mm-87 === But here I now read that people are indeed very attached to their>calculators, and so I am very curious: What do you use them for? Just to join in the general HP nostalgia: I used an HP-45 for many years, and it functioned fine for quite awhile, if a bit stickily,after a Pepsi spill that got into most of the keys. I nally ditchedit when some of the crucial buttons stopped responding.Around the same time I had an HP-65 programmable at work, and I wasprogramming it to do various numerical things I needed (like Lagrangeinterpolation of calibration data).I have bought one or two non-HP scientic calculators since then, but rarely used one. Almost every calculation I want to do these days Ijust do on the Matlab window I always have open on my computer. Andsince that includes a really lovely graphics capability, Ive neverhad occasion to want to use a graphic calculator. I have no idea howto use a TI-83, and I havent yet found enough incentive to want tolearn.One of these days IOll probably want to invest in a scienticcalculator for use away from the ofce, but Im not sure if I haveany need for programmability. It *would* be nice to have hexconversion at my ngertips; the last calculator purchased for me byan employer was an HP model that included lots of computer-relatedstuff like that. - Randy === > Ive always wondered about Mathematical Olympics. Dont the judges have> to be smarter than any of the contestants, to be able to judge them> accordingly? In normal Olympics, the judges can be any skinny little> wimps off the street, as they can merely watch the athletes without> competing themselves. Ignoring the time constraint --- in a lot of cases --- for makingthe judgement call, the most important qualications, IMHO, areintegrity, knowing the rules and knowing how to apply the rules. (Butmath rules are not just any rules.) Notwithstanding, getting a couple hundred of rainbow eyes andbroken noses and cut cheeks (analogically in boxing) may or may nothurt the qualications more than the injured.> But mathematics is different. Unlike athletics, its not plain to see> who is better than who. You have to be a mathematician yourself. Shall we start the denition of (the qualication of) amathematician? :) Math is not, IMHO, to gure out who is better than who, a pettyproposition. It is about truth. If your focus is math contests, ratherthan math in its pure form, then the result is no different fromanother type. A democratic view of the performance of thecontestants at the moment with a specic set of congurations. The lazy God (sorry for sometimes misspellings) by Minkowskycould have had pathetic performances in contests.> Theres *one* thing that suggest that the judges dont have to be the> smartest ones after all - in certain theoretical calculations, checking> whether something has been done is far easier than actually doing it. Wonder, if there can be exceptions (to easier).> For example consider the simple array sort. Actually sorting the array> cant be done faster than O(nlog n) time, this has been proven. But Practically, I sincerely hope that stays proven.> checking if the array has been sorted can be done in O(n) time with a> simple algorithm any child could come up with. Maybe this can be> expanded to more complicated calculations? === > Lets dene a sum: n 1> s(n)= S -------------> k=0 k! Now, we want to get this sum: oo> S(x)= S s(n) x^n ; x<1> n=0 Is there a closed form and if so how to evaluate itHow about this: Multiply by x and get the following:x S(x) = SUM (n=1,oo) s(n-1) x^nNow subtract S(x) and use the following fact: s(n) - s(n-1) = 1/n!I get S(x) = e^x/(1-x).-Michael. === > Lets dene a sum: n 1> s(n)= S -------------> k=0 k!Now, we want to get this sum: oo> S(x)= S s(n) x^n ; x<1> n=0Is there a closed form and if so how to evaluate it> Look at S(x) - xS(x).Rick === > Ah, but theres a difference for me. When I enter 6.67259 x 10 EE -11 and> press = (equal key), I get 6.67259 ^-10.This is the problem. If you enter 6...*10 EE -11 then you really compute6..*10*10^-11 since (EE z) is an abbreviation for doing *10^z.Hence I would suggest entering 6.. * 10 then pressing a button like y^x (itis so called on my TI 30X I think its also on TI 36X) and then entering-11. === >Ah, but theres a difference for me. When I enter 6.67259 x 10 EE -11 and>press = (equal key), I get 6.67259 ^-10.THERES your problem! You multiplied 6.67 by 10x10^(-11), which is6.67x10^(-10). You should have punched in 6.67 EE -11 . DONT punch in the 10!--Dan Grubb === 1/9 = 0.1111...> 2/9 = 0.2222...> ...> 8/9 = 0.8888...> 9/9 = 0.9999...But, 9/9 = 1, so 0.9999... equals 1This is just some funny stuff that crept into my head. Now, is there anything> wrong in this proof? If so, what is it?> If you know that each of these lines is the sum of a geometric series,that, then you need more to make this a proof. === > But here I now read that people are indeed very attached to their> calculators, and so I am very curious: What do you use them for?> I would be especially interested to discover any professional uses for> the graphing and symbolic calculators (my impression so far has been> that no one uses them except high school teachers and their students).You never forget the rst time you were in love....I cherish my HP 32S II so much that I wouldnt dream of bringing it with meto work. It might get lost or stolen or worse... That leaves me with astupid Windows Calculator, which doesnt do RPN. I must suffer to protect myloved ones...-Michael. === > Ellipse is produced if a plane intersects only one nappe of a cone.Is there a way to get the parameters of the ellipse as a function of> the angle of the plane with the axis of the cone?I know that they depend on the height of the cone and diameter of the> circle at the bottom of the cone.I couldnt nd exactly what youre asking for at Mathworld(http://mathworld.wolfram.com/Cone.html), but heres somethoughts from the geometry.(Plane that is perpendicular to the axis produces a circle.)Let theta be the angle the plane makes with the axis.Lets say the cone has height h and base diameter d.Let us suppose the plane cuts the axis at distancey from the apex, where the diameter is (d/h)y.Half of this, 0.5dy/h, is the semi-minor axisof the ellipse. The center point of the ellipseis the place where the plane cuts the axis.Now lets look at the semi-major axis. Considera planar slice through our plane and cone thatincludes the semi-major axis and the axis ofthe cone. (I suggest you draw a picture. Imreferring to a picture as I write this). Thepicture is something like this. Apex A | | cone | | B Axis / | / | / Cutting plane |/ CWe have a triangle formed by the apex A, theellipse center C, and the highest place Bwhere the ellipse cuts the cone. The angleACB is the angle theta at the bottom of thistriangle. The angle CAB at the top, the apexangle of the cone is arctan(0.5d/h). Call thisangle CAB = phi.BC is the semimajor axis of the ellipse. We cannd it from, for example, the law of sines: BC/sin(CAB) = AC/sin(ABC)I dened y = AC. sin(ABC) = sin(pi - (CAB + ACB)) = sin(CAB+ACB)So BC = semimajor axis = y*sin(phi)/sin(phi+theta)unless I made a mistake.Now you have the semiminor axis (0.5dy/h) and thesemimajor axis. Thats enough to calculate any otherparameters of the ellipse you want. - Randy <4a10pvcg1d5066dbuoa4s7s9pcus13rlui@no.spam> === >Ah, but theres a difference for me. When I enter 6.67259 x 10 EE -11 and>press = (equal key), I get 6.67259 ^-10. This is the problem. If you enter 6...*10 EE -11 then you really compute> 6..*10*10^-11 since (EE z) is an abbreviation for doing *10^z.> Hence I would suggest entering 6.. * 10 then pressing a button like y^x (it> is so called on my TI 30X I think its also on TI 36X) and then entering> -11.Or just 6.67259 EE -11.-- P.A.C. SmithIf the Apocalypse comes, beep me. <*> http://www.srcf.ucam.org/~pas51 === Im constructing an algorithm to generate the harmonic series of amusical instrument. Ive been butting my head with this equationwhich I have to invert, but Ive had no luck so far:y(x) = (1 + sqrt(1 + px)) / (1 + sqrt(1 + x))where 0 < p < 1.So the problem is to turn x into a function of y.I get the feeling this is either trivial and Ive missed somethingobvious, or it cannot be done analytically in a closed form.-Alistair === - 2y = px - 2y.sqr[(1+x)(1+px)] + xy^2-(2y + px + xy^2) = -2y.sqr[(1+x)(1+px)](2y + px + xy^2)^2 = 4y^2 (1+x)(1+px)4y^2 + p^2 x^2 + x^2 y^4 + 4pxy + 2px^2 y^2 + 4xy^3 = 4y^2 + 4xy^2 + 4pxy^2 + 4px^2 y^2p^2 x^2 + x^2 y^4 + 4pxy + 2px^2 y^2 + 4xy^3 = 4xy^2 + 4pxy^2 + 4px^2 y^2p^2 x^2 + x^2 y^4 + 4pxy - 2px^2 y^2 + 4xy^3 = 4xy^2 + 4pxy^2p^2 x^2 + x^2 y^4 + 4pxy - 2px^2 y^2 + 4xy^3 - 4xy^2 - 4pxy^2 = 0x^2 (p^2 + y^4 - 2py^2) + x(4py - 4py^2 + 4y^3 - 4y^2) = 0x^2 (p - y^2)^2 + 4x(py - py^2 + y^3 - y^2) = 0x^2 (p - y^2)^2 + 4x(py(1 - y) - y^2 (1 - y)) = 0x^2 (p - y^2)^2 + 4xy(p - y)(1 - y) = 0x = -4y(p - y)(1 - y) / (p - y^2)^2 >I get the feeling this is either trivial and Ive missed something >obvious, or it cannot be done analytically in a closed form. >Ive the unsettling notion you better check my work and my answer.y(0) = 1; x(1) = 0; well it does check for x = 0, but thatsa special case anyway as I divided by x for the nal solution.---- === Im constructing an algorithm to generate the harmonic series of a> musical instrument. Ive been butting my head with this equation> which I have to invert, but Ive had no luck so far: y(x) = (1 + sqrt(1 + px)) / (1 + sqrt(1 + x)) where 0 < p < 1. So the problem is to turn x into a function of y. I get the feeling this is either trivial and Ive missed something> obvious, or it cannot be done analytically in a closed form.> -Alistair> a computer equation solver (MATLAB symbolic toolbox) gave x = 4 * y * (1 - y) * (y - p) / (y^2 - p)^2but got confused when instructed to check by plugging in.Cause: ambiguity in taking square roots from squares.So, I applied common sense and got the same result.The method:First, realize that sqrt(p) < y < 1 (subtract and compare withzero). Then introduce an intermediate unknown t = 1 + sqrt(1 + x), so t > 0 and x = t * (t - 2) === >3 men went to a motel. Theres only 1 room left and it costs 30 bucks. So, each>man forked out 10 bucks.Later, the owner discovered he over-charged the 3 men. It should be 25 bucks>instead. So, he sent his runner to give 5 bucks back to the 3 men. The 3 men>think the owner is very honest, and thus each took 1 buck back, leaving 2 bucks>to the runner as tips.Now, since they took 1 buck back, each of them paid 9 bucks. 2 bucks to the>runner. So: (9*3)+2 = 29. Wheres the 1 dollar?>There is no one dollar to look for, i.e.:10 + 10 + 10 = 30 = 25 + 5 = 25 + 2 + 1 + 1 + 1--> 9 + 9 + 9 = 25 + 2so its 27 - 2 = 25 and not 27 + 2 = 30adam === >Once upon a time, the battery died and thought that Id>just use this other calculator until I had time to get>a new one. I could not function. RPN is so imbedded>into my thinking that I could not use a regular calculator.>I ended up doing the stuff on paper and the task of getting>a new battery jumped to number 1 priority.Its the same with me. I use a slide rule and/or paper and pencil if> an RPN calc isnt available. (Though my 48GX, 41CX and 16C are usually> within arms reach.)My laptop is often within arms reach and I use an RPN calculator on it, although I do also have a physical one somewhere. I, too, can not function with an algebraic-notation calculator, despite using algebraic notation often enough in various programming languages.-- David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine, School of Information & Computer Science === >Once upon a time, the battery died and thought that Id>>just use this other calculator until I had time to get>>a new one. I could not function. RPN is so imbedded>>into my thinking that I could not use a regular calculator.>>I ended up doing the stuff on paper and the task of getting>>a new battery jumped to number 1 priority.Its the same with me. I use a slide rule and/or paper and pencil if>> an RPN calc isnt available. (Though my 48GX, 41CX and 16C are usually>> within arms reach.)My laptop is often within arms reach and I use an RPN calculator on it, >although I do also have a physical one somewhere. I, too, can not >function with an algebraic-notation calculator, despite using algebraic >notation often enough in various programming languages.I dont see any relationship between those other kinds of calculationsand algebra or programming languages. Its because of my familiarityof programming that I cant use those other calculators.RPN is completely algebraic to me. When calculating any equation,you do the insides rst with a heirarchy of operation. Those non-RPN calculators are left to right with no push down list./BAH/BAH === >Once upon a time, the battery died and thought that Id>>just use this other calculator until I had time to get>>a new one. I could not function. RPN is so imbedded>>into my thinking that I could not use a regular calculator.>>I ended up doing the stuff on paper and the task of getting>>a new battery jumped to number 1 priority.>Its the same with me. I use a slide rule and/or paper and pencil if>> an RPN calc isnt available. (Though my 48GX, 41CX and 16C are usually>> within arms reach.)My laptop is often within arms reach and I use an RPN calculator on it,>although I do also have a physical one somewhere. I, too, can not>function with an algebraic-notation calculator, despite using algebraic>notation often enough in various programming languages. I dont see any relationship between those other kinds of calculations> and algebra or programming languages. Its because of my familiarity> of programming that I cant use those other calculators. RPN is completely algebraic to me. When calculating any equation,> you do the insides rst with a heirarchy of operation.> Those non-RPN calculators are left to right with no push down list.>You mean the ones with algebraic notation and no parentheses? (Yes, theyreally exist.) Or with the (seemingly always) insufciently deep nesting.Ive exceeded 9 levels of parentheses with algebraic (left-to-right) entry,but I vaguely remember *once* exceeding a 4-level stack on an RPNcalculator. I couldnt even tell you the circumstances. I might bemisremembering -- or remembering an error I made.I get around the algebraic problems by (drum roll, please) working from theinside out and either using stored memory locations or writing down theintermediate results. I also claim that anyone who can actually use analgebraic calculator for difcult calculations (like mortgage paymentcalculations) does the equivalent (well, maybe not, since it can be donewith about 5 levels of parentheses). But after a while, you forget whichlevel of parentheses youre in, so you have to be making notes of some sorton a sheet of paper.Oh, I forgot algebraic notation but no algebraic hierarchy. So 1+2*3=9.Aaaaaauuuuuuuggggggghhhhhhhhh! Usually with no parentheses.Jon Miller === rearrange it until you havesomething that depends on x & y and sqrt(1+x) = sqrt(1+px)square itthe l.h.s. will have sqrt(1+x) in it. Isolate it and square again. I thinkyou will then be left with a quadrative in x which you can solve. Im constructing an algorithm to generate the harmonic series of a> musical instrument. Ive been butting my head with this equation> which I have to invert, but Ive had no luck so far: y(x) = (1 + sqrt(1 + px)) / (1 + sqrt(1 + x)) where 0 < p < 1. So the problem is to turn x into a function of y. I get the feeling this is either trivial and Ive missed something> obvious, or it cannot be done analytically in a closed form.> -Alistair === hot-girl escribi.97 en el mensaje> a sphere : x^2 + (y-2)^2 + (z-3)^2 =1 let point p : when tangent line of sphere pass on (0,0,c) is meet x-y> plane, the point of contact is named p let c1,c2 : value c that trace of p satify parabola. solve that c1+c2 ?? ------------------------------------- i will regard it as difcult...... help me....my genius teacher.... i wait your ultra power advice. thank sir.Actually is is easy ...What surface descrive the tangent line throw (0, 0, c)? What type ofsections produce a plane in that surface? When this section ia a parabola?Note that the tangent line can meet the x-y plane on the side of the sphereor on the side of point (0, 0, c).-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === good advice....good teacher... === I am looking for some old HP calculators like HP 41CV, HP 41CX, HP>> 71B, HP 15C, HP 16C, HP 67 and any others in the 1980s era...If youI have an HP55 and no, Im not selling.>Bought it in 1975 and it still works :-)> http://www.dotpoint.com/xnumber/hp55.htm I would like to point out that this is a rarity in sci.math:> a thread in which people wax eloquent about calculators. In my personal experience, I have seen calculators be useful at work for 1. simple arithmetic which involves too many digits to be fun or interesting> 2. repetitive specialized calculations (e.g. a banker will use a calculator> which can compute effective interest rates on a loan or investment). Anything simpler people do by hand and anything more complex uses a computer.> In short: it always seemed to me that almost no one would use a calculator> as a regular part of their job except for situations #1 and #2. But here I now read that people are indeed very attached to their> calculators, and so I am very curious: What do you use them for?> I would be especially interested to discover any professional uses for> the graphing and symbolic calculators (my impression so far has been> that no one uses them except high school teachers and their students). In particular, if anyone has ever denied employment to an applicant> because he or she was found not to have the necessary competence with> calculators, that would be information I ought to have.My HP-55 was my rst exposure to programming ( I was 19back then in 1975). It was love at rst sight - although it hadonly 50 steps. It was the best toy I ever had.In university I used it for astronomy, geodesy and physicscourses. In the rst year we were forced to use log and trigtables for the exam problems. The next year we were allowedto use a calculator, but then the problems suddenly turned outto be purely theoretical.I used it to calculate graphs for functions for the drawingsI used to make on the left side of my notes. We were andstill are inseparable ;-)few years ago (with some extra features not present on theoriginal): http://users.pandora.be/vdmoortel/dirk/Stuff/HP-55.jpgDirk Vdm === Corolllary of:http://www.bearnol.pwp.blueyonder.co.uk/Math/psq.htm === Hmm.. You claim that hcf (pm, p(m+M)) = hcf (pm, M). Lets take n =8, m = 2, M = 12, pm = 3, p(m+M) = 2. Now hcf(pm, p(m+M)) = hcf(3, 2)= 1, and hcf(pm, M) = hcf(3, 12) = 3, and where I come from 1 != 3. As this statement is completely false, I dont see how the rest of theargument is to hold together. Perhaps you meant to make somestipulations on the choices of pm. If so, what are they? === sets of 10 as hard copy posters. However, I am now offering them on a CD.The images still measure 8 x 11-1/2 inches, same size as the posters.If you get a chance, please visit my website, where every poster isavailable to see (as a thumbnail). The URL is athttp://www.mathisradical.com/Catalog.htmIm not a professional sales person, just a math teacher who came up with aproduct that ts a need in a mathematics classroom. They are inexpensiveand useful. I also have a free newsletter that includes lesson plans,biographies of mathematicians and Black Line Masters foryour classroom.Sincerely,Kavon Rueterkavon@mathisradical.comhttp://www.mathisradical.com === >Im not a professional sales person, just a math teacher who came up with a>product that ts a need in a mathematics classroom. Of course youre not professional. Someone who was professional wouldrealize that posting the exact same message in the same newsgroupmere days apart would be spamming.Doug === How sweet of you to notice!KavonIm not a professional sales person, just a math teacher who came up witha>product that ts a need in a mathematics classroom. Of course youre not professional. Someone who was professional would> realize that posting the exact same message in the same newsgroup> mere days apart would be spamming. Doug <1xUcb.14254$O85.6040@pd7tw1no> <3f79e264$7$fuzhry+tra$mr2ice@news.patriot.net> <3f82ebb8$3$fuzhry+tra$mr2ice@news.patriot.net> <3f8c413c$18$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Terminate: SPA(GIS) === >You have a point, two points actually, but I did not say that proof>is a mathematical concept, talk about are Mathematical concepts.>Any that follow from the denition of natural number I give below.You dont give a denition. You list a few names of natural numbers,and mention the successor function. And you certainly havent answeredmy question as to what properties you believe they have that do notfollow from the Peano postulates.>In some cases it is easy to tell; Then tell.>In his paper G.9adel showed how, for any sound formal system for>natural numbers (that is, one in which all provable statements are>true),The word true in that context is a technical term quite differentfrom what is generally understood by true.>This is false. Incorrect.>G.9adel showed how to nd true statements about>natural numbers not provable from the (rst-order) Peano>Postulates.He showed to to prove additional theorems by adjoining to the PeanoPostulates an additional axiom, essentially asserting that the PeanoPostulates were consistent.>I do not know what you mean by a metaphysical concept. Things that you talk about without ever dening or characterising ina precise enough fashion to allow rational discourse.>The natural>numbers are the numbers>0, 1, 2, 3, 4, ....>When spelled out this becomes>0, s(0), s(s(0)), s(s(s(0))), s(s(s(s(0)))), ... .Thats not a denition. That doesnt even say that they obey thePeano Postulates.>Natural numbers are perfectly well-dened, despite the fact that>they cannot be axiomatised in a rst-order system. Then dene them.>There is nothing strange about natural numbers. There is nothing strange about oranges, either, but that doesnt makethem apples.>We encounter them every day.In isolation. We do not encounter The Naturals every day; inparticular, we do not encounter induction.>Your use of the word mathematics is non-standard.No.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === >In his paper G.9adel showed how, for any sound formal system for>natural numbers (that is, one in which all provable statements are>true),The word true in that context is a technical term quite different> from what is generally understood by true. Different how? === Working on a research project and cant nd the specic equationswere looking for.Given a hyperbolic paraboloid (HP) rotated 45 Deg. such that twoplanes of symmetry are y=+-x, the equation is z=xy:(1) Were looking for the equation of an HP rotated as described abovehaving a parabola with the equation 12x^2-3x+2 above the line y=x, anda parabola with the equation -12x^2 +2 below the line y=-x.(2) What method should be used to nd the equation of the hyperbolacreated by the intersecting plane z=k?(3) Were having trouble nding resources, text books andpublications of any type dealing with the specic equations ofrotated HPs...can you point us in the right direction? Specicsites, publications, or people you might know?Any and all help appreciated!Morgan and Brian === Suppose you have *any* three-dimensional gure given as arelationship between x, y, and z, and you want to rotate it, say,around the Z axis, counterclockwise, by an angle of t. Then all youhave to do is rotate each individual point, and the entire thing willbe rotated. Imagine a point (x, y, z). Rotating around the Z axis will notraise or lower the point, so just think about the (x, y) part for aminute. Imagine it on the plane, with the line segment from theorigin to (x,y) drawn in. Then imagine a circle centered at theorigin with this segment as its radius. Call the radius r (equal tosqrt(x^2 + y^2)), and the angle, measured counterclockwise from thepositive X axis (equal to arctan (y/x)) t1. Then from trigonometry wehave x = r cos t1 and y = r sin t1. Now you just want to replace xand y with their rotated equivalents, so if were rotating an angle oft to the left, we want x = r cos (t1 + t) and y = r sin (t1 + t). But recalling the trig identities:sin (s + p) = sin s cos p + cos s sin pcos (s + p) = cos s cos p sin s sin p We havex = r cos(t1 + t) = r (cos t1 cos t - sin t1 sin t) = (r cos t1) cos t - (r sin t1) sin t = x cos t - y sin ty = r sin(t1 + t) = r (sin t1 cos t + cos t1 sin t) = (r sin t1) cos t + (r cos t1) sin t = y cos t + x sin tz = z (since rotating around the Z axis doesnt raise or lower thepoint)So, saying that your equation was z = xy, then after rotating aroundthe Z axis by an angle of t, just plug in x for x, y for y and zfor z, and your equation becomesz = (x cos t - y sin t)(y cos t + x sin t)z = (sin t cos t) x^2 + (cos^2 t - sin^2 t) xy - (sin t cos t) y^2.Dont be scared by the trig functions; they all depend just on t, soonce you choose t the equation just looks like, say, z = ax^2 + bxy +ay^2 for two numbers a and b.Also, when the gure moves, it takes its lines of symmetry with it --If your lines of symmetry were y = x and y = -x, now they arey cos t + x sin t = x cos t - y sin t and y cos t + x sin t = y sin t- x cos tBut these can be reduced toy (sin t + cos t) = x (cos t - sin t) and y (sin t - cos t) = x (sin t+ cos t).Now, of course, there are two other kinds of rotation -- rotationabout the Y and Z axes. However, these have pretty much the sameequations, so you can gure them out yourselves. Now all rotationscan be made up of a rotation in the X direction, another in the Ydirection, and another in the Z direction.If you want to nd the intersection of a plane with a hyperbola, justset them equal to each other. I.e. if the plane is z = k and thehyperbola is z = xy, then k = xy, or in other words, y = k/x.Hyperbolas can be rotated in exactly the same manner asthree-dimensional shapes -- a rotation around the Z-axis in 3d is thesame as a rotation about the origin in 2d.Sorry if this is kind of hard to read, but its hard to teach whatwould basically be a couple of weeks in a high-school analyticinformation, look up information about rotating conics in any analyticgeometry book.> Working on a research project and cant nd the specic equations> were looking for.Given a hyperbolic paraboloid (HP) rotated 45 Deg. such that two> planes of symmetry are y=+-x, the equation is z=xy:(1) Were looking for the equation of an HP rotated as described above> having a parabola with the equation 12x^2-3x+2 above the line y=x, and> a parabola with the equation -12x^2 +2 below the line y=-x.(2) What method should be used to nd the equation of the hyperbola> created by the intersecting plane z=k?(3) Were having trouble nding resources, text books and> publications of any type dealing with the specic equations of> rotated HPs...can you point us in the right direction? Specic> sites, publications, or people you might know?Any and all help appreciated!Morgan and Brian === Suppose you have a device that will show a value M such thatM ~ N( 230, 0.023 ). What is the probability that we get valueof 230.3 or less?Well, simply P(M < 230.3) gives me around 0.500 (Mathematica).On the other hand, i tried to standarize the variable and got:x = (230.3-230)/sqrt(.023) = 1.978 P(Z < 1.978) = 0.976I cant explain the large difference between those two. Also, i can not really decide which one is correct (if any). what do you think?-- KindlyKonrad------------------------------------------------- --May all spammers die an agonizing death; have no burial places; their souls be chased by demons in Gehenna from one room to another for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sence to be lazy--------------------------------------------------- === > Suppose you have a device that will show a value M such that>M ~ N( 230, 0.023 ). What is the probability that we get value>of 230.3 or less?>Well, simply P(M < 230.3) gives me around 0.500 (Mathematica).>On the other hand, i tried to standarize the variable and got:>x = (230.3-230)/sqrt(.023) = 1.978 >P(Z < 1.978) = 0.976N(230, 0.023) means normal distribution with expectation 230 andvariance 0.023. Whatever Mathematica gives you is wrong as it appearsyou didnt give it the correct variance.You can estimate the sensibility of the result by looking at thestandard deviation, sqrt(0.023) ~= 0,152. Your value is about +2*sigmaaway from the expectation, so its more than 95% of the distribution.This agrees with your calculations. === >Suppose you have a device that will show a value M such that>M ~ N( 230, 0.023 ). What is the probability that we get value>of 230.3 or less?>Well, simply P(M < 230.3) gives me around 0.500 (Mathematica).>On the other hand, i tried to standarize the variable and got:>x = (230.3-230)/sqrt(.023) = 1.978 >P(Z < 1.978) = 0.976I cant explain the large difference between those two. Also, i can >not really decide which one is correct (if any). what do you think?>Havent done the calculations, but your second caculation is more likely to be correct. I suspect you did not enter the formula correctly in Mathematica.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === > Suppose you have a device that will show a value M such that> M ~ N( 230, 0.023 ). What is the probability that we get value> of 230.3 or less?> Well, simply P(M < 230.3) gives me around 0.500 (Mathematica).> On the other hand, i tried to standarize the variable and got:> x = (230.3-230)/sqrt(.023) = 1.978 > P(Z < 1.978) = 0.976I cant explain the large difference between those two. Also, i can > not really decide which one is correct (if any). what do you think?Basic sanity check here: can you *really* believe that the probability of a value greater than almost 2 standard deviations above the mean is nearly 50%? Thats not even mathematically possible (Chebyshevs inequality says that regardless of the distribution, at most 1/4 of it can lie two or more sds from the mean). === >> M ~ N( 230, 0.023 )>> P(M < 230.3) = 0.500 (Mathematica)>> x = (230.3-230)/sqrt(.023) = 1.978>> P(Z < 1.978) = 0.976> Can you *really* believe that the probability of a value greater > than almost 2 standard deviations above the mean is nearly 50%?Yes, i can, but thats depending on the fact that ive been studyingstatistics for 6 weeks this far. :)I didnt think of what you mentioned, due to my ignorance. I will, -- KindlyKonrad------------------------------------------------- --May all spammers die an agonizing death; have no burial places;their souls be chased by demons in Gehenna from one room toanother for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sence to be lazy--------------------------------------------------- === > And its raised to a power. Once you relax that you a) lose your> proof of the Erdos-Selfridge theorem and b) have a meaningless> result. Among all the (nitely many) prime factors, surely there> has to be a largest one. Thats not news.D. Cool === ) === Could you please check the following for me:What is the probability that a card selected from a deck is an ace orheart?The way I see it is 52/4 = 16 (hearts) plus 3 (the aces minus theheart) = 19Answer is 52/19 = 2.73682 probability. === >Answer is 52/19 = 2.73682 probability.Can a probability be greater than one?Doug === In sci.math, jelly:> Could you please check the following for me:What is the probability that a card selected from a deck is an ace or> heart?The way I see it is 52/4 = 16 (hearts) plus 3 (the aces minus the> heart) = 19Answer is 52/19 = 2.73682 probability.Youre walking backwards with a defective calculator, butyoure not off the path -- yet. :-)The standard way might be along the lines of:[1] Count the number of hearts. Thats 13.[2] Count the number of aces. Thats 4.[3] Count the number of aces of hearts. Thats 1.[4] Count the number of cards: 52.[5] Compute the desired number of events out of all events: ace or heart is obviously 13 + 4 - 1 (we dont want to doublecount ace of hearts), out of 52, or 16/52 = 0.3077 = 30.77%. (Note that the events must be of equal probability. Since were drawing a card from an unmarked deck this isnt a problem here, but some formulations of problems may not be so lucky.)Your next assignment: compute the probability of drawing aface card (JQK) or a heart. :-) Your answer should be 42.3%,with logic identical to the above except for changes to someof the cards to protect the innocent Queens. :-)-- #191, ewill3@earthlink.net -- which is why the Jacks always look excited :-)Its still legal to go .sigless. === > Could you please check the following for me: What is the probability that a card selected from a deck is an ace or> heart? The way I see it is 52/4 = 16 (hearts) plus 3 (the aces minus the> heart) = 19 Answer is 52/19 = 2.73682 probability.>First try to read your textbook where you will nd that probability of anevent by denition takes values in interval from 0 to 1.selecting heart is 13/52 and probability of ace heart is 1/52. What willbe probability of event ace or heart ?Goran === Could you please check the following for me:What is the probability that a card selected from a deck is an ace or> heart?The way I see it is 52/4 = 16 (hearts)Check your division here. plus 3 (the aces minus the> heart) = 19Answer is 52/19 = 2.73682 probability.Probabilities must be between 0 and 1. You did something(rather simple) wrong here.> Almost. Just two mistakes. === > Could you please check the following for me: What is the probability that a card selected from a deck is an ace or> heart? The way I see it is 52/4 = 16 (hearts) plus 3 (the aces minus the> heart) = 19 Answer is 52/19 = 2.73682 probability.>First of all, a suit has only 13 hearts, not 16. Sorry if I just broke threehearts; Im a heart breaker :-)Secondly, probability is less than one. I get something like 16/52.-Michael. === === There are many ways to develop Lebesgue integral, and one way is asfollows:f is lebesgue integrable if there exists a sequence {f_n} of step functionsuch that(a) sum int |f_n| < infty(b) f(x) = sum f_n(x) for every x such that sum |f_n(x)| < inftyThen the integral of f is dened as int f = sum int f_nMy question is: does anyone know any book that introduces Lebesgueintegral in this way?TC === There are many ways to develop Lebesgue integral, and one way is as> follows: f is lebesgue integrable if there exists a sequence {f_n} of step function> such that (a) sum int |f_n| < infty> (b) f(x) = sum f_n(x) for every x such that sum |f_n(x)| < infty Then the integral of f is dened as int f = sum int f_n My question is: does anyone know any book that introduces Lebesgue> integral in this way?No it doesnt look right. The function:f(x) = 1 if 0> follows:>f is lebesgue integrable if there exists a sequence {f_n} of step function>> such that>(a) sum int |f_n| < infty>> (b) f(x) = sum f_n(x) for every x such that sum |f_n(x)| < infty>Then the integral of f is dened as int f = sum int f_n>My question is: does anyone know any book that introduces Lebesgue>> integral in this way?No it doesnt look right. The function:f(x) = 1 if 0f(x) = 0 otherwiseis lebesgue integrable. And this function satises the condition above:Let the rationals in (0,1) be r_1, r_2, ... . Let f_n(x) = 0except for f_n(r_j) = 1, 1 <= j <= n. Then f_n is a stepfunction, the sum of the integrals of |f_n| is nite, and f(x) = sum f_n(x) for every x such that sum |f_n(x)| isnite.>I think you would have to approximate that using>simple functions (rather trivial since it one already) not step>functions. It is possible that step function might be dened>differently. Secondly the word sum seems to be occuring where there would>normally be sup. Thirdly for a non-negative measurable function f the>integral is usually dened as sup int phi dx, where sup is taken over all>measurable simple functions phi, such that phi <= f. The sup is in general>going to be over an uncountable set so any use of sequences of step>functions looks wrong.Comments on how this approach differ from the approach youknow (ok, from the more usual approach) dont show that thisapproach is wrong. The last comment about how that sup isover an uncountable set so any use of sequences seems wrongis way off: In spite of the fact that the sup is over anuncountable family of simple functions we certainly knowthat we can get a sequence of simple functions convergingto f, with the usual approach...Im not _certain_ offhand that the conditions above aretrue precisely when f is Lebesgue integrable, but Im certainly not certain that this is not so - I know thatthere _are_ approaches very much like this that dowork.************************David C. Ullrich === I was curious as to why an ordered pair (x,y) would be dened as{{x}{x,y}}. I understand that a set has no order but therefore, wouldntthat be the same as {{x,y} {x}}.If anyone has any good links to this just for some background information,that would be realy helpfuly. === >I was curious as to why an ordered pair (x,y) would be dened as>{{x}{x,y}}. I understand that a set has no order but therefore, wouldnt>that be the same as {{x,y} {x}}.Yes it is. So what? (Its not the same as (y,x), which is {{y},{x,y}}.)>If anyone has any good links to this just for some background information,>that would be realy helpfuly.************************David C. Ullrich === I was curious as to why an ordered pair (x,y) would be dened as> {{x}{x,y}}. I understand that a set has no order but therefore, wouldnt> that be the same as {{x,y} {x}}.Yes, it is the same. Its also the same as {{y,x},{x}} and{{x},{y,x}}. These are *all* the same set.And by the way, in the special case that y=x you can even writethis ordered pair as {{x}}. See why?The question is not whether this set has a unique representationwith curly braces; its whether the set unambiguously denes theordered pair. That is, if two different ordered pairs turned outto be the same set by denition, then the denition would be nogood. Can you nd any such case? For example, write out as manydifferent expressions for (y,x) as you please, and see if any ofthem matches an expression for (x,y). None will, except of coursein the special case that x=y.The ordering of the pair is preserved not in the ordering of theset (theres none!) but rather in the fact that the rst elementof the pair appears as the one and only singleton in the set. === >I was curious as to why an ordered pair (x,y) would be dened as>{{x}{x,y}}. I understand that a set has no order but therefore, wouldnt>that be the same as {{x,y} {x}}.Yes, it would. But the point is that {{x,y},{x}} = {{z,w},{z}} if andonly if x=z and y=w. So that (x,y) = (z,w) if and only if x=z and y=w,so that (x,y)=(y,x) if and only if x=y. So its not that the set isnow ordered, is that the set {{x}, {x,y}} can be used to say which ofx and y goes rst and which goes second.= Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)= Arturo Magidinmagidin@math.berkeley.eduX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9HFuwt08500; === 3 men went to a motel. Theres only 1 room left and it costs 30bucks. So, each man forked out 10 bucks.Later, the owner discovered he over-charged the 3 men. It should be25 bucks instead. So, he sent his runner to give 5 bucks back to the3 men. The 3 men think the owner is very honest, and thus each took 1buck back, leaving 2 bucks to the runner as tips.Now, since they took 1 buck back, each of them paid 9 bucks. 2 bucks to the runner. So: (9*3)+2 = 29. Wheres the 1 dollar?You are add/sub. unrelated values that should not be add/sub.Two of the 3 persons received $1.66 back and one took $ 1.67 which = a total of $5.00 returned.Then two of the persons pooled .66 cents each and the one whoreceived $1.67 pooled .67 cents giving the runner a $ 2.00 tip. Then you subtract $ 30.00 ( original cost) - $ 2.00 (tip) = $ 28.00( balance after tip) - $ 3.00 ( $ 1.00 each of the three men hadextra after tip) = $ 25.00 which is the nal total the three menhad to pay for the room excluding tip.Then again, forget all of the above and say the missing $ 1.00 wasleft as a tip for the maid. ;-)Portly X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9HFurN08456; === >>t(0)=x; t(n)=1-n*t(n-1); number in parenthesis is subscript>>dene R=t(20)>>show R(1-e^(-1))<1/21>a hint:prove by induction that, for n even,t(n) = n! * (sum (k=0,...,n) ((-1)^k / k!) - 1/e ).sorry, I forgot the x-argument:t(n)(1-1/e) = n! * (sum (k=0,...,n) ((-1)^k / k!) - 1/e ).Best wishes>Torsten. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9HIqq420222; === he must have meant latex to use latex if u r a beginner and have access to pc...i recommendTeXnic centeraravindI am new to here. I want to know if there are any softwares to write>equations.>One of my professor said something like letac or something like>that. I didnt ask for the spelling. I want to know if a software like>that exists.thnksSuresh === I recall what integration by parts means in a single variable, but Iam puzzled by a reference to it in rewriting a certain integral overdx,dt which occurs in a derivation of the Euler-Lagrange equation. Since there are no antiderivatives in more than one variable (so faras I can see?), the mnenomic about recognizing half the derivative ofthe product of two functions doesnt seem applicable.Is there a common meaning of integration by parts in severalvariables, is the author implying the technique is used in one of thesingle variable integrations composing the double integral, or is mybrain fried? Now that I think about it, it seems to me Ive seencasual references to integration by parts in other integrals overseveral variables -- possibly occuring in EM -- but I never reallytried to understand what was going on. === > I recall what integration by parts means in a single variable, but I> am puzzled by a reference to it in rewriting a certain integral over> dx,dt which occurs in a derivation of the Euler-Lagrange equation.> Since there are no antiderivatives in more than one variable (so far> as I can see?), the mnenomic about recognizing half the derivative of> the product of two functions doesnt seem applicable. Is there a common meaning of integration by parts in several> variables, is the author implying the technique is used in one of the> single variable integrations composing the double integral, or is my> brain fried? Now that I think about it, it seems to me Ive seen> casual references to integration by parts in other integrals over> several variables -- possibly occuring in EM -- but I never really> tried to understand what was going on.The idea in 2D is to write (del scalar).(vector) asdel.(scalar * vector) - scalar * (del.vector)The del.(scalar * vector) can then be converted into an integral ofscalar * vector . (outward normal) taken around the boundary in theright-hand screw rule sense.[More relevant to your specic problem, consider the abuse of notationdF/d(grad y) == (dF/d(dy/dx), dF/d(dy/dt))for scalars F(x,t,y, grad y) and y(x,t), all derivatives partial.]-- P.A.C. SmithIf the Apocalypse comes, beep me. <*> http://www.srcf.ucam.org/~pas51 boundary=----=_NextPart_000_0032_01C394ED.D1966400 === ------- ------------------------------------------------------------- -You have two rich aunts, Ruby is going to give you a $1000 a year until you are 17, Amber is going to start by giving you $100, and each subsequent year she will give you twice as much as the year before. If you assume you get exactly 17 payments, how much will you have in total from each aunt? Write an expression that shows how much money you have after x years for each scheme and determine when they are equal. Which is a better deal?How would I write the expression for this?Mark === ------------------------------------------------ --------------------- You have two rich aunts, Ruby is going to give you a $1000 a year until you are 17, Amber is going to start by giving you $100, and each subsequent year she will give you twice as much as the year before. If you assume you get exactly 17 payments, how much will you have in total from each aunt? Write an expression that shows how much money you have after x years for each scheme and determine when they are equal. Which is a better deal? How would I write the expression for this? MarkDidnt your teacher teach you about writing expressions before giving you this homework problem? === > You have two rich aunts, Ruby is going to give you a $1000 a year until > you are 17, Amber is going to start by giving you $100, and each > subsequent year she will give you twice as much as the year before. If > you assume you get exactly 17 payments, how much will you have in total > from each aunt? Write an expression that shows how much money you have > after x years for each scheme and determine when they are equal. Which > is a better deal?> How would I write the expression for this?You want to know how you would write an expression for this? I can anyof us tell?Besides, the subject of your post is misleading. The problem is notinteresting at all.Jose Carlos Santos === error or 0 ?===> error or 0 ?It depends on what you need for your application. As several other postshave pointed out, there is nothing that *always* works, no matter how 0/0arises. However, sometimes, its clear from context what 0/0 should be*in this one particular case*. Levinson and Redheffer (_Complex Variables_)stated this as all removable singularities are assumed to be removed,meaning that ne print details are omitted.It is my understanding that some computer languages dene it to beundened (or NAN) and some to be 0. I believe (but if its important,youd better check it yourself) that the relevant IEEE standard says itshould be undened, but I may be misremembering.Just do whatever works. But document it.Jon Miller === > error or 0 ?It might be dened in the limiting sense. E.g. f(x) = sin(x)/x. At x=0, this is 0/0. If we dene f(0) to be the limit of f(x) as x -> 0, it has the value 1, since sin(x) -> x as x -> 0.Gib === error or 0 ?It might be dened in the limiting sense. E.g. f(x) = sin(x)/x. At >x=0, this is 0/0. If we dene f(0) to be the limit of f(x) as x -> 0, >it has the value 1, And if you consider the function g(x) = sin(2x)/x then exactlythe same argument would give 0/0 = 2. Which is why 0/0is left undened.>since sin(x) -> x as x -> 0.Blech. Thats meaningless. You meant sin(x) ~ x as x -> 0.>Gib************************David C. Ullrich === >error or 0 ?It might be dened in the limiting sense. E.g. f(x) = sin(x)/x. At>x=0, this is 0/0. If we dene f(0) to be the limit of f(x) as x -> 0,>it has the value 1, And if you consider the function g(x) = sin(2x)/x then exactly> the same argument would give 0/0 = 2. Which is why 0/0> is left undened.since sin(x) -> x as x -> 0. Blech. Thats meaningless. You meant sin(x) ~ x as x -> 0.Gib ************************ David C. Ullrich0/0 is an emoticon for eyeballs with nerd glasses.Bob Pease === >> error or 0 ?> It might be dened in the limiting sense. E.g. f(x) = sin(x)/x. At > x=0, this is 0/0. If we dene f(0) to be the limit of f(x) as x -> 0, > it has the value 1, since sin(x) -> x as x -> 0.If 0/0 = 1 then 0/0 = 1*n for arbitrary n. Then 1 = 0/0 = n hence 1 = n. That is the kind of grief one gets if 0/0 is dened. Therefore it is not dened.Bob Kolker === error or 0 ?It might be dened in the limiting sense. E.g. f(x) = sin(x)/x. At >x=0, this is 0/0. If we dene f(0) to be the limit of f(x) as x -> 0, >it has the value 1, since sin(x) -> x as x -> 0.This doesnt say anything about the 0/0, though. Its not possible todene 0/0 as anything real because doing so will lead immediately tocontradiction. Assume 0/0 is dened as Z, where Z is in R. Now,Z + Z = (0 + 0)/0 = Z <=> Z = 0, but lim(n->0, n/n) = 1 = Z, so 0 = 1. === > error or 0 ?lim_(x,y)->(0,y) (x/y) = 0lim_(x,y)->(x,0) (x/y) = inf=> lim_(x,y)->(0,0) (x/y) = not dened. === > error or 0 ?Neither. It is undened.Bob Kolker === The world *is* a nite set of resources...>Since this is sci.math, perhaps you can dene the terms of the abovesentence and the assumptions about the physical nature of the world thatjusties this statement. After all, even the lowly hydrogen atom hasinnitely many energy levels according to some theories. === >Since this is sci.math, perhaps you can dene the terms of the above>sentence and the assumptions about the physical nature of the world that>justies this statement. After all, even the lowly hydrogen atom has>innitely many energy levels according to some theories.But its still only one hydrogen atom.The *usefulness* of that hydrogen atom may be increased as we learnhow to use its different energy levels for various purposes - and, asI noted, for a given level of technology, these resources have onevalue, and for a higher level, they have a greater value.However, statements about what technology may be able to do in thefuture are forward-looking, as they say in the prospecti, and so,just as you cant tell me that a cure for AIDS is going to beavailable by month X, one could make a case for making futureprojections for feeding and housing humanity *as if* technologicalprogress (despite the absence of any big disaster) comes immediatelyto a screeching halt.Myself, I think _thats_ going a bit far.John Savardhttp://home.ecn.ab.ca/~jsavard/index.html === I recall what integration by parts means in a single variable, but I>am puzzled by a reference to it in rewriting a certain integral over>dx,dt which occurs in a derivation of the Euler-Lagrange equation.>Since there are no antiderivatives in more than one variable (so far>as I can see?), the mnenomic about recognizing half the derivative of>the product of two functions doesnt seem applicable.Is there a common meaning of integration by parts in several>variables, is the author implying the technique is used in one of the>single variable integrations composing the double integral, or is my>brain fried? Now that I think about it, it seems to me Ive seen>casual references to integration by parts in other integrals over>several variables -- possibly occuring in EM -- but I never really>tried to understand what was going on.>Stuff thats analogous to int. by parts comes up when youredealing with several variable. E.g. if you are integrating a divergence over a volume you can change that into a surface term (i.e.a boundry term) just like how you get a boundry term when integratingby parts in 1-d (where the boundry is just 2 points):That is, in 1-d:Integral[u dv/dx, {x,a,b}]= Integral[d(u v)/dx - v du/dx, {x,a,b}]= Integral[d(u v)/dx,{x,a,b}] - Integral[v du/dx, {x,a,b}]= u(b)v(b) - u(a)v(a) - Integral[v du/dx, {x,a,b}]= (the boundry term) - Integral[v du/dx, (x,a,b}]Whereas, in 3-d (let v be a vector, f a scalar, del isthe vector (d/dx,d/dy,d/dz), and . means dot product):Integral[f del.v, {volume}]= Integral[del.(f v) - v.del(f), {volume}]= Integral[del.(f v),{volume}] - Integral[v.del(f),{volume}]= Integral[(f v).{surface}] - Integral[v.del(f),{volume}]= (the boundry term) - Integral[v,del(f),{volume}]hmm... this notation is looking pretty nasty and esoteric...but you get the point. In 1-d integration by parts you geta surface term but the surface is just two points (theendpoints of integration). In 3-d you get a surface term whereyou evaluate your integrand over the surface of the volume ontegration.hope that helps,adam === EG>I recall what integration by parts means in a single variable,but I>am puzzled by a reference to it in rewriting a certain integral over>dx,dt which occurs in a derivation of the Euler-Lagrange equation.>Since there are no antiderivatives in more than one variable (so far>as I can see?), the mnenomic about recognizing half the derivative of>the product of two functions doesnt seem applicable....> Stuff thats analogous to int. by parts comes up when youre> dealing with several variable. E.g. if you are integrating a > divergence over a volume you can change that into a surface term (i.e.> a boundry term) just like how you get a boundry term when integrating> by parts in 1-d (where the boundry is just 2 points):> That is, in 1-d:Integral[u dv/dx, {x,a,b}]= Integral[d(u v)/dx - v du/dx, {x,a,b}]= Integral[d(u v)/dx,{x,a,b}] - Integral[v du/dx, {x,a,b}]= u(b)v(b) - u(a)v(a) - Integral[v du/dx, {x,a,b}]= (the boundry term) - Integral[v du/dx, (x,a,b}]Whereas, in 3-d (let v be a vector, f a scalar, del is> the vector (d/dx,d/dy,d/dz), and . means dot product):Integral[f del.v, {volume}]= Integral[del.(f v) - v.del(f), {volume}]= Integral[del.(f v),{volume}] - Integral[v.del(f),{volume}]= Integral[(f v).{surface}] - Integral[v.del(f),{volume}]= (the boundry term) - Integral[v,del(f),{volume}]> hmm... this notation is looking pretty nasty and esoteric...> but you get the point. In 1-d integration by parts you get> a surface term but the surface is just two points (the> endpoints of integration). In 3-d you get a surface term where> you evaluate your integrand over the surface of the volume of> integration.Bingo! Got it!In several dimensions, analogues of the fundamental theorem ofcalculus are of the form:Int[DifOp(f),{inside}] = Int[f,{boundary}],Where DifOp is differential operator.Such things had occured to me before. Assuming there is also ananalogue to the product rule in the form DifOp(fg) = fDifOp(g) +gDifOp(f), you are all set to integrate by parts.made my day! Now the rest of the day may be spent in sloth. ;-)BTW: Ive adopted your notation. Is that Mathematica or some such, orsomething you invented on the y for this post? === >>Suppose that:>>Y(x)=(k_1 - k_2/x)^.5>>Where the ks are positive constants.>>Is it proper to claim that Y is proportional to x^-.5? In the>>presentation, the symbol alpha was used in place of the words is>>proportional to and I was wondering if that was a rigorous use of>>that symbol.Youve already been told the answer is no. However, coming from a>physics background Ill add that under some circumstances, one might>say something like Y varies as x^0.5 for small x, if that was the>regime of interest. The physicist would write something like> Y ~ x^-0.5 for small x >Coming from a physics background myself, I must disagree. A goodphysicist would not write something like y ~ x^-.05 _for small x_because it doesnt matter if x itself is small...In fact, since x probably is dimensionfull, smallness is relative.x is small relative to what scale? Also, the constantsprobably are dimensionful... at least one of them is if x is.So what _really_ matters (as I said in my rst post) is ifk_1 << k_2/xsince k_1 must have the same dimensions as k_2/x it is properto compare these quantities... and if k_2/x is much bigger thank_1, THEN y is approx. proportional to x^-.5whew... sorry about ranting. cheers,adam === >Suppose that:Y(x)=(k_1 - k_2/x)^.5Where the ks are positive constants.Is it proper to claim that Y is proportional to x^-.5? In the>presentation, the symbol alpha was used in place of the words is>proportional to and I was wondering if that was a rigorous use of>that symbol.>>[...]>since k_1 must have the same dimensions as k_2/x it is proper>to compare these quantities... and if k_2/x is much bigger than>k_1, THEN y is approx. proportional to x^-.5>If k1 << k2/x, then Y in this case is complex and multiply valued. Somehow, I suspect that this was not the intent in the rst place.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === The Lagrangian density ... so much Lagrangian L per unit volume ...>suggests we think of L = T - V as a kind of stuff, possibly with the>same justication as thinking of energy and momentum -- as much or as>little.Just what kind of stuff would it be? Is there any natural>interpretation of what Lagrangian is, or is it to simply be regarded>as a function which encodes the dynamics of the system in its>dependence on the variables, and magically recovers the ordinary>equations of motion via a rule set?>Nah, I dont think of L as a kind of stuff... Id saythat I think of H = T + V as a kind of stuff (energy). Butwhen I think of L I just think of that which, when integratedover time, is minimized to nd the classical path...orthat which gives the correct equation of motion when pluggedinto the Euler-Lagrange eqns.So, in that case, the lagrangian density is that which, whenintegrated over the _four-volume_, is minimized to nd theclassical path.I guess there is something nice about the lagragian density, becauseto work with it, we do not seperate out time as something specialto integrate over. The lagrangian density treats time and spaceequally, which is good. It helps a lot when you are trying tounify special relativity and quantum mechanics into QFT.adam === Let O be a bounded open set in R^n and f: O --> R^n a C^1 function withcompact support, i.e. it is 0 off a compact set. Is it true that theintegral of div f (the divergence of f) over O is 0?I know that if the boundary of O is nice, this follows from the Stokestheorem.My concern is what if the boundary of O is bad. Will the integral be still0?I believe that one can nd an open subset Q of O containing supp f, thathas a nice boundary so that Stoke theorem applies. But where can I nd sucha proof? === > Let O be a bounded open set in R^n and f: O --> R^n a C^1 function with> compact support, i.e. it is 0 off a compact set. Is it true that the> integral of div f (the divergence of f) over O is 0?> I know that if the boundary of O is nice, this follows from the Stokes> theorem.> My concern is what if the boundary of O is bad. Will the integral be still> 0?> I believe that one can nd an open subset Q of O containing supp f, that> has a nice boundary so that Stoke theorem applies. But where can I nd such> a proof?Urysohns Lemma should give you the set youre looking for, buttheres really no need for it. The integral of div f over O is thesame as the integral of div f over any set containing supp f. So youcan just ignore O and replace it by something nicer. === f can be seen has f:Q->R^n with Q containing O and Q with smooth boundary :D(for example Q=a ball)-GS-> Let O be a bounded open set in R^n and f: O --> R^n a C^1 function with> compact support, i.e. it is 0 off a compact set. Is it true that the> integral of div f (the divergence of f) over O is 0?> I know that if the boundary of O is nice, this follows from the Stokes> theorem.> My concern is what if the boundary of O is bad. Will the integral be still> 0?> I believe that one can nd an open subset Q of O containing supp f, that> has a nice boundary so that Stoke theorem applies. But where can I ndsuch> a proof? === I dont see why Q can be a ball.> f can be seen has f:Q->R^n with Q containing O and Q with smooth boundary:D> (for example Q=a ball) -GS-Let O be a bounded open set in R^n and f: O --> R^n a C^1 function with>compact support, i.e. it is 0 off a compact set. Is it true that the>integral of div f (the divergence of f) over O is 0?>I know that if the boundary of O is nice, this follows from the Stokes>theorem.>My concern is what if the boundary of O is bad. Will the integral bestill>0?>I believe that one can nd an open subset Q of O containing supp f,that>has a nice boundary so that Stoke theorem applies. But where can I nd> such>a proof? === >I dont see why Q can be a ball.Any compact set, for example the support of your function,is contained in some ball. This follows from the Heine-Boreltheorem...>> f can be seen has f:Q->R^n with Q containing O and Q with smooth boundary>:D>> (for example Q=a ball)>-GS->>Let O be a bounded open set in R^n and f: O --> R^n a C^1 function with>>compact support, i.e. it is 0 off a compact set. Is it true that the>>integral of div f (the divergence of f) over O is 0?>>I know that if the boundary of O is nice, this follows from the Stokes>>theorem.>>My concern is what if the boundary of O is bad. Will the integral be>still>>0?>>I believe that one can nd an open subset Q of O containing supp f,>that>>has a nice boundary so that Stoke theorem applies. But where can I nd>> such>>a proof?>************************David C. Ullrich === O is bounded. There exists r>0 such that B(0,r)=:Q contains O. f can be seenas a C^1 function Q->R^n, with f(x)=0 if x is in QO.And then you can apply Stokes.GS> I dont see why Q can be a ball.f can be seen has f:Q->R^n with Q containing O and Q with smoothboundary> :D>(for example Q=a ball)-GS-Let O be a bounded open set in R^n and f: O --> R^n a C^1 functionwith>> compact support, i.e. it is 0 off a compact set. Is it true that the>> integral of div f (the divergence of f) over O is 0?>> I know that if the boundary of O is nice, this follows from theStokes>> theorem.>> My concern is what if the boundary of O is bad. Will the integral be> still>> 0?>> I believe that one can nd an open subset Q of O containing supp f,> that>> has a nice boundary so that Stoke theorem applies. But where can Ind>such>> a proof?>> === >>Let A be the real matrix>> / 0 a b c >> A= / a 0 d e >> b d 0 f / >> c e f 0 / >>where a^2+b^2+c^2+d^2+e^2+f^2 = 6 .>>Consider that r_1 =< r_2 =< r_3 =< r_4 are eigenvalues of A.>>It is known that r_4= 10. >>Its true that all eigenvalues are integer numbers ?>>If A is non-singular, which is the inverse matrix A^{-1} ?>>It cant happen that r_4 = 10. The characteristic polynomial of>A is t^4 - (a^2+b^2+...+f^2) t^2 + c1 t + c0 for some c1 and c0.>So youd need r1 + r2 + r3 + r4 = 0 and >-6 = r1 r2 + r1 r3 + r1 r4 + r2 r3 + r2 r4 + r3 r4 > = 1/2 ((r1 + r2 + r3 + r4)^2 - r1^2 - r2^2 - r3^2 - r4^2)>i.e. r1^2 + r2^2 + r3^2 + r4^2 = 12. In particular >r4 <= 2 sqrt(3). Suppose we remove the requirement r_4=10. The only ways to >get 12 as a sum of four squares are 1^2 + 1^2 + 1^2 + 3^2 and >0^2 + 2^2 + 2^2 + 2^2, so the only integer solutions to>r1+r2+r3+r4=0 and r1^2+r2^2+r3^2+r4^2=12 with r1<=r2<=r3<=r4>are [-1,-1,-1,3] and [-3,1,1,1]. It is possible to get these>eigenvalues, e.g. with a=b=c=d=e=f=1 or a=b=c=d=e=f=-1 respectively.Robert Israel israel@math.ubc.ca>Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2> Robert Israel ,> According to your important remarks, condition> instead of r_4=10 we must have ,e.g. x_4=3 .=Other remarks.Let S(A)=a^2+b^2+c^2+e^2+f^2 . I have observed the following :1) Suppose S(A)=10*M^2 and min_{i=1,2,3}|r_i- r_{i+1}| >= 2M . Then [r_1,r_2,r_3,r_4]=[-3M,-M,M,3M]2) Let k=2 or k=3 . If r_k >= sqrt((4-k)*S(A)/(2k)) then eigenvalues are : [-T,-T,T,T] , T:= sqrt(S(A)/2) for k=3 , [-V,U,U,U ] , U:=sqrt(S(A)/6) , V=sqrt(3*S(A)/2) when k=2 . 3) If r_4 >= sqrt(2*S(A)/2) , then we have the eigenvalues [-W,-W,-W,Z] , W=sqrt(S(A)/6) , Z=sqrt(3*S(A)/2) .4) Let us suppose r_1 >= - sqrt(S(A)/6) . The eigenvalues are as in case 3) .5) If r_4-r_1 >= 2*sqrt(S(A)) we obtain eigenvalues [-sqrt(S(a)),0,0, sqrt(S(A)) ].6) When r_4-r_1 =< sqrt(2*S(A)) , then spectrum of A is [ -Q,-Q,Q,Q ] , Q= sqrt(S(A)/2) .=C:|z|=3f(z) = (z^2)*(z+i) / {(z-2i)^2}*(z-3i)when we use residue theorem,can z=3i ignore ??only z=2i.......apply??how do you solved it? === thank you... === > C:|z|=3f(z) = (z^2)*(z+i) / {(z-2i)^2}*(z-3i)when we use residue theorem,can z=3i ignore ??only z=2i.......apply??how do you solved it?Your function f is written in a strange way. If the expressionfor f is interpreted literally, then f is a rational functionwhose numerator is z^2.(z + i).(z - 3i) and the denominator is(z - 2i)^2. Then the integral is easily calculated by theresidue theorem. However, if f is a rational function whosenumerator is z^2.(z + i) and the denominator is(z - 2i)^2.(z - 3i), then your problem simply makes no sense,since your function f is not dened at one of the points ofthe circle {z : |z| = 3}.Jose Carlos Santos === When using Cauchys Residue Theorem, we need to choose asimple closed contour which does not pass through any singularities.Now, f has singularities at 2i and 3i.If you choose C: |z|=1, then the contour integral is 0.If you choose C: |z|=4, then the contour integral is 2ki(Res(f,2i)+Res(f,3i) ).Michael Leunghot-girl .b9.a6.b9g.97.a6l.97sD:bmqc97$q9q$1@news.hananet.net...> C:|z|=3 f(z) = (z^2)*(z+i) / {(z-2i)^2}*(z-3i) when we use residue theorem, can z=3i ignore ?? only z=2i.......apply?? how do you solved it? === Show that for each sequence of random variables {X_n : n = 1,2,...} thereexist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_nconverges to zero with probability one. Note : There is no hypothesiswhatsoever on the random variables.How can I prove this? This seems like a strange assertion to me that lacksintuition.Steve === > Show that for each sequence of random variables {X_n : n = 1,2,...} there> exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n> converges to zero with probability one. Note : There is no hypothesis> whatsoever on the random variables.How can I prove this? This seems like a strange assertion to me that lacks> intuition.First, I dont see why (as another poster mentioned) we would need theX_ns dened on the same probability space. The resulting randomsequence lives on the innite product of the probability spaces ofthe X_ns, which we can construct without needing the individualspaces to be the same.Now, for each n, choose a_n so large that P(|X_n| > a_n/n) < 2^{-n}.By the Borel-Cantelli Lemma, the probability that |X_n| > a_n/n forinnitely many n is zero. Thus, with probability one, |X_n/a_n| <=1/n for all n sufciently large, i.e. X_n/a_n --> 0.Steve === > Show that for each sequence of random variables {X_n : n = 1,2,...} there> exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n> converges to zero with probability one. Note : There is no hypothesis> whatsoever on the random variables.How can I prove this? This seems like a strange assertion to me that lacks> intuition.The intuition I can just possibly supply, if not the proof.Hmm... well, on second thought the intuition on second thought is a_little_ tricky! But essentially you are claiming that the randomvariables in your sequence are well-behaved random variables (aredundant claim), and so you can pick a sequence of real numbers whichdominate them in a well-dened sense.For example its clear (I hope) that we can pick a sequence ofconstants such that X_n/a_n converges to zero in expectation: i.e.,the real series E[X_n/a_n] converges to zero. Thats one kind ofdominance.A different kind of dominance would be picking the a_n such that, forsome other sequence of constants c_n which converge to zero, X_n/a_n exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n>converges to zero with probability one. Note : There is no hypothesis>whatsoever on the random variables.How can I prove this? This seems like a strange assertion to me thatlacks>intuition. The intuition I can just possibly supply, if not the proof. Hmm... well, on second thought the intuition on second thought is a> _little_ tricky! But essentially you are claiming that the random> variables in your sequence are well-behaved random variables (a> redundant claim), and so you can pick a sequence of real numbers which> dominate them in a well-dened sense. For example its clear (I hope) that we can pick a sequence of> constants such that X_n/a_n converges to zero in expectation: i.e.,> the real series E[X_n/a_n] converges to zero. Thats one kind of> dominance.>I guess one problem I have is that the above claim is not clear (the claimabout expectation). Why is this true?> A different kind of dominance would be picking the a_n such that, for> some other sequence of constants c_n which converge to zero, X_n/a_n <> c_n with probability one. That dominance would be stronger than the> rst kind of dominance, and _that_ assertion in fact is false: no> matter how much we try to squeeze the series with the a_n there> remains a positive probability that any given element of the sequence> of X_n will be sufciently large to break our claim, even before we> add for all n. We have a kind of Goldilocks dominance search: the rst kind was too> weak (easy to prove), the second kind was too strong, and maybe the> last kind is _just right_. ;-) You will have to carefully consider> the denition of convergence and maybe reach a proof by contradiction> -- what we are saying is that, while the sequences may have excursions> on their travel to zero, we will almost always be able to travel far> enough out on the sequence to nd that the excursions never again> rise beyond any given level. Good luck. === >>Show that for each sequence of random variables {X_n : n = 1,2,...}>there>>exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n>>converges to zero with probability one. Note : There is no hypothesis>>whatsoever on the random variables.>>How can I prove this? This seems like a strange assertion to me that>lacks>>intuition.>The intuition I can just possibly supply, if not the proof.>Hmm... well, on second thought the intuition on second thought is a>> _little_ tricky! But essentially you are claiming that the random>> variables in your sequence are well-behaved random variables (a>> redundant claim), and so you can pick a sequence of real numbers which>> dominate them in a well-dened sense.>For example its clear (I hope) that we can pick a sequence of>> constants such that X_n/a_n converges to zero in expectation: i.e.,>> the real series E[X_n/a_n] converges to zero. Thats one kind of>> dominance. >I guess one problem I have is that the above claim is not clear (the claim>about expectation). Why is this true?It doesnt matter. Its a fact that |X_n| is nite almost surely. It follows that P(|X_n| > a) -> 0 as a -> innity. Hence you canpick a_n such that P(|X_n| > a_n/n} < ___, and if you ll in theblank properly that shows that X_n -> in probability.>> A different kind of dominance would be picking the a_n such that, for>> some other sequence of constants c_n which converge to zero, X_n/a_n <>> c_n with probability one. That dominance would be stronger than the>> rst kind of dominance, and _that_ assertion in fact is false: no>> matter how much we try to squeeze the series with the a_n there>> remains a positive probability that any given element of the sequence>> of X_n will be sufciently large to break our claim, even before we>> add for all n.>We have a kind of Goldilocks dominance search: the rst kind was too>> weak (easy to prove), the second kind was too strong, and maybe the>> last kind is _just right_. ;-) You will have to carefully consider>> the denition of convergence and maybe reach a proof by contradiction>> -- what we are saying is that, while the sequences may have excursions>> on their travel to zero, we will almost always be able to travel far>> enough out on the sequence to nd that the excursions never again>> rise beyond any given level. Good luck.>************************David C. Ullrich === > Show that for each sequence of random variables {X_n : n = 1,2,...} there> exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n> converges to zero with probability one. Note : There is no hypothesis> whatsoever on the random variables.How can I prove this? This seems like a strange assertion to me that lacks> intuition.> Can you choose a_n so that Prob{X_n > a_n/n} goes to 0?Thats not enough in itself, but its a start. === > Show that for each sequence of random variables {X_n : n = 1,2,...} there> exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n> converges to zero with probability one. Note : There is no hypothesis> whatsoever on the random variables.How can I prove this? This seems like a strange assertion to me that lacks> intuition.Wait, you get to see X_n rst and then pick a_n? Analysis is*denitely* not my eld, but how about picking a_n = X_n * n; thisway,lim n->oo (X_n/a_n) = lim n->oo (1/n) = 0?If, on the other hand, youre asking for a sequence a_n that works forEVERY X_n, its clearly not possible; theres always a sequence X_n >=n*a_n. However, you can have probability 1 without having thestatement be true in general; i.e. if I am to pick an integer, theprobability that I will not pick 0 is 1, yet it is obviously notcertain that I will not pick 0. However, this statement still*sounds* untrue; I dont see offhand how to resolve it, but it cantbe too hard.Sorry if I sound like a moron; Im (a) just starting college and (b)going into algebra, not analysis. In fact I dont really like numbersanyway. Kinda weird to be going into math then, I guess, butwhatever. === >Show that for each sequence of random variables {X_n : n = 1,2,...} there>exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n>converges to zero with probability one. Note : There is no hypothesis>whatsoever on the random variables.How can I prove this? This seems like a strange assertion to me that lacks>intuition.Wait, you get to see X_n rst and then pick a_n? Analysis is> *denitely* not my eld, but how about picking a_n = X_n * n;but that a_n would not be constant, it would be random just like X_n is> this> way,lim n->oo (X_n/a_n) = lim n->oo (1/n) = 0?If, on the other hand, youre asking for a sequence a_n that works for> EVERY X_n, its clearly not possible; theres always a sequence X_n >=> n*a_n. However, you can have probability 1 without having the> statement be true in general; i.e. if I am to pick an integer, the> probability that I will not pick 0 is 1, yet it is obviously not> certain that I will not pick 0. However, this statement still> *sounds* untrue; I dont see offhand how to resolve it, but it cant> be too hard.Sorry if I sound like a moron; Im (a) just starting college and (b)> going into algebra, not analysis. In fact I dont really like numbers> anyway. Kinda weird to be going into math then, I guess, but> whatever. === >Show that for each sequence of random variables {X_n : n = 1,2,...} there>exist a sequence of constants {a_n : n = 1,2,...} such that X_n/a_n>converges to zero with probability one. Note : There is no hypothesis>whatsoever on the random variables.How can I prove this? This seems like a strange assertion to me that lacks>intuition.>First note that, for the problem to make sense, all the random varaibles must have as domain the same probability space. Heres an idea, though I am not sure it will work: Can you rst show it for discrete probability spaces, and then use some limiting (= approximation) arguments to get the general case?-- Stephen J. Herschkorn herschko@rutcor.rutgers.eduX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9I0ODn09862; === >> ...>>Yes. And so what?>>Readers should note that the poster admits that his position requires>>a function that is either a unit or has f as its only non-unit factor>>depending on certain conditions as m varies from 0 to positive>>innity. A condition follows.>>Yes, I still see no problem with such a function.Then give a *single* function in ALL of mathematics which behaves as>> you wish.Youre making up some wacky mathematics.Oh, come on. Do you know the Moebius function?>> mu(n) = 1 if n = 1>> 0 if n is divisible by a square>> (-1)^r if n = the product of r distinct primes.>> Wacky enough? But indeed, not the function we seek, see below.>> Ive found a simpler refutation of this posters position.Remember I haveP(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)>Variables: m, f, x, u E Ring of Algebraic integersand the factorization>P(m) = (a_1(m) x + uf)(a_2(m) x + uf)(a_3(m) x + uf)>Variables: a_1, a_2, a_3, roots of cubic dened as follows.>Cubic: a^3 + 3(-1+mf^2)a^2-f^2(m^3 f^4 - 3m^2 f^2 + 3m)The poster wishes for a variable function which has the property of>being a factor of f, so Ill use w_1(m) w_2(m) w_3(m) = f^2, where a_1>has w_1(m) as a factor for all integer m.Then a_1(m) x + uf must have w_1(m) as a factor, so dividing through>givesa_1(m)/w_1(m) + uf/w_1(m)where uf/w_1(m) cant be an algebraic integer for all integer m.It might help for me to put in actual numbers for u and f, which I can>do as the variables are independent of each other, so let u=2, f=7,>then itsa_1(m)/w_1(m) + 14/w_1(m)and clearly, if w_1(m) varies with m, then 14/w_1(m) is not an>algebraic integer for all integer m.>No, not clearly.>For those who STILL need help, consider that if you had 14/w_1(m) = r(m)introducing r(m) for the result of the division, thenw_1(m) r(m) = 14sow_1(m) r(m) - 14 = 0which would force zeroes for m.That is, you cant have algebraic integer functions, that is functions>that give algebraic integer results, and not have only certain values>of m that would work.>Patently false. Let w_1(m) = 14^{1/m} andr(m) = 14^{(m - 1)/m}. Both of these are nonconstantalgebraic integer functions of the integer argument m,and their product is 14. There are trillions(low estimate ...) of other examples. Remember,the ring of algebraic integers is extremely large.>That is, you can have something like 2m+ 7 = 21, that works for a>particular value of m, but you cant have functions in algebraic>integers that will multiply to give 14 for all integer m.To get such functions, you have to go outside the ring into a eld.>No indeedy; see up above.>That refutes the position of Dik T. Winter, and note that as Ive>pointed out that poster clearly either has limited mathematical>ability, or hes been lying now for some time.>Dik may have limited ability, but the limit is prettyhigh, and I have not known him to lie at all.One principle worth remembering here: when you haveeliminated almost every explanation you can think of, andthe only things that are left seem farfetched, whatever theyare, if they cannot be eliminated, they must be part of the truth.James Harris === > ...>>Yes. And so what?>>Readers should note that the poster admits that his position requires>>a function that is either a unit or has f as its only non-unit factor>>depending on certain conditions as m varies from 0 to positive>>innity. A condition follows.>>Yes, I still see no problem with such a function.Then give a *single* function in ALL of mathematics which behaves as>> you wish.Youre making up some wacky mathematics.Oh, come on. Do you know the Moebius function?>> mu(n) = 1 if n = 1>> 0 if n is divisible by a square>> (-1)^r if n = the product of r distinct primes.>> Wacky enough? But indeed, not the function we seek, see below.>> Ive found a simpler refutation of this posters position.Remember I haveP(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)Variables: m, f, x, u E Ring of Algebraic integersand the factorizationP(m) = (a_1(m) x + uf)(a_2(m) x + uf)(a_3(m) x + uf)Variables: a_1, a_2, a_3, roots of cubic dened as follows.Cubic: a^3 + 3(-1+mf^2)a^2-f^2(m^3 f^4 - 3m^2 f^2 + 3m)The poster wishes for a variable function which has the property of>being a factor of f, so Ill use w_1(m) w_2(m) w_3(m) = f^2, where a_1>has w_1(m) as a factor for all integer m.Then a_1(m) x + uf must have w_1(m) as a factor, so dividing through>givesa_1(m)/w_1(m) + uf/w_1(m)where uf/w_1(m) cant be an algebraic integer for all integer m.It might help for me to put in actual numbers for u and f, which I can>do as the variables are independent of each other, so let u=2, f=7,>then itsa_1(m)/w_1(m) + 14/w_1(m)and clearly, if w_1(m) varies with m, then 14/w_1(m) is not an>algebraic integer for all integer m. > No, not clearly.And the poster is incorrect, as below he gives the example of 14^{1/m}which blows up at m=0. >For those who STILL need help, consider that if you had 14/w_1(m) = r(m)introducing r(m) for the result of the division, thenw_1(m) r(m) = 14sow_1(m) r(m) - 14 = 0which would force zeroes for m.That is, you cant have algebraic integer functions, that is functions>that give algebraic integer results, and not have only certain values>of m that would work. > Patently false. Let w_1(m) = 14^{1/m} and> r(m) = 14^{(m - 1)/m}. Both of these are nonconstant> algebraic integer functions of the integer argument m,> and their product is 14. There are trillions> (low estimate ...) of other examples. Remember,> the ring of algebraic integers is extremely large.And his w_1(m) blows up at m=0.Math society is rather despicable in its weakness, and I fear itsbecause people are taught to believe that they make up math, as its not about logic but what makes them feel good.Rather than just accept the truth, math society continues to arguewith me, showing its base irrationality. >That is, you can have something like 2m+ 7 = 21, that works for a>particular value of m, but you cant have functions in algebraic>integers that will multiply to give 14 for all integer m.To get such functions, you have to go outside the ring into a eld. > No indeedy; see up above.Its actually trivial that for all integer m, you cant have arelation like14/w_1(m) = r(m)with functions varying with m, any more than you can have y=14/xin the ring of algebraic integers, for all integer x.Its really trivial but math society wishes to hide the error in coremathematics--that over hundred year old denition error--and incontinuing to ght the truth, they show their true disdain forconsistency in mathematics itself.It IS a despicable display. >That refutes the position of Dik T. Winter, and note that as Ive>pointed out that poster clearly either has limited mathematical>ability, or hes been lying now for some time. > Dik may have limited ability, but the limit is pretty> high, and I have not known him to lie at all.One principle worth remembering here: when you have> eliminated almost every explanation you can think of, and> the only things that are left seem farfetched, whatever they> are, if they cannot be eliminated, they must be part of > the truth.Which is actually something good for readers to consider when theynd it hard to believe that math society could really be thiscorrupt.But remember, just like you cant have xy=2, for all integer x, wherex and y are both integers, you cant have 14/w_1(m) = r(m) oruf/w_1(m) = r(m), where w_1(m) and r(m) are algebraic integerfunctions for all integer m.So when you accept what you know about mathematics, and consider thatmathematicians have been arguing with me for months, with whatconclusion are you left?Mathematicians have gone rogue, and are acting against world societyas they try to convince by lying about mathematics, and you can seejust how corrupt they have become, as now even when caught, theyrestill lying.James Harris === > ...>>Yes. And so what?>>Readers should note that the poster admits that his position requires>>a function that is either a unit or has f as its only non-unit factor>>depending on certain conditions as m varies from 0 to positive>>innity. A condition follows.>>Yes, I still see no problem with such a function.Then give a *single* function in ALL of mathematics which behaves as>> you wish.Youre making up some wacky mathematics.Oh, come on. Do you know the Moebius function?>> mu(n) = 1 if n = 1>> 0 if n is divisible by a square>> (-1)^r if n = the product of r distinct primes.>> Wacky enough? But indeed, not the function we seek, see below.>> Ive found a simpler refutation of this posters position.Remember I haveP(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)Variables: m, f, x, u E Ring of Algebraic integersand the factorizationP(m) = (a_1(m) x + uf)(a_2(m) x + uf)(a_3(m) x + uf)Variables: a_1, a_2, a_3, roots of cubic dened as follows.Cubic: a^3 + 3(-1+mf^2)a^2-f^2(m^3 f^4 - 3m^2 f^2 + 3m)The poster wishes for a variable function which has the property of>being a factor of f, so Ill use w_1(m) w_2(m) w_3(m) = f^2, where a_1>has w_1(m) as a factor for all integer m.Then a_1(m) x + uf must have w_1(m) as a factor, so dividing through>givesa_1(m)/w_1(m) + uf/w_1(m)where uf/w_1(m) cant be an algebraic integer for all integer m.It might help for me to put in actual numbers for u and f, which I can>do as the variables are independent of each other, so let u=2, f=7,>then itsa_1(m)/w_1(m) + 14/w_1(m)and clearly, if w_1(m) varies with m, then 14/w_1(m) is not an>algebraic integer for all integer m. > No, not clearly.And the poster is incorrect, as below he gives the example of 14^{1/m}which blows up at m=0. >For those who STILL need help, consider that if you had 14/w_1(m) = r(m)introducing r(m) for the result of the division, thenw_1(m) r(m) = 14sow_1(m) r(m) - 14 = 0which would force zeroes for m.That is, you cant have algebraic integer functions, that is functions>that give algebraic integer results, and not have only certain values>of m that would work. > Patently false. Let w_1(m) = 14^{1/m} and> r(m) = 14^{(m - 1)/m}. Both of these are nonconstant> algebraic integer functions of the integer argument m,> and their product is 14. There are trillions> (low estimate ...) of other examples. Remember,> the ring of algebraic integers is extremely large.And his w_1(m) blows up at m=0.Math society is rather despicable in its weakness, and I fear itsbecause people are taught to believe that they make up math, as its not about logic but what makes them feel good.Rather than just accept the truth, math society continues to arguewith me, showing its base irrationality. >That is, you can have something like 2m+ 7 = 21, that works for a>particular value of m, but you cant have functions in algebraic>integers that will multiply to give 14 for all integer m.To get such functions, you have to go outside the ring into a eld. > No indeedy; see up above.Its actually trivial that for all integer m, you cant have arelation like14/w_1(m) = r(m)with functions varying with m, any more than you can have y=14/xin the ring of algebraic integers, for all integer x.Its really trivial but math society wishes to hide the error in coremathematics--that over hundred year old denition error--and incontinuing to ght the truth, they show their true disdain forconsistency in mathematics itself.It IS a despicable display. >That refutes the position of Dik T. Winter, and note that as Ive>pointed out that poster clearly either has limited mathematical>ability, or hes been lying now for some time. > Dik may have limited ability, but the limit is pretty> high, and I have not known him to lie at all.One principle worth remembering here: when you have> eliminated almost every explanation you can think of, and> the only things that are left seem farfetched, whatever they> are, if they cannot be eliminated, they must be part of > the truth.Which is actually something good for readers to consider when theynd it hard to believe that math society could really be thiscorrupt.But remember, just like you cant have xy=2, for all integer x, wherex and y are both integers, you cant have 14/w_1(m) = r(m) oruf/w_1(m) = r(m), where w_1(m) and r(m) are algebraic integerfunctions for all integer m.So when you accept what you know about mathematics, and consider thatmathematicians have been arguing with me for months, with whatconclusion are you left?Mathematicians have gone rogue, and are acting against world societyas they try to convince by lying about mathematics, and you can seejust how corrupt they have become, as now even when caught, theyrestill lying.James Harris === >>As I have written already multiple times, it is trivially monic in other>>rings, the cubic (x - b1)(x - b2)(x - b3) serves just ne. But you>>want to go to a ring where your divisibility claims work. It includes>>the ring of algebraic integers. But which of the two numbers>> (-sqrt(7) + sqrt(15))/2 and (-sqrt(7) - sqrt(15))>>is a unit in that ring? And why precisely that number?>...>(B) In the Object ring, this marvellous construction that xes the>ring of algebraic integers by allowing numbers that should be in,>there are polynomias with integer coefcients, irreducible over Q,>which have ->SOME<- roots in the ring but not all of them.And in case (B), how do we know which one is there, as you ask?I think (B) is indeed precisely the case. James wants all those numbers>in his ring such that the cubics in the bs are monic with numbers in>his rings for all possible polynomials P(m). So when we look at the>polynomial in a for m=1 and f=2 we get (in the algebraic integers)>three roots that are obviously not coprime to f. To get at his claim>that exactly *one* of the as is coprime to f he has to make one of>those common factors (that I have written above) units. They are both>roots of an irreducible polynomial with integer coefcients of degree 4.>Actually sometime ago he made a remark that suggested this.The main problem is that he has to make this choice for each and every>polynomial that comes up. Now I am not sure, but it may be possible>to have a ring where only a single root of a quadratic is in his>ring,Depends on the quadratic and the factors involved (in order for thering A[w], with A the algebraic integers and w an algebraic number) tointersect Q exactly on the integers, certain conditions on w must bemet; not any old w will do. So he would have to prove that there isalways one of the factors in his situation where this condition ismet.>I very much doubt that a choice can be made across *all* polynomials>he comes up with without getting conicts. He has to show that a>consistent choice can be made.That is probably the most difcult part, if not impossible. If youllremember, back when he was factoring his cubic as(v^3+1)x^3 - 3vx + 2 = (sqrt(v+1)x + b1)(sqrt(v+1)x + b2)((v^2-v+1)x +b3),the issue of whether Q[sqrt(v+1),b1,b2,b3] contained 1/f arose. Andwhile the issue was never settled for that case, I did manage to showthat if you took the ring you got by adding ALL values of sqrt(v+1),b1, b2, b3, for all values of v different from -1, then that ringcontained all primes congruent to 1 modulo 6, even though it was notclear that they were all in any specic extensionQ[sqrt(v+1),b1,b2,b3].(Reference:http:// groups.google.com/groups?selm=9p2ada%241kva%241% 40agate.berkeley.eduSept 28, 2001)Taking possible values of v, even restricting to v=-1+mf^{2j}, with fand j xed, it may indeed (and probably is) impossible to chooseroots in a consistent way across all values of m (or even a restrictedbut innite set of values of m) that keep the resulting ringcontained in C - (Q-Z).== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of gures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indenite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan== Arturo Magidinmagidin@math.berkeley.edu === [.snip.]>Its actually trivial that for all integer m, you cant have a>relation like14/w_1(m) = r(m)with functions varying with m, any more than you can have y=14/xin the ring of algebraic integers, for all integer x.The two situations are completely dissimilar, and it betrays acomplete confusion on your part that you would try to conate them.In the rst situation, 14/w_1(m) = r(m), the denominator is notallowed to take ALL integer values, or ALL algebraic integer values;it is merely assumed to be a function whose DOMAIN is the integers,and whose image is not a singleton (you again forgot to mention youmean non-constant function).By contrast, in the latter situation you are demanding that a specicfunction, the identity, be used.If you wanted to make a similar situation, you would have to ask that14/w(x) be an integer, where w(x) is a function dened for allintegers and that takes integer values. In that case, you surely agreethat there are plenty of functions that will do the trick and whichare neither constant nor given by polynomials; for example, w(x) = 1if x is not divisible by 7, w(x) = 2 if x is divisible by 7 but not by2, and w(x) = 7 if x is divisible by 14.As to your original situation,>Its actually trivial that for all integer m, you cant have a>relation like14/w_1(m) = r(m)with functions varying with m, not only had you already been given several that are dened for allintegers, but here is one (I have a variant of a restriction elsewhere)which is dened for all complex numbers (and in particular, that isdened for all algebraic numbers, all algebraic integers, allintegers). It is trivial how one could extend it to any set largerthan the complex numbers, in any number of ways.(No, this is not the function we have here, but it makes the pointthat your claims that no such function can exist are just simplywrong): Let w_1(m) be dened as follows:(1) If m is a transcendental positive real number, then let w_1(m) = sqrt(7).(2) If m is a transcendental negative real number, let w_1(m) = sqrt(2).(3) If m is a transcendental complex number with Re(m)*Im(m)>0, let w_1(m) = 3-sqrt(2). If m is a transcendental complex number with Re(m)*Im(m)<0, let w_1(m)= 3+sqrt(2). If m is a transcendental purely imaginary number, then let w_1(m) = 7+sqrt(42).(4) If m is an algebraic integer, then let f(x) be the unique monic polynomial with rational coefcients which has m as a root and is irreducible over Q: f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1 x + a_0. Note that by irreducibility, a_0 is not equal to zero. Write a_0=r/s, where r and s are integers, s>0, and gcd(r,s)=1. We now dene an auxiliary prime p: If r = 1 or r=-1, let p=5 (or your favorite prime, if you dont like 5). If r is not equal to 1 or -1, let p be the largest rational prime that divides r. Consider all the roots of the monic polynomial with integer coefcients x^5 - 14px^4 + px - 14. (or your favorite monic polynomial with integer coefcients that has a divisor of 14 as constant term). Among the ve roots, let w_1(m) be the largest root in the lexicographic order. That is, we order the complex numbers by the rule a + bi <= c + di (a,b,c,d reals) if and only if ap then w_1(p) is a root of x^5 - 14px^4 + px - 14;and w_1(q) is a root of x^5 - 14qx^4 + qx - 14.Any common root of both polynomials would be a root of theirdifference, which is 14(q-p)x^4 - (q-p)x = (q-p)x ( 14x^3 - 1), andthe only algebraic integer root of this is x=0, so w_1(p) and w_1(q)are distinct.A complicated function, yes, but the point is that there are manyreally complicated functions in the world, and they can do a lot ofthings that you do not seem to believe are possible.>Its really trivial And really wrong.>But remember, just like you cant have xy=2, for all integer x, where>x and y are both integers, you cant have 14/w_1(m) = r(m) or>uf/w_1(m) = r(m), where w_1(m) and r(m) are algebraic integer>functions for all integer m.No, they are not similar situations at all. Any integer can be writtenas a product of two integers in only a nite number of ways. Anyalgebraic integer can be written as the product of two algebraicintegers in an innite number of ways; and if the algebraic integeris not a unit (like 14), then it can be written as a product of twoalgebraic integers in an innite number of ->essentially different<-ways. [. rant about mathematicians lying because they do not accept Jamess false claims as true removed.] === == [Gabriele Rossetti] has left a vast body of writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee xe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn away from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a ash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do not nd it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman =Arturo Magidinmagidin@math.berkeley.eduX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9I4Vds24432; === >Ellipse is produced if a plane intersects only one nappe of a cone.Is there a way to get the parameters of the ellipse as a function of>the angle of the plane with the axis of the cone?I know that they depend on the height of the cone and diameter of the>circle at the bottom of the cone.(Plane that is perpendicular to the axis produces a circle.)Here is an interesting reference about that:http://mathworld.wolfram.com/DandelinSpheres.html === With the function y = 8/x, gives 2^4 = 4^2, Im wondering if there are other === > With the function y = 8/x, gives 2^4 = 4^2, Im wondering if there are otherIf I had a dollar for every time those old chestnut has been posted to sci.math, I would be a rich young man.Bob Kolker === > With the function y = 8/x, gives 2^4 = 4^2, Im wondering if there are> other functions which would have more than this.Note: Perhaps David McAnally and I dont understand your question. You mayneed to clarify matters. But anyway, hoping that I do understand...Youre saying that the relations x^y = y^x and y = 8/x have in common thetwo points (x,y) = (2,4) and (4,2). Thats true. But those relations alsohave three other points in common: (-2,-4), (-4,-2), and (Sqrt(8),Sqrt(8))[where Sqrt denotes the nonnegative square root].As to your wondering if there are other functions [than y = 8/x] whichwould have more than this: Of course! There are functions whose graphshave an innite number of points in common with the graph of the relationx^y = y^x. The simplest such other function would be y = x.David Cantrell === >With the function y = 8/x, gives 2^4 = 4^2, Im wondering if there are otherYou could try the function from (1,innity) to (1,innity) implicitly dened by the solution for y in the equation log y/y = log x/x such that either x and y are distinct or x = y = e.David McAnally-------------- === I should clarify, I want to have the points (2,4) and (4,2) in the function.With the function y = 8/x, gives 2^4 = 4^2, Im wondering if there areother You could try the function from (1,innity) to (1,innity) implicitly> dened by the solution for y in the equation log y/y = log x/x such that> either x and y are distinct or x = y = e. David McAnally -------------- === >I should clarify, I want to have the points (2,4) and (4,2) in the function.The points (2,4) and (4,2) are in the function that I specied.David McAnally>>With the function y = 8/x, gives 2^4 = 4^2, Im wondering if there are>other>You could try the function from (1,innity) to (1,innity) implicitly>> dened by the solution for y in the equation log y/y = log x/x such that>> either x and y are distinct or x = y = e.>David McAnally>--------------I need to prove a theorem that states:If R is a number set and for any 2 numbers there exists a point of R betweenthem, then every number is a limit point of R.Is this basically the same proof as proving that there is a rational numberbetween any 2 irrationals and there is an irrational between any 2 rationals? Or am I way off base here? I guess Im not too sure how I could easily makethe transition from that to this theorem or if its possible at all. === > I need to prove a theorem that states: If R is a number set and for any 2 numbers there exists a point of R> between them, then every number is a limit point of R.>... then every number of R? is a limit ...R = {0} union (1,2) satises the condition,yet 0 isnt a limit point of R. === I need to prove a theorem that states:>If R is a number set and for any 2 numbers there exists a point of R>> between them, then every number is a limit point of R.>... then every number of R? No, every number.>is a limit ...R = {0} union (1,2) satises the condition,It does? What is an element of R between -2 and -1?>yet 0 isnt a limit point of R.Re-read the question - assume that she meant is very easy.************************David C. Ullrich === I need to prove a theorem that states:>If R is a number set and for any 2 numbers there exists a point of R>> between them, then every number is a limit point of R.>... then every number of R? is a limit ...R = {0} union (1,2) satises the condition,>yet 0 isnt a limit point of R.Do you mean {0} union ]1, 2]? I think the OP left out the conditionthat the number set be open/closed. === >> I need to prove a theorem that states:>If R is a number set and for any 2 numbers there exists a point of R>> between them, then every number is a limit point of R.>... then every number of R? is a limit ...R = {0} union (1,2) satises the condition,>yet 0 isnt a limit point of R. Do you mean {0} union ]1, 2]?>No, I said ]1,2[ , however ]1,2] , that is (1,2] , will sufcewhen 0 is added.>I think the OP left out the condition> that the number set be open/closed.>Im supposed to be psychic?Heck no, are students in such a hurry or so lazy that theywont bother to ask coherent questions?Alas too many students need to learn how to state questions.The rationals, which are neither open nor closed,satises both the condition and the conclusion.So premises other than open could be divined.As for closed, { 1/n | n in N } / {0} is closed setwith only one limit point, namely 0. So closed doesntsufce as 1/2, 1/3, ... arent limit points of the set.Its pointless to guess what poorly written posts intend.If its of worth, the OP will soon make amends. === I need to prove a theorem that states: If R is a number set and for any 2 numbers there exists a point of R> between them, then every number is a limit point of R....> Im supposed to be psychic?> Heck no, are students in such a hurry or so lazy that they> wont bother to ask coherent questions?The OPs statement, exactly as given, is (a) true, and(b) easy to prove: if x is any number and e>0, then R contains a point r between x and x+e, so |x-r| < e. === Suppose that n,p are positive integers, and x1,x_2,... (x_k=/=0) are real numbers.Let us denoteA(x_1,x_2,...,x_n)= (SUM_{k=1 to k=n}x_k)*(SUM_{k=1 to k=n} 1/x_k)B_p(x_1,x_2,...,x_n)= (SUM_{k=1 to k=n}x_k)^{p}*(SUM_{k=1 to k=n} 1/x_k^p)and assume that the following is true : === > Suppose that n,p are positive integers, and > x1,x_2,... (x_k=/=0) are real numbers.Let us denoteA(x_1,x_2,...,x_n)= (SUM_{k=1 to k=n}x_k)*(SUM_{k=1 to k=n} 1/x_k)B_p(x_1,x_2,...,x_n)= (SUM_{k=1 to k=n}x_k)^{p}*(SUM_{k=1 to k=n} 1/x_k^p)and assume that the following is true :> A(x_1,x_2,...,x_n)= 1 for all admissible x_k ,(x_k=/=0, x_k in R),> it follows> B_p(x_1,x_2,...,x_n)=1 for all odd integers p.Find n . Its true that n takes only one value , namely n=3 ?> [A Corrected form : Sorry !] Suppose that n,p are positive integers,(n>=2),and x1,x_2,... (x_k=/=0) are real numbers.DenoteA(x_1,x_2,...,x_n)= (SUM_{k=1 to k=n}x_k)*(SUM_{k=1 to k=n} 1/x_k)B_p(x_1,x_2,...,x_n)= (SUM_{k=1 to k=n}x_k)^{p}*(SUM_{k=1 to k=n} 1/x_k^p)and assume that the following is true :,, If A(x_1,x_2,...,x_n)= 1 ,then B_p(x_1,x_2,...,x_n)=1 for all odd integers p. Find n . Its true that n takes only one value , namely n=3 ?Alex/Proposer === >Sigh. Yes, the following, if the reasoning as (sic) actuallycorrect, can be easily formalized in ZF.... note that for example the in operator belowis not going to correspond to the in in ZF, itsgoing to be just some predicate, with axiomsinvolving it. > C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)>Classication> C4 AyAx[Az(z in y <-> z in x) -> {(set y & set x) <-> y=x}]> (Equi-membered classes are identical iff these are sets.)Put your money where your mouth is. Lets see your formalization.--John === >Sigh. Yes, the following, if the reasoning as (sic) actually>correct, can be easily formalized in ZF.... note that for example the in operator below>is not going to correspond to the in in ZF, its>going to be just some predicate, with axioms>involving it.>> C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)>>Classication>> >> C4 AyAx[Az(z in y <-> z in x) -> {(set y & set x) <-> y=x}]>> (Equi-membered classes are identical iff these are sets.)Put your money where your mouth is. Lets see your formalization.Its clear to anyone with half a clue that your reasoning, ifcorrect, can easily be formalized in ZFC. (Um: I should havesaid with some predicate in place of in and also somepredicate in place of =, instead of just pointing out thatyour in would not correspond to the in in ZFC.) Thisis because theres nothing non-standard about your_reasoning_, all thats non-standard is your _axioms_regarding in and =.In case you dont have half a clue, the part of theformalization corresponding to those two axiomsmight beC3 EyAx[ni(x, y) <-> Et(ni(x,t)) & A] (with y not free in A)Classication C4 AyAx[Az(ni(z,y) <-> ni(z,x)) -> {(S(y) & S(x)) <-> I(x,y)}] (Equi-membered classes are identical iff these are sets.)>--John************************David C. Ullrich === David Petry a .8ecrit dans le message deIm looking for a reference for the proof that the innite series> There is an entirely elementary proof of this relation,> which I suspect is well known but Ive never seen it. Factor (1+x)^N - (1-x)^N, for N an odd integer, as product ( w^k(1+x) - w^{-k}(1-x) ), k=-(N-1)/2..(N-1)/2> where w = exp( i pi / N ) Then nd the coefcient of x^3 for that expression, and> compare it to the same coefcient using the Binomial Theorem> for the rst expression, and take a limit.I nd :n*Binomial(n,3)=Sum((cotan(k*Pi/n))^2,k=1..(n-1)/2)and ....Please give me an hint for continue... === Dear All,I would like to know whether it is possible to dene the limit of afunction f as x tends to +oo, where f is dened on R^+, except on aninnite countable number of points of R^+ (for instance, except onthe set of integers).Paul === > Dear All, I would like to know whether it is possible to dene the limit of a> function f as x tends to +oo, where f is dened on R^+, except on an> innite countable number of points of R^+ (for instance, except on> the set of integers).or for instance, something like gamma(-x) ...I think you can safely use the standard limit denitions: For all P>0, there is a Q such that for all x of dom(f): x > Q ==> f(x) < -P For all e>0, there is a Q such that for all x of dom(f): x > Q ==> |f(x)-b| < e For all P>0, there is a Q such that for all x of dom(f): x > Q ==> f(x) > PI think these denitions work ne for the functions you have in mind.Of course the gamma(-x) would have no limit for x --> +ooDirk Vdm === The usual proofs condisider the open ball with the center point removed> (or the one point compactications). This set is homotopic to a> sphere, which is not contractible. This allows a proof that open sets> in R^2 and open sets in higher dimensions are not homeomorphic using> fundamental (rst homotopy) groups.If you know about higher homotopy groups, this will also give the general> result. However, the higher homotopy groups are not so easy to calculate.> Ok, correct. I would have no idea to calculate these homotopy groups,but it is not too hard to visualize so Ill believe thatpi_n(S^n) = Zpi_n(S^m) = 0 for m < n. === function g : R^2 -> Zlet Z : R*{0} U {0}*Rdene quotient topology of Z such that function g is quotient map.(hint : At this time, any two points dont separated to open set.)-------------------um...............i dont knowhelp me........my good teacher. === > function g : R^2 -> Z> let Z : R*{0} U {0}*R dene quotient topology of Z such that function g is quotient map.>The quotient topology for Z is { U subset Z | g^-1(U) open in R^2 }> (hint : At this time, any two points dont separated to open set.)> === continuous function fnlet {fn} is uniformly convergence fdene gn(x)=fn{x + (1/n)}show U.C gn -> f (U.C=uniformly convergence)-----------------------------because fn -> f : U.Cany E>0 , N existany x , n>=N => |fn(x)-f(x)| < Eand i must show gn -> f : U.C|gn(x)-f(x)| = |fn{x + (1/n)} -f(x)|= |fn{x + (1/n)} - fn(x) + fn(x) -f(x)|<= |fn{x + (1/n)} - fn(x)| + |fn(x) -f(x)|< 2/E + 2/E = Emy problem is |fn{x + 1/(n^2)} - fn(x)| U.C ??i think that it is pointwise convergence.because lim[fn{x + 1/(n^2)}] = fn(x)thus......my progress is wrong??how to solved it????....my best teacher.... === Can anyone please point me toward an implementation (in any language, even> pseudo ones) of Knuths spectral test? I have a random generator that I need to test, and I can not nd> implementations of the test... (I have found implementation specially geared for congrual tests, but this> is not what I need as my generator is not congrual).>Does this help?http://random.mat.sbg.ac.at/results/karl/spectraltest/ Also, are these tests (written in C) in DIEHARD?HTH, Flip === Does the following system have a solution:1024 o x1 mod y2048 o x2 mod y8192 o x3 mod y16834 o x4 mod yx1+x2+x3+x4+33 =yX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9ICLpP18575; === >Im doing some reading in abstract algebra (Dummit and Foote book) on tensor>products. Im not getting the idea behind it and its driving me nuts with>confusion. According to the book the motivation behind construction of>tensor products, from what I understood, is that given any R-module N is it>possible to extend/embed it to/in an S-module where R is a subring of S? I>will quote the book here: We begin the construction by returning to the>basic module axioms in order to examine whether we can dene products of>the form sn for s in S and n in N. These axioms start with the abelian group>N together with a map from S X N to N where the image of the pair (s, n) is>denoted by sn. It is therefore natural to consider the free Z-module on the>set S X N i.e. the collection of nite commuting sums of the elements of>the form (s, n) where s in S and n in N. This is an abelian group where>there are no relations between any distinct pairs (s, n) and (s, n), i.e.>no relations between the formal products sn, and in this abelian group the>original module has been thoroughly distinguished from the new>coefcients from S. The book the goes on to dene a subgroup H>generated by all the elements of the form(s1 + s2, n) - (s1, n) - (s2, n) .... (1)>(s, n1 + n2) - (s, n1) - (s, n2) .... (2)>(sr, n) - (s, rn) .... (3)and we take the quotient of the free abelain group by H to satisfy the>relations for a S-module and the compatibitly relation of the action of R on>N. The resulting quotient group S(X)R is called the tensor product of of S>and N over R. where s (X) n denotes the coset containing the (s, n).Now there is a natural map T: N --> S (X) N (over R). T(n) = 1 (X) nNow what is confusing me is the following: Since the group is free abelian>group, for any (s, n) there is no element (1, n) such that (s, n) = (1,>n), so N is not an S-Module if we identify it by its embedding into S (X) N>( T(n) = 1 (X) n ), because for N to be an S-module, then for every sn>there must exist an n such that sn = n.The book also says that the relations (1), (2) and (3) were the minimal>relations we had to impose in order to obtain an S-module, so it is the best>possible S-module to serve as a target for an R-module homomorphism from N.>This also confuses me since I thought that the R-module homomorphism from N>to S(X)N (call it Y) is onto, since if S(X)N is an S-module each>(s, n) must equal an element (1, n), so imposing an extra relationships>would mean that Y is not a homomorphism. This highly confuses me too. Is N a>subring of S(X)N? If, so how?? Also, why did we start with a free group and>take its quotient group by H to get the required relationships? We could>have imposed the relationships required for an S-module on the elements in>the rst place instead of making the group free and taking the quotient?>Tensor products seems to be a very interesting (and important) topic, so I>would really like to understand the idea behind it very well. If anyone>could help me I would really appreciate it.>X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9ICLp118589; === > Perplexing Patterns of Square Numbers> - B.S.Rangaswamy Observe closely the following patterns of nonzero square numbers. The squares read alike across and downwards in all these patterns. Perplex-5 Perplex-6 5 7 1 2 1 -239^2 1 5 1 3 2 1 -389^2> 7 2 3 6 1 -269^2 5 5 3 5 3 6 -744^2> 1 3 9 2 4 -118^2 1 3 3 2 2 5 -365^2> 2 6 2 4 4 -162^2 3 5 2 8 3 6 -594^2> 1 1 4 4 9 -107^2 2 3 2 3 2 4 -482^2> 1 6 5 6 4 9 -407^2 Perplex-7 Perplex-8 1 7 1 8 7 2 1 -1311^2 1 1 1 2 8 8 9 6 -3336^2> 7 3 3 3 2 6 4 -2708^2 1 6 3 5 3 9 3 6 -4044^2> 1 3 7 3 5 8 4 -1172^2 1 3 8 4 5 8 4 1 -3721^2> 8 3 3 4 7 6 9 -2887^2 2 5 4 6 2 1 1 6 -5046^2> 7 2 5 7 6 3 6 -2694^2 8 3 5 2 1 3 2 1 -9139^2> 2 6 8 6 3 2 1 -1639^2 8 9 8 1 3 5 2 9 -9477^2> 1 4 4 9 6 1 6 -1204^2 9 3 4 1 2 2 2 5 -9665^2> 6 6 1 6 1 9 5 6 -8134^2 There exist 15 more perplex-5 patterns> 13 perplex-6 > 09 perplex-7 > 02 Perplex-8 > and a scintillating Perplex-9 pattern It is thrilling and educating excercise to arrive at a few of these patterns. One should not expect to get all these patterns in ten days time. Hundred or its multiple days are reasonable, not with continous entanglement but with leisurely and positive attitude.I have taken much more number of days to arrive at all these 15+13+9+2+1 amazing patterns.> The two Perplex-8 patterns are prize catches and are to be arrived at, with immense analytical and choosy efforts. While, Perplex-9 with over 8 thousand nonzero 9 digit square numbers and more than a thousand pentillion(Pentillion=10^30)available combinations,demands Herculean efforts. If a second perplex-9 is discovered, it can be Wonder of WONDERS. I am still not able to exhaust all available combinations for perplex-9.> It is stimulating and exciting to discover these perplexing patterns. Formulate a few such patterns. It is certain that you get thrilled, elated and educated with the exploration of each of these patterns from the vivid galaxy of square numbers.> Wishing you all the best in your explorations.HINTS : A. Start with last column/row.> B. Refer to Perplex-5 presented in greater details under > Inquisitive problems in Teacher Exchange:13-16 at > http:// mathforum.org/te/higher.html. C. Utilize extensively the sorted out lists of nonzero > square numbers.> D. Forming interlocking square numbers on SCRABBLE BOARD with > Number Tiles can be helpful,as it completely eliminates > paper work.> E. Start with perplex-5 formations. -------------> Perplex-5 can be directly viewed at >> http: //mathforum.org/te/exchange/hosted/rangaswamy HINTS for formation of Perplex-6 patterns :>A.Compute a sorted out list of 6 digit nonzerosquares,numbering 376 Example-Squares ending with 16 are generated from > 346^2,396^2,446^2 - - - - 996^2 - in increments of 50> 354^2,404^2,454^2 - - - - 954^2 - & so on.> B. There exist 3 Perplex-6 patterns having 544644 in last row/column> 2 111556 > 2 166464 > 2 more 165649 > one Perplex-6 pattern each having 154449, 516961, > 594441,695556 in last row/column.> These hints can certainly assist in forming the Perplexing Patterns of Square Numbers. >- B.S.Rangaswamy>The THIRTEEN Perplex-6 patterns can be obtained by expanding the following sets of 6 squares each :1) 865^2 2) 885^2 3) 965^2 4) 379^2 5) 939^2 6) 369^2 7) 761^2 8)479^2 668^2 918^2 582^2 694^2 944^2 631^2 891^2 486^2 932^2 568^2 432^2 565^2 335^2 829^2 969^2 984^2 476^2 526^2 526^2 814^2 844^2 335^2 435^2 642^2 472^2 472^2 472^2 658^2 482^2 815^2 535^2 704^2 738^2 738^2 738^2 407^2 407^2 334^2 334^2 408^29) 981^2 10)715^2 11)965^2 12)344^2 13)661^2 806^2 409^2 627^2 357^2 615^2 544^2 424^2 372^2 935^2 828^2 608^2 477^2 462^2 585^2 962^2 796^2 536^2 478^2 585^2 478^2 408^2 719^2 771^2 834^2 393^2 Now, is the task of forming SEVEN Perplex-7 patterns out of 1061,7 digit nonzero square numbers. 3 Perplex-7 patterns end with 3114^2 One each of perplex-7 patterns end with 1204^2, 2538^2, 3108^2 & 3157^2 . Wishing you all happy and wonder-ful time in exploration of these perplex-7 patterns .- B.S.Rangaswamy. boundary=----=_NextPart_000_0064_01C39591.1FD16900 === ------- ------------------------------------------------------------- -Does the following system have a solution:1024 o x1 mod y2048 o x2 mod y8192 o x3 mod y16834 o x4 mod yx1+x2+x3+x4+33 =y === > [snip]> We have statement B:Two coins were ipped and at least one is a head.>> We have statement B:Two coins were ipped and at least one is a>> tail.>> What do you argue? What do you say now?>> I say that mathematically speaking, B and B are the same statement.It depends on what question youre going to ask. Youve taken half the> problem statement but havent said whether youre changing the other> half.If you make a swap of heads and tails THROUGHOUT THE ENTIRE QUESTION,> then they are equivalent questions.To whit: With a weighted coin which falls on heads with probability> 0.75, what is the probability of three heads in a row?Same question as: With a weighted coin which falls on tails with> probability 0.75, what is the probability of three tails in a row?But that does not mean that a weighted coin which falls on tails with> probability 0.75 is identical to a weighted coin which falls on> heads with probability 0.75. Here is a different question:With a weighted coin which falls on tails with probability 0.75, what> is the probability of three heads in a row?See how that works? Change the labelling in the whole question: same> question. Change the labelling in half the question, might be the same> or might be different. - Randy>>We have coins of equal density. Heads and tails are equally likely. I>>was trying to get Ullrichs opinion nailed down. He seems to be>>wa?g.No, he is not. You are misinterpreting his quite clear statement.>> He is trying to say your misinterpretation is not his original>> statement. The difference is quite clear when the two are>> laid side by side, as you did in your quote.>He chastised me for thinking that changing from heads to tails would>change that question. I didnt think it would and he didnt either.>Now he says he wasnt arguing that it wouldnt. No, Im not arguing that.>Thats wa?g.>>You and I seem to agree.I mostly avoid your threads because your silly obsession>> with this particular probability problem and your>> peculiar reading of it. So no, I would not say we agree.>We agree that changing from heads to tails does not alter the>question, or the answer. Or we dont agree? I say that it doesnt.>Surely you agree with that?>>When we ipped HH and said, Two coins were ipped and at least one>>is a head. What are the chances for two heads? or we ipped TT, and>>said Two coins were ipped and at least one is a tail. What are the>>chances for two tails? We would have the same question with the same>>answer.No. The question does not include what was ipped. Take the>> statements in quotes, i.e.>Two coins were ipped. It says that right in the question. WERE>FLIPPED, past tense. Were talking about a historical event. All we>know about the event is what we read in the statement.>> Two coins were ipped and at least one is a head. What >> are the chances for two heads?andTwo coins were ipped and at least one is a tail. What are the>> chances for two tails? And I will say these are the same question given a fair>> coin. Youve just relabelled heads and tails.>When they landed HH the rst statement was true, when they landed TT,>the second statement was true. At HT, or TH, either statement would>have been true.>Consider the second statement. We know that the coins landed TT, or>TH, or HT. Suppose that they landed TH. The other statement would have>been true. If it had been made, would we have had a different>question?>> When you add that there is a particular outcome given, its>> silly to ask about probability since theres only one>> point in the sample space. The probability is 1 in both>> of those cases.>>This is a probability question. The coins were ipped once. We can do>probability because we can re iterate the ip, umpteen thousands of>times. The tricky part is that we are re iterating a rst time ip.>We ip umpteen thousand times and each one was a rst time ip.>With HH, the heads statement was made, with TT, the tails statement>was made, and with HT, or TH, one, or the other. Either would have>been true, and would have been the same statement. OR, changing the>color would have changed the statement.>Likewise, with a HT, or a TH ip, either statement would be true.Huh? The statement includes a question: What are the>> chances for two heads (tails)? How can that question>> be true?>>Thats the point. Does changing from heads to tails alter the>question.>Either question would have the same answer. For one of them to have>>answer other than 1/2, they would have to have different answers.No, they have the same answer of 1/3.>Thats why I have this strange obsession. There are always qualied>people like you who will argue with me, and dont seem to want to>understand my argument.>> Case 1: I tell you two coins were ipped and one of the>> coins is a head. Therefore you know that I have either>> HH, TH, or HT with equal probability. I ask you what is>> the probability that I got HH? The answer is 1/3.>>I know that they landed HH, or HT, or TH. With HH, you made the heads>statement with probability one, but you made the tails statement with>probability 1/2.>There are now three possible outcomes. They are no longer equally>likely outcomes.Case 2: I tell you two coins were ipped and one of the>> coins is a head. Therefore you know that I have either>> TT, HT, or TH with equal probability. I ask you what is>> the probability that I got TT? The answer is 1/3.>I know that you have TT, HT, or TH. They are no longer equally>probably. Try it. You cant do it.Flip the coins but dont look. You cant make either statement, prior>to inspection. After inspection you can make one statement, or the>other. With HH, you can only make the heads statement. With TT, you>can only make the tails statement.With HT, or TH, you can make one, or the other, but not both. If you>make both, then I know that the probability is one that the two coins>are different. The four are equally probable, prior to inspection.>After inspection, you can only make one statement, or the other. After>the statement, the three that are left are not equally probable.>> You will see that in exchanging H and T in the>> two paragraphs throughout, I have not changed the>> meaning or the interpretation. There is nothing like>> a requirement that the probability be 1/2. Your nal>> statement is coming from somewhere outside of probability>> theory.You ipped the coins and got the answer wrong. Thats what keeps me>obsessed. My reading of the question is strange? My reading may be>strange, but you get the wrong answer, and you ipped the coins.Eldon>> - Randy************************David C. Ullrich === change that question. I didnt think it would and he didnt either.>Now he says he wasnt arguing that it wouldnt. No, Im not arguing that.>Thats wa?g.>>He chastised me for thinking that changing from heads to tails would>>change that question. I didnt think it would and he didnt either.>>Now he says he wasnt arguing that it wouldnt. No, Im not arguing that.>Thats wa?g.>> ************************David C. Ullrich>Okay, we have two questions:>B. Two coins were ipped and at least one is a head. What are the>chances for two heads?>B. Two coins were ipped and at least one is a tail. What are the>chances for two tails?Changing from heads to tails either changes the question, or it>doesnt.>I say no, Randy Poe says no.What do you say? You said you werent arguing this, or that. Does the>changing from heads to tails alter the question, or not?Yes, or no?The two questions are obviously equivalent. (That doesnt saythat Two coins were ipped and at least one is a headand Two coins were ipped and at least one is a tail are equivalent, and it _certainly_ doesnt say that Two coins were ipped and at least one is a headand Two coins were ipped and at least one is a tailare _the same statement_, which is the idiocy yousaid Id said.)>Eldon:~)Eldon************************David C. Ullrich === >>We have coins of equal density. Heads and tails are equally likely. I>>was trying to get Ullrichs opinion nailed down. He seems to be>>wa?g.No, he is not. You are misinterpreting his quite clear statement.>> He is trying to say your misinterpretation is not his original>> statement. The difference is quite clear when the two are>> laid side by side, as you did in your quote.>He chastised me for thinking that changing from heads to tails would>change that question.Correctly. Exchanging the words heads and tails THROUGHOUT thequestion changes nothing. Its just a relabeling of the sides of thecoins.> I didnt think it would and he didnt either.Good.>Now he says he wasnt arguing that it wouldnt.You quoted your change, the one that changes the situation, and theone that David said earlier does not. You have quoted them in the samepost. They are quite different. Study them. I have already explainedseveral times (see what is capitalized above) why your new change isdifferent from your old change. Im not going to berate this point anymore since Ive already said the difference over and over. I can onlyimagine how many times people whove been in this thread longer havetried to explain to you. - Randy === > You quoted your change, the one that changes the situation, and the> one that David said earlier does not. You have quoted them in the same> post. They are quite different. Study them. I have already explained> several times (see what is capitalized above) why your new change is> different from your old change. Im not going to berate this point any> more since Ive already said the difference over and over. I can only> imagine how many times people whove been in this thread longer have> tried to explain to you. - RandyThe truth is in the archives. But just study this thread.1 Virgil said that the compliment to at least one is a head is TT.2 Eldon says that is not true.3 With the statement at least one is a head, we know that TT didnthappen. We know that HH happened, or HT happened, or TH happened. Or,or, not all three.4 When HH happened and at least one is a head was generated, thenthe compliment would be, TT happened and at least one is a tail wasgenerated.5 When HT happened and at least one is a head was generated, thecompliment would be, HT happened and at least one is a tail wasgenerated.6 When TH happened and at least one is a head was generated, thecompliment would be, TH happened and at least one is a tail wasgenerated.7 This is only true if changing from heads to tails iscomplimentary.8 Eldon argues that it is.9 Eldon erred when he said that Doctor Ullrich has argued for that.Dr. Ullrich actually said that surely Eldon isnt dumb enough to thinkdifferent.10 Eldon should have said, surely Doctor Ullrich isnt dumb enough tothink thats wrong. In this, Eldon seems to have erred:)11 Eldon should have said, Randy Poe argues that heads and tailsare complimentary.12 Thats what I meant when I said that we agree. We agree thatheads and tails are complimentary.13 Now that we agree on that, we should be able to agree on the answerto our question.14 Flip two coins, they will land HH, or HT, or TH, or TT.15 We can say about our question that two coins were ipped and atrue statement was made.16 When those coins land HH, make the at least one is a headsstatement.17 When they land TT, make the at least one is a tails statement.18 When they land HT, or TH, make one statement, or the other, withoutbias.19 Then answer the question.Observations:There are similar questions which correctly answer 1/3. The key isnot, whether or not the statement was at least one is.The key????***The key is whether or not there was a prejudice toward one outcomeor the other, prior to inspection of the toss.***Another observation: Consider two different objects, such as an orange ball and a greenball.Randomly choose one.Announcing the winner does not communicate a prior prejudice.Eldon:) === Suppose the random variables X_i, i = 1,2,... are i.i.d. Showthat if E((X_1)^2) < innity and if M_n = max(X_1,X_2,...,X_n), then(M_n)/squareroot(n) converges to zero with probability one.I thought about altering the strong law of large numbers assuming secondmoments, then using a subsequence trick, but am getting nowhere.Can anyone give a proof for this please?Marc === >Suppose the random variables X_i, i = 1,2,... are i.i.d. Show>that if E((X_1)^2) < innity and if M_n = max(X_1,X_2,...,X_n), then>(M_n)/squareroot(n) converges to zero with probability one.I thought about altering the strong law of large numbers assuming second>moments, then using a subsequence trick, but am getting nowhere.Can anyone give a proof for this please?I dont have an actual proof, but you might want to note that youcan express P(M_n/sqrt(n) > eps) explicitly in terms of the distribution of X_1 (because P(M_n <= delta) = P(X_1 <= delta)^n.)>Marc>************************David C. Ullrich === Theres something called Kolmogorovs something that Im almostcertain will help. You can nd it in Gerald Follands Real Analysisbook, right before he proves the strong laws of large numbers. Itgives a good bound for M_n, which I think will do the job.> Suppose the random variables X_i, i = 1,2,... are i.i.d. Show> that if E((X_1)^2) < innity and if M_n = max(X_1,X_2,...,X_n), then> (M_n)/squareroot(n) converges to zero with probability one.I thought about altering the strong law of large numbers assuming second> moments, then using a subsequence trick, but am getting nowhere.Can anyone give a proof for this please?Marc === >:> What is the false implication?>:>:> : The false implication is that b is a non-unit factor of 2 like>: sqrt(2), or 2^{1/3}, when in fact, appropriately, its a factor of 1.>: Unfortunately the poster deleted out pertinent information, so Ill>: make the effort to put it back in so that it makes sense to readers>: without forcing them to go back to previous posts.>: I gave the example of xy = 2, where x and y are algebraic integers,>: where x=2a and y=b, so b is an algebraic integer, but a is not, but>: it is a member of what I call the object ring.>: The object ring is a commutative ring that includes all numbers such>: that -1 and 1 are the only members that are both a unit and an>: integer, where no non-unit member is a factor of any two integers that>: are coprime.>: Notice that denition for the object ring excludes possibilities like>: a=1/2 as then 2 would be a unit since 2(1/2) = 1.>: So, in fact, in the object ring, ab=1 and b IS a unit, but because a>: is inappropriately excluded from the ring of algebraic integers by the>: denition of algebraic integers as roots of monic polynomials with>: integer coefcients, you have the false implication that b is some>: other type factor of 2.>: Now then, say you follow algebra, and then you nd that you have 2 as>: a factor of x, well youve been pushed out of the ring of algebraic>: integers where x does NOT have 2 as a factor, even though x=2a.>:>: And you can see several problems that popped up by that exclusion as b>:>: is not a unit in the ring of algebraic integers, when it should be,>:>: and x does NOT have 2 as a factor in the ring, though x=2a.>:>>:>So, what is the problem? I havent seen anything that looks like >:>a contradiction, only an assertion that something is not a unit>:>which should be according to some unstated principle. Whats the>:>principle?>:> : Given that xy=2, with x=2a, y=b, with b an algebraic integer, x an>:> : algebraic integer, while a is not, it *should* be that y does not>:> : share non-unit factors with 2, since x *should* have 2 as a factor,>:> : but in the ring of algebraic integers, because the denition>:> : arbitrarily excludes a, neither of those is the case.>:>:> What is behind the shoulds, i.e. what are the bad consequences of >:> excluding a? Before you were talking about being able to prove two >:> contradictory results.>:>:> Mike>: The problem occurs because algebra does NOT recognize the arbitrary>: denition, so using algebra you can nd yourself in situations where>: you prove that x has 2 as a factor, but then again, because of this>: arbitrary denition, it does not, in the ring of algebraic integers.>: Since the assumption is that algebraic integers is an appropriate>: ring, it appears that youve proven two contradictory things:>>What do you mean by an appropriate ring?> : One where you cant appear to prove contradictory things.> : >: 1. x has 2 as a factor>: 2. x does NOT have 2 as a factor>>But arent 1. and 2. actually:>>1. x has 2 as a factor in the object ring> : Ill remind of the example of 2 and 6 in the ring of evens to help> : readers with context, as there 2 is not a factor of 6 because 3 isnt> : even.> : Now then, with my example, you have x=2a, where x is an algebraic> : integer, but as a is not, it doesnt have 2 as a factor, in the ring> : of algebraic integers, which falsely implies, given that xy=2, where y> : is an algebraic integer that x and y have non-unit factors of 2 in the> : same sense as like if you have sqrt(2) as a factor for both.> : But, in fact, *neither* has what you might call non-trivial factors of> : 2 in the ring of algebraic integers, which I have to say as trivially> : they have themselves as factors.> : That is, given that xy=2, x is trivially a factor of 2, in the ring of> : algebraic integers, as is y, when in fact x has a factor that IS 2, a> : non-trivial factor, in an appropriate ring, that is, one which does> : not allow contradictions.> : Its worth noting that proving that x doesnt have a factor of 2, in> : the ring of algebraic integers, requires going to the eld of> : algebraic numbers, which is itself, of course, not awed.>2. x does NOT have 2 as a factor in the algebraic integers> : And proving that requires going to the eld of algebraic numbers,> : which hasnt really been discussed at this point.> : So then *in the ring of algebraic integer* you can appear to prove> : *both* things, and only gure out that youve been pushed out of the> : ring of algebraic integers, by going to the eld of algebraic> : numbers.Im not following. You let x = 2a where x is an algebraic integer> and a is not. So x does not have a factor of 2 in the algebraic> integers. What is the contradictory proof that x *does* have a factor > of 2 in the algebraic integers?In my case I use a special polynomial P(m), which is special in that Ican rather easily factor it into non-polynomial factors to appear toprove within the ring of algebraic integers that x has 2 as a factor.So more fully, the problem with the ring of algebraic integers is that*in the ring of algebraic integers* you can prove with my example thatx has 2 as a factor, and then go to the eld of algebraic numbers andprove that x/2 cant be an algebraic integer.James Harris === [snip]> So more fully, the problem with the ring of algebraic integers is that> *in the ring of algebraic integers* you can prove with my example that> x has 2 as a factor, and then go to the eld of algebraic numbers and> prove that x/2 cant be an algebraic integer. James HarrisThere is no problem in the ring of algebraic integers. Your argument is fundamentally awed and your proofcontains fatal errors.--A fool and his proof are soon refuted.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Im trying to show the 1 - 1 correspondence between R and the power set of Nwithout resorting to the binary decimal argument. I know I can use theSchroeder-Bernstein Theorem. There is no problem nding an injection fromthe power set of N into R. Im having difculty nding an injection fromR into the power set of N.Help!L === > Im trying to show the 1 - 1 correspondence between R and the power set of N> without resorting to the binary decimal argument. I know I can use the> Schroeder-Bernstein Theorem. There is no problem nding an injection from> the power set of N into R. Im having difculty nding an injection from> R into the power set of N.Its probably easier to nd an injection from [0,1) into P(N). Thissufces, because its easy to see that each nondegenerate interval hasthe same cardinality as R.You say you dont want to use the binary decimal argument. Is a purebinary argument acceptable? For each x in [0,1) there is a binaryrepresentation, where to avoid ambiguity we avoid any representation thatends in all 1s. This corresponds in an obvious way with a unique memberof P(N), and the resulting map is an injection. Its not a bijection,because conite sets in P(N) are not in the range of this map.-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Let n>=4 and A=||a_{i,j}|| be a n x n real symmetric matrixwith a_{i,i}=0 , i =1,2,...,n . DenoteS(A)= SUM_{i=1 to i= n-1}SUM_{j=i+1 to j=n}a_{i,j}^2 .Suppose that r_1== 2*sqrt(S(A)) , then A is singular.2) If n is even and r_n -r_1 =< sqrt(8*S(A)/n) , then det(A)= (-2*S(A)/n)^{n/2} ; If n is odd and r_n -r_1 =< sqrt(8*S(A)/(n^2 -1)), then det(A)= ?3) Suppose that r_n = sqrt(2*(n-1)*S(A)/n) or r_n = sqrt(2*S(A)/(n(n-1))) . Its possible to nd A^{-1} ?If true , above questions are of interest ?Dear All,I would like to nd two one-variable functions, f and g, such that:* limit of f(x)/g(x) is an indetermination of the type 0/0 or oo/oo;* limit of f(x)/g(x) does not exist at 0;* limit of f(x)/g(x) exists at 0;* g(x) different of zero in some neighborhood of 0 (except possiblyat 0) is a false proposition.Paul === >Dear All,I would like to nd two one-variable functions, f and g, such that:* limit of f(x)/g(x) is an indetermination of the type 0/0 or oo/oo;>* limit of f(x)/g(x) does not exist at 0;>* limit of f(x)/g(x) exists at 0;>* g(x) different of zero in some neighborhood of 0 (except possibly>at 0) is a false proposition.??? If I understand what you mean by the fourth condition, itsays that there exists a sequence x_n -> 0 such that g(x_n) = 0.If so then the limit of f/g cannot exist.Paul************************David C. Ullrich === Given a quartic monic polynomial with integer coefcients and with four complex roots (two pairs of complex conjugate roots),eg: x4-10x3+47x2-90x+81how do I nd the min poly of |x|^2 (involving the complex absolute value)for the example above it is: x2-27x+81 (yeah I can get the 81 OK)without using a computer or Ferarris method of solving the quartic to get all the roots explicitly and then working back.I presume there is a generalization to certain even degree polynomials, but I reckon I can get that if someone can remind me how to do it for quartics.I feel stupid for having forgotten how to do this, but it *is* driving me insane.Bill.P.S: apologies if this appears twice or even thrice. I have been trying to get it to post all day, but my internet provider has a known news even after a day, and I can rarely read all responses. I am experimenting with other options. === Can somebody answer this question?Assume that V is a vector spaceAssume that S is a subspace of VAssume that O is the zero vector of VIs O also the zero vector of S? In other words, is the zero vector of Vinherited by S?In other words if the, zero vector in V is not in S then is S not asubspace?thank youMarch === >Can somebody answer this question?Assume that V is a vector space>Assume that S is a subspace of V>Assume that O is the zero vector of V>Is O also the zero vector of S? In other words, is the zero vector of V>inherited by S?Yes. By denition, a subspace must be a vector space with theoperations of V; that is, you dont get to dene the vector sum of Sany which way you want, you must add vectors in S as if they were inV, and the result must be in S.Since there is only one zero vector in V, and since for ANY vector vin V, if v+w = v then w must be the zero vector of V, then for anyvector s of S, if s+w = s, then w must be the zero vector of V.>In other words if the, zero vector in V is not in S then is S not a>subspace?Correct. =Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)= Arturo Magidinmagidin@math.berkeley.edu === >Yes. By denition, a subspace must be a vector space with the>operations of V; that is, you dont get to dene the vector sum of S>any which way you want, you must add vectors in S as if they were in>V, and the result must be in S.I was wondering about the minimal required denition. I was told theinclusion of the zero vector must be explicit to avoid a problem wherethe empty set could be a subspace of V, but 0 * v = 0 implies directlythat 0 must be in S so why should it be required explicitly? === >Yes. By denition, a subspace must be a vector space with the>>operations of V; that is, you dont get to dene the vector sum of S>>any which way you want, you must add vectors in S as if they were in>>V, and the result must be in S.I was wondering about the minimal required denition. I was told the>inclusion of the zero vector must be explicit to avoid a problem where>the empty set could be a subspace of V, but 0 * v = 0 implies directly>that 0 must be in S so why should it be required explicitly?No, it does not imply that directly. You must assume that S isnonempty: otherwise, where does v come from?A vector space V over the eld K is dened to be a set, which weusually call V as well, together with two operations: a vector sum,+, which takes pairs of elements of V as arguments and replies with anelement of V; and a scalar product, which takes as arguments anelement of K and an element of V, and returns an element of V. Theconditions that these set and operation must satisfy are:(1) For any three elements v, w, z of V, (v+w)+z = v+(w+z)(2) There exists an element in V, called 0, such that for every v in V, 0+v = v+0 = v.(3) For every v in V there exists an element in V, called -v, such that v+(-v) = (-v)+v = 0.(4) For every two elements v and w of V, v+w = w+v.(5) For every two elements v,w of V, and any k in K, k*(v+w) = (k*v) + (k*w).(6) For every element v of V, 1*v = v.(7) For every element v of V and every two elements k,m of K, (k+m)*v = (k*v) + (m*v) (km)*v = k*(m*v).If you remove condition (2), then just must remove condition (3) aswell (since it only makes sense in terms of condition 2), and in thatcase, the empty set together satises all conditions by vacuity(there is no element in V, so all statements of the form for allelements v of V are true).If you have a vector space V, then a subset S of V is a subspace ifand only if (a) S is nonempty; (b) For any r,s in S, r+s (added as elements of V) is also in S;and (c) For any s in S and k in K, k*s (multiplied as if s is an element of V) must also be in S.One can then prove that any S that satises (a), (b), and (c) willalso satisfy (1)-(7), by using the sum and scalar product from Vapplied only to elements of S.But you must assume that S is nonempty; if you do not assume S isnonempty, then the empty set satises conditions (b) and (c) byvacuity again, but S is not a vector space itself because it does notsatisfy condition (2). You can change (a) to (a) S contains 0, because (a) and (a) areequivalent in view of condition (c), as you note. However, in order toapply condition (c) to get the 0 vector in S, you must rst assumethat there is SOMETHING in S, in order to say: Ah, take some v in S;then 0*v is in S, so 0 is in S. What if there is nothing in S? Thenyou cannot conclude that 0 is in S.= =Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)= Arturo Magidinmagidin@math.berkeley.edu === >Ive noted a problem in algebraic number theory with the inclusiveness>of the denition of algebraic integers as roots of monic polynomials>with integer coefcients.Various posters have argued that in fact there is no problem, but>heres a short refutation of their primary objection.First I haveP(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)and the factorizationP(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)where the as are given by the following cubic:a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).My nding that only two of the as have f as a factor without regard>to the value of m has been vigorously disputed. > Its incorrect. See below.Mathematicians, you see, have other priorities than actual validity ofmathematics, as they have *social* issue. See below.>However, consider w_1(m), a factor of a_1 that is a factor of f, as>well as a function that varies with m, then it follows thata_1 x + uf has w_1(m) as a factor, so dividing through by w_1(m) givesa_1 x/w_1(m) + uf/w_1(m)but then uf/w_1(m) cannot in general be an algebraic integer as its>not representable as a polynomial with a nite number of terms if>w_1(m) varies with m. > A totally off-the-wall, unjustied statement, and, as it> so happens, incorrect. But for now, if you want to claim> it is true, the shoe is on your foot: try to prove it.Ive introduced r(m), to handle the result of uf/w_1(m).So the poster is requesting that I prove that r(m) w_1(m) = uf, does not exist over the ring of algebraic integers.Ive concluded that using numbers for u and f, as they are*independent* of m, helps, so let u=2, f=13.Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m), so you havexy=26and now Ill chat further.Then the question is, does their exist a multiplicative inverse in thering of algebraic integers for 26/w_1(m) for *all* algebraic integersm?The simple answer is that if w_1(m) varies with m, then it must varyover an innite number of algebraic integer values as m varies overalgebraic integers.But w_1(m) must vary from 0 to innity if it varies with m.Replies Ive seen have shied away from trying something like w_1(m) =m+1 because most readers can immediately realize that 26/(m+1) oranything like it, cant be an algebraic integer for all m.Instead Ive seen examples like 26^{1/m} or more trying, 26^{1/m^2+1},but notice that because m^2+1 can equal 0, *in the ring of algebraicintegers*, you can get 26^{1/0} for the ms that are the roots ofm^2+1. >My guess is that some may be assuming that f is replaceable by some>function of m, but in fact, its independent of the value of m, No, we are not assuming f is replaceable by some function of m. > We assume f itself is *constant* with respect to m. However, the factorization > of a polynomial P(m) whose coefcients are functions of m, hence > dependent on m, is also in general dependent on m. We do *not* assume> that the corresponding factorization of f is constant with respect> to m.So if you had any doubts about how low mathematicians could go,consider that after all, theyre trying to convince that you can haveone algebraic integer function dened by the multiplicative inverseof another algebraic integer function over all algebraic integer m,like with f=7, 14/w_1(m) = r(m).Its like saying that you can have xy=2, where x and y are integers,or algebraic integers, where x varies over the ring, and y remains init, though of course, I can just show that for x=5 that doesnt work.Hmmm...thats why posters have tried to pick w_1(m) using exponentialfunctions.Fascinating as theyre showing how much they understand.>so its>like 1/(x+1) which is also not representable by a polynomial if x is>an algebraic integer not equal to 0 or -2.So the objection is refuted by the impossibility of uf/w_1(m) being an>algebraic integer, > Also incorrect and false. Remember what you said above: ... consider w_1(m), a factor of a_1 that is a factor of f... If w_1(m) is a factor of f, that can only mean f/w_1(m) is> an algebraic integer, which of course implies that uf/w_1(m) > is an algebraic integer: this is your *assumption* here.The assumption is itself a contradiction, in that it requires theability to dene one algebraic integer function by the multiplicativeinverse of another algebraic integer function.Basic algebra.>for all algebraic integers m, if w_1(m) varies with>m.My hope is that posters who have been so successful in convincing>others that my argument is awed will post concessions. Absolutely not! You are inching closer to the truth in this. You are seeing > how this can work. You are correct that w_1(m) is a factor of> f that depends on m. As noted above this implies that > uf/w_1(m) is an A.I., and a1(m)/w_1(m) is also. I think you > are beginning to see how this can happen.Readers should note that attempt to push the impossible, one algebraicinteger function dened by the multiplicative inverse of anotheralgebraic integer function.My guess is that seeing something like uf/w_1(m) may confuse some whomight believe that u and f are functions of m, so I like to show theirindependence by tossing in values, like u=2, f=13, so you have26/w_1(m).> I note a couple of other things. First, the polynomial in a> that you mention above:[#] a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).Your claim is that two roots of this polynomial have a factor> that is f. This means that if r is one of those roots, then> r = f*c, where c is an algebraic integer. This implies thatf^3*c^3 + 3(-1 +m*f^2)*f^2*c^2 - f^2*(m^3*f^4 - 3*m*f^2 + 3*m) = 0,or, factoring out f^2, f*c^3 + 3*(-1 + m*f^2)*c^2 - (m^3*f^4 - 3*m*f^2 + 3*m) = 0.> The polynomial in c on the left is non-monic and primitive> (if f is not a multiple of 3). If it is irreducible, then> c cannot be an algebraic integer. When f = 5 and m = 1, this> equation is 5*c^3 + 72*c^2 - 553 = 0,and the polynomial in c is easily shown to be irreducible. > Therefore c cannot be an algebraic integer. This contradicts> your central claim: NONE of the roots of [#] can have a> factor that is 5 = f.> Second note: It may be worthwhile to see how irreducibility> is inextricably tied to factorization of roots. Assume that> Q(m) = x^2 + m*x + 30,where m is an integer. The constant term is divisible by 5 (also by 2 and 3, but I> will focus on 5 here). For certain values of m, this polynomial is reducible, and the> corresponding roots are not both divisible by 5: m = 11: r1 = 5, r2 = 6 m = 13: r1 = 10, r2 = 3 m = 17: r1 = 15, r2 = 2 m = 31: r1 = 30, r2 = 1. In all these examples, obviously one root is *divisible* by 5> and the other is *coprime* to 5. And of course in all these > examples, the polynomial is reducible. This is parallel to > Harriss cubic when m = 0: two of the roots are divisible> by f and one is coprime to f. In this particular case his polynomial> [#] is reducible. In general it is not. Now look at an other example. Say, m = 14. The > polynomial is irreducible, because the discriminant D^2 = m^ - 4*30 = 76, which is not a perfect square. The roots of the > polynomial are:> r1 = (-14 + sqrt(76))/2 = -7 + sqrt(19) and r2 = -7 - sqrt(19). Clearly both r1 and r2 are algebraic integers. Assume that r1 is a multiple of 5: r1 = 5*s1. Then 25*s1^2 + 70*s1 + 30 = 0. Factor out 5:[*] 5*s1^2 + 14*s1 + 6 = 0. This happens to have discriminant D^2 = 76. [This is> not a coincidence!]. Thus the polynomial in [*] is> *also* irreducible. Therefore s1 cannot be an algebraic> integer. Therefore r1 cannot be divisible by 5. The same can similarly be shown for r2. However, it is now easy to show that both r1 and r2 must both> be *non-coprime* to 5. For, suppose r1 is coprime to 5.> The fact that r1*r2 = 5*6would then imply that r2 is DIVISIBLE by 5, which > contradicts the result above that both r1 and r2 are> NOT divisible by 5. Conclusion: for m = 14, the polynomial is irreducible,> and both roots have a nonunit algebraic integer factor> in common with 5. This is true more generally. There is nothing special> about 14. Try, for example, m = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, > 14, 15, 16, 18, ... MOST values of m yield an irreducible > polynomial, and both of the roots will both share algebraic > integer factors in common with 5. In general it is hard to write down exactly what these > factors are, even in the quadratic case. The important > thing to know about them here is that, as suggested above > by the w_1(m) notation, *they will be dependent on m*.Whats fascinating here is how casually posters will just starttalking, which I think actually usually works with a lot of readerswho dont even bother to read carefully.Just remember that the base position here is that in the ring ofalgebraic integers you can have one function dened by themultiplicative inverse of another function.Remember xy=26.James Harris === >> Ive noted a problem in algebraic number theory with the inclusiveness>> of the denition of algebraic integers as roots of monic polynomials>> with integer coefcients.>Various posters have argued that in fact there is no problem, but>> heres a short refutation of their primary objection.>First I have>P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)>and the factorization>P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)>where the as are given by the following cubic:>a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).>My nding that only two of the as have f as a factor without regard>> to the value of m has been vigorously disputed.>> Its incorrect. See below. Mathematicians, you see, have other priorities than actual validity of> mathematics, as they have *social* issue. See below.However, consider w_1(m), a factor of a_1 that is a factor of f, as>> well as a function that varies with m, then it follows that>a_1 x + uf has w_1(m) as a factor,>so dividing through by w_1(m) gives>a_1 x/w_1(m) + uf/w_1(m)>but then uf/w_1(m) cannot in general be an algebraic integer as its>> not representable as a polynomial with a nite number of terms if>> w_1(m) varies with m.>> A totally off-the-wall, unjustied statement, and, as it>so happens, incorrect. But for now, if you want to claim>it is true, the shoe is on your foot: try to prove it. Ive introduced r(m), to handle the result of uf/w_1(m). So the poster is requesting that I prove that r(m) w_1(m) = uf, does not exist over the ring of algebraic integers. Ive concluded that using numbers for u and f, as they are> *independent* of m, helps, so let u=2, f=13. Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m), so you have xy=26 and now Ill chat further. Then the question is, does their exist a multiplicative inverse in the> ring of algebraic integers for 26/w_1(m) for *all* algebraic integers> m? The simple answer is that if w_1(m) varies with m, then it must vary> over an innite number of algebraic integer values as m varies over> algebraic integers. But w_1(m) must vary from 0 to innity if it varies with m. Replies Ive seen have shied away from trying something like w_1(m) => m+1 because most readers can immediately realize that 26/(m+1) or> anything like it, cant be an algebraic integer for all m. Instead Ive seen examples like 26^{1/m} or more trying, 26^{1/m^2+1},> but notice that because m^2+1 can equal 0, *in the ring of algebraic> integers*, you can get 26^{1/0} for the ms that are the roots of> m^2+1.My guess is that some may be assuming that f is replaceable by some>> function of m, but in fact, its independent of the value of m, No, we are not assuming f is replaceable by some function of m.>We assume f itself is *constant* with respect to m. However, thefactorization>of a polynomial P(m) whose coefcients are functions of m, hence>dependent on m, is also in general dependent on m. We do *not* assume>that the corresponding factorization of f is constant with respect>to m. So if you had any doubts about how low mathematicians could go,> consider that after all, theyre trying to convince that you can have> one algebraic integer function dened by the multiplicative inverse> of another algebraic integer function over all algebraic integer m,> like with f=7, 14/w_1(m) = r(m). Its like saying that you can have xy=2, where x and y are integers,> or algebraic integers, where x varies over the ring, and y remains in> it, though of course, I can just show that for x=5 that doesnt work. Hmmm...thats why posters have tried to pick w_1(m) using exponential> functions. Fascinating as theyre showing how much they understand.so its>> like 1/(x+1) which is also not representable by a polynomial if x is>> an algebraic integer not equal to 0 or -2.>So the objection is refuted by the impossibility of uf/w_1(m) being an>> algebraic integer,> Also incorrect and false. Remember what you said above: ... consider w_1(m), a factor of a_1 that is a factor of f... If w_1(m) is a factor of f, that can only mean f/w_1(m) is>an algebraic integer, which of course implies that uf/w_1(m)>is an algebraic integer: this is your *assumption* here. The assumption is itself a contradiction, in that it requires the> ability to dene one algebraic integer function by the multiplicative> inverse of another algebraic integer function. Basic algebra.IF THIS IS BASIC ALGEBRA, WHY CANT I UNDERSTAND YOUR ARGUMENT? I UNDERSTANDALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME.for all algebraic integers m, if w_1(m) varies with>> m.>My hope is that posters who have been so successful in convincing>> others that my argument is awed will post concessions.> Absolutely not! You are inching closer to the truth in this. You are seeing>how this can work. You are correct that w_1(m) is a factor of>f that depends on m. As noted above this implies that>uf/w_1(m) is an A.I., and a1(m)/w_1(m) is also. I think you>are beginning to see how this can happen. Readers should note that attempt to push the impossible, one algebraic> integer function dened by the multiplicative inverse of another> algebraic integer function. My guess is that seeing something like uf/w_1(m) may confuse some who> might believe that u and f are functions of m, so I like to show their> independence by tossing in values, like u=2, f=13, so you have> 26/w_1(m). I note a couple of other things. First, the polynomial in a>that you mention above:[#] a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).Your claim is that two roots of this polynomial have a factor>that is f. This means that if r is one of those roots, then>r = f*c, where c is an algebraic integer. This implies thatf^3*c^3 + 3(-1 +m*f^2)*f^2*c^2 - f^2*(m^3*f^4 - 3*m*f^2 + 3*m) = 0,or, factoring out f^2, f*c^3 + 3*(-1 + m*f^2)*c^2 - (m^3*f^4 - 3*m*f^2 + 3*m) = 0.>The polynomial in c on the left is non-monic and primitive>(if f is not a multiple of 3). If it is irreducible, then>c cannot be an algebraic integer. When f = 5 and m = 1, this>equation is 5*c^3 + 72*c^2 - 553 = 0,and the polynomial in c is easily shown to be irreducible.>Therefore c cannot be an algebraic integer. This contradicts>your central claim: NONE of the roots of [#] can have a>factor that is 5 = f.> Second note: It may be worthwhile to see how irreducibility>is inextricably tied to factorization of roots. Assume that> Q(m) = x^2 + m*x + 30,where m is an integer. The constant term is divisible by 5 (also by 2 and 3, but I>will focus on 5 here). For certain values of m, this polynomial is reducible, and the>corresponding roots are not both divisible by 5: m = 11: r1 = 5, r2 = 6 m = 13: r1 = 10, r2 = 3 m = 17: r1 = 15, r2 = 2 m = 31: r1 = 30, r2 = 1. In all these examples, obviously one root is *divisible* by 5>and the other is *coprime* to 5. And of course in all these>examples, the polynomial is reducible. This is parallel to>Harriss cubic when m = 0: two of the roots are divisible>by f and one is coprime to f. In this particular case his polynomial>[#] is reducible. In general it is not. Now look at an other example. Say, m = 14. The>polynomial is irreducible, because the discriminant D^2 = m^ - 4*30 = 76,which is not a perfect square. The roots of the>polynomial are:> r1 = (-14 + sqrt(76))/2 = -7 + sqrt(19) and r2 = -7 - sqrt(19). Clearly both r1 and r2 are algebraic integers. Assume that r1 is a multiple of 5: r1 = 5*s1. Then 25*s1^2 + 70*s1 + 30 = 0. Factor out 5:[*] 5*s1^2 + 14*s1 + 6 = 0. This happens to have discriminant D^2 = 76. [This is>not a coincidence!]. Thus the polynomial in [*] is>*also* irreducible. Therefore s1 cannot be an algebraic>integer. Therefore r1 cannot be divisible by 5. The same can similarly be shown for r2. However, it is now easy to show that both r1 and r2 must both>be *non-coprime* to 5. For, suppose r1 is coprime to 5.>The fact that r1*r2 = 5*6would then imply that r2 is DIVISIBLE by 5, which>contradicts the result above that both r1 and r2 are>NOT divisible by 5. Conclusion: for m = 14, the polynomial is irreducible,>and both roots have a nonunit algebraic integer factor>in common with 5. This is true more generally. There is nothing special>about 14. Try, for example, m = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12,>14, 15, 16, 18, ... MOST values of m yield an irreducible>polynomial, and both of the roots will both share algebraic>integer factors in common with 5. In general it is hard to write down exactly what these>factors are, even in the quadratic case. The important>thing to know about them here is that, as suggested above>by the w_1(m) notation, *they will be dependent on m*. Whats fascinating here is how casually posters will just start> talking, which I think actually usually works with a lot of readers> who dont even bother to read carefully. Just remember that the base position here is that in the ring of> algebraic integers you can have one function dened by the> multiplicative inverse of another function. Remember xy=26.> James Harris === In sci.physics, Virgil where the as are given by the following cubic:a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).>> The above cubic non-equation doesnt give anything.Presumably, it gives three roots when set to 0, whichwould have to be determined using standard methods.If one assumes f and m to be integers, then the roots arealgebraic integers, by denition. I do not know offhandwhether the same could be said of the roots if f and m arealgebraic integers (although I for one dont see why not).The product of these roots is divisible by f^2, althoughone has to be careful: I dont think one can conclude thattwo of the roots are divisible by f, as James appears toclaim elsewhere. Also, weird things happen in algebraicintegers regarding factorization, probably because of theproliferation of units. Taking f=3, we can factorize f^29 = 3^(2/3) * 3^(2/3) * 3^(2/3)as a perfectly valid product therein, and 3^(2/3) isa root of x^3 - 9 = 0 and therefore an algebraic integer.The factorization is not unique of course :-) ; one canalso write9 = 3^(1/2) * 3^(3/4) * 3^(3/4)9 = 3 * 3 * 19 = 3 * 3^(1/3) * 3^(2/3)etc.Mr. Harris does like to jump blindly over crevasses, it seems;the territory should instead be carefully negotiated. :-)-- #191, ewill3@earthlink.netIts still legal to go .sigless. === Ive noted a problem in algebraic number theory with the inclusiveness>of the denition of algebraic integers as roots of monic polynomials>with integer coefcients.Various posters have argued that in fact there is no problem, but>heres a short refutation of their primary objection.First I haveP(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) Cant you articulate your object in English? Such as, If you dene> algebraic integers as roots of monic polys with integer coefcients,> then the following odd thing happens: etc. etc. Just say exactly what> you think the contradiction or problem is.Yes! Thats what I would like to hear too.It seems like hes trying to prove some theorem, but unable to formulate this theorem. === Cc: [.snip.]>Various posters have argued that in fact there is no problem, but>heres a short refutation of their primary objection.First I haveP(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)and the factorizationP(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)where the as are given by the following cubic:a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).My nding that only two of the as have f as a factor without regard>to the value of m has been vigorously disputed.However, consider w_1(m), a factor of a_1 that is a factor of f, as>well as a function that varies with m, then it follows thatFactor where?I assume you mean: w_1(m) is, for each value of m, an algebraicinteger which is a factor of a_1(m), and also a factor of f, both inthe ring of algebraic integers.>a_1 x + uf has w_1(m) as a factor, Should be a_1(m)x + uf, but yes.>so dividing through by w_1(m) givesa_1 x/w_1(m) + uf/w_1(m)Should be a_1(m)x/w_1(m) + uf/w_1(m), so this is g_1(m)/w_1(m), yes.>but then uf/w_1(m) cannot in general be an algebraic integer (Yes, Im interrupting mid-sentence). Ill repeat the full sentence ina second)Why not? You said that w_1(m) was an algebraic integer which is both afactor of a_1 and a factor of f.Since w_1(m) is a factor of f, f/w_1(m) is an algebraic integer. Andif u is an algebraic integer, then u*[f/w_1(m)] is an algebraic integer.So this is an algebraic integer for every value of m for which thefunction is dened. Unless you are saying that the product of two algebraic integers isnot necessarily an algebraic integer? (Thats a theorem of Eisenstein,Gauss, and Dedekind)? >but then uf/w_1(m) cannot in general be an algebraic integer >as its>not representable as a polynomial with a nite number of terms if>w_1(m) varies with m.No, this is wrong. There does not have to be a single polynomial thatyou can plug m into and that gives you a polynomial for w_1(m). Soyou just need an expression with the property that for ->each<- valueof m, what you get is a monic polynomial with integer coefcientswhich has w_1(m) as a root.You want the VALUES uf/w_1(m) each to be an algebraic integer. Thatmeans that for EACH value of m there must be a polynomial, monic withinteger coefcients, that has uf/w_1(m) as a root. There does nothave to be a single polynomial, nor a single family of polynomialswith easy formulas depending on m. All there has to be is for ->each<-value of m, ->some<- polynomial with integer coefcients that worksfor ->that<- value. The polynomials dont have to be related to oneanother by any one nite formula.There is no nite expression with a nite number of terms with theproperty that for every value of m, you get a monic polynomial withinteger coefcients that has all of-|m|-1, -|m|, -|m|+1,....,0,....,|m|-2, |m|-1as roots. But there are expressions that for each value of m, give youa monic polynomial with integer coefcients that has all of them as roots:For example, let g(n,m) = { 1 if |n|<=|m| { 0 if |n|>|m|.Then takef(m,x) = prod_{i=-innity}^{innity} (g(i,m)x - [g(i,m)i - 1])where the index i ranges over all integers.Or, there is no single expression with a nite number of terms which,for each nonzero integer value of m gives you a polynomial that hasthe largest prime p that divides m, 1 if m=1 or -1, and 0 if m=0, as aroot. But there is a single (innite) expression that, for each valueof m, gives you a polynomial that has such a root. For instance,x*(Sum (from i=0 to innity) [m/i]prod_{j=1 to [m/i]}(x-j))where [m/i] is the oor of m/i. There is no reason to demand that there be a single nite expressionthat works; such an expression would be ->sufcient<-, but notnecessary, for the w_1(m) to be, each, an algebraic integer. In fact,there is no need to demand a single expression either, although it istrivial to construct one: for each value of m, let g_m(x) be amonic polynomial with integer coefcients that has w_1(m) as a root,and simply take a sum over all ms of xi_{n}*g_n(x), where xi_{n} isthe characteristic function of the singleton {n}: equal to 1 when theinput is n, and equal to zero otherwise. === === Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of gures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indenite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan== Arturo Magidinmagidin@math.berkeley.edu === > >>Cc: > [.snip.]>Various posters have argued that in fact there is no problem, but>heres a short refutation of their primary objection.First I haveP(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)and the factorizationP(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)where the as are given by the following cubic:a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).My nding that only two of the as have f as a factor without regard>to the value of m has been vigorously disputed.However, consider w_1(m), a factor of a_1 that is a factor of f, as>well as a function that varies with m, then it follows that>>Factor where?>>I assume you mean: w_1(m) is, for each value of m, an algebraic>>integer which is a factor of a_1(m), and also a factor of f, both in>>the ring of algebraic integers.> Mathematicians are disgustingly bad liars when caught, and this one is> no exception.Mathematicians are extraordinarily patient, even with those who blatantly do not deserve such consideration.Gib === [.ad hominem removed.]>> Should be a_1(m)x + uf, but yes.>so dividing through by w_1(m) gives>>a_1 x/w_1(m) + uf/w_1(m)Should be a_1(m)x/w_1(m) + uf/w_1(m), so this is g_1(m)/w_1(m), yes.>but then uf/w_1(m) cannot in general be an algebraic integer >> (Yes, Im interrupting mid-sentence). Ill repeat the full sentence in>> a second)Why not? You said that w_1(m) was an algebraic integer which is both a>> factor of a_1 and a factor of f.Thats the assertion that is shown to be impossible if w_1(m) varies>as m varies.No, you ->claim<- you show it to be impossible, but the reason yougive is invalid.>Its impossible because uf is not a function of m.Its not a question of uf, its a question of a_1(m). The factorw_1(m) depends on the value of a_1(m).>Basic, and easy stuff, but it takes a mathematician to try and lie>anyway, even when beaten.>> Since w_1(m) is a factor of f, f/w_1(m) is an algebraic integer. And>> if u is an algebraic integer, then u*[f/w_1(m)] is an algebraic integer.So this is an algebraic integer for every value of m for which the>> function is dened. Unless you are saying that the product of two algebraic integers is>> not necessarily an algebraic integer? (Thats a theorem of Eisenstein,>> Gauss, and Dedekind)? Rather than tell the truth this mathematician is trying to distract.No, Im trying to see why you claim it is not an algebraic integer,when it is clear that the hypothesis you had up to that point gaveimmediately that it ->was<-.>But it doesnt take a rocket scientist to know that given an integer>f, you cant have a function w_1(m) in algebraic integers such that>for *any* integer m, f/w_1(m) is an integer.Nonsense.>Some people may be confused by the symbol f, so as its independent>and I can give it a value, let f=13.Now then, Arturo Magidin needs you to believe that maybe there exists>a function in algebraic integers such that 13/w_1(m) is an algebraic>integer for *all* integer m.So, you are saying that there are no functions w_1(m), with domain theintegers and with algebraic integer values, other than constantfunctions, with the property that 13/w_1(m) is an algebraic integerfor each integer m?Not the function we have right now, but what is wrong with somethinglikew_1(m) = 13^{1/(m^2+1)}?That function is dened for every integer, has algebraic integervalues for every integer, is a divisor of 13 for every value of m, andis not constant.Or, since there are a countably innite number of proper divisors of13 in the algebraic integers, simply choose any well ordering ofthem and any well ordering of the integers, biject the twowell-orderings, and that gives you a function which is dened foreach integer m, has values in the algebraic integers, and such that13/w_1(m) is an algebraic integer for *all* integer m.Interesting that, when I gave you a mathematical response, here:>>but then uf/w_1(m) cannot in general be an algebraic integer >>as its>>not representable as a polynomial with a nite number of terms if>>w_1(m) varies with m.No, this is wrong. There does not have to be a single polynomial that>> you can plug m into and that gives you a polynomial for w_1(m). So>> you just need an expression with the property that for ->each<- value>> of m, what you get is a monic polynomial with integer coefcients>> which has w_1(m) as a root.You want the VALUES uf/w_1(m) each to be an algebraic integer. That>> means that for EACH value of m there must be a polynomial, monic with>> integer coefcients, that has uf/w_1(m) as a root. There does not>> have to be a single polynomial, nor a single family of polynomials>> with easy formulas depending on m. All there has to be is for ->each<->> value of m, ->some<- polynomial with integer coefcients that works>> for ->that<- value. The polynomials dont have to be related to one>> another by any one nite formula.>> There is no nite expression with a nite number of terms with the>> property that for every value of m, you get a monic polynomial with>> integer coefcients that has all of-|m|-1, -|m|, -|m|+1,....,0,....,|m|-2, |m|-1as roots. But there are expressions that for each value of m, give you>> a monic polynomial with integer coefcients that has all of them as roots:For example, let g(n,m) = { 1 if |n|<=|m|>> { 0 if |n|>|m|.Then takef(m,x) = prod_{i=-innity}^{innity} (g(i,m)x - [g(i,m)i - 1])where the index i ranges over all integers.Or, there is no single expression with a nite number of terms which,>> for each nonzero integer value of m gives you a polynomial that has>> the largest prime p that divides m, 1 if m=1 or -1, and 0 if m=0, as a>> root. But there is a single (innite) expression that, for each value>> of m, gives you a polynomial that has such a root. For instance,x*(Sum (from i=0 to innity) [m/i]prod_{j=1 to [m/i]}(x-j))where [m/i] is the oor of m/i. >> There is no reason to demand that there be a single nite expression>> that works; such an expression would be ->sufcient<-, but not>> necessary, for the w_1(m) to be, each, an algebraic integer. In fact,>> there is no need to demand a single expression either, although it is>> trivial to construct one: for each value of m, let g_m(x) be a>> monic polynomial with integer coefcients that has w_1(m) as a root,>> and simply take a sum over all ms of xi_{n}*g_n(x), where xi_{n} is>> the characteristic function of the singleton {n}: equal to 1 when the>> input is n, and equal to zero otherwise.to engage in an ad hominem, again:Readers on sci.physics and sci.logic note that the mathematicians have>been caught in a base lie trying to hide an over one hundred year old>error.>To believe them you need to believe in an impossible function which>would allow 13/w_1(m) to be an algebraic integer for ALL integer m.Why is such a function impossible? Because you couldnt think of one?>Anyone who disagrees need just give an algebraic integer function>w_1(m) such that 13/w_1(m) is an algebraic integer for all integer m.w_1(m) = 13^{1/(m^2+1)} seems to t the bill quite nicely. And noneof its values are units.>Of course, mathematicians are stepping into the problem that given>some thing like 13/w_1(m) you need, say, r_1(m) as an algebraic>integer function such that13/w_1(m) = r_1(m), so w_1(m) r_1(m) - 13 = 0which necessarily has zeroes. Yes, every value of m is a zero.> That is, only SOME values where it will>work.Huh? Are you under the impression that w_1(m)r_1(m) would be apolynomial in m? Why would you think that?== === Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of gures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indenite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan== Arturo Magidinmagidin@math.berkeley.edu === >Ive noted a problem in algebraic number theory with the inclusiveness>of the denition of algebraic integers as roots of monic polynomials>with integer coefcients.Various posters have argued that in fact there is no problem, but>heres a short refutation of their primary objection.First I haveP(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)and the factorizationP(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)where the as are given by the following cubic:a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).If the a_is depend on m, why dont you write it properly?P(m) = ( a_1(m) x + u f ) ( a_2(m) x + u f ) ( a_3(m) x + u f )>My nding that only two of the as have f as a factor without regard>to the value of m has been vigorously disputed.However, consider w_1(m), a factor of a_1 that is a factor of f, as>well as a function that varies with m, then it follows thata_1 x + uf has w_1(m) as a factor, i.e. w_1(m) | a_1(m) x + u f for all algebraic numbers m>so dividing through by w_1(m) givesa_1 x/w_1(m) + uf/w_1(m)but then uf/w_1(m) cannot in general be an algebraic integer as its>not representable as a polynomial with a nite number of terms if>w_1(m) varies with m.I hope you realize that the functions a_i are also not given by polynomialsin m (i.e., there does not exists a polynomial A_i(M) in A[M], thepolynomial ring in one variable over the algebraic numbers, such thatA_i(m) = A_i(m) for all algebraic numbers m).For the function w_1, well, the only thing youve mentioned about it isthat w_1(m) | a_1(m) and w_1(m) | f (for all algebraic numbers m, I think). So how could one expect a_1(m) x / w_1(m) + u f / w_1(m), or u f / w_1(m)to have any particular form at all? Of course, for u f / w_1(m)to be an algebraic integer for all algebraic integers m, it doesnthave to be expressible as a polynomial in m over the algebraic integers.With all you assumptions, it is of course trivial that u f / w_1(m)is an algebraic integer (since f is divisible by w_1(m) in the algebraicintegers).>My guess is that some may be assuming that f is replaceable by some>function of m, but in fact, its independent of the value of m,Everybody understands that, Im sure.>so its>like 1/(x+1) which is also not representable by a polynomial if x is>an algebraic integer not equal to 0 or -2.But I wonder who understands this. I denitely dont.(By the way, if x *is* an algebraic integer unequal to -1, then1/(x+1) *is* expressible as a polynomial over the algebraic integers -a constant polynomial. But you probably mean that there is nopolynomial F(X) over the algebraic integers such that F(x) = 1/(x+1)for all algebraic integers unequal to -1. I still wonder what 0 and-2 have to do with it - maybe just a mistake.)>So the objection is refuted by the impossibility of uf/w_1(m) being an>algebraic integer, for all algebraic integers m, if w_1(m) varies with>m.Can you repeat the denition of algebraic integer again forthe newsgroup and tell us how you conclude that u f / w_1(m)is not an algebraic integer?Peter van Rossum-- -- Peter van Rossum, | Universal law of linearity: for allDept. of Mathematics, New Mexico | f : R -> R and for all x, y in R:State University, Las Cruces, NM, USA. | f(x + y) = f(x) + f(y) === [.snip.]>Its not possible. Ill put in values to help those readers confused>by symbols (though algebra is BASED on symbols) by letting u=2, f=13,>then you have26/w_1(m)and you dont have to be a rocket scientist to know that no function>w_1(m) that actually varies with m can exist such that 26/w_1(m) is an>algebraic integer for *all* integer m.> w_1(m) = 13^{1/(m^2+1)}*2^{|m|/(m^2+1)}Im not saying thats the function in question (it is not), but there> is a function that actually varies with m such that 26/w_1(m) is an> algebraic integer for *all* integer m. So the argument that no such> function can exist is simply bogus.Well that example works, so now Ill say algebraic integer m, and itblows up at m^2+1 = 0.For those confused by the discussion, Ive pointed out thatmathematicians arguing with me have been arguing for a relationrelating the multiplicative inverse of one algebraic integer functionto another over all algebraic integers.For instance, with u=2, f=714/w_1(m) = r(m)which would relate w_1(m) to r(m) for all algebraic integer m, whichis like claiming that xy=2 for all integer x, with an integer y.Its that multiplicative inverse that blows away the objection as infact, w_1(m) is a constant with regard to m--independent of m--as Iverepeatedly shown.However, accepting basic algebra would mean that mathematicians acceptthat theres an over one hundred year old denition problem in coremathematics.Many of you may have believed that I was making something up, but youshould have enough mathematical knowledge to understand that therelations these people are arguing for, cannot exist.Unless youre lost on the example of xy=2, in the ring of integers, incomparison to 14/w_1(m) = r(m) in the ring of algebraic integers.Its that / that makes the difference. Math people are lying herebecause the information is so earthshaking, and they are corrupt.James Harris === [.snip.]>Its not possible. Ill put in values to help those readers confused>>by symbols (though algebra is BASED on symbols) by letting u=2, f=13,>>then you have>>26/w_1(m)>>and you dont have to be a rocket scientist to know that no function>>w_1(m) that actually varies with m can exist such that 26/w_1(m) is an>>algebraic integer for *all* integer m.>> w_1(m) = 13^{1/(m^2+1)}*2^{|m|/(m^2+1)}Im not saying thats the function in question (it is not), but there>> is a function that actually varies with m such that 26/w_1(m) is an>> algebraic integer for *all* integer m. So the argument that no such>> function can exist is simply bogus.Well that example works, so now Ill say algebraic integer m, and it>blows up at m^2+1 = 0.So let me see if I have this right:(1) You claimed something was impossible under a given set of conditions.(2) When I proved that your claim was simply false, you changed the set of conditions, and now argue that what I said was wrong.Great. So, rst of all, you were wrong. You claimed no such function couldexist, but theres a function that did everything you required it todo. And second of all, rather than admit you were wrong, you justchange the rules.Heres an example that works for EVERY algebraic NUMBER m:Step 1. Given an algebraic number m, let f(x) be the unique monicpolynomial with rational coefcients, irreducible over Q, which has mas a root. Write it as: f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0.Note that a_0 must be different from 0.Step 2. Write a_0, a rational number, as a_0 = r/s, with r and sintegers, r and s relatively prime.Step 3. If r=1 or -1, let q = 1.Step 4. If r is not equal to 1 or -1, then let q be the largest rational prime that divides r.Step 5. Let w(m) be a root of the polynomial x^4 + 13qx^3 + qx + 13.Then:(a) w(m) is an algebraic integer.(b) w(m) divides 13 in the ring of algebraic integers, since the product of all the roots is 13, and every root is an algebraic integer.(c) w(m) is not constant: it takes different values at different integers.(d) w(m) takes each value a countably innite number of times, so w(m) cannot be given by a polynomial. [.snip.]What now, James? Are you going to change your claim again in order toavoid the fact that you were just plain wrong?>For those confused by the discussion, Ive pointed out that>mathematicians arguing with me have been arguing for a relation>relating the multiplicative inverse of one algebraic integer function>to another over all algebraic integers.For those confused by the discussion, James claimed there could be NOfunction (other than a constant function), from the integers to thealgebraic integers, with the property that at each integer we got analgebraic integer which was a divisor of 13. James was wrong. He doesnot like to admit he is wrong, so hes changing the subject.>For instance, with u=2, f=714/w_1(m) = r(m)which would relate w_1(m) to r(m) for all algebraic integer m, which>is like claiming that xy=2 for all integer x, with an integer y.No, its not like claiming that at all. Because the integers are aUFD with a nite number of units, so there are only a nite numberof ways in which each integer can be expressed as a product ontegers.The algebraic integer are NOT a UFD, have an innite number of units,and any algebraic integer can be expressed as a product of algebraicintegers in an innite number of ways.So it is not at all like claiming that.>Its that multiplicative inverse that blows away the objection as in>fact, w_1(m) is a constant with regard to m--independent of m--as Ive>repeatedly shown.As youve repeatedly ->claimed<-, but as has repeatedly been shown,with EXPLICIT calculations, is not the case. [.rest deleted.] === ==Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of gures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indenite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan== Arturo Magidinmagidin@math.berkeley.edu === >> If the a_is depend on m, why dont you write it properly?P(m) = ( a_1(m) x + u f ) ( a_2(m) x + u f ) ( a_3(m) x + u f )Thats a style issue. I see it as a gesture of futility and anguish>at the reality of this easy refutation.More an issue of clarity. Its strange to have some dependencieson m (P(m), w_1(m) later on) explicitely and others not. But Iguess youre not concerened with presenting a clear argument.>> i.e. w_1(m) | a_1(m) x + u f for all algebraic numbers m>so dividing through by w_1(m) gives>>a_1 x/w_1(m) + uf/w_1(m)>>but then uf/w_1(m) cannot in general be an algebraic integer as its>>not representable as a polynomial with a nite number of terms if>>w_1(m) varies with m.I hope you realize that the functions a_i are also not given by polynomials>> in m (i.e., there does not exists a polynomial A_i(M) in A[M], theThats a rather stupid lie given that I put the polynomial in this>post.Again it isa^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)and its roots are a_1, a_2 and a_3.You seem to miss the point. Let me write it down more explicitely:there are no algebraic integers c_0, c_1, ..., c_n such thata_1(m) = c_0 + c_1 m + c_2 m^2 + ... + c_n m^n for all algebraic integersm. If you still think there are such algebraic integers, I wouldlike to see explicitely what they are.What you have is a polynomial Q(A,M) in two variables over the algebraicintegers (namely Q(A,M) = A^3 + 3(-1+Mf^2)A^2 - f^2(M^3f^4 - 3M^2f^2 + 3M))such that Q(a_1(m),m) = 0 for all algebraic integers m.>> With all you assumptions, it is of course trivial that u f / w_1(m)>> is an algebraic integer (since f is divisible by w_1(m) in the algebraic>> integers).Its not possible. Ill put in values to help those readers confused>by symbols (though algebra is BASED on symbols) by letting u=2, f=13,>then you have26/w_1(m)and you dont have to be a rocket scientist to know that no function>w_1(m) that actually varies with m can exist such that 26/w_1(m) is an>algebraic integer for *all* integer m.You have *assumed* that w_1(m) divides f (i.e. 13, in this case) forall algebraic integers m. So let me give a function w_1 that makes26/w_1(m) an algebraic integer, even an integer, for all algebraic numbersm. Here it is:w_1(m) = { 1 if m = 0 { 13 if m <> 0Of course, this is not a polynomial expression, but that is sort of thewhole point.>Its a show of how broken math society is that Peter van Rossum would>dare to make that stupid assertion.Yes! Ive been insulted by JSH. This nally makes my name as a>> Can you repeat the denition of algebraic integer again for>> the newsgroup and tell us how you conclude that u f / w_1(m)>> is not an algebraic integer?Mathematicians are pathetic liars.What is the lie in that question?>Ill use u=2, f=13 again, now then, NO function in algebraic integers>exists such taht 26/w_1(m) is an algebraic integer for all integers m,>if w_1(m) varies with m.You can scroll back to see a counterexample.>Its just not possible, but it takes a mathematician to lie about it.Another question, maybe easier. What is a function in algebraic integers?Peter van Rossum-- -- Peter van Rossum, | Universal law of linearity: for allDept. of Mathematics, New Mexico | f : R -> R and for all x, y in R:State University, Las Cruces, NM, USA. | f(x + y) = f(x) + f(y) === I have a degree in physics but a career in information technology. Ihave always had a desire to study physics or applied mathematics tothe graduate (Masters) level. I wouldnt expect this to increase myincome. My desire stems from a natural curiosity; I enjoy theseelds.Im in my mid-fties. Will my age be (perceived as) an impediment toeither undergraduate or graduate studies?Any suggestions as to - how one chooses between/among programs in applied mathematics andphysics?- the attitudes of people in each eld?- level of effort/time required? I would appreciate the advice of anyone in a similar situation.tom === > I have a degree in physics but a career in information technology. I> have always had a desire to study physics or applied mathematics to> the graduate (Masters) level. I wouldnt expect this to increase my> income. My desire stems from a natural curiosity; I enjoy these> elds.Im in my mid-fties. Will my age be (perceived as) an impediment to> either undergraduate or graduate studies?Most denitely. As Hardy noted in A Mathematicians Apology, math is a young mans game.That doesnt mean age is an impediment to your learning. But as far as whether age will be perceived as such, of course it will. === > I have a degree in physics but a career in information technology. I> have always had a desire to study physics or applied mathematics to> the graduate (Masters) level. I wouldnt expect this to increase my> income. My desire stems from a natural curiosity; I enjoy these> elds. Im in my mid-fties. Will my age be (perceived as) an impediment to> either undergraduate or graduate studies?I dont see why it would. You had to study a lot of math to get a degree inphysics so you must be capable. The only question is whether you have theconcentration to do so and are willing to devote the time. I assume by thispost that you are willing to devote the time?> Any suggestions as to> - how one chooses between/among programs in applied mathematics and> physics?What are your goals? What do you like/enjoy. What do you think youd be goodat? What parts have you been good at in the past?I was in a grad program but had to stop due to family illness. I may returnsoon depending on whether something pops up in the mean time. But things area bit different for me since my goals have changed and things are a bit inux as of now.Pete === > I have a degree in physics but a career in information technology. I> have always had a desire to study physics or applied mathematics to> the graduate (Masters) level. I wouldnt expect this to increase my> income. My desire stems from a natural curiosity; I enjoy these> elds.Im in my mid-fties. Will my age be (perceived as) an impediment to> either undergraduate or graduate studies?Any suggestions as to> - how one chooses between/among programs in applied mathematics and> physics?> - the attitudes of people in each eld?> - level of effort/time required?I would appreciate the advice of anyone in a similar situation.tomFollowing my retirement I earned my masters degree in mathematics (numbertheory) at age 57 and my doctorate at age 62. Everyone has been mostfriendly and helpful. This was a full time retirement project for me and Icontinue to go to the university a couple of days a week to attend seminarsand pursue my research interests. I count it as one of the most rewardingexperiences of my life.The main attitudes I met were friendly curiosity and some good humoured envythat I didnt have to go job hunting, apply for grants, teach, publish etcetc unless I wanted to.The effort required was considerable and I dont think I could have done iton a part time basis while holding down a full time job.Of course, many people are better organized (and less lazy :) ) than I am. Jack Fearnley === Ive been trying to understand this concept both as it applies to ourelementary understanding of (a,b), and the more general meaning intopology. I have reached the following conclusions:1.- The two concepts agree when the topology involved is that of allopen subsets of R. That is any open set in the topology is open inthe usual, elementary sense, that is the set is a neighborhood of allits points. Even here, however, the null set and R are both open andclosed.2.- For other topologies (say the collection of all subsets of R)a setmay be open according to one denition, closed according to theother, or even both open and closed. The 2 denitions are thereforeunrelated.Am I understanding this stuff correctly?There must be some general utility in dening open sets as theelements of a properly dened topology. Can someone please outlinethe reason? Many thanks. === >Ive been trying to understand this concept both as it applies to our>elementary understanding of (a,b), and the more general meaning in>topology. I have reached the following conclusions:>1.- The two concepts agree when the topology involved is that of all>open subsets of R. That is any open set in the topology is open in>the usual, elementary sense, that is the set is a neighborhood of all>its points. Even here, however, the null set and R are both open and>closed.2.- For other topologies (say the collection of all subsets of R)a set>may be open according to one denition, closed according to the>other, or even both open and closed. The 2 denitions are therefore>unrelated.Am I understanding this stuff correctly?Yes.Open sets in topology are an abstraction. The idea in topology is todene the notion of nearness or neighborhood. An open setcontaining x contains the points that are near x; how near dependson the open set.There are many ways of dening nearness, which translate intodifferent topologies. The usual topology in R is related to theconcept of nearness derived from a metric, which is a measure ofdistance: the distance between two real numbers x and y is |x-y|. Opensets are (ultimately) dened in terms of this distance.But this is not the only way to dene nearness. You could say thatyou do not want two real numbers to be thought of as near each otherunless they are the same real number: that gives you the discretetopology, where every set is open. Or you could say that any two realnumbers are near each other, which gives the indiscrete topology (theonly nonempty open set is the entire set).Or perhaps you are concerned with niteness in some sense, so youdene nearness in terms of being elements of a set whose complementis nite.>There must be some general utility in dening open sets as the>elements of a properly dened topology. What is a properly dened topology? A topology on a set is acollection T of subsets that satises the following properties: (1) The empty set is in T. (2) The full set is in T. (3) If A and B are in T then A intersect B is in T. (4) If {A_i}_i in I is a family of sets, and each A_i is in T, then the union of the A_i is also in T.Any collection of subsets that satises these 4 properties is aproperly dened topology on the set. === ==Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)= Arturo Magidinmagidin@math.berkeley.edu === For those of you trying to keep up with the mathematical facts in thediscussions about the error in core mathematics from a problem with adenition, this post will outline the important ones quickly andsuccinctly.1. First the problematic denition:Algebraic integers are dened to be roots of monic polynomials withinteger coefcient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, wheremonic refers to the leading coefcient.My assertion is that the over hundred year old denition excludesnumbers that have to be included to keep from having contradictioni.e. mathematical inconsistency.2. The important tool I use is a polynomial:P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)The form of the polynomial allows me to factor P(m) intonon-polynomial factors, and the factorization with those factors isP(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)where the as are roots of the following cubic:a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).3. Dispute centers around what happens when I divide P(m) by f^2,which youll note is a factor of the polynomial in the ring ofalgebraic integers.4. Mathematicians have argued that f^2 divides off as a function of mbecause if they concede that it divides off independent of m, then Ican show that only two of the roots ofa^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)have f as a factor.5. However, it turns out that if you go to the eld of algebraicnumbers you can prove that for *certain* values of m and f, the rootsof the cubic do not have f as a factor *in the ring of algebraicnumbers* which is the inconsistency.That is, for the math to be consistent, two of the roots *should* havef as a factor as long as m and f are algebraic integers, but while Ican show they do for a particular values like m=1, f=sqrt(2), thereare other values you can show they do not *in the ring of algebraicintegers* which results from the denition and its focus on monicpolynomials.Note: In the ring of algebraic integers you cant see the problem buthave to go to the eld of algebraic numbers as from within the ringof algebraic integers it appears that only two of the roots have afactor that is f.James Harris === > For those of you trying to keep up with the mathematical facts in the> discussions about the error in core mathematics from a problem with a> denition, this post will outline the important ones quickly and> succinctly. 1. First the problematic denition: Algebraic integers are dened to be roots of monic polynomials with> integer coefcient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where> monic refers to the leading coefcient. My assertion is that the over hundred year old denition excludes> numbers that have to be included to keep from having contradiction> i.e. mathematical inconsistency. 2. The important tool I use is a polynomial: P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) The form of the polynomial allows me to factor P(m) into> non-polynomial factors, and the factorization with those factors is P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where the as are roots of the following cubic: a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). 3. Dispute centers around what happens when I divide P(m) by f^2,> which youll note is a factor of the polynomial in the ring of> algebraic integers. 4. Mathematicians have argued that f^2 divides off as a function of m> because if they concede that it divides off independent of m, then I> can show that only two of the roots of a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) have f as a factor. 5. However, it turns out that if you go to the eld of algebraic> numbers you can prove that for *certain* values of m and f, the roots> of the cubic do not have f as a factor *in the ring of algebraic> numbers* which is the inconsistency. That is, for the math to be consistent, two of the roots *should* have> f as a factor as long as m and f are algebraic integers, but while I> can show they do for a particular values like m=1, f=sqrt(2), there> are other values you can show they do not *in the ring of algebraic> integers* which results from the denition and its focus on monic> polynomials. Note: In the ring of algebraic integers you cant see the problem but> have to go to the eld of algebraic numbers as from within the ring> of algebraic integers it appears that only two of the roots have a> factor that is f.> James HarrisIs it me or can anyone follow what JSH is ranting about?David Moran === Is it me or can anyone follow what JSH is ranting about?David MoranSee: http://www.google.com/search?q=%22James+Harris%22+site% 3Awww.crank.net http://www.crank.net/harris.html === >For those of you trying to keep up with the mathematical facts in the>discussions about the error in core mathematics from a problem with a>denition, this post will outline the important ones quickly and>succinctly.1. First the problematic denition:Algebraic integers are dened to be roots of monic polynomials with>integer coefcient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where>monic refers to the leading coefcient.My assertion is that the over hundred year old denition excludes>numbers that have to be included to keep from having contradiction>i.e. mathematical inconsistency.2. The important tool I use is a polynomial:P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)The form of the polynomial allows me to factor P(m) into>non-polynomial factors, and the factorization with those factors isP(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)where the as are roots of the following cubic:a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).3. Dispute centers around what happens when I divide P(m) by f^2,>which youll note is a factor of the polynomial in the ring of>algebraic integers.4. Mathematicians have argued that f^2 divides off as a function of m>because if they concede that it divides off independent of m, then I>can show that only two of the roots ofa^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)have f as a factor.Its not a question of conceding. It is simply not true that thefactors have the property you claim they do. Weve given explicitexamples with explicit calculations where they do not have theproperty you claim they do.Your argument about why they must divide off independent of m hasnot been understood or accepted by anyone that I am aware of otherthan you.>5. However, it turns out that if you go to the eld of algebraic>numbers you can prove that for *certain* values of m and f, the roots>of the cubic do not have f as a factor *in the ring of algebraic>numbers* which is the inconsistency.No, this is not a correct summary of the situation, for various reasons: (1) In the ring of algebraic numbers, every number other than zero is a factor of every other algebraic number: the algebraic numbers form a eld. So it is impossible to say the roots of the cubic do not have f as a factor *in the ring of algebraic numbers*. Nobody has made such a claim.(2) There is no need to go to the eld of algebraic numbers; for every value of m for which the cubic x^3 + 3(-1+mf^2)x^2 - f^2(m^3f^4 - 3m^2f^2 + 3m) is irreducible over Q, one can construct the minimal polynomial of a/f, where a is any root of f; this only requires you to go to the ring A[1/f], where A is the ring of all algebraic integers, and monic and irreducible with integer coefcients, it follows that cannot be a factor of any root in the ring of algebraic INTEGERS.>That is, for the math to be consistent, two of the roots *should* have>f as a factor as long as m and f are algebraic integers,ONLY if your argument were valid, which nobody but you believes to bethe case.> but while I>can show they do for a particular values like m=1, f=sqrt(2), there>are other values you can show they do not *in the ring of algebraic>integers* which results from the denition and its focus on monic>polynomials.Which is logical nonsense. A denition cannot introduce acontradiction, since a denition is simply shorthand for anexpression. If instead of dening algebraic integers we simplytalked about roots of monic polynomials with integer coefcients,there would be absolutely no change in mathematics (except that allnumber theory books would be longer). So IF there were acontradiction, it cannot be because of the DEFINITION, it would haveto be because some property that people have claimed the roots ofmonic polynomials with integer coefcients have, in actuality they donot have. You have not identied any such property, and therefore youhave not identied any problems (since you have not establish thevalidity of your argument about the divisibility of the coefcients). == === Arturo Magidinmagidin@math.berkeley.edu