mm-208 === Subject: Re: Bouncin Off the Walls (of a hexagon)>(This puzzle combines a couple related puzzles I have posted.)We have a regular hexagon whose sides are lables, clockwise from top,>A through F.the hexagons inner walls as if the walls were mirrored, but it also>affects the direction of *itself*; for whenever it crosses its own>path (as already drawn), it passes through the path, but is re§ected>as if a mirror has been placed perpendicularly to the previous path at>the point of intersection.if such a particular path actually exists, because the paths final>direction is heavily dependent upon the accuracy in which the path is>drawn and the accuracy of the angles re§ected.But I will give the order of the hexagons surfaces as visited by the>path (as drawn at the time of each particular crossing) below.(I know this, if my by-hand approximation was not too §awed.)(I have no idea. You best use exact-rational arithetic to get this, if>it is possible to figure out at all.)E, B , D, 2 crossings, F, E, 1 crossing, F, 3 crossing, D, B, 3 ^^^^^^^^^^^^^^^^^^^Problem: I tried to draw the path, and if the ray goes from F to E without acrossing, it cant get back to F with only one crossing, due to line 1 (from Ato E) and line 6 (which crosses line 1 and goes to F) being in the way.Did I just draw it differently, or is there a missing crossing?>crossings, F, 7 crossings, and back to its staring point.(If no crossings are listed between letters, than no crossings occur>between them.)(If someone solves this, they are going to have to post some link to Patrick Hamlyn posting from Perth, Western AustraliaWindsurfing capital of the Southern HemisphereModerator: polyforms group Does the series sin(1)/1 + sin(2)/2 + sin(3)/3 + sin(4)/4 + ... converge?> How does one prove convergence (or divergence)?. If it converges what > is a good way to estimate its value?No need to estimate it; the value can be computed exactly. Its(pi-1)/2.Note that sum_{n=1}^infty z^n/n = - Log(1-z) for complex z with |z| < 1 (and Log is the principal branch of thelogarithm). Replace z by r e^(it), for values of t other than pi/2 +2k pi, let r increase to 1, and appeal to Abels theorem to concludethatsum_{n=1}^infty e^(itn) / n = -Log(1-e^(it)) = -Log(e^(it/2) - Log(e^(-it/2) - e^(it/2)) = -i t/2 - Log(-2 i sin(t/2)) = -i t/2 - Log(2 e^(-i pi/2) sin(t/2)) = -i t/2 - ln(2) + i pi/2 - ln sin(t/2)if I havent made a total mess of the calculation. So we get ******************************************* * * * sum_n cos(n t)/n = -ln (2 sin(t/2)) * * * * sum_n sin(n t)/n = (pi - t)/2. * * * *******************************************In particular, taking t = 1, we get sum_n sin(n)/n = (pi -1)/2 approx 1.0707963267948966192,which agrees with the answer Mathematica returns. (But Mathematica 4.2insists that sum cos(n)/n diverges. Its wrong.)Note that you would deduce the sine sum if you computed the Fourierseries of f(x) = x. Not, perhaps, something you would have guessed;but hopefully something you would have computed sometime in your life.--Ron numbers> 4. Form the number P = p1*p2*...*pn + 1.Actually, he formed two numbers P = p1*p2*...*pn + 1 and Q =p1*p2*...*pn - 1, and insisted that any proof with just one was invalid.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ Walls (of a hexagon) [...] whenever it crosses its own path (as already drawn), it passes through the path, but is re§ected as if a mirror has been placed perpendicularly to the previous path at the point of intersection.[...] E, B , D, 2 crossings, F, E, 1 crossing, F, 3 crossing, D, B, 3> ^^^^^^^^^^^^^^^^^^^> Problem: I tried to draw the path, and if the ray goes from F to E> without a crossing, it cant get back to F with only one crossing,> due to line 1 (from A to E) and line 6 (which crosses line 1 and goes> to F) being in the way. Did I just draw it differently, or is there a missing crossing?> crossings, F, 7 crossings, and back to its staring point.The ray goes from F to E (no crossing), then bounces back betweenthe (just created) F-E ray and the original A-E ray. It then hitsthe last segment of the D-F ray and is re§ected back to F.[ It goes on to cross A-E, E-B, B-D, hits D, goes to B, then crossesB-D, E-B and A-E and reaches F, and then I have no idea what the lastseven crossings are. :( Re: DO MATHEMATICIANS READ WITH HALF A LIGHTBULB?> You want to prove that 1 + a + a^2 + a^3 . . . = 1 / (1 - a) FOR ALL> a.First of all notice that this statement does not make sense when a => 1, because the right side is undefined, hence you are trying to say> something using things without definitions -- i.e. potential nonsense.So you may say, fine, Ill DEFINE 1/0 for ya. Its infinity!> Then you have to invent your own infinity arithmetic. For you cannot> have infinity follow the same rules that we all agree regular numbers> follow. For example, 1 / 0 = infinity, 2 / 0 = infinity, so by> definition of division 0 * infinity has all the values of the rainbow.> So infinity * 0 is not unique. Incidentally, 0 / 0 is therefore not> unique either, since the answer is the number which multiplied by 0> gives 0 but this is true for all numbers. So you can replace the> with and and have yourself a nice little world with your infinity> and zero arithmetic.No one says you cant do that. It just has to be consistent and> logical. You can invent your own math. If its interesting and no one> has done it before, you can even publish it. :-)You obviously have great familiarity with the subject and have foughtwith it long enough to appreciate its difficulty.0 / 0 is the easiest case. The symbols ask us to count how many timeswe can subtract zero from zero to the limit of zero. We reach thelimit after the first required subtraction, and thus our answer is 1,a process 1.Any non-zero x / 0 = Infinity, and the objections to the rubbernature of the possible results can be answered in the Cartesian planewhere any two inverse x and y slopes multiply out to a slope of 1through very certain points on the x and y axes. Beginning with aninfinite-slope y value and a zero-slope x value, the resulting 1-slopeline will have a definate relation to any chosen beginning points. Although those beginning points can be any points, the beginningcontext chosen is real.Thus, 5 / 0 = Infinity, and, inversely in that particular beginningcontext, Infinity * 0 = 5. The fact that multiplications involvingInfinity and zero can lead to such results is not a contradiction ofnormal math, for Infinity and zero have logical properties that exceedthe properties of normal numbers. Specifically, Infinity includesboth every particular number AND all such numbers, which givesinfinity the logical power to exceed the algebraic limits of processeswith ordinary numbers. In other words, although a conclusion cannotexceed the import of its premisses, Infinity has the necessary extralogical elements to justify its results within normal algebra. Zeroobtains to the some of the same power with the extra logical elementof being both a normal point on the number line as well as,exotically, representing NO number.Although I havent studied the logic in detail, Ive read that when anirresistable force meets and unmovable object, what happens logicallyis EVERYTHING. Thus, infinity and zero math is ultimately consistentwith logic and algegra by virtue of the extra logical elementsinvolved.The difficult task is not to call such difficulties undefined, butrather to figure MATHEMATICIANS READ WITH HALF A LIGHTBULB?> Very nice, for hundreds of years we have erroneously assumed the> geometric series converges only for |x|<1, but you have proven> otherwise.If we had assumed such, it wouldnt be worth much. 63 28 EB DA E6 44 E5 5E EC F3 04 26 4E BF 1A 92X-Tom-Swiftie: I cant think of a good AI technique, Tom said searchingly> Although I still do not know what converges refers to, the form of> your objection is argumentum ad populum. Let us remember that for> many centuries, the world was most assuredly §at.I agree of course that argumentum ad populum is invalid, and haveseparately protested that. But let us remember that:1) The world has always been round, whatever people thought, and2) Nobody important has thought the world generation of large prime numbers> You were trying to conclude that, for any prime P, N is a larger> prime. No, I wasnt. I was saying that it might be a larger prime, but that it might be a composite of primes at least one of which is larger than P.Sorry, youre right. The *original poster* was (if I understand them> correctly) trying to make the conclusion noted above.Yes, the OP is indeed to blame. I wish Id made a better job of explaining Euclid, though.-- Richard Heathfield : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, MATHEMATICIANS READ WITH HALF A LIGHTBULB?You want to prove that 1 + a + a^2 + a^3 . . . = 1 / (1 - a) FOR ALL> a.First of all notice that this statement does not make sense when a => 1, because the right side is undefined, hence you are trying to say> something using things without definitions -- i.e. potential nonsense.So you may say, fine, Ill DEFINE 1/0 for ya. Its infinity!> Then you have to invent your own infinity arithmetic. For you cannot> have infinity follow the same rules that we all agree regular numbers> follow. For example, 1 / 0 = infinity, 2 / 0 = infinity, so by> definition of division 0 * infinity has all the values of the rainbow.> So infinity * 0 is not unique. Incidentally, 0 / 0 is therefore not> unique either, since the answer is the number which multiplied by 0> gives 0 but this is true for all numbers. So you can replace the> with and and have yourself a nice little world with your infinity> and zero arithmetic.No one says you cant do that. It just has to be consistent and> logical. You can invent your own math. If its interesting and no one> has done it before, you can even publish it. :-)You obviously have great familiarity with the subject and have fought> with it long enough to appreciate its difficulty.0 / 0 is the easiest case. The symbols ask us to count how many times> we can subtract zero from zero to the limit of zero. We reach the> limit after the first required subtraction, and thus our answer is 1,> a process 1.Any non-zero x / 0 = Infinity, and the objections to the rubber> nature of the possible results can be answered in the Cartesian plane> where any two inverse x and y slopes multiply out to a slope of 1> through very certain points on the x and y axes. Beginning with an> infinite-slope y value and a zero-slope x value, the resulting 1-slope> line will have a definate relation to any chosen beginning points. > Although those beginning points can be any points, the beginning> context chosen is real.Thus, 5 / 0 = Infinity, and, inversely in that particular beginning> context, Infinity * 0 = 5. By the same logic, 6/0 = infinity, therefore infinity * 0 = 6. We then For those who have the book, it is W. Rudin, Principles of>Mathematical Analysis, chapter 2 ( Basic Topology), problem 18:>[A set E is perfect if E is closed and every point of E is a limit>point of E] Is there a nonempty perfect set in R which contains no>rational number?> ... [ contructs open set containing rationals and with measure < 1, takex X as its complement ] ...Try showing (1) X is not countable (2) the isolated points of X constitute at most a countable set (3) X {isolated points of X} is closedI dont think this will work. Try considering points of X such that>every neighborhood contains an uncountable infinity of points of X>instead. By Cantor-Bendixson, (2) must be true. (3) seems clear since X, being closed, contains all its accumulation points, and these are also exactly the accumulation points of X{isolated points}. Is there a problem with (1) or C-B too much to take as known?> While statements (1)-(3) are correct, it seems that you are hinting> that X isolated points is perfect. That isnt true. [ Blush ]> Suppose X is the closed subset of the reals consisting of 0, {1/n: n>0},> and the closed interval [2,3]. Then X isolated points is 0 union> [2,3] is not perfect.Right, and you dont even need the interval; {0} isnt perfect. In order to make my idea work we need X1=X{isolated pts} X2=X1{isolated ... X(omega) = intersect[i I went to Indian Institute of Technology (IIT), Powai, Mumbai to> explain mechanism of my Action Device and to seek technical help. I> met Dr. Amitay Issac of Aerospace Engineering Department and I tried> to explain very basic component/idea of this action device. I have> given in my homepage what exactly I tried to convince him.http://www.geocities.com/actiondeviceBut he insisted that point B will shift its position along Y axis!.> I had to return in few minutes.Now I tried to convince again to Dr. G Arvind Rao of Aerospace> Engineering Department by email, but he also said that point B will> shift its position along Y axis !.> Hmmm... did you consider that they could be right, and you could be wrong?Laura, where from you suddenly dropped in this mess? You just dont> know, what is going on. I thought about this thousands of times in> last 13 months. I had posted idea of whole device in many newsgroup.> This is just one of the basic component or idea behind this invention.> At least this problem was not arised. And now suddenly this problem> propped up. It appears that the problem has popped up every time you haveshown it to anyone familiar with physics. > Indian Institute of Technology is most prestigious college in India.> This institute gives people for Aviation Industry around the world.> And I just wonder, why so highly educated people fail to understand> such simple problem.> Maybe, just maybe, they do understand it.Have you done elementary Geometry Laura? Take a look at my homepage.http://www.geocities.com/actiondevice What is on that page is silly. Point B is pulled downward byany downward force applied to it unless support is provided tohold it up. You provide no support, so point B is pulled down. > In fact, this is not problem at all. But what a tragedy, I am> facing such ridiculous problems.I can end my all problems anytime, but I am following the rules of> this battle, waiting game.> Build a working model and submit it to them for examination. Doesnt matter how much force it produces, as long as it proves that your idea works.No, not yet. You just dont know what is going on around me. Things> are under absolute control. You will never believe it. Do it anyway. The parts will cost you less than $US 5, and itwill take you less than four hours to build and test it. > I am just watching how the minds of highly educated people around> the world are controlled by that Supreme Force named God.> Let me get this straight.... *God* doesnt want this device discovered? Why not? And if not, whats stopping him from destroying you to make sure you stay quiet?He does want this device to be discovered. This is exactly why> He controlled absolutely everything in my personal life. He> navigated things in last 17 years in such a way that my thought> process moves only in one direction. He trained me to gain> absolute power of imagination. God and I have an agreement. I will not say anything untrueabout God or deliberately do anything to harm anyone, and inexchange, God will not control me in any way. So far, God hashonored our agreement perfectly. I hope that Im upholding myside as well, but, unlike God, Im not perfect. I promise you that God will not stop you from making andoperating a model of your device, and that God will notinterfere in any way with its operation. However, God willnot make it work for you, either. If you are right, it willwork. If you are wrong, it will not work. God cares whetherit works or not, and knows whether it works or not, but Godwill not change you, your invention, or anything else to makeit work or prevent it from working. > This device is very simple. But there is no victory without> sufferings. And He has discovered His own ways to trap me. You misunderstand Gods purpose. I do not understand Godspurpose, either, but God has made it clear to me that yourunderstanding is incorrect. God is not trapping you. > Things are being controlled very cleverly. Dont believe me? People in this NG will not answer clearly the question I have> posed. Will point B move along Y axis in XY plane? It needs> just yes/no. But they will remain silent(or they will be> humorous). They will ignore me. Because they are controlled. Most people ignore you because they have no interest in you. Other people use humor with you because they believe you arementally ill. You need to try to show them that you are not. And some people have answered your question. The person whoposts under the handle Bored Huge Krill said on November 9that point B will fall, and showed you why. George Dishmanexplaind to you where the §aw in your device is on August 24.Phil Holman told you on August 1st why it will not work.Rodney Long also told you on August 1st why it will not work. Other people are not being controlled. You are being controlled, but *not* by God. If you are man enough to fasten a couple of springs to aframework and test it, you can do it in less than four hours.I guarantee it. Are info, . Google found this as well:> http://www.math.uga.edu/~mbaker/pdf/pnt2.pdf> Its a 17-page pdf, covering the background about zeta and the Laplace> transform, as well as Newmans proof of PNT.> LHSorry for the delayed reply; I wanted to point out that between thenand now Ive updated my online analytic/combinatorial number theorynotes to include an exposition of Hildebrands elementary proof of theprime number theorem. Whether this is the best proof of the PNT isleft for the reader to judge... The notes are online athttp://www.princeton.edu/~ppollack/notes/Im sure there are still plenty of errors lurking in this chapter andthe others; corrections and comments are more than welcome!BTW, I feel a bit sheepish about not including the reference to theBaker-Clark paper myself. As the link above points out, Baker was theadvisor for my senior thesis at UGA (which was an earlier version ofmy notes). Speaking as objectively as I can, his expositions are amodel of clarity, so Im glad someone had NEED HELP BADLY (sorry, maths not psych)Expires: 28 daysHenriWilson skrev i melding Now consider the case of a charge of mass m being accelerated in a field. Using a=F/m, and m=Mo.gamma, dv/dt=(F/c.Mo)[sqrt(c^2-v^2)], which is:... nonsense.You have screwed this up before, and I have shown>you the correct equation before.F = dp/dt = d/dt (m*gamma*v)>gamma =1/sqrt(1 - v^2/c^2)dv/dt = (1/((v^2/c^2)*gamma^3 + gamma))*F/m((v^2/c^2)*gamma^3 + gamma)*dv = (F/m)*dtgamma*v = (F/m)*t + CYes OK.>if the charge is accelerated from standstill, v = 0 when t = 0, C = 0>v = c*t/sqrt(t^2+T^2) where T = m*c/F>v approaches c asymptotically(If you assume F is constant). I still cannot see the philosophical reason for using F=dp/dt rather thanF=m.dv/dt where m=f(v)> dv/[sqrt(c^2-v^2)]=kdt. Integrating: arcsin(v/c)=kt+B or v=c.sin(kt+B) if t=0 then v=0, so B=0 So we have v=c.sin(kt), where k is definitely finite. How come? Even in its exponential form this doesnt make any sense. Does it mean the SR equation is wrong or that matter becomes anti matter as c is exceeded?It means that you are unable to do the math correctly.The math is correct.Its just that the you a simpler approach before.>If we assume the charge is accelerated from standstill>in a constant electric field excerting the force F, we get:>p(t) = F*t = m*gamma*v>(the derivation isnt necessary since we integrate it back in the next step!)Ill think about it.>Henri Wilson. See the Stupidity of Criteria for Complete Metrizability?A Tychonov (or T_3.5, or completely regular) space is topologically completeiff it is a G_delta (countable intersectionof open sets) in its Cech-Stone compactification (which exists because ofcomplete regularity).A topologically complete metrisable space is completeley metrisable.For proofs, see eg. Engelking, General Topology.Henno Brandsma--> Could anyone point me towards some theorems which give criteria for> topological completeness, HELP BADLY (sorry, maths not psych)Expires: 28 days> as it enters a static electric field.Parse the bloody sentence in quotes, Henry. It doesnt say that. Isnt>English your first language? - Randy, grabbing his popcorn and going back to watch the>entertainment>Moron. Stop kissing Andsernons arse.Youre irritating me with your inability to read English.I finally reached threshold.Field already existed.Different from: Distant source of field changes.Situation 1: Field already present. Sources not changing. Field not>changing.>Situation 2: Field changing. Sources changing. Different.Static. Dynamic. Close. Far.Moron. - Randy - RandyRandy, you are so ignorant, you dont even realise that the introduction of anelectron (actually lots) will affect the field.Henri Wilson. See the Stupidity of cleverly. Dont believe me?> People in this NG will not answer clearly the question I have posed.> Will point B move along Y axis in XY plane? It needs just yes/no.> But they will remain silent(or they will be humorous). They will> ignore me. Because they are controlled.No, Abhi, you have been answered. Repeatedly. Read my reply from a couple of> weeks back. Asking whether the point B moves is a little meaningless given> your defined frame of reference. But there will be a net force acting on> point B towards point D, exactly the same as the net force acting on point D> towards point B. So the rod BD is subject to a compressive force.KrillI am really finding myself in con§agration. Things are still underHis absolute command including your mind.I went to Indian Institute of Technology and started expalaining verybasic idea of this device. But they raised problem which does notexist at all. What I tried to convince them is given in my homepage.Please note that I have changed my homepage from previous one.http://www.geocities.com/actiondeviceWhat those few of very brilliant people in India insisted that point Bwill move along Y axis. I agree that due to forces acting at pointB, third vector will be produced and point B will move in space inperpendiculer direction to XY plane. But those people insist that itwill move along Y axis !Now I posed the same problem before you people and what you said that,Asking whether the point B moves is a little meaningless given yourdefined frame of reference. This is definitely not answer of myquestion.(1) Those people in IIT said that point B will move along Y axis.(2) You said that it is meaningless in given frame of reference.What exactly is going on around us, Krill? Perhaps you dont know, but now I know that HE has absolute commandover everything in this universe including conscious mind of humanbeing. You dont know, but HE knows very well magnitude of thisdevice. It is going to change course of history, physics and it isgoing to open gateway to whole universe.This device is very simple to build but HE knows very well whichthings in my personal life can crash me and He has done it in mosthorrible, brutal way with perfect timing leaving me struggling withmyself.People in this NG will not answer my question clearly. I know WHY?He is in absolute to generate unidirectional force.http://www.geocities.com/actiondevice> Abhi,> What is a solid angle?> Can you give an example of a solid angle?> Solid angle means angle which does not change due to forces acting atpoint B in this V-shaped spring ABC.Wall of your room makes an angle of 90 degree with roof of your room.This angle does not change if you push the wall by your hand. This isexample of solid angle.How is weather out there ProblemAssuming this is true, will some kind soul prove it for me?Let A be a nonempty set under a total ordering R and let D be adenumerable subset of A. Then there is a one-one correspondence SbetweenD and the (set of) natural numbers such that for any two (distinct)elements x and y of D, S(x) < S(y) iff xRy. (Where < signifies theusual less-than relation on the natural numbers.)If perchance it is false, a disproof will be this is true, will some kind soul prove it for me?Let A be a nonempty set under a total ordering R and let D be a>denumerable subset of A. Then there is a one-one correspondence S>between>D and the (set of) natural numbers such that for any two (distinct)>elements x and y of D, S(x) < S(y) iff xRy. (Where < signifies the>usual less-than relation on the natural numbers.)If perchance it is false, a disproof will be appreciated.>As stated it is false: there is no order-preserving bijection between the set of integers and the setof natural numbers.KP-- E-MAIL: K.P.Hart@EWI.TUDelft.NL PAPER: Faculty EWIPHONE: +31-15-2784572 TU DelftFAX: +31-15-2786178 Postbus 5031URL: http://aw.twi.tudelft.nl/~hart n=1..infinity) < infinity ?The sequence tan(n)/n does not converge to 0; I imagine this is true, since Israel says hes proved it.>But its not all that obvious (its not clear to me whether>you were meaning to say it was obvious or not...)Its obvious if youre willing to accept that infinitely manycontinued fraction convergents to 2/pi have an odd numerator...Im not sure you need to appeal to the reference Robert Israel nntp.itservices.ubc.ca>It seems to me that a moments re§ection on the recurrence relation forp_{n+1}/q_{n+1} shows that the density of odd numerators in the sequenceof convergents is at least 1/2 (since you can never have two consecutiveeven numerators), and of course the same is true for denominators. Ordid I miss something glaringly obvious? -- === ErickSubject: Re: Bottom line on prime counting you may have noticed frenetic activity from posters trying toconvince you that theres nothing sinister about mathematicians doingtheir best to downply my find of a way to count prime numbers byintegrating a partial difference equation, but whats the bottom line?Does what I found work or not?Now that is an interesting question, isnt it! > pssst, hey Harris... your stuff doesnt work.... but dont tell anybody...It works, it just isnt what he claims it is: new or interesting. Hes just thinking about it differently.-- Will polygons of 5 corners which havethe maximum number or parallel/perpendicular different directionsamong lines drawn from their corners are:1) An irregular pentagon based in the square and the intersection ofthe parallels to the diagonals drawn from 2 adjacent corners.2) An irregular pentagon based in a 1-sqrt(3) parallelogram with alsothe parallel to the diagonals marking the 5th point.As an example, the polygon with maximum number of DIFERENT paralleldirections would be a regular pentagon, with 5 different pairs think youre referring to /infinity > | ----- |> 1 1/2 | (4 n)! (1103 + 26390 n)|> ---- = 2/9801 2 | ) -----------------------|> Pi | / 4 (4 n) (4 n) |> | ----- (n!) 4 99 |> n = 0 />I must have done something wrong with Google; found it easily enough on asecond attempt:http://elephant.linux.net.cn/pi.en.html(9801 = 99^2 and 396 = 4*99)It looks like a formula from another planet. And each summand of the seriesadds EIGHT significant digits. If sin(n)/n , n=1..infinity) < infinity ?> I imagine this is true, since Israel says hes proved it.> But its not all that obvious (its not clear to me whether> you were meaning to say it was obvious or not...)No, it is not obvious, at least not to me. I was in a hurry when Iwould post one today. Now, I wont have to, since what I had in mindwas about the same thing that not know what>convergence means in the above infinite series. Basically it means that (1 + a + a^2 + a^3 + ...) has a finite answer.Example: a = 0.5 means that the outcome is 2.>I do know that>infinity has the property of both containing each particular number>that it comprizes as well as having no end of such numbers, and thats>it.If 2 + 3 + 4 + . . . ad infinitum is subtracted from 1 + 2 + 3 + . . . ad infinitum, then it seems obvious that you will be subtracting the equally>infinite tails of two infinite series at point 2 on the number line,>and thus leave the non-infinite 1 that makes the only difference>between the two series, i.e., the non-infinite difference of 1 in each>series starting points on the number line.As obvious as this might seem to you, it is also wrong. Think of thefollowing ïnumber:a = 1 + 2 + 3 + ...Now multiply this by two so it becomes:b = 2 + 4 + 6 + ...Side by side for clarity:a = 1 + 2 + 3 + 4 + 5 + 6 + 7 + ...b = 2 + 4 + 6 + ... You might think that a and b contain an equal number of ïelements(and infinitely many of them). However, you can also see that allelements in b are also in a, yet a has elements that b doesnt (alluneven numbers). So a appears to have more elements than b, but thetwo must also have the same number of elements. This is obviouslynonsense... the reason is that more, equal or less simply doesnt workwell with infinite. If you have an infinite number of elements and youadd one element, you will still have an infinite number of elements.In math:infinity + 1 = infinitybut also:infinity + 2 = infinityThis obviously leads to problems as:Infinity - infinity = ?It can be anything (from the above alone it can either be 1 or 2).Infinity simply doesnt abide by ïnormal rules.In the proof you gave you did this (or something similar):(a + a^2 + a^3 + ...) - (a^2 + a^3 + ...) = aHowever, for a in outside the range <-1, 1> this boils down toinfinity - infinity and the result of that operation is unknown (soyou cant say it is a). For an a in the range <-1,1>, this will resultto {some number}-{some other number). This operation has a result andtherefore your proof is valid for this range of having difficulties solving these two limits.> I mustnt use Lhospital rule:> a) lim(x*(2^(1/x))-x) where x increases to infinite.> b) lim(cosh(x)-1)/(x^2) where x approaches 0.Yesterday i have send the solution of the first limit!!! but i haventsee it in the group!!Now i post you the solutions of 1/x=u then u-->0 and lim((2^u)-1/u)/u =lim(u*2^u-1)/(u^2) u-->0 === => lim-->-inf because u*2^u-->0 and-1/(u^2)-->-infSubject: Re: so-called associative property of the addition> a + ( b + c ) = ( a + b ) + c> allows us to write both sides of the equality as> a + b + c.> This property is *not* valid for so-[badly]-called> infinite sums. You cannot write something like> a1 + ( a2 + a3 + ... ) = (a1 + a2) + a3 + ...> In fact, the thing> a1 + a2 + a3 + ....> is not even a sum to begin with!> It is a so-called limit of a series of partial sums:> s1 = a1> s2 = a1 + a2> ...> sn = a1 + a2 + ... + an> If this series has a limit for n -> infinity, then one is> allowed to use the abbreviation> limit(sn; n -> infinity) = a1 + a2 + a3 + ...> The property of having a limit in the previous sentence> is something that can be verified by other means. Okay, now please give an example of where s1 = a1 and s2 = a1 + a2 but> (a1 + a2) - a2 DOES NOT = a1.It is *always* true that (a1 + a2) - a2 = a1These are finite sums and they have the associativeproperty.Infinite sums do not have this property, so writing a1 + ( a2 + a3 + ... ) = (a1 + a2) + a3 + ...is nonsense.O.t.o.h writing a1 + ( a2 + a3 + ... + an ) = (a1 + a2) + a3 + ... + anis perfectly okay since these are finite sums.But see below... Although the associative property of addition is important in the end> of my proof, the most important property is the distributive property> of multiplication over addition where the common factor of an infinite> series (a + a^2 + a^3 . . . ad infinitum) gets distributed and> multiplied over 1-a and leaves two equal infinities with one starting> at a and the other at a^2, but where the infinite portions of both> infinities can be exactly leaving the non-infinite point of a on the number> line.The distributive properties (b+c)*(a1+a2) = b*(a1+a2) + c*(a1+a2) b*(a1+a2) = b*a1 + b*a2are always valid.For infinite sums however one cannot write: b*(a1+a2+...) = b*a1 + b*a2 + ... (b+c)*(a1+a2+...) = b*(a1+a2+...) + c*(a1+a2+...)simply because the thing a1 + a2 + ...is not a sum to begin with. It is a limit of a series.If and only if the limit exists and is a real number, there isno problem because then you can use properties of realnumbers.Now, suppose that limit( sn, n -> infinity )exists and is real, then it can be abbreviated to S,or to the symbolic (!) form: a1 + a2 + ...but it *is* a real number, so then one can write (b+c)*S = b*S + c*Sor, intruducing the symbolic (!) notation S = a1 + a2 + ...one can write symbolically (!): (b+c)*(a1+a2+...) = b*(a1+a2+...) + c*(a1+a2+...).In the same circumstances one can also prove that limit( b*sn, n -> infinity )exists and is real and thus can be symbolically (!) written as b*a1 + b*a2 + ...One can also prove that this limit has the value b*limit( sn, n -> infinity )which is therefore equal to b*Sor symbolically (!) to b*(a1+a2+...)so that you finally can write, symbolically (!): b*(a1+a2+...) = b*a1 + b*a2 + ... However, Im listening, so please show my WHY the associative> process of addition no longer works with an infinite sum series. If your> proof depends on the seemingly illogical form of various outcomes with> some substitutions, then I invite you to explore those forms deeper> rather than risk a circular argument that my math is wrong because it> gives wrong outcomes.Taken together:One is only allowed to write b*(a1+a2+...) = b*a1 + b*a2 + ... (b+c)*(a1+a2+...) = b*(a1+a2+...) + c*(a1+a2+...). a1 + ( a2 + a3 + ... ) = (a1 + a2) + a3 + ...if one knows *in front* that the symbolic quantity (a1+a2+...)exists and has a real (finite) value.So using these ïproperties in a proof to show thatit exists is not allowed.N.O.Way has given the correct proof that the limitonly exists when -1 < a < 1. In his proof he did useassociativity and distributivity of finite sums only.Your math has given an example of what happenswhen you apply these ïproperties where you arenot allowed to use HALF A LIGHTBULB?>Although I havent studied the logic in detail, Ive read that when an>irresistable force meets and unmovable object, what happens logically>is EVERYTHING. That does not logically follow. The question ïWhat happens when anirresistable force (a force that can move any object) meets anunmovable object? can simply not be answered. The only thing thatlogically follows from the concepts of an unmovable object and anirresistable force is that they cannot both exist.Premisse 1: A force that can move all objects is an irresistableforce.Premisse 2: There exists an object that cannot be moved.conclusion: No force can move all objects, so no force can be anirresistable force.OrPremisse 1: There exists a force that can move all objects.Premisse 2: An object that cant be moved by any force is an unmovableobject.Conclusion: No object can be an unmovable object.Since logically both an irresistable force and an unmovable objectcannot exist, any argument that starts out with stating that they doboth exist leads to an unsound i first looked at integral[cos(ax)/(b^2+x^2)], limits -infinityto +infinity, i thought i could solve it by some substitution. Butafter trying all my wits on that, i decided to consult an table ofintegrals and found the result as ci(ab)sin(ab) + si(ab)cos(ab)where ci(x) is integral[cos(t)/t] and si(x) is integral[sin(t)/t].Well... that made me curious and i started looking up ways to solveintegrals in general. I had by now collected a lot of terms, like,local and global analysis, perturbation theory, Liovelle expansionsand so on. I wnt to know, if these are the ways in which the integralsthat cannot be normally solved are being integrated and put into thosetables.And something interesting popped up. I came across something calledRiccatti equation(i am not sure if i spelled it right) and it hasapplication in signal processing. i dont know what kind ofapplications...can someone enlighten me on that too.Also, please do suggest me some readable books in which i can learnabout the local and global analysis and kind of Apocalypse NOW! I went to Indian Institute of Technology and started expalaining very> basic idea of this device. But they raised problem which does not> exist at all. What I tried to convince them is given in my homepage.> Please note that I have changed my homepage from previous one. http://www.geocities.com/actiondevice What those few of very brilliant people in India insisted that point B> will move along Y axis. I agree that due to forces acting at point> B, third vector will be produced and point B will move in space in> perpendiculer direction to XY plane. But those people insist that it> will move along Y axis !http://www.dishman.me.uk/George/Abhi/abhi.gifIf you stretch the springs by moving A to E andC to F, the forces exerted by the springs are shownby the thick red arrows. To add the forces, the thinred lines make them into a parallelogram and the greenarrow shows the resulting total. Dont take my wordfor it though, check how to add forces in any mechanicstext book.To stop B moving, Newtons law says you need an equaland opposite reaction which is shown by the blue arrow.Without that, B will move in the direction of the greenarrow. That is along the Y axis so they are rightaccording to the standard rules of mechanics.> People in this NG will not answer my question clearly.They all said B will move down the Y axis. That is avery clear answer.Think of the line EF as a bow and the springs as thestring. Which way of errorSuppose that g(x) is proposed as an approximation of f(x) on [a,b].> What are the most popular ways of measuring how well g approximates f> over that interval?> [...]If f and g are probability densities then theres the asymmetricKullback-Leibler divergence D(f,g) = int f(x) cos(ax)/(b^2+x^2)> hi, when i first looked at integral[cos(ax)/(b^2+x^2)], limits -infinity> to +infinity, i thought i could solve it by some substitution. But> after trying all my wits on that, i decided to consult an table of> integrals and found the result as ci(ab)sin(ab) + si(ab)cos(ab)> where ci(x) is integral[cos(t)/t] and si(x) is integral[sin(t)/t].> Well... that made me curious and i started looking up ways to solve> integrals in general. I had by now collected a lot of terms, like,> local and global analysis, perturbation theory, Liovelle expansions> and so on. I wnt to know, if these are the ways in which the integrals> that cannot be normally solved are being integrated and put into those> tables.Just a quick thought: You can view the result as a function of a and b, anddo some integral transformations. I mean, consider e.g. the Laplacetransform or the Fourier transform with respect to a and/or b. Thesetransforms are expressed as integrals, and the trick is then to interchangethe two integrals. I cant work out the details, but its a trick Ive seenmore than done elementary Geometry Laura? Take a look at my homepage.http://www.geocities.com/actiondevice> What is on that page is silly. Point B is pulled downward by> any downward force applied to it unless support is provided to> hold it up. You provide no support, so point B is pulled down.When I say minds, thinking ability, intelligence of people around theworld is being controlled, this is the reason.I am not applying any downward force. The force is applied to point Aalong the direction of line BE which makes 60 degree angle with X axisand the force is applied to point C along the direction of line BFwhich also makes 60 degree angle with X axis.Again I am talking about V-shaped SPRING which can Re: Approximating Pi by Rationals>I feel that this is probably a well-known subject for number theorists,>but Ive never read anything about it. The question is how closely can>we approximate pi by rationals. I would think as close as you wanted to, because:Let a, b, c and d be positive integers.0 < (a/b) < pi0 < pi - (a/b)e = pi - (a/b), so e>0. For all e there exists a ïc and a ïd so: 0 < (c/d) < eSo:a/b + c/d < pi(a*d)/(b*d) + (c*b)/(b*d) < pi(a*d + c*b)/(b*d) < pi(a*d + c*b)/(b*d) is a rational that is closer to pi than mathematics.> Isnt arithmetic a part of mathematics?> Hardly, no (strange as that may sound). It does sound strange. And also unbelievable. Especially since> Arithmetic is taught as a part of Mathematics in school.The only mathematics in arithmetic is when you ask yourself thingslike why does this method work, is there a quicker one, etc.Applying the methods is then no longer maths, it is arithmetic.Just like you can use mathematical thoughts in a new piece ofmusic. Then, after composing, playing the music is not math, is it?And of course, the anglosaxon world uses ïmathematics as asynonym of arithmetics, adding to the confusion.> That may also explain> why the math teaching world is not very fond of tricks like this.> It is not a trick, it is a sound method for multiplication.> Yes. But ive seen a site about Vedic maths showing some other things. We are talking about multiplication here, and nothing else.But it is used as pr for ïthe rest of Vedic maths...> They really are tricks, giving a student no ïbasis or ïinsight at all. Certainly the multiplication method gives us a terrific insight into> the terrific advantages of place value. > Nice tricks, mind you. But tricks. Fine. All of mathematics is art, or a bag of tricks. All> civilisation comes from trickery with nature, and its subsequent> manipulation. Those who do it best, are the winners. You need to> have great basis and insight to do any trick properly.But math is more; it is tricks + insight why.What is vital in math education is to give the children a basis,something to fallback on, when it gets less directly intuitive.> It gives> a very thorough insight to the whole process, once properly> understood.> It is a welcome addition to arithmetic education.> But when it comes to insight, the distributive law is much better. Explain the distributive law, and show us how it is much better! With> respect to what?With respect to insight. The definition of the number sequence,the definition (=the meaning) of addition and multiplication,*those* are the mathematical aspects of arithmetic.The rest is ttrivial calculus. A computer can do it.> Children are likely to like this method, and teachers> will find it useful to teach.> Yes, i also definitely think so. We agree on one thing. Good. > Mathematics teachers are desperately fighting the general prejudice> that mathematics is ïjust arithmetic.> Well, that is because they are not doing the arithmetic right. Once> that is done right, following proper understanding of what numbers> really are and what they really mean, then things will be better for> all concerned.> Here i disagree. Worshipping vedic maths as a potential redemption for> maths education would be fatal. Not for most Indians and all non-bigots, I hope. It is not a question> of worshipping anything; it is to use some wonderful methods to do our> arithmetic better, and thus improve the entire basis of mathematics.Doing arithmetic better will not help mathematics a single bit.Arithmetic is *not* the basis of math. How about formula for where I can measure the speed of a car when> it hits a still object based on cars weight, objects weight, and total> distance the object was thrown. I realize that there are many other factors> such surface friction, in this case road, but im just looking for an> estimate, not to be as exact as posible.Relevant variables: Cars mass m1 Objects mass m2 Gravitiational acceleration g Total distance thrown: d Cars speed: vBy dimensional analysis,[m1] = [m2] = M; [g] = LT^-2; [d] = L; [v] = LT^-1Looks like v = (d/g)^(1/2) * f(m2/m1) for some function f.-- P.A.C. SmithThe vast majority of Iraqis want to live in a peaceful, free world.And we will cleverly. Dont believe me? People in this NG will not answer clearly the question I have> posed. Will point B move along Y axis in XY plane? It needs> just yes/no. But they will remain silent(or they will be> humorous). They will ignore me. Because they are controlled.> Most people ignore you because they have no interest in you.I said that I am trapped. My every action, word, thought wascontrolled in this NG or outside world. Now circumstances, through mypast postings, are created in this NG in such a way that even if I amtalking truth, they will not take interest in me. This is just one ofthe part of trap.> Other people use humor with you because they believe you are> mentally ill. You need to try to show them that you are not.Not me. It is their job to show that their intelligence, thinkingability, conscious minds are not being controlled. They just need toanswer simple question. Whether point B will move along Y axis ornot?They can refer figure on my homepagehttp://www.geocities.com/actiondeviceNow I am going to explain again what I am trying to say.In above figure AB and CB are V-shaped spring of same length andstiffness and both springs are in relaxed state initially. Angle ABCis solid angle . Let angle ABC be 60 degree in this figure(for thesake of explanation only, in actual Action Device this angle will bevery small). At t = 0, we apply same magnitude of force on point A and C and wepull point A of spring AB towards point E in the direction of line BEwhich makes 60 degree angle with X axis and we pull point C ofspring CB towards point F in the direction of line BF which also makes60 degree angle with X axis. And we are pulling point A and C insuch a way that these points must stretch or extend to point E and Fresp. on x axis.Please note that we are pulling points A and C in the direction whichmakes 60 degree angle with X axis. We are NOT pulling point A and C indownward direction.At t = t, spring AB is stretched and point A of spring AB reaches topoint E on X axis. Also at t = t, spring CB is stretched and point Cof spring reaches to point F on X axis.We will find that point B has not shifted its position along Y axis.It remains where it was at t = 0. Because if point B shifts itsposition along Y axis, to say, point B, angle ABC (or EBF, becausepoint A coincides with point E and point C coincides with point F)will be different from angle ABC i.e. greater than 60 degree. But asstated above, angle ABC is solid angle which does not change due toforces acting at point B.Yes, point B will shift its position in space but certainly not alongY axis or in XY plane. Third vector will be produced at point Bdirection of which will be perpendiculer to XY plane. Am I right or formula that looks not very sympathetic and I would like toprove that it can never be an integer... The main problem being thatit includes some powers of 2 and 3. I found no way to do this, so Imasking for your help.Heres the monster :24 * [ 24 * ^(2p+n) - 3^(n+1) * (6 * 3^p + 4^p) ] / [27 * 3^(p+n) - 16* 2^(2p+n)]where p and n are integers greater than psych) You say that the force on a charge due to an electric field acts> instantaneously. Correct?>Why do you ask what I am saying, when what I amsaying is quoted right above?>I am saying: as it enters a static electric field. So there is also an opposite force acting on the electrodes. even if the electrodes are light years apart. IS THAT WHAT YOU ARE SAYING?>I am saying:> as it enters a static electric field.>We have two electrodes - say 1 km apart.>(Or a light year apart - if you insist)>The potential difference is 1 million volts.>(Or a zillion volts - if the distance is a light year)>There is a small hole in the negative electrode.>We inject an electron through this hole.>When will a force act on the electron?>I am still saying:> as it enters a static electric field.>But what are YOU saying?>Not untill the electrode 1 km away feels the opposing force?> IS THAT WHAT YOU ARE SAYING? NO PAUL. I am saying that your claim infers that the effect of an injected> electron will be felt INSTANTLY at the far electrode.I have understood that you think my answer is wrong.I am asking you: WHAT IS YOUR ANSWER?Since you claim that the force cannot act instantly,I am asking you, WHEN will the force act?You claim that the distance to the other electrode is relevant.WHY is it relevant?How does this distance affect the delay before the forceon the electron starts acting?This is a concrete scenario.Please state what you think will happen.> Of course, since the> electron existed BEFORE it was injected, the effect would have already been> there even though the near electrode was in the way. The only way this> experiment can even be hypothesized is by either ïannihilating a very large> number of electrons or by monitoring the force on the far electrode with> movement of the electron mass towards it.Say, what the hell are you babbling about?Dont tell fairytales about suddenly disappearing charges!ADDRESS THE SCENARIO GIVEN!It is a concrete scenario which in principle could be made.(And which EASILY could and IS made with shorter distancebetween the electrodes. You have it in your TV set!)Let it be a hot wire in the hole in the electrode.Thermionic emission of electrons.When one of these electrons gets out of the holeand into the static field between the electrodes,how long time will it take before a force start acting on it?My answer is: as it Re: DO MATHEMATICIANS READ WITH HALF A LIGHTBULB?Very nice, for hundreds of years we have erroneously assumed the> geometric series converges only for |x|<1, but you have proven> otherwise.Although I still do not know what converges refers to, the form of> your objection is argumentum ad populum. Let us remember that for> many centuries, the world was most assuredly §at.Very Respectfully,> RayConvergence is how infinite series are formalised so that we candispense with the very confusing infinity associated with them.Infinity has confused philosophers and mathematicians alike throughouthistory, and so it was necessary to replace the notion with somethingwhich is very definite and can be reasoned about - convergence.As Guass put it:Infinity is only a figure of speech, meaning a limit to which certainratios may approach as closely as desired, when others are permittedto increase indefinitely.It means that when you say the sum of an infinite series you arereally just referring to a limit, if it exists, of a sequence offinite partial sums. Theres nothing illogical about this.Also bear in mind that your proof makes implicit use of several lawsof the real numbers (though applying them where they are notguaranteed to apply) which you then ignore by redefining division andthen claiming that divisions by zero exist. If you are going toreinvent mathematics to cope with whatever you want infinite sums tobe, you are going to need to go back to such foundations and work fromthere. One of your problems is getting round the contradiction statedthatinfinity * 0 = 5infinity * 0 = 6therefore 5 = 6. Youre not getting round this one unless you redefineequality and then the whole of mathematics. Personally, Im happy nothaving to think of infinity as a real number, and Im happy not havingto think of infinite sums as literally being the sum of an infinitenumber of hi, when i first looked at integral[cos(ax)/(b^2+x^2)], limits -infinity to +infinity, i thought i could solve it by some substitution. But after trying all my wits on that, i decided to consult an table of integrals and found the result as ci(ab)sin(ab) + si(ab)cos(ab) where ci(x) is integral[cos(t)/t] and si(x) is integral[sin(t)/t]. Well... that made me curious and i started looking up ways to solve integrals in general. I had by now collected a lot of terms, like, local and global analysis, perturbation theory, Liovelle expansions and so on. I wnt to know, if these are the ways in which the integrals that cannot be normally solved are being integrated and put into those tables.Just a quick thought: You can view the result as a function of a and b, and> do some integral transformations. I mean, consider e.g. the Laplace> transform or the Fourier transform with respect to a and/or b. These> transforms are expressed as integrals, and the trick is then to interchange> the two integrals. I cant work out the details, but its a trick Ive seen> more than once.Or you can observe that cos(ax)/(b^2+x^2) is the real part ofexp(iax)/(b^2 + x^2) and then to use <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h}YL`!h_XXr5Q>_nGsY2_> I feel that this is probably a well-known subject for number theorists,> but Ive never read anything about it. The question is how closely can> we approximate pi by rationals. More specifically:For integer n>0, let f(n) be the largest integer m such m/n < pi.> Let d(n) = pi - f(n)/n.Then d(n) measures how accurately we can approximate pi by a rational> with denominator n.How small can d(n) be? Clearly, d(n) < 1/n. But can we make d(n) much> smaller than that? Q1: Can we find arbitrarily large values of n such that d(n) < 1/n^2? Q2: Can we find arbitrarily large values of n such that d(n) < 1/n^3? Q3: In general, for each p>1, can we find arbitrarily large values of> n such that d(n) < 1/n^p?NO! Surprisingly, K. Mahler showed that d(n) > 1/n^(42). Thatexponent 42 has been improved subsequently. I guess the best currentresult is 8.0161, due to M. Hata.-- G. A. Edgar probability 2......> we cuts a two place into a wire at random.let length of wire is 1We got the wire of a three piece.when we made a triangle from three wire piece,find that probability of possibility.----------------------------------very difficult......um....help....me please.......Put the wire on a number line with left end-point at 0 and right > end-point at 1. Let c and d be the points at which the wire is> cut such that 0 < c < d < 1. Now use the fact that in a triangle> the sum of the lengths of any two sides must be greater than the> third. For example, we must have c + (d - c) > 1 - d. When> you get the appropriate inequalities between c and d, plot them> in the Cartesian plane with horizontal axes labelled c and d.> Then find the area of the solution region. This will be your> probability.hmm, this looks a lot like a homework/tutorial question (it certainlyappeared in examples sheets for the cambridge tripos over the last fewyears), and if so you really should think about conditionalprobability. someone offered a pictorial proof above involving asquare. there is one involving an equilateral triangle, whose side isthe length of the stick/piece of wire. divide the triangle into 4smaller equilateral triangles in the obvious way. pick a point atrandom in the triangle. draw the perpendiculars from each side tomeet at that point. the lengths of these segments add up to the length of the piece of wire again. the only points where by you canmake a traingle are those lying in the central triangle, and, asabove, the probability of picking quaternionists,It is a pleasure to observe that quaternions and 3D and 4D rotation continue to be of interest, at least for a group of afficionados.Some years ago I developed a PC-based multibody system for visualisation and animation of 4D wireframe figures. It runs in the Windows MSDOSbox as well under the bare MSDOS (version 5 or later).For me it was of great help to understand 4D geometry, algebraically as well as geometrically and kinematically.Unfortunately I have no FTP site, but whoever is interested in seeing 4D in action isinvited to request a copy, which will be sent at no cost as EMail attachments to a cover EMail letter. The usual freeware / sharewareconditions apply.Please feel free to visit my site at http://www.xs4all.nl/~plast/Jempub.htm for somematrix-oriented proof that leads directly to computer Fundamental Reason for High Achievements of JewsThe word ïSatan means ïaccuser in hebrew.> As can be seen by looking at the headers,> I did not make the post above.Technically, we do not know that.Virtually anyone can create a header.And this is a good thing. Afterall, you all murdered Jesus.Nearly everyone inwardly loves that which is evil, and hates thatwhich is good.There is nothing that offends more, than what is right and what isgood.Nontheless, Satan will vouch for you. You did not make the postabove.Obviously, some dishonest person> who cannot address the issues I raise,> is taking the dishonest approach to> promoting his agenda.As the best measure of a person, or group,> are the tactics they use,> I suggest that the poster re§ect on his actions,> use his real name, and address the issues raised,> rather than try to obscure someone elses messages> using dishonest, immoral tricks.In the long run, honesty is the best policy.Maybe, maybe not. The world is ruled by lies. People attack anddestroy the truth, and instinctively wish to destroy it. Many people,do not deserve the truth. They will un§inchingly seek to putobstacles in the way of anything that they find to be good. This isbecause they are jealous, and wish to steal that which they think isvirtuous and ïtrue for themselves, and wish to deny it for everyoneelse. Truth is only ïtruth when they benifit from it. No one trulybelieves in it, however.Most people have been trained from birth, to issue filth from theirmouths, and that is the only true form of enjoyment for them.Do not cast your pearls unto swine. Are all swine? Probably so,but once every now and then one comes across strange phenomenon, andone might probably even doubt that.The truth is a pearl, and at least most men, are processes X_t and sigma_t be defined as:dX_t = (mu+betasigma_t^2)dt + sigma_t dW_t + rho dZ_{lamda t}dsigma_t^2 = -lamda sigma_t^2 dt + dZ_{lamda t}with sigma_0^2 >0. ( mu, beta, lambda >0 and rho<0 are constants).W is a Brownian motion, Z an independent Levy process with no gaussiancomponent and positive increments (a subordinator).Assuming Z has no deterministic drift and the cumulant k(theta) = logE(e^{theta Z_1}), where it exists, takes the form:k(theta)=int_{x>=0}(e^{theta x}-1)w(x)dxwhere w(x) is the density of the Levy measure of Z_1. (There are some otherassumptions on Z but theyre just technical making sure w(x) is reasonableso that sigma^2 has a stationary distribution)The problem is to find the dynamics of S_t=e^{X_t}using Itos formula. Ihave an answer, which looks right but I cant see how it has been derived.The value I have is:dS_t=S_{t-}(b_t dt _ sigma_t dW_t + dM_t)where b_t is the process given byb_t=mu + lambda k(rho) + (beta + 1/2)sigma_t^2and M_t is the martingale Levy processM_t = Sigma_{0 prove that it can never be an integer... The main problem being that> it includes some powers of 2 and 3. I found no way to do this, so Im> asking for your help. Heres the monster : 24 * [ 24 * ^(2p+n) - 3^(n+1) * (6 * 3^p + 4^p) ] / [27 * 3^(p+n) - 16> * 2^(2p+n)]>Is this 24*[24*???^(2p+n)...> where HELP BADLY (sorry, maths not psych) as it enters a static electric field.>Parse the bloody sentence in quotes, Henry. It doesnt say that. Isnt>English your first language? - Randy, grabbing his popcorn and going back to watch the>entertainment>Moron. Stop kissing Andsernons arse.Youre irritating me with your inability to read English.>I finally reached threshold.>Field already existed.>Different from: Distant source of field changes.>Situation 1: Field already present. Sources not changing. Field not>changing.>Situation 2: Field changing. Sources changing. Different.>Static. Dynamic. Close. Far.>Moron.> - Randy> Randy, you are so ignorant, you dont even realise that the introduction of an> electron (actually lots) will affect the field.Amazing how much verbiage you can introduce without answering ïs simple question.How long after the electron enters the field willit be affected?Imagine two plates generating a static electricfield. An electron passes through a hole in the plates.It is moving at 100 m/sec. How long before it is actedon by the field? One second? One microsecond? One nanosecond?When does the velocity start changing from 100 m/sec? Howfar does the electron get before this happens? - Indian Institute of Technology and started expalaining very> basic idea of this device. But they raised problem which does not> exist at all. What I tried to convince them is given in my homepage.> Please note that I have changed my homepage from previous one.> http://www.geocities.com/actiondevice> What those few of very brilliant people in India insisted that point B> will move along Y axis. I agree that due to forces acting at point> B, third vector will be produced and point B will move in space in> perpendiculer direction to XY plane. But those people insist that it> will move along Y axis !http://www.dishman.me.uk/George/Abhi/abhi.gifIf you stretch the springs by moving A to E and> C to F, the forces exerted by the springs are shown> by the thick red arrows. To add the forces, the thin> red lines make them into a parallelogram and the green> arrow shows the resulting total. Dont take my word> for it though, check how to add forces in any mechanics> text book.To stop B moving, Newtons law says you need an equal> and opposite reaction which is shown by the blue arrow.> Without that, B will move in the direction of the green> arrow. That is along the Y axis so they are right> according to the standard rules of mechanics.People in this NG will not answer my question clearly.They all said B will move down the Y axis. That is a> very clear answer.Think of the line EF as a bow and the springs as the> string. Which way will the green arrow go?GeorgeRoger that.But my illusion will come to an end if few other people in this NGconfirm what you have said. I will repeat zillion times that in thisV-shaped spring, angle ABC is solid angle. Please read it === Subject: Re: Selecting the correct graph> I would like to know some important points when selecting a certain> graph for the data I have i.e. Using a Pie graph for displaying the> largest % of Annual sales, in a five-year period, for a business.OK. naturals Adjunct Assistant Professor at the University of Montana.Theres clearly something wrong somewhere in part of my argument,since I had concluded S(a+b+c) to be no more than 15, and PeterMontgomery exhibited an example with S(a+b+c)= 24...>Write x = a_0 + 10a_1 + 10^2a_2 + ... + 10^n a_n> y = b_0 + 10b_1 + 10^2b_2 + ... + 10^n b_n>with 0<= a_i,b_i < 10, and at least one of a_n,b_n nonzero. So S(x) = a_0 + a_1 + ... + a_n> S(y) = b_0 + b_1 + ... + b_nDefine d_0,....,d_n recursively as follows:d_0 = 0 if a_0+b_0 < 10>d_0 = 1 if a_b+b_0 > 9.d_{i+1} = 0 if a_{i+1} + b_{i+1} + d_i < 10>d_{i+1} = 1 if a_{i+1} + b_{i+1} + d_i > 9(The d_i are the carries)Then (x+y) + (a_0+b_0 - 10d_0) + 10(a_1+b_1+d_0-10d_1) + ... +>10^n(a_n+b_n+d_{n-1}-10d_n) + 10^{n+1}d_n. Then S(x+y) = a_0+b_0 + a_1 + b_1 + ... + a_n+b_n + > +(d_0+...+d_n)- 10(d_0+...+d_n)> = S(x) + S(y) - 9(d_0+...+d_n).>Now, you assume that S(a+b) <= 4> S(b+c) <= 4> S(a+c) <= 4>Say we take 2(a+b+c) = (a+b) + (b+c) + (a+c).Now, lets consider the carries involved: Each of a+b, a+c, b+c has at>most 4 nonzero digits, and each nonzero digit is at most 4. How many>carries can there be? For there to be carries when we add a+b, a+c,>and b+c together, there must be corresponding entries, at least one of>which is a 4, and the other 2 are either 4s or 3s. But that means>that there is at most 1 carry. That is:S(2(a+b+c)) = S(a+b) + S(a+c) + S(b+c) orS(2(a+b+c)) = S(a+b) + S(a+c) + S(b+c) - 9.I think this is right, although the best way to do it is to say thatin adding a+b and a+c there can be no carries, etc...>Now we want to relate S(2(a+b+c)) to S(a+b+c)Again, let x = a_0 + 10a_1 + 10^2a_2 + ... + 10^n a_n>with 0<= a_i <10, a_n>0So S(x) = a_0 +...+ a_n.Define e_0,....,e_n by lettinge_i = 0 if 0<=a_i <5>e_i = 1 if 4 - 10(e_0 + ... + e_n)> = 2(a_0+...+a_n) - 9 (e_0+...+e_n)> = 2*S(x) - 9 (e_0+...+e_n)Where e_0+...+e_n = number of digits larger than 4 in x.Let x = (a+b+c). S(a+b+c) = (1/2)[ S(2(a+b+c)) + 9(e_0+...+e_n)]This is:S(a+b+c) = (1/2)[ S(a+b)+S(a+c)+S(b+c) + 9(-1+e_0+...+e_n)]orS(a+b+c) = (1/2)[ S(a+b)+S(a+c)+S(b+c) + 9(e_0+...+e_n)]Now, S(a+b) + S(a+c) + S(b+c) <= 12.Since the digits of a+b, a+c, and b+c add up to 4 each, it is easy to>verify that e_0+...+e_n <=2 (that is, the sum (a+b)+(a+c)+(b+c) cannot>have more than 2 digits larger than 4). This is the mistake, of course. The example Peter gave hada=b=c=5050505, so a+b = a+c = b+c = 10101010. the e_i are the number of digits greater than 4 in a+b+c, not in(a+b)+(a+c)+(b+c), which is what I calculated. I think this can still work to show no such number exist; the obviousupper bound given by Peter of 60 is not tight enough, but perhaps abit more careful study will I accept as reality. --- Calvin (Calvin and simple problemTiago Paix.8bo grava .88 la saucisse et au marteau:> Does anyone has an ideia how can this be solved?> I have two differential equations:> X(t)=a-b*X(t)> Y(t)=c-d*Y(t)> I want to know when the solutions intersect. By doing it straight forward> (just X(t)=Y(t) ) I end up with the transcendental equation:(Y-X)(t) = c - a - (d-b)(Y-X)(t)(Y-X)(t) = K exp((b-d)t) + (a-c)So (Y-X)(t) = 0 <=> [ln((c-a)/k)]/[b-d]Therefore, it depends on the K which can be any real and must bedeterminer by any condition (have you some properties of graphs of 0/1-polytopes> To every polytope, we associate a graph in the following way: take its> vertices as nodes. The nodes are joined by an edge if and only if the> corresponding vertices are adjacent.How can we decide, given any graph, whether it is the graph of some> 0/1-polytope or not? > If there is no exact criterion known, is there a good sufficient one?Google for realizability of polytopes. Its a quite lively areaof research.A 0/1-polytope is a polytope where all its vertices have coordinates in {0,> 1}.Hmm, I fail to see why you need fixed coordinates in the firstplace. If you happen to find a realizable polytope for some givengraph, then you should be able to name the vertices with some0/1-coordinates in some suitable space. Maybe I am missing THE BOOKobviously has a bias towards content related to Erdos. And of course beauty is relative (some of the proofs I find ...er... inaccessible).But I cannot fault the authors at all; deep, readable, meaty, true to purpose (they put in beautiful -proofs-, not necessarily beautiful -truths-, though of course there is overlap).Now to my issue: of the sections in the book (number theory, geometry, analysis, combinatorics, graph theory), the authors seem to cover the range of modern mathematics fairly well, a bit lopsided, but still a good cover (probability is in a few of the proofs, even algebra (a finite division ring is a field), topology is really just a fancy word for geometry, right?).-But- there is one field which is entirely missing: Logic.Are there no beautiful proofs in the field of logic?Goedels first incompleteness theorem has a large part to it that is totally ugly algebra/number theory/encoding, but the gist of it is simple.What about the related halting problem or (almost identical) nondenumerability of reals?Sure there are lots of important theorems of logic, Im just trying to think of those that have Big Number Game> | A quote from: | The New Fowlers Modern English Usage, Third Edition, Edited by R. W. | Burchfield, The acknowledged authority on English usage | [all of that from the front of the dust jacket...] | | Under the topic billion: | It is best now to work on the assumption that the word means ïa | thousand millions in all English-speaking areas...I suggest that the word million, billion and trillion be abolished,>and usage adapted around SI prefixes.>Eg: Bill Gates is not worth billions of dollars; he prefixes also havemultiple meanings. The SI 10-based ones. And the 2-based ones.The attempt by the SI people to force gibibytes and other suchabortions on the computer crowd isnt going to succeed, BitTorrentnotwithstanding. -- Matthew T. Russotto mrussotto@speakeasy.netExtremism in defense of liberty is no vice, and moderation in pursuitof justice is no virtue. But extreme restriction of liberty in pursuit of a DO MATHEMATICIANS READ WITH HALF A LIGHTBULB?X-DMCA-Notifications: http://www.giganews.com/info/dmca.html Im listening ... so tell me in English why Im wrong.> Do you know what it means for a series to converge? Are you talking about a series when you write a + a^2 + a^3 + ... ? If not, know what>convergence means in the above infinite series. Then the person reading with half a lightbulb is _you_.>I do know that>infinity has the property of both containing each particular number>that it comprizes as well as having no end of such numbers, and thats>it.Uh, no, thats not it, thats meaningless nonsense.To say that a_1 + ... converges to s means this: For ever eps > 0there exists an integer N such that |s - (a_1 + ... + a_n)| < eps for every n > N.Thats the _definition_ - thats what mathematicians _mean_ byinfinite sums. If youre saying something about infinite sumswhich cannot be proved from that definition (which you are)then whatever youre talking about has nothing to do withthe infinite sums that mathematicians are talking about.(NOTE that theres no mention of infinity whatever in thedefinition! Hence no need to figure out whats meant bystatements like infinity contains each particular number,etc.)>If 2 + 3 + 4 + . . . ad infinitum is subtracted from 1 + 2 + 3 + . . . ad infinitum, then it seems obvious that you will be subtracting the equally>infinite tails of two infinite series at point 2 on the number line,>and thus leave the non-infinite 1 that makes the only difference>between the two series, i.e., the non-infinite difference of 1 in each>series starting points on the number line.Please explain how ïconvergence refutes that logic.There is no _logic_ in the previous paragraph! You saysomething _seems_ _obvious_. Saying something seems obviousdoes not prove that its true.If you want to _prove_ what you just said the proof willhave some reference to the definition above. And thatwill be impossible, because according to the definitionabove there is simply no such thing as 2 + 3 + ... grava .88 la saucisse et au marteau: Does anyone has an ideia how can this be solved? I have two differential equations:> X(t)=a-b*X(t) Y(t)=c-d*Y(t)> I want to know when the solutions intersect. By doing it straight forward (just X(t)=Y(t) ) I end up with the transcendental equation:(Y-X)(t) = c - a - (d-b)(Y-X)(t)> Im sorry, but I think you made a mistake (unless Im missing something).c-a-(d-b)(Y-X) = c - a - d Y + d X + b Y - b XWhich is not Y(t) - X(t)You have to complement that change of n=1..infinity) < infinity ?X-DMCA-Notifications: http://www.giganews.com/info/dmca.html>The sequence tan(n)/n does not converge to 0; I imagine this is true, since Israel says hes proved it.But its not all that obvious (its not clear to me whetheryou were meaning to say it was obvious or not...)Its obvious if youre willing to accept that infinitely many>continued fraction convergents to 2/pi have an odd numerator...Im not sure you need to appeal to the reference Robert Israel nntp.itservices.ubc.ca>It seems to me that a moments re§ection on the recurrence relation for>p_{n+1}/q_{n+1} shows that the density of odd numerators in the sequence>of convergents is at least 1/2 (since you can never have two consecutive>even numerators), and of course the same is true for denominators. Or>did I miss something glaringly obvious?I have no idea. Whats obvious depends on the reader; since I knownothing about the continued fraction expansion of 2/pi nothingabout it is obvious to me.> -- ErickDavid C. I went to Indian Institute of Technology and started expalaining very> basic idea of this device. But they raised problem which does not> exist at all. What I tried to convince them is given in my homepage.> Please note that I have changed my homepage from previous one.> http://www.geocities.com/actiondevice> What those few of very brilliant people in India insisted that point B> will move along Y axis. I agree that due to forces acting at point> B, third vector will be produced and point B will move in space in> perpendiculer direction to XY plane. But those people insist that it> will move along Y axis !http://www.dishman.me.uk/George/Abhi/abhi.gifIf you stretch the springs by moving A to E and> C to F, the forces exerted by the springs are shown> by the thick red arrows. To add the forces, the thin> red lines make them into a parallelogram and the green> arrow shows the resulting total. Dont take my word> for it though, check how to add forces in any mechanics> text book.To stop B moving, Newtons law says you need an equal> and opposite reaction which is shown by the blue arrow.> Without that, B will move in the direction of the green> arrow. That is along the Y axis so they are right> according to the standard rules of mechanics.People in this NG will not answer my question clearly.They all said B will move down the Y axis. That is a> very clear answer.Think of the line EF as a bow and the springs as the> string. Which way will the green arrow go?GeorgeIt is no more physics or Newtonian mechanics, Pythagoras theorem,George. For me, it is war between negative forces and positive forcesof nature. Logic, math, physics, intelligence, thinking ability ofwell educated, brilliant people like you is succumbed to dark, evilforces of nature. People on this earth are being controlled by darkforces of nature.What you fail to understand that if point B shifts its position alongY axis, angle ABC or EBF will change and as I am saying again andagain, angle ABC is solid angle.But I will not allow dark, evil forces of nature around me tooverpower me. I am talking about changing course of history, physicsand opening gateway to entire universe.Why you people are being An apparently simple problemTiago Paix.8bo grava .88 la saucisse et au marteau:> Im sorry, but I think you made a mistake (unless Im missing something).> c-a-(d-b)(Y-X) = c - a - d Y + d X + b Y - b XHuh? I just substracted the first line to the mathematics.> Isnt arithmetic a part of mathematics?> Hardly, no (strange as that may sound).> It does sound strange. And also unbelievable. Especially since> Arithmetic is taught as a part of Mathematics in school.The only mathematics in arithmetic is when you ask yourself things> like why does this method work, is there a quicker one, etc.> Applying the methods is then no longer maths, it is arithmetic.> Just like you can use mathematical thoughts in a new piece of> music. Then, after composing, playing the music is not math, is it?And of course, the anglosaxon world uses ïmathematics as a> synonym of arithmetics, adding to the confusion.We also had the word reckoning (with an typical variety of spellings over the centuries) for that purpose. Certainly the meaning of ïmathematics has become diluted by its applicationto simple acts of calculation.I believe that the English phrase reading, writing and reckoning predates the, more common nowadays, reading, writing and ïrithmetic. (Those are the so-called three ïRs.) (Alas, I cant justify that belief with any hard facts.)Phil-- Unpatched IE vulnerability: mhtml wecerr CAB §ipDescription: Delivery and installation of an executableReference: === Subject: Re: Vedic Mathematics --- Myth and RealityHerman Jurjus grava .88 la saucisse et au marteau:> It does sound strange. And also unbelievable. Especially since> Arithmetic is taught as a part of Mathematics in school.> The only mathematics in arithmetic is when you ask yourself things> like why does this method work, is there a quicker one, etc.> Applying the methods is then no longer maths, it is arithmetic.Huh? Just to know, why does include arithmetic for you? Are Galoisfields arithmetic? Are congruence rigns arithmetic? In that case, howcan you say that those arent maths? In the other case, why dont like Gates, those prefixes also have) multiple meanings. The SI 10-based ones. And the 2-based ones.And the mixed-based ones.A 3,5 §oppy disk holds 1,44 megabytes. (1,44 * 1000 * 1024 bytes)SaSW, Willem (at stack dot nl)-- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No Im not paranoid. You all think Im paranoid, dont you consider positive integers k and j with m and n digits> respectively. Assume moreover that neither k nor j is equal to 0. > Can we determine exactly the number of digits in the product kj? If you are permitted to take logarithms, its trivial.> Its> either m+n-1 or m+n, but which one it is changes depending on the> nature of k and j. I wanted to use this to determine the number of> digits in, say 2^64. Now, this one might not be so difficult because> many people know 2^32 off the top of their head so they know it has 10> digits. So they would know immediately that 2^64 either has 20 or 19,> but still its not clear (to me anyway) if its 20 or 19 without> multiplying it out. But still, what about 3^64? And what about the> number of digits of a^b where a and b are any positive integers?Youre permitted to calculate all the partial products largest first, and stop as soon as you know that the remaining terms can no longer cause the product to change the number of digits due to propagating carries.e.g. I know 2^16 is 65536. Therefore 2^32 = 65536*65535 = 6xxxxx * 6xxxxx = 3600000000 + y therefore 2^32 has 10 digits as it cant have 9 as y is positive.If I know that 2^32 is 4294967296, then 2^64 = 4294967296*4294967296 = 4xxxxxxxxx * 4xxxxxxxxx = 16000000000000000000 + ytherefore 2^64 has 20 digits as it cant have 19 as y is positive.> Secondly, is there an easy way to determine exactly the first r digits> of the product of an m-digit and n-digit number, without computing the> whole number? It seems like you can estimate the first r digits by> multiplying together some chunk of the beginning of the first number> with some other chunk of the beginning of the second number, and> looking at the first r digits of the result. But is there some> formula in terms of m and n that gives you exactly the size of the> chunk you should take so that the result is guaranteed to be exact? > Or do you have to multiply the whole numbers together?There are methods of performing multiplication that can help here.Namely what I learnt as the French Peasant, or Par Gelosia, method, which is also known as the Vedic method.Basically, if you lay the digits of the two values at the sides of a 2D grid, perform the calculations diagonal by diagonal.e.g. If multiplying 4-digit numbers abcd * efgh, and you only wantthe to 3 digits of the answer.Calculate: e f g h----------------+--a*e a*f a*g a*h | ab*e b*f b*g | bc*e c*f | cd*e | dThe 4th diagonal (d*e)+(c*f)+(b*g)+(a*h) might not even be needed, depending on whether the value from the first 3 diagonals can beaffected by a carry of 4.Phil-- Unpatched IE vulnerability: Click hijackingDescription: Pointing IE mouse events at non-IE/system windowsReference: http://safecenter.net/liudieyu/HijackClick/ HijackClick-Content.HTMExploit: http://safecenter.net/liudieyu/HijackClick/HijackClick2- learned a lot, not the least of which was thatsin(1) + sin(2) + ... + sin(n) is bounded independent of n.Now this thing about tan(n)/n raises another question, for which I will start a new thread.--Jim BuddenhagenTo reply copy jbuddenh@REMOVEtexas.net Halton ArpRather naively I thought that if you put out something simple enoughthat most people could understand it, then theyd be willing to atleast question authority long enough to see when authority figures getit wrong.Then once they got a better sense that some people, even with a title,will just say things for their own benefit, theyd want to know thetruth, and be willing to look outside the mainstream.So I have my math results, which I think are rather simple, and nowits a matter of presenting them to the world, right?But then theres Dr. Halton Arp.You see, Dr. Arp is a scientist, a world renowned scientist and he has*data*, real, hard astronomical data, which is more substantive indisproving the commonly taught Big Bang Theory, than the data used tosupport that theory.So it should be a no-brainer, the theory has to be changed to fit thedata.Thats not what has happened.Notice, for those who know of Dr. Arps personal theories that Ididnt mention them. I said *data*, as I dont agree with much of Dr.Arps cosmological theories.Whats not in doubt though is the fact that the data he has takes awaythe possiblity of the commonly taught Big Bang Theory with moresubstantive DATA than there is in support of that theory.James HarrisMy math discoveries, found for cleverly. Dont believe me? People in this NG will not answer clearly the question I have> posed. Will point B move along Y axis in XY plane? It needs> just yes/no. But they will remain silent(or they will be> humorous). They will ignore me. Because they are controlled.> Most people ignore you because they have no interest in you.I said that I am trapped. My every action, word, thought was> controlled in this NG or outside world.You are not trapped. You have free will. Nobody forcedyou to post these words. You are free to post or not topost, and you are free to post whatever words you want whenyou do post. You are also free to post to other newsgroups.> Now circumstances, through my> past postings, are created in this NG in such a way that even if I am> talking truth, they will not take interest in me.You are posting in a physics newsgroups. Your posts will beevaluated on the correctness of the physics content, not onyour personality. Your physics is incorrect, but you seemunwilling to accept that. You have gone to eminent Indianphysicists and been given a correct analysis. You didntlike the analysis.> Other people use humor with you because they believe you are> mentally ill. You need to try to show them that you are not.Not me. It is their job to show that their intelligence, thinking> ability, conscious minds are not being controlled. They just need to> answer simple question. Whether point B will move along Y axis or> not?Many people have answered this question: Point B willmove along the Y axis.They can refer figure on my homepagehttp://www.geocities.com/actiondeviceNow I am going to explain again what I am trying to say.In above figure AB and CB are V-shaped spring of same length and> stiffness and both springs are in relaxed state initially. Angle ABC> is solid angle.Apparently this means you have some rigid structureforcing this angle to be approximately constant.> Let angle ABC be 60 degree in this figure(for the> sake of explanation only, in actual Action Device this angle will be> very small).> At t = 0, we apply same magnitude of force on point A and C and we> pull point A of spring AB towards point E in the direction of line BE> which makes 60 degree angle with X axis and we pull point C of> spring CB towards point F in the direction of line BF which also makes> 60 degree angle with X axis. And we are pulling point A and C in> such a way that these points must stretch or extend to point E and F> resp. on x axis.Please note that we are pulling points A and C in the direction which> makes 60 degree angle with X axis. We are NOT pulling point A and C in> downward direction.But they have a downward component.Systems exactly like this have been used to two boatsalong canals (B is a boat, A and C are mules onshore, the whole point is to move the boat forward)or to plow land (B is a plow, A and C are oxen) forcenturies. They work. B moves forward, as it isdesigned to do.> At t = t, spring AB is stretched and point A of spring AB reaches to> point E on X axis. Also at t = t, spring CB is stretched and point C> of spring reaches to point F on X axis.We will find that point B has not shifted its position along Y axis.Of course it has. What is preventing it from shifting?Try it. Put yourself at point B and have two people pullyou with ropes at angles. You think you wont move?> It remains where it was at t = 0. Because if point B shifts its> position along Y axis, to say, point B, angle ABC (or EBF, because> point A coincides with point E and point C coincides with point F)> will be different from angle ABC i.e. greater than 60 degree. But as> stated above, angle ABC is solid angle which does not change due to> forces acting at point B.Oh, OK. I see what youre saying. ABC is a rigid structurewhich does not allow points A and C to separate horizontally.Nothing is completely rigid, so B will move a little bitas the structure is stretched and pulled tight. At thatpoint youll reach static equilibrium, so long as theforces are indeed confined to pull at angles but thestructure prevents the horizontal separation fromgrowing.Unfortunately, you put the forces on springs ratherthan confine them to the steel structure. So they canhappily move apart.> Yes, point B will shift its position in space but certainly not along> Y axis or in XY plane.If the forces are confined to the xy plane, so will the motion be.> Third vector will be produced at point B> direction of which will be perpendiculer to XY plane.> Am I right or Suppose that g(x) is proposed as an approximation of f(x) on [a,b]. What> are the most popular ways of measuring how well g approximates f over that> interval? Here are several measures. In each case, the smaller the measure is, the> better the approximation is considered to be. AInf. The maximum of |absolute error| over the interval,> where absolute error = g(x) - f(x).> RInf. The maximum of |relative error| over the interval,> where relative error = (absolute error)/f(x). A2. The root-mean-square of |absolute error| over the interval.> R2. The root-mean-square of |relative error| over the interval. A1. The average of |absolute error| over the interval.> R1. The average of |relative error| over the interval. All of these measures may be thought of as power means (also calledHoelder> means). They have the form * ( Integral( |error|^p ) / (b-a) ) ^ (1/p) where the integral is taken with respect to x from a to b, and error is> either absolute or relative. Obviously, in the cases of A1 and R1, p = 1,> and in the cases of A2 and R2, p = 2. The value of p is not so obvious,> however, in the cases of AInf and RInf. But in the limit as p increases> without bound, the power mean gives simply the maximum, as needed in AInf> and RInf. As such, for those cases, we may say that p = +oo.> Here are some questions of mine. Are there any important measures of error in form * which use values of p> other than 1, 2, and +oo? Are there any important measures of error which are not in form * ? Clearly, using p = 2 yields a measure which is intermediate between those> with p = 1 and p = +oo. In that sense, p =2 represents a nice compromise.> But is there anything really special about p = 2 (say, as opposed to p = 4> or p = 3/2) ? (Of course, I grant that the integral is typically fareasier> to evaluate analytically when p = 2 than when p = 4 or 3/2. But Im> wondering if p = 2 is special for a more fundamental reason than that.)>I dont know if youll consider this relevant or not.Sometimes errors of less than 0 are considered unimportant. This isusually when risk rather than error is being measured. In particular,in analyzing warranty data, lifetime in excess of the warranty period is notconsidered a risk that needs to be measured, but failures before thewarranty period expires are.Similarly, when actual financial results are worse than predicted, errorsare analyzed, but when they are better than predicted, that is taken asevidence of financial or managerial acumen. This means that efforts must bemade to minimize underperformance, but overperformance is just fine.I have also seen theoretical discussions that p other than 1, 2 or ooshould at least be considered (gives different weight to different sizeerrors), but I have not seen it done in practice. I have seen complaintsthat absolute error is more important than r.m.s. error, and why do all youmathematical types keep using this complicated r.m.s. error? (My usualanswer is that it makes the calculations doable. My response to theinevitable Huh? is Just take my word for arose (n=1,2,3...). The discussion showed that this sequence does not approach zero. Now |tan(11)/11| is about 20. So I wondered: does tan(n)/n (or |tan(n)/n|) take on arbitrarily large values? So far the largest value I have found is 556.3, but n is very large.I didnt see anything in the other thread that answered this. Does anyone know? Jim Buddenhagen------------To reply copy jbuddenh@REMOVEtexas.net grava .88 la saucisse et au marteau: Im sorry, but I think you made a mistake (unless Im missing something).> c-a-(d-b)(Y-X) = c - a - d Y + d X + b Y - b XHuh? I just substracted the first line to the LIGHTBULB? > The ïproof you did must be wrong somewhere as the equation doesnt> work with any value outside the range <-1,1>. I believe the error is> that the operations shown can only be applied to absolute convergent> series (ïabsoluut convergente reeksen in dutch). Since the series is> not convergent at all for a being outside <-1,1>, the whole proof is> nonsense.> Im sure someone else can explain this better... I posted virtually the same arguments at the www.johnpatrick.com> message board and recieved a similar response from The Truth.> ______________________________________________________________ _______Essential in your derivation is the step [(a + a^2 + a^3 . . .) -> (a^2 + a^3 + a^4 . . .)] = a.> But this equivalence only holds if the series a + a^2 + a^3 . . .> converges, and it only converges for certain a, not for any a.<< There is only one element of infinity that is not common to both> infinite series and that is a to the first power, and thus the> difference between the two infinite series is simply a to the first> power. If you will take the time to explain in English why the a Ive> isolated is not the certain a that you require, Ill listen, but> youve otherwise said nothing.> -------------------------------------------------------------- ---------- Im listening ... so tell me in English why Im wrong.>Better yet.I dont even care what multiplication and division mean. I do know whataddition means.Since 1 + 2 + 4 + 8 + . . . = -1, I propose that I give you $1 today. (The - means that it goes from me to you.) In exchange, you will give me$1 today, $2 tomorrow, $4 the following day and so on. (The lack of a -[sometimes called a +] means the money goes from you to me.) Since they areequal, you will surely agree to 2^30 < 1 + 2 + 4 + . . . (going onforever), you can stop after just a month. Then you must be making a profiton the deal, so n=1..infinity) < infinity ?[...]> So we get *******************************************> * *> * sum_n cos(n t)/n = -ln (2 sin(t/2)) *> * *> * sum_n sin(n t)/n = (pi - t)/2. *> * *> ******************************************* In particular, taking t = 1, we get sum_n sin(n)/n = (pi -1)/2 approx 1.0707963267948966192, which agrees with the answer Mathematica returns. (But Mathematica 4.2> insists that sum cos(n)/n diverges. Its wrong.)It might be noted, however, that the current version of Mathematica doesnot make that mistake:In[1]:= FullSimplify[Sum[Sin[n]/n, {n, 1, Infinity}]]Out[1]= (-1 + Pi)/2In[2]:= FullSimplify[Sum[Cos[n]/n, {n, 1, Infinity}]]Out[2]= -Log[2 - an end if few other people in this NG> confirm what you have said. I will repeat zillion times that in this> V-shaped spring, angle ABC is solid angle. Please read it again.http://www.geocities.com/actiondeviceSo, what do you do to *make* it a solid angle? Nail all three pointsto the board or what?If B is attached to nothing but the springs (and A/C are attached to nothing buttheir respective springs) the shape of the whole contraptionat time t will be determined by the characteristics of the springs andthe friction of the springs and of point B on the surface youre dragging sci.physics, James Harris<3c65f87.0311170550.5c7f0ae3@ and mathematics, but while> finding roots of polynomials is typically the aim of the average> researcher, polynomials themselves can be used as powerful tools for> analyzing the roots of *other* polynomials.The concepts are advanced, but can be approached by first considering> a basic example.The basic factorization to start is(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > 49(x^3 + 5x^2 + 3x + 1)What about your original polynomial? Or even a polynomialwhich doesnt have a_3 = 1 and a_0 = 1, havingnon-unit roots?Followups.[rest snipped]-- #191, ewill3@earthlink.netIts still legal to go James Harris<3c65f87.0311170548.272ff192@ and mathematics, but while> finding roots of polynomials is typically the aim of the average> researcher, polynomials themselves can be used as powerful tools for> analyzing the roots of *other* polynomials.The concepts are advanced, but can be approached by first considering> a basic example.The basic factorization to start is(c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > 49(x^3 + 5x^2 + 3x + 1)Why this particular polynomial? Why not x^3 - 8,with (c_3 x + 8) as a factor?Or your original polynomial with (c_3 x + 22) as a factor?[rest snipped]-- #191, ewill3@earthlink.netIts still legal to go thread the sequence {tan(n)/n} arose (n=1,2,3...). The> discussion> showed that this sequence does not approach zero. Now |tan(11)/11| is> about> 20. So I wondered: does tan(n)/n (or |tan(n)/n|) take on arbitrarily> large> values? So far the largest value I have found is 556.3, but n is very> large.The poiunt is that tan(n) is big in absolute value iff n is close toan odd multiple of pi/2, that is if pi/2 is approximately n/mwhere m is odd. Now there are certainly infinitely many (m,n) with|pi/2 - n/m| < 1/m^2. Were m odd then |tan n| would be about|n - m pi/2|^{-1} > m which is approx 2n/pi. So |tan n/n| would bebounded away from zero. Its not obvious that there would be infintelymany cases with m odd, and to get a better result we need a betterDiophantine approximation result on pi/2. I dont know if thereis such a result.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 Note that both are a special case of L^Hopitals rule> (note spelling):What are we supposed to notice? That you dont have a circum§ex? That youdont know that the (usual) substitution for a circum§ex o is os? Thatslepping falmes always contain question > What role does the Jacobson radical> ab = a+b - ab That is not the Jacobson radical. The Jacobson radical of a ring is the intersection> of it proper left (or right) ideals.You need some conditions on the left ideals that youare taking the intersection of. In the case of a finitedimensional algebras the Jacobson radical is the intersectionof the nilpotent left ideals. In general the left ideals shouldbe, for example, right-quasi-regular, which means every elementin the ideal has a right inverse with respect to the circle operationmonoid given by a o b = a + b - ab. This is, no doubt, whyMr Dement confused the circle operation with the Jacobson radical.A counter-example to your characterization is the two dimensionalalgebra on the plane R^2 with multiplication (x,y)(a,b) = (0,xa).The only proper non-zero ideal is {0}xR. But in this case the Jacobsonradical is the whole algebra.Exterior algebras have non-trivial Jacobson radicals, and Im think theyhave some applications in differential geometry and physics, but I doubtthat knowing the definition of Re: another quaternion question (oops) > What role does the Jacobson radical> ab = a+b - ab> That is not the Jacobson radical.> The Jacobson radical of a ring is the intersection> of it proper left (or right) ideals. You need some conditions on the left ideals that you> are taking the intersection of. In the case of a finite> dimensional algebras the Jacobson radical is the intersection> of the nilpotent left ideals. In general the left ideals should> be, for example, right-quasi-regular, which means every element> in the ideal has a right inverse with respect to the circle operation> monoid given by a o b = a + b - ab. This is, no doubt, why> Mr Dement confused the circle operation with the Jacobson radical.>Actually it should be the intersection of the regular maximal left ideals ofthe ring (or the intersection of the left primitive ideals of the ring).It is the _union_ of the left quasi-regular left ideals or in the finitedimensional (or Artinian) case the union of the nilpotent left ideals.I hope Ive got it right this time.Edwin LIGHTBULB?X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 08:15 === PM, raydpratt@hotmail.com (raydpratt) said:>Subject: DO MATHEMATICIANS READ WITH HALF A LIGHTBULB?No, but you do.>Rudy ticked me off yesterdayDo you always get angry at what you do not understand?>Prove that 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a)It would have helped had you had some comprehension of what he menatby the left hand side, the right hand side, or prove.>So I did:No, because you failed to grasp the issue of convergence.>However, before we discuss it, lets give a step-by-step definition>of division so that everybody can follow along:Too bad that you didnt do that.>A / B asks as to countNo.>plus count any fractional units of BWhat is a fractional unit of B? Answer without resort to handwavingor circular definitions.>The brain stops workingWell, yours does.have taken it as a sign that you were beyond your depth.>Ill finish reading the book, and I might even look at all the>problems, and perhaps Ill be enriched by it,I doubt it. Theres too much that you know that isnt so.>but it really makes me wonder how much logic mathematicians really >study.Youre in the position of somebody who is tome depth wondering whypeople go to concerts. Its not so much that you havent got a clue,but that you dont even realize that you havent got a clue.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to unbounded?> In another thread the sequence {tan(n)/n} arose (n=1,2,3...). The> discussion showed that this sequence does not approach zero. Now> |tan(11)/11| is about 20. So I wondered: does tan(n)/n (or |tan(n)/n|)> take on arbitrarily large values? So far the largest value I have found> is 556.3, but n is very large. I didnt see anything in the other thread that answered this. Does> anyone know?You might look at the last part of.The last sentence of that says The sequences of maxima and minima arealmost certainly unbounded, but it is not known how to prove this fact.Disclaimer: The wording in that last phrase is Erics, not mine. (Ofcourse I suspect that the sequences are unbounded. But I certainly wouldntcall it a Conformal GroupI am still looking for a good complete pedagogical discussion of this andwhat the 4 special conformal generators mean physically. I know what allthe others mean.The other 4 generators are generators of boosts to uniformlyaccelerated motions. SU(2,2) acts singularly on the Minkowski space.It acts smoothly, in a non-singular way, on the compactifiedMinkowski space, that is with added light cone review hyperbolic motion in MTW.So there is a connection to the equivalence principle.I still need to read a definitive pedagogical reviewof the details and also look again at the Utiyama &Kibble papers to make sure I remember them correctly.Since Einsteins gravity field for curved space-timecomes from locally gauging the translation subgroup T4of the conformal group, and the torsion field, absentin GR comes from locally gauging the Lorentz subgroupO(1,3) what new compensating fields do we get fromlocally gauging the entire 15 parameter conformal group?Thats the question here. If the hyperspaceis globally not oriented like a one-sided Mobius strip or a Klein Bottle then Trivial(x) locallyunwraps the global twists.Ark corrected:Jack,Not quite true. Imagine space-time that consists of just one point -call it x, and let the fiber over it, f(x), is a Klein bottle. Thehyperspace itself is then just the Klein bottle. It is non-oriented.Yet local trivialization gives you {x} X f(x) - which is the same -still non-oriented.What local trivialization does - it allows us to untwist twistsalong the base, but not along the fibre.arkOK ...What do they locally gauge to?My hunch is /zpf,u ...Yes, I think so too, and have written some stuff about iton my web page athttp://www.innerx.net/personal/tsmith/coscongraviton.htmlThe basic reference for that work is a paper by Aldrovandi and Pereira athttp://xxx.lanl.gov/abs/gr-qc/9809061which describes in some detail how the special conformal groupgives rise to cosmological constant type terms.My contribution, built on their nice math foundation,is to count degrees of freedom and get a result thatat a critical time in the past the ratio of matter forms inour universe should have been:67% Dark Energy27% Dark Matter 6% Ordinary Matterand if you follow evolution in ways that seem reasonable to me,you get a present-day content in a range of (depending on whetherCold Dark Matter is in the form of Primordial Black Holes,or MOND, or a mixture thereof)Remember Tony it is an essential falsifiable precise prediction of my MACRO-QUANTUM vacuum emergent gravity model that dark matter and dark energy are essentially the same exotic vacuum zero point energy density w = -1 field/zpf = (alpha)^-1[(alpha)^3/2|Vacuum Coherence|^1 - 1]where alpha = 1/(variable string tension) is a generalized Witten parameter(h = c = 1 convention that Witten likes to use)/zpf > 0 is anti-gravitating dark energy/zpf < 0 is gravitating dark matterWeak curvature Poisson eq isGrad^2(Exotic Vacuum Potential) = c^2/zpfLarge clumps of w = -1 /zpf < 0 like what keeps galaxies stable mimics w = 0 CDM like you pointed out in our e-mail discussions.The same idea works on the micro-scale keeping the spatially extended electron stable by holding in the self-electric charge and compensating the spin centrifugal inertial force.Indeed Brian Greenes vibrating strings of pure energy in NOVAs The Elegant Universe are in reality these Type II superconducting vortex cores of exotic vacuum where Vacuum Coherence --> 0 with trapped EM §ux.Universal Regge Slopes follow trivially.Therefore my precise prediction is that no dark matter detector will click with the right stuff. Only false positives. All dark matter detectors will remain dark if correctly designed. This is a null experiment like the Michelson-Morley experiment that was asking the wrong question. Dark matter, like dark energy is a virtual off mass space to trigger detector clicks.If experiment shows otherwise my theory is falsified in the Popper sense.Tony continued:68-75% Dark Energy28-21% Dark Matter 4% Ordinary MatterThe observed composition by WMAP:73% Dark Energy23% Dark Matter 4% Ordinary Matteris between the 75-21-4 result of assuming Cold Dark Matter ismade up of Primordial Black Holes,This I disagree with as given above. A large glob of exotic vacuum with /zpf < 0need not be a black hole with a singularity.and to the 71-25-4 result of assuming that Cold Dark Matter isa (reasonable to me) mixture of Primordial Black Holes and MOND.Unfortunately from my point of view, I cannot put these resultson the Cornell arXiv because I am blacklisted.It is especially unfortunate because they indicate that yourmodel is on the right track (except that I dont like the naiveform of supersymmetry that exists in superstring theory).TonyAlso from TonySaul-, you ask about more details on the OD and TD structures,but I havent really done much more than I described in my previouse-mail message. It is pretty clear in my mind, but I havent writtendown details yet (and being blacklisted I am not too enthusiastic aboutdoing much more writing down of stuff -...As to supersymmetry, if any reasonably interpreted experimentalresults clearly show the existence of one of the naive supersymmetryand should be thrown in the trash can.My model does, as you suggest in a remark about my 27-dim spaces,have a subtle form of supersymmetry that does appear at veryhigh energies (when the spacetime goes to 8-dim and the Higgslevel, there is only one generation of fermions (8 of them)and all 28 D4 generators are on the same footing.In my 8-dim Lagrangian,the gauge boson term has 28 x 1 = 28 degrees of freedomandthe fermion term also has 8 x 7/2 = 28 degrees of freedomthe factor of 7/2 coming from the effective dimensionality ofTherefore, the boson degrees of freedom energies,and we get to see nice symmetries down here at low energiesdue to residual effects of the high-energy subtle from THE BOOK>What about the related halting problem or (almost identical) >nondenumerability of reals?That second one Arp> You see, Dr. Arp is a scientist, a world renowned scientist and he has> *data*, real, hard astronomical data, which is more substantive in> disproving the commonly taught Big Bang Theory, than the data used to> support that theory.Arps data is presented in the _Atlas of Peculiar Galaxies_, Astrophysical Journal Supplement Number 123, Volume 14, November 1966.In the Atlas Arp presents galaxies that appeared abnormal. Follow-upobservations showed that some, not all, of the galaxies were in facttwo galaxies that are apparently interacting. What caused the doubt about the Big Bang was that some of these pairs have very differentred-shifts. If the galaxies are close to each other the differentred-shifts would sound the death knell for expansion and the BB.However, as observing technique has improved weve determined thatmost of these pairs are simply close in the line of sight and areat very different distances. There are a few cases that have notbeen elucidated, the necessary obsrvations are, at best, difficult.These remaining cases do not constitute an overthrow of the BB, to do that would require high quality observations of difficult objects; big results require big data, obscure, difficult cases do not provide that.The Atlas was and remains an important astronomy tool, the carefullwork on its members has provided important insights and data aboutgalaxy formation and interaction. The value of a scientific ideais not whether it is right or wrong, but rather the questions it leadsus to ask. - Dont remember who said that, wish it was me.For a usefull review of the state of BB theory have a look atAstronomy and Astrophysics_, _Standard Cosmology and Alternatives:anti-BBers ) admit that the reason most cosmologists like the BBis that the thoery makes many predictions that both occour naturalyin the theory and have been confirmed observationaly.As we say in sci.astro,Dark skies,tom-- We have discovered a therapy ( NOT a cure ) for the common cold. Play tuba for an *******************************************> * *> * sum_n cos(n t)/n = -ln (2 sin(t/2)) *> * *> * sum_n sin(n t)/n = (pi - t)/2. *> * *> *******************************************> In particular, taking t = 1, we get> sum_n sin(n)/n = (pi -1)/2 approx 1.0707963267948966192,> which agrees with the answer Mathematica returns. (But Mathematica 4.2> insists that sum cos(n)/n diverges. Its wrong.)It might be noted, however, that the current version of Mathematica does> not make that mistake:In[1]:= FullSimplify[Sum[Sin[n]/n, {n, 1, Infinity}]]Out[1]= (-1 + Pi)/2In[2]:= FullSimplify[Sum[Cos[n]/n, {n, 1, Infinity}]]Out[2]= -Log[2 - 2*Cos[1]]/2Theyve improved FullSimplify too, then; in 4.2, Mathematica returnsthe sum of the sine series as -1/2 I (Log[1-E^-I] - Log[1-E^I])(approximately--this is from memory), and FullSimplify doesnt help.Furthermore, FullSimplify wouldnt touch ArcTan[Sin[t]/(1-Cos[t])]either, although it has an obvious simplification. Ill have to trythat in 5.0 (which I have at work, but not at home).--Ron illusion will come to an end if few other people in this NG>confirm what you have said. I will repeat zillion times that in this>V-shaped spring, angle ABC is solid angle. Please read it again.http://www.geocities.com/actiondevice>The problem in your reasoning is that although the springs at point Bare welded in angle ABC, you are forgetting that the springs will bend(the only way to avoid this is by using completely rigid ïspringswhich arent obviously not really springs anymore). Now you could usesprings by placing the whole in a rigid V-shaped tube. That way youcould keep the angle between the complete springs constant. You willfind that point B tries to move but cant, because the rigid tube iskeeping it from moving (it is supplying the force to ïcounter theforce on point B).The best way to figure out that you are mistaken, is by doing a littletest (as you obiously are not going to trust newtons laws). Get twosprings and connect them together in a ïsolid angle. Now see whathappens if you pull on the end of the strings (if you place yourfinger under point B to keep point B in the same place, you canactually feel it pushing).BTW you are aware that you are trying to disprove the is no more physics or Newtonian mechanics, Pythagoras theorem,> George. For me, it is war between negative forces and positive forces> of nature. Logic, math, physics, intelligence, thinking ability of> well educated, brilliant people like you is succumbed to dark, evil> forces of nature. People on this earth are being controlled by dark> forces of nature.Then dont believe us. Build it.What you fail to understand that if point B shifts its position along> Y axis, angle ABC or EBF will change and as I am saying again and> again, angle ABC is solid angle.Build it and see.But I will not allow dark, evil forces of nature around me to> overpower me. I am talking about changing course of history, physics> and opening gateway to entire universe.Build it and see. - math results, which I think are rather simple, and now>its a matter of presenting them to the world, right?One would think so... however, what is important in that presentationis that you make no jumps in the proof of your argument. For example,you have repeatedly claimed this:>Let(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where the as are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)But you havent shown us once what the roots of that latter equationactually are. My advise to you is that you show us. Give us a_1(x),a_2(x) and a_3(x). If you can present the roots then we can verifythat those as do indeed ArpRather naively I thought that if you put out something simple enough> that most people could understand it, then theyd be willing to at> least question authority long enough more than a dozen people have pointed out graveprocedural errors and §at out empirical wrong results in your mostrecent spews. You responded like a diversity hire - turning yourback, screaming ïDISCRIMINATION!, shuf§ing down the same incompetentroad some more... all the while demanding a larger paycheck.http://www.crank.net/harris.html Its not every braying jackass that gets a whole page at crank.netHey stooopid, we enjoy cleaning up your messes like we enjoy scrapingdog off our shoes on a curb, distasteful but expedient. Heystooopid, the whole world is against you for good and just objectivereasons. Take a pooper scooper to yourself.-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos (oops) What role does the Jacobson radical ab = a+b - ab That is not the Jacobson radical. The Jacobson radical of a ring is the intersection of it proper left (or right) ideals. You need some conditions on the left ideals that you are taking the intersection of.Doh! I meant maximal left ideals ...-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 LIGHTBULB?Ray, you write:> You obviously have great familiarity with the subject and have fought> with it long enough to appreciate its difficulty.0 / 0 is the easiest case. The symbols ask us to count how many times> we can subtract zero from zero to the limit of zero. We reach the> limit after the first required subtraction, and thus our answer is 1,> a process 1.You may think of it that way, although a more general way is to saythat A divided by B gives the number that, when multiplied by B, givesA. Thus division is the inverse of multiplication. The reason I saythis is because multiplication is not always repeated addition (andthus division is not always repeated subtraction).Like with all inverses, though, Multiplication can be a single-valuedoperator and Division, its inverse, can be multivalued. As is the casewith division by zero. In this case, there is no unique multiplicativeinverse you can use. The notation a / b is usually meant to symbolizea * b^-1, but b^-1 this time is a SET of multiple values.If you want to find out moer about this, you should probably read alittle about Group Theory and Divisors of Zero. These things are partof Abstract Algebra, which shows what we can derive if we assume somevery abstract (unconnected from particular implementations) things.Multiple-valued inverses abound in things like Complex Analysis, wherethey deal with them (see branches). What you describe below is sort o§ike branches:> Thus, 5 / 0 = Infinity, and, inversely in that particular beginning> context, Infinity * 0 = 5. This is sort of like choosing a branch of the function a / 0.> The fact that multiplications involving> Infinity and zero can lead to such results is not a contradiction of> normal math, for Infinity and zero have logical properties that exceed> the properties of normal numbers. Just be careful that you dont go into hand-waving, which basicallyassumes that you have all the details worked out, but in fact theymight prove very difficult, if not impossible, to work out. If theyare impossible to work out, then the hand waving, while certainlyeasier to do than rigorous mathematical proof, may turn out to havebeen a waste of time (since you were stating incorrect results).> Specifically, Infinity includes> both every particular number AND all such numbers, which gives> infinity the logical power to exceed the algebraic limits of processes> with ordinary numbers. Id say basically all of modern math is based on Set Theory. If youwant to find out the foundations, check out:Zermelo-Fraenkel set theory (say, Wikipedia.org)Peanos Construction of the natural numbersSee how the real number sare CONSTRUCTED to know what they ARE. How doyou think you get all those axioms about the real numbers? Theyarent axioms, theyre postulates because its possible to constructthe system from something more basic.One way to do so is with Naturals -> Integers -> Rationals -> Reals,the last part can be done with Dedekind Cuts or a number of otherdifferent ways.> In other words, although a conclusion cannot> exceed the import of its premisses, Infinity has the necessary extra> logical elements to justify its results within normal algebra. Zero> obtains to the some of the same power with the extra logical element> of being both a normal point on the number line as well as,> exotically, representing NO number.Like I said, there might be something in this, but I dont knowwhether its worth the effort to just prove that the geometric seriesconverges to the expression on the right. After all, youll have todefine the basics of an entire system of mathematics just to fit thatfact. Sometimes its worked, thoguh, look at complex numbers: try toshow there exists a number i s.t. i^2 + 1 = 0 . Notice youll have todefine your own operations so that ^ and + make sense. :-)Take a look ou there. Ive heard of a few systems realted toinfinity-arithmetic. Theres a whole big field though, with set theoryand transfinite numbers... Cantors theory... etc. > Although I havent studied the logic in detail, Ive read that when an> irresistable force meets and unmovable object, what happens logically> is EVERYTHING. Thus, infinity and zero math is ultimately consistent> with logic and algegra by virtue of the extra logical elements> involved.Pick a particular subject you are interested in and study it in moredepth. You will see that there are a lot of issues involved in thisstuff. :-)> The difficult task is not to call such difficulties undefined, but> rather to figure them out.I also want to remark that sometimes generalization is undesirable.For example, the generalization of the reals to the complex variablesgives us many new thihgs, and is quite desirable. Generalizingfurther, to Quaternions, and introducing Cayler numbers, makes usunable to say things we were able to say about the more specialcases i.e. complex numbers and real numbers. Thus the theorems aredifferent, and focus on different things.So, often, its useful to not define things, especially if we dontneed them, just for the sake of generality. Besides, according toGodels second incompleteness theorem, given any developed system o§ogic with axioms and so forth, there will ALWAYS be statements whichare undecidable -- whether they are true or not. (See the Matrix :-).If you define those you will have MORE undecidable statements. Sodefining stuff you dont need is an unnecessary exercise. When HELP BADLY (sorry, maths not psych) And you have not provided any theory of E&M that allows any suchthing as a reverse field. Nor why there should be any kind ofspeed limit involved. Nor why it should follow any such thingas the kinetic energy formula observed in accelerators. Nor haveyou provided a relation between energy and mass if you dontaccept relativity.Socks radiation from an acceleraed charge! fields associated with a moving charge! The ïBack EMF concept. I would be most amazed if a moving charge DID NOT alter the field around itself, wouldnt you?Quite.>are accelerated.>You KNOW the following, Henry.>In an accelerator going at full efficiency, we KNOW that>because it looses this energy as synchrotron radiation in the bends>of the circuit.(Very obvious and easily measurable.)>So we - and YOU - know that the RF-cavities never ceases>is only few mm/s below the speed of light.>So why do you keep pretending that the E-field is not>speed approaches c, when you KNOW that isnt true? I DID NOT SAY THAT.Indeed you did.> The question is how much energy?Forgotten that too?> You are making no attempt to answer that one. In typical fashion, you pretend> the relevant & Penroses Twistor Conformal GroupErratum/zpf = (alpha)^-1[(alpha)^3/2|Vacuum Coherence|^1 - 1]should be/zpf === Re: A NOTy problem.> This post not CCd by email>Subject: A NOTy problem.> I have been puzzling over what ought to be a simple logic problem>A crime is committed by one of the following;>Alan, Bob, Chris or Dave.>Each makes a statement to the police but three of them lie.>Alan says, I didnt do it.>Bob says, Alan is lying.>Chris says,Bob is lying.>Dave says, Bob did it.>If Chris told truth: Alan told truth; two soothe sayers.>Thus Chris lied; Bob told truth; Alan did it.>Alan lied; Dave is wrongGday Gday Alan, Well that is brilliant but I am not. Somehow you must have> eliminated other entry points to the problem. That is quite a skill> backtrack because of further reductio absurdum. Best wishes,Im not brilliant either. Thats why I usually try to use formallogic in puzzles like this. Lets write A,B,C,D for the four guysinvolved, and t.X for X tells the truth and g.X for X is guilty. Then what we know can be formalized as (1) t.A == not g.A (2) t.B == not t.A (3) t.C == not t.B (4) t.D == g.B (5) exactly one of t.A, t.B, t.C, t.D is trueWe see that we dont know much about guiltiness, but that (2), (3),and (5) say a lot about truthfulness: (5)== using (2) on the second part, and (3) then (2) on the thirdpart exactly one of t.A, not t.A, t.A, t.D is true== t.A is false since the first and third part cannot both be true (not t.A) / (not t.D)== using (1) for the left part and (4) for the right part g.A / (not g.D)So Alan is guilty, and Dave isnt.Putting this into words, and ignoring Dave, we get the following: Chris says that Alan tells the truth, so either they both tell the truth, or they both lie. But three of them lie, so they cant both tell the truth. So they both lie. So Alan lied when he said he didnt do it, so he did it.Groetjes, <> Two questions:> First, consider positive integers k and j with m and n digits> respectively. Assume moreover that neither k nor j is equal to 0.> Can we determine exactly the number of digits in the product kj? Its> either m+n-1 or m+n, but which one it is changes depending on the> nature of k and j. I wanted to use this to determine the number of> digits in, say 2^64. Now, this one might not be so difficult because> many people know 2^32 off the top of their head so they know it has 10> digits. So they would know immediately that 2^64 either has 20 or 19,> but still its not clear (to me anyway) if its 20 or 19 without> multiplying it out. But still, what about 3^64? And what about the> number of digits of a^b where a and b are any positive integers?> log base 10 (denoted log10) is a good measure of the number of digits of any> number in base 10.> Let §oor(x) denote the integer less than or equal to x, then the number of> digits of a number x is> 1+§oor(log10(x))For 2^64, we get 1+§oor(64*log10(2))HTHAmer,Well, I was hoping for something that we could do by inspection,otherwise you may as well simply multiply the two numbers togetherusing a problem in a book to try and determine if there was apolynomial p(x) with at least 2 nonzero terms such that p(x)^2 hadexactly the same number of nonzero terms as p(x). I proved itcouldnt happen for linear, quadratic, or cubic polynomials, and Ithink I proved it couldnt happen for quartic polynomials also. ThenI found a quintic where it is true: Namely, p(x)=4x^5+4x^3-2x^2+2x+1,[p(x)]^2=16x^10+32x^9+28x^6+4x^3+x^2. Now the begging question iswhether or not there is a polynomial q(x) such that q(x)^2 has FEWERnon-zero terms than q(x). I believe the answer is most likely yes,but I have not the patience to start multiplying out 6th and 7thdegree polynomials with pencil and paper. The next begging question,assuming the answer to the previous question is yes, is the following: Let Z(p) denote the number of non-zero terms of a polynomial p. Consider the set P={p| p an arbitrary polynomial and Z(p^2) I have a question. For a randomly moving object in two-dimensional plane, the object has to> move from point X to point Y. During the movement, there are two random> processes posing on the object. For example, one process is the irregular> geograph and the other process is the varying weather. The two processesmay> be correlated. Plz give some suggestions on where can I find the related reference and which Yan> http://www.ntu.edu.sg/home5/pg01308021Look for the random walk topic in any advanced mathematical statisticsbook. You just have a random walk in 2 dimensions here. Im not going tosuggest any of the old standard books here.David Institute of Technology and started expalaining very> basic idea of this device. But they raised problem which does not> exist at all. What I tried to convince them is given in my homepage.> Please note that I have changed my homepage from previous one.> http://www.geocities.com/actiondevice> What those few of very brilliant people in India insisted that point B> will move along Y axis. I agree that due to forces acting at point> B, third vector will be produced and point B will move in space in> perpendiculer direction to XY plane. But those people insist that it> will move along Y axis !> http://www.dishman.me.uk/George/Abhi/abhi.gif> If you stretch the springs by moving A to E and> C to F, the forces exerted by the springs are shown> by the thick red arrows. To add the forces, the thin> red lines make them into a parallelogram and the green> arrow shows the resulting total. Dont take my word> for it though, check how to add forces in any mechanics> text book.I think you missed my point here. Since the red arrowsare along the springs, they are in the XY plane. Whenyou add them using the parallelogram, the result mustalso be in the XY plane. There is no perpendicularforce, the movement will be along the Y axis.> To stop B moving, Newtons law says you need an equal> and opposite reaction which is shown by the blue arrow.> Without that, B will move in the direction of the green> arrow. That is along the Y axis so they are right> according to the standard rules of mechanics. What you fail to understand that if point B shifts its position along> Y axis, angle ABC or EBF will change and as I am saying again and> again, angle ABC is solid angle.I know but I am responding to your diagram and there isnothing in that to keep the angle the same. What I amsaying first like everyone else is that, as you havedrawn it, point B will move and the angle will change.Something extra will be needed to prevent it movingthat is not shown in the diagram at the moment, and Icannot speculate on what you might add.Whatever method you choose, if it stops point B movingthen it can only do so by providing the reaction forceshown by the blue arrow because the sum of the blue andgreen must be zero if the acceleration of point B is tobe zero. F = m * a, remember? If a=0, F must be zero too.> Why you people are being controlled, this is the reason behind it.It is not control but common education. I learnt mechanicsfrom teachers and textbooks and so did everyone else. Wehave all learnt the same rules so we polynomial> I ran across a problem in a book to try and determine if there was a> polynomial p(x) with at least 2 nonzero terms such that p(x)^2 had> exactly the same number of nonzero terms as p(x). I proved it> couldnt happen for linear, quadratic, or cubic polynomials, and I> think I proved it couldnt happen for quartic polynomials also. Then> I found a quintic where it is true: Namely, p(x)=4x^5+4x^3-2x^2+2x+1,> [p(x)]^2=16x^10+32x^9+28x^6+4x^3+x^2.Make that 16 x^10 + 32 x^8 - 16 x^7 + 32 x^6 - 8 x^5 + 20 x^4 + 4x + 1.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 http://www.dishman.me.uk/George/Abhi/abhi.gif> To stop B moving, Newtons law says you need an equal> and opposite reaction which is shown by the blue arrow.> Without that, B will move in the direction of the green> arrow. That is along the Y axis so they are right> according to the standard rules of mechanics. Roger that. But my illusion will come to an end if few other people in this NG> confirm what you have said. I will repeat zillion times that in this> V-shaped spring, angle ABC is solid angle. Please read it again.I have explained in another reply that, although you saythis, there is nothing in the diagram to make it happen.However, I also wanted to mention that your use of thephrase solid angle will be confusing to many people.This web page explains what solid angle means: http://en.wikipedia.org/wiki/Solid_angleYou might VC5UG2ZG8e4wGnjsueOcEHhvG5S2fvUiW6mj3Hxc6npc3fqcNONfgVGoogle for realizability of polytopes. Its a quite lively area> of research.> Unfortunately, nothing to find about realizability concerning graphs...> A 0/1-polytope is a polytope where all its vertices have coordinates in {0, 1}.Hmm, I fail to see why you need fixed coordinates in the first> place. If you happen to find a realizable polytope for some given> graph, then you should be able to name the vertices with some> 0/1-coordinates in some suitable space. Maybe I am missing something.> It took me quite a long time to find a counterexample, but in fact it isvery easy: take n>4 arbitrary. Obviously there is no 0/1-polytope that hasthe same graph as a polygon with n vertices, because if would be >=3dimensional, so that its graph has higher connectivity.-- Phyics is much too hard for physicists.reverse my forename for saying. ABC is a rigid structure> which does not allow points A and C to separate horizontally. Nothing is completely rigid, so B will move a little bit> as the structure is stretched and pulled tight. At that> point youll reach static equilibrium, so long as the> forces are indeed confined to pull at angles but the> structure prevents the horizontal separation from> growing. Unfortunately, you put the forces on springs rather> than confine them to the steel structure. So they can> happily move apart.I think he envisages something like putting a coathangerbent into a V shape into the springs so the A and C endsare free to move but the other ends are fixed at the bend.> Yes, point B will shift its position in space but certainly not along> Y axis or in XY plane. If the forces are confined to the xy plane, so will the motion be.Think of a dowsing rod §ipping with slight changes ofstrain across the points and you might get an insightinto his thinking. (You may not want an insight MATHEMATICIANS READ WITH HALF A LIGHTBULB?> Im listening ... so tell me in English why Im wrong.Do you know what it means for a series to converge?> Are you talking about a series when asking -- and the answer is no, I do not know what> ïconvergence means in the above infinite series. I do know that> infinity has the property of both containing each particular number> that it comprizes as well as having no end of such numbers, and thats> it.If 2 + 3 + 4 + . . . ad infinitum is subtracted from 1 + 2 + 3 + . . . ad infinitum, then it seems obvious that you will be subtracting the equally> infinite tails of two infinite series at point 2 on the number line,> and thus leave the non-infinite 1 that makes the only difference> between the two series, i.e., the non-infinite difference of 1 in each> series starting points on the number line.Lets consider these and some other sums.S1 = 1 + 2 + 3 + ...S2 = 2 + 3 + 4 + ...S3 = 2 + 4 + 6 + 8 + ...S4 = 3 + 5 + 7 + 9 + ...Now, you say it seems obvious that S2 - S1 = 1.Notice that S3 can be obtained by doubling S1 termfor term. That is, S3 = 2*S1.Notice also that S3 + S4 = S2.Therefore S2 = S3 + S4 = 2*S1 + S4. S2 is more thandouble S1 (which is already infinite). In fact itsS4 more than double S1, and S4 is clearly infinite.So are you really so sure that the difference betweenS2 and S1 is only 1?> Please explain how ïconvergence refutes that logic.Because since none of these sums converge, you cant dothings like S2 = oo S1 = oo + 1 S2 - S1 = oo + 1 - oo = 1By suitable manipulation of symbols, I can obtain oo - Apocalypse NOW!X-DMCA-Notifications: http://www.giganews.com/info/dmca.html<<< Why you people are being controlled, this is the reason behind it. -Abhi.Why dont you seek a close mental health facilityand get a grip on the paranoid delusions you display?They have all kinds of new drugs to modulatethe delusional periods.... R. Mays---------------------------------------------------------- -------------------Some where within the Quantum StateHttp://.Mays.Com/story.htmlhttp://paul.mays.com/ mayday.htmlhttp://paul.mays.com/rainy.htmlPhysics is experience, arranged in economical order. -Ernst first looked at integral[cos(ax)/(b^2+x^2)], limits -infinity> to +infinity, i thought i could solve it by some substitution. But> after trying all my wits on that, i decided to consult an table of> integrals and found the result as ci(ab)sin(ab) + si(ab)cos(ab)> where ci(x) is integral[cos(t)/t] and si(x) is integral[sin(t)/t].For the INDEFINITE integral, perhaps. The DEFINITE integral which youproposed is integrated by the method of residues, and the answer (for a > 0, b > 0) is e^(ab) pi/b. It suffices to be able to integratecos(cx)/(x^2+1) for c = ab.This really bugs me--people post integrals in pseudo-generality whichcan be reduced by a simple linear change of variable.> Well... that made me curious and i started looking up ways to solve> integrals in general. I had by now collected a lot of terms, like,> local and global analysis, perturbation theory, Liovelle expansions> and so on. I wnt to know, if these are the ways in which the integrals> that cannot be normally solved are being integrated and put into those> tables.And something interesting popped up. I came across something called> Riccatti equation(i am not sure if i spelled it right) and it has> application in signal processing. i dont know what kind of> applications...can someone enlighten me on that too.Also, please do suggest me some readable books in which i can learn> about the local and global analysis and kind of stuff.One question at a CosmologyDo not reply to mindspring which is a dummy address so that I can send mail on my regular MacMail program out from a WiFi Caffe. Use sarfatti@well.comalso I will be moving soon so sarfatti@pacbell.net will be defunct. sarfatti@well.com is the one to use.Jack, you ask:... 4 special conformal generators ...What do they locally gauge to?My hunch is /zpf,u ...Yes, I think so too, and have written some stuff about iton my web page athttp://www.innerx.net/personal/tsmith/coscongraviton.htmlThe basic reference for that work is a paper by Aldrovandi and Pereira athttp://xxx.lanl.gov/abs/gr-qc/9809061which describes in some detail how the special conformal groupgives rise to cosmological constant type terms.Yeah that paper is interesting.They seem to say that the deSitter group limits to 15 parameter conformal group when cosmological constant limits toinfinity! Also there is an interesting stringy duality between infinite and zero cosmological constant. It is the intermediate cases that is of interest. Also localizing - not just a space of constant curvature, i.e. locally gauge the Lie algebra of the De Sitter group which is more general than the conformal group?infinity hence gravity as curvature is no longer possible.BTW the world hologram ideaLp*^2 = Lp^4/3(c/Ho)^1/3is alluring, but has real problems of consistent interpretation such as cosmological time increase in the Regge slope i.e.Wittensalpha = Lp*^2 = 1/(string tension)means the string tension decreases as the universe 3D space expands.My guess is that current astrophysics falsifies that?The electron rest mass from Higgs field part of Vacuum Coherence is (h = c = 1)m ~ e^2/zpf*^1/2 ~ e^2/(alpha)^1/2What does this do to e/m?If /zpf* = 1/Lp*^2 = 1/G* = (alpha)^-1And if Blackett relation for quantized trappedEM §ux in the Wheeler micro wormhole of Mass without mass and Charge without charge:e = G*1/2 m = (alpha)^1/2mm ~ (alpha)m^2(alpha)^-1/2 = (alpha)^1/2m^2Ignoring m = 0 root.m ~ (alpha)^-1/2e is then invariant, but e/m is not.Ho = R(now),t/R(now)in the FRW metric.We want G* on large scale > 10^-3 cm to be G(Newton) and it is also thought to be G(Newton) at the Planck scale10^-33 cm, yet we want G* = Approximating Pi by Rationals >I feel that this is probably a well-known subject for number theorists,but Ive never read anything about it. The question is how closely canwe approximate pi by rationals. More specifically:For integer n>0, let f(n) be the largest integer m such m/n < pi.Let d(n) = pi - f(n)/n.Then d(n) measures how accurately we can approximate pi by a rationalwith denominator n.How small can d(n) be? Clearly, d(n) < 1/n. But can we make d(n) muchsmaller than that? Q1: Can we find arbitrarily large values of n such that d(n) < 1/n^2? Q2: Can we find arbitrarily large values of n such that d(n) < 1/n^3? Q3: In general, for each p>1, can we find arbitrarily large values of n such that d(n) < 1/n^p?>NO! Surprisingly, K. Mahler showed that d(n) > 1/n^(42). That>exponent 42 has been improved subsequently. I guess the best current>result is 8.0161, due to M. Hata.>What if we redefine d(n) as min (pi - f(n)/n, [f(n)+1]/n - pi)?-- Stephen J. Herschkorn Cosmology, De Sitter Group Lie AlgebraLet R be a stringy Kaluza-Klein compactification scale of an extra space dimensionSuppose(c/H(t))Lp*(t) = R^2(t)Use the holographicLp*(t)^2 = Lp^4/3(c/H(t))^2/3with the world hologram on the surface of a Planck sphere.Therefore,(c/H(t))^4/3Lp*(t)^2/3 = R^2(t)This one puts the world hologram on the past light cone wave front of thickness Lp*(t) back c/H(t) in time from t = now.The first one is a dual relation.Do not reply to mindspring which is a dummy address so that I can send mail on my regular MacMail program out from a WiFi Caffe. Use sarfatti@well.comalso I will be moving soon so sarfatti@pacbell.net will be defunct. sarfatti@well.com is the one to use.Jack, you ask:... 4 special conformal generators ...What do they locally gauge to?My hunch is /zpf,u ...Yes, I think so too, and have written some stuff about iton my web page athttp://www.innerx.net/personal/tsmith/coscongraviton.htmlThe basic reference for that work is a paper by Aldrovandi and Pereira athttp://xxx.lanl.gov/abs/gr-qc/9809061which describes in some detail how the special conformal groupgives rise to cosmological constant type terms.Yeah that paper is interesting.They seem to say that the deSitter group limits to 15 parameter conformal group when cosmological constant limits toinfinity! Also there is an interesting stringy duality between infinite and zero cosmological constant. It is the intermediate cases that is of interest. Also localizing - not just a space of constant curvature, i.e. locally gauge the Lie algebra of the De Sitter group which is more general than the conformal group?infinity hence gravity as curvature is no longer possible.BTW the world hologram ideaLp*^2 = Lp^4/3(c/Ho)^1/3is alluring, but has real problems of consistent interpretation such as cosmological time increase in the Regge slope i.e.Wittensalpha = Lp*^2 = 1/(string tension)means the string tension decreases as the universe 3D space expands.My guess is that current astrophysics falsifies that?The electron rest mass from Higgs field part of Vacuum Coherence is (h = c = 1)m ~ e^2/zpf*^1/2 ~ e^2/(alpha)^1/2What does this do to e/m?If /zpf* = 1/Lp*^2 = 1/G* = (alpha)^-1And if Blackett relation for quantized trappedEM §ux in the Wheeler micro wormholeof Mass without mass and Charge without charge:e = G*1/2 m = (alpha)^1/2mm ~ (alpha)m^2(alpha)^-1/2 = (alpha)^1/2m^2Ignoring m = 0 root.m ~ (alpha)^-1/2e is then invariant, but e/m is not.Ho = R(now),t/R(now)in the FRW metric.We want G* on large scale > 10^-3 cm to be G(Newton) and it is also thought to be G(Newton) at the Planck scale10^-33 cm, yet we want G* = quaternion question (oops) > What role does the Jacobson radical ab = a+b - ab> That is not the Jacobson radical.> The Jacobson radical of a ring is the intersection of it proper left (or right) ideals. You need some conditions on the left ideals that you are taking the intersection of. Doh! I meant maximal left ideals ...>Just maximal left ideals is not enough. Take the nilpotent algebraA = R^2 with multiplication (x,y)(a,b) = (0,xa). A has only three ideals:A , {0}xR, and {(0,0)}. But in this case the Jacobson simple problemWhat are you guys talking about? The original system was>X(t)=a-b*X(t)>Y(t)=c-d*Y(t)>Nicolas subtracted to get>(Y-X)(t) = c - a - (d-b)(Y-X)(t)>which is wrong! The correct result would be(Y-X)(t) = c - a - (d Y- b X)(t)which is not helpful.Solving the equations, you want to find t such thatX(0) exp(-b t) + a/b = Y(0) exp(-d t) + c/dLamberts W function might help.-- Stephen J. Herschkorn random process - I have great trouble reading this post. It is always hard figure out the statistical content whenthe context is super-abstract. It becomes impossible to figure, when idiosyncratic terminologyis combined with bad spelling -I have a question. For a randomly moving object in two-dimensional plane, the object has to move from point X to point Y. - *has* to move? Or, you want to assume that does...During the movement, there are two randomprocesses posing on the object. - random forces, pushing on the object?For example, one process is the irregular geograph - geography?and the other process is the varying weather. The two processes may be correlated. Plz give some suggestions on where can I find the related reference and which book specify such problem.Okay. Concretely, as I imagine it: You have a mountain goat whose wandering depends on the hilliness and weather (whileexposure depends partly on hills, too). Where will he go, how fast? Look up animal husbandry.Concretely as you want: What is the problem or question?-- Rich Ulrich, wpilib@pitt.eduhttp://www.pitt.edu/~wpilib/index.htmlTaxes are Reconsidering Halton Arp> So it should be a no-brainer, the theory has to be changed to fit the> data.In addition to potential problems with a theory, the availabledata may be incomplete (e.g. Neptune was not known whenirregularities were found in the path of Uranus) or theobservations may be wrong (e.g. Schiaparellis canals on Mars.)Please note that I have made no comment on the work of Arp noron the Big Bang theory. I am merely pointing out that there areother possibilities Argument with professor <3fb93260$1$fuzhry+tra$mr2ice@news.patriot.net> <874qx2o9u0.fsf@becket.becket.net>h X-Treme: C&C,DWS at 03:26 PM, tb+usenet@becket.net ( Bushnell, BSG) said:>For what its worth, there *is* an extension of the cross product for>vectors in higher dimensions.FSVO extension. The exterior product for a vector space V does not mapVxV->V, but into the exterior algebra of V.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to <3fb6ed2f$2$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-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWSsaid:> off you cunting little Jew.Is that what su madre la puta taught you, feygeleh? A fongul! Are youjealous over how I used kus emak, tonto? A bas! -- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to READ WITH HALF A LIGHTBULB? X-Treme: C&C,DWS at 08:43 PM, raydpratt@hotmail.com (raydpratt) said:>Let us remember that for many centuries, the world was >most assuredly §at.Let us not remember things that never happened. That story is as bogusas the cherry-tree story. Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to Proofs from THE BOOKh X-Treme: C&C,DWS at 04:13 PM, Mitch Harris said:>topology is really just a fancy word for geometry, right?Wrong. There are many issues in Geometry beyond the scope of Topology.Before the last few centuries, hardly any work in Geometry hadanything to do with Topology.>-But- there is one field which is entirely missing: Logic.Of course there are. But, as you noted, the book was inspired by Erdosand you should have expected it to cover the fields in which heworked.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to (sorry, maths not psych)HenriWilson skrev i melding> I still cannot see the philosophical reason for using F=dp/dt rather than> F=m.dv/dt where m=f(v)Its Newtons second law of motion in its original form.The change of motion is proportional to the force impressed; and is made in the direction of the straight line in which the force is impressed.What did Newton mean by motion?Those who have Logic missing from Proofs from THE BOOK> -But- there is one field which is entirely missing: Logic.Are there no beautiful proofs in the field of logic?Goedels first incompleteness theorem has a large part to it that is > totally ugly algebra/number theory/encoding, but the gist of it is simple.What about the related halting problem or (almost identical) > nondenumerability of reals?screw the stuff thats on the border of math n logic - theres plentyof awesome stuff smack in the middle of logic:(1) the Henkin proof of semantic completeness of 1st order logic -beautiful algebraic proof, prolly the most common used for new logicalsystems....(2) most any proof made by Gentzen, but most awesome and mostfundamental of course would be his cut-elimination proof. (hisconsistency of arithmetic proof is killer too, but it takes us out ofpure logic....)(3) whatever proof Frege gave at the end of his Begriffsschrift - somemathematical sentence or other, if I recall. Included here because ofits powerful and instant proof-of-concept role o§ogic-as-we-know-it-and-as-invented-by-frege.(4) derivation of the least-fixpoint concept of a truth predicate, andKripkes intuitive discussion. (I never learned tarskis contraryresult for expressively rich languages well enough to include ithere..... :( )(5) the lowenheim-skolem construction.Im sure theres more, but these are just off the top of my head.....sci.logic might give you some more focussed Jeff Root: > Action Device to generate unidirectional force.http://www.geocities.com/actiondevice> Abhi,> What is a solid angle?> Can you give an example of a solid angle?Solid angle means angle which does not change due to forces> acting at point B in this V-shaped spring ABC. Okay. Everyone else uses the term solid angle to meansomething completely different and unrelated to your use ofthe term. I will interpret your use of it according to yourdefinition. > Wall of your room makes an angle of 90 degree with roof of> your room. This angle does not change if you push the wall> by your hand. This is example of solid angle. Whether the angle changes or not depends on the constructionof the room and on how hard I push. If I press on the windowglass I can press quite gently and see a small change in theangle, because the glass bends fairly easily and conditionsfor seeing even a small change are optimal. If I press on amore solid part of the wall, where it is built of wood framecovered with lath and plaster, or where it is concrete blocks,I would have to either press much harder or use some kind ofsensitive measuring instrument to see a change in the angle.I have pounded on the wall with a hammer, and when I did so,the ceiling nearby vibrated noticeably. That means the anglechanged slightly and brie§y when I applied a large force. In your device, what prevents the angle ABC from changing whenforces are applied through the springs? What makes the anglesolid? How solid is it? Did you realize that your invention is an application of theclassic question, What happens when an irresistable forcemeets an immoveable object? The answer to that classic question is: The object moves.In the real world, all forces are irresistable, and there isno such thing as an immoveable object. Any force, even asmall, weak force, applied to any object, causes some motionof the object, even if it is only a very small motion. > How is weather out there in Menneapolis, Jeff? Very pleasant today. Yesterday it was cloudy and a bit rainy,but now the sky is beautiful blue. The humidity has been highthe last couple of days, which made the temperature seem mild.I could see my breath even though the air did not feel know any mathematicians. Mathematicians always delete my e-mails.Ten days ago I posted the link to introduce my new proof of the FTCand a funny thing happened. Besides a couple of replies about mystyle, there were no replies about the substance of what I have to sayfrom the sci.math area. I had to go to alt.sci.physics.new-therories off&th=1ea9a3a962a95979to get any response at all. This is strange because calculus shouldbe of interest to mathematicians. I suppose analytical thinkers tendtowards math and visual thinkers tend towards physics. You should beinterested because even though you may be expert at the art ofcalculus, you dont know why it works until you see what I have found.Calculus had no proof for the fundamental theorem for two hundredyears. Infinitesimals were used to explain the practice but bothLeibniz and Newton knew there was something wrong with them and theycouldnt be proved. Limits were developed to improve upon theinfinitesimal and avoid awkward terms like estimate and approximate inthe calculus. Although proving the calculus is valid, limits do notenlighten you as to how the calculus works its magic.One problem I have is that anybody who already knows the calculus willsee the connection but not the benefit of my proof. They wont haveto do anything different or see the big value in understanding. Newstudents will appreciate it because all the new rules they learn willfall into place and the objectives of the calculus will be understood.So here is the challenge. If you think youre smart read the abovethread and go to: www.precalculus.netfirms.com/#3 But you cant because its right. So come back and say its right soothers will know. Spread the word, this stuff belongs in a coordinate system in four dimensions and consider the linesegment starting from the point (0,0,0,0) and ending at the point(1,1,1,1) the lenght of this line segment is 2 so that in general thelenght of a line segment similar to this is the square root of thenumber of dimensions. In infinite dimensional space we have a linesegment starting from the point (0,0,0...) and ending at the point(1,1,1...) that has an infinite lenght. go to the end of this line andadd one units lenght to it. What are the coordinates of this newpoint? It cant be a real number greater than one because the distancebetween any two real points would result in an infinite lenght.therefore the coordinates of this point are an naturalsPeter L. Montgomery escribi.97 en el You can get at least 24: a = b = c = 5050505.> a+b = a+c = b+c = 10101010> a+b+c = 15151515> S(a+b+c) = 24 An upper bound is 60. S(a + b + c) = S(10a + 10b + 10c) <= S(5a + 5b) + S(5a + 5c) + S(5b + 5c)> <= 5S(a + b) + 5S(a + c) + 5S(b + c) <= 15*4 = 60.A solution is a = 4554554555 b = 5455455455 c = 5545545545S(a+b) = S(a+c) = S(b+c) = 4; S(a+b+c) = 51(By Larrosa Ca.96estroA Coru.96a categorical definition of a group, (instead of the Re: Apocalypse NOW!Abhi replied to Jeff Root: > Have you done elementary Geometry Laura? Take a look at my homepage.http://www.geocities.com/actiondevice> What is on that page is silly. Point B is pulled downward by any downward force applied to it unless support is provided to hold it up. You provide no support, so point B is pulled down.When I say minds, thinking ability, intelligence of people around the> world is being controlled, this is the reason.I am not applying any downward force. The force is applied to point A> along the direction of line BE which makes 60 degree angle with X axis> and the force is applied to point C along the direction of line BF> which also makes 60 degree angle with X axis. What makes you think that that is not a downward force? Instead of saying downward, I will use your terms. You are applying forces in the direction of BE and in thedirection of BF. So point B is pulled toward point E andtoward point F unless some opposing force holds it back. > Again I am talking about V-shaped SPRING which can store elastic> potential energy. Yes. But in order to stretch a spring, force must beapplied to both ends. If you pull on only one end of aspring, the whole spring will move, and will not stretch. > Think again.... I did. Now you act. Did you make and test your device, yet?If you havent, do it *now*! It is challenge Adjunct Assistant Professor at the University of Montana. [.snip.]>Ten days ago I posted the link to introduce my new proof of the FTC>and a funny thing happened. Besides a couple of replies about my>style, there were no replies about the substance of what I have to say>from the sci.math area. I had to go to alt.sci.physics.new-therories> off&th=1ea9a3a962a95979>to get any response at all. This is strange because calculus should>be of interest to mathematicians. I suppose analytical thinkers tend>towards math and visual thinkers tend towards physics. You should be>interested because even though you may be expert at the art of>calculus, you dont know why it works until you see what I have found.Calculus had no proof for the fundamental theorem for two hundred>years. Infinitesimals were used to explain the practice but both>Leibniz and Newton knew there was something wrong with them and they>couldnt be proved. Limits were developed to improve upon the>infinitesimal and avoid awkward terms like estimate and approximate in>the calculus. Although proving the calculus is valid, limits do not>enlighten you as to how the calculus works its magic.That may be ->your<- experience. I found limits not only veryenlightening, but they even made clear why the notion of continuityis important.By the way, it is ->perfectly possible<- to develop calculus usinginfinitesimals. >One problem I have is that anybody who already knows the calculus will>see the connection but not the benefit of my proof. They wont have>to do anything different or see the big value in understanding. New>students will appreciate it because all the new rules they learn will>fall into place and the objectives of the calculus will be understood.Again, that may be your impression, because it worked for->you<-. Whether students will appreciate it because all the newrules they learn will fall into place is not a judgement you can makebased on your experience and your experience alone. It is your guess,and to establish it you would need to perform some controllededucational experiments.>So here is the challenge. If you think youre smart read the above>thread and go to:> www.precalculus.netfirms.com/#3 >But you cant because its right. So come back and say its right so>others will know. Spread the word, this stuff belongs in textbooks. I disagree. Your approach is limited to functions which are one-to-one(what you term non-reversing) in the region in question, which makesit annoying to apply to a general function: you would first need tofigure out the intervals on which it is one-to-one before yourapproach can be used. That makes the approach certainly less usefulthan the standard approach. Why should it be in a textbook, absent->actual evidence<- that it is useful for a significant proportion ofstudents?(No, I did not check it very carefully, though in general it seemsabout right; your function M(x) is closely related to Newtons Method,and a formula for it could be developed directly by using thederivative. Without the derivative, you end up having to doapproximations: your claim that you can find it graphically reallyamounts to claiming that you can draw accurate tangents byhand. Thats really hardly true in practice, so your claims towardsthe end really end up being that you can find M(x) and R(x) by usingthe derivative, which means that you are really just running around incircles. If you have any objective evidence that the approach will bebeneficial for a significant number of students, then you should bepresenting your findings to educators and to writers of about what I accept as reality. --- Calvin (Calvin and infinitesimals exist>Set up a coordinate system in four dimensions and consider the line>segment starting from the point (0,0,0,0) and ending at the point>(1,1,1,1) the lenght of this line segment is 2 so that in general the>lenght of a line segment similar to this is the square root of the>number of dimensions. In each of these cases, the coordinate system you set up is,implicitly, imposing a metric structure on your n-dimensional vectorspace which makes it a Euclidian space. So far so good;your claims are sound, to this point.>In infinite dimensional space But here we begin to slip into imprecision (which will soonlead to nonsense). *What* infinite dimensional space?If you are (somewhat sloppily, but with at least a bit oftradition behind you) thinking of each of your standard Euclidian spaces being included in the next as a hyperplane(so that the point with coordinates (x_1,...,x_n) in then-dimensional standard space is identified with the pointwith coordinates (x_1,...,x_n,0) in the (n+1)-dimensionalstandard space, then indeed the union of the countablesequence of finite-dimensional Euclidian spaces can bethought of as an infinite-dimensional Euclidian space (to wit, R^infty with a standard metric)--but then its not true that >we have a line>segment starting from the point (0,0,0...) and ending at the point>(1,1,1...) because the point (1,1,1,...) (1 in every position) isnt *in*that infinite dimensional space. Even if you completeR^infty in the usual way, to obtain the standard model ofseparable Hilbert space little-ell-two, (1,1,1...) wont be there. So asserting that the line segment in question>has an infinite lengthis false. Now, of course, there *are* sequence spaces whichinclude (1,1,1...), for instance, little-ell-infinity. Butunfortunately for your program, these spaces arent Euclidian(and with the little-ell-infinity metric, that line segmenthas length 1).Thus, the rest of your post:>go to the end of this line and>add one units lenght to it. What are the coordinates of this new>point? It cant be a real number greater than one because the distance>between any two real points would result in an infinite lenght.>therefore the coordinates of this point are an infinitesimal distance>past one.is unjustified (and in fact it is unjustifiable, and either falseor nonsensical depending on how its interpreted).It was a try, though.Lee many of the opinions are emotionally> bound, maybe resorting to just BITCH will heat the debate enough> to carry the meaning and sentiment. We can all practice... YOU> BITCH! Its the capitalising that does it, if GG switched> to *emphasising* like that, but then it kills the §ow.... but> were over a lot of hurdles already. People can in§ict more abuse> without obscenity.HercHow true!Didnt Kronecker drive Cantor insane without the help offour-letter words? Wasnt Pertti Lounesto--one of thefew credentialled mathematicians to post with any frequencyon sci.math--felled by a campaign whose brutality bearswitness to the sanguinary nature of ïlesprit NUT POST ROTE LIE TOOL UP INTEREST? SPENT LOT IS TRUE!B: how about LONE PROSTITUTE?Pertti Lounesto: ...I have neved poster anagrams of your name. Why notattack on my mathematics, instead of my person, orrather my name, or rather permutations of lettersof my name? Why not you, and your countryman ..., whoalso posted anagrams of my name, explore my mathematics,which develops further the works of your countryman, William K. Clifford? C: I have received the following message concerninga once-frequent participant in sci.math.THIS IS WITH GREAT REGRETS THAT WE LET YOU KNOW THATPERTTI LOUNESTO HAS DIED WHEN SWIMMING IN CRETE ONJUNE 21. HE WAS A FRIEND AND INSPIRATION OF MANYB: > Pertti Lounesto > - born in Finland > - invented quadratic nature of triality > - will die in Finland I am sorry to hear that he has been defenition of a Group Adjunct Assistant Professor at the University of Montana.>What is the categorical definition of a group, (instead of the usuall>definition based on binary operations).Not sure if this is what you mean, but a group is a category with onlyone object in which every arrow is about what I accept as reality. --- Calvin (Calvin and engineer study analysis?Hello...I want to make a postgraduate study in analysis, because I like mathsa lot and my undergraduate final work was about iterative methods forregularization problems.I am an engineer and in the faculty, they are requesting me an essayabout why should I study analysis. I must make a good writing becausemy admission depends on it.I would like you to help me and give me some potentially rewarding challenge [...] you might want to see the text file at> to Simpson - and his should, reasonably, be the standardinterpretation - Goodsteins Theorem is seen as a number-theoretictruth that appeals necessarily to transfinite reasoning. If so, theTheorem would involve a concept of intuitive mathematical truth thatis of a higher order than that required to see that the G.9adeliansentence is an effectively (verifiably) true number-theoreticassertion under the standard interpretation.This follows since G.9adels reasoning can be carried out even in a weakArithmetic such as Robinsons system Q; thus the truth of theG.9adelian sentence under the standard interpretation could, arguably,be considered even more intuitive than the truth of anynumber-theoretic assertion that appeals to mathematical induction.Accordingly, the choice seems to lie between abandoning the concept ofintuitive truth that forms the bedrock of Peanos axioms, and - ifGoodsteins Theorem is, indeed, a true number-theoretic assertion -finding a non-standard interpretation under which Goodsteinsreasoning can be mirrored constructively.As I argue elsewhere, I see a serious problem in admitting the former;hence, my interest in invalidating the standard interpretation ofGoodsteins reasoning stems from a belief in the XcWf4GZUgeu9GNrFC8fVJ4Emwgm+ada+CCLyqhCwRFgbND3UKN5RocGiven a C^infty manifold M, a lagrangian L:TM-->R, is there a way toformulate the Euler-Lagrange equation without explicitly using charts?-- Phyics is much too hard for physicists.reverse group, (instead of the usuall> definition based on binary z : M -> M such that d 1 x z M ---> M x M -------> M x M | | | | m V V 1 --------------------> Mand v |---> |---> _ _ | | V V 0 |--------------> u = v*v^-1commute. A monoid is a set M with maps m : M x M ---> M and e : 1 ---> Msuch that 1 x m M x M x M -------> M x M | | | m x 1 | m V V M x M -----------> M m e x 1 1 x e 1 x M -------> M x M -------> M x 1 | | | | ell | m | s V V V M = M = Mcommute.-- dont know any mathematicians. >Mathematicians always delete my e-mails.Maybe its because youre condescending to them. very cleverly. Dont believe me? People in this NG will not answer clearly the question I have> posed. Will point B move along Y axis in XY plane? It needs> just yes/no. But they will remain silent(or they will be> humorous). They will ignore me. Because they are controlled.> Most people ignore you because they have no interest in you.I said that I am trapped. My every action, word, thought was> controlled in this NG or outside world. Now circumstances,> through my past postings, are created in this NG in such a way> that even if I am talking truth, they will not take interest> in me. This is just one of the part of trap. About as many people here take interest in you as take interestin Ralph Sansbury, Jack Sarfatti, Brad Guth, or others who postideas which are contradicted by observation. You are not beingignored more than others who do the same kind of things you aredoing. Several people have replied to you in this and otherthreads. But most people have no interest in what Ralph orJack or Brad or you have to say. Most people have no interestin what I have to say. That is not part of a trap. Differentpeople are just interested in different things. Other people use humor with you because they believe you are mentally ill. You need to try to show them that you are not.Not me. Yes, you! Nobody can to do it for you. You must do it. > It is their job to show that their intelligence, thinking> ability, conscious minds are not being controlled. They can do so if they want to, but it is not their job todo so. > They just need to answer simple question. Whether point B will> move along Y axis or not? And as several people have told you, according to yourdescription, it is completely and unambiguously clear thatpoint B will move along the Y axis. Also, whatever it isthat is keeping angle ABC solid will either bend or break,or both. > They can refer figure on my homepagehttp://www.geocities.com/actiondeviceNow I am going to explain again what I am trying to say.In above figure AB and CB are V-shaped spring of same length and> stiffness and both springs are in relaxed state initially. Angle> ABC is solid angle . Let angle ABC be 60 degree in this figure> (for the sake of explanation only, in actual Action Device this> angle will be very small). Why does the angle need to be very small? Why not use theactual angle in the explanation? > At t = 0, we apply same magnitude of force on point A and C and> we pull point A of spring AB towards point E in the direction of> line BE which makes 60 degree angle with X axis and we pull> point C of spring CB towards point F in the direction of line BF> which also makes 60 degree angle with X axis. And we are> pulling point A and C in such a way that these points must> stretch or extend to point E and F resp. on x axis.Please note that we are pulling points A and C in the direction> which makes 60 degree angle with X axis. We are NOT pulling> point A and C in downward direction. Yes you are. You are not pulling them *straight* down, but you*are* pulling them down. When you go down a stair at a 60 degree angle, are you goingdown? Yes, obviously you are. > At t = t, spring AB is stretched and point A of spring AB reaches> to point E on X axis. Also at t = t, spring CB is stretched and> point C of spring reaches to point F on X axis.We will find that point B has not shifted its position along Y axis.> It remains where it was at t = 0. Because if point B shifts its> position along Y axis, to say, point B, angle ABC (or EBF,> because point A coincides with point E and point C coincides with> point F) will be different from angle ABC i.e. greater than> 60 degree. But as stated above, angle ABC is solid angle which> does not change due to forces acting at point B.Yes, point B will shift its position in space but certainly not> along Y axis or in XY plane. Third vector will be produced at> point B direction of which will be perpendiculer to XY plane.Am I right or wrong? Wrong, obviously. Brad Guth thinks he sees buildings and roads and bridges inthe Magellan radar images of the surface of Venus. Nobodyelse sees them when they look at the same images. He thinkseverybody else is wrong. The guy who posts here with the handle Oriel36 thinks thatthe fact that Earth is 360 degrees around means that Earthrotates exactly 360 degrees in 24 hours, and that astronomersare wrong when they say it rotates approximately 361 degreesin 24 hours. They are unable to see certain things as they really are. Your error proof that infinitesimals existNope.You can add something of ïlength one to (1,1,1...), however it will besomething like sqrt(6/pi^2 )*(1/2, 1/3, 1/4, 1/5, ...). So the coordinatesare (1+1/2*sqrt(6/pi^2), 1+ 1/3*sqrt(6/pi^2), ...).In fact, if we say that a vector X = (x_0, x_1, x_2, ...) where the x_nsare real,vectors that have finite length are those such that,summing over the naturalnumbers,Sum(x_n^2) is finite. All finite length vectors of are of this form, andall vectors of this form have finite length.All sums in the following are infinite sums.Another problem with this proof is that it assumes that if you cant add avector of length one along the line determined by (1,1,1,...), that thisimplies the existence of infinitesimals. Since no such vector exists, thisonly implies that no such vector exists. Suppose such a vector exists.Then (a,a,a,...) has length 1. So sum(a^2) = a^2 + a^2 + a^2 + ... = 1^2 =1. Subtracting a^2 from both sides we have a^2 + a^2 +a^2 +... = 1 = 1-a^2.1 = 1-a^2 => 1-1 = -a^2 = 0. But then sum(a^2) = 0. This is acontridiction, so no such vector exists. Additionally, no real numberexists such that a + a + a + ... = 1.Justin Van Winkle> Set up a coordinate system in four dimensions and consider the line> segment starting from the point (0,0,0,0) and ending at the point> (1,1,1,1) the lenght of this line segment is 2 so that in general the> lenght of a line segment similar to this is the square root of the> number of dimensions. In infinite dimensional space we have a line> segment starting from the point (0,0,0...) and ending at the point> (1,1,1...) that has an infinite lenght. go to the end of this line and> add one units lenght to it. What are the coordinates of this new> point? It cant be a real number greater than one because the distance> between any two real points would result in an infinite lenght.> therefore the coordinates of this point are an infinitesimal distance> past sha1:uaW4RDErZw5TI89Kun84aC3nODs=>What is the categorical definition of a group, (instead of the usualldefinition based on binary operations). Not sure if this is what you mean, but a group is a category with only> one object in which every arrow is invertible.He may also mean group in a category.A group in a category C with binary products and terminal object is atriple G, m:G x G -> G, i:G -> G> satisfying ...For ..., draw commutative diagrams which express the usual equations,such as G ------> G x G | |! | | m v v 1 --------> G eThis diagram expresses that g x g^{-1} = e. Ill bet hes asking for what you said, but just for completenesssake I mention it.-- A set having three members is a single thing wholly constituted byits members but distinct from them. After this, the theologicaldoctrine of the Trinity as ïthree in one should be childs play. --Max Black, having the following equation :S = { [(x%n) + 1] % p for (x/n) even { [(x%n) + 1 + p/2] % p for(x/n) oddand p>1Its easily proven that S(a) = S(a+2n) because of the fact they are both evenor both odd for starters and x%n = (x+2n)%n.Problem is I wish to prove that S(a) != S(a+1) but i have no idea on how tohandle that equation due to the multiple modulos in it. I know you cantjust get rid of the modulo signs and need to add an extra factor (say b*p)but even then im still stuck. Any crank alert]James Harris> Rather naively I thought that if you put out something simple enough> that most people could understand it,www.crank.net/harris.htmlor (more fun) do your own googling with shrewd combinations of keywords. Itried moron factorization bulland Google found James Harris in Please!X-DMCA-Notifications: http://www.giganews.com/info/dmca.htmlA phase transition is when a system resides atthe critical point between its static and chaoticattractors. Such as a cloud residing at the criticalpoint between water and air.To create a Grand Unified Theory we mustfind a way to unite all the realms of the universeinto one model of understanding. Uniting thefour forces is not the answer. We must unitethe realms of quantum motion, classical motionand evolution into one seamless view, explainablewithin one frame of reference and mathematics.We must unite Darwin, Heisenberg and Einstein.This can be done by replacing a model that examinesthe specific nature of things such as matter, lightand life, with a model that uses the ...behavior ofthese realms at their phase transitions.As matter, light and life all experience this behavior.Which is when they stand poised at the boundarybetween its static and chaotic tendency.For example. In a system comprising matter, lightand energy; matter would reside in the staticattractor basin, while energy defines the chaoticattractor. When those two are at the boundary orcritical point between each other, when one cant tell i§ight is generated. Light would fill the dynamicattractor that results from the critical interactionbetween static and chaotic states. Much asa cloud is the dynamic attractor that results fromthe critical interaction of water and air.Another example. In a system comprising gravity, classicalmotion and cosmic expansion; gravity would reside in thestatic attractor basin, while cosmic expansion woulddefine the chaotic attractor. When those two are at theboundary between each other, when one cant tellif its contracting or expanding...matter or energy....thenthe dynamic attractor of classical motion is generated.These two examples provide the smallest and largestscale phase transition states in the universe. Withlight being the observable result of the smallest transitionstate, and classical motion the observable result ofthe largest scale phase transition.In complexity science, ïcomplex is defined to be when asystem resides midway between its own extremesin possible behavior. Or simply at the phase transition statewithin its own possibility space.Since quantum motion defines one extreme, and classical motionthe opposite extreme in possibility space, the two compriseendpoints of a single system. The midpoint, or the phasetransition state between the resulting dynamic attractorsof light and motion, would be order. Or self-organization.When using phase transitions to model the universe, we cannow treat quantum motion, evolution and classical motionwithin the same system or frame of reference. And use themathematics of phase transitions for all three.See links below for this math.One model can explain the quantum and classical endpointsin possible motion, while also defining the structure withinwhether material or living order. Nothing in the universeis left out.Self-organization or evolution occur at the critical pointbetween ...light and motion.Darwin, Heisenberg and Einstein United!JonathanAn Introduction to Complex SystemsTorsten Reil, Department of Zoology, University of Oxfordhttp://users.ox.ac.uk/~quee0818/complexity/ complexity.htmlSelf-Organizing Systems (SOS) good, inexpensive math software?> For various obscure reasons Ive needed an inexpensive shareware> software which would calculate basic arithmetic functions, ln(n),> e^n, roots, and the basic trig. functions,...all out to pretty> high decimal points, say 50 or so (or more depending on how playful I get).Maxima (http://maxima.sourceforge.net) has the capability to set the number of decimal places for calculations, and supportsthe functions mentioned. Maxima is a general symbolic algebra program, a descendant ofDOE Macsyma (but not commercial Macsyma). It is quirky but hasmany useful weaponsCommentary 3The hyperspace H consists of fibers f(x) that areeither copies of or representations of the symmetrygroup G.>Well, that convinces me... It should. They are bible fibers. Wouldnt want to blaspheme.> Why not? I guess you have never be surrounded by a pack of ravenous bible thumpers.-- Oddly Syria will not declare its Golan Heights a militarybombing and artillery range and advise everyone to leavethe area for their own safety. -- The Iron Webmaster, Harris can tell us any of Harps theories, butcn you reprise the basic elements?... anyway,he was, according to his memoir, cut out of a lotof dark skies time by the establishemnt,who seem to really love the Hubble interpretation. my understanding of Arps correlations was that a)they were statistically unlikely, and b)there are obvious/apparent *connections* between many of them. in any case,teh primary argument is just that the redshift doesnt haveto be interpreted as a Doppler shift; does it? > Arps data is presented in the _Atlas of Peculiar Galaxies_, > Astrophysical Journal Supplement Number 123, Volume 14, November 1966. > These remaining cases do not constitute an overthrow of the BB, > to do that would require high quality observations of difficult > objects; big results require big data, obscure, difficult cases > do not provide that. > For a usefull review of the state of BB theory have a look at> Astronomy and Astrophysics_, _Standard Cosmology and Alternatives:> anti-BBers ) admit that the reason most cosmologists like the BB> is that the thoery makes many predictions that both occour naturaly> in the theory and have been confirmed observationaly.--ils duces I NEED HELP BADLY (sorry, maths not psych)Expires: 28 days>But what are YOU saying?Not untill the electrode 1 km away feels the opposing force? IS THAT WHAT YOU ARE SAYING? NO PAUL. I am saying that your claim infers that the effect of an injected electron will be felt INSTANTLY at the far electrode.I have understood that you think my answer is wrong.I am asking you: WHAT IS YOUR ANSWER?>Since you claim that the force cannot act instantly,>I am asking you, WHEN will the force act?>You claim that the distance to the other electrode is relevant.>WHY is it relevant?>How does this distance affect the delay before the force>on the electron starts acting?This is a concrete scenario.>Please state what you think will happen.You keep talking about the force on the electron. Im more concered about tehforce at the far electrode. > Of course, since the electron existed BEFORE it was injected, the effect would have already been there even though the near electrode was in the way. The only way this experiment can even be hypothesized is by either ïannihilating a very large number of electrons or by monitoring the force on the far electrode with movement of the electron mass towards it.Say, what the hell are you babbling about?>Dont tell fairytales about suddenly disappearing charges!What happens after this: (e+) + (e-) = ?ADDRESS THE SCENARIO GIVEN!It is a concrete scenario which in principle could be made.>(And which EASILY could and IS made with shorter distance>between the electrodes. You have it in your TV set!)Let it be a hot wire in the hole in the electrode.>Thermionic emission of electrons.>When one of these electrons gets out of the hole>and into the static field between the electrodes,>how long time will it take before a force start acting on it?I could probably write a whole book answring that.However I would conficently say that, before it starts to move, the force isinstant. My answer is:> as it enters a static electric field.WHAT IS YOUR ANSWER?I just gave it to you.; Now please answer MY question.If a highly charged sphere is moved, say, backwards and forwards between twoelectrodes, what happens to the force on those electrodes. Does it changeinstantly or is there a time lag?Hint: neither you nor anyone else knows the answer. Henri Wilson. See the Stupidity Re: Numerical integration, prime counting>Ive repeatedly pointed out that I found a partial difference equation>which when integrated over a certain range gives a count of prime>numbers.And people have repeatedly pointed out that integration is the> wrong word here. But why should you care about that? It should> be up to us to figure out what you mean - silly requirements like> saying precisely what you mean only apply to lesser minds. Right.>Now I want to talk about going from the partial difference equation to>a partial differential equation.Its also been repeatedly pointed out that that thing you get> is _not_ (quite) a partial differential equation. But never mind> that...>Here again is the difference equation and the instructions for the>integration:>[...]>Thats easy to program for those interested in doing the calculations.>That is, now using non-discrete variable types like doubles, you can>program that in and let delta y approach 0.>However, it does take a *tremendous* amount of computing power or>time, as delta y drops.>Ive done a few calculations and found interesting results.>Basically, summing the partial differential apparently gives you a>close approach to the prime counting distribution, which is closer>than li(x) itself!And its been repeatedly pointed out that youve given _no_> evidence that the solution to the pde should have anything> whatever to do with counting primes.Hint: The fact that you _say_ something _appears_ to be so does not> count as evidence, because you have zero credibility left after> making so many false statements.Why should JSH have zero credibility, while--after your blunders,you, a Ph.D. in mathematics and professor of same--retain yours?Could it be that that what commands respect on sci.math and sci.logicis not mathematical prowess, but a stop-at-nothing determinationto silence or neutralize dissenters from the canon?>Im curious about replies from posters with experience doing numerical>integrations of partial differential equations.James Harris>My math discoveries, found for Correy:Dont *all* of the following indicate that you had *no idea* (and*still* have no idea) wh Ex~(x=x) follows from C1-C4?C1 AxAy[x=y -> Az(z in x <-> z in y)]C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] C3 EyAx[x in y <-> Et(x in t) & A] (with y not free inA)ClassificationC4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] (WeakExtensionality)David Ullrich:If you meant NGB set theory then no, C1-C4 arenot inconsistent with set theory. It does _not_follow that C1-C4 give an example of somethingwhich is not equal to itself, or an example ofsomething which does not exist. It is correct that I have no idea why Ex~(x=x)follows from C1-C4. This is because (assumingthat NBG is consistent) NBG has a model inFOL=. In that model everything is equal Correy:You have no idea why Ex~(x=x) follows from C1-C4because you are brain-dead analysis teacher whocan only work with the routines he has memorized.David Ullrich:Could be. Now show us why Ex~(x=x) _does_ followfrom Ullrich:Exhibit of proof of Ex~(x=x) from C1-C4 and someone (sorry, maths not psych)Expires: 28 daysStatic. Dynamic. Close. Far.Moron. - Randy Randy, you are so ignorant, you dont even realise that the introduction of an electron (actually lots) will affect the field.Amazing how much verbiage you can introduce without >answering ïs simple question.How long after the electron enters the field will>it be affected?Randy, havent you noticed that whenever is in trouble, he always attemptsto hijack the converstaion by diverting the subject down a side track.Imagine two plates generating a static electric>field. An electron passes through a hole in the plates.>It is moving at 100 m/sec. How long before it is acted>on by the field? One second? One microsecond? One nanosecond?>When does the velocity start changing from 100 m/sec? How>far does the electron get before this happens?That is not the question I have raised.See my latest answer to . - RandyHenri Wilson. See the Stupidity of Online Calculus III (MultiVariable) classEveryone;Im trying to find sources for reasonably priced, good math coursesonline. Im specifically looking for Calculus III and beyond. Im notinterested in any of the pseudo schools that want your money to sellyou a degree, I want the information.. Im limited from pursuing thisin a formal school environment due to disabilty although over theyears I did manage to struggle through the first of the Physicssequence and Calculus II. Is there an instructor out there that might consider a self-study withme? I am slow so a self-paced program would work best. A dimensional analysis.The cars mass 1134 KgObjects Mass 68 Kglength 22.861 M> Im trying to find a formula for where I can measure the speed of a carwhen> it hits a still object based on cars weight, objects weight, and total> distance the object was thrown. I realize that there are many otherfactors> such surface friction, in this case road, but im just looking for NEED HELP BADLY (sorry, maths not psych)Expires: 28 daysHenriWilson skrev i melding I still cannot see the philosophical reason for using F=dp/dt rather than F=m.dv/dt where m=f(v)Its Newtons second law of motion in its original form.>The change of motion is proportional to the force impressed;> and is made in the direction of the straight line in which the force> is impressed.What did Newton mean by motion?>Those who have studied his texts think he meant momentum.>It is interesting because the term ïv.dm/dt is explains the energy increasethat is normally associated with ïrelativistic mass increase.It actually supports my argument that this energy really goes into the ïreversefield bubble that forms around a moving charge.Henri Wilson. See the Stupidity of I NEED HELP BADLY (sorry, maths not psych)Expires: 28 days>And you have not provided any theory of E&M that allows any such>thing as a reverse field. Nor why there should be any kind of>speed limit involved. Nor why it should follow any such thing>as the kinetic energy formula observed in accelerators. Nor have>you provided a relation between energy and mass if you dont>accept relativity.>Socks radiation from an acceleraed charge! fields associated with a moving charge! The ïBack EMF concept. I would be most amazed if a moving charge DID NOT alter the field around> itself, wouldnt you?Quite.are accelerated.You KNOW the following, Henry.In an accelerator going at full efficiency, we KNOW thatbecause it looses this energy as synchrotron radiation in the bendsof the circuit.(Very obvious and easily measurable.)So we - and YOU - know that the RF-cavities never ceasesis only few mm/s below the speed of light.So why do you keep pretending that the E-field is notspeed approaches c, when you KNOW that isnt true? I DID NOT SAY THAT.Indeed you did.> The question is how much energy?Forgotten that too?Come on , you are being silly now. You know what ïasymptote means.> You are making no attempt to answer that one. In typical fashion, you pretend the relevant question does not exist.A blatant lie.>Henri Wilson. See the Stupidity of is tan(n)/n unbounded?> In another thread the sequence {tan(n)/n} arose (n=1,2,3...). The> discussion showed that this sequence does not approach zero. Now> |tan(11)/11| is about 20. So I wondered: does tan(n)/n (or |tan(n)/n|)> take on arbitrarily large values? So far the largest value I have found> is 556.3, but n is very large.> I didnt see anything in the other thread that answered this. Does> anyone know?You might look at the last part of> .The last sentence of that says The sequences of maxima and minima are> almost certainly unbounded, but it is not known how to prove this fact.> Disclaimer: The wording in that last phrase is Erics, not mine. (Of> course I suspect that the sequences are unbounded. But I certainly wouldnt> call it a fact.)David, unless Im going crazy, the n values in the sequence run out before the tanc() values. What are the two missing n values that correspond to the two highest tanc()s? (OEIS doesnt have them either).Phil-- Unpatched IE vulnerability: Web Archive buffer over§owDescription: Possible automated code execution.Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0303/107.html== === =Subject: Re: Reconsidering Halton Arp But then theres Dr. Halton Arp.You see, Dr. Arp is a scientist, a world renowned scientist and he has> *data*, real, hard astronomical data, which is more substantive in> disproving the commonly taught Big Bang Theory, than the data used to> support that theory.So it should be a no-brainer, the theory has to be changed to fit the> data.Thats not what has happened.Nor should it have happened, simply because one person came up withdata one time that appears to contradict established theories. Thisreally touches upon an interesting and important set of principles.Because physics, astronomy, and many other sciences have strongexperimental (that is, observational) components as well astheoretical components, there exists that possibility thatobservations are incorrect or misinterpreted. When there arethousands of observations that support a theory, then one observationthat contradicts it will not be used to throw it out. It is used as acaution. It is used to say probably, but maybe not. It is used totry to push back the frontiers of knowledge. And these are done onlywhen that observation can be replicated.I understand that somebody of the stature of Lord Kelvin, possiblyKelvy himself, came up with a theoretical calculation that proved thatthe Earth could not be as old as the geologists were claiming becauseit would have lost all of its heat long before and would no longerhave supported life. This calculation had some emotional appeal(positive and negative) as well as theoretical appeal, since it tendedto support certain popular religious beliefs of the time (whichhavent actually entirely gone away). At the time this calculationwas made, nobody knew about radioactivity, and so the calculationswere not appropriate to the real situation and therefore effectivelywrong. Thus they didnt end up overthrowing established geology, butthey must have scared a lot of people.And then there is the fact that the theories of scientists dontprovide us complete and total understanding of the universe. Therefore data which serve to delimit and expand the frontiers ofknowledge are treasured (mostly), but they dont generally serve tothrow out all of established physics. That is unlikely given theoverwhelming experimental evidence for so much of it. They can causesome theories to change; they can cause some theories to be tossedout; They can cause things that were simply not understood to be fitwithin a framework so that they are finally understood.It is extremely naive to think that discoveries such as Arps shouldcause a rapid upheaval in astronomy. They should be and are takenseriously, examined, and fit into the Reconsidering Halton ArpSo I have my math results, which I think are rather simple, and nowits a matter of presenting them to the world, right?One would think so... however, what is important in that presentation> is that you make no jumps in the proof of your argument. For example,> you have repeatedly claimed this:Let(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where the as are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)But you havent shown us once what the roots of that latter equation> actually are. My advise to you is that you show us. Give us a_1(x),> a_2(x) and a_3(x). If you can present the roots then we can verify> that those as do indeed lead to both equations.> Its only a bit tedious to find the roots. First, let t = -1 + 49*xto keep matters simpler and also denote (-1 + sqrt(-3))/2 by theta.Then, the three roots are-t + P + Q,-t + P * (theta) + Q * (theta)^2-t + P * (theta)^2 + Q * (theta)whereP = ((1 - t^3)/2 + (1/2) * sqrt((1 - 3*t^3) * (1 + t^3)))^{1/3}Q = ((1 - t^3)/2 - (1/2) * sqrt((1 - 3*t^3) * (1 + t^3)))^{1/3}Now take sundry values for x and see whether 7 divides exactly two ofthe roots in the ring of algebraic integers, as James claims. ArpRather naively I thought that if you put out something simple enough> that most people could understand it, then theyd be willing to at> least question authority long enough to see when authority figures get> it wrong.Arp has done much good work. Heres some of it: Atlas of Peculiar Galaxies http://nedwww.ipac.caltech.edu/level5/Arp/frames.htmlCatalogue of Southern Peculiar Galaxies & Association http://nedwww.ipac.caltech.edu/level5/SPGA_Atlas/ frames.htmlCrank Information http://www.crank.net/harris.html http://www.crank.net/usenet.html another thread the sequence {tan(n)/n} arose (n=1,2,3...). The discussion > showed that this sequence does not approach zero. Now |tan(11)/11| is about > 20. So I wondered: does tan(n)/n (or |tan(n)/n|) take on arbitrarily large > values? So far the largest value I have found is 556.3, but n is very large.Well, tan(n), which is always defined, can take arbitrarilylarge values, and I think it is easy to prove (please, *actualmathematicians* correct me when Im wrong :-))First, given any real number x that is not an rational number,then it is true that for every epsilon > 0, there exists arational number r such that |x - r| < epsilonThe proof is straightforward: Pick any two fractions r1 andr2, with r1 < x, r2 > x. Take r = (r1 + r2) / 2. r isobviously a rational number, and |x - r| < |r2 - r1|/2.Given that x is irrational, r can not be x. So, eitherr > x, or r < x. Pick r and either r1 or r2 to have twovalues surrounding x (if r < x, then pick r and r2; ifr > x, pick r1 and r). Repeat the process.After N times, |x - r| < |r2 - r1|/2^N.With N sufficiently large, the difference can be madearbitrarily small, and thus less than epsilon.Now, we take a rational approximation of pi. Lets call itp/q Then, there exists integer numbers N and k such that|N - p/q (k + 1/2)| is arbitrarily small|N - p/q (k + 1/2)| = |2qN - p(2k+1)| / 2|q|So, pick q arbitrarily large (by multiplying both numeratorand denominator by whatever number is necessary), and thenpick N = p and k = q, such that the above expression isequal to 1 / 2|q|But since we can pick a rational number that is arbitrarilyclose to pi, and given that rational number, we can find Nand k such that (k+1/2) times that rational number is aninteger number N, then we conclude that tan(n) can bearbitrarily large -- since there is at least one n thatcan be arbitrarily close to pi(k+1/2) for some k.Now, does that mean that tan(n)/n takes arbitrarily largevalues as well?? (intuitively, it seems to me that it does,since tan(x) goes very steeply to infinity; but its notclear, from the way to construct the N required to bearbitrarily close, if that N has to be so large that itwill counter the effect of the large === value of tan(N))Carlos--Subject: Re: Hello Sci.* NG, Do You went to Indian Institute of Technology and started expalaining very>basic idea of this device. But they raised problem which does not>exist at all. What I tried to convince them is given in my homepage.>Please note that I have changed my homepage from previous one.http://www.geocities.com/actiondeviceWhat those few of very brilliant people in India insisted that point B>will move along Y axis. I agree that due to forces acting at point>B, third vector will be produced and point B will move in space in>perpendiculer direction to XY plane. But those people insist that it>will move along Y axis !http://www.dishman.me.uk/George/Abhi/abhi.gifIf you stretch the springs by moving A to E andC to F, the forces exerted by the springs are shownby the thick red arrows. To add the forces, the thinred lines make them into a parallelogram and the greenarrow shows the resulting total. Dont take my wordfor it though, check how to add forces in any mechanicstext book.To stop B moving, Newtons law says you need an equaland opposite reaction which is shown by the blue arrow.Without that, B will move in the direction of the greenarrow. That is along the Y axis so they are rightaccording to the standard rules of mechanics.>People in this NG will not answer my question clearly.They all said B will move down the Y axis. That is avery clear answer.Think of the line EF as a bow and the springs as thestring. Which way will the green arrow go?George> Roger that.But my illusion will come to an end if few other people in this NG> confirm what you have said. I will repeat zillion times that in this> V-shaped spring, angle ABC is solid angle. Please read it again.I interpret this to mean you have somehow stapled the springs at point B. If you have not secured the springs at be, the ends *will* move, no matter what you say to the contrary.I recommend that you take a math course that will teach you how to work with vectors. It seems clear from your responses that you do not understand how vectors Reconsidering Halton ArpRather naively I thought that if you put out something simple enough> that most people could understand it, then theyd be willing to at> least question authority long enough to see when authority figures get> it pointed out grave> procedural errors and §at out empirical wrong results in your most> recent spews. You responded like a diversity hire - turning your> back, screaming ïDISCRIMINATION!, shuf§ing down the same incompetent> road some more... all the while demanding a larger paycheck.http://www.crank.net/harris.html> Its not every braying jackass that gets a whole page at crank.netHey stooopid, we enjoy cleaning up your messes like we enjoy scraping> dog off our shoes on a curb, distasteful but expedient. Hey> stooopid, the whole world is against you for good and just objective> reasons. Take a pooper scooper to yourself.James,Tom Kirkes response above is an accurate summary of the status ofArps work. There are a few instances where two galaxies withdifferent redshifts appear to be dynamically interacting, which wouldof course be impossible under the accepted interpretation of redshift.This occurence is an artifact of looking through a three dimensionalvolume --- sometimes objects at different distances will lie close toeach other (or even on top of each other) on the sky plane. I dontthink that there is any evidence that this happens more frequentlythan expected. There is no big conspiracy here ---- it has a verysimple and expected geometrical explanation. By the way, most of Arpspeculiar galaxies really are interacting systems where the constituentmembers have the same naively I thought that if you put out something simple enough> that most people could understand it, then theyd be willing to at> least question authority long enough to see when authority people have pointed out grave> procedural errors and §at out empirical wrong results in your most> recent spews. You responded like a diversity hire - turning your> back, screaming ïDISCRIMINATION!, shuf§ing down the same incompetent> road some more... all the while demanding a larger paycheck.And sci.phy-sickies call Uncle Al a cultural analysis, because I like maths> a lot and my undergraduate final work was about iterative methods for> regularization problems. I am an engineer and in the faculty, they are requesting me an essay> about why should I study analysis. I must make a good writing because> my admission depends on it. I would like you to help me and give me some useful comments. question as I am an engineer by trade (MS), butalso possess an MS in Mathematics.I have found both disciplines to be quite complementary, but I must say thatmy mathematics training has greatly enriched my everyday work life.Mathematics training has taught me a very disciplined approach to problemsolving in general and this is useful in my profession.An area like analysis really allows you to develop different ways ofapproaching problem solving than is typical in an engineering curriculum.In fact, universities are finally realizing that mixing of differentprograms can be very beneficial to both students and industry. For example,look at Computational Science programs and it is clear that they will beuseful and may replace the typical engineering only versus math onlycurriculums around the world.In todays world, we need people who can attack problem solving from manyangles and having the ability to apply tools that you will learn in analysiscan only help to aid you. I think I would try to focus on the fact thatthis analysis curriculum may help to provide a foundation for research andhow to apply some of these tools to the engineering profession itself.Anyway my 2 cents, HTH, is a scientist, a world renowned scientist and he has> *data*, real, hard astronomical data, which is more substantive in> disproving the commonly taught Big Bang Theory, than the data used to> support that theory.Arps data is presented in the _Atlas of Peculiar Galaxies_, > Astrophysical Journal Supplement Number 123, Volume 14, November 1966.In the Atlas Arp presents galaxies that appeared abnormal. Follow-up> observations showed that some, not all, of the galaxies were in fact> two galaxies that are apparently interacting. What caused the doubt > about the Big Bang was that some of these pairs have very different> red-shifts. If the galaxies are close to each other the different> red-shifts would sound the death knell for expansion and the BB.However, as observing technique has improved weve determined that> most of these pairs are simply close in the line of sight and are> at very different distances. There are a few cases that have not> been elucidated, the necessary obsrvations are, at best, difficult.These remaining cases do not constitute an overthrow of the BB, > to do that would require high quality observations of difficult > objects; big results require big data, obscure, difficult cases > do not provide that.The Atlas was and remains an important astronomy tool, the carefull> work on its members has provided important insights and data about> galaxy formation and interaction. The value of a scientific idea> is not whether it is right or wrong, but rather the questions it leads> us to ask. - Dont remember who said that, wish it was me.For a usefull review of the state of BB theory have a look at> Astronomy and Astrophysics_, _Standard Cosmology and Alternatives:> anti-BBers ) admit that the reason most cosmologists like the BB> is that the thoery makes many predictions that both occour naturaly> in the theory and have been confirmed observationaly.As we say in sci.astro,Dark skies,tomWish there were more like you around. Unfortunately, the Uncle Al typespredominate in these the sequence {tan(n)/n} arose (n=1,2,3...). Thediscussion> showed that this sequence does not approach zero. Now |tan(11)/11| isabout> 20. So I wondered: does tan(n)/n (or |tan(n)/n|) take on arbitrarilylarge> values? So far the largest value I have found is 556.3, but n is verylarge. Well, tan(n), which is always defined, can take arbitrarily> large values, and I think it is easy to prove (please, *actual> mathematicians* correct me when Im wrong :-)) First, given any real number x that is not an rational number,> then it is true that for every epsilon > 0, there exists a> rational number r such that |x - r| < epsilon The proof is straightforward: Pick any two fractions r1 and> r2, with r1 < x, r2 > x. Take r = (r1 + r2) / 2. r is> obviously a rational number, and |x - r| < |r2 - r1|/2. Given that x is irrational, r can not be x. So, either> r > x, or r < x. Pick r and either r1 or r2 to have two> values surrounding x (if r < x, then pick r and r2; if> r > x, pick r1 and r). Repeat the process. After N times, |x - r| < |r2 - r1|/2^N. With N sufficiently large, the difference can be made> arbitrarily small, and thus less than epsilon. Now, we take a rational approximation of pi. Lets call it> p/q Then, there exists integer numbers N and k such that |N - p/q (k + 1/2)| is arbitrarily small |N - p/q (k + 1/2)| = |2qN - p(2k+1)| / 2|q| So, pick q arbitrarily large (by multiplying both numerator> and denominator by whatever number is necessary), and then> pick N = p and k = q, such that the above expression is> equal to 1 / 2|q| But since we can pick a rational number that is arbitrarily> close to pi, and given that rational number, we can find N> and k such that (k+1/2) times that rational number is an> integer number N, then we conclude that tan(n) can be> arbitrarily large -- since there is at least one n that> can be arbitrarily close to pi(k+1/2) for some k. Now, does that mean that tan(n)/n takes arbitrarily large> values as well?? (intuitively, it seems to me that it does,> since tan(x) goes very steeply to infinity; but its not> clear, from the way to construct the N required to be> arbitrarily close, if that N has to be so large that it> will counter the effect of the large value of tan(N))>My hunch is that if n mod 2pi is random with a uniform distribution on[0,2pi) and dense then tan(n)/n can take arbitrarily large values becausethe slope of tan(x) tend to infinity. Like you I have a gut feeling but cantprove it.Alternatively maybe the recurrence relation for tan(n)/n can help:tan(n+1)/(n+1) = that infinitesimals existThe barbarians are at the gate being controlled very cleverly. Dont believe me?> People in this NG will not answer clearly the question I have posed.> Will point B move along Y axis in XY plane? It needs just yes/no.> But they will remain silent(or they will be humorous). They will> ignore me. Because they are controlled. No, Abhi, you have been answered. Repeatedly. Read my reply from acouple of> weeks back. Asking whether the point B moves is a little meaninglessgiven> your defined frame of reference. But there will be a net force acting on> point B towards point D, exactly the same as the net force acting onpoint D> towards point B. So the rod BD is subject to a compressive force.> Krill I am really finding myself in con§agration. Things are still under> His absolute command including your mind.>I dont think so, Abhi. Were just trying to explain why your device doesntwork as you believe. Maybe you could consider that maybe, just maybe, allthese people who know about this stuff (I have two engineering degrees, bythe way) and are trying to tell you it wont work are right and it is youwho is wrong. Not some dark conspiracy or anything like that.> I went to Indian Institute of Technology and started expalaining very> basic idea of this device. But they raised problem which does not> exist at all. What I tried to convince them is given in my homepage.> Please note that I have changed my homepage from previous one. http://www.geocities.com/actiondevice What those few of very brilliant people in India insisted that point B> will move along Y axis. I agree that due to forces acting at point> B, third vector will be produced and point B will move in space in> perpendiculer direction to XY plane. But those people insist that it> will move along Y axis !where is the force acting anywhere other than in the XY plane that wouldcause this to happen? The system you describe is entirely planar. Now I posed the same problem before you people and what you said that,> Asking whether the point B moves is a little meaningless given your> defined frame of reference. This is definitely not answer of my> question. (1) Those people in IIT said that point B will move along Y axis.> (2) You said that it is meaningless in given frame of reference. What exactly is going on around us, Krill?I said it is meaningless unless you specify your frame of reference. You areusing some unusual terms which may be confusing. When you say that ABC is asolid angle I interpret that to mean that the angle is somehow made rigidby means of spring devices that do not §ex. Maybe others assume somethingdifferent.What is certainly the case is that there is a force exterted on the rod BDat point B towards D. So B is being pulled along the Y axis. But withinthe reference frame of your device, it wont move. Costraining the lengthBD, the length DC and the angle ABD to be invariant means the shape cantchange. Perhaps you dont know, but now I know that HE has absolute command> over everything in this universe including conscious mind of human> being. You dont know, but HE knows very well magnitude of this> device. It is going to change course of history, physics and it is> going to open gateway to whole universe.>Abhi, just go and build the damn thing already.> This device is very simple to build but HE knows very well which> things in my personal life can crash me and He has done it in most> horrible, brutal way with perfect timing leaving me struggling with> myself. People in this NG will not answer my question clearly. I know WHY?>How clear an answer do you need Abhi?Krill> He is in absolute command. psych)How long after the electron enters the field willit be affected?Randy, havent you noticed that whenever is in trouble, he always attempts>to hijack the converstaion by diverting the subject down a side track.I have noticed this in conversations between you and . Between youand me, too. However, it isnt who does not. Nor is it me.When does the velocity start changing from 100 m/sec? Howfar does the electron get before this happens?That is not the question I have raised.I know. Its the question he was asking you, that you didnt answer.At least you admit that rather than answer the question, you changedthe subject. You raised this question Off the Walls (of a hexagon) The ray goes from F to E (no crossing), then bounces back between> the (just created) F-E ray and the original A-E ray. It then hits> the last segment of the D-F ray and is re§ected back to F.While this is the topological restriction from the problem data, Idid the angle chase last night and this just does not work -- afterleaving E the ray cannot intersect the last segment of the D-F ray.So I dont think there is a solution after all.This leaves four possibilities:1) I stuffed up the angle chase, and it is possible after all;2) There is an alternate topology that I have missed;3) The information provided about the edges and crossings was wrong; or4) The original diagram was just too inaccurate.Im leaning towards 4 at the moment, with a side order of 3. But Imwilling to be convinced otherwise. === Subject: Re: Apocalypse NOW! stiffness and both springs are in relaxed state initially. Angle ABC> is solid angle . Let angle ABC be 60 degree in this figure(for the> sake of explanation only, in actual Action Device this angle will be> very small).Im assuming that by solid angle, this angle is somehow made rigid.Slightly odd, but it can be done. You could make each of the members AB andCB out of two concentric tubes, able to slide in and out, with an internalspring. Is that kind of what you have in mind? At t = 0, we apply same magnitude of force on point A and C and we> pull point A of spring AB towards point E in the direction of line BE> which makes 60 degree angle with X axis and we pull point C of> spring CB towards point F in the direction of line BF which also makes> 60 degree angle with X axis. And we are pulling point A and C in> such a way that these points must stretch or extend to point E and F> resp. on x axis. Please note that we are pulling points A and C in the direction which> makes 60 degree angle with X axis. We are NOT pulling point A and C in> downward direction.>But you are, as everybody has tried to point out ad nauseam. Using yourcartesian system, and some notation I introduced in an earlier post, theforce exterted by the springs AB and CB on the point B can be decomposedinto an {x,y} vector as:{ -F sin(alpha), -F cos(alpha) } and{ F sin(alpha), -F cos(alpha) }respectively. The two springs exert no net force along the x axis on B,but -2Fcos(alpha) along the y axis (the minus sign indicates that the forceis towards the origin, or downwards in your chosen orientation.> At t = t, spring AB is stretched and point A of spring AB reaches to> point E on X axis. Also at t = t, spring CB is stretched and point C> of spring reaches to point F on X axis.>yes, we got that.> We will find that point B has not shifted its position along Y axis.> It remains where it was at t = 0. Because if point B shifts its> position along Y axis, to say, point B, angle ABC (or EBF, because> point A coincides with point E and point C coincides with point F)> will be different from angle ABC i.e. greater than 60 degree. But as> stated above, angle ABC is solid angle which does not change due to> forces acting at point B.this implies that in order to maintain your solid angle (which seems to bevalidated by your explanation to mean rigid) there is a torque exterted atB on each of the members to maintain that.Nonetheless, just because point B is constrained not to move in your frameof reference, doesnt mean the force goes away. That was what I was talkingabout before. The fact that B doesnt move relative to E and F just meansthat you have redefined the boundary of your closed system. You are relyingon an external force at E and F to prevent *the whole thing moving*.When you throw in the other components of your device needed to complete theclosed system with no external forces applied, it all space but certainly not along> Y axis or in XY plane. Third vector will be produced at point B> direction of which will be perpendiculer to XY plane. Am I right or wrong?wrong. Where is a force perpendicular to the XY plane in your Bundle Physics/Consciousness 2)] B L U R ...> http://www.astrocentral.co.uk/beagle.html Constraining the Topology of the Universe> http://arxiv.org/abs/astro-ph/0310233 Commentary 2> [...]> One can also imagine a fiber of strings of qubits.> 1 qubit is a parallel infinity of c-bits.> i.e.> |qubit> = |1 c-bit><1c-bit|qubit> + |0 c-bit><0 c-bit|qubit > Where there is a continuous infinity of different c-bit bases> or orthonormal frames each corresponding, for example,> the the angular orientation of an inhomogeneous field> magnet in a Stern-Gerlach filter for spin qubits> in the DARPA spintronics project or like the billion billion> Single Electron Transistors inside the human brain at the> sub-microtubular protein dimer hydrophobic cage level forming> the hardware interface with external world whose software is our stream> of inner consciousness.> [[[[(({}}}]]]]> Each possible orientation is a primitive parallel quantum universe.> The quantum computer computes in all possible> orientations simultaneously like a continuous> infinity of classical Turing machines in a> distributed network working on the same problem> - or so the folklore goes.> as one fifth of all species known to exist> today. In recent centuries, hundreds of> species have disappeared, almost always> as a result of human activities.> http://www.worldwildlife.org/news/pubs/specieslist.html ~^~> What our Dreams Tell Us> Do our dreams give us messages from our bodies about health problems> we may not be aware of? The ancient Greeks thought that dreams> contained information that could be used to diagnose disease. With> some diseases, specific dreams are more likely to occur; however,> people who have the most severe cases of these diseases often say> they dont dream at all.Dr. Trisha Macnair> female heart patients dream of separation. Migraine sufferers have> dreams containing extreme fears (perhaps because they fear the onset> of another headache). People whose brain scans show signs of dementia> or brain shrinkage often dream about losing something, especially> money or food (ie. something essential).Victims of stroke, epilepsy or Parkinsons disease have noted changes> in the amount of time they spend dreaming and in the quality of their> dreams, which have fewer visual images. Theyre also less able to> remember their dreams. People with high blood pressure have dreams> filled with hostility (one of the causes of their problem?)Patients with narcolepsy, who find it hard to stay awake, dream> about strange and frightening events. People under the in§uence> of alcohol and drugs (including sedatives and antidepressants)> have nightmares when the drugs are stopped. Asthma patients have> very emotional dreams, perhaps because not being able to breathe> is such an emotional experience.People with psychosomatic illnesses (who tend to think theyre> sick when theyre not) have dreams filled with aggression, fear> and helplessness, which are probably the underlying causes of> this condition.Dreams occur during REM (Rapid Eye Movement) sleep. This is when> the brain is most active and our sleep is the deepest. People who> are deprived of REM sleep dont feel as if theyve slept enough.> It occurs roughly every hour to 1 hours, several times a night.> REM is tied to bodily changes in temperature, pulse rate, and> blood pressure, so the dreams that are produced can actually set> off heart attacks, migraine and asthma attacks. In South East> Asia there is a rare disorder where men die mysteriously in their> sleep, called Pok-Kuri, which may be caused by abnormal heart> rhythms during REM sleep. ~^~ Stuart Hameroffs Home Page:http://www.consciousness.arizona.edu/hameroff/ index.htmlThe Elegant Universe homepagehttp://www.pbs.org/wgbh/nova/elegant/ _________The Religious Experience of Philip K. Dick by R. Crumb http://www.philipkdick.com/weirdo.htmTHE POWER of NOWby Eckhart Tollehttp://www.eckharttolle.com/ ... It is finding your true nature beyondname and form. The inability to feel this connectednessgives rise to the illusion of separation, from yourselfand from the world around you. You then perceive yourself,consciously or unconsciously, as an isolated fragment.Fear arises, and con§ict within and without becomesthe norm. ... -- Eckhart Tolle http://www.eckharttolle.com/http://adidam.org/Yes! There is no religion, no Way of God, no Way of DivineRealization, no Way of Enlightenment, and no Way ofLiberation that is Higher or Greater than Truth Itself..... Therefore, Reality (Itself) Is Truth,and Reality (Itself) Is the Only Truth. -- Adi Da Samraj, a.k.a. Bubba Free John, a.k.a. Da Free John, a.k.a. Da Kalki, a.k.a. the Ruchira Avatar, Adi Da Love-Ananda Samraj, a.k.a. Franklin Jones http://adidam.org/ http://www.daplastique.com/home.htmlLet me share with you this little model Ive worked outabout who we are as human beings. I call it theThree-Plane Consciousness Model. If I were to take apicture of who I see you to be, the picture would showthree Is ---three different levels of who you are,planes on which you have an identity.Number One is what I call ego, thats the I we allknow very well, the plane of the body, mind, andpersonality; of all those things we think we are.Number Two I call the soul; the soul measures time notin days and years but in incarnations, and its the Ithat was around before we as egos were born and thatwill be around after we as egos die.And Number Three is ... just Number Three. We all havedifferent names for it, and wars are fought over whatto call it, so I avoid all that by just calling itNumber Three.I see our task as learning to live on more than one ofthose planes simultaneously, experiencing ourselves asegos and souls at the same time. And since you gottabe one to see one, once we are resting in our souls,then we will see others as souls as well. Then when welook into another persons well say, Are you in there?Im in here. Far out!When we are able to look behind even that identity assoul, well see that we have still another identitybecause we are also Number Three.Thats the mystic I, because in Number Three theresactually only one of us. Your Number Three isnt merelylike my Number Three---Theyre the same thing. -- Baba Ram Das, a.k.a. Richard Alpert http://ramdasstapes.org/index.htm _________Committee for Surrealist Investigation of Claims of the Normal [ CSICON ] http://www.rawilson.com/csicon.shtml < C O N T A C T >Uppaluri Gopala Krishnamurti (Born 9 July 1918)http://www.well.com/user/jct/mystiq1.htmTHE MYSTIQUE OF ENLIGHTENMENTPart One [Excerpt]U.G. KrishnamurtiPeople call me an ïenlightened man -- I detest that term -- theycant find any other word to describe the way I am functioning.At the same time, I point out that there is no such thing asenlightenment at all. I say that because all my life Ive searchedand wanted to be an enlightened man, and I discovered that thereis no such thing as enlightenment at all, and so the questionwhether a particular person is enlightened or not doesnt arise.I dont give a hoot for a sixth-century-BC Buddha, let alone allthe other claimants we have in our midst. They are a bunch ofexploiters, thriving on the gullibility of the people. There isno power outside of man. Man has created God out of fear. So theproblem is fear and not God. ______________I discovered for myself and by myself that there is no self torealize -- thats the realization I am talking about. It comesas a shattering blow. It hits you like a thunderbolt. You haveinvested everything in one basket, self-realization, and, inthe end, suddenly you discover that there is no self to discover,no self to realize -- and you say to yourself What the hell haveI been doing all my life?! That blasts you. _______________All kinds of things happened to me -- I went through that, you see.The physical pain was unbearable -- that is why I say you reallydont want this. I wish I could give you a glimpse of it, a touchof it -- then you wouldnt want to touch this at all. What you arepursuing doesnt exist; it is a myth. You wouldnt want anythingto do with this.UG: You see, I maintain that -- I dont know, whatever you callthis; I dont like to use the words ïenlightenment, ïfreedom,moksha or ïliberation; all these words are loaded words, theyhave a connotation of their own -- this cannot be brought aboutthrough any effort of yours; it just happens. And why it happensto one individual and not another, I dont know.Questioner: So, it happened to you?UG: It happened to me.Q: When, Sir?UG: In my forty-ninth year.But whatever you do in the direction of whatever you areafter -- the pursuit or search for truth or reality -- takesyou away from your own very natural state, in which you alwaysare. Its not something you can acquire, attain or accomplishas a result of your effort -- that is why I use the word ïacausal.It has no cause, but somehow the search come to an end.Q: You think, Sir, that it is not the result of the search? I askbecause I have heard that you studied philosophy, that you wereassociated with religious people ...UG: You see, the search takes you away from yourself -- it is inthe opposite direction -- it has absolutely no relation.Q: In spite of it, it has happened, not because of it?UG: In spite of it -- yes, thats the word. All that you do makesit impossible for what already is there to express itself. That iswhy I call this ïyour natural state. Youre always in that state.What prevents what is there from expressing itself in its own wayis the search. The search is always in the wrong direction, so allthat you consider very profound, all that you consider sacred, isa contamination in that consciousness. You may not (Laughs) likethe word ïcontamination, but all that you consider sacred, holyand profound is a contamination.So, theres nothing that you can do. Its not in your hands.I dont like to use the word ïgrace, because if you use theword ïgrace, the grace of whom? You are not a specially chosenindividual; you deserve this, I dont know why.If it were possible for me, I would be able to help somebody.This is something which I cant give, because you have it.Why should I give it to you? It is ridiculous to ask for a thingwhich you already have.Q: But I dont feel it, and you do.UG: No, it is not a question of feeling it, it is not a questionof knowing it; you will never know. You have no way of knowingthat at all for yourself; it begins to express itself. There isno conscious.... You see, I dont know how to put it. Never doesthe thought that I am different from anybody come into myconsciousness. [...]((({})))Continued at: The Archetype and the Beast ï98http://pw1.netcom.com/~mthorn/arcbeast.htm [~][^][~]Disingenuous Demagogues Deteriorate DailyAll Politicians are Demagogues, yet not allDemagogues are Politicians... ~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~ Sagittarius assimilated by OUR Milky Way: Resistence Is Futile! http://www.astro.virginia.edu/~mfs4n/sgr/ ~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~^~ T h e Y e z i d i s o f K u r d i s t a n> http://www.songsouponsea.com/Promenade/GnosisE.htmlEARTHlights (by NASA):> http://antwrp.gsfc.nasa.gov/apod/image/0011/earthlights_ dmsp.jpg ~0~ Were like a few bacteria> waiting for the next §ush.> -- Peter Prehn> www.contemposcribe.com/cbcwash/centwashgeology.htm ~0~ICE MEMORY by ELIZABETH KOLBERT> Does a glacier hold the secret of how> civilization began--and how it may end?Excerpt:> ... Over the past decade or so, there has been a> shift--inevitably labelled a ïparadigm shift--in> the way scientists regard the Earths climate. The> new view goes under the catchphrase ïabrupt> climate change, although it might more> evocatively be called neo-catastrophism, after the> old, Biblically inspired theories of §ood and> disaster. Behind it lies no particular theoretical> insight--scientists have, in fact, been> hard-pressed to come up with a theory to make> sense of it--but it is supported by overwhelming> empirical evidence, much of it gathered in> Greenland. The Greenland ice cores have shown> that it is a mistake to regard our own, relatively> benign experience of the climate as the norm. By> now, the adherents of neo-catastrophism include> virtually every climatologist of any standing.> Abrupt climate changes occurred long before> there was human technology, and therefore have> nothing directly to do with what we refer to as> global warming. Yet the discovery that for most> of the past hundred thousand years the Earths> climate has been in §ux, changing not gradually, or> even incrementally, but violently and without> warning, cant help but cast the global-warming> debate in new terms. It is still possible to imagine> that the Earth will slowly heat up, and that the> landscape and the weather will gradually evolve in> response. But it is also possible that the change> will come, as it has in the past, in the form of> something much worse. ~0~See also: North Greenland Ice core Project (NGRIP)> http://www.glaciology.gfy.ku.dk/ngrip/index_eng.htmVOLUNTEERS NEEDED For 180 Light Year Round Trip Excursion!> [Estimated time of departure & arrival still under review]Scientists Discover Planetary System Similar to Our Own:> http://www.nsf.gov/od/lpa/news/03/pr0373.htm The small so-called grays ... are simply mature> human fetuses grown and tailored within artificial> wombs. I would assume that the reptilian and> insectoid humanoid entities reportedly seen by> some abductees are just other examples of> genetically engineered life forms culled> from reptilian and insect life forms on Earth.> -- Raymond Fowler> http://www.nicap.dabsol.co.uk/bio-fowler.htm(Everybody ïwonders about Raymond) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~B L U R ...> http://www.astrocentral.co.uk/beagle.htmlAn Atlas of The Universehttp://www.anzwers.org/free/universe/SDSS: Sloan Digital Sky Surveyhttp://www.sdss.org/WMAP: Wilkinson Microwave Anisotropy Probehttp://map.gsfc.nasa.gov/http://lambda.gsfc.nasa.gov/The Origins of the Fear of Death and Dyinghttp://primal-page.com/death.htmStuart Hameroffs Home Page:http://www.consciousness.arizona.edu/hameroff/ index.htmlhttp://www.mazepath.com/uncleal/ message constitute permission for an emailed lifetime of regret.> at 03:26 PM, tb+usenet@becket.net ( Bushnell, BSG) said:>For what its worth, there *is* an extension of the cross product for>vectors in higher dimensions.FSVO extension. The exterior product for a vector space V does not map> VxV->V, but into the exterior algebra of V.Right. My point is that there is a bijection of the exterior Apocalypse NOW!> Oh, OK. I see what youre saying. ABC is a rigid structure> which does not allow points A and C to separate horizontally. Nothing is completely rigid, so B will move a little bit> as the structure is stretched and pulled tight. At that> point youll reach static equilibrium, so long as the> forces are indeed confined to pull at angles but the> structure prevents the horizontal separation from> growing. Unfortunately, you put the forces on springs rather> than confine them to the steel structure. So they can> happily move apart.>this is where I believe Abhi is getting confused. Hes mixing two things.One can assume for the purposes of analysis that the angle ABC really isreally rigid and has no elasticity at all. Two things are true when youconsider the entire closed system:One : all the forces at each point can be vector summed of all the confusion. Torques are extertedat ABC by the other components of the system (which as described are free tomodify angles). For instance, there is a compressive force in rod AD whichresults in a clockwise torque being applied at the joint ABD. However, bydefinition, the angle ABD is rigid, meaning that the device is able toresist that torque and exert an equal and opposite torque at that point.Thats what being a rigid angle means, by definition.Abhis confusion is in assuming that the torque causes the force to go away.It Re: Iterated Map: x <- x - r/y; y <- y - r/xFor those less interested in plotting potentially pretty pictures, Iam reposting this with my conjecture (previously at bottom of originalpost) at the top, as so to have those plot-disinterested people bemore likely aware of (and possibly therefore prove/disprove) theconjecture.(New stuff in this reply as well.)I conjecture that, if sum{m=1 to oo} r(m) converges (absolutely anyway), then the x and y sequences bothconverge, and the x and y sequences diverge if the r-sum diverges.(And the x and y sequences are as defined immediately below, ofcourse.)By the way, years ago I asked about the sequencex(m+1) = x(m) - 1/x(m),which is a specific case of this situation here.(r(m) = 1; x(1) = y(1))This sequence of xs varied greatly depending upon the exact value ofx(1).Leroy> I am curious, since I am unable to currently plot anything now,> if the following iterated plot produces an interesting graph:x(m+1) = x(m) - r(m)/y(m);y(m+1) = y(m) - r(m)/x(m).({r(m)} is some predetermined sequence, preferably causing the> sequences x and y to converge, and r(m) never is x(m)*y(m) for any m.)Now, {x(1),y(1)} is the coordinate of the iterated-upon point;and, say, we plot a color at this coordinate which corresponds tolimit {m -> oo} sqrt(x(m)^2 + y(m)^2),as an example, if r is such that the sequences converge.Any {r(k)} lead to any interesting plots??> Also....If r is nonzero,x(2+m) > = x(1+m) + (x(1+m) - x(m)) x(m) r(1+m)/(x(1+m) r(m));y(2+m) > = y(1+m) + (y(1+m) - y(m)) y(m) r(1+m)/(y(1+m) r(m)).So {x(k)} and {y(k)} are each dependent on the first 2 terms of their> sequences only.> But x(2) = x(1) -r(1)/y(1) and y(2) = y(1) -r(1)/x(1), so x(1) still> affects the y sequence and conversely.> Now, if> limit {k ->oo} > r(k+1)/r(k) = R exists, and R < 1,then both the x sequence and y sequence converge.Furthermore, if R > 1, both sequences diverge.But I do not know how to determine what happens if R = 1 or R does not> exist.I conjecture that, if sum{m=1 to oo} r(m) converges (absolutely anyway), then the x and y sequences both> converge, and the x and y sequences diverge if the r-sum diverges.Can anyone prove/disprove this Walls (of a hexagon) [...] whenever it crosses its own path (as already drawn), it passes through the path, but is re§ected as if a mirror has been placed perpendicularly to the previous path at the point of intersection.[...] E, B , D, 2 crossings, F, E, 1 crossing, F, 3 crossing, D, B, 3> ^^^^^^^^^^^^^^^^^^^> Problem: I tried to draw the path, and if the ray goes from F to E> without a crossing, it cant get back to F with only one crossing,> due to line 1 (from A to E) and line 6 (which crosses line 1 and goes> to F) being in the way.> Did I just draw it differently, or is there a missing crossing? crossings, F, 7 crossings, and back to its staring point.The ray goes from F to E (no crossing), then bounces back between> the (just created) F-E ray and the original A-E ray. It then hits> the last segment of the D-F ray and is re§ected back to F.[ It goes on to cross A-E, E-B, B-D, hits D, goes to B, then crosses> B-D, E-B and A-E and reaches F, and then I have no idea what the last> seven crossings are. :( ]> I believe you have it correct so far ...:)(Yeah, the last 7 crossings are combinationsHiI have a problem I would like to solve but the answer eludes meIn a series of symbols (0s and 1s) I would like to calculate how many possible combinations there are.eg; 4 symbols with 3 zeroes and 1 one has 4 possible combinations1000010000100001a slightly more complex example4 symbols with 2 zeroes and 2 ones has 6 possible combinations110010101001011001010011in each series the number of 0s and 1s are fixed0s always out number 1sI would like to know how to solve this for n zeroes and n onesin particular problem I would like to be able to solve has 240 zeroes and 16 ones (but not always)Any clues how I could go about this, it is not as logical as it first seems eg 240 * 239 * 238 ... (at least to me)I expect the result to be somewhere around 2^64 question(unrelated newsgroups removed)ConsiderK = a+bL =ab (I use K for K1 and L for K2 for clarity).Solve for a,b as follows:L = a(K-a)L = aK - a^2a^2 - Ka + L = 0This is just a quadratic eqn in a; the solutions area= (K +/- sqrt(K^2 - 4L))/2Same deal for b.A similar approach should work for your equations, but will (presumably)lead to a cubic equation instead of a quadratic. Unfortunately, the generalsolution to a cubic is an ugly looking thing and your equations will not Suppose: K2 = a + b + c> K1 = a b + b c + c a> K0 = a b c I want to acheive 2a - b - c, 2b - a - c, 2c - a - b using any> combinations of K2, K1, K0 using any operations (+, -, *, , roots, logs,> etc.) under real numbers. 1) How do I know if such combinations of K2, K1, K0 will get me to 2a -b -> c?> 2) analysis?Because it will teach you that math is not a collection of formulas, butrather a way of thinking.> Hello... I want to make a postgraduate study in analysis, because I like maths> a lot and my undergraduate final work was about iterative methods for> regularization problems. I am an engineer and in the faculty, they are requesting me an essay> about why should I study analysis. I must make a good writing because> my admission depends on it. I would like you to help me and give silence the discoveries made my non-mathematicians.I refer you to the posts of my good friend, James Harris.MB> Im not a mathematician. I dont know any mathematicians.> Mathematicians always delete my e-mails. Ten days ago I posted the link to introduce my new proof of the FTC> and a funny thing happened. Besides a couple of replies about my> style, there were no replies about the substance of what I have to say> from the sci.math area. I had to go to dq=&hl=en&lr=&ie=UTF-8&safe=off&th=1ea9a3a962a95979> to get any response at all. This is strange because calculus should> be of interest to mathematicians. I suppose analytical thinkers tend> towards math and visual thinkers tend towards physics. You should be> interested because even though you may be expert at the art of> calculus, you dont know why it works until you see what I have found. Calculus had no proof for the fundamental theorem for two hundred> years. Infinitesimals were used to explain the practice but both> Leibniz and Newton knew there was something wrong with them and they> couldnt be proved. Limits were developed to improve upon the> infinitesimal and avoid awkward terms like estimate and approximate in> the calculus. Although proving the calculus is valid, limits do not> enlighten you as to how the calculus works its magic. One problem I have is that anybody who already knows the calculus will> see the connection but not the benefit of my proof. They wont have> to do anything different or see the big value in understanding. New> students will appreciate it because all the new rules they learn will> fall into place and the objectives of the calculus will be understood. So here is the challenge. If you think youre smart read the above> thread and go to:> www.precalculus.netfirms.com/#3> But you cant because its right. So come back and say its right so> others will know. Spread the word, this stuff belongs in textbooks.> posting this ahead of the particular response from Fred that Iam responding to appearing on Google.]> The ray goes from F to E (no crossing), then bounces back between> the (just created) F-E ray and the original A-E ray. It then hits> the last segment of the D-F ray and is re§ected back to F. While this is the topological restriction from the problem data, I> did the angle chase last night and this just does not work -- after> leaving E the ray cannot intersect the last segment of the D-F ray.> So I dont think there is a solution after all. This leaves four possibilities: 1) I stuffed up the angle chase, and it is possible after all;> 2) There is an alternate topology that I have missed;> 3) The information provided about the edges and crossings was wrong; or> 4) The original diagram was just too inaccurate. Im leaning towards 4 at the moment, with a side order of 3. But Im> willing to be convinced at F, and then on E, such that itcrossesthe D-F line at the segment between the point the D-F line crosses theinitial A-E segment and side F, before returning to F (making a loop).Looking at my sketch, it looks like I have the angle of the firstF-side bounce way off (having the path go towards E too quickly), butnot so much so that the topology could work arithmetic, rather than mathematics.> Isnt arithmetic a part of mathematics?> Hardly, no (strange as that may sound).> It does sound strange. And also unbelievable. Especially since> Arithmetic is taught as a part of Mathematics in school.The only mathematics in arithmetic is when you ask yourself things> like why does this method work, is there a quicker one, etc.Fair enough, so you do agree that arithmetic is a part of mathematics.> Applying the methods is then no longer maths, it is arithmetic.One may think of arithmetic as the computational arm of mathematics. Without arithmetic, mathematics becomes extremely wooly, andrestricted to a closed few. Like, very few people can understand amathematical theorem, just by itself. When there is a working-out ofthe theorem using arithmetic, people understand the theorem a lotbetter. To divorce arithmetic from mathematics is to rarefy the scopeof mathematics, and while it may put mathematicians on a high pedestal(like Einsteinian physicists) there is the risk that the public willbe alienated. So such elevation may well be temporary.> Just like you can use mathematical thoughts in a new piece of> music. One can also use gardening or dog-training thoughts in a new piece ofmusic. Or any other thoughts. I dont understand what you are tryingto say.>Then, after composing, playing the music is not math, is it?In a keyboard where you can store the music, there is a lot of maths! Such as whatever is involved in digitising the music, controlling theSNR with what bit quantisation, etc.> And of course, the anglosaxon world uses ïmathematics as a> synonym of arithmetics, adding to the confusion.No, it is not a synonym. It is a part. We learn arithmetic, thenalgebra, then geometry, then trigonometry, then co-ordinate geometry,then calculas and finally probability and statistics in anglosaxonschools, and they are all classified as mathematics subjects. It sohappens that arithmetic questions are not asked in maths exams in thehigher classes. Then, when we study computer science in engineeringcolleges, and how to make the computer do arithmetic, we have to learnthe basics once again in greater depth.> That may also explain> why the math teaching world is not very fond of tricks like this.> It is not a trick, it is a sound method for multiplication.> Yes. But ive seen a site about Vedic maths showing some other things.> We are talking about multiplication here, and nothing else.But it is used as pr for ïthe rest of Vedic maths...Is it? Have you read the book in question? Is it marketed in Westernworld? Exactly who is doing the PR? I just happened to chance on thebook while I was in Kolkata, buying the works of Kalidas in Sanskrit. Nobody forced it on me! And now, here I find that book denounced as ahoax, fraud, etc. by certain people. I am only trying to present whatI learnt from that book.> They really are tricks, giving a student no ïbasis or ïinsight at all.> Certainly the multiplication method gives us a terrific insight into> the terrific advantages of place value.> Nice tricks, mind you. But tricks.> Fine. All of mathematics is art, or a bag of tricks. All> civilisation comes from trickery with nature, and its subsequent> manipulation. Those who do it best, are the winners. You need to> have great basis and insight to do any trick properly.But math is more; it is tricks + insight why.You cannot do clever tricks successfully without possessing insight. If you try to do tricks without possessing insight into the details,you will soon be shown a silly bungler.> What is vital in math education is to give the children a basis,> something to fallback on, when it gets less directly intuitive.The Vedic multiplication method gives a great deal of insight into therelevance of place value. The current method is certainly very clumsyin contrast. It has been agreed that my definition for that method isthe same as for the advanced method of convolution, used formultiplying very large numbers. If primary schoolchildren use suchadvanced tricks at their very tender age, surely their capabilitieswill develop fast.> It gives> a very thorough insight to the whole process, once properly> understood.> It is a welcome addition to arithmetic education.> But when it comes to insight, the distributive law is much better.> Explain the distributive law, and show us how it is much better! With> respect to what?With respect to insight. The definition of the number sequence,> the definition (=the meaning) of addition and multiplication,> *those* are the mathematical aspects of arithmetic.As I said, the Vedic multiplication method gives terrific insight. > The rest is ttrivial calculus. A computer can do it.Indeed, with the Vedic multiplication method we can multiply tenthousand digit numbers on a ordinary pc, without needingmultiprecision routines, and with only a few lines of code notrequiring any great genius to write.> Children are likely to like this method, and teachers> will find it useful to teach.> Yes, i also definitely think so.> We agree on one thing. Good.> Mathematics teachers are desperately fighting the general prejudice> that mathematics is ïjust arithmetic.> Well, that is because they are not doing the arithmetic right. Once> that is done right, following proper understanding of what numbers> really are and what they really mean, then things will be better for> all concerned.> Here i disagree. Worshipping vedic maths as a potential redemption for> maths education would be fatal.> Not for most Indians and all non-bigots, I hope. It is not a question> of worshipping anything; it is to use some wonderful methods to do our> arithmetic better, and thus improve the entire basis of mathematics.Doing arithmetic better will not help mathematics a single bit.No, I do not agree. Children will be freed from the torture o§earning arithmetic the modern bad way, and they should have a morepositive learning attitude towards mathematics as a result ofincorporating Vedic arithmetic in primary schools. Improvingarithmetical methods will increase computing efficiencies, and modernlife is about increasing efficiencies. Also, there are profoundphilosophical depths is arithmetic, such as 1/0 that is quite lostupon very many, I am afraid.> Arithmetic is *not* the basis of math. How about geometry?Geometry is very logical, and deals with space. To put it intopractice, we need arithmetic. My work is on mathematical modelling,and the models are so complex, geometry becomes quite useless torepresent them. We have to represent them with abstract classes,using arithmetic. Solutions are obtained by number-crunching, veryefficiently. Prior efforts such as Petri-nets failed, they were toogeometrical.Arindam akin to Brouwer, new?Let S be the set of (x,y,z) in R^3 such thatx + y + z = 1andx>=0 and y>=0 and z>=0.Let f be a continous map S to S. Define three closed subsets of S by:A = set of (x,y,z) such that f(x)>=xB = set of (x,y,z) such that f(y)>=yC = set of (x,y,z) such that f(z)>=zand a fourth by:D = (A cap B) cup (B cap C) cup (C cap A).Show that D contains a connected subset T which meets all three sides of S,ie. T contains the three points(0,a,b)(d,0,c)(e,f,0)for some a,b,...,f, not necessarily nonzero.I think I can grind out a proof, but maybe the result is already familiar ArpYou see, Dr. Arp is a scientist, a world renowned scientist and he has> *data*, real, hard astronomical data, which is more substantive in> disproving the commonly taught Big Bang Theory, than the data used to> support that theory.Arps data is presented in the _Atlas of Peculiar Galaxies_, > Astrophysical Journal Supplement Number 123, Volume 14, November 1966.In the Atlas Arp presents galaxies that appeared abnormal. Follow-up> observations showed that some, not all, of the galaxies were in fact> two galaxies that are apparently interacting. What caused the doubt > about the Big Bang was that some of these pairs have very different> red-shifts. If the galaxies are close to each other the different> red-shifts would sound the death knell for expansion and the BB.However, as observing technique has improved weve determined that> most of these pairs are simply close in the line of sight and are> at very different distances. There are a few cases that have not> been elucidated, the necessary obsrvations are, at best, difficult.These remaining cases do not constitute an overthrow of the BB, > to do that would require high quality observations of difficult > objects; big results require big data, obscure, difficult cases > do not provide that.I went to Google, and found a relevant link.In figure A below, there are four objects. NGC 7603 is a spiral galaxywith a redshift value of 0.029. Object 1 is a quasar with z = 0.057.Objects 2 and 3 are quasar-like objects with z values of 0.243 and0.391 respectively. As L.97pez-Corredoira and Guti.8errez noted:Everything points to the four objects being connected amongthemselves, but how to explain the different redshifts? (p. L17). Howto explain indeed? Gribbin lamented: That strikes at the foundationstone of received cosmological wisdom (p. 65). It certainly does! Ascase where we once again are experiencing a situation where data getthrown out if they dont fit the theory. Big Bang cosmology simplycannot explain Arps anomalies.date November 1966, in what I see as a not subtle attempt to skewreader opinion.As Ive said, Dr. Arp projection a closed map?Hello group,I am trying to decide wether the canonical projectionp_i: X -> X_i, X = X_1x...xX_n,which is an open map, is also closed or not. I already found out thatthe closed subsets of the product space X are infinite intersectionsof sets like X-U_l, whereU_l = U^l_1 x ... x U^l_n with all U^l_i open in X_i.But the problem ist the intersection of all of these subsets:p_i( intersection(l element L)(X-U_l)) subset intersection(lelement L)p_i(X-U_l)Between the left and right side of the above statement there notalways exists equality. The right side seems to be a closed subset,as p_i(X_U_l) = X_i, but this doesnt help me.Can anyone give me a hint?Rene.-- Ren.8e MeyerStudent of Physics & MathematicsZhejiang University, Nodes?Consider the Binary Node tree of naturals. 0 / 0 1 / / 0 1 0 1 . . .(I hope it is lined up.)Obviously if you consider the omegath row, it will have beta-1 (C) elements,but since omega is the first limit ordinal, it is the first row with aninfinite number of elements.However, consider the previous row. What ordinal does it represent? omega?How many elements does it have? beta-1? Let us move up the tree until wefind a row with beta-0 (aleph-0) elements. What ordinal corresponds to thisrow?Furthermore, if P=NP, wouldnt this mean that combinationsAre you familiar with combinatorial and permutations? If so, your answeris, if z=zeros and o=ones, (z+o)C(o) or (z+o)C(z). They are equal.> Hi I have a problem I would like to solve but the answer eludes me In a series of symbols (0s and 1s) I would like to calculate how many> possible combinations there are. eg; 4 symbols with 3 zeroes and 1 one has 4 possible combinations 1000> 0100> 0010> 0001 a slightly more complex example 4 symbols with 2 zeroes and 2 ones has 6 possible combinations> 1100> 1010> 1001> 0110> 0101> 0011> in each series the number of 0s and 1s are fixed> 0s always out number 1s I would like to know how to solve this for n zeroes and n ones in particular problem I would like to be able to solve has 240 zeroes and> 16 ones (but not always) Any clues how I could go about this, it is not as logical as it first> seems eg 240 * 239 * 238 ... (at least to me) I expect the result to be somewhere around 2^64 but could Halton ArpSo it should be a no-brainer, the theory has to be changed to fit the> data.In addition to potential problems with a theory, the available> data may be incomplete (e.g. Neptune was not known when> irregularities were found in the path of Uranus) or the> observations may be wrong (e.g. Schiaparellis canals on Mars.)Please note that I have made no comment on the work of Arp nor> on the Big Bang theory. I am merely pointing out that there are> other possibilities than §awed theories.Which is why science is about keeping an open mind, but the problemhere is that certain people made up their minds years ago.Remember the shocking nature of the dispute with Halton Arp is howhes been treated by other scientists whove fought to pushalternative theories off the radar screen.They say its settled. The data says that its not.I want to emphasize that point! The issue here is that certain peopleare working to dismiss alternative theories, other explanations infavor of what is now the current Big Bang Theory.The problem is that the data refutes their model, but theyve decidednot to give in to the data.As a side reference consider the entropy problem of the typical blackhole model.However, some physicists mentioned an alternative theory whichinvolves gravastars.What happened?Physics is becoming dogma dogged, as certain people make their careersbased on particular theories or notions that have been sold to thepublic, and then figure out that they can push their theories againstthe data because the current system lets them.Or hey, maybe no one wants to piss off the people who might bebothered if scientists start admitting that maybe the Big Bang isbunk, or that black holes dont exist?Maybe they think the public would be too upset to Ordinals, Binary Nodes?> Consider the Binary Node tree of naturals.> 0> / > 0 1> / / > 0 1 0 1 . (I hope it is lined up.)> Obviously if you consider the omegath row, I dont know what that means. I can certainly write down the n-th row for any natural number n. But I have no way of knowing what you mean by the omegath row. Its not well-defined.it will have beta-1 (C) elements,beta than none at all I guess.> but since omega is the first limit ordinal, it is the first row with an> infinite number of elements.> You havent defined what you mean by the omegath row. For example if I ask you what is the 47th node of row 2000, you could tell me. But you cant tell me whats the 47th node of row omega.> However, consider the previous row. Well clearly if you intend the omegath row to be the result of some type of limiting process, there is no immediately previous row. You can define omega to be the first ordinal that follows all the finite ordinals, but you cannot then speak of the ordinal immediately preceding know where I can find any referene(paper, book, etc)for G/MMPP/1.Ive been trying, but I cant find it.Any reply would define an actionSO(n+1) x S^n ----> S^nwhere G = SO(n+1) and X = S^n,then what is the stabilizer of an element x in S^n? Is it SO(n+1) in any way?That is, can I mod out SO(n+1) by a unbounded? > In another thread the sequence {tan(n)/n} arose (n=1,2,3...). The> discussion showed that this sequence does not approach zero. Now> |tan(11)/11| is about 20. So I wondered: does tan(n)/n (or> ||tan(n)/n|)> take on arbitrarily large values? So far the largest value I have> found is 556.3, but n is very large.> I didnt see anything in the other thread that answered this. Does> anyone know?> You might look at the last part of> .> The last sentence of that says The sequences of maxima and minima are> almost certainly unbounded, but it is not known how to prove this> fact. Disclaimer: The wording in that last phrase is Erics, not mine.> (Of course I suspect that the sequences are unbounded. But I certainly> wouldnt call it a fact.) David, unless Im going crazy, the n values in the sequence run out> before the tanc() values. What are the two missing n values that> correspond to the two highest tanc()s? (OEIS doesnt have them either).Below is part of what I sent to Eric. It should answer your question.David-------------------------The program below is fairly efficient. It lists successive maxima and minimain the sequence, preceded by the value of n giving that extremum.$MaxExtraPrecision = 400; min = 0; max = 0; i = 0;While[i < 350, i = i + 1;v = N[Tan[n]/n, 70];If[v < min, min = v; Print[{n, N[min, 6]}]];If[v > max, max = v; Print[{n, N[max, 6]}]]]{1,1.55741}{2,-1.09252}{11,-20.541}{122925461,2.65934}{ 534483448,3.58205}{3083975227,4.3311}{214112296674652,18.0078} {1317811389848379909481978463177998812826691414678853402757616 ,-54.5197}{ 140278322695820130839421972690860545853324377324794798150266034 6275822,-74.7721}{ 656663628873183465002720935077228312202585023816648634619965107 978725953831633144141594852184603577778946951537005011,18.0566 }{ 237245119117113587257974180846905994200670445824975057251225928 091938884235884204270870335868832984721112397578700716369398842 Mathematics --- Myth and RealityArithmetic is indeed very important. And, being from India, wehave a special fascination for the decimal calculations and thedifferent methods. In fact, there is quite a bit of research toimplement decimal computing and storage using electronic devices.For example, at one point of time I got fascinated by the possibleapplication of the very interesting resonant tunneling devices forOne of those was published in the journal electronics letters (UK): New static storage scheme for analogue signals using four-state resonant-tunneling devices, Electronics Letters, 29, 1435--1437 (1993).Those interested in electon devices and multivalued memories mayfind the paper interesting. But, at the same time, I always likesuch arithmetic methods where the learner can clearly see why themethod is working. For example, the basic square root findingmethod that we were taught is not clear to me. Can anyone explain?For example, to find the square root of 225, 225 | 15 <---- 1 -- 25|125 125 ---Arjoe> mathematics.>Isnt arithmetic a part of mathematics?>Hardly, no (strange as that may sound).>It does sound strange. And also unbelievable. Especially since>Arithmetic is taught as a part of Mathematics in school.>The only mathematics in arithmetic is when you ask yourself thingslike why does this method work, is there a quicker one, etc.Fair enough, so you do agree that arithmetic is a part of mathematics.Applying the methods is then no longer maths, it is arithmetic.One may think of arithmetic as the computational arm of mathematics. > Without arithmetic, mathematics becomes extremely wooly, and> restricted to a closed few. Like, very few people can understand a> mathematical theorem, just by itself. When there is a working-out of> the theorem using arithmetic, people understand the theorem a lot> better. To divorce arithmetic from mathematics is to rarefy the scope> of mathematics, and while it may put mathematicians on a high pedestal> (like Einsteinian physicists) there is the risk that the public will> be alienated. So such elevation may well be temporary.Just like you can use mathematical thoughts in a new piece ofmusic. One can also use gardening or dog-training thoughts in a new piece of> music. Or any other thoughts. I dont understand what you are trying> to say.Then, after composing, playing the music is not math, is it?In a keyboard where you can store the music, there is a lot of maths! > Such as whatever is involved in digitising the music, controlling the> SNR with what bit quantisation, etc.And of course, the anglosaxon world uses ïmathematics as asynonym of arithmetics, adding to the confusion.No, it is not a synonym. It is a part. We learn arithmetic, then> algebra, then geometry, then trigonometry, then co-ordinate geometry,> then calculas and finally probability and statistics in anglosaxon> schools, and they are all classified as mathematics subjects. It so> happens that arithmetic questions are not asked in maths exams in the> higher classes. Then, when we study computer science in engineering> colleges, and how to make the computer do arithmetic, we have to learn> the basics once again in greater depth.That may also explainwhy the math teaching world is not very fond of tricks like this.>It is not a trick, it is a sound method for multiplication.>Yes. But ive seen a site about Vedic maths showing some other things.>We are talking about multiplication here, and nothing else.>But it is used as pr for ïthe rest of Vedic maths...Is it? Have you read the book in question? Is it marketed in Western> world? Exactly who is doing the PR? I just happened to chance on the> book while I was in Kolkata, buying the works of Kalidas in Sanskrit. > Nobody forced it on me! And now, here I find that book denounced as a> hoax, fraud, etc. by certain people. I am only trying to present what> I learnt from that book.They really are tricks, giving a student no ïbasis or ïinsight at all.>Certainly the multiplication method gives us a terrific insight into>the terrific advantages of place value.Nice tricks, mind you. But tricks.>Fine. All of mathematics is art, or a bag of tricks. All>civilisation comes from trickery with nature, and its subsequent>manipulation. Those who do it best, are the winners. You need to>have great basis and insight to do any trick properly.>But math is more; it is tricks + insight why.You cannot do clever tricks successfully without possessing insight. > If you try to do tricks without possessing insight into the details,> you will soon be shown a silly bungler.What is vital in math education is to give the children a basis,something to fallback on, when it gets less directly intuitive.The Vedic multiplication method gives a great deal of insight into the> relevance of place value. The current method is certainly very clumsy> in contrast. It has been agreed that my definition for that method is> the same as for the advanced method of convolution, used for> multiplying very large numbers. If primary schoolchildren use such> advanced tricks at their very tender age, surely their capabilities> will develop fast.It gives>a very thorough insight to the whole process, once properly>understood.>It is a welcome addition to arithmetic education.But when it comes to insight, the distributive law is much better.>Explain the distributive law, and show us how it is much better! With>respect to what?>With respect to insight. The definition of the number sequence,the definition (=the meaning) of addition and multiplication,*those* are the mathematical aspects of arithmetic.As I said, the Vedic multiplication method gives terrific insight. The rest is ttrivial calculus. A computer can do it.Indeed, with the Vedic multiplication method we can multiply ten> thousand digit numbers on a ordinary pc, without needing> multiprecision routines, and with only a few lines of code not> requiring any great genius to write.>Children are likely to like this method, and teachers>will find it useful to teach.>Yes, i also definitely think so.>We agree on one thing. Good.Mathematics teachers are desperately fighting the general prejudicethat mathematics is ïjust arithmetic.>Well, that is because they are not doing the arithmetic right. Once>that is done right, following proper understanding of what numbers>really are and what they really mean, then things will be better for>all concerned.>Here i disagree. Worshipping vedic maths as a potential redemption formaths education would be fatal.>Not for most Indians and all non-bigots, I hope. It is not a question>of worshipping anything; it is to use some wonderful methods to do our>arithmetic better, and thus improve the entire basis of mathematics.>Doing arithmetic better will not help mathematics a single bit.No, I do not agree. Children will be freed from the torture of> learning arithmetic the modern bad way, and they should have a more> positive learning attitude towards mathematics as a result of> incorporating Vedic arithmetic in primary schools. Improving> arithmetical methods will increase computing efficiencies, and modern> life is about increasing efficiencies. Also, there are profound> philosophical depths is arithmetic, such as 1/0 that is quite lost> upon very many, I am afraid.Arithmetic is *not* the basis of math. How about geometry?Geometry is very logical, and deals with space. To put it into> practice, we need arithmetic. My work is on mathematical modelling,> and the models are so complex, geometry becomes quite useless to> represent them. We have to represent them with abstract classes,> using arithmetic. Solutions are obtained by number-crunching, very> efficiently. Prior efforts such as Petri-nets failed, they were too> geometrical.Arindam