes æ æ æ abs(R - 2*k^2) is k = 2038. æThis yields x = k/2 = 1019 and > z = 3x = 3057. æThis in turn gives æ æ æ 1019^2 = y^2 + R - 2*k^2, which produces æ æ æ y^2 = 1019^2 - R + 2*k^2 or æ æ æ y^2 = 1038361 - 8302859 + 8306888 æ æ æ æ æ = 1043290, and æ y = 1020.975024, that is, y is not an integer. æIt is true that æ æ æ (3x - y)*(3x + y) = R, but this is not an integer factorization. æThe > CORRECT values of x and y are not 3057 and > 1020.975024 respectively, but rather æ æ æ æx = 2922, y = 485. æ You might start with a 'y' in the neighborhood of > 1021 and search above and below this value until you > found y = 485, but this would take a while. æFor T > very large (e.g. T ~ 10^100), a lot of such searching > would likely be needed. æPlus in an actual problem, you > would not have the advantage of knowing in advance > that T factored in the form æ æ æ(3x - y)*(3x + y). æ Conclusion: If you are lucky and R is of the form > above, the Harris approach might give the right > answer, but very likely will not do so if R is large, and > you will need to search. æIf R is not of the expected form, > you have more searching to do over primes > 3. æIn general > this method has some resemblance to Fermat's method, > but is likely less efficient. Marcus. Well, in general the minimum value for k will be given by the maximum > k such that abs(T-(a^2+1)k^2) is a minimum where 'a' is given as I've > talked about before, and you can use other primes p than 3. æI was > using 3 as an example as I found it intriguing that it might factor an > RSA public key. In general, the maximum number of steps from your k found using the > minimum is given by k/2p in the positive direction. Fascinating math as I figure it all out. æStill making some missteps > but it's getting a lot clearer as time goes on. That k/2p number of steps can be HUGE so I'm having to back down a bit > as I contemplate more. James Harris- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - === Subject: Re: Surrogate factoring and helper primes posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Numbers for which this factoring scheme works > when p = 3 are of the form (3k - y) * (3k + y). > If R is NOT of the form > (3x - y)*(3x + y), but rather, say, of the form R = (17x - y)*(17x + y), then the search based on p = 3 will not succeed at all. Actually James is aware of this; though he doesn't mention it in his post, his blog entry on the subject states that z defined by z = (f + g)/2 should be divisible by 3, which implies that R = f*g can be written in the form (3k - y)*(3k + y). Since an RSA composite is presumably not divisible by 3 its factors have residues of 1 or 2 mod 3, so you would expect that roughly half of them can be written in this form (unless there is some reason to prefer composites which can't, I really wouldn't know) so it would be quite something if he had an efficient way to factor numbers which can. That's a big if of course; as usual I can't make sense of his proof, and couldn't get his method to work on the first example I tried, which was 1021*1031. Though it's possible that I misunderstood the method - James, can you demonstrate how it is supposed to work for R = 1021*1031 = 1052651? === Subject: Re: Surrogate factoring and helper primes posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) Numbers for which this factoring scheme works > when p = 3 are of the form (3k - y) * (3k + y). > If R is NOT of the form > (3x - y)*(3x + y), but rather, say, of the form R = (17x - y)*(17x + y), then the search based on p = 3 will not succeed at all. Actually James is aware of this; though he doesn't mention it in his > post, his blog entry on the subject states that z defined by z = (f + > g)/2 should be divisible by 3, which implies that R = f*g can be > written in the form (3k - y)*(3k + y). Since an RSA composite is > presumably not divisible by 3 its factors have residues of 1 or 2 mod > 3, so you would expect that roughly half of them can be written in > this form (unless there is some reason to prefer composites which > can't, I really wouldn't know) so it would be quite something if he > had an efficient way to factor numbers which can. Agreed. If. > That's a big if of > course; as usual I can't make sense of his proof, There is no proof. But in his description, one unclear part is, why should k be chosen to minimize abs(T - 2*k^2) ? > and couldn't get his > method to work on the first example I tried, which was 1021*1031. Yes, that's one where I do not get it to work either (and where Fermat would give the answer almost instantly). It is more likely to work for smaller numbers. Marcus. > Though it's possible that I misunderstood the method - James, can you > demonstrate how it is supposed to work for R = 1021*1031 = 1052651? === Subject: Re: Surrogate factoring and helper primes posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) That's a big if of > course; as usual I can't make sense of his proof, There is no proof. I suspect as much, of course, though I'm currently attempting a period of giving James more benefit of the doubt than usual. I expect that this period will end abruptly the next time he accuses me of being a space alien working to block his research in a bid to bring forward the EEP, or something. === Subject: Re: Surrogate factoring and helper primes posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) > Misprints below at *****: > The real story is that I was finishing out the theory on surrogate > factoring. > z^2 = y^2 mod T > I wondered a few years ago if there weren't another way where you > could connect factorizations, and after years of research I've found > that with > z^2 = y^2 + nT, > where n is some non-zero integer I can add a few more variables using > z = x+.87k, so > (x+.87k)^2 = y^2 + nT > and if you have k and .87, and if k = 2.87x then you find > x^2 = y^2 + nT - (1 + .87^2)k^2. > You may recall I talked a lot recently about factors mod p, where p is > an odd prime, as now is where helper primes come in allowing you to > solve for k modulo p: > k^2 = (.87^2+1)^{-1}(nT) mod p. > So how did I introduce prime p? Actually with > 2.87x = k + pr 2 > so the full equation is actually > x^2 = y^2 + nT - (1 + .87^2)k^2 + kpr 2 > but if you find k such that the absolute value of nT - (1 + .87^2)k^2 is > a minimum, then you minimize the absolute value of r 2 as well, and > make it 0. That maximal k for the minimum must also have 2.87 as a factor, since > then 2.87x = k so k will always be even. > So the primes help you find the factorization and then vanish without > a trace, which is why I call them helper primes. > To see one in operation, let T=119 and n=1, and p = 3. > Then > k^2 = (.87^2+1)^{-1}(nT) mod p = (.87^2+1)^{-1}(119) mod 3. > And .87 = 1 works giving k^2 = 1 mod 3, so I can use any integer k > coprime to 3. > And finding k such that the absolute value of 119 - 2k^2 is a minimum > gives k=8. And that was with k=1 mod 3. If that didn't work, then there is k=-1 > mod 3 to check with as well. The main correction here is emphasizing that x must be an integer. > Then I have x = 4, just like that, and substituting > 16 = y^2 + 119 - 128 = y^2 - 9 > so y = 5, and > z = x + .87k = 4 + 8 = 12, and z-y = 12-5 = 7, and z+y = 12+5 = 17. > Notice that p=3 just helped and then vanished without a trace. > And yes, it IS possible than an RSA public key could be factored by > p=3. That may be surprising but the math now is elementary since you can > just look back with the surrogate factoring answer. Consider R and RSA public key, such that R mod 3 = 2. Then you have from the surrogate factoring theory that k = 1 mod 3, or > k = -1 mod 3 and .87 = 1, and k is found such that R - 2k^2 is a minimum and k is even, as that is required to make x an integer. Now here's the scarily simple proof that it will work. Let z^2 = y^2 + R. Then if z = x + .87k, and 2.87x = k, it is true then that x = z/3 so if z is divisible by 3, then x exists as an integer. Now substituting x^2 + 2.87kx + (.87k)^2 = y^2 + R, so x^2 = y^2 + R - 2.87kx - (.87k)^2 and I can get rid of x, just on the right side to have x^2 = y^2 + R - (1 + .87^2)k^2 Now if k is coprime to 3 and .87 = 1, R - 2k^2 = 0 mod 3, so x^2 = y^2 + R - 2k^2 leaving just the result that k is the maximal values such that the > absolute value of R - 2k^2 is a minimum. Remember k exists as an integer and x exists as an integer if z has 3 > as a factor, and R mod 3 = 2. Since R is a public key it is odd with > only two prime factors. To finish the proof use 2.87x = k + 3j, instead above, and get to x^2 = y^2 + R - 2k^2 - 3kj and note when j is 0. The approach I used was to note that if you > assume you have an absolute minimum and increase or decrease k, which > has to move mod 3, then j must increase, regardless of which direction > you move in, so the minimum absolute value gives the correct k, as > long as that k is even, and you have two possibles: k = 1 mod 3 and k > = -1 mod 3, so BOTH must be checked. I don't know if that's subtle or not but I'll leave it for now. Note then that ANY RSA public key that has a residue of 2 mod 3 can be > attacked with this approach to factor it with p=3, where if z is > divisible by 3, it WILL fall. But regardless, other primes can be used if those conditions aren't > met. The main point was to emphasize the power of this technique. It CAN factor an RSA public key with p = 3, and very trivial > equations. James Harris Numbers for which this factoring scheme works > when p = 3 are of the form > *****> (3k - y) * (3k + y). Should be: *****> (3x - y)*(3x + y). For example: R = 11 * 31 = 341. The value of k > which minimizes abs(R - 2k^2) and is even and not > divisible by 3 is: k = 14. This yields x = 7 and > z = 21 and x^2 = y^2 + R - 2k^2 reduces to 49 = y^2 + 341 - 2*196, or y^2 = 49 + 51 = 100, so y = 10, and the factors of R are (z - 10)*(z + 10) = (21 - 10)*(21 + 10) = 11 * 31. So in this case again, the method works. But: does the algorithm ALWAYS work for T of the > form (3*x - y)*(3*x + y)? Let R = 11 * 61 = 671. This is of the form (3*x - y)*(3*x + y) for x = 12 and y = 25. However, the value of k which > is not divisible by 3 and which minimizes abs(R - 2k^2) is k = 20. Thus Harris's method would give x = 10 and > z = 30, and 100 = y^2 + 671 - 2*400, or y^2 = 229, which is not a perfect square. So no integer > factorization in this case. Note that > k = 20 = 2 mod 3. Choosing the optimal value of k > which is even and congruent to 1 mod 3 yields k = 16. > Thus x = 8, z = 24, and 64 = y^2 + 671 - 2*256 or y^2 = 64 - 159 = -95 which is negative, and abs(y^2) is not a > perfect square anyway. So this does not give an > integer factorization either. What is not clear here is why Harris wants > to choose k such that k is not equal to 0 mod 3 > and abs(R - 2k^2) is a minimum. As just shown, in general this does > not provide an integer factorization. Chances are that > y^2 as defined above is not a perfect square of an integer. > Thus in general even if you know that R is of the form > (3x - y)*(3x + y), you will have to search for y. > If R is very large, this search is may need to cover > a wide range. If R is NOT of the form > (3x - y)*(3x + y), but rather, say, of the form R = (17x - y)*(17x + y), then the search based on p = 3 will not succeed at all. Here is another example: R = 2437 * 3407 = 8302859 This product is of the form (3x - y)*(3x + y). If you use the Harris algorithm to try to find x > and y, here is how it works: The value of k (even integer, coprime to 3) which > minimizes abs(R - 2*k^2) is k = 2038. This yields x = k/2 = 1019 and > z = 3x = 3057. This in turn gives 1019^2 = y^2 + R - 2*k^2, which produces y^2 = 1019^2 - R + 2*k^2 or y^2 = 1038361 - 8302859 + 8306888 = 1043290, and y = 1020.975024, that is, y is not an integer. It is true that (3x - y)*(3x + y) = R, but this is not an integer factorization. The > CORRECT values of x and y are not 3057 and > 1020.975024 respectively, but rather x = 2922, y = 485. You might start with a 'y' in the neighborhood of > 1021 and search above and below this value until you > found y = 485, but this would take a while. For T > very large (e.g. T ~ 10^100), a lot of such searching > would likely be needed. Plus in an actual problem, you > would not have the advantage of knowing in advance > that T factored in the form (3x - y)*(3x + y). Conclusion: If you are lucky and R is of the form > above, the Harris approach might give the right > answer, but very likely will not do so if R is large, and > you will need to search. If R is not of the expected form, > you have more searching to do over primes > 3. In general > this method has some resemblance to Fermat's method, > but is likely less efficient. Marcus. === Subject: Re: Surrogate factoring and helper primes posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) > ... ANOTHER misprint, at: #######: Misprints below at *****: > The real story is that I was finishing out the theory on surrogate > factoring. > z^2 = y^2 mod T > I wondered a few years ago if there weren't another way where you > could connect factorizations, and after years of research I've found > that with > z^2 = y^2 + nT, > where n is some non-zero integer I can add a few more variables using > z = x+.87k, so > (x+.87k)^2 = y^2 + nT > and if you have k and .87, and if k = 2.87x then you find > x^2 = y^2 + nT - (1 + .87^2)k^2. > You may recall I talked a lot recently about factors mod p, where p is > an odd prime, as now is where helper primes come in allowing you to > solve for k modulo p: > k^2 = (.87^2+1)^{-1}(nT) mod p. > So how did I introduce prime p? Actually with > 2.87x = k + pr 2 > so the full equation is actually > x^2 = y^2 + nT - (1 + .87^2)k^2 + kpr 2 > but if you find k such that the absolute value of nT - (1 + .87^2)k^2 is > a minimum, then you minimize the absolute value of r 2 as well, and > make it 0. > That maximal k for the minimum must also have 2.87 as a factor, since > then > 2.87x = k > so k will always be even. > So the primes help you find the factorization and then vanish without > a trace, which is why I call them helper primes. > To see one in operation, let T=119 and n=1, and p = 3. > Then > k^2 = (.87^2+1)^{-1}(nT) mod p = (.87^2+1)^{-1}(119) mod 3. > And .87 = 1 works giving k^2 = 1 mod 3, so I can use any integer k > coprime to 3. > And finding k such that the absolute value of 119 - 2k^2 is a minimum > gives k=8. > And that was with k=1 mod 3. If that didn't work, then there is k=-1 > mod 3 to check with as well. > The main correction here is emphasizing that x must be an integer. > Then I have x = 4, just like that, and substituting > 16 = y^2 + 119 - 128 = y^2 - 9 > so y = 5, and > z = x + .87k = 4 + 8 = 12, and z-y = 12-5 = 7, and z+y = 12+5 = 17. > Notice that p=3 just helped and then vanished without a trace. > And yes, it IS possible than an RSA public key could be factored by > p=3. > That may be surprising but the math now is elementary since you can > just look back with the surrogate factoring answer. > Consider R and RSA public key, such that R mod 3 = 2. > Then you have from the surrogate factoring theory that k = 1 mod 3, or > k = -1 mod 3 and .87 = 1, and k is found such that > R - 2k^2 > is a minimum and k is even, as that is required to make x an integer. > Now here's the scarily simple proof that it will work. > Let z^2 = y^2 + R. > Then if z = x + .87k, and 2.87x = k, it is true then that > x = z/3 > so if z is divisible by 3, then x exists as an integer. > Now substituting > x^2 + 2.87kx + (.87k)^2 = y^2 + R, so > x^2 = y^2 + R - 2.87kx - (.87k)^2 > and I can get rid of x, just on the right side to have > x^2 = y^2 + R - (1 + .87^2)k^2 > Now if k is coprime to 3 and .87 = 1, > R - 2k^2 = 0 mod 3, so > x^2 = y^2 + R - 2k^2 > leaving just the result that k is the maximal values such that the > absolute value of R - 2k^2 is a minimum. > Remember k exists as an integer and x exists as an integer if z has 3 > as a factor, and R mod 3 = 2. Since R is a public key it is odd with > only two prime factors. > To finish the proof use 2.87x = k + 3j, instead above, and get to > x^2 = y^2 + R - 2k^2 - 3kj > and note when j is 0. The approach I used was to note that if you > assume you have an absolute minimum and increase or decrease k, which > has to move mod 3, then j must increase, regardless of which direction > you move in, so the minimum absolute value gives the correct k, as > long as that k is even, and you have two possibles: k = 1 mod 3 and k > = -1 mod 3, so BOTH must be checked. > I don't know if that's subtle or not but I'll leave it for now. > Note then that ANY RSA public key that has a residue of 2 mod 3 can be > attacked with this approach to factor it with p=3, where if z is > divisible by 3, it WILL fall. > But regardless, other primes can be used if those conditions aren't > met. > The main point was to emphasize the power of this technique. > It CAN factor an RSA public key with p = 3, and very trivial > equations. > James Harris Numbers for which this factoring scheme works > when p = 3 are of the form *****> (3k - y) * (3k + y). Should be: *****> (3x - y)*(3x + y). For example: R = 11 * 31 = 341. The value of k > which minimizes abs(R - 2k^2) and is even and not > divisible by 3 is: k = 14. This yields x = 7 and > z = 21 and x^2 = y^2 + R - 2k^2 reduces to 49 = y^2 + 341 - 2*196, or y^2 = 49 + 51 = 100, so y = 10, and the factors of R are (z - 10)*(z + 10) = (21 - 10)*(21 + 10) = 11 * 31. So in this case again, the method works. But: does the algorithm ALWAYS work for T of the > form (3*x - y)*(3*x + y)? Let R = 11 * 61 = 671. This is of the form (3*x - y)*(3*x + y) for x = 12 and y = 25. However, the value of k which > is not divisible by 3 and which minimizes abs(R - 2k^2) is k = 20. Thus Harris's method would give x = 10 and > z = 30, and 100 = y^2 + 671 - 2*400, or y^2 = 229, which is not a perfect square. So no integer > factorization in this case. Note that > k = 20 = 2 mod 3. Choosing the optimal value of k > which is even and congruent to 1 mod 3 yields k = 16. > Thus x = 8, z = 24, and 64 = y^2 + 671 - 2*256 or y^2 = 64 - 159 = -95 which is negative, and abs(y^2) is not a > perfect square anyway. So this does not give an > integer factorization either. What is not clear here is why Harris wants > to choose k such that k is not equal to 0 mod 3 > and abs(R - 2k^2) is a minimum. As just shown, in general this does > not provide an integer factorization. Chances are that > y^2 as defined above is not a perfect square of an integer. > Thus in general even if you know that R is of the form > (3x - y)*(3x + y), you will have to search for y. > If R is very large, this search is may need to cover > a wide range. If R is NOT of the form > (3x - y)*(3x + y), but rather, say, of the form R = (17x - y)*(17x + y), then the search based on p = 3 will not succeed at all. Here is another example: R = 2437 * 3407 = 8302859 This product is of the form (3x - y)*(3x + y). If you use the Harris algorithm to try to find x > and y, here is how it works: The value of k (even integer, coprime to 3) which > minimizes abs(R - 2*k^2) is k = 2038. This yields x = k/2 = 1019 and > z = 3x = 3057. This in turn gives 1019^2 = y^2 + R - 2*k^2, which produces y^2 = 1019^2 - R + 2*k^2 or y^2 = 1038361 - 8302859 + 8306888 = 1043290, and y = 1020.975024, that is, y is not an integer. It is true that (3x - y)*(3x + y) = R, but this is not an integer factorization. The #######> CORRECT values of x and y are not 3057 and Should be: #######> CORRECT values of 3x and y are not 3057 and > 1020.975024 respectively, but rather > #######> x = 2922, y = 485. Should be: #######> 3x = 2922, y = 485. > You might start with a 'y' in the neighborhood of > 1021 and search above and below this value until you > found y = 485, but this would take a while. For T > very large (e.g. T ~ 10^100), a lot of such searching > would likely be needed. Plus in an actual problem, you > would not have the advantage of knowing in advance > that T factored in the form (3x - y)*(3x + y). Conclusion: If you are lucky and R is of the form > above, the Harris approach might give the right > answer, but very likely will not do so if R is large, and > you will need to search. If R is not of the expected form, > you have more searching to do over primes > 3. In general > this method has some resemblance to Fermat's method, > but is likely less efficient. Marcus. === Subject: Re: Surrogate factoring and helper primes posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) > Misprints below at *****: > The real story is that I was finishing out the theory on surrogate > factoring. > z^2 = y^2 mod T > I wondered a few years ago if there weren't another way where you > could connect factorizations, and after years of research I've found > that with > z^2 = y^2 + nT, > where n is some non-zero integer I can add a few more variables using > z = x+.87k, so > (x+.87k)^2 = y^2 + nT > and if you have k and .87, and if k = 2.87x then you find > x^2 = y^2 + nT - (1 + .87^2)k^2. > You may recall I talked a lot recently about factors mod p, where p is > an odd prime, as now is where helper primes come in allowing you to > solve for k modulo p: > k^2 = (.87^2+1)^{-1}(nT) mod p. > So how did I introduce prime p? Actually with > 2.87x = k + pr 2 > so the full equation is actually > x^2 = y^2 + nT - (1 + .87^2)k^2 + kpr 2 > but if you find k such that the absolute value of nT - (1 + .87^2)k^2 is > a minimum, then you minimize the absolute value of r 2 as well, and > make it 0. That maximal k for the minimum must also have 2.87 as a factor, since > then 2.87x = k so k will always be even. > So the primes help you find the factorization and then vanish without > a trace, which is why I call them helper primes. > To see one in operation, let T=119 and n=1, and p = 3. > Then > k^2 = (.87^2+1)^{-1}(nT) mod p = (.87^2+1)^{-1}(119) mod 3. > And .87 = 1 works giving k^2 = 1 mod 3, so I can use any integer k > coprime to 3. > And finding k such that the absolute value of 119 - 2k^2 is a minimum > gives k=8. And that was with k=1 mod 3. If that didn't work, then there is k=-1 > mod 3 to check with as well. The main correction here is emphasizing that x must be an integer. > Then I have x = 4, just like that, and substituting > 16 = y^2 + 119 - 128 = y^2 - 9 > so y = 5, and > z = x + .87k = 4 + 8 = 12, and z-y = 12-5 = 7, and z+y = 12+5 = 17. > Notice that p=3 just helped and then vanished without a trace. > And yes, it IS possible than an RSA public key could be factored by > p=3. That may be surprising but the math now is elementary since you can > just look back with the surrogate factoring answer. Consider R and RSA public key, such that R mod 3 = 2. Then you have from the surrogate factoring theory that k = 1 mod 3, or > k = -1 mod 3 and .87 = 1, and k is found such that R - 2k^2 is a minimum and k is even, as that is required to make x an integer. Now here's the scarily simple proof that it will work. Let z^2 = y^2 + R. Then if z = x + .87k, and 2.87x = k, it is true then that x = z/3 so if z is divisible by 3, then x exists as an integer. Now substituting x^2 + 2.87kx + (.87k)^2 = y^2 + R, so x^2 = y^2 + R - 2.87kx - (.87k)^2 and I can get rid of x, just on the right side to have x^2 = y^2 + R - (1 + .87^2)k^2 Now if k is coprime to 3 and .87 = 1, R - 2k^2 = 0 mod 3, so x^2 = y^2 + R - 2k^2 leaving just the result that k is the maximal values such that the > absolute value of R - 2k^2 is a minimum. Remember k exists as an integer and x exists as an integer if z has 3 > as a factor, and R mod 3 = 2. Since R is a public key it is odd with > only two prime factors. To finish the proof use 2.87x = k + 3j, instead above, and get to x^2 = y^2 + R - 2k^2 - 3kj and note when j is 0. The approach I used was to note that if you > assume you have an absolute minimum and increase or decrease k, which > has to move mod 3, then j must increase, regardless of which direction > you move in, so the minimum absolute value gives the correct k, as > long as that k is even, and you have two possibles: k = 1 mod 3 and k > = -1 mod 3, so BOTH must be checked. I don't know if that's subtle or not but I'll leave it for now. Note then that ANY RSA public key that has a residue of 2 mod 3 can be > attacked with this approach to factor it with p=3, where if z is > divisible by 3, it WILL fall. But regardless, other primes can be used if those conditions aren't > met. The main point was to emphasize the power of this technique. It CAN factor an RSA public key with p = 3, and very trivial > equations. James Harris Numbers for which this factoring scheme works > when p = 3 are of the form > *****> (3k - y) * (3k + y). Should be: *****> (3x - y)*(3x + y). For example: R = 11 * 31 = 341. The value of k > which minimizes abs(R - 2k^2) and is even and not > divisible by 3 is: k = 14. This yields x = 7 and > z = 21 and x^2 = y^2 + R - 2k^2 reduces to 49 = y^2 + 341 - 2*196, or y^2 = 49 + 51 = 100, so y = 10, and the factors of R are (z - 10)*(z + 10) = (21 - 10)*(21 + 10) = 11 * 31. So in this case again, the method works. But: does the algorithm ALWAYS work for T of the > form (3*x - y)*(3*x + y)? Let R = 11 * 61 = 671. This is of the form (3*x - y)*(3*x + y) for x = 12 and y = 25. However, the value of k which > is not divisible by 3 and which minimizes abs(R - 2k^2) is k = 20. Thus Harris's method would give x = 10 and > z = 30, and 100 = y^2 + 671 - 2*400, or y^2 = 229, which is not a perfect square. So no integer > factorization in this case. Note that > k = 20 = 2 mod 3. Choosing the optimal value of k > which is even and congruent to 1 mod 3 yields k = 16. > Thus x = 8, z = 24, and 64 = y^2 + 671 - 2*256 or y^2 = 64 - 159 = -95 which is negative, and abs(y^2) is not a > perfect square anyway. So this does not give an > integer factorization either. What is not clear here is why Harris wants > to choose k such that k is not equal to 0 mod 3 > and abs(R - 2k^2) is a minimum. As just shown, in general this does > not provide an integer factorization. Chances are that > y^2 as defined above is not a perfect square of an integer. > Thus in general even if you know that R is of the form > (3x - y)*(3x + y), you will have to search for y. > If R is very large, this search is may need to cover > a wide range. If R is NOT of the form > (3x - y)*(3x + y), but rather, say, of the form R = (17x - y)*(17x + y), then the search based on p = 3 will not succeed at all. Here is another example: R = 2437 * 3407 = 8302859 This product is of the form (3x - y)*(3x + y). If you use the Harris algorithm to try to find x > and y, here is how it works: The value of k (even integer, coprime to 3) which > minimizes abs(R - 2*k^2) is k = 2038. This yields x = k/2 = 1019 and > z = 3x = 3057. This in turn gives 1019^2 = y^2 + R - 2*k^2, which produces y^2 = 1019^2 - R + 2*k^2 or y^2 = 1038361 - 8302859 + 8306888 = 1043290, and y = 1020.975024, that is, y is not an integer. It is true that (3x - y)*(3x + y) = R, but this is not an integer factorization. The > CORRECT values of x and y are not 3057 and > 1020.975024 respectively, but rather x = 2922, y = 485. You might start with a 'y' in the neighborhood of > 1021 and search above and below this value until you > found y = 485, but this would take a while. For T > very large (e.g. T ~ 10^100), a lot of such searching > would likely be needed. Plus in an actual problem, you > would not have the advantage of knowing in advance > that T factored in the form (3x - y)*(3x + y). Conclusion: If you are lucky and R is of the form > above, the Harris approach might give the right > answer, but very likely will not do so if R is large, and > you will need to search. If R is not of the expected form, > you have more searching to do over primes > 3. In general > this method has some resemblance to Fermat's method, > but is likely less efficient. Marcus. === Subject: THE VED CONTAIN REFERENCES TO INFINITY Vedas contain references to infinity Special Correspondent THE HINDU Tuesday, February 19, 2008 [Caption] Frits Staal delivering a lecture at Indian Institute of Mathematical Science, Taramani, in Chennai on Monday. - Photo: R.Ragu Chennai - Ancient India had recognised importance of the concept of infinity whereas other civilisations frowned upon it, Frits Staal, Professor Emeritus of Philosophy and South and Southeast Asian Studies, University of California, Berkeley, said on Monday. Noting that Yajur Ved had used very large numbers, Prof. Staal, regarded as an authority on linguistics and the study of Vedic ritual and mantras, said there were many references to infinity in the Ved and Upanishads. Large numbers were extrapolated from the number of bricks used for the construction of altars. To a query on the largest number used in Indian traditions, he replied that Buddhists and the Ramayana went up to the number of ten to the power of 60. Prof. Staal, a scholar of Greek and Indian logic and philosophy and Sanskrit grammar, was delivering a talk at the Institute of Mathematical Science here on language and method in Indian science. He said the Rig Ved was familiar with the distinction between cardinal numbers and ordinal numbers. Sound system Describing the sound system of language as a major discovery in linguistics, Prof. Staal, whose book on Ved will be published by Penguin India in April, said this happened in India around 7th century BCE, not despite the absence of writing but because of it. The writing system came to India after the Asoka era (around 300 BCE). Till then, all these complex concepts, discussed in the Ved, were carefully and orally transmitted. More at: http://www.hindu.com/2008/02/19/stories/2008021958340200.htm Jai Maharaj http://tinyurl.com/24fq83 http://www.mantra.com/jai http://www.mantra.com/jyotish Om Shanti Hindu Holocaust Museum http://www.mantra.com/holocaust Hindu life, principles, spirituality and philosophy http://www.hindu.org http://www.hindunet.org The truth about Islam and Muslims http://www.flex.com/~jai/satyamevajayate DISCLAIMER AND CONDITIONS o Not for commercial use. Solely to be fairly used for the educational purposes of research and open discussion. 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Does anyone here *not* believe me, instead believe I'm lying about that fact? === Subject: Re: Tommy, don't be so cranky ;-) > response. > If I said I won an Olympiad or a Putnam competition or had a Ph.D > in Math, certain others here wouldn't believe it, so there is no > value in saying such things. I actually *did* win the Putnam competition. (I placed top five.) > Do you believe me? Does anybody here believe me? > Does anyone here *not* believe me, instead believe I'm lying about that > fact? === Subject: ***Solutions Manuals and Test Banks AVAILABLE - Accounting, Finance, Taxation*** posting-account=Oef_qQoAAAC_nhbXm5pHPNx-3IJ2uJp1 1.1.4322; .NET CLR 2.0.50727; InfoPath.1; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) I have the COMPLETE SOLUTIONS MANUALS in PDF and WORD format for the following textbooks: 1. Intermediate Accounting, 12th edition by Kieso, Weygandt, Warfield; ISBN-10: 0471749559; ISBN-13: 978-0471749554 For this one the TEST BANK is available too. 2. 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Juvinall, Fundamentals of machine component design, 4th ed, John Wiley& Son === Subject: Re: What would the end velocity be if...? posting-account=HXfuyAoAAADfxdPuI920odl79Rgch8Vx Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > light, which is 300 000 000 km/s. How's that again? === Subject: Re: What would the end velocity be if...? The accelerator would have to be about 4.59*10^13 m long, > or 307 AU, extending into the inner Oort cloud. And if you think that's too fantastic for anybody to propose with a straight face, check out a book by Marshall Savage named The Millennial Project. -- Mike Combs ---------------------------------------------------------------------- By all that you hold dear on this good Earth I bid you stand, Men of the West! Aragorn