mm-146 Ok, I'm trying to work on my homework and am stuck on 4.9 #10 of Vector>Analysis by Davis.The question states By means of Stokes' theorem, 'nd S F*dR around the>ellipse x^2+y^2=1, z=y, where F=xi+(x+y)j+(x+y+z)k.I got the curl of F and that equalled i-j+k but I'm not really sure how to do>the rest of the problem. Any help would be appreciated. I've wasted a lot of>time and gotten almost nowhere.ahh, z=y means the ellipse is tilted in the z-y plane and theprojection of it on the x-y plane is x^2+y^2=1so the area vector of the ellipse is pointing in the(0,-1,1) direction so that you get two contributionsfrom curlF dot dA I.e. integrate (-j+k)dot(-j+k)= 2 over the circle to get 2 PiAlso, Check that the back of book is right by doing it explicitly:x=costy=sintz=sintt==0..2PidR=i(-sint)+j(cost)+k( cost)F=i(cost)+j(cost+sint)+k(cost+2sint)FdotdR=(-costsint)+( cos^2t+costsint)+cos^2t + 2costsintintegrate from 0 to 2pi = 0 + Pi + 0 + Pi + 0 = 2 Piadam = at 03:12 PM, habshi@anony.com (habshi) said:ROTF,LMAO! You list a bunch of tripe that has nothing to do withMathematics, But you forget G. H. Hardy's greatest discovery,Ramanujuan! Plus, of course, you misrepresented some and you neglectedChinese and Greek work much earlier than some things you list.-- Shmuel (Seymour J.) Metz, SysProg and = INVENTIONS & DISCOVERIES> INVENTION OF NUMERALS> Numerals are found in the inscriptions of Ashoka The Great in the 3rd> Century BC. This knowledge traveled from there to Europe and West. In> Arab countries even now numerals are known as HINDSE: from India. La> time, It is India that gave us the ingenious method of expressing all> numbers by means of ten symbols - Prof. O.M. Mathew in Bhavan's> Journal INVENTION OF ZERO> Brahmagupta was the 'rst mathematician to treat ZERO (0) as a number> and showed its mathematical operations INVENTION OF ARITHMETIC> Arithmetic was discovered by Indians in about 2nd Century BC.> Bhaskaracharya's book Lilavathi is regarded as the 'rst book on> modern arithmetic. The Arabs learnt and adopted it from India and> spreaded it to Europe. In 499 AD Aryabhatta 'nished his work> Aryabhatt, giving rules of Arithmetic (Encyclopedia Britannica) INVENTION OF ALGEBRA> In Western Europe the knowledge of Algebra was borrowed, not from> Greece but from Arabs, who acquired this from India. Algebra is the> only Arabic name for Bijaganitha. Aryabhatta was one of the 'rst to> use Algebra (Encyclopedia Britannica) INVENTION OF GEOMETRY AND TRIGNOMETRY> The brick work of Harappa and Mohenjodaro excavations show that people> of ancient India (2500 BC) possessed knowledge of Geometry. Aryabhatta> formulated the rules for 'nding the area of a ?triangle', which led> to the origin of Trignometry. DISCOVERY OF ASTRONOMY> The knowledge of the motion of heavenly bodies was discovered by> Aryabhatta (499 AD), Latadeva (505 AD) and Brahmagupta (628 AD), for> calculating the timing of eclipses. In Surya Sidhanta' Latadeva,> talked about the earth's axis and called it SUMERU. That the earth is> a sphere and it rotates on its own axis, was known to Varahamihira> and other Indian astronomers much before Copernicus published this> theory. (Jewish Encyclopedia) INVENTION OF CALENDAR MAKING> Discovery of measurement of time and discovery of nomenclature of> days, month and years and invention of calendar making was made in> India. In his book ?Surya Sidhanta' Latadeva (505 AD) divided the year> into 12 months. Seven planets of the solar system effect the earth's> atmosphere and their names were added to the seven days of the week,> which was accepted all over the world. DISCOVERY OF THEORY OF GRAVITATION> In his book ?Sidhanta Shiromani' Bhaskaracharya mentions about force> of attraction resembling gravity, discovered centuries later by> Newton. (Jewish Encyclopedia) INVENTION OF IRON PRODUCTS IN 3000 BC> The word AYAS occurs in the four Vedas which denotes iron. Ashoka> pillar at Mehrauli, New Delhi and another iron pillar in Karnataka> stand proof of India's metallurgical heritage (A study published in> the magazine ?The Current Science'). INVENTION OF COPPER, BRONZE AND ZINC> The copper and bronze artifacts dates back to Indus Valley> Civilization (2500 BC). According to treatise RASARATNAKAR, zinc was> made in around 50 BC at Zawar in Rajasthan (India). INVENTION OF CHEMICALS PROCESSES, DYES AND CHEMICAL COLORS> Chemistry known as RASAYAN SHASTRA was invented in India. Elphinstone> to prepare the sulphate of copper, zinc and iron and carbonates of> lead and iron. RASAVIDYA or Indian alchemy made its appearance around> 5th Century AD (National Science Centre, New Delhi)At http://ancienthistory.about.com/gi/dynamic/offsite.htm?site= http%3A%2F%2Fmembers.aol.com%2Fbbyars1%2F'rst.html is the title The First Mathematicians.At http://www.stormloader.com/ajy/zero.html is the title Zero. =Psycho-hypocrite Antonio once again resorts to the same ad-hom he sowhiningly accuses others of! How the mighty have fallen!.Look at this hilarious bullshit:> oh by the way since you are in an informative mood.. tell them how you> have been caught lying so much.. that nobody belives you anymore..Nobody ia a bigger liar than you, as I have repeatedly demonstrated. BTW, you have now proven evolution by evolving into a weasel.Budikka =>> See, exactly the same as your method.>Not at all! It's 4 things to combine, no matter how it's done. Oh, it matters a lot how it's done. Because it is done differently, it> is a different method. Because it is done more easily and quickly> both in the mind and on paper, it is a better method. Absolutely. But:> 1. is it Vedic?That's what I too would like to know. I think it is, for I suspect myfather knew this method. If a number of Indian grandparents from allover India jump up saying that they knew this method, that it wastaught to them traditionally, then it would be Vedic (and not a fraudderived from say Tractenberg or someone and put in a book and passedoff as Vedic) unless one showed that they were all lying.> 2. is it really such a big deal?Oh yes. The idea is that it should be taught in primary schools,replacing the current multi-line method. If that is indeed done allover the world, then certainly is is a big deal.> Ad 1: The expression crosswise vertical doesn't say anything about the order of addition.That's the beauty of it.> Ad 2: Math (at least western math) is much more than arithmetic. It sure is a cool thing> to do 345 x 167 in the mind. Especially for school children. But it's not an advance> in the 'eld of mathematics. It's simply arithmetic.And isn't arithmetic really wonderful? Especially when you know whatit is all about? Vedic arithmetic shows you that. If implemented, itwill be an advance in arithmetic, and arithmetic was taught to me inschool as a part of mathematics. We started with arithmetic, then wehad algebra, then geometry, then trigonometry, then co-ordinategeometry and 'nally calculas.Look, let us do the following sum using Vedic arithmetic:(1234*5678 + 5659873 - 9987654)*345 just like that. Well, we get( (5)(16)(34)(60)(61)(52)(32) + 5659873 - 9987654)*345or ((10)(22)(39)(69)(69)(59)(35) - 9987654)*345 or ((1)(13)(31)(62)(63)(54)(31))*345or (3)(43)(150)(379)(632)(724)(624)(394)(155)or 925010495.I'm not sure if I have worked it out correctly, for I was only tryingto show the Vedic method, how beautifully and elegantly it works, thatis, without bothering about all the carries from one operation toanother, as done in current methods taught in school. Concentratingon the place values, and not the carries, makes life lot easier forthe working-out. If the beauty of all this does not strike you oranyone, I have nothing further to say.And still I don't know the one-line division and square rootextraction method. As far as I can see, that is also unknown toanyone around. But these methods have been published in the book onVedic mathematics, that is hailed as a fraud.Arithmetic is the basis of all computation. And computation is soimportant, in business, 'nance, engineering, etc. Faster computationwill naturally lead to greater ef'ciencies.> And for the purpose of math education, learning people calculation 'tricks'> is generally considered poor education of math.We are talking about teaching the fundamentals of arithmetic tochildren using better methods. After 44 years of dealing withmathematics, I am of the 'rm conviction that mathematics is an art,and like all arts, it is a bag of tricks. I am con'dent that peoplewill learn to love maths, once they get over their dread ofarithmetic, taught the usual way. Anyway, the best judges for thiswill not be experienced mathematicians, but primary school teachersand their charges, oce the green signal has been given by theeducators.Arindam Banerjee. Herman Jurjus =>> Equivalently, M*N is the same as M*N mod (M + N - 1).>> Sorry, this should be multiplication of M digits with N digits, base b,>> is equivalent to multiplication modulo b^(M + N - 1), i.e. M+N-1 digits.Oh well, so FFT or not, looks like multiplying M by N by any method means>MN multiplications! Well, when these multiplications are hardwired (as in>human memory for single digits) the computational issues (On*n) becomes>really irrelevant, for they all are done in no time at. Like, the video>extraction for radar data processing is done by NAND gates - its all done in>real time! You are in error. The number of multiplications required for> multiplying two numbers with the FFT method is O(n*log(n) where n is> the larger of the two numbers; it is not m*n.Fine, just multiply 12345 by 67809 using FFT with less than 25multiplications. Do it here. > Richard Harter, cri@tiac.net> http://home.tiac.net/~cri, http://www.varinoma.com> We have people from every planet on the earth in this State.> -- California Governor Gray Davis =>> Equivalently, M*N is the same as M*N mod (M + N - 1).>> Sorry, this should be multiplication of M digits with N digits, base b,>> is equivalent to multiplication modulo b^(M + N - 1), i.e. M+N-1 digits.Oh well, so FFT or not, looks like multiplying M by N by any method means>MN multiplications! Well, when these multiplications are hardwired (as in>human memory for single digits) the computational issues (On*n) becomes>really irrelevant, for they all are done in no time at. Like, the video>extraction for radar data processing is done by NAND gates - its all done in>real time! You are in error. The number of multiplications required for> multiplying two numbers with the FFT method is O(n*log(n) where n is> the larger of the two numbers; it is not m*n. Fine, just multiply 12345 by 67809 using FFT with less than 25> multiplications. Do it here.Do not misinterpret this beahaviour as typical arrogance of brahmins.So he demands explanations in his own ways. If he wants to learn howto calculate squres and if you teach him how to 'nd cubes, he may getconfusedand may stop learning mathematics. Richard Harter, cri@tiac.net> http://home.tiac.net/~cri, http://www.varinoma.com> We have people from every planet on the earth in this State.> -- California Governor Gray Davis =>> Equivalently, M*N is the same as M*N mod (M + N - 1).>> Sorry, this should be multiplication of M digits with N digits, base b,>> is equivalent to multiplication modulo b^(M + N - 1), i.e. M+N-1 digits.Oh well, so FFT or not, looks like multiplying M by N by any method means>MN multiplications! Well, when these multiplications are hardwired (as in>human memory for single digits) the computational issues (On*n) becomes>really irrelevant, for they all are done in no time at. Like, the video>extraction for radar data processing is done by NAND gates - its all done in>real time! You are in error. The number of multiplications required for> multiplying two numbers with the FFT method is O(n*log(n) where n is> the larger of the two numbers; it is not m*n. Fine, just multiply 12345 by 67809 using FFT with less than 25> multiplications. Do it here.Do not misinterpret this beahaviour as typical arrogance of brahmins.So he demands explanations in his own ways. If he wants to learn howto calculate squres and if you teach him how to 'nd cubes, he may getconfusedand may stop learning matheamatics. Richard Harter, cri@tiac.net> http://home.tiac.net/~cri, http://www.varinoma.com> We have people from every planet on the earth in this State.> -- California Governor Gray Davis => 1. When a student (or anyone) asks for a letter, it is > implicit that they are asking Will you write me a _good_> recommendation. Although I do have one example to the contrary. Many years ago, an undergraduate asked me to write her a letter of recommendation for mathematics graduate school. I told her that I wouldn't be able to say very much beyond she completed most of the homework assignments and she passed the course, but she still wanted me to write the letter, which I did. The story turned out to be that she was an overseas student whose government had been paying her expenses. The catch was that she had to apply to graduate school. Didn't have to go to grad school, but had to apply. She didn't want to go, and knew as well as I did that it wasn't a good idea for her to go, so what she actually wanted was a letter that would get her rejected! Unfortunately for her, one of the places she applied to was so desperate for students that it accepted her despite the lukewarm letters of recommendation she presented. I regret that I don't know how things turned out in the end.-- => letters of recommendation are about the most ludicrous mechanism in>> place in academia - it's a shame. institutional admissions exams>> would be far superior.>> Letters of recommendation may well be ludicrous. I have no reason>> to believe, and much reason to disbelieve, that institutional>> admissions exams would be any better at deciding who would be>> a good graduate student.perhaps you do not believe that a solid undergraduate>preparation is key component in graduate studies success.>is that it? I suppose I believe that, for a value of key that may be> lower than your value of key. What I'm sure I believe is that> (1) institutional admissions exams probably aren't particularly > good at measuring that component in graduate studies success> (since I don't have much faith in exams), and thatwell, your faith regarding institutional exams has to be the weakestdefense you have for your belief. perhaps if you made a statement, andjusti'ed it, regarding subjectivity, perhaps you would have betterluck.> (2) as Bart Goddard has pointed out, letters of recommendation do> have at least some chance of indicating that other (for me, much more> key) component in graduate studies success, namely, the > ability to *do mathematics*, while (3) institutional admissions> exams surely couldn't do anything at all towards indicating that> other component.the ability to solve mathematical problems, say in an admissions exam,is without a doubt, a much better indicator of ability to domathematics. in fact, the ability to solve mathematics problems is theonly accurate indicator of ability in mathematics. do you understandwhere i am coming from?>> Go into detail on how you'd make such an exam, if you don't mind.i do not need to go into any detail to ascertain that a>resounding reason for the lack of institutional exams in the>united states is correlated with the economic and logistics>required to put such systems in place. You certainly do not need to go into any detail to do that> on *my* behalf, because I didn't express (and don't have) any> interest in your opinion on that question, which I didn't raise.> The question I raised was, How would you make such an exam (with> some reasonable hope that it would do what you want it to do)?you propose the question, you seek the answer. not from me though.> Lee Rudolph =>> ...>> (2) as Bart Goddard has pointed out, letters of recommendation do>> have at least some chance of indicating that other (for me, much more>> key) component in graduate studies success, namely, the >> ability to *do mathematics*, while (3) institutional admissions>> exams surely couldn't do anything at all towards indicating that>> other component.the ability to solve mathematical problems, say in an admissions exam,>is without a doubt, a much better indicator of ability to do>mathematics. Scarcely without a doubt! The conditions of an exam, admissionsexam or otherwise, are entirely different from the conditions under which (real, practicing) mathematicians (of my broadpersonal acquaintance--which, I admit, does not include you,who may for all I know work along entirely different lines)actually do mathematics. And such problems as must,necessarily, be posed on an exam (admissions exam orotherwise), are just that--problems (with known solutions)to solve, not new mathematics to do (and possibly fail at).>in fact, the ability to solve mathematics problems is the>only accurate indicator of ability in mathematics. I would trust the seriously considered opinion of almost anyof my professional acquaintances, expressed in a letter ofrecommendation, that a third party had--or had not--the ability to do mathematics, far more than I would trust anadmissions exam. I wouldn't trust a stranger's opinionas much, but I'd still trust it more than an exam. >do you understand>where i am coming from?Either (if I am to be charitable) from somewhere far from whereI am, or (otherwise) from a position of profound ignorance ofwhat it means to do mathematics.Lee Rudolph = ...>> (2) as Bart Goddard has pointed out, letters of recommendation do>> have at least some chance of indicating that other (for me, much more>> key) component in graduate studies success, namely, the >> ability to *do mathematics*, while (3) institutional admissions>> exams surely couldn't do anything at all towards indicating that>> other component.the ability to solve mathematical problems, say in an admissions exam,>is without a doubt, a much better indicator of ability to do>mathematics. Scarcely without a doubt! The conditions of an exam, admissions> exam or otherwise, are entirely different from the conditions > under which (real, practicing) mathematicians (of my broad> personal acquaintance--which, I admit, does not include you,> who may for all I know work along entirely different lines)> actually do mathematics. And such problems as must,> necessarily, be posed on an exam (admissions exam or> otherwise), are just that--problems (with known solutions)> to solve, not new mathematics to do (and possibly fail at).if you were wishing to establish that the conditions between takingexams and being a mathematician are different, congratulations, youhave done a superv job! i wonder though, how does that damage mystatement?>in fact, the ability to solve mathematics problems is the>only accurate indicator of ability in mathematics. I would trust the seriously considered opinion of almost any> of my professional acquaintances, expressed in a letter of> recommendation, that a third party had--or had not--the > ability to do mathematics, far more than I would trust an> admissions exam. I wouldn't trust a stranger's opinion> as much, but I'd still trust it more than an exam. i reckon that is a problem you and many others share.>do you understand where i am coming from? Either (if I am to be charitable) from somewhere far from where> I am, or (otherwise) from a position of profound ignorance of> what it means to do mathematics.perhaps now you will reject the notion that a mathematician is de'nedby his ability to solve math problems?no, i am not referring to ones with known solutions...> Lee Rudolph =>> The question I raised was, How would you make such an exam (with>> some reasonable hope that it would do what you want it to do)? you propose the question, you seek the answer. not from me though.> a general knowledge one would expect a math major to have mastered at a pretty good school. I downloaded the practiceexam a few months ago and worked it. It is, of course, multiple choice, but one has to evaluate contour integral,say which of the following is not a basis for a vector space,do arithmetic in a dihedral group, and so on. No proofs,but there are a couple questions from every normal coursein the undergraduate sequence: topology, number theory,algebra, analysis, probability, statistics, etc. And Calculus,just to check.Some graduate school require the test. The reason I downloadedthe test was that I was designing a new math major, and wantedto go through the questions and how many of the them a potentialgraduate from my new program would have a chance of answering.The practice test was good for pointing out holes in my program. (Namely, complex analysis and topology.) Not thatI have to kowtow to the test designers, but the test designersare trying to 'nd what's typical for an American math major,and so I thought the test would be at least somewhat normative.Anyway, if one is curious, the practice test is downloadablefrom the GRE website.Bart =algebra> system,> which runs in any computer with Java. You can play it online. www.SymbMath.com = It plots and analyses any x-y data for peak location, peak height,> peak> width, semi-derivative, derivative, integral, semi-integral,> convolution,> deconvolution, curve 'tting, and separating overlapped peaks> and> background.> www.chemSoftware.com =system. www.mathHandbook.com =I'm trying to get the deadwood off my reading shelf, and I just'nished BITCH, by Elizabeth Wurtzel, which turned out to be worthreading for her powerful, insightful, and soul-baring epilogue, buttoo much of the rest of the book was just too much, over and over,about O.J. Simpson. Worse, Ms. Wurtzel is a cinema-holic whodismisses the woes of real-life victims who don't play their partswell, as if we should all have just the right life scripts for ourtragedies and have the necessary acting abilities to earn her respectand sympathy, or at least a tepid encore?And, one of the next space-hoggers on my shelf is INFINITY AND THEMIND: THE SCIENCE AND PHILOSOPHY OF THE INFINITE, by Rudy Rucker.Rudy ticked me off yesterday -- I had hoped to get a good start on'nishing the rest of the book, but then he stopped writing his barelycomprehensible Math-eze and added some stinking number questions,including the following:Prove that 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a)So I did:(1 - a) [1 + (a + a^2 + a^3 . . .)] = (1 - a) (1 / (1 - a)) {multiplying both sides by (1 - a)}(1 - a) + [(1 - a) (a + a^2 + a^3 . . .)] = 1 {interim result}(1 - a) + [(a + a^2 + a^3 . . .) - (a^2 + a^3 + a^4 . . .)] = 1{interim result}(1 - a) + a = 1 {interim result}1 = 1 {'nal proof of equality}This proves that the original equation is true no matter what value ofa we use, but Rudy argues that the original formula makes no sensewhen we substitute certain values for a.However, before we discuss it, let's give a step-by-step de'nition ofdivision so that everybody can follow along:A / B asks as to count how many times we can subtract all of B fromA to the limit of A, plus count any fractional units of B that we cansubtract from A, and to state the total as our answer.For example, 9 / 5 asks us to count how many times we can subtract 5from 9 (once), plus how many fractional units of 5 we can subtractfrom the remainder of 4 (and our total answer is 1 & 4/5ths).Now, in the above formula:1+ (a + a^2 + a^3 . . .) = 1 / (1 - a), Rudy notes that if we use 2 for a, then the formula becomes:1 + (2 + 4 + 8 . . .) = 1 / (1 - 2), or1 + (2 + 4 + 8 . . .) = 1 / -1The brain stops working when we see an ever-increasing series on theleft side and a simple negative fraction on the right side, and weconclude that the two expressions are not equal. However, if we forcethat supposedly simple negative fraction through the true processrequired by division, as explained above, we will see how complex thatsimple fraction really is, and that it exactly equals the expressionon the left side:1 / -1 says to state how many times we can subtract -1 from 1 tothe limit of 1 plus state the fractional units.The 'rst time that we subtract -1 from 1 we get 2 and move away fromour limit of 1 instead of towards it (which makes sense as a reverseprocess from a division process using numbers with the same sign), andwe immediately have a whole count of 1 subtraction:1 + (. . .)After subtracting -1 from 1, the 1 now has a remainder of 2 afterbeing subtracted from, and thus our division limit is now 2 instead of1, and the fractional units of -1 that can subtracted from the limitof 2 becomes 2 / -1, or -2. As such, the ïfractional' units to beadded to our count comes from the next division of 2/ -2., which saysto subtract -2 from 2, which equals 4 (to the limit of 2), and to addthe interim answer to our series:1 + (2 + . . .)And so on as above:1 + (2 + 4 + 8 . . .) = 1 / -1The rote learners among us will immediately object that 1 / -1 = -1,and thus that -1 cannot possibly equal 1 + (2 + 4 + 8 . . .). Theargument ignores the fact that transforming 1 / -1 into -1 is analgebraic process (-1 = [-1] * 1 = [-1] * 1 / 1 = etc.), and is not agiven fact, and any such transformation should be performed on bothsides of the equation to be truly and logically valid in all cases,regardless that it is irrelevant in most cases. Thus, whenever thedivision process itself is crucial to the meaning of an equation, itcannot simply be dispensed with or collapsed on one side of theequation alone.Thus, the above idea is not contrary to algebra, but rather states apoint that is not being taught. It's a distinction, not acontradiction.Rudy gives a more interesting but nonetheless complying example of asubstitution for a where the substitution is -1:If a = -1 in: 1 + (a + a^2 + a^3 + a^4 + a^5 . . .) = 1 / (1 - a),then all the even powers of a will equal 1 and all the odd powers ofa will equal -1:1 + (-1 + 1 - 1 + 1 - 1 . . .) = 1 / (1 - [-1]) {interim}1 + (-1 + 1 - 1 + 1 - 1 . . .) = 1 / 2 {'nal}What Rudy notes is that at any point in the in'nite series on theleft, the total can be either 0 or 1, and thus that it can be arguedthat the evidence shows that the sum is ultimatlely either 0 or 1. Rudy then 'nds it amusing that the right side of the equation seemsto average this mind bender by declaring the sum to be the average of0 and 1, which is 1/2.No such metaphysics are needed to understand the formula. Although 1/2 is in its simplest form, it is still in the form of a requireddivision process that continues ad in'nitum. Although I haven'tworked out the exact mechanics of the math process on the right sidethat produces exactly the sequence on the left side, I imagine that itis not much more dif'cult than a 2-cycle engine the produces +1 atthe start and alternately procuces -1 every other cycle. (Perhapssomething like the following: 1 / 2 = zero 2's subtractable (leaving +1 untouched) + 1 / 2, and 1 - 2 = -1 (forced past the limit of 1),and -1 - (-2) = 3, and 3 / 2 = +1 with a remainder of 1 / 2, adin'nitum.)1 + (-1 + 1 - 1 + 1 - 1 . . .) = 1 / 2 {ad in'nitum}Rudy argues that the left side of the above equation represents aThompson Lamp where it can be argued that the 'nal state of thelamp is on, or off, however one chooses to argue the case. Theargument is spurious since it is plainly stated that the light, if itwere such, is being turned on and off at speci'c points in themathematical process, ad in'nitum.I'll 'nish reading the book, and I might even look at all theproblems, and perhaps I'll be enriched by it, but it really makes mewonder how much logic mathematicians really study.Very Respectfully,Ray => I'm trying to get the deadwood off my reading shelf, and I just> 'nished BITCH, by Elizabeth Wurtzel, which turned out to be worth> reading for her powerful, insightful, and soul-baring epilogue, but> too much of the rest of the book was just too much, over and over,> about O.J. Simpson. Worse, Ms. Wurtzel is a cinema-holic who> dismisses the woes of real-life victims who don't play their parts> well, as if we should all have just the right life scripts for our> tragedies and have the necessary acting abilities to earn her respect> and sympathy, or at least a tepid encore? And, one of the next space-hoggers on my shelf is INFINITY AND THE> MIND: THE SCIENCE AND PHILOSOPHY OF THE INFINITE, by Rudy Rucker. Rudy ticked me off yesterday -- I had hoped to get a good start on> 'nishing the rest of the book, but then he stopped writing his barely> comprehensible Math-eze and added some stinking number questions,> including the following: Prove that 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a)> for all a?Looks like you're better off sticking with E. Wurtzel. =>Prove that 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a)So I did:(1 - a) [1 + (a + a^2 + a^3 . . .)] = (1 - a) (1 / (1 - a)) >{multiplying both sides by (1 - a)}And what if a = 1?>(1 - a) + [(1 - a) (a + a^2 + a^3 . . .)] = 1 {interim result}(1 - a) + [(a + a^2 + a^3 . . .) - (a^2 + a^3 + a^4 . . .)] = 1>{interim result}Ah, a bit of shuf§ing the terms of a conditionally convergent series,that's always a neat trick in any proof.>(1 - a) + a = 1 {interim result}1 = 1 {'nal proof of equality}Very nice, for hundreds of years we have erroneously assumed thegeometric series converges only for |x|<1, but you have provenotherwise. =Look, math is basically a formalization of common sense and logic. So,if something produces nonsense, that means along the way you have madean erroneous step. Otherwise logic is illogical, so you haveundermined the foundations of logic, a feat similar to what BertrandRussel and later Kurt Godel were able to do.Now, as to your question...You want to prove that 1 + a + a^2 + a^3 . . . = 1 / (1 - a) FOR ALLa.First of all notice that this statement does not make sense when a =1, because the right side is unde'ned, hence you are trying to saysomething using things without de'nitions -- i.e. potential nonsense.So you may say, 'ne, I'll DEFINE 1/0 for ya. It's in'nity!Then you have to invent your own in'nity arithmetic. For you cannothave in'nity follow the same rules that we all agree regular numbersfollow. For example, 1 / 0 = in'nity, 2 / 0 = in'nity, so byde'nition of division 0 * in'nity has all the values of the rainbow.So in'nity * 0 is not unique. Incidentally, 0 / 0 is therefore notunique either, since the answer is the number which multiplied by 0gives 0 but this is true for all numbers. So you can replace thewith and and have yourself a nice little world with your in'nityand zero arithmetic.No one says you can't do that. It just has to be consistent andlogical. You can invent your own math. If it's interesting and no onehas done it before, you can even publish it. :-)So, notice:You want to prove that 1 + a + a^2 + a^3 . . . = 1 / (1 - a) FOR ALLa.This means you have to de'ne what the above statement MEANs for alla. We've just treated the case a = 1. Now let's see the case a > 1.What does the . . . mean in your express? if it means add in'nitelymany numbers together I say there is still a lot of things unde'ned.What do you mean by add in'nitely many numbers together? Instead, Iwant to use the de'nition that is used by all mathematicians, and iswell de'ned:SERIES [i=0...in'nity] a^i = lim [n->in'nity] SUM[i=0...n] a^iand I want to prove that = 1/1-a)The standard proof goes like this:for any partial sum,a * SUM[i=0...n] a^i = SUM[i=1...n+1] a^i = a^(n+1) - 1 + SUM[i=0...n]a^iSo we solve for SUM[i=0...n] a^i and we getSUM[i=0...n] a^i = (a * SUM[i=0...n] a^i) + (1 - a^n+1)(1 - a) * SUM[i=0...n] a^i) = (1 - a^n+1)SUM[i=0...n] a^i = (1 - a^n+1) / (1-a) // IF a != 1Taking the limit as n goes to inity NOW givesSUM[i=0...in'nity] a^i = 1 / (1-a) // IF |a|<1SUM[i=0...in'nity] a^i = a - INFINITY / (1 - a) // IF |a|>1This is why limits were invented, also. You don't want to deal within'nities straight out. You will get confused about what the heck isgoing on. TO make things precise, you usually want to deal with FINITErepresentations of the same thing.For example, the SERIES (a_n) is de'ned as the limit of the partialsums s_n = SUM [i=1...n] (a_i). If this limit is in'nity, we stopcaring about its value.If two series both diverge to in'nity but we want to see whether oneapproximates the other very well, we take the limit of the RATIOS ofthe partial sums, and that's how we reach our conclusion.All this rests on the concepts of FOR EVERY and THERE EXISTS. Insteadof talking about in'nity and FOR EVERY, you can use De Morgan's Laws,which are simply logical to our minds, and prove the negativestatement about THERE EXISTS. People who have dif'culty understandingin'nite constructions (I also do, sometimes), should realize thatthey can replace FOR EVERY x, P(x) with NOT [THERE EXISTS x, NOTP(x)]. This is useful for example for Cantor's DiagonalizationArgument.Some things are just impossible to de'ne if you want them to 'texisting standards. For example, it's impossible to de'ne an orderrelation between complex numbers without violating the axioms oforder. In the same way, you can make up your own arithmetic with zeroand in'nity, just make clear what, if anything, you are violating innormal arithmetic (as I did above with dividing by zero). Once youhave a consistent system, and you're sure it works, it might give youuseful results, but you always have to be careful about translatingthe results into regular arithmetic.Enjoy,Grisha =>Rudy ticked me off yesterday -- I had hoped to get a good start on>'nishing the rest of the book, but then he stopped writing his barely>comprehensible Math-eze and added some stinking number questions,>including the following:Prove that 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a)Suppose a=2 (you did somewhere). Insert this value in the aboveequation. The equation is false for a=2, hence you can't prove thatthe right side and the left side of the equation are equal for allvalues of a. Any ïproof' that does do this must therefore be incorrect(as logic dictates).So maybe the equation given isn't the whole story. Let's start withthis:1 + a + a^2 + ... + a^(n-1) + a^nMultiply by (1-a):(1 - a) * (1 + a + a^2 + ... + a^(n-1) + a^n) =(1 + a + a^2 + ... + a^(n-1) + a^n) - (a * (1 + a + a^2 + ... +a^(n-1) + a^n)) =(1 + a + a^2 + ... + a^(n-1) + a^n) - (a + a^2 + a^3 + ... + a^n +a^(n+1)) =1 - a^(n+1)So:1 + a + a^2 + ... + a^(n-1) + a^n = (1 - a^(n+1)) / (1 - a)When n goes to in'nity, the right hand side of this equation onlybecomes equal to the right hand side of the equation given above byRudy, if a has a value between -1 and 1. So the complete statementshould have been:1 + (a + a^2 + a^3 . . .) = 1 / (1 - a) with -1 < a < 1>So I did:(1 - a) [1 + (a + a^2 + a^3 . . .)] = (1 - a) (1 / (1 - a)) >{multiplying both sides by (1 - a)}(1 - a) + [(1 - a) (a + a^2 + a^3 . . .)] = 1 {interim result}(1 - a) + [(a + a^2 + a^3 . . .) - (a^2 + a^3 + a^4 . . .)] = 1>{interim result}(1 - a) + a = 1 {interim result}1 = 1 {'nal proof of equality}This proves that the original equation is true no matter what value of>a we use, but Rudy argues that the original formula makes no sense>when we substitute certain values for a.The ïproof' you did must be wrong somewhere as the equation doesn'twork with any value outside the range <-1,1>. I believe the error isthat the operations shown can only be applied to absolute convergentseries (ïabsoluut convergente reeksen' in dutch). Since the series isnot convergent at all for a being outside <-1,1>, the whole proof isnonsense.I'm sure someone else can explain this better... => The ïproof' you did must be wrong somewhere as the equation doesn't> work with any value outside the range <-1,1>. I believe the error is> that the operations shown can only be applied to absolute convergent> series (ïabsoluut convergente reeksen' in dutch). Since the series is> not convergent at all for a being outside <-1,1>, the whole proof is> nonsense. I'm sure someone else can explain this better...I posted virtually the same arguments at the www.johnpatrick.commessage board and recieved a similar response from The Truth._______________________________________________________ ______________>>Essential in your derivation is the step [(a + a^2 + a^3 . . .) -(a^2 + a^3 + a^4 . . .)] = a.But this equivalence only holds if the series a + a^2 + a^3 . . .converges, and it only converges for certain a, not for any a.< work with any value outside the range <-1,1>. I believe the error is> that the operations shown can only be applied to absolute convergent> series (ïabsoluut convergente reeksen' in dutch). Since the series is> not convergent at all for a being outside <-1,1>, the whole proof is> nonsense. I'm sure someone else can explain this better... I posted virtually the same arguments at the www.johnpatrick.com> message board and recieved a similar response from The Truth.> ______________________________________________________________ _______>>Essential in your derivation is the step [(a + a^2 + a^3 . . .) -> (a^2 + a^3 + a^4 . . .)] = a.> But this equivalence only holds if the series a + a^2 + a^3 . . .> converges, and it only converges for certain a, not for any a.<< There is only one element of in'nity that is not common to both> in'nite series and that is a to the 'rst power, and thus the> difference between the two in'nite series is simply a to the 'rst> power. If you will take the time to explain in English why the a I've> isolated is not the certain a that you require, I'll listen, but> you've otherwise said nothing.> -------------------------------------------------------------- ---------- I'm listening ... so tell me in English why I'm wrong.The so-called associative property of the addition a + ( b + c ) = ( a + b ) + callows us to write both sides of the equality as a + b + c.This property is *not* valid for so-[badly]-calledin'nite sums. You cannot write something like a1 + ( a2 + a3 + ... ) = (a1 + a2) + a3 + ...In fact, the thing a1 + a2 + a3 + ....is not even a sum to begin with!It is a so-called limit of a series of partial sums: s1 = a1 s2 = a1 + a2 ... sn = a1 + a2 + ... + anIf this series has a limit for n -> in'nity, then one isallowed to use the abbreviation limit(sn; n -> in'nity) = a1 + a2 + a3 + ...The property of having a limit in the previous sentenceis something that can be veri'ed by other means.hth.Dirk Vdm => I'm listening ... so tell me in English why I'm wrong.Do you know what it means for a series to converge?Are you talking about a series when you write a + a^2 + a^3 + ... ?If not, what? =[SNIP]I hope it's a bad, ugly troll. Otherwise, it's a bad, uglymathematician.-- Nicolas, who wonders why there is always such people posting on Usenetnewsgroups. =Commentary 2Given coordinate patch C(x) in the base space M in a neighborhood of point x and 'ber f(x)form the local Cartesian product C(x)f(x) with ordered pair X = (x,fo).Take the union C(x)f(x)/C(x')f(x')/... of all such local products.There are redundant ordered pairs X because the coordinate patches C(x) and C(x') as sets overlapwith non-vanishing intersection C(x)/C(x')=/= Empty Set.Identify the redundant multiple images of the same actual point of the base space M usingthe symmetry group G as an equivalence relation. That is, two ordered pairs X and X' areidenti'ed or equivalent if x = x' < C(x)/C(x') and if fo' = gfo where g < G to form disjointequivalence classes {f(x)} that are the distinct points of the 'ber in hyperspace H.This is all local at a 'xed base point x like in an internal gauge force symmetry.g is also called a transition function.The hyperspace H is the factor space of the union C(x)f(x)/C(x')f(x')/ ... mod G.The projection map P:(x,{fo}) -> xWhen M is the curved space-time of Einstein's gravity theory in addition to the G equivalencein the extra space dimensions of the 'ber, x'(E) = Diff(4)x(E) at 'xed event Eto make disjoint equivalence classes {x(E)} mod Diff4(E).One can imagine a hybrid where the 'ber is a discrete space of strings of c-bits.One can also imagine a 'ber of strings of qubits.1 qubit is a parallel in'nity of c-bits.i.e.|qubit> = |1 c-bit><1c-bit|qubit> + |0 c-bit><0 c-bit|qubit>Where there is a continuous in'nity of different c-bit basesor orthonormal frames each corresponding, for example,the the angular orientation of an inhomogeneous 'eldmagnet in a Stern-Gerlach 'lter for spin qubitsin the DARPA spintronics project or like the billion billionSingle Electron Transistors inside the human brain at thesub-microtubular protein dimer hydrophobic cage level formingthe hardware interface with external world whose software is our stream of inner consciousness.Each possible orientation is a primitive parallel quantum universe.The quantum computer computes in all possibleorientations simultaneously like a continuousin'nity of classical Turing machines in adistributed network working on the same problem - or so the folklore goes.to be continued.Commentary 1The 'ber bundle as an idea has 4 parts.1. A structure symmetry group G.2. The total hyperspace H or, in some applications Wheeler's BIT.3. The projection map P.4. The base space M or, in some applications. Wheeler's IT.The hyperspace H consists of 'bers f(x) that areeither copies of or representations of the symmetrygroup G.The projection map P collapses a 'ber f(x) in the hyperspace H toa point x in the base space M.All of these objects are continuum differential manifoldsdepending on the continuum of real numbers which itsassociated issues of Cantor's in'nity of in'nities ofCabalistic Aleph's in an ascending Jacob's Ladder.This is not a discrete combinatoric mathematics althoughsuch a skeletal structure is associated with it as inHerman Weyl's Theory of Groups and Quantum Mechanicsand as in Saul-Paul Sirag's presentation of V.I. Arnold'sA-D-E mathematics of everything.The base space is covered by an atlas of local coordinate patcheswith all important overlap transition functions sewing thepatches together like a quilt.M is space-time in local micro-quantum 'eld theory of pointThe extra-dimensions of hyperspace formthe Calabi-Yau space of vibrations of thesuperstring beyond space-time.The connection on the total hyperspace H is the potentialof a local gauge force.Examples of connections is the 4 potential Au(x) inMaxwell's electromagnetism with G as U(1).There are similar connections for the Yang-Mills weak forcewith G = SU(2) and the strong force with G = SU(3).Classical general relativity, as distinct from local micro-quantum'eld theory, has the torsion-free symmetric three-index non-tensorLevi-Civita connection with G as the Diff(4) group.The latter comes from locally gauging the 4 parameter translation subgroup(generated by the 4-momentum Pu of globally §at special relativity )of the 15 parameter conformal group of Roger Penrose's massless twistors.Bottom -> Up: Given base space M and symmetry group G construct thehyperspace H as a quilt patchwork.Top -> Down: Given hyperspace H and symmetry group G construct thebase space M as the non-overlapping partition of hyperspace into G-orbitscalled the quotient space of H mod G in the principal bundle.Micro-quantum source renormalizable local 'elds of spin 1/2 lepto-quarks are associated vector bundles.Micro-quantum force renormalizable local 'elds of spin 1 gauge force bosons (electro-weak and strong) arefrom the principal bundle.There is no renormalizable quantum gravity in this precise sense.This is because classical Einstein gravity is a More is different (P.W. Anderson)emergent collective effect as in Andrei Sakharov's metric elasticity of aninstability in the globally §at false vacuum of the interacting lepto-quark source/electroweak-strong force.Einstein's gravity + uni'ed exotic vacuum dark energy/matter with Andrei Linde's chaotic in§ationary cosmology are the result of the continual phase transitions from globally §at false high entropy micro-quantum vacua to locally curved macro-quantum low entropy metastable vacua.to be continued: =>I'm looking for a solution to a set of linear equations which has a very >speci'c form. Given k>0, and real vectors a_0^{k-1} and b_0^{k-1} (all >>= 0, if it matters), I need to 'nd a real vector c_0^{k-1} such that: a_0 c_0 + a_{k-1} c_1 + a_{k-2} c_2 + ... + a_1 c_{k-1} = b_0> a_1 c_0 + a_0 c_1 + a_{k-1} c_2 + ... + a_2 c_{k-1} = b_1> a_2 c_0 + a_1 c_1 + a_0 c_2 + ... + a_3 c_{k-1} = b_2> a_3 c_0 + a_2 c_1 + a_1 c_2 + ... + a_4 c_{k-1} = b_3> ...> a_{k-1} c_0 + a_{k-1} c_1 + a_{k-3} c_3 + ... +> a_0 c_{k-1} = b_{k-1}If I'm reading this correctly, you want to solve (sum a_m J^m) c = bfor c, where coef'cients a_i and the vector b are given, and J isthe 0-1 matrix whose 1's are precisely in the (i+1,i) entries (includingthe (1,k) entry). This matrix J is easily diagonalized (if w = exp(2pi i / k) then the vector (1,w^m,w^(2m),...) is an eigenvaluecorresponding to the eigenvalue w^(-m) ). In this new basis thecoordinates of c are found simply by dividing those of b by the valuesof P(x) = sum a_m x^m at x = 1, w, support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAFLMLK13048; => how can i get a better approximation?you have to 'rst say in what sense you want the polynomial to 'tthe function (least square 't maybe?) i.e. provide a norm in function space and then solve the resulting minimizing problem with respect to the coef'cients support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAFLMIm12988; => I see, thx. I wasn't clear on who was 'ring at who :) But JSH does> sometimes threaten people with litigation, and he had somebody in court one> time, for calling him a crank. The case was quickly tossed.>>I don't think there was any such incident.>> He has threatened on occasion; but I think Mr. Hammick may be getting>> confused with the case where Underwood Dudley ->was<- sued for calling>> someone a crank. Dilworth v. Dudley:>> http://www.law.emory.edu/7circuit/jan96/95-2282.html The suit was dismissed. It wasn't just one suit. The 'rst, suing me, my publisher, and>the president of my school for, among other things, conspiracy to deny>Mr. Dilworth's civil rights (mathematicians wouldn't talk to him>because I called him a crank, he said), was in federal court in>Wisconsin where it was indeed dismissed. Mr. Dilworth then appealed>to the Seventh Circuit Court of Appeals where it was again dismissed. >I was pleased that the decision was written by Judge Posner,>well-known for his many books, who had clearly read some of my>_Mathematical Cranks_.> By the way, my respect for the legal profession went up as a>result of this process. Mr. Dilworth represented himself, I assume>because no lawyer would take his case. The lawyers for the>Mathematical Association of America also did a lot of work, 'nding>all sorts of precedents where a person was called a scab, a traitor,>and other nasty things in print and the courts let the authors get>away with it.> This wasn't enough for Mr. Dilworth, who started the legal>process all over again by suing in Wisconsin _state_ court. Once>again the case was dismissed, but this time with the requirement that>Mr. Dilworth pay $7000 or so for his opponents' legal fees. That>'nally stopped him, though he continued to send me letters.> Earlier this year he died, so now we can all call Mr. Dilworth>whatever we want.> However, the Law of Conservation of Cranks still operates, and>his kind has not died out. Presumably, the kind of crank who sues has>not died out either, so everyone be careful. Woody Dudley Dilworth's ghost is still cackling, however - see http://www.rcfp.org/news/1996/0422k.html,that the suit was dismissed in Dudley's favor, and thatan appeals court upheld the dismissal. Professor loses libel suit over label as ïcrank'and clearly identi'es Dudley as the professor! (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAG2ohQ02519; =If the last seven digits on n! are 8000000, support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAG3cXV05345; =>1. I've posted a lot on sci.math over a long period of time, to your>knowledge, have I *ever* been right?You know what they say, even a broken clock is right twice a day...Seriously though, some of the things you say are right. Your proofs usually have one or two major errors in them, the rest is 'ne. But in math one error is all it takes to produce completely incorrect results.If you're asking whether one of your major proclamations have ever been right, for instance I have a short proof for FLT, I've ïdiscovered' the prime counting function, The algebraic integers are §awed, etc, then to my knowledge all of them have been wrong.2. Do I have *any* correct results, or do you think I just talk and>never say anything that is mathematically correct?Some of your results are correct, though I believe you are greatly over-estimating their importance. Again, non of the major results you have claimed to achieve are correct, to my knowledge.If instead of over-estimating your own importance you would simply post your result you wouldn't be treated as an arrogant prick.3. To your knowledge, has ANYONE ever posted agreement with me on>*anything*?Of course. Most people who review your proofs agree that they are right up to the 'rst mistake. Also there are the occasional cranks and nuts. To my knowledge no competent mathematician has ever agreed with one of your major results.4. In your opinion, have I ever won an argument on the newsgroup?I don't know what winning an argument means, the term is obviously subjective. Of the serious arguments you have had with critics such as Arturo Magidin and Nora Baron, my personal impression was that your argument was hopelessly lost because you were simply wrong. I can't talk about arguments I didn't read.5. To your knowledge, have I ever caught other posters in errors?I can't remember, but I must say that even if you did I probably wouldn't remember. It simply doesn't seem to matter that much.>No approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAG5H7H11133; =>1. I've posted a lot on sci.math over a long period of time, to your>knowledge, have I *ever* been right?2. ...do...I just talk and never say anything that is mathematically correct?...>James HarrisNow you have been right. (In part 2 support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAGDFMa11708; =This is a (elementary?) geometry problem and I'm looking for a simplesolution.In a unit circle a chord is drawn.The distance of the centerfrom the chord is x (0<=x<=1).What is the length of the chord?Is it proportional to sqrt(1-x^2)?Olivio =In sci.math, Olivio This is a (elementary?) geometry problem and I'm looking for a simple> solution. In a unit circle a chord is drawn.The distance of the center> from the chord is x (0<=x<=1).What is the length of the chord?> Is it proportional to sqrt(1-x^2)?Half of the chord, the radius to the end of the chord half,and the line from the circle center to the bisection pointof the chord results in a right triangle. Therefore:hypotenuse = 1side adjacent = xside opposite = sqrt(1 - x^2)Bear in mind that this is only half of the chord, but theother half is symmetrical; the answer is 2 * sqrt(1 - x^2).Or one can do it analytically. Place the chordperpendicular to the X axis and to the right of the origin;therefore the circle point above the X axis is (x,y). Whatis y? Well, x^2 + y^2 = 1 as we're hypothesizing a unitcircle; the length of the chord is then 2y = 2* sqrt(1-x^2). Olivio> -- #191, ewill3@earthlink.netIt's approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAGHk5s28159; support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAGIeEQ31949; =This is a (elementary?) geometry problem and I'm looking for a simple>solution.In a unit circle a chord is drawn.The distance of the center>from the chord is x (0<=x<=1).What is the length of the chord?>Is it proportional to sqrt(1-x^2)?OlivioYes it support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAH2tNx32757; =>Can you please let me know if there is/was a Roman Numeral U and>>what amount it stands for.>>The Romans used the letter ïV' to represent the number 5. In some>medieval and earlier text the letter ïU' was frequently substituted>for ïV'. The Romans had both the letters ïU' and ïV' in their >alphabet, but sometimes (always? We dont know.) substituted ïV'>for ïU' when carving words into marble (or any other material they>had at hand). For example S.P.Q.R = SENATUS POPULUSQUE ROMANUSin carved form readsSENATVS POPVLVSQVE ROMANVS.Some say this stems from the fact that a ïV' is so much easier to>carve into marble (or whatever) than a ïU'. Maybe this is the>reason for the resubstitution of the numeral ïV' to ïU' in some >texts. Well, some people like to pay tribute to their >conjectures.Sincerly, ND> =Why do these old posts keep showing up here throughsome address at mathforum.org? Is there some sort ofglitch at the Math Forum approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAH2tR900344; =>> How long has the relation between complex numbers and particular 2x2>> matrixes been known?>> Are there any future plans to ïphase out' complex numbers and replace>> them with 2x2 matrixes which work exactly the same?>Complex numbers are isomorph 2x2-rotation-matrices are isomorph>ordinary 2D-vectors (with a certain multiplication added) -You calculate>them exactly the same - and you can calculate them in mixed mode too:>see the mixed-mode calculator:>http://i-z.eu.tt or http:// i-is-no-longer-imaginary.gmxhome.dewhich demonstrates, that you don't need this imaginarystuff any longer,>nor the Argand diagramm nor the imaginary axis, just like Caspar>Wessel started 2oo years ago without these.>Have fun >HeroThis is all very well, but personally I think the nice thing about complex numbers is that you can usually just treat them as real numbers anyway, so why go to the trouble of working out the corresponding matrices etc. Also take a little function like f(z)=z^2, if you wanted to differentiate it you'd have to separate the real and imaginary parts, 'nd the Jacobian etc, when you can just treat it like a real-valued function and just say f'(z)=2z as support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAH2tVY00372; =I still believe I discovered the fractal spirals described at:http://fractalspiral.zxcvb.orgThe formula is:x = sum (i = 1 to n) (r/i) sin (i*theta + i^2*pi*variable)y = sum (i = 1 to n) (r/i) cos (i*theta + i^2*pi*variable)I was told they were done by Michel Mendes France in the midnineties, but I am not able to verify this as I am outside the mathworld. The only web link I have found is:http://hypo.ge-dip.etat-ge.ch/www/math/html/node56.htmlIt is not the same thing. I would gladly reward the person whodonate to the charity of your choice $100.00 (US). I know this isn'ta lot but I don't have a lot. I need to know the correct representation of the formula. In itthere is a variable, and I would like to know what letter wasassigned to it. I would also enjoy reading technical jargon andactually understading the math being described. But mostimportantly, I need resolution so I stop wasting time thinking Iinvented this.k at symbl zxcvb dot org- K evin D support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAH2tbv00410; =Really, I am serious. I think this has great potential for usefull applications.Based on the given Quantitative-Qualitative model to assess intelligence I feel there must co-exist a Stupidity Constant in all assesed intelligence, otherwise there would be no basis for said intelligence.All assessed and thus measured intelligence must be also based on stupidity.Usefull applications could include: Finding limits to existing systems outlined by purely positive criteria. Knowing there must exist negative criteria and that the intelligence is based as much on stupidity as anything else. In other words there are as many good reasons against supporting these existing systems (previously supported only by positive criteria) as those supporting them. This would give new meaning to Mathematical Applications and new meaning to their limitidness and 'niteness.Zim Olsonhttp://www.zimmathematics.com All assessed and thus measured intelligence must be also based on> stupidity.I'll concede that yours is.--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAH2tjS00455; =>If the last seven digits on n! are 8000000, compute the value of n.Since the number ends with exactly six 0's, it must contain 5^6 as a factor, but not 5^7. =In sci.math, Dale ShoultsIf the last seven digits on n! are 8000000, compute the value of n. Since the number ends with exactly six 0's, it must contain 5^6 as a > factor, but not 5^7.> Given that hint, it's trivial; it has to be at least 25!, andless than 30!.27! = 10888869450418352160768000000-- #191, ewill3@earthlink.netIt's by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hAH36Dp02273; =I thought of another application of my 30 minute prior post A Mathematical Stupidity Constant to assessed IntelligenceI have come up with a System Transformation used to determine any system functionality which is outlined at:http://www.zimmathematics.com/app.htmIn the system transformation outlined here it contains a non-functional component. A usefull application can be found in explaining the Quantum Mechanic Slit experiment phenomena where exhibiting non causal or non rational behavior as my System Transformation said must be included in determining any system functionality.Zim Olsonhttp://www.zimmathematics.com => What role does the Jacobson radical> ab = a+b - abThat is not the Jacobson radical.The Jacobson radical of a ring is the intersectionof it proper left (or right) ideals.> play in the quaternion algebra> (lets say, over the reals), if any?None---literally! The Jacobson radical of H is {0}since H is a division ring.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) =Discrete Spaces Eventually Bore Converging Sequences. Proof:Should debonair (xj) dare to converge upon demure x, then for any nhood of x, (xj) would eventually squeeze into the nhood.As X is so discrete, a mere {x} will do for x's nhood.Thus eventually (xj) would be squeezed into {x}.Now when (xj) has eventually been con'ned to {x}, it will 'nd it's values severely restricted.Some may even consider it solitary con'nement.Yet there (xj) will be for the rest of it's eternal life, allowed the company of none but x. Thus boredom, QED. ;-)---- =JOn> When I did A level maths at school, about 30+ years ago, in my> analytical geometry course I was taught a subject called ïinversion'. => When I did A level maths at school, about 30+ years ago, in my> analytical geometry course I was taught a subject called ïinversion'.> .... > I have never come across this again, and the subject seems to heve> disappeared from text books.... You'll 'nd it in Chapter 5 of the recent text-book by Brannan,Esplen & Gray, Geometry, Cambridge U.P., 1998. They show how inversionis connected with Moebius transformations of the complex plane.> .... Does anyone know how/when this concept came about?.... I'd like to know more about that, too. Can anyone help? Ken Pledger. => .... Does anyone know how/when this concept came about?.... > I'd like to know more about that, too. Can anyone help?It goes back to Apollonius (262-190 BC), at least. I say this because thereand inversion relative to the ellipse, the hyperbola, and the parabola in the'conic sections' of Apollonios, by B.A. Rozenfeld (Istoriko-MatematicheskieIssledovaniya, 30 (1986), 195-199), but I haven't read it (I don't knowwhere to 'nd it and, besides, I can't read russian).Jose Carlos Santos =To every polytope, we associate a graph in the following way: take itsvertices as nodes. The nodes are joined by an edge if and only if thecorresponding vertices are adjacent.How can we decide, given any graph, whether it is the graph of some0/1-polytope or not? If there is no exact criterion known, is there a good suf'cient one?A 0/1-polytope is a polytope where all its vertices have coordinates in {0,1}.TIA,Tobias-- Phyics is much too hard for physicists. => To every polytope, we associate a graph in the following way: take its> vertices as nodes. The nodes are joined by an edge if and only if the> corresponding vertices are adjacent. How can we decide, given any graph, whether it is the graph of some> 0/1-polytope or not? > If there is no exact criterion known, is there a good suf'cient one?I have no idea, but an obvious necessary condition is that G is (log_2 |V(G)|)-connected.-- David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine, School of Information & Computer Science =What is the order of the orthogonal group of order n over the 'nite 'eld GF(q) , i.e O(n,GF(q)) ? =HenriWilson skrev i melding> Now consider the case of a charge of mass m being accelerated in a 'eld.> Using a=F/m, and m=Mo.gamma, dv/dt=(F/c.Mo)[sqrt(c^2-v^2)], which is:... nonsense.You have screwed this up before, and I have shownyou the correct equation before.F = dp/dt = d/dt (m*gamma*v)gamma =1/sqrt(1 - v^2/c^2)dv/dt = (1/((v^2/c^2)*gamma^3 + gamma))*F/m((v^2/c^2)*gamma^3 + gamma)*dv = (F/m)*dtgamma*v = (F/m)*t + Cif the charge is accelerated from standstill, v = 0 when t = 0, C = 0v = c*t/sqrt(t^2+T^2) where T = m*c/Fv approaches c asymptotically> dv/[sqrt(c^2-v^2)]=kdt.> Integrating: arcsin(v/c)=kt+B or> v=c.sin(kt+B) if t=0 then v=0, so B=0 So we have v=c.sin(kt), where k is de'nitely 'nite. How come? Even in its exponential form this doesn't make any sense. Does it mean the SR equation is wrong or that matter becomes anti matter as c> is exceeded?It means that you are unable to do the math correctly.But I have shown you a simpler approach before.If we assume the charge is accelerated from standstillin a constant electric 'eld excerting the force F, we get:p(t) = F*t = m*gamma*v(the derivation isn't necessary since we integrate it back in the next step!)Paul =>> as it enters a static electric 'eld.Let us consider a charged sphere somewhere in the universe. It exerts a force>on every other charge. If we can arrange for it to lose that charge somehow,>you are claiming that all those forces disappear INSTANTLY.Parse the bloody sentence in quotes, Henry. It doesn't say that. Isn'tEnglish your 'rst language? - Randy, grabbing his popcorn and going back to watch theentertainment =>> as it enters a static electric 'eld.>>Let us consider a charged sphere somewhere in the universe. It exerts a force>>on every other charge. If we can arrange for it to lose that charge somehow,>>you are claiming that all those forces disappear INSTANTLY.Parse the bloody sentence in quotes, Henry. It doesn't say that. Isn't>English your 'rst language? - Randy, grabbing his popcorn and going back to watch the>entertainment>Moron. Stop kissing Andsernon's arse.Henri Wilson. See the Stupidity of Relativity.www.users.bigpond.com/hewn/index.htm => Remember the old vacuum tube triodes.>> Acccording to you, a signal on the grid would be instantly felt at the>anode.>> If that's not instantaneous communication, what is?>> Henri Wilson.his way of giving up. Paul wants a static 'eld at the grid. That's ok,>until the electron reaches the grid, then that static charge has to reverse>sign, and is no longer static. Leaving aside the slew rate of the ampli'er>driving the grid (which in any real experiment we could not do), if the grid>reversed from positive (attracting the electron) to negative as it passed>through (propelling the electron on), it would still take a 'nite amount of>time for the 'eld beyond the grid where the electron is heading to change,>because c is 'nite. Hence at the instant the electron reaches the grid,>even if the grid potential is zero, there is a negative 'eld ahead of the>electron to slow it down. Hence the electron cannot attain c, by this>method. However, a positron coming down the other way would have a closing>velocity with the electron that was greater than c. Precisely. Paul Andsernon doesn't know what he's talking about.I note with interest that you think Androcles' incoherent babbleshows that I don't know what I am talking about. :-)Did you actually read it, Henry?in accelerators, isn't it? :-)Hatch or Anrocles - doesn't matter.They are 100% correct as long as they disgree with relativity.Right? :-)Paul, always amused when Henry agrees to nonsense => Consider an object being accelerated by a idealistic jet of water or a> continuous ïstream of elastic ping pong balls'. What is its subsequent velocity> pattern?>> [snippage] > M.dv/dt=m(Vo-v) or: >>[more snippage]>>Henri, you've got major problems with this. It assumes>>Newtonian momentum transfer, and so implies a Newtonian>>equation for kinetic energy. And there's a real problem>>most assuredly not 1/2 mv^2. So your model has gross>>disagreement with observation.>>Socks>> What an idiot!>> Naturally, if one includes assumptions that a theory is correct in an attempt>> to prove it wrong, that attempt is likely to fail.Yes you are, and you did, and it did. You assumed Newton, and>you got Newton, and those are clearly not similar to what is>obverved in accelerators.> A moving charge surrounds itself with a volume of ïreverse 'eld' because the>> main 'eld takes time to operate. The energy associated with this ïback 'eld'>> is equivalent to a mass increase.And their speeds are also not observed to go above c.It is very hard to accelerate anything to c let alone beyond it.The experiments are probably incapable of measuring superluminal speeds anyway.And you have not provided any theory of E&M that allows any such>thing as a reverse 'eld. Nor why there should be any kind of>speed limit involved. Nor why it should follow any such thing>as the kinetic energy formula observed in accelerators. Nor have>you provided a relation between energy and mass if you don't>accept relativity.>Socksradiation from an acceleraed charge!'elds associated with a moving charge!The ïBack EMF' concept.I would be most amazed if a moving charge DID NOT alter the 'eld arounditself, wouldn't you?Henri Wilson. See the Stupidity of Relativity.www.users.bigpond.com/hewn/index.htm =And you have not provided any theory of E&M that allows any such>thing as a reverse 'eld. Nor why there should be any kind of>speed limit involved. Nor why it should follow any such thing>as the kinetic energy formula observed in accelerators. Nor have>you provided a relation between energy and mass if you don't>accept relativity.>Socks radiation from an acceleraed charge! 'elds associated with a moving charge! The ïBack EMF' concept. I would be most amazed if a moving charge DID NOT alter the 'eld around> itself, wouldn't you?Quite.are accelerated.You KNOW the following, Henry.In an accelerator going at full ef'ciency, we KNOW thatbecause it looses this energy as synchrotron radiation in the bendsof the circuit.(Very obvious and easily measurable.)So we - and YOU - know that the RF-cavities never ceasesis only few mm/s below the speed of light.So why do you keep pretending that the E-'eld is notspeed approaches c, when you KNOW that isn't true?Another case of selective memory loss?What you admit knowing in one posting,you have forgotten in the next, eh?Paul =>And you have not provided any theory of E&M that allows any such>>thing as a reverse 'eld. Nor why there should be any kind of>>speed limit involved. Nor why it should follow any such thing>>as the kinetic energy formula observed in accelerators. Nor have>>you provided a relation between energy and mass if you don't>>accept relativity.>>Socks>> radiation from an acceleraed charge!>> 'elds associated with a moving charge!>> The ïBack EMF' concept.>> I would be most amazed if a moving charge DID NOT alter the 'eld around>> itself, wouldn't you?Quite.>are accelerated.You KNOW the following, Henry.>In an accelerator going at full ef'ciency, we KNOW that>because it looses this energy as synchrotron radiation in the bends>of the circuit.(Very obvious and easily measurable.)>So we - and YOU - know that the RF-cavities never ceases>is only few mm/s below the speed of light.So why do you keep pretending that the E-'eld is not>speed approaches c, when you KNOW that isn't true?I DID NOT SAY THAT.The question is how much energy?You are making no attempt to answer that one. In typical fashion, you pretendthe relevant question does not exist.Another case of selective memory loss?>What you admit knowing in one posting,>you have forgotten in the next, eh?Paul>Henri Wilson. See the Stupidity of Relativity.www.users.bigpond.com/hewn/index.htm => Let's get this straight.>> You say that the force on a charge due to an electric 'eld acts>> instantaneously. Correct?>> Let us consider a charged sphere somewhere in the universe. It exerts a>force>> on every other charge. If we can arrange for it to lose that charge>somehow,>> you are claiming that all those forces disappear INSTANTLY.>> I will continue when I receive your answer (if one is forthcoming).> Interesting question, Henry. I've used it myself, on the discussion of>gravity propagation. If a star were to convert *all* it mass to radiation in>one super-supernova, how long would it take for us to detect its loss of>gravity? I don't mean watch its planet suddenly §y away, that would be a>distant observation and would reach us at c. I mean detect the gravity loss>right here. This would be the biggest gravity pulse (negative going) that we>could possibly detect. But in order to detect the loss, where would have to>be aware of it in the 'rst place. As it turns out, the nearest star to us>(other than the sun) is 3.9 light-years away, and that is just too far to>detect it's gravity directly. So I fail to understand why anyone would>attempt an experiment to search for gravity waves, other than to give>themselves some funding on a futile attempt. There's a lot of money to be>made out of relativity, and very few people are altruistic.>Androcles>Exactly.There is one problem that I'm sure Paul will pounce on. That is, how to makecharge suddenly disappear.Here is another interesting question. The ïMass' of an object is made up of a major proportion, the total mass of allenergy.Presumably, the two portions play an equal role in Newton's gravitation Law. If there is no distinction between the two, is it possible that all MATTER isnothing but some kind of manifestation of various ï'elds'?Also, since mass is lost/gained in a Nuclear explosion, would we not expectmeasurable gravity waves to be produced. Even though the amount of matter isrelatively small, the sum total of all the potential energy associated withthat matter is quite large. (If the nuclear bomb is considered to be the centre of the universe, what isthe energy required to raise all the other matter in the universe to itspresent distance from that point. Might it just turn out to be mc^2?). Henri Wilson. See the Stupidity of Relativity.www.users.bigpond.com/hewn/index.htm => Let's get this straight.>> You say that the force on a charge due to an electric 'eld acts>> instantaneously. Correct?>> Let us consider a charged sphere somewhere in the universe. It exerts a>force>> on every other charge. If we can arrange for it to lose that charge>somehow,>> you are claiming that all those forces disappear INSTANTLY.>> I will continue when I receive your answer (if one is forthcoming).> Interesting question, Henry. I've used it myself, on the discussion of>gravity propagation. If a star were to convert *all* it mass to radiationin>one super-supernova, how long would it take for us to detect its loss of>gravity? I don't mean watch its planet suddenly §y away, that would be a>distant observation and would reach us at c. I mean detect the gravityloss>right here. This would be the biggest gravity pulse (negative going) thatwe>could possibly detect. But in order to detect the loss, where would haveto>be aware of it in the 'rst place. As it turns out, the nearest star tous>(other than the sun) is 3.9 light-years away, and that is just too far to>detect it's gravity directly. So I fail to understand why anyone would>attempt an experiment to search for gravity waves, other than to give>themselves some funding on a futile attempt. There's a lot of money to be>made out of relativity, and very few people are altruistic.>Androcles> Exactly. There is one problem that I'm sure Paul will pounce on. That is, how tomake> charge suddenly disappear. Here is another interesting question.> The ïMass' of an object is made up of a major proportion, the total massof all> energy. Presumably, the two portions play an equal role in Newton's gravitationLaw.> If there is no distinction between the two, is it possible that all MATTERis> nothing but some kind of manifestation of various ï'elds'?Since matter is mostly empty space anyway, why can't I pass through a brickwall?And of course the answer is that I am repelled by the electrons. So what areelectrons?> Also, since mass is lost/gained in a Nuclear explosion, would we notexpect> measurable gravity waves to be produced. Even though the amount of matteris> relatively small, the sum total of all the potential energy associatedwith> that matter is quite large.As I understand it, a mountain is just detectable. An nuclear device issmall enough to carry on a plane. Therefore you would need a *Very*sensitive device to measure the gravity of the weapon, and that would implyproximity. If you detonate the device, I think you are likely to destroy theinstrument before it could react :) (If the nuclear bomb is considered to be the centre of the universe, whatis> the energy required to raise all the other matter in the universe to its> present distance from that point. Might it just turn out to be mc^2?). Henri Wilson.> See the Stupidity of Relativity.> www.users.bigpond.com/hewn/index.htm =>> Of course you are funny.> If electric and magnetic 'elds acted instantaneously, light would travel at> in'nite speed.>>I note with interest that your statement is so ambiguously put>>that it may be right as well as wrong.>>But I take for granted that your statement is meant to be>>relevant to my claim, which was:>> as it enters a static electric 'eld.>>So assuming that acting instantly is to be understood>>in this sense, an unambiguous version of your statement>>becomes:>> travel at in'nite speed.>>Was this what you meant to say, Henry?>>If not, what DID you mean to say, and what is the relevancy>>of what you meant to say to your action time of the static>>accelerating 'eld in an accelerator?>>Paul, 'nding the acrobatic show a little monotonous>> Remember the old vacuum tube triodes.>> Acccording to you, a signal on the grid would be instantly felt at the anode.So if:> as it enters a static electric 'eld.>it follows:>A signal on the grid would be instantly felt at the anodeExplain why, please.> If that's not instantaneous communication, what is?Henry, you know of course that you are babbling nonsense.>There simply isn't possible to be as stupid as you pretend.Paul> Let's get this straight. You say that the force on a charge due to an electric 'eld acts> instantaneously. Correct?Why do you ask what I am saying, when what I amsaying is quoted right above?I am saying: as it enters a static electric 'eld.> Let us consider a charged sphere somewhere in the universe. It exerts a force> on every other charge. If we can arrange for it to lose that charge somehow,> you are claiming that all those forces disappear INSTANTLY.And why the hell do you think I would claim something as stupid as that?You MUST know this is stupid nonsense, Henry.Why the hell do you pretend that it is possible to interpret mystatement above to mean that the electric 'eld will act instantlyelectric 'eld?This is actually too bloody stupid even for you, Henry.Why do you pretend to be such a moron?> I will continue when I receive your answer (if one is forthcoming).Answer what?You are desperated to evade the point, are you?Lets take this from the beginning.Of course there are really no static electric'elds in an accelerator, and I never said it was.that is a resonance cavity with a very powerful EM-'eld.The typical frequency is in the order of 100MHz (or higher).where the E-'eld is longitudinal.The crucial point is that the phase of this resonator isadjusted such that the E-'eld peaks (in the same direction)enters the accelerating stretch.That is the point. Why is that so hard to get?So what I am claiming is still:it enters the (quasi) static electric 'eld. And the forceit speed.Can you please explain how you can use this forinstant communication, Henry?Of course you cannot.The only decent thing you can do is to admit thatthis claim is wrong.But you won't do that, will you?You will rather keep insisting that when I say thatthe RF-cavity in an accelerator, then I really have statedthat a force acts on an electron in the Andromeda galaxyin the accelerator.Won't you?Paul => Of course you are funny.>> If electric and magnetic 'elds acted instantaneously, light would travel at>> in'nite speed.>>I note with interest that your statement is so ambiguously put>that it may be right as well as wrong.>>But I take for granted that your statement is meant to be>relevant to my claim, which was:> as it enters a static electric 'eld.>>So assuming that acting instantly is to be understood>in this sense, an unambiguous version of your statement>becomes:> travel at in'nite speed.>> Remember the old vacuum tube triodes.>> Acccording to you, a signal on the grid would be instantly felt at the anode.>>So if:>> as it enters a static electric 'eld.>>it follows:>>A signal on the grid would be instantly felt at the anode>>Explain why, please.> If that's not instantaneous communication, what is?>>Henry, you know of course that you are babbling nonsense.>>There simply isn't possible to be as stupid as you pretend.>>Paul>> Let's get this straight.>> You say that the force on a charge due to an electric 'eld acts>> instantaneously. Correct?Why do you ask what I am saying, when what I am>saying is quoted right above?I am saying:> as it enters a static electric 'eld.So there is also an opposite force acting on the electrodes.even if the electrodes are lightyears apart. IS THAT WHAT YOU ARE SAYING?> Let us consider a charged sphere somewhere in the universe. It exerts a force>> on every other charge. If we can arrange for it to lose that charge somehow,>> you are claiming that all those forces disappear INSTANTLY.And why the hell do you think I would claim something as stupid as that?Because you just DID, above, (even though you probably do not realise what yousaid).You MUST know this is stupid nonsense, Henry.>Why the hell do you pretend that it is possible to interpret my>statement above to mean that the electric 'eld will act instantly>electric 'eld?The bloody 'eld can be from one side of the universe to the other for all Icare. This is actually too bloody stupid even for you, Henry.>Why do you pretend to be such a moron?You just do not understand. You are very confused.> I will continue when I receive your answer (if one is forthcoming).Answer what?>You are desperated to evade the point, are you?Lets take this from the beginning.>Of course there are really no static electric>'elds in an accelerator, and I never said it was.>that is a resonance cavity with a very powerful EM-'eld.>The typical frequency is in the order of 100MHz (or higher).>where the E-'eld is longitudinal.>The crucial point is that the phase of this resonator is>adjusted such that the E-'eld peaks (in the same direction)>enters the accelerating stretch.>That is the point. Why is that so hard to get?That is obvious, irrelevant and NOT the point we are discussing.The question is, does a moving charge feel the ïfull strength' of theaccelerating 'eld? (similarly, does a falling object feel the full ïforce of gravity'? Tom Robertsonce told me they didn't)The evidence shows that charges do NOT accelerate in ïNewtonian fashion'. SRclaims they behave as though their mass increases by gamma. I am suggesting alternative explanations based on two possibilities. Firstly,the 'eld DOES take time to act and therefore the 'eld gradient at the pointof the moving charge IS NOT the same as that which would act on a charge ATREST. Secondly, the movement of the charge itself creates an opposing 'eldthat effectively reduces the magnitude of the 'eld gradient around the movingcharge.I repeat that if what you claim is true then you have invented instantaneouscommunication.So what I am claiming is still:>it enters the (quasi) static electric 'eld. And the force>it speed.How do you de'ne KE?Can you please explain how you can use this for>instant communication, Henry?I told you, above.Of course you cannot.>The only decent thing you can do is to admit that>this claim is wrong.It is NOT wrong.Fill a box with electrons. The box attracts every positive charge in theuniverse. What happens to that attraction if the box suddenly disappears? You say it goes to zero instantly. So if you can make the box appear and disappear in an intelligent manner, youwill be able to communicate instantly with anyone, ANYWHERE!!!!Congratulations Paul. You will certainly be awarded a Nobel Prize.But you won't do that, will you?>You will rather keep insisting that when I say that>the RF-cavity in an accelerator, then I really have stated>that a force acts on an electron in the Andromeda galaxy>in the accelerator.Won't you?No Paul. You are considerating only 'elds between electrodes and not spherical 'eldsemanating from point charges. Paul>Henri Wilson. See the Stupidity of Relativity.www.users.bigpond.com/hewn/index.htm => You say that the force on a charge due to an electric 'eld acts>> instantaneously. Correct?Why do you ask what I am saying, when what I am>saying is quoted right above?I am saying:> as it enters a static electric 'eld. So there is also an opposite force acting on the electrodes.> even if the electrodes are light years apart. IS THAT WHAT YOU ARE SAYING?I am saying: as it enters a static electric 'eld.We have two electrodes - say 1 km apart.(Or a light year apart - if you insist)The potential difference is 1 million volts.(Or a zillion volts - if the distance is a light year)There is a small hole in the negative electrode.We inject an electron through this hole.When will a force act on the electron?I am still saying: as it enters a static electric 'eld.But what are YOU saying?Not untill the electrode 1 km away feels the opposing force? IS THAT WHAT YOU ARE SAYING?Come on, make your point.What is the action time of the force on the electron?Why do you think the distance to the other electrode is relevant?How does the distance to the other electrode affectthis action time?Please don't say something like we don't know.Because we DO know.Do YOU know?The rest is =>> You say that the force on a charge due to an electric 'eld acts> instantaneously. Correct?>>Why do you ask what I am saying, when what I am>>saying is quoted right above?>>I am saying:>> as it enters a static electric 'eld.>> So there is also an opposite force acting on the electrodes.>> even if the electrodes are light years apart.>> IS THAT WHAT YOU ARE SAYING?I am saying:> as it enters a static electric 'eld.We have two electrodes - say 1 km apart.>(Or a light year apart - if you insist)>The potential difference is 1 million volts.>(Or a zillion volts - if the distance is a light year)>There is a small hole in the negative electrode.>We inject an electron through this hole.>When will a force act on the electron?>I am still saying:> as it enters a static electric 'eld.>But what are YOU saying?>Not untill the electrode 1 km away feels the opposing force? IS THAT WHAT YOU ARE SAYING?NO PAUL. I am saying that your claim infers that the effect of an injectedelectron will be felt INSTANTLY at the far electrode. Of course, since theelectron existed BEFORE it was injected, the effect would have already beenthere even though the near electrode was in the way.. The only way thisexperiment can even be hypothesized is by either ïannihilating' a very largenumber of electrons or by monitoring the force on the far electrode withmovement of the electron mass towards it.If the effects are instantaneous as you claim, you will have achievedinstantaneous communication.Come on, make your point.>What is the action time of the force on the electron?>Why do you think the distance to the other electrode is relevant?>How does the distance to the other electrode affect>this action time?Please don't say something like we don't know.>Because we DO know.>Do YOU know?OK I will agree with you. Assume it is instant. Therefore I can send messagesto the far electrode instantaneously.The Paul Andsernon snips again!Paul>Henri Wilson. See the Stupidity of Relativity.www.users.bigpond.com/hewn/index.htm =>:> 18. Let G be the group of all real 2-by-2 matrices>:> (a b)>:> (c d), with ad - bc non-zero, under matrix multiplication, and let N>:> be the subgroup consisting of those elements of G with ad - bc = 1.>:> Prove that N contains G', the commutator subgroup of G.>:> 19. In problem 18 show, in fact, that N = G'.: I assume, that your matrices have their coef'cients in some commutative>: 'eld k. If k* are the nonzero elements of k (with multiplication),>: you can verify that: (a b)>: (c d) --> ad-bc: does indeed give a surjective homomorphism det: G ---> k*.: This will exhibit N = ker(det) and G/N = k* commutative.All true, but this only shows that N contains G'; the hard part is the other >direction (problem 19, show that N = G'), which is not coming to me right now.Ted> Find a few classes of matrices in G'. For example, all rotation matrices (cos(t) sin(t)) (-sin(t) cos(t))for real t are in G'. So are upper triangular matrices with ones on the diagonal.Choose enough classes that you generate all of G' when they are merged.Then try to show that every member of your classes is in N.-- After California's recall election, wild'res Schwartz-en-ed the Bush-lands on its geographic right (when we wanted the forests to be Green). pmontgom@cwi.nl Home: San Rafael, California Microsoft Research and CWI = No. 18 is OK, I think. Any matrix of the form A B A^(-1) B^(-1) has> determinant 1, and thus so do all products of elements of this form.> Thus G' is contained in N. To show that G' actually equals N, I 'rst> thought of working out the form of a general element of G' and solving> the resulting equations, but these turned out to be rather horrible.> The next problem in the book is a simpler one of the same form, and I> was able to show you could actually get away with quite simple A and B> to get the general element. My attempts to do the same thing with this> one have been fruitless so far.G' is a normal subgroup of G = GL(2,R). If one could show that onenontrivial unipotent matrix A was in G' you would be done. By thisI mean a matrix A =/= I with (A-I)^2 = 0. These matricesform a single conjugacy class in G. If one is in G', they allare: in particular (1 x // 0 1) and (1 0// y 1) are in G' forall x and y, and these elementary matrices generate SL(2,R).I suppose it's just a question of playing with commutators until touget such a matrix.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) =:> Thus G' is contained in N.::> 19. In problem 18 show, in fact, that N = G'.: I assume, that your matrices have their coef'cients in some commutative: 'eld k. If k* are the nonzero elements of k (with multiplication),: you can verify that: (a b): (c d) --> ad-bc: does indeed give a surjective homomorphism det: G ---> k*.: This will exhibit N = ker(det) and G/N = k* commutative. >All true, but this only shows that N contains G'; the hard part is >the other direction (problem 19, show that N = G'), which is not >coming to me right now.Apply theorem: Let G be a group with commutator subgroup G'.(a) The subgroup G' is normal in G, and the factor group G/G' is abelian.(b) If N is any normal subgroup of G, then the factor group G/N is abelian if and only if G' is contained in N.-- G metabelian?Now from N = G' is it possible to show N is Abelian?In general, when G is invertible nxn matrices with *,then N = G' as shown above.Is it still true that N is Abelian?That is do rigid rotations commute?It seems so. How's it shown and does N = G' help?---- => :> 18. Let G be the group of all real 2-by-2 matrices> :> (a b)> :> (c d), with ad - bc non-zero, under matrix multiplication, and let N> :> be the subgroup consisting of those elements of G with ad - bc = 1.> :> Prove that N contains G', the commutator subgroup of G.> :> No. 18 is OK, I think. Any matrix of the form A B A^(-1) B^(-1) has> :> determinant 1, and thus so do all products of elements of this form.> :> Thus G' is contained in N.> :> 19. In problem 18 show, in fact, that N = G'.> : I assume, that your matrices have their coef'cients in some commutative> : 'eld k. If k* are the nonzero elements of k (with multiplication),> : you can verify that> : (a b)> : (c d) --> ad-bc> : does indeed give a surjective homomorphism det: G ---> k*.> : This will exhibit N = ker(det) and G/N = k* commutative.>All true, but this only shows that N contains G'; the hard part is>the other direction (problem 19, show that N = G'), which is not>coming to me right now. Apply theorem: Let G be a group with commutator subgroup G'.> (a) The subgroup G' is normal in G, and the factor group G/G' is abelian.> (b) If N is any normal subgroup of G, then the factor group G/N is> abelian if and only if G' is contained in N.I don't think I follow you here. We already know that G' is containedin N. The problem is the reverse inclusion, namely that N is containedin G'. -- G metabelian?> Now from N = G' is it possible to show N is Abelian?> In general, when G is invertible nxn matrices with *,> then N = G' as shown above. Is it still true that N is Abelian?> That is do rigid rotations commute?> It seems so. How's it shown and does N = G' help? ----Again, I think you've lost me here. Certainly N is not abelian in theoriginal question. For example, the matrices(3 1)(8 3)and(2 5)(1 3) do not commute.John Harrison =Consider a system of DE: 1/c dx_i/dt + x_i = sum_{jk} a^i_{jk}x_j*x_k, where a^i_{jk} are non-negative real numbers and a^i_{jk}=a^i_{kj} (i.e. the matrix a^i is symmetric). Given the initial values of x_i, can the system be analytically solved: a) in general case; b) under the condition sum_i a^i_{jk}=1; c) under the condition b) and sum_i x_i=1? =>Consider a system of DE:> 1/c dx_i/dt + x_i = sum_{jk} a^i_{jk}x_j*x_k, where a^i_{jk} are> non-negative real numbers and a^i_{jk}=a^i_{kj} (i.e. the matrix a^i> is symmetric).> Given the initial values of x_i, can the system be analytically> solved:> a) in general case;> b) under the condition sum_i a^i_{jk}=1; > c) under the condition b) and sum_i x_i=1?Probably not in general. Maybe for the case of a 2 x 2 system, althoughthe solution might be horrendously complicated. In case (b), ifX = sum_i x_i, you have 1/c dX/dt + X = X^2 which can be solved (andX=1 is a solution). So this effectively reduces the dimension of the system by 1.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 = > Action Device to generate unidirectional force. http://www.geocities.com/actiondevice Abhi, What is a solid angle? Can you give an example of a solid angle? -- Jeff, in Minneapolis . =Just for record.I went to Indian Institute of Technology (IIT), Powai, Mumbai toexplain mechanism of my Action Device and to seek technical help. Imet Dr. Amitay Issac of Aerospace Engineering Department and I triedto explain very basic component/idea of this action device. I havegiven in my homepage what exactly I tried to convince him.http://www.geocities.com/actiondeviceBut he insisted that point B will shift its position along Y axis!.I had to return in few minutes.Now I tried to convince again to Dr. G Arvind Rao of Aerospaceshift its position along Y axis !.Indian Institute of Technology is most prestigious college in India.This institute gives people for Aviation Industry around the world.And I just wonder, why so highly educated people fail to understandsuch simple problem.In fact, this is not problem at all. But what a tragedy, I am facingsuch ridiculous problems.I can end my all problems anytime, but I am following the rules ofthis battle, waiting game.I am just watching how the minds of highly educated people around theworld are controlled by that Supreme Force named God.-Abhi. => Just for record. I went to Indian Institute of Technology (IIT), Powai, Mumbai to> explain mechanism of my Action Device and to seek technical help. I> met Dr. Amitay Issac of Aerospace Engineering Department and I tried> to explain very basic component/idea of this action device. I have> given in my homepage what exactly I tried to convince him. http://www.geocities.com/actiondevice But he insisted that point B will shift its position along Y axis!.> I had to return in few minutes. Now I tried to convince again to Dr. G Arvind Rao of Aerospace> shift its position along Y axis !.Hmmm... did you consider that they could be right, and you could be wrong? Indian Institute of Technology is most prestigious college in India.> This institute gives people for Aviation Industry around the world.> And I just wonder, why so highly educated people fail to understand> such simple problem.Maybe, just maybe, they do understand it. In fact, this is not problem at all. But what a tragedy, I am facing> such ridiculous problems. I can end my all problems anytime, but I am following the rules of> this battle, waiting game.Build a working model and submit it to them for examination.Doesn't matter how much force it produces, as long as it proves that youridea works. I am just watching how the minds of highly educated people around the> world are controlled by that Supreme Force named God.Let me get this straight.... *God* doesn't want this device discovered? Whynot? And if not, what's stopping him from destroying you to make sure youstay quiet? => Just for record. I went to Indian Institute of Technology (IIT), Powai, Mumbai to> explain mechanism of my Action Device and to seek technical help. I> met Dr. Amitay Issac of Aerospace Engineering Department and I tried> to explain very basic component/idea of this action device. I have> given in my homepage what exactly I tried to convince him. http://www.geocities.com/actiondevice But he insisted that point B will shift its position along Y axis!.> I had to return in few minutes. Now I tried to convince again to Dr. G Arvind Rao of Aerospace> shift its position along Y axis !. Hmmm... did you consider that they could be right, and you could be wrong?Laura, where from you suddenly dropped in this mess? You just don'tknow, what is going on. I thought about this thousands of times inlast 13 months. I had posted idea of whole device in many newsgroup.This is just one of the basic component or idea behind this invention.At least this problem was not arised. And now suddenly this problempropped up.> Indian Institute of Technology is most prestigious college in India.> This institute gives people for Aviation Industry around the world.> And I just wonder, why so highly educated people fail to understand> such simple problem. Maybe, just maybe, they do understand it.Have you done elementary Geometry Laura? Take a look at my homepage.http://www.geocities.com/actiondevice> In fact, this is not problem at all. But what a tragedy, I am facing> such ridiculous problems. I can end my all problems anytime, but I am following the rules of> this battle, waiting game. Build a working model and submit it to them for examination.> Doesn't matter how much force it produces, as long as it proves that your> idea works.No, not yet. You just don't know what is going on around me. Thingsare under absolute control. You will never believe it.> I am just watching how the minds of highly educated people around the> world are controlled by that Supreme Force named God. Let me get this straight.... *God* doesn't want this device discovered? Why> not? And if not, what's stopping him from destroying you to make sure you> stay quiet?He does want this device to be discovered. This is exactly why Hecontrolled absolutely everything in my personal life. He navigatedthings in last 17 years in such a way that my thought process movesonly in one direction. He trained me to gain absolute power o'magination.This device is very simple. But there is no victory withoutThings are being controlled very cleverly. Don't believe me? People in this NG will not answer clearly the question I have posed.Will point B move along Y axis in XY plane? It needs just yes/no.But they will remain silent(or they will be humorous). They willignore me. Because they are controlled.Laura, Watch Out Apocalypse In Action....-Abhi. =Please, can someone help me with this (dif'cult) exercise ?Let u_k be a positive real sequence, such that the series sum( 1/u_k,k=1..in'nity) converges.Let T_n = u_1 + ... + u_n. Prove that the series sum (n/T_n,n=1..in'nity) converges and that sum (n/T_n, n=1..in'nity) <= 2 *sum( 1/u_k, k=1..in'nity). => Let u_k be a positive real sequence, such that the series sum( 1/u_k,> k=1..in'nity) converges.> Let T_n = u_1 + ... + u_n. Prove that the series sum (n/T_n,> n=1..in'nity) converges and that sum (n/T_n, n=1..in'nity) <= 2 *> sum( 1/u_k, k=1..in'nity).>D.8esol.8e je t'avais oubli.8e !! (Solution du Monier, 3.2.23):By Cauchy-Schwarz inequality applied to the vectors (sqrt(u_1), ...,sqrt(u_n)) and (1/sqrt(u_1),...,n/sqrt(u_n)),(1+2+...+n)^2 <= (u_1+...+u_n)*(1/u_1+2^2/(u_2)^2+...+n^2/(u_n)^2).It follows that: (2n+1)/(u_1+...+u_n) <= 4*(2n+1)/(n^2*(n+1)^2)*sum(k^2/u_k,k=1..n); summing for N>0,sum((2n+1)/(u_1+...+u_n), n=1..N) <= 4*sum((2n+1)/(n^2*(n+1)^2),n=1..N)*sum(k^2/u_k, k=1..n) = 4*sum(k^2/u_k*sum((2n+1)/(n^2*(n+1)^2),n=k..N), k=1..N) <= 4*sum(k^2/u_k*1/k^2, k=1..N)because:sum((2n+1)/(n^2*(n+1)^2), n=k..N) = sum(1/n^2-1/(n+1)^2, n=k..N) = 1/k^2 -1/(N+1)^2 <= 1/k^2,whence:sum((2n+1)/(u_1+..+u_n), n=1..N) <= 4*sum(1/u_k, k=1..N)<=4*sum(1/u_k,k=1..+oo).@+Julien Santini =God save Monier =I've written at least twice about this subject in the past, withoutreceiving any feedback. I'd be glad to read any kind of comment!>>Exponentials, and logarithms of invertible elements, exist in any Banach >>algebra. Look up holomorphic functional calculus.I should qualify that. The holomorphic functional calculus exists in>complex Banach algebras, while the usual quaternions are an algebra>over the reals. Of course there's no trouble complexifying to get an expand to someextent on the relationships between C (the complex *'eld*) and thethe *algebra* M(2,R) of 2x2 real matrices.Of course nothing of what I'm saying has a sound mathematical meaning,but IMO my observations yield a very natural point of view in somerespects, e.g. when dealing with some particular problems. (HoweverI'll give as a brief account as possible!)The point is that M(2,R) can be thought of (being isomorphic to) a2-dimensional complex algebra with a basis given by {1,chi}satisfying chi^2=1, ichi + chi i=0.On the other hand (the algebra isomorphic to) M(2,R) is a4-dimentional real algebra with a basis given by {1,i,chi,ichi}: inparticular these elements are not (the images of) the standard basisvectors of M(2,R).Note that in this sense the elements of M(2,R) are (numbers thatconstitute) another hypercomplex extension of C.Now, you can work abstractly with this extension of the ring ofcomplexes, and it is not important how you do intepret them. But ifyou want to have a direct expression of z+wchi (z,w in C) in terms ofsquare matrices, then a *possible choice* of i and chi for thetranslation is: i=[0 -1] chi=[1 0] [1 0], [0 -1].It's worth to notice that it is not a mere chance that the secondmatrix acts on a column vector as complex conjugation.Now, it is straightforward to realize that for A=z+wchi det(A)=|z|^2-|w|^2, A^{-1}=det(A)^{-1} (z*-wchi) if det(A)neq 0.(the latter identity works also if you abstractly *de'ne* det(A) asabove). Interestingly the operatorial norm of A is ||A||=|z|+|w|.Now, as an example, let's 'nd the solutions of the equation x^2=-1.Let x=a+bchi, a,b in C: the following two equations must besatis'ed: a^2+|b|^2=-1, 2Re(a)b=0.If (i) b=0 then a^2=-1 => a=+i or a=-i; if (ii) bneq 0, then a=ik, kin R, |b|^2=k^2-1 => |k|>1. By allowing |k|>=1 one can express all thesolutions including those found at point (i) as x=ik+sqrt(k^2-1)e^{itheta}chi,period!If do the similar calculation for x^2=1, then you 'nd that thesolutions found at the corresponding point (i) cannot be incorporatedin a general expression and one has x=1 or x=-1 or x=ih+sqrt(h^2+1)e^{iphi}chi, h in R.Another interesting exercise is to look for numbers/matrices I,X thatsatisfy the same identities as i,chi. (Since 1,-1 commute with everyelement of the algebra, then X must be chosen of the latter form!)Hope this was a TEASER!!Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc =Polynomials are well-known in science and mathematics, but while'nding roots of polynomials is typically the aim of the averageresearcher, polynomials themselves can be used as powerful tools foranalyzing the roots of *other* polynomials.The concepts are advanced, but can be approached by 'rst consideringa basic example.The basic factorization to start is(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1)with the c's algebraic integers, notice that only two of the c's have7 as a factor.It might help to go the *other* way, and start with (d_1 x + 1)(d_2 x + 1)( d_3 x + 1) = x^3 + 5x^2 + 3x + 1and now multiply by 49.In the 'rst example you're looking at a product and realizing thatfrom the distributive property a(b+c) = ab + ac, you know there's*one* way it could be produced, which is to multiply something likethe second example by 49.The distributive property is key here. Understanding it thoroughly,is of prime importance.Now notice that you can abstract from here as you're looking at*functions* of x, as introducingf_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, you have(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1).Notice that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1as long as you're in a ring where 7 is not a factor of 1.Which is consistent with what was found before, as only two of thefunctions have the property that 7 is a factor.Now I'll move on to a more complicated example.Let(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)so they are functions of x, and since one of the roots equals 3 atx=0, I haveb_3(x) = a_3(x) - 3, so that all the functions in(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)equal 0, when x=0.Those of you who 'nd it hard to use the distributive property withthe *product* can imagine the factorization from *before* 49 beingmultiplied.It's harder to show here as the polynomial which de'nes the functionin that factorization is not displayable in general.So I started at the end, with 49 already multiplied because then I cangivea^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).That slight change, starting at the end, means that you have tounderstand the distributive property fully and *trust* it.Now notice that I have the result that only two of the roots of thecubica^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)can have factors in common with 7, so the 49 splits between those two.What's so startling is that the result is for a *family* ofpolynomials as it applies for any algebraic integer x.James HarrisMy math discoveries, found for pro'thttp://mathforpro't.blogspot.com/ => Polynomials are well-known in science and mathematics, but while> 'nding roots of polynomials is typically the aim of the average> researcher, polynomials themselves can be used as powerful tools for> analyzing the roots of *other* polynomials. The concepts are advanced, but can be approached by 'rst considering> a basic example. The basic factorization to start is (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > 49(x^3 + 5x^2 + 3x + 1) with the c's algebraic integers, notice that only two of the c's have> 7 as a factor. It might help to go the *other* way, and start with (d_1 x + 1)(d_2 x + 1)( d_3 x + 1) = > x^3 + 5x^2 + 3x + 1 and now multiply by 49. In the 'rst example you're looking at a product and realizing that> from the distributive property a(b+c) = ab + ac, you know there's> *one* way it could be produced, which is to multiply something like> the second example by 49. The distributive property is key here. Understanding it thoroughly,> is of prime importance.> I like this example. Of course if you want to multiply (d_1 x + 1)(d_2 x + 1)( d_3 x + 1)by 49 so that you have 7 as the constant term in two ofthe factors, you are right, there is basically one way to do it (modulo permutations of d1, d2, and d3). However you can distributefactors of 49 through the three linear factors above in in'nitely many *other* ways, and the resulting polynomialis still the same. For example, (7^{4/3}*d1*x + 7^{4/3})*(7^{1/2}*d2*x + 7^{1/2})*(7^{1/6}*d3*x + 7^{1/6}).If you do this, you still get 49(x^3 + 5x^2 + 3x + 1)just as before. The thing is, there is no particular reason you needto get 7 as the constant term for two of the factors. In theexample you give, when x = 0, the constant term P(0) is 49,and of course 49 = 7^2 = 7^{12/6} = 7^{4/3} * 7^{1/2} * 7^{1/6},right? So the constant terms multiply together as they should. And this is only *one* way to use the distributive law to distribute the factors of 49 across the three factors. You caneven de'ne the factorization in three parts as a function of x if you want.> Now notice that you can abstract from here as you're looking at> *functions* of x, as introducing f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, you have (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). Notice that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 as long as you're in a ring where 7 is not a factor of 1. Which is consistent with what was found before, as only two of the> functions have the property that 7 is a factor. Now I'll move on to a more complicated example. Let (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) so they are functions of x, and since one of the roots equals 3 at> x=0, I have b_3(x) = a_3(x) - 3, so that all the functions in (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) equal 0, when x=0. Those of you who 'nd it hard to use the distributive property with> the *product* can imagine the factorization from *before* 49 being> multiplied. It's harder to show here as the polynomial which de'nes the function> in that factorization is not displayable in general. So I started at the end, with 49 already multiplied because then I can> give a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> Yes - note that now the a's are indeed functions of x. When youconsider 5*a1(x) + 7,and you want to factor a factor of 7 out of it, there are many ways that could be done, depending on the value of a1(x). When x = 0, a1(x) = 0. You know that a1(1), for example, is not 0. You argue, for no visiblereason, that a1(1) must be divisible by 7 because the constant termof the product is 7*7*22, and you think that if you divide 7*7 outof the whole expression, you cannot divide any piece of 7 outof the constant term 22 because 7 and 22 are relatively prime in the algebraic integers. You are indeed right that 7 and 22 are relatively prime. However, you don't *need* to have 22 itself divisible by any part of 7. The only thing you need to have divisible by some part of 7 is 5*b3 + 22.Now: you know that 22 is coprime to 7 and you know that 5 is coprimeto 7. Suppose b3 is also coprime to 7. Can I conclude from thatthat 5*b3 + 22 is also coprime to 7? Isn't it possible under the hypotheses just In fact, there is a decomposition of 7*7 of the form r*s*t suchthat: 1. 7*7 = r*s*t 2. 7/r and 7/s are algebraic integers 3. a_1/r and a2_/s are algebraic integers 4. (b_3*x + 22)/t is an algebraic integer. You may well shriek, WHAT ABOUT X ? SEE THAT X IN THERE?IT'S A VARIABLE! WHAT'S IT DIVISIBLE BY? Here's the key thing. The numbers r, s, and t, just like a_1,a_2, and b_3, are *** functions of x ***. That's exactly how it works.Divisibility of the constant terms is not important. What is important is that when you multiply everything out after youhave distributed the parts of 49 among the three factors, theproduct of the constant terms is still 22. Here's what you get: (7/r) * (7/2) *(22/t) = (7*7/(r*s*t))*22 = (49/(r*s*t))*22 = 22,just as it should. ***IT IS NOT IMPORTANT THAT 22/t IS NOT ANALGEBRAIC INTEGER***. WHAT IS IMPORTANT THAT (5*b_3 + 22)/t IS AN ALGEBRAIC INTEGER. The other key idea here is the functions-of-x idea. How 49splits into three parts depends on the value of x. You may say,How can r, s, and t know about x? They know about x becausethey are intimately connected to a_1, a_2, and b_3, and clearlythese coef'cients are functions of x because a_1 and a_2 anda_3 = b_3 + 3 are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0 See those x's in there? There is NO REASON to think that because, when x = 0,r, s, and t are respectively 7, 7, and 1, that the same is truefor all values of x. That is *not* a requirement which one candeduce from the fact that the constant terms are 7, 7, and 22.> That slight change, starting at the end, means that you have to> understand the distributive property fully and *trust* it.> No problem with the distributive law. The problem is, you seemto be able to imagine using it in only one way. However 7*7 can be split up as a product of three algebraic integers in in'nitely manyways, and thus distributed among the three factors of your polynomial inin'nitely many ways. Not all of those ways give algebraic integercoef'cients. As we have shown MANY TIMES, a_1 and a_2 cannot bedivisible, in the algebraic integers, by 7. Assuming they are leadsto a contradiction when x <> 0. Bottom line: for x <> 0, 49 does not split up the way you think itdoes. There is *another way* to split it up which does not resultin the contraction just mentioned. You appear to be making the mistakeof thinking that 22/t has to be an algebraic integer. > Now notice that I have the result that only two of the roots of the> cubic a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) can have factors in common with 7, so the 49 splits between those two.> No - you have seen the proof *many* times. No root of the polynomial above can be divisible by 7, and no root can be relatively prime to 7. If you assume so you get a contradiction.It's inescapable! Nora B.> What's so startling is that the result is for a *family* of> polynomials as it applies for any algebraic integer x. > James Harris My math discoveries, found for pro't> http://mathforpro't.blogspot.com/ => Polynomials are well-known in science and mathematics, but while> 'nding roots of polynomials is typically the aim of the average> researcher, polynomials themselves can be used as powerful tools for> analyzing the roots of *other* polynomials. The concepts are advanced, but can be approached by 'rst considering> a basic example. The basic factorization to start is (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > 49(x^3 + 5x^2 + 3x + 1) with the c's algebraic integers, notice that only two of the c's have> 7 as a factor. It might help to go the *other* way, and start with (d_1 x + 1)(d_2 x + 1)( d_3 x + 1) = > x^3 + 5x^2 + 3x + 1 and now multiply by 49. In the 'rst example you're looking at a product and realizing that> from the distributive property a(b+c) = ab + ac, you know there's> *one* way it could be produced, which is to multiply something like> the second example by 49. The distributive property is key here. Understanding it thoroughly,> is of prime importance. Now notice that you can abstract from here as you're looking at> *functions* of x, as introducing f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, you have (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). Notice that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 as long as you're in a ring where 7 is not a factor of 1.> This is the only way to divide both side by 49 if the f_i arelinear functions of x. If you use more complicated functionsall bets are off.> Which is consistent with what was found before, as only two of the> functions have the property that 7 is a factor. Now I'll move on to a more complicated example. Let (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)> More complicated functions. All bets are off. - William Hughes don't you try and get a life instead of repeatedly swamping thisnewsgroup with endless variations of the same material? Is this what youmeant when you said you were going to turn up the volume? If so, why notpost in all caps? Maybe that will prove the contents are correct.--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com =Are there any techniques that can ef'ciently do polynomial division? (I am looking for techniques similar to Karatsuba used formultiplication ...)Jaco = Are there any techniques that can ef'ciently do polynomial division? > (I am looking for techniques similar to Karatsuba used for> multiplication ...) JacoHave you checked Knuth? = Are there any techniques that can ef'ciently do polynomial division?> (I am looking for techniques similar to Karatsuba used for> multiplication ...)long divisionSuppose that g(x) is proposed as an approximation of f(x) on [a,b]. Whatare the most popular ways of measuring how well g approximates f over thatinterval?Here are several measures. In each case, the smaller the measure is, thebetter the approximation is considered to be.AInf. The maximum of |absolute error| over the interval, where absolute error = g(x) - f(x).RInf. The maximum of |relative error| over the interval, where relative error = (absolute error)/f(x).A2. The root-mean-square of |absolute error| over the interval.R2. The root-mean-square of |relative error| over the interval.A1. The average of |absolute error| over the interval.R1. The average of |relative error| over the interval.All of these measures may be thought of as power means (also called Hoeldermeans). They have the form* ( Integral( |error|^p ) / (b-a) ) ^ (1/p)where the integral is taken with respect to x from a to b, and error iseither absolute or relative. Obviously, in the cases of A1 and R1, p = 1,and in the cases of A2 and R2, p = 2. The value of p is not so obvious,however, in the cases of AInf and RInf. But in the limit as p increaseswithout bound, the power mean gives simply the maximum, as needed in AInfand RInf. As such, for those cases, we may say that p = +oo.Here are some questions of mine.Are there any important measures of error in form * which use values of pother than 1, 2, and +oo?Are there any important measures of error which are not in form * ?Clearly, using p = 2 yields a measure which is intermediate between thosewith p = 1 and p = +oo. In that sense, p =2 represents a nice compromise.But is there anything really special about p = 2 (say, as opposed to p = 4or p = 3/2) ? (Of course, I grant that the integral is typically far easierto evaluate analytically when p = 2 than when p = 4 or 3/2. But I'mwondering if p = 2 is special for a more fundamental reason than that.)David Cantrell => Suppose that g(x) is proposed as an approximation of f(x) on [a,b]. What> are the most popular ways of measuring how well g approximates f over that> interval? Here are several measures. In each case, the smaller the measure is, the> better the approximation is considered to be. AInf. The maximum of |absolute error| over the interval,> where absolute error = g(x) - f(x).> RInf. The maximum of |relative error| over the interval,> where relative error = (absolute error)/f(x). A2. The root-mean-square of |absolute error| over the interval.> R2. The root-mean-square of |relative error| over the interval. A1. The average of |absolute error| over the interval.> R1. The average of |relative error| over the interval. All of these measures may be thought of as power means (also called Hoelder> means). They have the form * ( Integral( |error|^p ) / (b-a) ) ^ (1/p) where the integral is taken with respect to x from a to b, and error is> either absolute or relative. Obviously, in the cases of A1 and R1, p = 1,> and in the cases of A2 and R2, p = 2. The value of p is not so obvious,> however, in the cases of AInf and RInf. But in the limit as p increases> without bound, the power mean gives simply the maximum, as needed in AInf> and RInf. As such, for those cases, we may say that p = +oo. > Here are some questions of mine. Are there any important measures of error in form * which use values of p> other than 1, 2, and +oo? Are there any important measures of error which are not in form * ?How about the geometric mean, A0 exp ( integral ( log |absolute error| ) / (b - a) ).-- => Here are some questions of mine. Are there any important measures of error in form * which use values of p> other than 1, 2, and +oo?I wouldn't dare try to answer this - I haven't learned enough yet (as ifdoing so is ever possible), but everything I do learn usually teaches methat I should have paid more attention to subjects I had previouslythought were unimportant. ;-) > Are there any important measures of error which are not in form * ?The error norms in higher order Sobolev spaces aren't quite in that form;but at least all the integral order spaces have norms in the form(Integral( Sum_i (|error_i|^p) ) ^ (1/p))which isn't much more general.> Clearly, using p = 2 yields a measure which is intermediate between> those with p = 1 and p = +oo. In that sense, p =2 represents a nice> compromise. But is there anything really special about p = 2 (say, as> opposed to p = 4 or p = 3/2) ?With any p, the absolute error norms are metrics on appropriately de'nedfunction spaces, so everything you can prove about normed vector spacesapplies to them.It's only for p = 2 that the metric comes from an inner product, however,upper bounds for 'nite element method errors in p=2 norms, for example.---Roy Stogner => Suppose that g(x) is proposed as an approximation of f(x) on [a,b].> What are the most popular ways of measuring how well g approximates f> over that interval?As pointed out by another responder, the answer depends on the application.For example, in time-dependent analysis, where x represents time, one mightwant to compare two time series. The comparison is made more complicated ifthe two signals look similar to the eye, but differ more in phase than inamplitude. For example, compare two sine waves of equal amplitude, butslightly different frequency. => Are there any important measures of error in form * which use values of p> other than 1, 2, and +oo?There are some harmonic means that, I think, involve expressions with p=-1.$.02 -Ron Shepard => Suppose that g(x) is proposed as an approximation of f(x) on [a,b]. What> are the most popular ways of measuring how well g approximates f over that> interval?I guess it depends on the 'eld of application, and also in theparticular application.In engineering, I believe the one used most frequently is themean-square error, since it is related to energy.> Here are some questions of mine. Are there any important measures of error in form * which use values of p> other than 1, 2, and +oo?I'm not familiar with any that are commonly used (I'm an electricalengineer -- maybe in other 'elds there might be)> Are there any important measures of error which are not in form * ?Not that I'm familiar with.> or p = 3/2) ? (Of course, I grant that the integral is typically far easier> to evaluate analytically when p = 2 than when p = 4 or 3/2. But I'm> wondering if p = 2 is special for a more fundamental reason than that.)Calculating the integral is usually irrelevant. What you want it'nd conditions that guarantee that the error is minimized, not'nding out what the error is.For instance, when you solve an overdetermined set of linearequations, you are calculating the optimal solution; you arenot calculating the error (though you know that it is minimum,and you could calculate it after you have the solution, if youwanted).HTH,Carlos-- => Maybe so, but in the mathematics community, truth takes precedence over> ego, which makes the 'eld very different from business, politics, and> many other human endeavors. Well, I think that the major difference is that in mathematics it is muchmore dif'cult to get away with bullshit, because of the nature of thediscipline. Hand waving can get you very in business, politics,philosophy, etc.; not so in mathematics. Now my opinion is that ego is just as strong in the mathematics communityas elsewhere, as can be seen in lots of heads of departments world over. X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft =>Sometimes a student would post an innocent question but he or she>would get scold for posting such a stupid question and would be>called a moron. What does that have to do with integrity? >It seems like these dese days, some mathematicians would use>profanity against others mathematicians.What does that have to do with integrity?>Where is the purity and integrity in mathematics?Neither purity nor integrity has anything to do with civility or withthe ability to suffer fools gladly.>Anyhow, I still love math by heart and I would like to give my>respect to many great math fanatics in this forum.You could best show your respect by not referring to them as fanatics.-- Shmuel (Seymour J.) Metz, SysProg and =|>Where is the purity and integrity in mathematics?||Neither purity nor integrity has anything to do with civility or with|the ability to suffer fools gladly.They are separate issues, but they are not unrelated. I 'nd thatthe people I've gotten acquainted with, who have a reputation fornot suffering fools gladly or who are proud of not doing so,nearly all can be seen erring on the side of prematurely concludingthat some idea or person was to be rejected. Incivility conduces tohostility, and hostility clouds the mind. On a crude level, it ispossible to be accurate about people who annoy one, but I 'nd ithard, and I think most people do too, to be very observant in sucha frame of mind. One feels a much greater temptation to slack offon efforts at describing these fools accurately, which is a lapsein integrity.For example, one mathematician I met had considered whetherit would work well to use an alternative treatment of a certainissue in functional analysis. He looked for anyone who mighthave addressed the issue, and found that of all the places helooked, it was solely in a textbook by a second mathematicianthat a reason was given why we should do it the usual way. Butthe 'rst mathematician had a counterargument, though, whichhe wanted to present to the second mathematician. The secondone (who gave me an impression of being usually conscientiousabout addressing mistakes in his work and so on) had thisdon't suffer fools gladly attitude, and when approached by the'rst, stormed angrily away declaring he had no time to discusssuch a thing. I realize that I'm expecting the reader to believe mewhen I indicate that this was an overreaction, although it's hardto demonstrate. It wasn't a matter of the second mathematicianbeing interrupted during something important and the like. Andthe discussion was not simply postponed to a more convenienttime.It is sometimes said that this kind of stern response is needed tokeep at bay the large volumes of nonsense that we are all exposedto. I think the value of such hostility is much overrated, however.Plenty of people are civil and relatively nonjudgemental withoutletting very much of their time be wasted on pointless pursuits.My opinion is that society in the U.S. is generally erring on theside of glibness, and that there are many involved in politics whoare keen to encourage us to err still farther in that direction.Rather than reasoning things out, we get discussions in whichpeople express shock and horror that the plain, obvious answeris not being accepted immediately. I see pundits trying to get menot only to agree with them, but to agree that it's so painfullyobvious that I needn't bother considering the other point of view,at least not seriously. This goes beyond incivility, but it's the kindof problem that incivility encourages.Keith Ramsay =Very well said. I agree.Have a tolerable existence. Eli. |>Where is the purity and integrity in mathematics?> |> |Neither purity nor integrity has anything to do with civility or with> |the ability to suffer fools gladly. They are separate issues, but they are not unrelated. I 'nd that> the people I've gotten acquainted with, who have a reputation for> not suffering fools gladly or who are proud of not doing so,> nearly all can be seen erring on the side of prematurely concluding> that some idea or person was to be rejected. Incivility conduces to> hostility, and hostility clouds the mind. On a crude level, it is> possible to be accurate about people who annoy one, but I 'nd it> hard, and I think most people do too, to be very observant in such> a frame of mind. One feels a much greater temptation to slack off> on efforts at describing these fools accurately, which is a lapse> in integrity. For example, one mathematician I met had considered whether> it would work well to use an alternative treatment of a certain> issue in functional analysis. He looked for anyone who might> have addressed the issue, and found that of all the places he> looked, it was solely in a textbook by a second mathematician> that a reason was given why we should do it the usual way. But> the 'rst mathematician had a counterargument, though, which> he wanted to present to the second mathematician. The second> one (who gave me an impression of being usually conscientious> about addressing mistakes in his work and so on) had this> don't suffer fools gladly attitude, and when approached by the> 'rst, stormed angrily away declaring he had no time to discuss> such a thing. I realize that I'm expecting the reader to believe me> when I indicate that this was an overreaction, although it's hard> to demonstrate. It wasn't a matter of the second mathematician> being interrupted during something important and the like. And> the discussion was not simply postponed to a more convenient> time. It is sometimes said that this kind of stern response is needed to> keep at bay the large volumes of nonsense that we are all exposed> to. I think the value of such hostility is much overrated, however.> Plenty of people are civil and relatively nonjudgemental without> letting very much of their time be wasted on pointless pursuits. My opinion is that society in the U.S. is generally erring on the> side of glibness, and that there are many involved in politics who> are keen to encourage us to err still farther in that direction.> Rather than reasoning things out, we get discussions in which> people express shock and horror that the plain, obvious answer> is not being accepted immediately. I see pundits trying to get me> not only to agree with them, but to agree that it's so painfully> obvious that I needn't bother considering the other point of view,> at least not seriously. This goes beyond incivility, but it's the kind> of problem that incivility encourages. Keith Ramsay =My research so far in using optic §ow for aerial robot navigation hasbeen mainly experimental (seehttp://www.pages.drexel.edu/~weg22/research.html if interested). However, I want to get more involved in the theoretical aspects of itby creating an optic §ow sensor numerical model. When the same inputis fed into my model and a sensor (www.centeye.com for example), theoutput should be the same. I have never developed any numericalmodels before and I was just wondering if anyone out there could offersome advice or suggestions on where to begin? Also, I wouldappreciate any information that can be shared about models that arealready in existence (possibly 1D).-Bill => Polynomials are well-known in science and mathematics, but while> 'nding roots of polynomials is typically the aim of the average> researcher, polynomials themselves can be used as powerful tools for> analyzing the roots of *other* polynomials. The concepts are advanced, but can be approached by 'rst considering> a basic example. The basic factorization to start is (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > 49(x^3 + 5x^2 + 3x + 1) with the c's algebraic integers, notice that only two of the c's have> 7 as a factor. It might help to go the *other* way, and start with (d_1 x + 1)(d_2 x + 1)( d_3 x + 1) = > x^3 + 5x^2 + 3x + 1 and now multiply by 49. In the 'rst example you're looking at a product and realizing that> from the distributive property a(b+c) = ab + ac, you know there's> *one* way it could be produced, which is to multiply something like> the second example by 49. The distributive property is key here. Understanding it thoroughly,> is of prime importance. Now notice that you can abstract from here as you're looking at> *functions* of x, as introducing f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, you have (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). Notice that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 as long as you're in a ring where 7 is not a factor of 1. Which is consistent with what was found before, as only two of the> functions have the property that 7 is a factor.> This part so far is OK.> Now I'll move on to a more complicated example. Let (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where the a's are roots of[***] a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) so they are functions of x, and since one of the roots equals 3 at> x=0, I have b_3(x) = a_3(x) - 3, so that all the functions in (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) equal 0, when x=0. Those of you who 'nd it hard to use the distributive property with> the *product* can imagine the factorization from *before* 49 being> multiplied.> As in the 'rst example you gave above, the 49 can bedistributed among the three factors in several differentways. There is no justi'cation for your implied claim below that two constant terms 7 should be error.> It's harder to show here as the polynomial which de'nes the function> in that factorization is not displayable in general. So I started at the end, with 49 already multiplied because then I can> give a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). That slight change, starting at the end, means that you have to> understand the distributive property fully and *trust* it. Now notice that I have the result that only two of the roots of the> cubic a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) can have factors in common with 7, so the 49 splits between those two.> When you were discussing the simpler polynomial above, yousaid that the coef'cients c_1, c_2, c_3 were algebraic integers,so presumably you are talking about the same thing here. Thatis, you are saying the a's are algebraic integers and two of themare divisible by 7 in the ring of algebraic integers. Thus assume a_1/7 is an algebraic integer; equivalently a_1 = 7*b_1,where b1 is an algebraic integer. But since you are saying thata_1 is a root of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) we must have that b_1 is a root of 7^3*b^3 + 3*(-1 + 49*x)*7^2*b^2 - 7^2*(2401*x^3 - 147*x^2 + 3*x) = 0.Now divide through by 7^2: 7*b^3 + 3*(-1 + 49*x)*b^2 - (2401*x^3 - 147*x^2 + 3*x) = 0.Finally, let x = 1: 7*b^3 + 144*b^2 - 2257 = 0.This polynomial is primitive, irreducible, and non-monic. Thereforenone of its roots can be algebraic integers. Therefore a_1/7 isnot an algebraic integer. Therefore a_1 is not divisible by 7. You continue to think that your proof that a_1 *is* divisible by7 is valid. If it were, it would imply a mathematical contradiction.Mathematics would be inconsistent. There is no sense in claimingthat the algebraic integers are incomplete. They are a perfectly well-de'ned subset of the real numbers. Something cannot both be inthat subset and not in that subset. Please explain your own conclusions on this. > What's so startling is that the result is for a *family* of> polynomials as it applies for any algebraic integer x.> For example, x = 1, which is what I used above. Your proof implies a contradiction, so it is either incorrect or math is inconsistent. I have identi'ed exactlyabove where you have made an unjusti'ed assumption. Your argument here is wrong. Nora B. James Harris My math discoveries, found for pro't> http://mathforpro't.blogspot.com/ = so that all the functions in (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) equal 0, when x=0.> We have been here many times. The problems is that a_1(x), a_2(x) andb_3(x) are not polynomials. Therefore, we do not know thatthe way in which the 49 distributes itself among thethree factors on the LHS is independent of x. Thus we cannotconclude that 7 divides (5 a_1(x)+ 7) for all x. Thus wecannot conclude that 7 divides a_1(x) for all x. -William Hughes =posted. I did comment on it already, but you did not answer. I thinkyou have not seen my response. So I will try again here. > Polynomials are well-known in science and mathematics, but while > 'nding roots of polynomials is typically the aim of the average > researcher, polynomials themselves can be used as powerful tools for > analyzing the roots of *other* polynomials.Oh, well, It appears you have modi'ed it a bit. Yes, polynomials arepowerful tools to analyse roots of other polynomials. Yup, thisparagraph was rewritten, as is the next. > The concepts are advanced, but can be approached by 'rst considering > a basic example. > The basic factorization to start is > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > 49(x^3 + 5x^2 + 3x + 1) > with the c's algebraic integers, notice that only two of the c's have > 7 as a factor.A comment I add this time and did not add in the last version. The c'scan *only* be algabraic integers because the constant factor of thepolynomial is 1. When that constant factor is 2, there is *no* suchdecomposition. > It might help to go the *other* way, and start with > (d_1 x + 1)(d_2 x + 1)( d_3 x + 1) = > x^3 + 5x^2 + 3x + 1 > and now multiply by 49.And again the same, newly added, comment: the d's are *only* algebraicintegers because the constant term of the cubic is 1. > In the 'rst example you're looking at a product and realizing that > from the distributive property a(b+c) = ab + ac, you know there's > *one* way it could be produced, which is to multiply something like > the second example by 49. > The distributive property is key here. Understanding it thoroughly, > is of prime importance.Hrm, an added paragraph. Does not add much information. > Now notice that you can abstract from here as you're looking at > *functions* of x, as introducing > f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, > you have > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). > Notice that dividing both sides by 49 gives > (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 > as long as you're in a ring where 7 is not a factor of 1.A repeat:This is independent of 7 being a factor of 1. Note however that itis *not* the only way to distribute 49 amongst the three factors onthe left hand side. This is the only way *only* if you require thatthe three factors on the left hand side are polynomials, if you havenot such an requirement it can be done differently. However, you canhave polynomials on the left hand side *only* because the roots of thepolynomial on the right hand side are units, i.e. divisors of 1. Itwill *not* work if that is not the case.An addition:If you do not have the requirement that the factors on the left handside are polynomials, the following factorisation is also possible(and gazillion others), and assuming x is integer: De'ne: w3(x) = gcd(f3(x) + 1, 7) { this can be 1 or some other divisor of 7. } w2(x) = 7 / w3(x).Now: [ (f1(x) + 7)/7 ][ (f2(x) + 7)/w2(x) ][ (f3(x) + 1)/w3(x) ]is a perfectly valid factorisation in the algebraic integer valuedfunctions on the integers of: x^3 + 5 x^2 + 3x + 1Note that the distributive property dictates: (a + b) * c = a * c + b * cin a particular ring. Not that (a + b) / c = a / c + b / cin a ring. That is, that when (a + b)/c is in a ring, there is norequirements that a/c and b/c are also in that ring. You keep onassuming that. > Which is consistent with what was found before, as only two of the > functions have the property that 7 is a factor.This is true *only* in some special cases... > Now I'll move on to a more complicated example. > Let > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so they are functions of x, and since one of the roots equals 3 at > x=0, I haveNote, that they are *not* polynomials. So the situation is differentfrom the one above. And indeed, in general *none* of the factors isdivisible by 7. Moreover, you can not even have the requirement thatwhen you divide by 49 the factors on the left must be polynomials,because you do not start with polynomials on the left. So there areother ways to distribute 49 amongst the three factors, however theway you distribute is a function of x.Remainder of repetition skipped.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ =Bullshit. Flush!Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ... converge?How does one prove convergence (or divergence)?. If it converges what is a good way to estimate its value?Jim Buddenhagen------------To reply copy jbuddenh@REMOVEtexas.net to address bar and edit out REMOVE => Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ... converge?> How does one prove convergence (or divergence)?. If it converges what > is a good way to estimate its value?> Recognize sin(x)/1 + sin(2x)/2 + sin(3x)/3 + sin(4x)/4 + ...as the Fourier series of a certain function, then evaluate it at x=1.Answer: (pi-1)/2 = 1.0708, approx.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ =James Buddenhagen says...Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ... converge?>How does one prove convergence (or divergence)?. If it converges what >is a good way to estimate its value?I'm not sure whether it converges, but if it does, I know what it convergesto 8^)Note that the in'nite series S = sin(1)/1 + sin(2)/2 + ...is the imaginary part of the series (using the relation exp(ix) = cos(x) + isin(x)). E = exp(i)/1 + exp(2i)/2 + ...This is a special case of the series x/1 + x^2/2 + ...(To see this, let x = exp(i))If this converges, it converges to the function log(1/(1-x)), so E = log(1/(1-exp(i)))To take the imaginary part, we need to rewrite1/(1-exp(i)) in the for A exp(iB) where A andB are real. To get it in this form, note 1/(1-exp(i)) = exp(-i/2)/(exp(-i/2) - exp(i/2)) = i exp(-i/2)/(2 sin(1/2)) = exp(i pi/2) exp(-i/2)/(2 sin(1/2)) = exp(i (pi/2 - 1/2))/(2 sin(1/2))Taking the log gives E = i (pi/2 - 1/2) - log(2 sin(1/2))Taking the imaginary part gives: S = (pi/2 - 1/2)--Daryl McCulloughIthaca, NY => Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ... converge?> How does one prove convergence (or divergence)?. If it converges what> is a good way to estimate its value?>Abel's rule =Julien Santini a .8ecrit dans le message de> Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ...converge?> How does one prove convergence (or divergence)?. If it converges what> is a good way to estimate its value?> Abel's ruleOK what about tan(n)/n ?Philippe => Julien Santini a .8ecrit dans le message de>> Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ...> converge?>> How does one prove convergence (or divergence)?. If it converges what>> is a good way to estimate its value?>> Abel's rule OK what about tan(n)/n ?The sequence tan(n)/n does not converge to 0; therefore, the seriestan(1)/1 + tan(2)/2 + tan(3)/3 + ... diverges.Jose Carlos Santos => Julien Santini a .8ecrit dans le message de> Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ...>> converge?> How does one prove convergence (or divergence)?. If it converges what> is a good way to estimate its value?> Abel's rule>> OK what about tan(n)/n ?The sequence tan(n)/n does not converge to 0; I imagine this is true, since Israel says he's proved it.But it's not all that obvious (it's not clear to me whetheryou were meaning to say it was obvious or not...)>therefore, the series>tan(1)/1 + tan(2)/2 + tan(3)/3 + ... diverges.>Jose Carlos SantosDavid C. Ullrich =>> OK what about tan(n)/n ?>The sequence tan(n)/n does not converge to 0; therefore, the series>tan(1)/1 + tan(2)/2 + tan(3)/3 + ... diverges.Correct. And for a proof that tan(n)/n does not converge to 0, you canA limit problem with explanation.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 => 1. I've posted a lot on sci.math over a long period of time, to your> knowledge, have I *ever* been right?Only on minor issues.> 2. Do I have *any* correct results, or do you think I just talk and> never say anything that is mathematically correct?Only on minor issues.> 3. To your knowledge, has ANYONE ever posted agreement with me on> *anything*?Only on minor issues.> 4. In your opinion, have I ever won an argument on the newsgroup?Only on minor issues.> 5. To your knowledge, have I ever caught other posters in errors?For the last time, only on minor issues. > James Harris => 1. I've posted a lot on sci.math over a long period of time, to your> knowledge, have I *ever* been right?Yes. 2. Do I have *any* correct results, or do you think I just talk and> never say anything that is mathematically correct?Yes ! 3. To your knowledge, has ANYONE ever posted agreement with me on> *anything*?Yes. 4. In your opinion, have I ever won an argument on the newsgroup?At least arguments of typographical errors. 5. To your knowledge, have I ever caught other posters in errors?At least typographical errors. You are welcome, We Pretty James Harris => 1. I've posted a lot on sci.math over a long period of time, to your> knowledge, have I *ever* been right?> Yes, but mostly not on the important things. You were right thatyou discovered a way to count primes that appears to be differentfrom other ways. > 2. Do I have *any* correct results, or do you think I just talk and> never say anything that is mathematically correct?> Mostly the same answer as above. You have been consistentlyand repeatedly wrong about all your signi'cant claims connected with Fermat's Last Theorem or a core error in mathematics. Howeveryou have found a different way to count primes. > 3. To your knowledge, has ANYONE ever posted agreement with me on> *anything*?> Yes.> 4. In your opinion, have I ever won an argument on the newsgroup?> Not a big one. A few little ones.> 5. To your knowledge, have I ever caught other posters in errors?> Yes. But far less often than you have been caught. Nora B. > James Harris = > 1. I've posted a lot on sci.math over a long period of time, to your > knowledge, have I *ever* been right?Yes. You have been right on occasion. > 2. Do I have *any* correct results, or do you think I just talk and > never say anything that is mathematically correct?Yes. You have been right on occasion. > 3. To your knowledge, has ANYONE ever posted agreement with me on > *anything*?Yes, agreement has been posted on occasion. > 4. In your opinion, have I ever won an argument on the newsgroup?No. > 5. To your knowledge, have I ever caught other posters in errors?I have no idea. Probably.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ = I'm having dif'culty proving that ||v||_p <= ||v||_2 for p>=2, here v> is a vector in R^n. Prove it when ||v||_p = 1 and then remember these are norms. =Let S(n) the decimal sum of n, this isS(17)=1+7=8,S(98)=9+8=17I have the following question: There are a, b and c, natural numbers, suchthatS(a+b)<5S(a+c)<5S(b+c)<5 andS(a+b+c)>50 ?I think that there are not.(Exuse-me I write english very bad)Pepe Bosch => Let S(n) the decimal sum of n, this is S(17)=1+7=8,> S(98)=9+8=17 I have the following question: There are a, b and c, natural numbers, such> that S(a+b)<5this is true then :a < 50 and b < 50> S(a+c)<5this also :a < 50 and c < 50> S(b+c)<5 andand this, of course :b < 50 and c < 50> S(a+b+c)>50 ?Uhm..if this statement were true, this would be true as well : a+b+c>599999 (as this is smallest number n for which S(n) > 50)And..that can't be true? I'm not sure if this is right since it's awfullysimple..perhaps your de'niton of decimal sum is different..-- Quaternion => Let S(n) the decimal sum of n, this is>> S(17)=1+7=8,>> S(98)=9+8=17>> I have the following question: There are a, b and c, natural numbers, such>> that>> S(a+b)<5this is true then :>a < 50 and b < 50No. Say a=1010, b= 200. Then a+b = 1210, and S(a+b) = 4 < 5.Write x = a_0 + 10a_1 + 10^2a_2 + ... + 10^n a_n y = b_0 + 10b_1 + 10^2b_2 + ... + 10^n b_n with 0<= a_i,b_i < 10, and at least one of a_n,b_n nonzero. So S(x) = a_0 + a_1 + ... + a_n S(y) = b_0 + b_1 + ... + b_nDe'ne d_0,....,d_n recursively as follows:d_0 = 0 if a_0+b_0 < 10d_0 = 1 if a_b+b_0 > 9.d_{i+1} = 0 if a_{i+1} + b_{i+1} + d_i < 10d_{i+1} = 1 if a_{i+1} + b_{i+1} + d_i > 9(The d_i are the carries)Then (x+y) + (a_0+b_0 - 10d_0) + 10(a_1+b_1+d_0-10d_1) + ... +10^n(a_n+b_n+d_{n-1}-10d_n) + 10^{n+1}d_n. Then S(x+y) = a_0+b_0 + a_1 + b_1 + ... + a_n+b_n + +(d_0+...+d_n)- 10(d_0+...+d_n) = S(x) + S(y) - 9(d_0+...+d_n).Now, you assume that S(a+b) <= 4 S(b+c) <= 4 S(a+c) <= 4Say we take 2(a+b+c) = (a+b) + (b+c) + (a+c).Now, let's consider the carries involved: Each of a+b, a+c, b+c has atmost 4 nonzero digits, and each nonzero digit is at most 4. How manycarries can there be? For there to be carries when we add a+b, a+c,and b+c together, there must be corresponding entries, at least one ofwhich is a 4, and the other 2 are either 4's or 3's. But that meansthat there is at most 1 carry. That is:S(2(a+b+c)) = S(a+b) + S(a+c) + S(b+c) orS(2(a+b+c)) = S(a+b) + S(a+c) + S(b+c) - 9.Now we want to relate S(2(a+b+c)) to S(a+b+c)Again, let x = a_0 + 10a_1 + 10^2a_2 + ... + 10^n a_nwith 0<= a_i <10, a_n>0So S(x) = a_0 +...+ a_n.De'ne e_0,....,e_n by lettinge_i = 0 if 0<=a_i <5e_i = 1 if 4>Let S(n) the decimal sum of n, this is>>S(17)=1+7=8,>>S(98)=9+8=17>>I have the following question: There are a, b and c, natural numbers, such>>that>>S(a+b)<5 > this is true then :> a < 50 and b < 50Surely not? What about a = b = 100? a+b = 200 and s(200) = 2 + 0 + 0 < 5. >>S(a+b+c)>50 ? > Uhm..if this statement were true, this would be true as well : a+b+c>599999 (as this is smallest number n for which S(n) > 50) And..that can't be true? I'm not sure if this is right since it's awfully> simple..perhaps your de'niton of decimal sum is different..> Doesn't work. You can get arbitrarily big values of a + b + c with S(a+b), etc < 5. I can't see whether or not you can make S(a+b+c) bigger than 50 or not, but will think about it. (Probably won't have too much success - number theory isn't my strength)David =>Let S(n) the decimal sum of n, this is>>S(17)=1+7=8,>S(98)=9+8=17>>I have the following question: There are a, b and c, natural numbers, such>that>>S(a+b)<5>> this is true then :>> a < 50 and b < 50Surely not? What about a = b = 100? a+b = 200 and s(200) = 2 + 0 + 0 < 5.>>S(a+b+c)>50 ?>> Uhm..if this statement were true, this would be true as well : >> a+b+c>599999 (as this is smallest number n for which S(n) > 50)>> And..that can't be true? I'm not sure if this is right since it's awfully>> simple..perhaps your de'niton of decimal sum is different..Doesn't work. You can get arbitrarily big values of a + b + c with >S(a+b), etc < 5. I can't see whether or not you can make S(a+b+c) bigger >than 50 or not, but will think about it. (Probably won't have too much >success - number theory isn't my strength)David> You can get at least 24:a = b = c = 5050505.a+b = a+c = b+c = 10101010a+b+c = 15151515S(a+b+c) = 24 An upper bound is 60.S(a + b + c) = S(10a + 10b + 10c) <= S(5a + 5b) + S(5a + 5c) + S(5b + 5c) <= 5S(a + b) + 5S(a + c) + 5S(b + c) <= 15*4 = 60. -- After California's recall election, wild'res Schwartz-en-ed the Bush-lands on its geographic right (when we wanted the forests to be Green). pmontgom@cwi.nl Home: San Rafael, California Microsoft Research and CWI =>Let S(n) the decimal sum of n, this is>>S(17)=1+7=8,>S(98)=9+8=17>>I have the following question: There are a, b and c, natural numbers,>such that>>S(a+b)<5>> this is true then :>> a < 50 and b < 50 Surely not? What about a = b = 100? a+b = 200 and s(200) = 2 + 0 + 0 < 5.Right.. of course!..I already thought there was somethign wrong.There's really nothing you can get for S(a) or S(b) out of S(a+b)<5.. So thesame counts for the relationship between S(a+b) and S(a+b+c)Sounds like a very complex problem then, unless you perhaps put a reasoningbehind ïcarries' when summating a+b with c and a with b+c (2*(a+b+c) =(a+b)+c + a+(b+c)), since the carries working in the sum can't have anincreasing effect on the decimal sum?.. So for the decimal sum to increasewhen adding two numbers, you'd need to have values that 'll each other upand don't cause carries..which eerily revolves around the boundary of 5.Well I'll stay out of this..stuff for the big boys-- Quaternion =I think you are right but I can't demonstrate thatPepe Bosch ha scritto nel messaggio> Let S(n) the decimal sum of n, this is S(17)=1+7=8,> S(98)=9+8=17 I have the following question: There are a, b and c, natural numbers, such> that S(a+b)<5> S(a+c)<5> S(b+c)<5 and> S(a+b+c)>50 ? I think that there are not. (Exuse-me I write english very bad) Pepe Bosch =>What experimental evidence?Transverse Doppler effect; Relativistic corrections to the spectrumof the Hydrogen atom; all experimental evidence for QED, which con§atespermanently settling the issue; relativistic momentum and energyhalf life of 15 minutes (neutrons) to travel light years for thousands ofyears across the cosmos to reach Earth; Michelson-Morley experiment;the constitutive relations D = epsilon_0 E, B = mu_0 H.search.yahoo.com/search?p=Evidence For Relativity>Moving clocks running slow? They don't.Directly observed to do so, in fact.>The GPS clocks run fast.... precisely as predicted by Relativity, and in the very amount predictedto do so. =>assumption, and is intuitive as well. Making the assumption that the time it>takes for a signal to reach an object is the same as the time it take for>the signal to return, when in the meantime [sic] you've moved away or>toward the object [...]Motion is relative. You were standing still. It was the object thatwas moving. So, there. => I provided proof for my claim. But of course you *conveniently*> snipped the link. That proof consists of fairly straightforward> reasoning.> LOL!!> I provided the proof that Einstein makes assumption, AFTER you denied it,> and called me a liar.You did no such thing.You are a disgusting lying pig.> Then YOU snipped my response, windbag.I did no such thing.You are a disgusting lying pig.> Not only that,> but you gave no proof at all.I did.You are a disgusting lying pig.> Assertion isn't proof, even if you think it is. I never claimed it were.You, however, have provided nothing but assertion for the subject line.> The subject line remains true, whether you like it or not, Ostrich.The subject line has been solidly disproven.You are a disgusting lying pig. Henri. I've been debating with him for years, and he's just one of> those relativists that refuse to see what is right under their nose. It's a> psychological hang-up that most people have. The don't want to admit they've> been made a fool of by Einstein, they have to appear in the eyes of others> as the one who has all the answers, but that's provided the question suits> them, of course. When it doesn't, they ignore it.> 1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v) )] = tau(x',0,0,t+x'/(c-v))> ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/> Where does he get that 1/2 from? I send you a letter on the 10th. I get a reply from you on the> 16th. So I assume you received the letter and replied to it on> the 13th:> 1/2( 10 + 16) = 13 Right... but perhaps still a little too far out. > I will try to improve the analogy, who knows, it may just make the> difference: Without mentioning it, Dirk also calculated that the total return time> was 16-10 =6 days.> Therefore, if we don't know anything about the speed of either> trajectory, we have no other option but to assume - or declare by> convention - that the one way delivery time is 1/2 * 6 = 3 days. You may have noticed that at that time Einstein was careful to declare> the total average return speed to be constant, in full agreement with> measurements and LET. No assumption there!> The later SRT assumption that the one-way speed is truly c in one's> own frame of reference, can not be proved if the Lorentz equations are> correct. ;-) Haraldxein: How simple. Bravo. When was the last OWLS-TWLS discussion? Was it decided by vote or physics? <8pr7rvojo5acmso735p31bo58d7g0rq43v@4ax.com> <4C0tb.4282$_g6.527@news-binary.blueyonder.co.uk>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft said:Duh! So is any scienti'c theory. But it has nothing to do withMathematics. Please take it to sci.physics.relativity.crackpots.Oh, BTW, welcome to my 'lters.*PLONK*-- Shmuel (Seymour J.) Metz, => All theories, in physics and elsewhere, are based on assumptions. No theory can bootstrap itself out of nothing. So what?> ïSo what' depends on the validity of the assumption. The PoR is an> assumption, and is intuitive as well. Making the assumption that the time it> takes for a signal to reach an object is the same as the time it take for> the signal to return, when in the meantime you've moved away or toward the> object, is a rather silly assumption that I will not accept. That's ïso> what'.> Androcles Then ignore relativity. But unless you can come up with something that agrees with the experimental evidence better than relativity, everyone else will ignore you. = All theories, in physics and elsewhere, are based on assumptions. No theory can bootstrap itself out of nothing. So what?> ïSo what' depends on the validity of the assumption. The PoR is an> assumption, and is intuitive as well. Making the assumption that thetime it> takes for a signal to reach an object is the same as the time it takefor> the signal to return, when in the meantime you've moved away or towardthe> object, is a rather silly assumption that I will not accept. That's ïso> what'.> Androcles Then ignore relativity. But unless you can come up with something that agrees with the> experimental evidence better than relativity, everyone else will> ignore you.What experimental evidence? Moving clocks running slow? They don't. The GPSclocks run fast. MMX? It's pretty obvious to anyone with half a brain thatthe result is what you would expect if the speed of light were sourcedependent. I would say that agreed with the experimental evidence muchbetter than relativity.Androcles = All theories, in physics and elsewhere, are based on assumptions. No theory can bootstrap itself out of nothing. So what?> ïSo what' depends on the validity of the assumption. The PoR is an> assumption, and is intuitive as well. Making the assumption that the> time it> takes for a signal to reach an object is the same as the time it take> for> the signal to return, when in the meantime you've moved away or toward> the> object, is a rather silly assumption that I will not accept. That's 'so> what'.> Androcles Then ignore relativity. But unless you can come up with something that agrees with the> experimental evidence better than relativity, everyone else will> ignore you.> What experimental evidence? Moving clocks running slow? They don't. The GPS> clocks run fast. MMX? It's pretty obvious to anyone with half a brain that> the result is what you would expect if the speed of light were source> dependent. I would say that agreed with the experimental evidence much> better than relativity.> AndroclesAndrocles1. Are you saying that the speed of light is source dependent? 2. What is your de'nition or explanation of ïsource dependency'?Peter Riedt = All theories, in physics and elsewhere, are based on assumptions. No theory can bootstrap itself out of nothing. So what?> ïSo what' depends on the validity of the assumption. The PoR is an> assumption, and is intuitive as well. Making the assumption that the> time it> takes for a signal to reach an object is the same as the time ittake> for> the signal to return, when in the meantime you've moved away ortoward> the> object, is a rather silly assumption that I will not accept. That's'so> what'.> Androcles Then ignore relativity. But unless you can come up with something that agrees with the> experimental evidence better than relativity, everyone else will> ignore you.> What experimental evidence? Moving clocks running slow? They don't. TheGPS> clocks run fast. MMX? It's pretty obvious to anyone with half a brainthat> the result is what you would expect if the speed of light were source> dependent. I would say that agreed with the experimental evidence much> better than relativity.> Androcles Androcles> 1. Are you saying that the speed of light is source dependent?Yes.> 2. What is your de'nition or explanation of ïsource dependency'? Peter RiedtJust throw a ball forward from your car window as you move.It should be obvious, and no different from throwing a photon from a star.The velocity of light in interstellar space is c, with respect to the star.If the star moves, then it still c with respect to the star. To an observerthe velocity is c+v, where v is the velocity of the star.Now create a mathematical model of a star in elliptical orbit accordingto Kepler's laws, determine when the light will arrive, calculate thevariation in brightness, calculate a spectrum, and see if it matchesempirical data. If it does, then Einstein's second postulate hasno foundation. A match to empirical data has been found.There are three models to consider.1) Newton-Galileo - photons and PoR, SoL is source dependent.2) Maxwell-Lorentz - wave and PoR, SoL is aether dependent3) Einstein - photon or wave, SoL is observer dependent.I don't know of any sensible fourth choice.MMX thows out 2), but cannot differentiate between 1) and 3).Therefore 1) should be given due consideration, someone has to do it,and that someone is me.Androcleshttp://www.androc1es.pwp.blueyonder.co.uk/ index.html => 1. Are you saying that the speed of light is source dependent? Yes. 2. What is your de'nition or explanation of ïsource dependency'?> Peter Riedt> Just throw a ball forward from your car window as you move.> It should be obvious, and no different from throwing a photon from a star.> The velocity of light in interstellar space is c, with respect to the star.> If the star moves, then it still c with respect to the star. To an observer> the velocity is c+v, where v is the velocity of the star.And for waves ... ? => Ray Steiner> I came up with another way of showing that x/tan x has no elementary> antiderivative.> We need only one result from Wiener's 1997 paper:> arcsin(x)/x does not have an elementary antiderivative. Let I = int(x/tan x dx)= int (x cot x dx)> Use integration by parts to get> I = x ln(sin x) - int( ln(sin x) dx)> Let I2= int( ln(sin x) dx) Let u= sin x, x = arcsin(u), dx = 1/sqrt(1-u^2) du> Then> I2= int ( ln(u)/sqrt(1-u^2) du) Finally, use parts again to get> I2= ln(u) arcsin(u) - int(arcsin(u)/u du).> So, by Wiener's result, the original integral is not elementary. More results:> By exactly the same method one can show that> I3 = int (x tan x dx) is not elementary.> Now, let's substitue u= tan x, x = arctan u, dx= 1/ (u^2 + 1) du in I3.> Then it reduces to> I4 = int( u*arctan(u)/(1+u^2) du).> so the second integral of my previous post is non-elementary.> Finally, consider> I5= int ( (arctan(x))^2 dx).> By parts, one can reduce it to integrating I4, so I5 is also> non-elementary.> LHAfter I posted this, I found an even simpler way of doing it.Again We need the result from Wiener's 1997 paper:arcsin(x)/x does not have an elementary antiderivative.In int(arcsin(x)/x dx) substitute x= sin u, dx= cos u du; arcsin x= uThen it becomesint( u cos u/sin u du) which is int(u/tan u).If the latter were elementary thenint(arcsin(x)/x dx) would be elementary, which is not the case.So the result follows easily.BTW, if we substitute x = cos u in the same integral, we also'nd that int(x tan x dx) is not elementary.Ray Steiner = Just as the > answer to a mathematical question is either right or right, I'd like a few of those questions!Some of us learned three-valued logic in high school. There are threeways to solve a problem: the right way, the wrong way, and the way youwere told to do it (which may have nothing to do with right or wrong).David Ames =>> Am I too dumb for math?>De'ne dumb. Math is a tool, and for some a pleasure>in its own right Ahem. It's not polite to talk about matherbation in public.... except among people of the same persuasion.David Ames =I have taught math for many years, both in a classroom setting and oneon one, and more often than not, the main problem lies in students notunderstanding the basic ideas behind the formulas, rather than justmemorizing the formulas. For many students, there is the mistakenbelief that knowing the formulas is enough.> [added sci.edu, comp.edu] >> I dont get it.Im a perfect 0 at Am I too dumb for math?You're asking the wrong question -- you're not>dumb for math; you *might* have a brain/mind that>works in a way that is not compatible with the way>of thinking that you need for understanding maths>(emphasis on the *might* -- you claim to be dumb>for maths, but who knows, maybe you're much better>newsgroup and just don't know it, or maybe you're a>high-expectation kind of person, and then anything>below Newton, or Gauss, or Fourier's brains means>too dumb for math in your mind? :-)) I've come across various students who viewed that they were missing the> mathematics gene (or programming gene, or whatever the particular> subject happened to be). In those cases it was uniformly the case that> their dif'culty was emotional/attitudinal, rather than cognitive. As you> mention, one unhelpful attitude is perfectionism, especially in hard-edged> subjects where some answers are clearly objectively *wrong* and thus the> student has no wiggle room to avoid the conclusion that they made an> error.Anyway, this, plus many of the things that have>been already said (mainly about math being genuinely>hard -- the more sophisticated level of math, the>harder, of course) Several of the missing gene students had the unhelpful attitude that> they expected maths to be easy, since they had found their schooling easy> so far. ISTM that cognitive issues do kick in when dealing with high levels of> abstraction where there are no readily-accessible concrete models. For> example, my brain hit the wall trying to visualise non-Hausdorff spaces,> and my painful memory of the rest of that topology course is of generally> mindless memorizing and proof cranking. >HTH,Carlos =>I have taught math for many years, both in a classroom setting and one>on one, and more often than not, the main problem lies in students not>understanding the basic ideas behind the formulas, rather than just>memorizing the formulas. For many students, there is the mistaken>belief that knowing the formulas is enough.This is not surprising, considering that this is essentiallyhow everything is taught in the elementary and high schools,and also in the courses through calculus.As just about all textbooks take that approach, and the greatbulk of teachers believe in teaching that way, it is hard tosee what can be done. We are NOT going to be able to teachthe teachers; this failed for the new math, and it was veryde'nitely tried.It is not even learning the basic ideas behind the formulas;this might get through to a percentage of the teachers. Itis knowing the properties and the concepts of the mathematicalobjects used, and these are not going to be learned by themistaken methods of the educationists.We can teach variables in the general sense (do NOT limit them to numbers) as linguistic entities with beginning reading. We can teach sound mathematical logic to at leasthalf of those in elementary school; it has been done. But we cannot succeed if learning mathematics is measuredby the multiple choice computational stuff on the exams nowbeing mandated. Nobody learns what multiplication means bymemorizing the tables, and one can understand what it meanswithout having any facility in computing answers.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University =Does anyone know an approximation for Stirling Numbers of the Secondkind, S(n,k), for very large values of n?S(n,1) = S(n,n) = 1 are easy, as are S(n,2) and S(n,n-1). It's theintermediate values of k that are dif'cult. I can't use thealternating series because of =>Does anyone know an approximation for Stirling Numbers of the Second>kind, S(n,k), for very large values of n?Graham, Knuth and Patashnik Concrete Mathematics refers to David and Barton, Combinatorial Chance, Hafner 1962, chap. 16, for asymptoticsof Stirling numbers.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =What will be a good reference to understand topological properties of L^p spaces ? =>says...>2. In your experience, is math quirky?>De'ne quirky. > Idiosyncratic. >>Sometimes the results one sees are not what one would intuitively think were the>>case but are nonetheless correct. Is that what you mean by quirky? > No, I guess I mean, like where sometimes there are rules that follow> in one instance, that you aren't sure still apply in another given> instance, and at times it can be so dif'cult to 'gure out if that> happens that you just have to rely on experts *telling* you that the> rule no longer applies.> By this de'nition, math is *not* quirky.-- Will Twentyman = > Well, well, well! If it ain't (Suk My) Dik! What's up, dude? > ... > My, oh my. You make a joke on my name, and in response I make one on your > The relevance is Jesse Hughis inability to distinguish between word > play and an insult--a distinction you were (apparently) able to make.Well, I agree with Jesse. You posted a wordplay that was intended tobe insulting (and childishly so), see the surrounding wording. I sawit as an insult. That I did not retaliate in kind was deliberate. Irarely insult people on Usenet. I think I have done that about two timesin 20 years of posting history.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ =Jack, you ask:... Tony what do you mean by D5 describing Gravity?What that would mean to me is that start from the D5 groupand end up with Einstein's local 'eld equationGuv = (superstring tension)^-1Tuvcan you actually do that?Similarly, start from D4 and get the U(1)xSU(2)xSU(3) principal 'berbundle of theelectroweak-strong gauge forces with the associated vector bundle ofthe lepto-quarksources. Can you do that? ....Yes, to both questions (although I don't use the conventionalsuperstring structure that you mention).I am trying to see if there is a precise mathematical connection between what you are doing andwhat I am doing in http://qedcorp.com/APS/EmergentGravity.pdfIn my theory Einstein's gravity 'eld plus exotic vacuum w = -1 dark energy/matter 'elds are all ODLRO c-number collective emergent low energy effective MACRO-QUANTUM 'elds in which Diff(4) is an emergent symmetry from the spontaneous breaking of the U(1) EM symmetry at the unstable false vacuum micro-quantum level of the lepto-quark sources/electroweak-strong gauge forces level.More speci'cally, I get an emergent c-number LOCAL giant MACRO-QUANTUM VACUUM COHERENT WAVEWorld Crystal Lattice Distortion Field = Lp*^2(Goldstone Phase Field),uEinstein's guv 'eld is the §at Minkowski metric + the Strain Tensor of the Distortion FieldTherefore, Einstein's Gravity Field emerges as modulation of the Goldstone Phase Field consistent with Andrei Sakharov's metric elasticity, which is the complementary view of P.W. Anderson's generalized phase rigidity in his More is different paradigm. That is, the basic emergent gravity coupling is precisely Ed Witten'salpha' = 1/(string tension) = 8piG*/c^4where I allow G* to be a scale-dependent variable that is 10^40 G(Newton) at the 1 fermi scale.Uni'ed Exotic Vacuum Dark Energy/Matter Field is from the modulation of lepto-quarks asmc^2 ~ e^2/zpf^1/2where /zpf ~ -1/(1 fermi)^2 at the 1 fermi scale in the vibrating dark matter quantized vortex string core where theWhere with h = c = 1 conventionalpha' is a scale-dependent variable not 'xed at 10^-66 cm^2.BTW I have come to the tentative conclusion that the alleged Holographic Universe formulaLp* = Lp^2/3L^1/3found in the LNL e-prints has serious problems of interpretation.My basic theory does not essentially depend on that additional assumption.While I see no basic problem in you getting the micro-quantum U(1)xSU(2)xSU(3) from your group theory, I donot understand how you get Einstein's Gravity. The Ed Witten argument that one gets a spin 2 quantum is not good enough for me since in the More is different emergence the consensus quantum gravity idea is not correct at all and that is what non-renormalizability of Einstein's GR in the low-energy sector is telling us.There may be some linear spin 2 perturbative random gravitons as micro-quantum noise or normal §uid in the emergent curved spacetime super§uid background from the modulation of the c-number Goldstone phase.So I am asking where is the MACRO-QUANTUM EMERGENCE in your group theory approach to deriving Einstein's Gravity from a more fundamental level?Tony:Those things (the forces of Gravity and the Standard Model) allcome from the 28-dim adjoint rep of the D4 subalgebra of D5.More particularly:D4 has a 15-dimensional D3 subalgebra that is the conformalalgebra SU(2,2) = Spin(2,4). It has:OK4 special conformal generatorsWhat do they locally gauge to?My hunch is /zpf,u10 anti-deSitter generators.The 4 Pu of T4 locally gauge to Einstein's guv.The 6 Muv of O(1,3) locally gauge to a Torsion Field.By a modi'ed conformal MacDowell-Mansouri mechanism,Is this where ODLRO MACRO-QUANTUM EMERGENCE is buried?All this is at conventional textbook level, for example,section 14.6 of Uni'cation and Supersymmetry, 2nd edition,by Rabindra Mohapatra, Springer-Verlag 1992.If you want a prominent establishment name dropped, Frank Wilczekmentions the MacDowell-Mansouri mechanism inhttp://xxx.lanl.gov/abs/hep-th/9801184where he notes that the mechanism was also independentlyformulated by Chamseddine and West.The mechanism was invented to make it possible to get gravityfrom the anti-deSitter part of Lie superalgebras used insupergravity theories.As to the Standard Model,look at the 28 generators of the D4 Lie algebra.Use 15 of them as above, and 1 moreto complete the SU(2,2) to U(2,2) = SU(2,2)xU(1).Then you have 12 generators left.The form the 12-dim Standard Model SU(3) x SU(2) x U(1).Here is how you can see that structure geometrically,and unambiguously:If you look at things (here I was inspired by Saul-Paul) interms of the Weyl re§ection group of the root vector space,and take away the 12 root vectors of the U(2,2) and the 4 Cartanalgebra elements of the 16-dim rank 4 U(2,2), that leaves youwith 28-12-4 = 12 root vectors.Since the root vectors of D4 form a 24-cell, which can be seenas a 12-vertex cuboctahedron plus two 6-vertex octahedra,you see that the remaining 12 root vectors form a pair of octahedra.Line the two octahedra up so that they share a common axis,and project the two octahedra into a space perpendicular to that axis,so that the 4 axis vertices fall on a line that forms theroot vector diagram (including Cartan origin vertices) ofthe 4-dim Lie algebra U(2) = SU(2) x U(1).The remaining 8 can be seen as the vertices of a cube: tb----xb | | | zb----yb | | | | yr-|--zr | | | xr----trNow look at the cube along its tb-tr diagonal axis,and project all 8 vertices onto a plane perpendicularto the tb-tr axis, giving the diagram yb xb zb tb tr zr xr yrwith two central points surrounded by two interpenetrating triangles,which is the root vector diagram of SU(3).Therefore,conformal,andthe remaining 12 give us the Standard Model SU(3) x SU(2) x U(1).Consideration of the relevant geometries and combinatorics givelevel calculations, so such higher order things as neutrino masses(they are in my model tree-level massless) remain to be fully calculated.Details are in my papers, including a paper athttp://www.innerx.net/personal/tsmith/TQ3mHFII1vNFadd97. pdfthat contains material barred from the Cornell arXiv due tome being blacklisted by them.Tony =By a modi'ed conformal MacDowell-Mansouri mechanism,I asked:Is this where ODLRO MACRO-QUANTUM EMERGENCE is buried?All this is at conventional textbook level, for example,section 14.6 of Uni'cation and Supersymmetry, 2nd edition,by Rabindra Mohapatra, Springer-Verlag 1992.If you want a prominent establishment name dropped, Frank Wilczekmentions the MacDowell-Mansouri mechanism paper seems very relevant in accord with what I amdoing in a simpler way independently. Wilczek is one of the besttheoretical physicists around for sure. I heard him speak twice nowthis year.where he notes that the mechanism was also independentlyformulated by Chamseddine and West.The mechanism was invented to make it possible to get gravityfrom the anti-deSitter part of Lie superalgebras used insupergravity theories. =Commentary 3The hyperspace H consists of 'bers f(x) that areeither copies of or representations of the symmetrygroup G.Jack, this is not quite correct. They are homogenous spaces onwhich the group operate transitively. Example, for the group SU(2),you can take as the 'bre a copy of SU(2) itself (3-dimensional), oryou can take sphere S^2, on which SU(2) operate (2-dimensional).Notice that S^2 is not a representation of SU(2). It is a quotientSU(2)/SO(2).Early Kaluza-Klein theories were operating with group Manifolds.Souriau, and later Witten, suggested more realistic theories where'bers could be of lesser dimensions. Thie rigorous mathematics andexamples of this latter approach have been developed in themonograph:Riemannian Geometry, Fibre Bundles, Kaluza-Klein Theories and AllThat... (World Scienti'c Lecture Notes in Physics, Vol 16)by Robert Coquereaux, Arkadiusz Jadczyk :-)arkPrincipal 'ber bundles for the local gauge forces:1. A transformation g of the symmetry group G acts on the ordered pair X = (x, fo) in hyperspace H with output gX.Question: Can gx = x' =/=x i.e. can one move the base point in this operation or must G always be the identity in the base space? That is, we always need, in addition to G a connection and a path in order to change location in the horizontal base space and the vertical 'ber space that is beyond space-time. G certainly moves fo up and down the vertical 'ber for every element g =/= identity. Does it also move x -> x' = gx =/= x horizontally along the base manifold without a connection 'eld and a path speci'ed? Clearly the answer must be NO. See below.The modern understanding of gauge invariance, as a symmetry under transformations ofquantum-mechanical wave functions, was reached by Weyl himself and also by London veryshortly after the new quantum mechanics was 'rst proposed. In this understanding ofabelian gauge invariance, and in its nonabelian generalization [2], the space-time aspect islost. The gauge transformations act only on internal variables. This formulation has hadgreat practical success. Still, it is not entirely satisfactory to have two closely related, yetde'nitely distinct, fundamental principles, and several physicists have proposed ways tounite them.One line of thought, beginning with Kaluza [3] and Klein [4], seeks to submerge gaugesymmetry into general covariance. Its leading idea is that gauge symmetry arises as a re§ec-tion in the four familiar macroscopic space-time dimensions of general covariance in a largernumber of dimensions, several of which are postulated to be small, presumably for dynam-ical reasons.Here we should take the opportunity to emphasize a point that is somewhatconfused by the historically standard usages, but which it is vital to have clear for whatfollows. When physicists refer to general covariance, they usually mean the form-invarianceof physical laws under coordinate transformations following the usual laws of tensor calculus,including the transformation of a given, preferred metric tensor. Without a metric tensor,one cannot form an action principle in the normal way, nor in particular formulate the ac-cepted fundamental laws of physics, viz. general relativity and the a purely mathematical point of view one might consider doing without the metric tensor;in that case general covariance becomes essentially the same concept as topological invari-ance. The existence of a metric tensor reduces the genuine symmetry to a much smaller one,in which space-times are required not merely to be topologically the same, but congruent(isometric), in order to be considered equivalent. In the Kaluza-Klein construction, for thisreason, the gauge symmetries arise only from isometries of the compacti'ed dimensions.Another line of thought proceeds in the opposite direction, seeking to realize generalcovariance [CapitalEth] in the metric sense [CapitalEth] as a gauge symmetry. arXiv:hep-th/9801184 v4 23 Apr 1998IASSNS-HEP-97/142Riemann-Einstein Structure from Volume and Gauge SymmetryFrank WilczekBTW Wilczek shows that Gennady Shipov's torsion theory is closely related toRoger Penrose's spinors in curved spacetime with the anti-symmetricspin connection as the locally induced compensating torsion 'eld.It all comes from locally gauging the O(3,1) subgroup of the Conformal Groupas I said previously based on Utiyama's and Kibble's papers from the mid-1960's.Whether or not Akimov's claims from Moscow that torsion waves from O(1,3) ofsuf'cient intensity to have psychotronic weapons bio-toxic effects can easily be generated when,in contrast, gravity waves from T4 are so hard to 'nd is another issue not considered here.The gravity wave T4 coupling parameter is essentially Ed Witten's alpha' = (superstring tension)^-1.What is the corresponding O(1,3) spin connection coupling parameter? Akimov's claims hangon the answer to that question. Is it easier to make propagating torsion dislocation topological string defectsthan to make propagating curvature disclination topological string defects in the MACRO-QUANTUMVacuum Coherence Field's Goldstone Phase? That's what Akimov's claims come down to in terms ofmy new theoretical paradigm for the emergence of Einstein's Gravity and the Uni'ed Exotic Vacuum Field ofw = -1 Dark Energy/Matter.2. The action of the symmetry group G on the total hyperspace H induces an equivalence relation ~ .That is, if X' = gX, g < G, then X' ~ X.3. ~ partitions hyperspace H into disjoint non-overlapping equivalence classes called G-orbitsG(X) = {gX, for all g < G}Remember that in this principal bundle fo is also a g < G.All G-orbits have identical structure and are diffeomorphic to G.4. This disjoint partition of hyperspace H gives the quotient space H/G that is the base space M with points x.Every point x of the base space M is really an equivalence class or G-orbit of a continuous in'nity of points of a larger dimensional Hermetic or occult hidden hyperspace implicate inside it. Worlds within worlds. Wheels within wheels. Shades of Bohm's Implicate Order?5. The Projection Map P is simply P:G-orbit -> x.This means that each individual G-Orbit is really associated with a single vertical 'ber at a single horizontal base space event. The G-orbit is the vertical 'ber beyond, in the usual physics applications, a localized spacetime event x, although we can have delocalized base spaces of twistors whose intersections are points. We can also perhaps have base spaces of 'nite strings both open and closed and even base spaces of higher dimensional brane worlds?Commentary 2Given coordinate patch C(x) in the base space M in a neighborhood of point x and 'ber f(x)form the local Cartesian product C(x)f(x) with ordered pair X = (x,fo).Take the union C(x)f(x)/C(x')f(x')/... of all such local products.There are redundant ordered pairs X because the coordinate patches C(x) and C(x') as sets overlapwith non-vanishing intersection C(x)/C(x')=/= Empty Set.Identify the redundant multiple images of the same actual point of the base space M usingthe symmetry group G as an equivalence relation. That is, two ordered pairs X and X' areidenti'ed or equivalent if x = x' < C(x)/C(x') and if fo' = gfo where g < G to form disjointequivalence classes {f(x)} that are the distinct points of the 'ber in hyperspace H.This is all local at a 'xed base point x like in an internal gauge force symmetry.g is also called a transition function.The hyperspace H is the factor space of the union C(x)f(x)/C(x')f(x')/ ... mod G.The projection map P:(x,{fo}) -> xWhen M is the curved space-time of Einstein's gravity theory in addition to the G equivalencein the extra space dimensions of the 'ber, x'(E) = Diff(4)x(E) at 'xed event Eto make disjoint equivalence classes {x(E)} mod Diff4(E).One can imagine a hybrid where the 'ber is a discrete space of strings of c-bits.One can also imagine a 'ber of strings of qubits.1 qubit is a parallel in'nity of c-bits.i.e.|qubit> = |1 c-bit><1c-bit|qubit> + |0 c-bit><0 c-bit|qubit>Where there is a continuous in'nity of different c-bit basesor orthonormal frames each corresponding, for example,the the angular orientation of an inhomogeneous 'eldmagnet in a Stern-Gerlach 'lter for spin qubitsin the DARPA spintronics project or like the billion billionSingle Electron Transistors inside the human brain at thesub-microtubular protein dimer hydrophobic cage level formingthe hardware interface with external world whose software is our stream of inner consciousness.Each possible orientation is a primitive parallel quantum universe.The quantum computer computes in all possibleorientations simultaneously like a continuousin'nity of classical Turing machines in adistributed network working on the same problem - or so the folklore goes.to be continued.Commentary 1The 'ber bundle as an idea has 4 parts.1. A structure symmetry group G.2. The total hyperspace H or, in some applications Wheeler's BIT.3. The projection map P.4. The base space M or, in some applications. Wheeler's IT.The hyperspace H consists of 'bers f(x) that areeither copies of or representations of the symmetrygroup G.The projection map P collapses a 'ber f(x) in the hyperspace H toa point x in the base space M.All of these objects are continuum differential manifoldsdepending on the continuum of real numbers which itsassociated issues of Cantor's in'nity of in'nities ofCabalistic Aleph's in an ascending Jacob's Ladder.This is not a discrete combinatoric mathematics althoughsuch a skeletal structure is associated with it as inHerman Weyl's Theory of Groups and Quantum Mechanicsand as in Saul-Paul Sirag's presentation of V.I. Arnold'sA-D-E mathematics of everything.The base space is covered by an atlas of local coordinate patcheswith all important overlap transition functions sewing thepatches together like a quilt.M is space-time in local micro-quantum 'eld theory of pointThe extra-dimensions of hyperspace formthe Calabi-Yau space of vibrations of thesuperstring beyond space-time.The connection on the total hyperspace H is the potentialof a local gauge force.Examples of connections is the 4 potential Au(x) inMaxwell's electromagnetism with G as U(1).There are similar connections for the Yang-Mills weak forcewith G = SU(2) and the strong force with G = SU(3).Classical general relativity, as distinct from local micro-quantum'eld theory, has the torsion-free symmetric three-index non-tensorLevi-Civita connection with G as the Diff(4) group.The latter comes from locally gauging the 4 parameter translation subgroup(generated by the 4-momentum Pu of globally §at special relativity )of the 15 parameter conformal group of Roger Penrose's massless twistors.Bottom -> Up: Given base space M and symmetry group G construct thehyperspace H as a quilt patchwork.Top -> Down: Given hyperspace H and symmetry group G construct thebase space M as the non-overlapping partition of hyperspace into G-orbitscalled the quotient space of H mod G in the principal bundle.Micro-quantum source renormalizable local 'elds of spin 1/2 lepto-quarks are associated vector bundles.Micro-quantum force renormalizable local 'elds of spin 1 gauge force bosons (electro-weak and strong) arefrom the principal bundle.There is no renormalizable quantum gravity in this precise sense.This is because classical Einstein gravity is a More is different (P.W. Anderson)emergent collective effect as in Andrei Sakharov's metric elasticity of aninstability in the globally §at false vacuum of the interacting lepto-quark source/electroweak-strong force.Einstein's gravity + uni'ed exotic vacuum dark energy/matter with Andrei Linde's chaotic in§ationary cosmology are the result of the continual phase transitions from globally §at false high entropy micro-quantum vacua to locally curved macro-quantum low entropy metastable vacua.to be continued: => Commentary 3 > The hyperspace H consists of 'bers f(x) that are> either copies of or representations of the symmetry> group G.> Well, that convinces me...X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft Schoenfeld) said:>I am not a mathematician by trade,That's obvious. More important, you lack the ability to listen.>but I was talking to a maths>professor and he absolutely refused to acknowledge the concept of a>cross product for two vectors that are not 3 dimensional.No doubt he also refused to acknowledge the concept of a pair o'ntegers i and j such that i+j>j+i. Im sure that there are all sortsof impossible concepts that he refused to acknowledge.>For me,>Let A be vector in N space>Let B be vector in N space>A x B = CYou haven't de'ned anything. What is C? Let's be concrete: ifA=(1,0,0,0) in R^4 with the L2 metric and B=(0,1,0,0), what is C=AxB? >Now, I can prove that C has the propertyHow, when you haven't de'ned it. Worse, one of the properties of thecross product in R^3 is that AxB is orthogonal to A and B, so ifC=AxB, C.A=C.B=0. Thus your>C.A / |C||A| = C.B / |C||B|Doesn't say much.>So, one could de'ne the cross product between two n-vectors as an>n-vector with the propertySure: just de'ne AxB=0 for every A and B. It would not, however, havethe properties of the cross product in R^3. If you don't want to dothat, then you would have to de'ne *WHICH* vector with thoseproperties, and there you run into dif'culties.>But, am I wrong?Of course you're wrong. Google for Exterior or Grassmann.>Or is the professor wrong? Not on this he isn't. at 10:29 PM, j.schoenfeld@programmer.net (John Schoenfeld) said:>What is there was no professor? Then you work it out a step at a time, without any handwaving, and yousee where you went wrong. Or you don't, and you remain deluded.>Don't you look stupid, believerSomebody does, but it isn't him. You look stupid for presenting as ade'nition something that in fact did not de'ne anything, and youlook stupid for rejecting good advice instead of thinking thingsthrough.But tell me, tonto, why are you paying tuition if you believe that theprofessor doesn't understand things as elementary as that?-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to =For what it's worth, there *is* an extension of the cross product forvectors in higher dimensions. The problem is that the cross productis a cheat. The generic operation is a tensor outer product, and it so happensthat the outer product of two 3D vectors has three components, so youcan map it into a vector easily. But this is a cheat of sorts, and itleads to confusions in many areas (especially when people start usingit in studying electrodynamics).Thomas => A cross product is a special case of the Clifford wedge product. Really this is an exterior algebra construction, not a Cliffordconstruction.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) => A cross product is a special case of the Clifford wedge product. The> wedge product of two vectors is called a bi-vector. It just so> happens that in R^3 bi-vectors are dual to vectors, so we can> interpret the three dimensional bi-vector as corresponding to a> special type of vector (usually referred to as an axial vector). This> correspondence breaks down in higher dimensions, since the ranks of> dual objects must add up to the number of dimensions. Hence, in R^4,> bi-vectors are dual to themselves and cannot be interpreted as vectors> at all. In even higher dimensions, the situation becomes all the more> hopeless.Did you insult me sometime recently? I can't remember.However, even if it was you, all is forgiven for that nice precis. =I would like to know some important points when selecting a certaingraph for the data I have i.e. Using a Pie graph for displaying thelargest % of Annual sales, in a 've-year period, for a business.Thanx,cSaviour =interesting.I am claiming to derive Einstein's gravity with the cosmological term as a large scalelimit of the exotic vacuum 'eld from an instability at least in the QED sector ofthe micro-quantum vacuum. Wilczek in contrast never does any micro-quantumtheory that I can see in his paper? He already starts with classical 'elds andmakes no attempt to derive gravity as emergent from the micro-quantum standardmodel.By a modi'ed conformal MacDowell-Mansouri mechanism,I asked:Is this where ODLRO MACRO-QUANTUM EMERGENCE is buried?All this is at conventional textbook level, for example,section 14.6 of Uni'cation and Supersymmetry, 2nd edition,by Rabindra Mohapatra, Springer-Verlag 1992.If you want a prominent establishment name dropped, Frank Wilczekmentions the MacDowell-Mansouri This paper seems very relevant in accord with what I amdoing in a simpler way independently. Wilczek is one of the besttheoretical physicists around for sure. I heard him speak twice nowthis year.where he notes that the mechanism was also independentlyformulated by Chamseddine and West.The mechanism was invented to make it possible to get gravityfrom the anti-deSitter part of Lie superalgebras used insupergravity theories. =My professor and my TA skipped their of'ce hours and Im totally lost.looking for guidance since Im confused.Consider relation schema R(A,B,C) and the set of functionaldependenciesF={B->A,A->C}1. All non-trivial relations. Is this correct? Im just guessing.B->A, A->C,B->C,AB->BC,AC->C, BC->CA,AB->C2. Find a non-empty instsance of R(give a number of rows) thatsatis'es every Functional Dependency in F.Is this correct?A B C2 1 33 2 44 3 53. Find an instance in R that satis'es every FD in F accept A->BHow do you get A->B???? I dont see it. 4. Possible to 'nd an instance an instance that satis'es every FD inF, but does not satisfy the FD AB->C. I have no idea. Im totally lost.Can someone point in the right direction? => Take the in'nite series expansion for e and put it into the in'nite> series expansion of e to the power of x and multiply the terms and you> end up with a series of aleph 1 terms>> This is nonsense. Even when you multiply everything out, there are>> still only countably many terms.>> Yes, compare countability of the rational numbers.> that sum to a 'nite result e to> the power of e. If you repete this process for e to the power of e to> the power of e do you end up with aleph 2 terms?>> For series with an arbitrary number of terms, look up the term summable>> family. But a necessary condition for convergence is that at most>> countably many terms are nonzero, this follows from the archimedean axiom>> of the real numbers.>If aleph 1 is the set of all subsets of aleph 0 No, aleph1 is 'rst uncountable ordinal. You must mean bet1, which is usually written c. Whether or not aleph1=c is called the Continuum Hypothesis, and is undecidable in ZFC.> Anyway, your argument is incorrect for c as well. then take the counting>numbers the 'rst subset is the empty set next come the sets with only>one member that is 1,2,3 etc next is the sets with two members these>can be arranged in a two dimensional array 1 2,1 3,1 4 etc 2 3,2 4,2>5 etc 3 4,3 5,3 6 etc etc all the way up to those with an in'nite>number of terms which can be arranged in an array with an in'nite>number of dimensions. Now if you want each one of these sets has a>complement however I think when you get to the sets with an in'nite>number of terms then each subset is duplicated. which subset did I>miss? You missed in'nite subsets. Your counting scheme simply won't work for the in'nite number of dimensions case. You did not describe the counting scheme for that case.As for e to the power of e the 'rst term is 1 next comes 1 + x>+ (x^2)/2 + then comes 1/2 + x/2 + (x^2)/4 + , x/2 + (x^2)/2 + (x^3)/4>+ , (x^2)/4 + (x^3)/4 + (x^4)/8 + etc arranged in a two dimensional>array all the way up to those with an in'nite number of terms that>can be arranged an array with an in'nite number of dimensions. you>have a one to one correspondence or at least a one to two>correspondence. CronOK lets review e^e=1+e+(e^2)/2+(e^3)/6+...= 1+ 1+1+1/2+1/6+1/24+...+ 1/2+1/2+1/4+1/12+1/48+ 1/2+1/2+1/4+1/12+1/48+ sorry about the messthis is the best I can do 1/4+1/4+1/8+1/24+1/48+ 1/6+1/6+1/12+1/36+1/144+ 1/6+1/6+1/12+1/36+1/144 1/12+1/12+1/24+1/72+ 1/6+1/6+1/12+1/36+1/144+ 1/6+1/6+1/12+1/36+1/144 1/12+1/12+1/24+1/72+ 1/12+1/12+1/24+1/72+1/288+ 1/12+1/12+1/24+1/72+1/2881/24+1/24+1/48+1/144+ this series has 1+aleph 0+(aleph 0)^2+(aleph0)^3+...epsilon 0 terms now I know that aleph 0 =epsilon 0 but thisseries contains more than epsilon 0 terms. the value of an in'niteseries depends on how you add up the terms this series is even worsehowever I want to leave it how it is to see if it can de'ne curvesthat a simple series can't => OK lets review e^e=1+e+(e^2)/2+(e^3)/6+...= 1+ 1+1+1/2+1/6+1/24+...> + 1/2+1/2+1/4+1/12+1/48+ > 1/2+1/2+1/4+1/12+1/48+ sorry about the mess> this is the best I can do > 1/4+1/4+1/8+1/24+1/48+ > 1/6+1/6+1/12+1/36+1/144+ 1/6+1/6+1/12+1/36+1/144 > 1/12+1/12+1/24+1/72+ 1/6+1/6+1/12+1/36+1/144+ > 1/6+1/6+1/12+1/36+1/144 1/12+1/12+1/24+1/72+ > 1/12+1/12+1/24+1/72+1/288+ 1/12+1/12+1/24+1/72+1/288> 1/24+1/24+1/48+1/144+ this series has 1+aleph 0+(aleph 0)^2+(aleph> 0)^3+...epsilon 0 terms now I know that aleph 0 =epsilon 0 but this> series contains more than epsilon 0 terms. the value of an in'nite> series depends on how you add up the terms this series is even worse> however I want to leave it how it is to see if it can de'ne curves> that a simple series can'tYou are mixing ordinals with cardinals. Aleph_0 is a cardinal, butepsilon_0 is an ordinal. The least trans'nite ordinal is called omega(or sometimes omega_0). Both omega and epsilon_0 are countable ordinals,which is to say that their cardinality is aleph_0.If A is any countably in'nite set, then AxA is likewise countable. Byinduction, we can conclude that A^k is countable for each natural numberk > 0. The union of { A^k : k > 0 } is likewise countable. Thecardinality of each of these sets is aleph_0.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft =>Take the in'nite series expansion for e and put it into the in'nite>series expansion of e to the power of x and multiply the terms and>you end up with a series of aleph 1 terms No, you end up with a series of Aleph_0 terms.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to = > Shmuel (Seymour J.) Metz, SysProg and JOAT> Fuck off you cunting little Jew.-- G.C. => Working in binary after the decimal point 'rst you have .000... and> its compliment .111... next comes those with one one .1 .01 .001 etc> and their compliment .0111... .00111... .000111... then those with> two ones .11 .101 .1001 etc .011 .0101 .01001 etc .0011 .00101 > .001001 etc etc and their compliments these can be arranged in two two> dimensional arrays all the way up to those with an in'nite number of> ones that can be arranged in two in'nite dimensional arrays. Which> real number did I miss? So now we have a one to two correspondence> between my original construction and the reals between zero and one> therefore a series with a continium of nonzero terms sum to a 'nite> result. Mr fritz says this is in impossible. So does e^(e^e) have c> members or 2^c members.Your compliments seems to mean, based on your examples above, representations of reals betweeen 0 and 1 having two binary representations. There are only countably many reals with such dual binary representation (or dual representation in any larger integral base), as all such numbers are rational.I do not see from the above too brief description how you intend to count those reals whose binary representations contain both in'nitely many zeros and in'nitely many ones.Indeed, a trivial translation between base 2 and base 4 allows a direct application of Cantor's diagonal proof in base 4 that there is no surjection from the naturals to reals between 0 and 1.Each such real, between 0 and 1, has a base 4 representation a = sum[i=1..oo. a_i/4^1], with all a_i in {0,1,2,3}And any dual representations are all 0's or all 3's from some point onwards, so 1's and 2's do not involve the dual representation uncertainties. For any listingFf: N -> [0,1]: n -> f(n), construct x as follows:let the n'th digit of x equal 1 if the nth base four digit of f(n) is not 1 and let it be 2 otherwise: x_n = 1 if f(n)_n <> 1 x_n = 2 if f(n)_n = 1Then x is between 0 and 1 but is not equal to any of the f(n).Since no assumptions were made about f except that its domain is N and its codomain is [0,1], and there is shown to be some member of the codomain not in the image of f, such f cannot ever be surjections.In fact, similar explicit constructions can produce a least countably many members of [0,1] not in the image of any such f. =>>Take the in'nite series expansion for e and put it into the in'nite>>series expansion of e to the power of x and multiply the terms and you>>end up with a series of aleph 1 terms that sum to a 'nite result e to>>the power of e. If you repete this process for e to the power of e to>>the power of e do you end up with aleph 2 terms?>> You still only get aleph 0 terms.>> I think the OP is making a very common error: mistaking the set of>> all subsets of N with the set of all *'nite* subsets of N> Working in binary after the decimal point 'rst you have .000... and> its compliment .111... next comes those with one one .1 .01 .001 etc> and their compliment .0111... .00111... .000111... then those with> two ones .11 .101 .1001 etc .011 .0101 .01001 etc .0011 .00101 > .001001 etc etc and their compliments these can be arranged in two two> dimensional arrays all the way up to those with an in'nite number of> ones that can be arranged in two in'nite dimensional arrays. Which> real number did I miss? Each number in your 'rst list has a 'nite number of 1's, and eachnumber in your second list has a 'nite number of 0's. You never reach anumber such as 1/3 = (.01010101...)_2 or 2/3 = (.10101010...)_2, in whichboth the 0's and the 1's are in'nite.>So now we have a one to two correspondence> between my original construction and the reals between zero and one> therefore a series with a continium of nonzero terms sum to a 'nite> result. Mr fritz says this is in impossible. So does e^(e^e) have c> members or 2^c membersIt has aleph_0 members. -- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. arithmetic mean and HM is harmonicmean of two consecutive primes.By plotting the order n (index n in p_n) of prime like spectral co-linear equations.Q: Resonance theory of allowed quantum numbers?Tapio