mm-201 === Subject: Re: recurrence relationselma recurrence relations:> x[n+1]=a.x[n]-b.y[n]> y[n+1]=c.y[n]+d.x[n]> I am interested in the long term behaviour of this model, for> different values of a,b,c,d and x[0] and y[0].You can get easily recurrence relations for x[n] and y[n] isolateds.x[n+1] = ax[n] - by[n]y[n+1] = cy[n] + dx[n]x[n+2] = ax[n+1] - by[n+1] = ax[n+1] - bcy[n] - bdx[n] = ax[n+1] -acx[n] + cx[n+1] - bdx[n] = (a+c)x[n+1] - (ac + bd)x[n]Initial values: x[0] = x[0], x[1] = a[x0] - by[0]Likely,y[n+2] = cy[n+1] + dx[n+1] = cy[n+1] + dax[n] - bdy[n] = cy[n+1] -acy[n] + ay[n+1] - bdy[n] = (a+c)y[n+1] - (ac + bd)y[n]Initial values: y[0] = y[0], y[1] = cy[0] + dx[0]So, you transform a linked pair of recurrence relations of order 1, in toequals requrrence relations of order 2, althougt with distincts initialvalues.The characteristic equation of both isr^2 - (a+c)r + (ac+bd) = 0 (#1)If r and r are distincts solutions of that equation, you havex[n] = Ar^n + Br^ny[n] = Cr^n + Dr^nwhere A, B, C and D are constants that you can determine with the initialvalues.If #1 has a doble root r, thenx[n] = Ar^n + Bnr^ny[n] = Cr^n + Dnr^n-- (statistics)how to make date more like Laplacian distribution? ...> Now for a simple example. Lets say you have a fair die. Each time you toss> it, the probabily of each outcome is 1/6. So theory says we have a uniform> distribution. Now toss the die 100 times and count how many times each value> shows up. Now theory says on average each value shows up 1/6 of the time.> However in any given sample (set of measurements), you cant expect this> perfect a distribution every time. Why you ask? That is because each trial> is independent of the prior ones. And in this situation I only tossed it 100> times and 100 is not a multiple of six, so not all bins can have the same> number of counts. Since the counts must be integers and 100/6 = 16.6666666,> you can see the one problem. ...This can raise an unrealistic expectation. If we toss the die 102 times,can we then ask that each face of an honest die to turn up 17 times?Jerry-- Engineering is the art of making what you want from things you can 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.>The poiunt is that tan(n) is big in absolute value iff n is close to>an odd multiple of pi/2, that is if pi/2 is approximately n/m>where 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 be>bounded away from zero. Its not obvious that there would be infintely>many cases with m odd, and to get a better result we need a better>Diophantine approximation result on pi/2. I dont know if there>is such a result.Yes there is. See my posting on the Subject A limit problem with explanation in sci.math.symbolic on 17 October 2001. But to get|tan n/n| to take on arbitrarily large values, I think youll need the continued fraction of pi to have unbounded elements. Almostcertainly its true, but we cant prove it. OTOH if x is a quadraticirrational, since the continued fraction of x has bounded elementsthe sequence tan(n pi x)/n will be bounded.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israelUniversity of Reconsidering Halton Arp> > The problem I have with that is that in looking at actual pictures you> can see material curling from one galaxy to another that is clearly> pulling the matter over.> > The problem is that you are looking at one messy chaotic system> in the line of sight of another messy chaotic system ( this is true> no matter how far apart or close they are ). It is quite true that these are distant objects. Readers can look for themselves athttp://perso.wanadoo.fr/lempel/red_shift_NGC_7603_ uk.htmwhich has some pictures of one of the so-called anomalies. > Having been burned a few times astronomers no longer accept simple > it looks like arguments. These arguments are used to start or> inform other studies of the objects. That sounds reasonable.> For these next cases I dont have references in front of me, an > examination of the archives of _Astronomy & Astrophysics_ will > locate these and other studies. > > For example Vera Rubin reports a case where a small spiral galaxy is > close to but behind an elliptical, there appears to be a bridge between> the two. As the red shifts are different this could be a counter case> to the BB. Dr Rubin measured the rotation velocities in the spiral> and through application of the Tully-Fischer(sp?) relation ( maximum > rotational velocity is correlated with galaxy mass and size ) showed> that the spiral was large and approximately at the red shift distance;> ie: much further than the elliptical.> > Radio astronomy was used in another study, a bridge was shown to> have two velocity components, one at low red shift, the other at> high red shift. There is nothing between them. > > These are both difficult measurements, the objects are small and> faint. The variation in signal is small compared to the total.> We should not be surprised that it has taken decades to elucidate> these cases, when Dr Arp first published the necessary measurements> were impossible.That sounds reasonable as well.> What JSH and others seem to think is that there is a large quasi-religous> hierarchy of priest-astronomers that impose a rigid orthodoxy on> the field. This ignores generations of intelligent, hard-working,> ambitious grad students ( and full professors for that matter ) who > would like nothing better than to upset the apple cart. That would> be a professional meal-ticket; lots of big telescope time and a> department head-ship.Id prefer the issue settled myself, and think it simpler if you canjust look at red-shift to tell how far away something in the sky is.However, despite the pleasing simplicity of the notion, dataincreasingly shows that it is *too* simple of an idea and the realworld or real Universe I should say, is more complicated. That meansthat researchers like Dr. Arp need to get a fair and complete hearing,and other scientists do not have the right to claim its settled.And dont make claims about graduate students as at the end of the daythey depend on the authorities, the full professors who make the finaldecisions.Grad students are important mind you, as Ive often found them helpfulin my mathematical research, but when it comes down to decision time,they go with authority.> date November 1966, in what I see as a not subtle attempt to skew> reader opinion.> > people who agree with Arp ). 1966 is the date that started things. > Subsequent reports have found a few other cases, Arp was good enough > to find most, but the new cases have not altered the argument.Ok thats not a big deal then, my apologies for misreading you.> To date there have been *NO* cases where physicaly close ( not line > of sight close ) objects have been shown to have markedly different> red shifts. Arp and others have provided a list of where this may> happen, in the cases we have examined it does not.> > Dark skies,> > tomAnd, Im confident that here youre just saying what you want tobelieve is true. And I have one case in particular then to mentionagain:See http://perso.wanadoo.fr/lempel/red_shift_NGC_7603_uk.htmIts not the only evidence Ive seen either, and its quite recent.Of course theres also the word of Dr. Arp as well.Are you challenging his interpretation of the data?Not that Id have a problem with that, but Id like to make sure thatyoure *stating* clearly that youre challenging his interpretation Group for Dark Energy/MatterKaisers paper 81JMP.pdf on his websitePhase Space Approach to Relativistic Quantum Mechanics IIIhttp://www.wavelets.com/vita.html#papersis of particular interest for several reasons.The Lie algebra for the 10 charges of the globally §at false Minkowski space-time vacuums Poincare group is[Ji,Jj] = Jk[To,Kr] = Tr[Ki,Kj] = -c^-2Jk[Ji,Kj] = Kk[Ji,Tj] = Tk[Tr,Ks] = -c^-2&rsTowhere i,j,k = 1,2,3 (or x,y,z in Cartesian frame of high school analytic geometry)(ijk) above in sense of cyclic permutationr,s = 1,2,3To is total energy generating time translations in globally §at Minkowski spacetimeTk = total linear momentum generating space translationsJk are space-components of total angular momentum generating space rotations along the k space axis in the ij space plane.Kk are the Lorentz boosts between two Global Inertial Frames along the k axis in space, which, by definition, are not accelerated frames. These are space-time rotations (0,k) in Minkowski space-time.The Poincare group is a semi-direct product of the translation subgroup Lie algebra {To, T1,T2,T3} with the Lorentz O(1,3) subgroup Lie algebra {J1,J2,J3,K1,K2,K3}. Since the commutators across the two subgroups do not commute.Note that the Galilean limit of Newtonian mechanics is c -> infinity where[Ki,Kj] = -c^-2Jk --> 0and[Tr,Ks] = -c^-2&rsTo --> 0Now Kibble showed (mid 1960s) that Einsteins 1915 theory of gravity without torsion fields comes from locally gauging the 4 generators {T0, T1, T2, T3}. This can be seen heuristically in the large scale L >> Lp limit of Hagen Kleinerts world crystal lattice elastic-plastic model of General Relativity.Globally §at 4D Minkowski space-time is simply a perfectly tiled lattice with unit cells at the Planck scale Lp. The symmetry of these unit cells is washed out in the large scale L >> Lp.The compensating gauge force fields for {T0, T1, T2, T3} are simply du(x), i.e. the distortion field of the world crystal at point x.Hagen Kleinert then shows that Einsteins 1915 theory comes from the strain tensor of the distorted world crystalguv = Minkowski metric + (1/2)[du(x),v + dv(x),u]with string defects of disclination whose large scale density is precisely Einsteins tidal force curvature tensor.The global translation group symmetry is broken, but is replaced by the Diff(4) local symmetry group. EEP is obeyed in the tetrad eu^a(x) formulation of the theory where u is in the curved base spacetime and a is in the locally almost §at tangent spacetime. Almost means we are far enough from a spacetime black hole singularity and larger than Lp so that tidal curvature g-force differentials and quantum gravity §uctuations are ignorable. Note, that the g-force is eliminated in LIFs on timelike geodesics, but the tidal differential curvature force from geodesic deviation is not eliminated. However, EEP only holds in the weak curvature non micro-quantum domain. That Einstein may not have realized this limitation on his theory in the early days is of no great consequence IMHO. I am not even sure that he did not realize it at least after 1925. I am alluding, of course, to Paul Zielinskis thesis that may be a valid footnote in the history of the emergence of Einsteins thought. If one wants a local stress energy density tensor for geometry it is simplyTuv(Geometry) = (String Tension)GuvDefine the non-exotic vacuum asTuv(matter/radiation/near EM fields) = 0/zpf = 0And in that caseTuv(Geometry) = 0.However, in the exotic w = -1 zero point energy density dark energy/matter vacuumTuv(Geometry) +(String Tension) /zpfguv = 0Metric engineering in the exotic vacuum isTuv(Geometry)^;v =/= 0Tuv(Geometry)^;v + /zpf^,vguv = 0assuming metricity and zero torsion in this simplest of cases./zpf = (alpha)^-1[(alpha)^3/2|Vacuum Coherence|^2 - 1]h = c = 1 convention temporarilyalpha = (String Tension)^-1 = Witten parameterThere is no need for the Yilmaz theory.See below on Vacuum Coherence quieting the random chaos of the micro-quantum field vacuum zero point §uctuations. This insight is entirely missing from current string theory as well as the program of Haisch and Puthoff. In the former case we have Ed Witten admitting that the smallness of the Cosmological Constant is very serious for his M Theory. In contrast Hal Puthoff simply hand waves away the problem, which simply will not do IMHO.Kleinert does not have any connection to quantum theory in his model.I have added that.My MACRO-QUANTUM Coherent Ansatz is:du(x) = Lp^2(Goldstone Phase(x)),u,u = partial derivativewhereVacuum Coherence Field = (Higgs Amplitude Field)e^i(Goldstone Phase Field)I also have a micro-quantum BCS type argument why this macro local field emerges from the QED sector of quantum field theory at least as a low energy approximate effective ODLRO c-number smooth field theory.This Vacuum Coherence Field is the scalar field of in§ationary cosmology in the FRW limit.The Lp^2 is consistent with Black Hole Thermodynamics of Bekenstein et-al.Back to the Lie Algebra, we see that there are non-vanishing commutators between the translation and the Lorentz subgroup algebras. Therefore, completeness and consistency (not in Godels sense) suggest that we must also locally gauge the Lorentz group. Utiyama showed, before Kibble, that locally gauging the entire Poincare group gives a larger theory than Einsteins with the additional torsion field as the compensating gauge force field from the 6 charges{J1,J2,J3,K1,K2,K3}. The torsion field appears as an antisymmetric Diff(4) 3rd rank tensor additional piece of the connection field for parallel transport of Diff(4) tensors along world lines in curved spacetime.The De Sitter group has a curvature parameter that is really Einsteins cosmological constant / in the sense of the FRW metric. Now apparently / -> infinity has Penroses 15 parameter conformal group of massless twistors as the limit and there is a kind of string duality/ ~ Lp^4//between zero and infinite cosmological constants which I suspect are at best metastable vacua.So the issue is additional compensating gauge force fields by locally gauging the entire Lie Algebra of the De Sitter group, which obviously will give the local /zpf(x) unified dark energy/matter field and perhaps something extra?Where can I find the Lie algebra of the De Sitter Group fully written out?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.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):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,andto 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 are a couple recurring results in maths which tend to be used as a testing ground for different theories. Generally quite simple things that you can use to get a feel for the theory. I can only think of a few examples off the top of my head, but Im sure Ive seen more of them in real and complex analysis. Things like the following:There are infinitely many primes. There are probably literally hundreds of these - a number of purely number theoretical ones, topological, analytic, etc. Its something that people often go back to as a neat ïhuh. Look. Ive got another proof of the infinitude of primes. Not that exciting, but a neat way to apply new theories.The reals are uncountable. Similar to the last one. I cant think of nearly as many, but at least three come to mind - baire category, binary expansion and the theorem that every countable dense totally ordered set without endpoints is isomorphic to Q (and the reals arent isomorphic to Q). Im sure Ive seen a few more, but they escape me at the moment.Fundamental theorem of algebra. There are proofs involving compactness, complex variable theory (several distinct ones involving complex variable theory in fact), algebraic topology, galois theory and half a dozen or so others. Its a nice result that is sufficiently complex (no pun intended) that it involves interesting mathematics, but simple enough that one can get an easy handle on it.So, this is basically a poll. Have you got any good examples of such results? What are your favourite proofs of these and others? That sort of thing. I know the questions are a bit vague, but thats mostly because Im interested in a wide variety of depends on the eigenvalues of the matrix [[a -b], [c >> d]]. One good introduction is Luenberger, Introduction to Dynamic >> Systems.>By the way, if the maximal modulus of the eigenvalues is less than one, >the system tends to the origin in the long term. Greater than one, it >generally goes off to infinity.>If your four coefficients are positve, the matrix has only complex >eigenvalues. This implies that the path of the system is almost periodic.Huh? Try[ 1 -2 ][ 1 4 ]which has eigenvalues 2 and 3.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of Numerical integration, prime countingX-DMCA-Notifications: http://www.giganews.com/info/dmca.html>> > >[...]>>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.logic>>is not mathematical prowess, but a stop-at-nothing determination>>to silence or neutralize dissenters from the canon?>> >> Tee-hee. Yeah, thats exactly right. Of course youre not going to>> get anyone to _admit_ that thats what commands respect here, because>> after all that would be giving the game away.>> >> Giggle.>>David Ullrich basically tries to tease posters.>>He latches on to certain people and then will not let go.>>He tracks them across Usenet, and then claims its their fault.> I especially like the part about stop-at-nothing determination>> to silence dissenters, coming from the only person Ive ever seen>> _explicitly_ threaten to _kill_ people who disagreed with him on>> usenet...>>And David Ullrich, a math professor at Oklahoma State University, lies>a lot.> *****************************>> John Correy says:>> >> Christian (what an oxymoron!): Degrade, demean, goad and bait me>> as Ullrich and the Boyz have done to JSH, and I wont tri§e with>> writing your employer: Ill come after you with an AK-47!>> >> >> David C. Ullrich>>While John Corrys use of hyperbole might seem questionable to some,>its David Ullrich who pushes the worst case interpretation in an>attempt to aggravate.>>If he really were worried, I doubt hed be teasing John Correy.>>What I found when I talked to Ullrichs boss at Oklahoma State>University, is that he lies even to his university, as I heard a>rather fascinating tale, second-hand through his boss, justifying>Ullrichs odd posting behavior, which is a tale his boss apparently>believed.Uh-huh. If you dont want everyone to find this as amusing asthey find everything else you say you should explain exactlywhat lie youre referring to - just saying you heard about a lieI told without saying what the lie was sounds silly.Exactly what was this fascinating tale?>Make no mistake, David Ullrich is a dedicated and persistent liar.>>He also is a math professor at Oklahoma State University.>James HarrisDavid C. coefficients are positve, the matrix has only complex >>eigenvalues. This implies that the path of the system is almost periodic.>> >>Huh? Try>>[ 1 -2 ]>[ 1 4 ]>>which has eigenvalues 2 and 3.>My bad. For some reason, was thinking in my head that det (A - u I) = (a-u)^2 + b d-- Stephen J. forward a copy of ALL headers otherwise we will be unable to in analysing the recurrence relations:>> x[n+1]=a.x[n]-b.y[n]>> y[n+1]=c.y[n]+d.x[n]>> I am interested in the long term behaviour of this model, for>> different values of a,b,c,d and x[0] and y[0].>>You can get easily recurrence relations for x[n] and y[n] isolateds.>>x[n+1] = ax[n] - by[n]>y[n+1] = cy[n] + dx[n]>>x[n+2] = ax[n+1] - by[n+1] = ax[n+1] - bcy[n] - bdx[n]>> = ax[n+1] -acx[n] + cx[n+1] - bdx[n]>> = (a+c)x[n+1] - (ac + bd)x[n]>>Initial values: x[0] = x[0], x[1] = a[x0] - by[0]>>Likely,>>y[n+2] = cy[n+1] + dx[n+1] = cy[n+1] + dax[n] - bdy[n]>> = cy[n+1] -acy[n] + ay[n+1] - bdy[n]>> = (a+c)y[n+1] - (ac + bd)y[n]>>Initial values: y[0] = y[0], y[1] = cy[0] + dx[0]>So, you transform a linked pair of recurrence relations of order 1, in to>equals requrrence relations of order 2, althougt with distincts initial>values.>>The characteristic equation of both is>>r^2 - (a+c)r + (ac+bd) = 0 (#1)>>If r and r are distincts solutions of that equation, you have>>x[n] = Ar^n + Br^n>y[n] = Cr^n + Dr^n>>where A, B, C and D are constants that you can determine with the initial>values.>>If #1 has a doble root r, then>>x[n] = Ar^n + not used to> using Excel. I usually use Mathematica, or something. I am just trying to> learn Excel so that I can add it to my resume.>> Lurch>> > I am also getting something strange when I try to calculate cos(pi/2).> I> > keep getting the result 6.12574E-17. Is that suppossed to be> essentially 0?> > Cant Excel give you 0? I know that from programming, but I thought MS> > could have just programmed that to represent 0. Whats up?> > You dont have exactly pi/2. You have a §oating-point> number which approximates pi/2 to 16 places, and thus> the cosine of that number approximates cos(pi/2) up> to 16 places.>Just select all cells on the page packing in a square, given radius, find maximum number of circles>Any pointers to a solution of the follwoing inverse circle packing>problem will be greatly appreciated:> Given a unit square, and a radius R, how does one find the maximum> number of circles, each of radius R, that can be packed into the square?Google for packing circles square. Let r(n) be the largest radius of n equal circles that can be packed in a unit square. If r(n+1) < R <= r(n) then your answer is n. AFAIK there is noreally effective way to answer these questions for large n (or equivalently small R), although there are methods that will probablycome close to the correct answer. Large in this context couldbe a couple of dozen.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia combinations>>can you point to a site that perhaps has a good description> > No, but heres a page with a couple examples: http://members.aol.com/mensanator666/fun/playing.htm> Hi are well written and illuminating.using your example....C(m.n) = m!/((n!)*(m-n)!)How many 8-bit binary numbers have exactly 4 ones?C(8,4) = 8!/(4!)*(4!) = 8*7*6*5*4*3*2 / 4*3*2 * 4*3*2 = 8*7*6*5 / 4*3*2 = 7*6*5 / 3 = 7*2*5 = 70 I used the following to calculate the possible combinations of 16 bits out of 256. in other words How many 256-bit binary numbers have exactly 16 ones?.from your example I got256!/(16!*(256-16)!)I plugged this into maxima (only downloaded yesterday :)and the answer was 10078751602022313874633200my final step was to calculate how many bits was required to store this numberlog2(10078751602022313874633200) = 83.05951932 bits( had to use excel as I couldnt find how to change the base in Maxima )So I can store the original 256 bit pattern into 84 bitsDoes system of quadratic equations?Amusingly, if youve got two quadratics in three variables, such as,ax^2 + by^2 = cdx^2 + ez^2 = f,then the underlying geometric object is an elliptic curve (assuminganonsingularity condition). So there are only finitely many solutionsin integers, and the solutions in rational numbers form a finitelygenerated abelian group (with the addition of an extra point or two).Similarly, the set of real solutions form a group, and ditto for theset of complex solutions.JoeS> >I ran into a problem where I would need to solve for a system (specifically>2) of quadratic equations. My search on the net lead me to systems of>polynomial useful. I have aquestion this time that is somewhat similiar to a question a gentlemanwho posted about books that have problems that are more difficult thanyour ordinary textbook problems. However, he was concerned withundergrad material. My question is, do any of you know of any booksthat contain particularly hard and challenging College Algebra andTrigonometry problems? Ive noticed and heard from my math teacherthat problems just arent written like they used to because theauthors didnt have to worry about the solutions and therefore put invery tough and challenging problems that I just cant seem to find intodays textbooks. Anyways, I would appreciate it if anyone could shedsome light on to where I can find textbooks (probably older ones) thatyou may have used that were challenging, or any other advice for thistype of thing.Also, what are some good books that would help to teach problemsolving in general? And my last question, Ive seen on some mathcompetitions problems like How many integers between 123,456,789 and987,654,321 are divisible by both 3 and 5? or How many digits does100! have?. How does one go about solving problems like these, isthis a special type of math I should look into (discrete math maybe?)? These are obviously not an everyday problem and I think it would befun to learn how to solve problems similar to invention is an application of the> classic question, What happens when an irresistable force> meets an immoveable object?>> The answer to that classic question is: The object moves.Err.. I can apply an irresistible force quite easily to an immovable object,such as my living room wall, by pushing against it.The wall doesnt move.I do.For every action there is an equal and opposite reaction.In a more extreme situation, the blast from the tail of a rocket is anirresistible force, attempting to move the launching pad and the entireEarth Approximating Pi by Rationals> Consider the set of positive numbers x with the following property. > For all positive c and m, there exists an integer n > m such > that n x - §oor(n x) < n^-c. Does this set have measure 0?What has been posted here before: Given an arbitrary fixed c > 2. The set of positive numbers x with the property that for every m > 0 there exists an n > m such that nx - §oor (nx) < n^(-c) has Reality> Try multiplying> 349123402913412309482130498213409821340982134908123094821394012 3434234-> 234234234234123094823109482310948123094812304981230948120394812 3342342-> 942309234233423423423914230491283409213840123984012398401239423 4234234> by> 234234213412340921384021934802193480213948012398402139840231984 0123433-> 234213412342134231412304981230498213412342134123421342314231423 4234233-> 985095823409852034958340295802983450943850934803498503498503495 8454434> Well, dc doesnt hesitate to do so: ( I wrapped the lines )> 817766456652629050901008344284490254577627250171455980161202844 97781> 581332926241857587759650622403785245998354970716445468160520692 73902> 046184389942519979041425889047625868811616312463493390880337712 69107> 608902144066228087296715611544924186507226247761867313135959246 69744> 836595048721823753507192726034874829926964706377031617648727702 10340> 267851139136628652454737676922270814471739889498365746896900300 52117> 92171893556Nor does C-BChttp://www.csd.uwm.edu/~whopkins/cbc/ index.html81776645665262905090100834428449025457762725017145598 016120284497781581332926241857587759650622403785245998354970716 445468160520692739020461843899425199790414258890476258688116163 124634933908803377126910760890214406622808729671561154492418650 722624776186731313595924669744836595048721823753507192726034874 829926964706377031617648727702103402678511391366286524547376769 222708144717398894983657468969003005211792171893556I know, this is way too simple, small and trivial for FFT.I should have put down 2 10,000 digit numbers. But still,lets see Mr. Indian Magic Numbers crank this trivialmultiplication out.Heres the RIGHT way to do it by hand.Step 1: Enumerate349123402913412309482130498213409821340982134908123094 8213940123434234- 234234234234123094823109482310948123094812304981230948120394812 3342342- 942309234233423423423914230491283409213840123984012398401239423 4234234(writing up the columns top to bottom, right to left by simplecounting). 1: 3491234029134123094821304982134098213409821349081230948.. .34234 2: 6982468058268246189642609964268196426819642698162461896.. .68468 3: 10473702087402369284463914946402294640229464047243692844.. .02702 4: 13964936116536492379285219928536392853639285396324923792.. .36936 5: 17456170145670615474106524910670491067049106745406154741.. .71170 6: 20947404174804738568927829892804589280458928094487385689.. .05404 7: 24438638203938861663749134874938687493868749443568616637.. .39638 8: 27929872233072984758570439857072785707278570792649847585.. .73872 9: 31421106262207107853391744839206883920688392141731078533.. .0810610: 34912340291341230948213049821340982134098213490812309482.. .42340(Item #10 is only a redundancy check and is otherwise not necessary)Step 2: Line up, indexed by the 2nd number .23423421341234092138402193480219348021394801239840213984023198 40123433- 234213412342134231412304981230498213412342134123421342314231423 4234233- 985095823409852034958340295802983450943850934803498503498503495 84544342: 6982468058268246189642609964268196426819642698162...3: 1047370208740236928446391494640229464022946404724...4: 139649361165364923792852199285363928536392853963...2: 6982468058268246189642609964268196426819642698...3: 1047370208740236928446391494640229464022946404...4: 139649361165364923792852199285363928536392853...2: 6982468058268246189642609964268196426819642...1: 349123402913412309482130498213409821340982...3: 104737020874023692844639149464022946402294......Step 3: Add >> >>[...]>>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.logic>is not mathematical prowess, but a stop-at-nothing determination>to silence or neutralize dissenters from the canon?> > Tee-hee. Yeah, thats exactly right. Of course youre not going to> get anyone to _admit_ that thats what commands respect here, because> after all that would be giving the game away. Answer my question: Why should JSH have zero credibility, while--after your blunders, you, a Ph.D. in 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.> > Lets not forget that the Vedic principle crosswise and vertical> does also lie at the heart of this modern, bad, clumsy way...> Because the Vedic principle does not talk about the order of> the have fully grasped the subtleties of the method. Isuppose you have to be a primary school teacher or a primary schoolchild to understand that! Not talking about the order of theadditions, that being implicit, is the real strength of the method,where there is independence of columns and thus avoidance of carryingtill the very last stage where it is done in one go. Thus, thismethod is admirable for parallel processing, and can be implementedwith very few lines of code. I could easily multiply two ten thousanddigit numbers using simple C code with the definition I have given ofthis algorithm. I could never do such a thing so simply with othermethods.Please remember that very learned people can also be shown to be veryfoolish. Like, all those who believed in Aristotle, and those todaywho believe in the Special Theory of --- Myth and Reality> Arithmetic is indeed very important. And, being from India, we> have a special fascination for the decimal calculations and the> different methods.Different you say? Well, okay! Have some fascination withthis:Addition Table: + | P E S C 0 D 2 3 9 --+------------------------------------- P | 0 D 2 3 9 C0P C0E C0S C0C E | C 0 D 2 3 9 C0P C0E C0S S | S C 0 D 2 3 9 C0P C0E C | E S C 0 D 2 3 9 C0P 0 | P E S C 0 D 2 3 9 D | D09 P E S C 0 D 2 3 2 | D03 D09 P E S C 0 D 2 3 | D02 D03 D09 P E S C 0 D 9 | D0D D02 D03 D09 P E S C 0Multiplication Table: x | P E S C 0 D 2 3 9 --+------------------------------------- P | 202 D0E D0D P 0 9 C0C C03 S0S E | D0E D00 D03 E 0 3 C0E C00 C03 S | D0D D03 P S 0 2 9 C0E C0C C | P E S C 0 D 2 3 9 0 | 0 0 0 0 0 0 0 0 0 D | 9 3 2 D 0 C S E P 2 | C0C C0E 9 2 0 S P D03 D0D 3 | C03 C00 C0E 3 0 E D03 D00 D0E 9 | S0S C03 C0C 9 0 P D0D D0E 202Examples: 0.3 C.0 D.0 C.0 0.0 x 0.3 x C.0 x D.0 + D.0 + 0.0 ----- ----- ----- ----- ----- C.00 D.00 D.00 0.0 0.0The implications are both clear and striking to the perceptive:something far in advance of anything the likes of Vedic Mathematics --- Myth and RealityAddition Table: + | P E S C 0 D 2 3 9 --+-------------------------------------Multiplication Table: x | P E S C 0 D 2 3 9 --+-------------------------------------should be numbered the READ WITH HALF A LIGHTBULB?> My answer is that a bird is more than its parts and that we cannot> mangle a bird to prove that the concept of a §ying bird is invalid. > As to the present problem [ 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a) ],> we cannot mangle the left side without equally mangling the right> side, and although algebra generally allows such mangling and> collapsing of processes into ïequivalent expressions, we have to be> careful when dealing with infinity processes since the processes> generate a specific series of sums, not just any series of sums.> > What does exactly mangling mean to you in this context? What is> extra logical element for that matter? Could you please provide> rigorous definitions for these terms?As an example, lets take the taboo division by zero by substituting 1for a.1 + (a + a^2 + a^3 . . .) = 1 / (1 - a), where a = 1:1 + (1 + 1 + 1 . . .) = 1 / 0The right side of the equation asks us to count the number of timesthat we can subtract zero from one until we reach the limit of one. Every time we go through the process, we add one to our count: 1 + 1 +1 . . . ad infinitum. Thus, we can see that the right side of theequation is a short description of a specific, infinite process thatwill more or less exactly produce the output of the left side of theequation.My hypothesis is that this is true for all substitutions of a andthat we cannot place parentheses around portions of the left side thatdo not agree with the output created by the right sides actualprocess, and I believe that someone more rigorously trained than mecan figure out the rules for finding the logical processes in all theright side processes for all substitutions fields>Hey, does anyone know what the difference between a ring and a field>is? > > A field must be commutative, AND every nonzero element of the ring> must have a multiplicative inverse.> > A ring need not be commutative, nor must every nonzero element> have a multiplicative inverse.And, according to some definitions, not even a multiplicative identity.> > >As far as I can tell they are subject to the same axioms,> > There are two axioms that a field satisfies but which a ring need not> satisfy: a field F is a ring, which, in addition to all the ring> axioms, satisfies:> > x*y = y*x for all x, y in F> > and For all x in F, if x is not 0, then there exists y in F such that xy=1.> > So the axioms of a ring are a proper subset of the axioms of a> field. Every field is a ring, but not every ring is a field.> > yet I>have been told that the set of integers are members of the set of>rings, but not members of the set of fields.> > This is nonsense as written. What you mean to say, presumably, is that> the integers, with the usual addition and multiplication, are a ring> but not a field.> > (There is no set of rings and there is no set of fields)> > Yes, that is correct. Because it is not true that for every nonzero> integer x there exists an integer y such that x*y = 1. For instance,> there is no integer y such that 2*1 = 1, but 2 is a nonzero integer.> > > Also, if there is a>difference, is it true that rings are a subset of fields?> > Every field is a ring, but not every ring if a field. So you would be> trying to say that the collection of all fields is contained in the> collection of all rings. Not the other way around.> > I accept as reality.> --- Calvin (Calvin and Hobbes)> (MI-Modified) Roman Numerals (was: Vedic Mathematics...)> Ironically, multiplication is far easieror more interesting, to say the least> -- almost combinatoric in fact -- in Roman numerals. And with> a minor adjustment, the system can be made positional (even> without a zero!), with the algorithm suitable extended.The generalized IV rule is rendered as such: For sequences of M,D,C,L,X,V,I of the form S = z0 a1 z1 a2 z2 ... an zn with z0, z1, z2, ... non-increasing sequence of single digits ai (i=1,...,n) each a string of 0,1,2, or more digits all of value less than zi, the value of S is z0 + z1 + z2 + ... + zn - the sum of the values of a1,a2,...,anThe abovementioned adjustments are1) Everything under a bar counts 1000-fold2) Everything after a comma counts 1000-fold3) Everything after a dash counts 10-fold.4) Sequences are interpreted only between dashes, commas and under bars.3) Dashes not used when ligatures used for the digits I,II,III,IV,V,VI, VII,VIII,IX, denoted respectively by ASCII equivalents by i,n,m,u,v,a,z,w,y and x.>Try writing 123987739383883938989874 in [the above-mentioned>modification of the] Roman numerals [this reply is in reference to].CXXIII,XIIIM,DCCXIL,CXXDIII,CXXMIII,XIILM,XIM,CXXXMIVIn Standard FormI-II-III-IX-VIII-VII-VII-III-IX-III-VIII-III-VIII-VIII-III- IX-III-VIII-IX-VIII-IX-VIII-VII-IVor in ligature form: i n m y w z z m y m w m w w m y m w y w y w z uAnother example: 1,000,001 = IX-IX-IX-IX-X-I = y y y y x iin Standard Form, along with numerous equivalents such as:1,000,001 = XCIX-IX-IX-X-I = IC-IX-IX-X-I = IC-IX--XCX-I = IC-IX--C-I = XMIX--C-I = IM--C-I = IM,C-I = IM,MI = IMI,I = M,I = I,,I1,000,000 = y y y y y x = IX-IX-IX-IX-IX-X = XCIX-IX-IX--XCX = CMXCIX--XC-XCX = XMIX--XC-C = IM---CMC = IM---M = IM,M = IMI, = M, = I,,Multiplication 3429 by 9231 is: III,CDIXXX by IX,CCXXXI.These are snapshots of the combinatorial process which, in fact,goes on very quick on the §y and is actually very easy to do on paper: III,CDIXXX by IX,CCXXXI.The formatting is for illustration only. You can actually write everythingdown as in step 2 directly.(1) Replication [negatives go on lower line; IIIII <-> V, VV <-> X, etc. conversions done on the §y]: (III,DXXX) by (X,CCXXXI) ( ,CI ) (I, ) -> (XXX,DCLXXXXIII,) (V,CXV,D) (C , ) (CCCVI,DCCCCXXX) (I, ) (III, ,) ( ,D , ) (MXXIII,C) ( XXX, ) (XXXV,MCCXV,MCCCCXXX) (III,MDLXIII,CCCXXXI) -> (XXXII,II ,MC) ( ,CCCL,I) -> (XXXI,MIII,C) ( ,CCCL,I) = XXXI,CCCLMIII,ICwhich is the answer, also equivalently expressed in any of the forms: = XXXI,LDCCIII,IC = XXXI,DCLCIII,IC = XXXI,DCLIII,ICor 31,653,099 in decimal equivalent. Also, are the other equivalents: XXXI,DCLIII,IC = XXXI,DCLIII,XCIX = XXXI---DCLIII---XCIX = III-I--LXV-III--IX-IX = III-I-VI-V-III--IX-IX = III-I-VI-V-II-X-IX-IX = m i a v n x y ywith the final two equivalents in Standard Form and Standard square, given radius, find maximum number of circles> > >Any pointers to a solution of the follwoing inverse circle packing>problem will be greatly appreciated:> > Given a unit square, and a radius R, how does one find the maximum> number of circles, each of radius R, that can be packed into the square?> > Google for packing circles square. Let r(n) be the largest> radius of n equal circles that can be packed in a unit square.> If r(n+1) < R <= r(n) then your answer is n. AFAIK there is no> really effective way to answer these questions for large n (or> equivalently small R), although there are methods that will probably> come close to the correct answer. Large in this context could> be a couple of dozen.> > Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2See also the OEIS entryhttp://www.research.att.com/projects/OEIS?Anum=A084616and the links given there. AFAIK E. Specht has extended his numericalsearch to several 100 circles, with results stable up College Algebra/Trig Problems, misc.With the problems you put forth, I would recommend a book on Number theory.You are in a college prep school, right? Then, I would look on Amazon.comfor some good books.Try:http://www.amazon.com/exec/obidos/tg/detail/-/ 0486409171/qid=1069290624/sr=8-1/ref=sr_8_1/102-0071628- 6208175?v=glance&n=507846http://www.amazon.com/exec/obidos/tg/ detail/-/0387982191/qid=1069290624/sr=8-2/ref=sr_8_2/102- 0071628-6208175?v=glance&n=507846You can also try the books by George Polya. He has some interestingproblems and methods to this group before and found it very useful. I have a> question this time that is somewhat similiar to a question a gentleman> who posted about books that have problems that are more difficult than> your ordinary textbook problems. However, he was concerned with> undergrad material. My question is, do any of you know of any books> that contain particularly hard and challenging College Algebra and> Trigonometry problems? Ive noticed and heard from my math teacher> that problems just arent written like they used to because the> authors didnt have to worry about the solutions and therefore put in> very tough and challenging problems that I just cant seem to find in> todays textbooks. Anyways, I would appreciate it if anyone could shed> some light on to where I can find textbooks (probably older ones) that> you may have used that were challenging, or any other advice for this> type of thing.>> Also, what are some good books that would help to teach problem> solving in general? And my last question, Ive seen on some math> competitions problems like How many integers between 123,456,789 and> 987,654,321 are divisible by both 3 and 5? or How many digits does> 100! have?. How does one go about solving problems like these, is> this a special type of math I should look into (discrete math maybe?)> ? These are obviously not an everyday problem and I think it would be> fun 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.> > The _calculus part_ of elementary school-arithmetic can> _hardly_ be called mathematics, thats all im saying.You are entitled to your opinion. > > 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. Thats an important point, yes. But restricting to numbers only> doesnt make maths very exciting for laymen, im afraid.> Personally, i would never have gotten any interest in mathematics> if not for geometry, reasoning from axioms, etc. Historically,> mathematical progress always ended fairly soon, in all cultures> that restricted maths to ïdoing calculations.When they do calculations properly, they succeed. The success ofmodern Western cultures lies very largely in their ability to docalculations very fast, using computers for number crunching, andterrific use of calculations in all aspects of engineering, finance,banking, selling, 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.> > Yes. But i meant something different, namely the everyday life> use of the word, not the order of education.Our education in mathemtics makes for our comprehension of the word.> To add: the word mathematics is used with two meanings:> one denotes the academic discipline, the other a synonym of> arithmetic. (Anyway thats the way it seems to me.)That is not the way it seems to me. To me arithmetic is what youfirst learn in mathematics. It is the first and most basic subject.> > > > 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 Western> world? Exactly who is doing the PR?> > Years ago i was present in a lecture about Vedic maths, already.> Someone visited universities with it. Recently, ive seen> a website with the same story. The impression they leave with me> is that they are plain PR, so i take the liberty to call it PR.That way everything that is demonstrated or talked about is PR. > (Your own posts are also PR for vedic maths. BTW, I dont> consider PR to be a dirty word.)Fair enough. But to me PR is something you do when you want to sellsomething. You are saying it is not a dirty word now, but in earlierposts your tone was that it was PR, a cheap selling trick, and thosewho promoted vedic maths were not trustworthy at all. Pleaseremember, I am not trying to sell you anything, or convince you ofanything. It does not matter very much to me, what you or anyone maypersonally think of Vedic maths. Ultimately, the best product in afraud-free environment needs no selling effort - that is why it is thebest product. What does matter to me, is whether the book is a fraudor not, derived from Western sources. Because this is what is beingclaimed by certain parties. While you and others have pointed outsome Western sources that may have described the ancient method, noone has claimed that they are absolutely original. One may find out,after consulting them, that such methods in Tractenberg were derivedfrom Vedic maths, as he may have acknowledged same! Even if themethod is original from Tractenberg, and not Vedic-Indian at all, itwould still remain a very good method that should be taught in primaryschools.> > 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.> > I wasnt talking here about the multiplication method,> but about some of the other tricks, as far as ive seen them.But the whole of Vedic Mathematics (including the multiplicationmethod) was being described as a fraud, of no use at all. Now that Ihave explained the multiplication, that is being left out of theaccusation! Who knows, when I may explain the other methods, howthinking might change!> 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.> > The alternative multiplication method is absolutely welcome.I dont know the rest of Vedic arithmetic, and I dont think I willknow till I get that book. Which may take a few years, I am afraid,unless someone gives me a copy. In the meantime I would be happy ifyou and others state that you dont know any one-line division andone-line square root extraction methods, of similar elegance as theVedic multiplication method. Maybe those who championed Vedic mathsdid not know what they were talking about. Like most of 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.> > Ah... now we suddenly are talking about more than the> alternative multiplication method?Apart from the multiplication, I have showed by example how well thatwill go with addition and subtraction. All the places areindependent, and the carry will be done at the last stage. Extremelyelegant. Now, if there is a division scheme that good, allcomputation could be revolutionised for speed and efficiency.> As said, the multiplication method is welcome. But what> ive seen of the rest makes me hesitate a lot.All depends, upon who shows what, and how.Arindam Banerjee.> > graphThanx Randy! You pointed me in the right direction with MATHEMATICIANS READ WITH HALF A LIGHTBULB?> In the proof you gave you did this (or something similar):> > (a + a^2 + a^3 + ...) - (a^2 + a^3 + ...) = a> > However, for a in outside the range <-1, 1> this boils down to> infinity - infinity and the result of that operation is unknown (so> you cant say it is a). For an a in the range <-1,1>, this will result> to {some number}-{some other number). This operation has a result and> therefore your proof is valid for this range of a.We commonly separate out finite elements from disparate infinitieswhen we perform math with inequalities. Wheres the problem then?Your other argument (not quoted above) that a denser infinity and aless dense infinity have equal elements is true, but here we aretalking about two infinities with exactly paired elements at the samepoints on the number line (except one point extra in one infinity inthe non-infinite direction) whose elements are exactly the same(powers of a), and thus we can match absolutely equal portions ofthe two infinities at any common point on the number line. Thisallows us to compare a pair of apples, one with a stem, and onewithout. This is not an apple asking -- and the answer is no, I do 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 > 0> there exists an integer N such that> > |s - (a_1 + ... + a_n)| < eps for every existIm not sure that anyone is following this thread, but I just realized thatthere was something I meant to add. This post is kindof silly, but Illpost it anyway:Infinitesimals dont make sense in the standard def. of the real numbers.So it is pointless to try to prove they exist.What I mean is, an infinitesimal would have the following property.if b is a real s.t. b>1, 1+infinitesimal < b . Also, 1+infinitesimal > 1.This implies that (1+infinitesimal) is not a real number, since the realsare hausdorff. For example, using axiom 4 from the link, there would haveto be 2 disjoint open sets which contain 1 and 1+infinitesimalrespectivally.http://mathworld.wolfram.com/ TopologicalSpace.htmlJustin Van Winkle> Nope.> You can add something of ïlength one to (1,1,1...), however it will be> something like sqrt(6/pi^2 )*(1/2, 1/3, 1/4, 1/5, ...). So thecoordinates> are (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_ns> are real,> vectors that have finite length are those such that,summing over thenatural> numbers,> Sum(x_n^2) is finite. All finite length vectors of are of this form, and> all 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 a> vector of length one along the line determined by (1,1,1,...), that this> implies the existence of infinitesimals. Since no such vector exists,this> only 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 a> contridiction, so no such vector exists. Additionally, no real number> exists 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 LIGHTBULB?> 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 that> it exists is not allowed.> N.O.Way has given the correct proof that the limit> only exists when -1 < a < 1. In his proof he did use> associativity and distributivity of finite sums only.> Your math has given an example of what happens> when you apply these ïproperties where you are> not allowed to use them.Im listening, and I have to leave open the possibility that you arecorrect.However, your argument boils down to: ïyou cant do it, and theresults of trying to do so give results we call invalid, and thatsour proof.I still believe that those invalid forms hide a discernible underlyinglogic that is consistent and goes deeper than weve looked at. As Islowly learn math and more logic, Ill keep my mind and eyes Ironically, multiplication is far easier -- almost combinatoric in fact --> in Roman numerals. And with a minor adjustment, the system can be> made positional (even without a zero!), with the algorithm suitable> extended.> ...> The generalized IV rule is rendered as such:> For sequences of M,D,C,L,X,V,I of the form> S = z0 a1 z1 a2 z2 ... an zn> with z0 > z1 > z2 > ... > zn each single digits;> ai (i=1,...,n) each a string of 0,1,2, or more digits> all of value less than zi,> the value of S is> z0 + z1 + z2 + ... + zn - the sum of the values of a1,a2,...,an> > The abovementioned adjustments are> (1) Everything under a bar counts 1000-fold> (2) Everything after a comma counts 1000-fold> (3) Everything after a dash counts 10-fold.> (4) Sequences are interpreted only between dashes, commas and > under bars.> > That makes the Roman system a superset of a positional 0-less > system, with an interesting combinatorial addition and multiplication > a lgorithm.> ...> I generally go 36*49 = 1254 + 90*13 - 660 = 1764,> or 57*28 = 1056 + 120*10 - 660 = 1596,> or 73*39 = 2127 + 100*12 - 480 = 2847.Interesting. Can you actually do any serious arithmetic with thisextended Roman system? Say calculating portfolio returns or elementary LIGHTBULB?> By the same logic, 6/0 = infinity, therefore infinity * 0 = 6. We then > have> > 5 = Infinity * 0 = 6> > How do you deal with that?They do not arise from the same beginning process, so they are not equal.Very HALF A LIGHTBULB?> That does not logically follow. The question ïWhat happens when an> irresistable force (a force that can move any object) meets an> unmovable object? can simply not be answered. The only thing that> logically follows from the concepts of an unmovable object and an> irresistable force is that they cannot both exist.That seems reasonable, and I wont argue with it directly. I need tostudy my copy of INTRODUCTION TO LOGIC, by Irving Copi, up through thetruth tables and symbolic logic that comes to that discussion. Ienjoy Aristotelian syllogisms and sorites as a pattern of thinking andargument, which I studied and used for legal work, and Ive begunstudying a limited amount of symbolic logic, but I havent got arational clue about how Copis statement about an irresistable forceand an unmovable object leads to EVERYTHING is rational. However,Copi is no idiot.Very existJustin Van Winkle grava .88 la saucisse et au marteau:> Im not sure that anyone is following this thread, but I just realized that> there was something I meant to add. This post is kindof silly, but Ill> post it anyway:> > Infinitesimals dont make sense in the standard def. of the real numbers.> So it is pointless to try to prove they exist.If you want to play with infinitesimals, you can search Mathematics --- Myth and Reality> > > > > > > The idea that the historicity of antiquities/artifacts is to be > > determined by a political process and not by independent > > research is as revolutionary as the design of your > > perpetual motion machine.> > My research on perpetual motion machines was an independent process.> > Research is an independent process but claims without > proper verification may amount to fraud with the society. Putting up new ideas that challenge old ones, and designs, are notclaims. They are new ideas that are not proven.> You have CLAIMED to have designed perpetual motion machine.I have given the design for same. I have not said that it works,since it has not been made. No one has said that it will not work,after going through the design. The mathematical theory no one hasproved wrong.To say that I am claiming anything, when all I have actually done isto publish new ideas and designs on a website and defend them inUsenet, is to be a LIAR which of course is exactly what you are, andalways have been, as is all too well know. A shameless LIAR and anot-so-cunning TWISTER.> You have CLAIMED to have proved Einsteins theories wrong.they are completely wrong. I will claim they are wrong when weactually produce energy without destroying mass, and when we will gofaster than light.Again, you are LYING and TWISTING, following your usual fashion.> These are extremely tall claims which are not supported > by any independent verification process.They are not *claims* at all. It is only you who is saying that theyare claims, in your familiar lying and twisting style. They are aninvitation to further research and thought in certain new directions. If no one is interested, that is not *my* fault. All I can see isthat no one is interested in helping me in any way. Fair enough, Iwill do what I can on my own, at my own pace, as and when time andopportunity permit.People who say I am wrong just make statements that I am wrong, in thesilly obtuse manner which is also your style. They do not counteranything I have stated. Naturally, I cannot take them seriously.> > It still is so. However, confirming historicity is a political> process. Historians are paid by kings/presidents/ministers/etc. via> universities.> > Not only historians but scientists too are paid by > kings/presidents/ministers/etc.. What is your point?They are forced to suck up to kings/presidents/ministers/etc. and so,pander to their biases. Arp [JSH 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.html> or (more fun) do your own googling with shrewd combinations of keywords. I> tried> moron factorization bull> and Google found James Harris in 0.24 seconds.> > This is just great. Now JSH can claim to have Google against him in> this massive international conspiracy to silence him. Eventually,> hell be the lone hero battling an entire universe controlled by Satan> or something else of similar magnitude. Dont fan the 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. Now the begging question is> whether or not there is a polynomial q(x) such that q(x)^2 has FEWER> non-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 7th> degree 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) the set D={Z(p)-Z(p^2)|p in P}. Is D finite? If so, what is the> maximum? Or at least an upper bound.MR0393387 (52 #14197)Freud, R.97bertOn the minimum number of terms in the square of a polynomial. (Hungarian. English, Russian summary)Mat. Lapok 24 (1973), 95--98 (1975).26A75 (10K20)Let $scr P$ be the class of polynomials $P_k$ with $k$ terms and integral coefficients. Let $Q(k)$ be the smallest number of terms in $P_k{}^2$ as $P_k$ varies over $scr P$. The author shows that $Q(k)/krightarrow 0$ as $krightarrow+infty$. In fact, he presents an upper bound on $Q(k)$ implying the above limit. This extends a result of P. Erd.9as [Nieuw Arch. Wiskunde (2) 23 (1949), 63--65; MR 10, 354] who obtained the above result when the coefficients of $P_k$ are assumed to be real numbers instead of integers.Reviewed by J. GalambosMR0027779 (10,354b)Erd.9as, P.On the number of terms of the square of a polynomial.Nieuw Arch. Wiskunde (2) 23, (1949). 63--65.10.0XLet $Q(k)$ be the least number of terms occurring among the squares $f_k(x)^2$ of all polynomials $f_k(x)$, where $f_k(x)$ has rational coefficients and exactly $k$ nonzero terms. It is proved that there exist positive constants $c_1,c_2$ such that $c_1<1$ and $Q(k) >> 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.> > >The poiunt is that tan(n) is big in absolute value iff n is close to>an odd multiple of pi/2, that is if pi/2 is approximately n/m>where 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 be>bounded away from zero. Its not obvious that there would be infintely>many cases with m odd, and to get a better result we need a better>Diophantine approximation result on pi/2. I dont know if there>is such a result.> > > Yes there is. See my posting on the Subject A limit problem with > explanation in sci.math.symbolic on 17 October 2001. But to get> |tan n/n| to take on arbitrarily large values, I think youll need > the continued fraction of pi to have unbounded elements. Almost> certainly its true, but we cant prove it. OTOH if x is a quadratic> irrational, since the continued fraction of x has bounded elements> the sequence tan(n pi x)/n will be bounded.Nice thread (nice diversion at 13 too). I dont think I saw anything in it that said either way whether tan(n)/ns bevaviour can be modelled by a random variable or not. If so, whats the distribution, and what kind of growth of the maxima can be expected? A parallel question is what happens if we strictly limit ourselves to the contfrac convergents?Yeah, I know, this is just numeric noodling, and not mathematically significant, but numerical noodling is fun. Ive pushed the extrema but due to the increased precision required Im suffering from a bit of a slow-down.conv# tan(n)/n n (first few digits)8872 -829.712489 449204777921823865995697961987799391078140834412081313226801 $9768 -958.007133 429202408867171028407029086630096030297054817541194937713460 $11282 -2534.645599 172419407891290112240403334491844273665014554087732835463547 $12284 -5430.634611 784868065223015048621954304085669560818256728978805648108777 $15503 24263.751532 693634101852993659879763497130205978431745222404157619617265 $24604 -12702.238257 789589514919363342054150946805086978105605233603412704914213$ checked to contfrac convergent #30000Thats only about 3 GHz hrs of CPU time, which implies that if someone were interested in taking it to convergent #100000 its only a day or sos work on a fast box. Do we have enough data-points to conjecture how far the extrema could be pushed in those few days? Just visually it looks like its not wildly far from linear in n at the moment, but there are so few data points. David (C.) - got a few spare GHz days? (Im stopping there, as Ive got primes to find.)Heres the (prettified!) GP script for the run I just did:p40000 need high precision.cf=contfrac(Pi/2,,30000); how far do you want to go?mint=0;maxt=0; should now be -12702 and 24263n=11;d=7;on=3;od=2; start at convergent 5. Dont ask why.for(i=5,30000, again how far do you want to go? m=cf[i]; t=n*m+on;on=n;n=t; t=d*m+od;od=d;d=t; find the next convergent if(i>5990, dont do stuff Ive already done t=tan(n)/n; if(((t>maxt)&&(maxt=t))||((t Putting up new ideas that challenge old ones, and designs, are not> claims. They are new ideas that are not proven.> I have given the design for same. I have not said that it works,> since it has not been made. No one has said that it will not work,> after going through the design. The mathematical theory no one has> proved wrong.> To say that I am claiming anything, when all I have actually done is> to publish new ideas and designs on a website and defend them in> Usenet, is to be a LIAR which of course is exactly what you are, and> always have been, as is all too well know. A shameless LIAR and a> not-so-cunning TWISTER.> they are completely wrong. I will claim they are wrong when we> actually produce energy without destroying mass, and when we will go> faster than light.> > Again, you are LYING and TWISTING, following your usual fashion. ...and more ranting in this fashion.What a ing lunatic.-- -------------------------------------------------------------- ----Got to get behind the mulein the morning and rid of embarrassing age spots Well it takes care of business TRUST YOURSELF swamp gas PAK CHOOIE UNF ultra-violent light Jump The Shark illuminati IYKWIM I believe Scientific Proof Of === God I believe in Jesus testifySubject: Mathematical >n}5I!mZ_u;dzR:@ZProhS[~Thanx Randy! You pointed me in the right direction with the first>link. Much appreciated...That last ones pretty good. I bookmarked it for myself. I especiallylike their examples of bad data presentation. One is a magazine coverfrom Ithaca, NY (home of Cornell Univ) with scary graphs that showCornell tuition going up at the same time as Cornell quality rankingapparently going down.The graphs on the cover are reprints (with no axes) of graphs thatappear inside (with axes). When the axes are labelled, you find that: (a) they are on completely different time scales, and (b) the sharp dip in Cornell ranking at what purports to bethe same time as the tuition rise is because previously it was ranked#15, and now Maybe some scientist (of MATHEMATICS) can explain why mathematicians > are spending their time developing mathematical ways to generate pr0n?Mostly because they dont Re: (statistics)how to make date more like Laplacian my question? Oh, its my problem> that I did not clearly present the background...Partly. Partly also you posed the question in a slightly troll-likeway. :-)> Here is the story: in deblocking of block DCT coded JPEG images, it was> known that the DCTed coefficients are Laplacian distributed... But now I am> looking at low bit rate JPEG images, so there are someking of artifacts...> in order to reconstruct the original images... many algorithms have been> devised... one possibility is to make the image coefficients more Laplcian> like...These dudes appear to use the fact that the coefficients are Laplacianto a reasonable level of accuracy:http://bmrc.berkeley.edu/research/publications/1996/ 110/imdsp.pshttp://citeseer.nj.nec.com/smoot96study.htmlSo what makes you say your coefficients are not? Are you looking atthe right coefficients?> So that came my question: how to make data more Laplican like...Please give> me some detailed explanation as I am not veteran in statistics...Well, what variations do you have available? Does your modificationhave to fit into JPEG? Or can it be seen as a completely separatestep?Usually, if you want to convert a random variable with onedistribution to another distribution you need to find a function thatrelates the two random variables. A first, naive, approach might be to use the histogram of the data youhave to find a piece-wise linear function that rescales the data sothat its histogram is Laplacian.This technique is sometimes used in image processing to make greyscaleimages have a uniform brightness distribution (see e.g. http://www.cs.tcd.ie/Fergal.Shevlin/courses/4d4/4BA10/ PixelBrightnessTransformations.pdf)Ciao,Peter K.-- Peter J. KootsookosI will ignore all ideas for new works [..], the invention of which has reached its limits and for whose improvement I see no further hope.- Julius Frontinus, c. AD you angry or something? Why are you throwing some cryptictables? How does the square root method work - that was my question.Arjoe> >>Arithmetic is indeed very important. And, being from India, we>>have a special fascination for the decimal calculations and the>>different methods.> > Different you say? Well, okay! Have some fascination with> this:> > Addition Table:> + | P E S C 0 D 2 3 9> --+-------------------------------------> P | 0 D 2 3 9 C0P C0E C0S C0C> E | C 0 D 2 3 9 C0P C0E C0S> S | S C 0 D 2 3 9 C0P C0E> C | E S C 0 D 2 3 9 C0P> 0 | P E S C 0 D 2 3 9> D | D09 P E S C 0 D 2 3> 2 | D03 D09 P E S C 0 D 2> 3 | D02 D03 D09 P E S C 0 D> 9 | D0D D02 D03 D09 P E S C 0> > Multiplication Table:> x | P E S C 0 D 2 3 9> --+-------------------------------------> P | 202 D0E D0D P 0 9 C0C C03 S0S> E | D0E D00 D03 E 0 3 C0E C00 C03> S | D0D D03 P S 0 2 9 C0E C0C> C | P E S C 0 D 2 3 9> 0 | 0 0 0 0 0 0 0 0 0> D | 9 3 2 D 0 C S E P> 2 | C0C C0E 9 2 0 S P D03 D0D> 3 | C03 C00 C0E 3 0 E D03 D00 D0E> 9 | S0S C03 C0C 9 0 P D0D D0E 202> > Examples:> > 0.3 C.0 D.0 C.0 0.0> x 0.3 x C.0 x D.0 + D.0 + 0.0> ----- ----- ----- ----- -----> C.00 D.00 D.00 0.0 0.0> > The implications are both clear and striking to the perceptive:> something far in advance of anything the likes of all who replied.> I learned a lot, not the least of which was that> sin(1) + sin(2) + ... + sin(n) is bounded independent of n.That statement intrigued me. I searched throughout this thread, to no avail,for information about that. So I investigated sin(1) + sin(2) + ... + sin(n)myself. For anyone whos interested, I found the lub and glb to be precisely 1/2*cot(1/4) and -1/2*tan(1/4)respectively.David equations?>> I ran into a problem where I would need to solve for a system>(specifically>> 2) of quadratic equations.>> There is a method of Newton for finding the zeros of>> arbitrary functions. In one dimension, you draw the tangent>> line at your current value of x, then go to where that>> tangent line intersects the x-axis.>> This idea is easily extended to functions of multiple>> variables and to simultaneous nonlinear equations>> like yours: make a linear approximation near your>> current value of x, then solve that linear system>> for your next value (or at least for the direction>> to move).>>Hmm. As far as I know there is no Convergence proved for the case of>arbitrary functions.Right. Thats why I included the paragraph talking about secondderivatives and how they affected convergence, and how far away fromsolutions it often diverges. I believe I also mentioned that there aremore robust methods out there. If you download code from a place likeNetlib, it will likely be fairly modern in its algorithms.> Is this different in my case?No, you have to worry about second derivatives as well. In your case(I said this already, but you snipped the paragraph) the question iswhether the matrices A and B are positive definite. If they are not,you have nonconvex functions and simple-minded algorithms will havetrouble or fail.>What about retrieving all roots? Doesnt Newton converge (if it does) to>just one root?If there are more than one, then which one it converges to will dependon the starting point and the sizes of your steps. That is true of allsearch methods. >> Heres one discussion:>> you, I will try this one.>> Im not clear if you want the theory or just pre-canned>> software. Both are widely available on the web.>>Both of course ;-)>In the meantime I am more interested in the theory, because I have solved my>problem at hand because I made succesful use of additional properties of my>equations. But it might be the case, that I will need to solve for other>similar problems, where these specific structure is not available. I just>want to be prepared.>Do>> a search for nonlinear simultaneous equations. For>> software, Netlib around a lot before posting to this group.Did you browse through Netlibs repository? Many top-of-the-linealgorithms end up there, and theyre free.>What I was thinking of was something like a theory about quadratic>simultaneous systems extending of what I know about linear systems.The link I gave Randy! I thought it was something like that, but I am not used to>using Excel.Its a property of any system doing numerical computation. Matlab doesthe same thing. So would a math library in FORTRAN or C++.> I usually use Mathematica, or something.Systems like Mathematica and Maple escape round-off error by doingcalculations symbolically, as we do.> I am just trying to>learn Excel so that I can add it to my resume.I recommend learning the optimizer. Its a powerful solver that doesinteger programs, linear programs, and nonlinear optimization, andmany people dont even know its there. - questionDoes anyone here know why the Lindenberg version of the Central LimitTheorem implies the traditional Central Limit Theorem?I.e.Lindenberg version :Let X_{n,k} be independent, E(X_{n,k}) = 0 and Sum (from k = 1 to n) ofVariance(X_{n,k}) = (s_n)^2Then lim P[ (X_{n,1} + .... + X_{n,n}) / s_n <= x] = phi(x) where phiis the cumulative normal distribution and <= denotes less than or equal toprovided thatlim Sum (from k = 1 to n) of E[ (X_{n,k})^2 * 1(|X_{n,k}| > e*s_n) ] = 0for all e > 0and E is expectationTradictional version :If X_k are i.i.d and E(X_k) = 0 and Var(X_k) = 1, thenlim P [ (X_1 + ... + X_n) scientist (of MATHEMATICS) can explain why mathematicians>are spending their time developing mathematical ways to generate pr0n?> http://perpetualocean.com/amgallery8.html>Whats wrong with human pr0n? Why are mathematicians taking the first>step on the slippery slope towards ROBOT PR0N??? WHERE WILL IT END??http://www.sixsixfive.com/195.htmlAlthough Bjorks All Is Full of Love video has to be theBest. Robot Pr0n. Ever. ïxcept that its so lovely and good.-- Chimes peal joy. Bah. Joseph Michael Bay Icy colon barge Cancer Biology Frosty divine Saturn Stanford University www.stanford.edu/~jmbay/ least of which was that> sin(1) + sin(2) + ... + sin(n) is bounded independent of n.> > That statement intrigued me. I searched throughout this thread, to no avail,> for information about that. So I investigated sin(1) + sin(2) + ... + sin(n)> myself. For anyone whos interested, I found the lub and glb to be precisely> > 1/2*cot(1/4) and -1/2*tan(1/4)> > respectively.Yes, thats right. To sum sin(x) + sin(2x) + ... + sin(nx), regardthis as being the imaginary part of sum_{k=1}^n (e^{ix})^n and equateto the imaginary part of the sum of this geometric series.This used to be taught in courses on trigonometric series (e.g. tocalculate the Fejer kernel). But are those taught anymore (to anybodybut engineers)?--Ron relationship between Goldbachsconjecture and the great prime number theorem. Let G(n) be the numberof prime pairs per n such that the sum of each pair equals n. ThusG(10)=2 because of 5+5 and 7+3 etc.Plotting G(n) the graph displays a definite bottom curve that looks like C(n)=f(n)/ln(f(n)). The only problem is what is f(n). See my pageathttp://www.geocities.com/dirkie6/goldbach.PDF to see my very useful. I have a> question this time that is somewhat similiar to a question a gentleman> who posted about books that have problems that are more difficult than> your ordinary textbook problems. However, he was concerned with> undergrad material. My question is, do any of you know of any books> that contain particularly hard and challenging College Algebra and> Trigonometry problems? Ive noticed and heard from my math teacher> that problems just arent written like they used to because the> authors didnt have to worry about the solutions and therefore put in> very tough and challenging problems that I just cant seem to find in> todays textbooks. Anyways, I would appreciate it if anyone could shed> some light on to where I can find textbooks (probably older ones) that> you may have used that were challenging, or any other advice for this> type of thing.> > Also, what are some good books that would help to teach problem> solving in general? And my last question, Ive seen on some math> competitions problems like How many integers between 123,456,789 and> 987,654,321 are divisible by both 3 and 5? or How many digits does> 100! have?. How does one go about solving problems like these, is> this a special type of math I should look into (discrete math maybe?)> ? These are obviously not an everyday problem and I think it would be> fun to learn how to solve problems similar to these.> There are many many many books on stuff like this. The best way toget started is to find a book geared towards teaching basic problemsolving skills. Most books you will find will probably be tooadvanced, however. I recommend one called Polynomials, by Barbeau. This one is not geared specifically towards problem solving, but theauthor does claim that to be one of the target audiences of the book. This book is VERY readable and the exercises range from extremelysimple to very challenging. Even most graduate students could learnalot of interesting things from this book, all presented in a veryeasy manner. I think this will develop your mind towards problemsolving in general, and you will be able to take on a more advancedbook after that.Now, about your question about being disivible by 3 and 5. A numberis divisible by both 3 and 5 if and only if it is divisible by 15. Sothe problem is equivalent to figuring out the number of integers inthat range which are divisible by 15. Now, long division shows that123,456,789 = 8230452*15 + 9987,654,321 = 65843621*15 + 6But what does this say really? It says that it takes 8230452multiples of 15 to get to 987,654,321, and 65843621 multiples of 15 toget to 987,654,321. So the difference of these two numbers is thenumber of multiples in between. Then you need to subtract 1, sincethe 8230452th multiple of 15 is right BEFORE 123,456,789. So theanswer is 65843621 - 8230452 - 1 = 57613168.To answer your other question, how does one go about solving problemssuch as this, theres really no good answer to this question. Yearsof hard work practicing problems of this nature, reading solutions(but only after alot of thought), and making conjectures of your ownand trying to prove them is the only way I know of. Besides, theresno general strategy for every problem. Every problem is different,and especially when it comes to competition type problems, theyregoing to throw everything in the book at you, and youll have to usetechniques from all over, plus alot of insight and intuition(which is developed through years of hard work, as mentioned above).If you finish the Polynomials book, Id recommend that a second readbe something in the realm of algorithmIm working on implementing Dixons character table algorithm,more-or-less as a matter of personal interest at this point. Myintent is to get the basic (and, I gather, slower) algorithmdown, and possibly implement the Dixon-Schneider algorithm downthe road.The algorithm, which at first seemed entirely opaque andsubsequently seemed pretty straightforward, now seems a littlemore subtle than my second impression had led me to believe...Gotta love that.Im hoping somebody here can lead me in the direction of anIn particular, a couple of things seem ambiguous to me rightnow... First, the algorithm requires the determination of thenullspace of an integer matrix mod p. Although Im familiarwith modular arithmetic at a basic level, its not clear to mewhether: (1) the matrix is operated on by the mod p, then thenullspace found, or (2) the whole operation is done mod p. Ifthe latter, I am (apparently) entirely unclear on how its done.Second, Dixon refers to considering the action of a matrix on aset in refining the eigenvectors. What is meant by consideringis also unclear to me...Any help you guys could offer would be great; Im at wits endtrying to make sense of this myself... ---------- . . . Except when they dont, Because sometimes they wont. - Dr. Seuss--------------------------------------------------------- (statistics)how to make date more like Laplacian distribution?>> ...>> Now for a simple example. Lets say you have a fair die. Each time youtoss> it, the probabily of each outcome is 1/6. So theory says we have auniform> distribution. Now toss the die 100 times and count how many times eachvalue> shows up. Now theory says on average each value shows up 1/6 of thetime.> However in any given sample (set of measurements), you cant expect this> perfect a distribution every time. Why you ask? That is because eachtrial> is independent of the prior ones. And in this situation I only tossed it100> times and 100 is not a multiple of six, so not all bins can have thesame> number of counts. Since the counts must be integers and 100/6 =16.6666666,> you can see the one problem.>> ...>> This can raise an unrealistic expectation. If we toss the die 102 times,> can we then ask that each face of an honest die to turn up 17 times?Jerry,Yes, we can ask but this will only happen about once every 49300 times.I was trying to illustrate two different points, the first was the casewhere the number of die values doesnt divide evenly into the sample size.And the other which is the more important is the case where we dont expecteach value (face value on a die) to show up in perfectly matched counts intrial runs. I know that you know this, but I think Walala now gets thepoint - I hope.Clay>> Jerry> -- > Engineering is the art of making what you want from things you can get.> pr0n>> Maybe some scientist (of MATHEMATICS) can explain why mathematicians> are spending their time developing mathematical ways to generate pr0n?You see, when a theoretician loves a set of axioms VERY much...[ CENSORED ]...Oh, yes, baby, yes, YES *YES*! ...Q.E.D.! Queue Eeeee Deeeeeeeeeeeee!> Whats wrong with human pr0n? Why are mathematicians taking the first> step on the slippery slope towards ROBOT PR0N???Hold on there! Once you use one slippery slope argument, its tempting totry to apply them everywhere!Soon therell be nothing to eat but boiled libertarian frogs!> WHERE WILL IT END??Well, I dont know, but when it comes to ROBOT PR0N, I think that HajimeSorayama was the beginning of the end.[ Long pause, chirping crickets ]See, because his name is Hajime, and he...and were talking aboutthe...and in Japanese...oh, never mind.> Will all pr0n stars be thrown onto the scrap heap as no longer> required once mathematicians have developed the ultimate pr0nographic> algorithm? Is there no hope for Pamela Anderson to ever work again?Hmm...theres a remake of John Henry crying out to be made here.(Obvious Bag: But my margin is too small to contain it)Still, I think Pamela Anderson might not be the best choice if yourelooking for all-natural Certified Organic Pr0n. I mean, what with theimplants, and the Tool Time, and the Barb Wire, mwa-hey, and oh lifeguardlady, and Woody Allen is running away, from the bouncing, mw-hurgn-whey.-- .:*~*:. .:*~*:. .:*~*:. .:*~*:. .:*~*:. `*._.:* :// `*._.:* ers. `*._.:* net/ `*._.:*.:*~*:. http .:*~*:. memb .:*~*:. cox. .:*~*:. Re: Vedic Mathematics --- Myth and Reality> Arithmetic is indeed very important. And, being from India, we> have a special fascination for the decimal calculations and the> different methods. In fact, there is quite a bit of research to> implement decimal computing and storage using electronic devices.> For example, at one point of time I got fascinated by the possible> application of the very interesting resonant tunneling devices for> One 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).Decimal storage sounds interesting, but I dont know exactly how itcan be any kind of improvement to memory storage. Unless you can packfour state devices in around the same real estate as binary, ofcourse. Then memory capacity will increase by a reasonable factor. So far as computation is concerned, yes, there should be a definitegain with four-level logic instead of binary. But I dont see how 4state will lead to decimal computing. If you have states 0 1 2 3, youneed 3 devices to write 10. In binary logic, you need 4 binarydevices to write 10.> Those interested in electon devices and multivalued memories may> find the paper interesting. But, at the same time, I always like> such arithmetic methods where the learner can clearly see why the> method is working. For example, the basic square root finding> method 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> ---> > ArjoeThat is the very same method I learnt! We did it blindly, like we didmultiplication blindly, and had no end of trouble with the carry insubtraction.See if you can get that book on Vedic Mathematics. That should givesome insight. I === Subject: Fields in Complex Space-TimeKaiser shows how dynamical fields F are extended to complex space-time using an Analytic Signal Transform whose AST Fw is a wavelet, i.e. eq 14 p. 6 of 48 whereFw = F(x [CapitalEth] sy/u) i.e. multi-scale resolution WAVELET!Z = x [CapitalEth] iyThe Kernel in the AST integral is (s [CapitalEth] iu)^-1s is integrated from [CapitalEth] to + infinityu is imaginary part of complex y.Note that Glauber coherent states are MACRO-QUANTUM states when |Z| >> 1 where generalizing from a single mode quantum harmonic oscillatora|Z> = Z|Z>[a,a*] = 1You can squeeze these states.Much more generally in phase space not confined to a single mode quantum harmonic oscillator there is, as shown by Kaiser, a generalized version of the above simple toy model that applies to solutions of Klein-Gordon, Dirac and Maxwell equations in 4Dim space-time generalized to complex space-time, which in the special virtual off mass shell BEC ODLRO MACRO-QUANTUM VACUUM yields: --> Higgs(z) e^iGoldstone(z)z = x - iy is an event in complex spacetime.Some provocative excerpts fromPhysical wavelets and their sources:Real physics in complex spacetime_Gerald KaiserCenter for Signals and Waveswww.wavelets.comWaves, wavelets, and complex spacetimeI begin with a brief review of my past efforts to extend classical and quantum theories tocomplex spacetime and interpret the results physically. By that I mean that the imaginaryspacetime coordinates, and any other extras associated with analyticity, are to be understood directly in terms of common observable attributes and not merely as a technical device for proving theorems or exotic higher dimensions inaccessible to mortals stuck in the real world like the poor souls in Platos cave. I have tried not to impose an a priori grand vision but, rather, interpret the imaginary coordinates in each theory by understanding their effects within that theory. Consequently, the interpretations vary somewhat from one theory to another. But they all have in common the following theme. In the extended theory, certain singular points (evaluation maps on fields or wave functions, source points, etc.) become in§ated to extended objects. This transformation is determined by analyticity and the particular theory. In every case, the structure of the objects is shaped by the equations of the theory and their degrees of freedom are specified precisely by the complex spacetime coordinates. The real coordinates give the center, and the imaginary coordinates the extent and orientation of the object in space and time. These ideas are similar in spirit to wavelet analysis, where a function of one variable (time, say) is expressed in terms of an additional variable describing the scale or resolution in the first. This analogy goes farther in the treatment of massless than massive fields, since the latter have an intrinsic scale and thus cannot be scaled arbitrarily. For relativistic fields with mass, spacetime orientation includes velocity, and this makes the complex spacetime an extended phase space. The relativistic coherent-state representations for massive Klein- Gordon and Dirac fields constructed in [K77, K78] interpolate between time-frequency and wavelet descriptions, behaving like the former in the nonrelativistic regime and like the latter in the ultrarelativistic one. In fact, there is a very close correspondence between the nonrelativistic limit in physics and the narrow-band approximation in signal theory; see [K90, K94, K96]. Although the results cited in this section are not new, I believe they have acquired some currency because of substantial progress recently in the understanding of the sources associated with retarded holomorphic fields. The new results focus on massless fields, but it is likely that similar computations exist for massive fields where the integrals are more difficult. Sources describe the breakdown of analyticity due to natural singularities and physically necessary branch cuts. What I find especially fascinating is that such branch cuts behave much like real matter. Depending on the theory, they carry charge, mass and spin, and they emit and absorb radiation. In spite of their simple origins, they turn out to have surprising and complex (pardon the expression) properties, the pursuit of which has the feeling of exploring hitherto unknown forms of matter and not merely the mathematical properties of branch cuts. The results of this search have intrigued and inspired me, and I hope to share this excitement with the reader. At any rate, the dismissal of ict in [MTW73] may have been premature. Even in general relativity, analytic continuations in time and space have borne some rich fruit, even if the physical basis of the procedure is often ill-understood. For example, an exquisitely simple geometric derivation of the Hawking temperature for Schwarzschild black holes is obtained [HI79] by analytically continuing the metric in time, interpreting the Euclidean time coordinate as an angle, and choosing its period to make the horizon a coordinate singularity like the origin in polar coordinates. The reciprocal of the imaginary time period is interpreted in the usual (KMS) way as a temperature, and this turns out to be nothing but the Hawking temperature! I confess that I do not understand this derivation in more than a formal way, but analytic continuation has, in any case, become the main strategy of black-hole thermodynamics, as explained in[Kr03]. The analytic continuation of spatial coordinates also has an honorable history in relativity, having played a major role (and conceptually an equally obscure one) in the discovery of charged spinning black holes by Newman et al. [N65]; see also [N73, NW74, K01a, N02]. And then there are the theories of twistors and H-spaces 11 ConclusionsAnalytic continuations to complex time and complex spacetime abound in physics, although the terminology of Wick rotations is, in my opinion, sometimes used too casually, without any mathematical justification or even any basis for justification (not even wrong). There have been times while reading papers (or even books) on string theory, for example, when was unable to tell whether the author was working in a Euclidean or Lorentzian signature. But even when justified, the extensions are usually regarded as mathematical methods without any particular physical significance. Here is a non-exhaustive list of examples known to me. In the correspondence between quantum field theory and statistical mechanics, the imaginary time (more precisely, its period) is related to the reciprocal temperature. But this is regarded as an analogy between the two theories, albeit a precise and very useful one. To make it more than analogy one might, for example, interpret the complex time as a combination of evolution and thermal parameters for a system in a local equilibrium state, something like the complex combination of the (also incompatible) position and momentum observables occurring in coherent-state representations. (These need not be eigenstates of a corresponding combinations of operators, as they are in the Bargmann-Segal representation. For example, the relativistic coherent states ez (29) do not depend on the existence of covariant spacetime operators, which do not in fact exist within the usual framework. In Wightman field theory, n-point functions are extended to tube domains in their difference variables and powerful methods of complex analysis are used to prove theorems like PCT and the connection between spin and statistics about the original fields in real spacetime [SW64]. There is no attempt to interpret the complex coordinates z = x - iy, although the interpretation of y as (proportional to) an expected energy-momentum in relativistic coherent states, proved for free fields in [K77, K78, K87], extends to general axiomatic fields [K90, Section 5.3]. In constructive quantum field theory [GJ87], the Euclidean region is used to correlate the n-point functions of a given theory by rigorous (Feynman-Kac) path integral methods. Then they are continued back to real spacetime and used to construct interacting fields. Again there is no attempt to interpret complex spacetime because the quantized field exists only in the Minkowskian region while the random field exists only in the Euclidean region. In between, there are only n-point (Wightman) functions. There have been various efforts to represent spacetime as a Shilov boundary of a complex domain (see [G01] and references therein), but I am not aware of any claiming to do physics directly inside these domains. Complex spacetime plays a prominent role in twistor theory [PR86, P87] and the theories of Heaven or H-spaces [HNPT78, BFP80], but again no direct interpretation is generally given to the complex coordinates. To the best of my knowledge, the only examples (aside from the relativistic coherent states and physical wavelets covered here) where complex spacetime coordinates are given a direct physical significance have appeared in the works of Newman et al. [N73, NW74], who have proved the following very intriguing result. Consider an isolated classical relativistic system in §at spacetime with positive total mass The total angular momentum splits into orbital and spin parts L + s, and L is made to vanish by translating to the center of mass. Similarly, if the system has total charge e =/= 0 and a magnetic moment [Micro], then its dipole tensor is reduced to [Micro] by translating to the center of charge. However: With a further imaginary translation by is/mc, the spin can be made to vanish. Thusspin may be identified with an imaginary center of mass. With an imaginary translation by i[Micro]/e, the magnetic moment can be made to vanish.Thus magnetic moment may be identified with an imaginary center of charge. If the centers of mass and charge coincide, then the spin and magnetic moment can betransformed away simultaneously by an imaginary translation. The necessary and sufficient condition for that is that the gyromagnetic ratio of the system have the Dirac value: [Micro] = (e/mc)s.In the massless case, the world lines with complex center of mass are replaced by a totally null complex plane if the spin (in real Minkowski space) is nonzero. This idea, although proved in §at spacetime, was inspired by the Kerr-Newman solution to the Einstein equation [N65], which is the universal model for spinning, charged black holes. It was discovered by performing a somewhat mysterious complex coordinate transformation on the spherically symmetric solution with mass and charge (Reissner-Nordstrom) which is, roughly, a general-relativistic version of extending the Newtonian potential from R3 to C3 The Kerr-Newman solution was soon realized to have the Dirac gyromagnetic ratio. Recently, an old debate was re-ignited with A. Trautman whether the Dirac value necessarily depended on the nonlinear character of the equations. Newman settled the question by showing that the Dirac ratio was obtained as well for the linearized solution [N02]. In the related work [K01a], the charge-current distribution for a (real, static) electromagnetic field defined as in [N73] by a holomorphic Coulomb potential was computed and shown to represent a rigidly spinning disk so that the rim moves at the speed of light. This is consistent with the fact that (191) represents the electromagnetic part of the linearized Kerr-Newman black === combinations>Subject: Re: (Q) How many possible combinations>Message-id: <32b184adb623d3a65c5b5a6733fa3ad8@news.teranews.com>>can you point to a site that perhaps has a good description>> >> No, but heres a page with a couple examples:>> >> http://members.aol.com/mensanator666/fun/playing.htm>> >>Hi examples are well written and >illuminating.>>using your example....>C(m.n) = m!/((n!)*(m-n)!)>>How many 8-bit binary numbers have exactly 4 ones?>>C(8,4) = 8!/(4!)*(4!)> = 8*7*6*5*4*3*2 / 4*3*2 * 4*3*2> = 8*7*6*5 / 4*3*2 > = 7*6*5 / 3> = 7*2*5> = 70 >>I used the following to calculate the possible combinations of 16 bits >out of 256. in other words How many 256-bit binary numbers have exactly >16 ones?.>>from your example I got>256!/(16!*(256-16)!)>>I plugged this into maxima (only downloaded yesterday :)>>and the answer was 10078751602022313874633200>>my final step was to calculate how many bits was required to store this >number>>log2(10078751602022313874633200) = 83.05951932 bits>( had to use excel as I couldnt find how to change the base in Maxima )>>So I can store the original 256 bit pattern into 84 bits>Does that sound reasonable ?Yep.C:Python23user>python dec2bin1.py10078751602022313874633200 in base 10 ( 26 digits ) is10000101011001000010001110000001100110111100101101010000001001 1101110011000111110000 in base 2 ( 84 digits )Although keep in mind that the worse case for an m-bit number is to have m/2 1s(the middle column of Pascals Triangle). Your exaple is column 17 which is along way away from column 128 (which would need 252 bits to hold its value).>>Chris--MensanatorAce of countingarent you splitting hairs, John --doesnt he really have some largely *infinitesimal* credibility,thus far in the 10-year programme of saying hes right? > Its also been repeatedly pointed out that that thing you get> is _not_ (quite) a partial differential equation. But never mind > Why should JSH have zero credibility, while--after your blunders, > to silence or neutralize dissenters from the canon? My FLT proof, done for life, liberty & the purfuit of happineff ...but Usenet was just too small to hold it! http://mathforprofit.blogspot.com/ > 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 in> A)Classification> C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] (Weak> Extensionality)--les eigenvalues of real symmetric matrix A is greater than a,all the eigenvalues of real symmetric matrix B is greater than b. Proof: All the eigenvalues of matrix A+B is greater than (a+b) Halton Arpoverlapping messy systems cannot exclude every anomaly,just by saying it. are you claiming that every single oneof Arps anomalies have been satisfactorily circumscribedby the Doppler-shift interpretation? or, as I think, were you saying thatnone of them have been proven to be physically close, becauseI didnt go there on the USS Enterprooz? > Having been burned a few times astronomers no longer accept simple > it looks like arguments. These arguments are used to start or> inform other studies of the objects. > > For these next cases I dont have references in front of me, an > examination of the archives of _Astronomy & Astrophysics_ will > locate these and other studies. > Subsequent reports have found a few other cases, Arp was good enough > to find most, but the new cases have not altered the argument.> > To date there have been *NO* cases where physicaly close ( not line > of sight close ) objects have been shown to have markedly different> red shifts. Arp and others have provided a list of where this may> happen, in the cases we have examined it does physicist (a Nobel prize winner) on TV many years ago who said he spent all of his time after he retired studying the statistics of Shakespeares writing.I think Im a very determined deterministic Eddington monkey sometimes. Take a 26 by 26 keyboard; the height of each key is proportional to the letter pair frequencies in the first ten thousand characters of Hamlet. I hit the highest key which stays down and it prints the letter t. Next I hit the highest key in the h row because the first key was the letter pair th, and it prints an h. Then I print an e, etc.. It will take another kind of monkey to interpret what I type.Here is the function g to simulate the mon-key.g[n_, rowStart_, y_] := Module[{t, k, an, ma, ne}, an = {rowStart}; t = y; k = rowStart; Do[ne = First[Flatten[Position[t[[k]], Max[t[[k]]]]]]; t[[k, ne]] = -1; k = ne; an = Append[an, ne], {n}]; an]The 26 by 26 table c of the letter pair frequencies from the first ten thousand characters of Hamlet in the statement below is from the book Mr. Babbages Secret, The Tale of a Cypher and APL, by Ole Immanuel Franksen, 1985.alphabet = CharacterRange[a, z]alphabet[[g[48,First[Flatten[Position[c,Max[c]]]],c]]]{t,h,e ,r,e,s,t,i,n,d,i,s,e,a,n,t,o,n,g,e,n,o,r,i,t,e,d,a,t,a,r,t,t,s ,a,l,e,t,r,a,s,o,u,r,o,f,t,u,t}I got a hit on an Internet search for the meaning of the word trasour and it is probably a tracery, something you might find in king tuts tomb (which was broken into twice in antiquity) and sold at an art sale.The 28 by 28 letter pair correlation matrix s from the book Scientific and engineering problem-solving with the computer, by William Ralph Bennett, Jr., 1976 based on the dialog from Act III of Hamlet with an alphabet that includes the space and the apostrophe: it is obvious the first letter must be a q because that row has a sum of one more than the sum of the corresponding column.alphabetTwo = Join[CharacterRange[a, z], { , }]alphabetTwo[[g[26,17,s]]]{q,u,r, ,t,h,e, ,a,n, ,m,e,r,e,n,d, ,w,h,a,t, ,s, ,i,n}They spell the Moslem religious book the Quran, sometimes.Does anybody know what this kind of characteristic run through a frequency table is called? Can anybody give a probability based estimate of what to expect Mathematics --- Myth and Reality> >>Arithmetic is indeed very important. And, being from India, we>>have a special fascination for the decimal calculations and the>>different methods. In fact, there is quite a bit of research to>>implement decimal computing and storage using electronic devices.>>For example, at one point of time I got fascinated by the possible>>application of the very interesting resonant tunneling devices for>>One 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).> > Decimal storage sounds interesting, but I dont know exactly how it> can be any kind of improvement to memory storage. Unless you can pack> four state devices in around the same real estate as binary, of> course. Then memory capacity will increase by a reasonable factor. > So far as computation is concerned, yes, there should be a definite> gain with four-level logic instead of binary. But I dont see how 4> state will lead to decimal computing. If you have states 0 1 2 3, you> need 3 devices to write 10. In binary logic, you need 4 binary> devices to write 10.The idea was to achieve a 10-peak-9-valley resonant-tunnel device (RTD),that when biased with an FET would give 10 stable states - enablingdecimal storage. But, at the time of the experiment, we had only4-peak-3-valley devices available - enabling 4 states. So, we used thatto essentially make a proof of concept - as to how to read and write thememory. In principle, the scheme should lead to space saving. Let ustake the case of decimal storage. If we normalize the voltage level, andaccept three places after the decimal point, say, 0.567 V, we can place5 in one of the 10-state devices (roughly the size of a transistor,these are vertically stacked heterojunctions), 6 in the next, and 7 inthe last. So, we need 3 FETs and 3 RTDs for this static storage. Now, ifinto acceptable levels of quantizations (10) and then need 10*6/10*4transistors to store them.> > > >>Those interested in electon devices and multivalued memories may>>find the paper interesting. But, at the same time, I always like>>such arithmetic methods where the learner can clearly see why the>>method is working. For example, the basic square root finding>>method 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> > That is the very same method I learnt! We did it blindly, like we did> multiplication blindly, and had no end of trouble with the carry in> subtraction.> > See if you can get that book on Vedic Mathematics. That should give> some insight. I left my copy in Kolkata!I am expecting to learn about the method from the other proof: Given: All the eigenvalues of real symmetric matrix A is greater than a,: all the eigenvalues of real symmetric matrix B is greater than b.: Proof: All the eigenvalues of matrix A+B is greater than (a+b)Hint: What is the relationship between the eigenvalues of a symmetric matrix A and the proof that infinitesimals existX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 02:08 PM, currentresident@veloemail.com (charles ramsey) === said:>Subject: proof that infinitesimals existAlas, no: you havent proven anything. Instead, youve madeunsubstantiated statements about things that you havent defined.>In infinite dimensional spaceWhat do you mean by infinite dimensional space?>the point (1,1,1...)What do you mean by the point (1,1,1...)?>that has an infinite lenghtWhat do you mean by length?>go to the end of this lineWhat do you mean by the end of a line?>add one units lenght to it.What do you mean by add one units lenght to it?>What are the coordinates of this new point? You tell me? You havent defined your space, your coordinate system oryour notion of length, much less proven any of their properties, so itis not possible to draw any infewrence about them. You havent evengiven enough definitions to prove that such a point either exists oris unique.>It cant be a real number greater than one because the distance>between any two real points would result in an infinite lenght.You havent proven that. Further, if the point doesnt exist then nosuch issue arises.>therefore the coordinates of this pointYou havent proven anything about the point, or even that it exists.>are an infinitesimal distance past one.What does an infinitesimal distance past one mean?-- 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 NOW!Androcles replied to Jeff Root: >> Did you realize that your invention is an application of the>> classic question, What happens when an irresistable force>> meets an immoveable object?>> >> The answer to that classic question is: The object moves.> > Err.. I can apply an irresistible force quite easily to an> immovable object, such as my living room wall, by pushing> against it.> The wall doesnt move. Yes it does. > I do. I now pronounce you man and wife. Ooops! Sorry, wrong context! > For every action there is an equal and opposite reaction. Yes. That fact is how I know that your wall moves when youpush on it. It may not move enough for you or me to detectwithout a micrometer gauge or a laser rangefinder or a verysensitive electronic motion detector, but it moves. > In a more extreme situation, the blast from the tail of a> rocket is an irresistible force, attempting to move the> launching pad and the entire Earth beneath it. The rocket> moves. I saw Space Shuttle Columbia launch on its next-to-last §ightfrom less than six miles away. I was certainly moved by it. -- questionIve searched for simplifications to arctan(x/2) to no avail. Doesanyone know of one, or know that it does not http://www.giganews.com/info/dmca.html> Androcles replied to Jeff Root:>Snipped cause I had nuten to say bout dat part......>> I saw Space Shuttle Columbia launch on its next-to-last §ight> from less than six miles away. I was certainly moved by it.>> -- Jeff, in Minneapolis>> .Aint it a hoot Jeff !!!I was working on the Cape for the STS 26 launch and watchedwith my escort from about 3.3 miles out at the Doppler station..And yes its a emotional rush to see a launch and until someonesees one live they will never know what a test your device, yet?> If you havent, do it *now*! It is your idea. *You* have to> make it work!>> > But that would make it impossible for him to revel in his current position> as unrecognized genius / savior / conspiracy victimWhen coincidence happen again and again, some point of time, you willbe forced to think that someone is executing well constructed planin order for specific purpose. Still, you try to avoid thinking insuch way. But suddenly, you find that coincidence frequency hasincreased, then you will have to admit that someone is controllingthe things around you.Let me tell you just one coincidence.Though I have a Time Theory which explain origin of universe, in myeveryday life, I seldom look at sky, at stars. But in the evening ofmany years, listening songs on walkman. I saw two satellites goingsouth to north, one from west to east. And I was just thinking, howbeautiful this starry sky is looking.And I saw bright §ash right up their among stars very clearly. You can explain it this way. That §ash was originated due to frictionof solar wind with atmosphere of earth. You can confirm the timingfrom media around the world.But what I am saying that, I was looking at starry sky in last manyyears. And how the hell on this earth I did it at perfect time?You will think that this is coincidence. But what about those lights§ashing before my eyes in my room, at different places source ofwhich I can not explain. There are many things happened in my personallife which I can not talk in public.And those things are not just coincidence, Laura. I was trapped, Iam trapped. Things around me are being controlled, navigated verycleverly.He know what 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 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?>> Come on Paul, you are being silly now. You know what ïasymptote means.>>Indeed, and that whats wrong.>>Here is what I have told you many times,>and what you once more have forgotten:>>time it passes through the accelerating field, regardless of its speed.Hmm!>The gained energy does NOT approach zero (or any other value)>asymptotically when the speed approached zero, because it is constant!Hmm!>>That is how much energy!>>The proof of that is that when the accelerator is in steady state,>the lost energy in the bends is equal to the energy gained in>the RF-cavities. The lost energy is radiated as synchrotron>radiation, which is easy to measure.There is nothing sensational about that.>This lost energy does NOT decrease when the speed of>the electrons increases, quite the contrary.Does that mean ïit increases.>Thus the gained energy does NOT decrease when the speed>of the electrons approaches c.>>So whatever you think happens to the field in the RF-cavities>when the speed approaches c, we KNOW for certain thatBut where does that energy go? >>Which proves you WRONG.>The radiation from an accelerated charge!>or the fields associated with a moving charge!>or The ïBack EMF concept.>any less when the speed approaches c.that does not con§ict with what I said.mean. Is it converted to 1/2mv^2? Or does some of it go into the ïrelativisticmass increase? If so, how and why? >>You know this, and have admitted to know this.You have misinterpreted my statements again.>>But you keep forgetting what you know,>and restate the claims you know are wrong over and over.>> You are making no attempt to answer that one. In typical fashion, you pretend> the relevant question does not exist.>>A blatant lie.>>... because you know I have answered that one many times.No you answer many questions but usually ones you make up.>>Paul>Henri Wilson. See the Stupidity of Logic missing from Proofs from THE BOOK>>What about the related halting problem or (almost identical) >>nondenumerability of reals?> > > That second one is in there, on page 88.Oh.Hmmm.. Ive got to start working on that reading comprehension thingy. And following the proof of |Z| < |R| they give |R|=|R^2|. They also have the Schroeder-Bernstein thm which I from THE BOOK> 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 had> anything to do with Topology.I meant by my short comment about topology and geometry that, by whatever route one is lead to topology (category theory, point-set analysis, differential manifolds, etc) that one is doing generalized geometry.Or at spiders> > Been trying to read Conant and Vogtmanns on a theorem of Kontsevich.> Now if you have two spiders S and T (as defined in the paper) there is> a mating operation o which requires the specification of one leg,> y, of S and one leg, x, of T giving a new spider (S,y)o(T,x). Further> development of the paper indicates that this operation should be> commutative i.e. we should have that (T,x)o(y,S) = (S,y)o(x,T).> However, for simple cases involving the associative operad this> doesnt seem to be the case. Does anyone know if the mating operation> as given in the paper in fact is commutative?In the absence of the definitons of the terms youre using,a reference to the paper might be nice.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 sci.physics, OrTiMoN:> Now its easy to stray from the actual point of the posting isnt it?> Well, OK, suppose you clarify on what it means for a number to exist,or not exist.Im assuming you wish to be mathematical about it so followupsreset, but Ill continue anyway...We can assume ï1 exists. ï1 is the first, the starting point.We can also assume that the successor of a number is greaterthan the number -- in fact, we might as well go whole hog andtake all of Peanos axioms, just for the fun of it. :-) (Admittedly,the fifth axiom, relating to a variant of weak induction, is morelike a meta-axiom, which bothers me a bit.)We assume 0. If one defines subtraction thats easy enough. Wealso get negative numbers for mostly free. Now we have J, theset of all integers.Define multiplication and division and we now get Q, exceptfor such oddities as 1/0, which doesnt exist (sincea * 0 = 0 for any rational, real, or complex a), and 0/0,which is mostly meaningless without a lot more context.(For example, lim{x->0} x^2 / x = 0, but one can alsoconstrue it as 0/0. Of course lim{x->0} ax/x = 0/0 = a,for any a. Or one can take lim{x->0} x/x^2 = 0/0 =larger than any number one can think of.)Now it gets a little tricky. The standard methods might bealong the lines of Dedekind cuts or Cauchy sequences. AFAICT,both work to define numbers which are provably not in Q, butwhich can be well-ordered. sqrt(-1) comes into play, theso-called imaginary numbers are created, and one eventuallyproves everythings complete, at least using the standardmethods of addition, subtraction, multiplication, and division.Along the way one gets some beautiful results such asexp(i * pi) = -1. Unfortunately, the complex numbers arentwell-ordered although its usually not a problem. (Of coursesubsets thereof can be: the real line, for example.)If one rescinds s > s, then we get into finite fields. Idont know the notation offhand but any prime p can generatesuch. (Non-primes dont do too well; if one assumes {0,1,2,3,4,5},for example, 2 * 3 = 0. Not good for a field, at leastusing standard arithmetic operations, modulo n. Otheroperations with similar properties are possible, though.)If one throws away multiplicative commutativity, one can dosome interesting stuff: quaternions, for example.One can also wander into statistics. For example, one couldcount the number of left-handed people who write with theirright hand. Since this is generally considered a contradiction,one could make a case here that the number of left-handed peoplewho write with their right hand would not exist -- but one canalso make a case that the number of such people is 0.Ditto for the number of squared circles and the number of peoplewho think that computers are less useful than the ancientconcept of phlogiston. :-)Or one can stick to my original point, unproductive asit is, in that a number doesnt exist as such except asas a small piece of metal with adhesive backing that canbe stuck on a mailbox, as a sequence of longitudinal wavesimpacting ones eardrum, or other such representations;if one has 5 chickens, for example, one can of courseenumerate them 1,2,3,4,5 (and even tag or brand them),but at the end of the day what does one have? A bunch ofchickens which can be paired off with 5 cows (if one has5 cows, of course), but does the number 5 exist per se?Of course not -- its a concept, albeit a very useful one.So please do clarify this question of non-existent numbers.-- #191, ewill3@earthlink.netIts still limits - Need Help!In sci.math, Jonathan Miller:>> In sci.math, Jonathan Miller>> :> 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 you dont know that the (usual) substitution for a circum§ex>> o is os? That slepping falmes always contain mispleddings?>> Jon Miller>> Search engines are notorious for missing the point when a rule>> name is misspelled. I did not intend insult. :-)> > Im sorry, I misinterpreted your remark.SOK.> >> Are you suggesting it should be spelled LH.99pitals Rule? Id>> go along with that, from the limited amount of French I know>> (oui, non, je tadore, je ne parle pas Fran.8dais, etc.)> > Well, either that or LHospital (he spelled it both ways himself, as the> Academie Francais [note mispellings] hadnt yet standardized the language).> But sometimes correct spellings miss web pages, because the page sponsor has> misspelled things.> > Ive been avoiding accents and cedillas because I was under the impression> that ASCII didnt support them, but I now realize that that is incorrect.> Theyre there, so theyll display correctly on everyones machine. Today I> avoided them out of sheer laziness. Theyre not on my keyboard.> > Jon Miller> This isnt ASCII but ISO-8859-1 (or ISO-10646-1; Im not entirelysure which). .99 = 244 or hex 0xF4, for example. But yourpoint is interesting; Ill admit Im not that familiar withLHospital/LH.99pital. (Come to think of it, when was thelast time I actually used *calculus*? :-) )ASCII proper is 0-127, and Im not entirely certain about that;the printable ASCII is 32-126 and the rest is control characters.Be careful, as theres at least one other coding out there. Imnot entirely certain, for example, what old BASIC programs aretrying to show when I get weird characters; the IBM character seton a PC had horizontal and vertical lines, but ASCII/ISO-8859-1 doesntprovide for that in its list of characters except for such thingsas ASCII 124 (|), the ASCII underscore 95 (_), and 175 (), whichon my font at least looks like it might be useful as an overbar.You can tell what Im expert in, Im sure. :-) Ill admit to someinterest in whether such things as neural networks might help insuch fuzzy endeavors -- but thats taking us a bit far afield. :-)-- #191, ewill3@earthlink.net -- but hey, this *is* Usenet :-)Its still legal to go Charlie Johnson:> >> Hi all,>> I am just starting to play a little with MS-Excel for Math purposes. Does>> anyone know of a way to calculate several values for a given function,> i.e.>> Cos(1, pi/2, pi/3, ...n) ? And, can a cell reference be given instead of>> putting in the values? Such as, Cos(A1:A10). (I have also cross-posted> to>> two something strange when I try to calculate cos(pi/2). I> keep getting the result 6.12574E-17. Is that suppossed to be essentially 0?> Cant Excel give you 0? I know that from programming, but I thought MS> could have just programmed that to represent 0. Whats up?Its a general problem in §oating point.For example, one can compute 1/3. In mathematics,1/3 = 0.333333333....ad infinitum. However, in §oating point the computer sees1/3 = 0x3fd5555555555555or in slightly more conventional notation1/3 = 2^(-3) * 2.AAAAAAAAAAAAA [base 16]whose exact value is in fact6004799503160661/18014398509481984or 0.333333333333333314829616256247390992939472198486328125000... pi, being a transcendental number, has a similar problem; it cannotbe exactly represented in the computer. Since most §oating-pointunits use polynomial approximations small errors can creep inthere, as well; small wonder that cos(pi/2) != 0 in such a system.Fortunately 6.1E-17 is very close to 0. :-)[.sigsnip]-- #191, ewill3@earthlink.netIts still legal to BOOK >What about the related halting problem or (almost identical) >>nondenumerability of reals?> > Nondenumerability of the reals is set theory, not logic.> There are lots of such proofs in set theory; many of these> elegant and somewhat paradoxical proofs are quite accessible> to the non-expert.Er... yes specifically set theory. Some people tend to put set theory in as one part of logic (namely heads of mathematics departments, and the that the thms/proofs/subject matter in set theory (at least in the elegance/accesibility area) have very different §avors, enough to separate them. Them thms/proofs pointed out by cdj are what I was asking for. And on re§ection, of the three logic thms I mentioned above, though the proofs have a similar §avor, the halting problem and uncountability of the reals dont feel logical (that is, spiders> >> Been trying to read Conant and Vogtmanns on a theorem of Kontsevich.>> Now if you have two spiders S and T (as defined in the paper) there is>> a mating operation o which requires the specification of one leg,>> y, of S and one leg, x, of T giving a new spider (S,y)o(T,x). Further>> development of the paper indicates that this operation should be>> commutative i.e. we should have that (T,x)o(y,S) = (S,y)o(x,T).>> However, for simple cases involving the associative operad this>> doesnt seem to be the case. Does anyone know if the mating operation>> as given in the paper in fact is commutative?>>In the absence of the definitons of the terms youre using,>a reference to the paper might be nice.Im sorry. Here it is:3. math.QA/0208169 [abs, ps, pdf, other] :Title: On a theorem of KontsevichAuthors: James Conant, Karen maths not psych)>>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 is>> instant.>>Its settled then.> Paul, it is by no means settled.>>We agree:> as it enters a static electric field.>>One can but wonder why it took you so long to realize>the obvious.>> Read what I said! Can you see the words before it starts to move?> Do you know what they mean?> Do you also know what the electron is doing for the rest of the time? IT IS> MOVING!OK.So lets change the scenario a little.We still have two electrodes 1 km apart, with a millionvolts potential difference between them.Behind the cathode, there is an electron gun shootingelectrons with an initial speed v_o through a small holein the cathode.The question is still:When does a force start to act on the electron?My answer is:as it enters a static electric field.What is your answer?[..]>> If a highly charged sphere is moved, say, backwards and forwards between two>> electrodes, what happens to the force on those electrodes. Does it change>> instantly or is there a time lag?>>Why do you have to ask?>There is obviously a time lag.>> Why didnt you say so earlier. One can but wonder why it took you so long to> realize the obvious.>> But are you quite sure of your answer?> After all, you DID say that electrostatic forces acted instantaneously. Have> you changed your mind?I find this line of arguing rather irritating.Are you really so desperate that you have to try to makeit appear that I have said something I never did in orderto refute statements I never made?Isnt this technique below you?Have you no personal integrity to take care of?> Can you refer to an experiment that shows the time lag?Henry, the physical consequence of moving a charged spherebackwards and forwards is that an EM-wave is emitted.Do you have to ask if there are experiments showing HELP BADLY (sorry, maths not psych)>How long after the electron enters the field will>it be affected?>>Randy, havent you noticed that whenever Paul 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 Paul. Between you>and me, too. However, it isnt Paul who does not. Nor is it me.>>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.>>I know. Its the question he was asking you, that you didnt answer.>At least you admit that rather than answer the question, you changed>the subject. You raised this question in lieu of giving an answer.>> - Randy> BULL!But you know Randy is right, dont you?You know bloody well the subject of the conversation wasspecifically:How long time does it take before a force act on a chargedis already there?cannot feel the force at the same instant as it enters the field,but that there must be some action time.You know bloody well that in order to defend this claim,YOU tried to hijack the converstaion by diverting the subject down a side track, namely the following:| Let us consider a charged sphere somewhere in the universe. It exerts a force| on every other charge. If we can arrange for it to lose that charge somehow,| you are claiming that all those forces disappear INSTANTLY.You know bloody well that my only claim was: as it enters a static electric field.You know bloody well that I NEVER claimed what youin your side track said I had claimed, namely:If a charge somewhere in the universe somehow suddenlydisappears, all charges in the universe will instantly be affectedIn light of this, your statement: Randy, havent you noticed that whenever Paul is in trouble, he always attempts to hijack the conversation by diverting the subject down a side track.appears rather pathetic.It is so obvious who is in trouble and is desperate to divertthe subject down a side track, that you can be sure thatRandy is not the only person NEED HELP BADLY (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 studied his texts think he meant momentum.>>Paul>> It is interesting because the term ïv.dm/dt is explains the energy increase>> that is normally associated with ïrelativistic mass increase.>>Of course.>The question is what is the momentum?>If the momentum is mv, then the mass _must_ increase with the speed.>In 1905 this was the accepted definition, and was why Einstein>said that the mass increased with speed.>>However, the modern approach is to say that>momentum = m*f(v) where m is the invariant mass>and f(v) =v/sqrt(1 - v^2/c^2)> It actually supports my argument that this energy really goes into the ïreverse>> field bubble that forms around a moving charge.>>Why not call your enigmatic bubble a fairy?>>more invisible but massive fairies clings to it.>When the fairies loose their grip in the bends of the accelerator,>their mass is transformed to synchrotron radiation.>>Make perfect sense, doesnt it?>> It makes just as much sense as saying ïmass increases for no apparent reason.> Where is your supporting PHYSICAL evidence? How can mass simply appear to> increase? What does ïmass increase mean Paul?> You people are really funny.You are not very good at reading, are you? :-)I am not saying mass is increasing.I am saying the mass is invariant.I am saying the momentum is m*v/sqrt(1-v^2/c^2)And you dont really have to ask for the physical evidencefor that, do you? Of course not.You know that this is verified all the time.And know what, Henry?Physical evidence doesnt go away by stating:I see no reason why Nature should behave as it does, maths not psych)>Why not call your enigmatic bubble a fairy?>>more invisible but massive fairies clings to it.>When the fairies loose their grip in the bends of the accelerator,>their mass is transformed to synchrotron radiation.>>Make perfect sense, doesnt it?>> Neat! Hot fairies.Indeed.And since synchrotron radiation may appear bluish,it also supports the notion that invisible fairies are RealityNntp-Posting-Host: hera.cwi.nl... > But all I am asking is to understand others points. I want to know > how you can use FFT to multiply 12345 by 67809 with less than 25 > multiplications. If really FFT does O(nlogn), then you should do it > in 5log5 (that is, 5*1.6 or 8) multiplications.Wrong. You do not understand what O(n log n) means. It means thatthere is a constant C such that for n large enough FFT can do it inless than C.n.log(n) multiplications.You assumed two things:1. 5 is sufficiently large.2. C = 1.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; TtZ7vmZpwee7u5tOeUiNirA4LS6eE-Rt-EHQCtx8lHYeqbXw6Mgq02Complex images work only as long the system is linear.That fails immediately, if non-linear parts appear,as usual in GR.Ulrich === Subject: Re: I NEED HELP BADLY (sorry, maths not psych)>>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 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?>> Come on Paul, you are being silly now. You know what ïasymptote means.>>Indeed, and that whats wrong.>>Here is what I have told you many times,>and what you once more have forgotten:>>time it passes through the accelerating field, regardless of its speed.>> Hmm!>>The gained energy does NOT approach zero (or any other value)>asymptotically when the speed approached zero, because it is constant!>> Hmm!>That is how much energy!>>The proof of that is that when the accelerator is in steady state,>the lost energy in the bends is equal to the energy gained in>the RF-cavities. The lost energy is radiated as synchrotron>radiation, which is easy to measure.>> There is nothing sensational about that.>>This lost energy does NOT decrease when the speed of>the electrons increases, quite the contrary.>> Does that mean ïit increases.Yes.per cycle in the RF-cavities regardless of the speed.But it will loose more and more energy per cycle inthe bends as the speed increases.When the two are equal, the accelerator is in steady state.when it is going at peak efficiency.>Thus the gained energy does NOT decrease when the speed>of the electrons approaches c.>>So whatever you think happens to the field in the RF-cavities>when the speed approaches c, we KNOW for certain that>> But where does that energy go?Into kinetic energy, of course.Einstein says: KE = m*c^2*(((1/sqrt(1 - v^2/c^2)) - 1)Newton says: KE = 0.5*m*v^2Why do you find the second of these equations more naturalthan the first?Two different theories, the first is experimentally confirmed,the second is experimentally falsified.Why do you have a problem with that?Why not simply accept it?>Which proves you WRONG.>The radiation from an accelerated charge!>or the fields associated with a moving charge!>or The ïBack EMF concept.>any less when the speed approaches c.>> that does not con§ict with what I said.Uh? :-)> mean. Is it converted to 1/2mv^2? Or does some of it go into the ïrelativistic> mass increase? If so, how and why?Answered above.>You know this, and have admitted to know this.>> You have misinterpreted my statements again.Another step in Henry Wilsons eternal cycle of §eeingstatements §ed 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.>> Lets not forget that the Vedic principle crosswise and vertical> does also lie at the heart of this modern, bad, clumsy way...> Because the Vedic principle does not talk about I dont think you have fully grasped the subtleties of the method.I dont think you have fully grasped the meaning of what i said.The methods are different, but both are based on crosswise vertical.The expression crosswise vertical _could_ lead to your method,but just as well to the usual one.In short, i didnt say the methods are the same, but that they are both basedon crosswise vertical.> I suppose you have to be a primary school teacher or a primary school> child to understand that! Not talking about the order of the> additions, that being implicit, is the real strength of the method,> where there is independence of columns and thus avoidance of carrying> till the very last stage where it is done in one go. Thus, this> method is admirable for parallel processing, and can be implemented> with very few lines of code. I could easily multiply two ten thousand> digit numbers using simple C code with the definition I have given of> this algorithm. I could never do such a thing so simply with other> methods.Yes, the gist of it is single register you only have to remember onenumber, all along the process.But is _that aspect_ of the method Vedic? The expression crosswise verticaldoesnt say much about it, does it?> Please remember that very learned people can also be shown to be very> foolish. Like, all those who believed in Aristotle, and those today> who believe in the Special Theory of Relativity!>> Arindam Banerjee.It is certainly true that the stupidest mistakes are oftenmade by numbers only> doesnt make maths very exciting for laymen, im afraid.> Personally, i would never have gotten any interest in mathematics> if not for geometry, reasoning from axioms, etc. Historically,> mathematical progress always ended fairly soon, in all cultures> that restricted maths to ïdoing calculations.>> When they do calculations properly, they succeed. The success of> modern Western cultures lies very largely in their ability to do> calculations very fast, using computers for number crunching, and> terrific use of calculations in all aspects of engineering, finance,> banking, selling, etc.The progress of modern mathematics in the past 300 yearsis due mainly to its advancement beyond calculus. And alternative multiplication method is absolutely welcome.>> I dont know the rest of Vedic arithmetic, and I dont think I will> know till I get that book. Which may take a few years, I am afraid,> unless someone gives me a copy.If you didnt read the book, how do you know that your multiplicationmethod is Vedic? Crosswise vertical is not much of a description,is polynomial~hello.......matrix A =2 1 0 00 2 0 00 0 1 10 0 -2 4find minimal polynomial of A---------------------------------my solving process is........characteristic polynomial of A isf(x)={(x-2)^2}(x-1)(x-4)thusminimal polynomial of A ism(x)=(x-2)(x-1)(x-4) or m(x)={(x-2)^2}(x-1)(x-4)thusi want find m(x) such that m(A)=0 or m(A)=0but i dont find that.please.....point out an my error. sir.....thank Re: Problem from Herstein 4> :> 18. Let G be the group of all real 2-by-2 matrices> :> (a b)> :> (c d), with ad - bc non-zero, under matrix multiplication, and let N> :> be the subgroup consisting of those elements of G with ad - bc = 1.> :> Prove that N contains G, the commutator subgroup of G.> :> 19. In problem 18 show, in fact, that N = G.> > : I assume, that your matrices have their coefficients in some commutative> : field k. If k* are the nonzero elements of k (with multiplication),> : you can verify that> > : (a b)> : (c d) --> ad-bc> > : does indeed give a surjective homomorphism det: G ---> k*.> > : This will exhibit N = ker(det) and G/N = k* commutative.> > All true, but this only shows that N contains G; the hard part is the other > direction (problem 19, show that N = G), which is not coming to me right now.Yes, you are right. I wonder were I had my head.Fortunately, Robin saved the day with the remark about the unipotentmatrices.For a given matrix(1 s)(0 1)one might as well try, to build the desired commutator from uppertriangualar matrices.Writing d for 1/d just calculate the commutator(d 0) (1 t) (d 0) (1 -t)(0 d) (0 1) (0 d) (0 1)this boils down to the equation s = t(d+1)(d-1)So all you need, is a d in k*, not equal to 0,1 or -1.The only case where the above does not work is k=Z/2Z;actually I wonder if the result really holds G be the group of all real 2-by-2 matrices>> :> (a b)>> :> (c d), with ad - bc non-zero, under matrix multiplication, and let N>> :> be the subgroup consisting of those elements of G with ad - bc = 1.>> :> Prove that N contains G, the commutator subgroup of G.>> :> 19. In problem 18 show, in fact, that N = G.>> >> : I assume, that your matrices have their coefficients in some>> : commutative field k. If k* are the nonzero elements of k (with>> : multiplication), you can verify that>> >> : (a b)>> : (c d) --> ad-bc>> >> : does indeed give a surjective homomorphism det: G ---> k*.>> >> : This will exhibit N = ker(det) and G/N = k* commutative.>> >> All true, but this only shows that N contains G; the hard part is the>> other direction (problem 19, show that N = G), which is not coming to me>> right now.> > Yes, you are right. I wonder were I had my head.> Fortunately, Robin saved the day with the remark about the unipotent> matrices.> > For a given matrix> > (1 s)> (0 1)> > one might as well try, to build the desired commutator from upper> triangualar matrices.> Writing d for 1/d just calculate the commutator> > (d 0) (1 t) (d 0) (1 -t)> (0 d) (0 1) (0 d) (0 1)> > this boils down to the equation s = t(d+1)(d-1)> > So all you need, is a d in k*, not equal to 0,1 or -1.> The only case where the above does not work is k=Z/2Z;> actually I wonder if the result really holds then.It doesnt: GL(2,2) is isomorphic to S_3, with its commutatorsubgroup corresponding to A_3. The unipotent matriceslie outsise this subgroup. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 who have the book, it is I.N.Herstein, Topics in Algebra (> 2nd edition), chapter 2 ( Groups), section 2.9 ( Homomorphisms),> problem 19 ( I include problem 18 for reference purposes)> 18. Let G be the group of all real 2-by-2 matrices> (a b)> (c d), with ad - bc non-zero, under matrix multiplication,I.e., G = GL(2,R)> and let N> be the subgroup consisting of those elements of G with ad - bc = 1.I.e., N = SL(2,R)> Prove that N contains G, the commutator subgroup of G.> 19. In problem 18 show, in fact, that N = G.I dont think Ive seen anyone point out that 19 is a trivialcorollary to the fact that N is perfect, i.e., N = N, as istrue for all SL(2,F) except for |F| in {2,3}.Not that I remember how to prove this, but maybe it will pointyou in the right direction.-- Jim solve a puzzle.>> xzy +> xyz> ---> yzx> I have to find the only leaves z = 5.-- Using M2, Operas revolutionary e-mail news story, hard to say whether its overrated or notThe story is at:http://www.af.mil/stories/story.asp?storyID= 123006043Basically, a 12 year old girl figures out thatA - (B/C) = (A-1) + (C/C) - (B/C),and uses it as an improved way to subtract fractions.However, the artical comes off making it sound like she has inventedcalculus or something. Obviously if a grown adult were to discoverthe time.On the other hand, the fact that she is 12 does lend her some credit. Im not sure how much it lends, though. Certainly if she were 3 oreven 6, it would lend a lot more, but 12 is pushing it. When I was 12I discovered that you could do those damned division problems whereyoure required to give the remainder, using a calculator:A/B = §oor(A/B) with a remainder of B*[A/B-§oor(A/B)](I used the QBasic int rather than §oor though)When I showed that to my math teacher her reply was youre trying todo algebra that you dont understand. She was an ignorant person toone extreme, but the teacher of the girl in this story seems to be anignorant person to the opposite extreme: his claim that this girlsdiscovery is some kind of revolution does a lot to make him look likean idiot. The fact that it got so far (the principal called it a dayof mathematical rejoicing and some ignorant USAF reporter put it onthe Air Force news) without anyone noticing that it is not really allthat big a deal, shines a very poor light on all parties involved.Of course it would be cruel and heartless to express such sentimentsto the girl, but on the other hand, is it no less cruel or heartlessto decieve her so utterly? Suppose, hypothetically, that she hadindependantly discovered the quadratic equation: certainly she woulddeserve great praise and honors, but to tell her you are the firstperson to ever solve this problem would be an outright black lie.It seems as though in the entire journey from the classroom to thenews, noone ever once thought of actually consulting a mathematician. (I guess a lot of ignorant folk would think the Interesting math news story, hard to say whether its overrated or not> > Basically, a 12 year old girl figures out that> A - (B/C) = (A-1) + (C/C) - (B/C) [...]The example presented in the Air Force news story (below) is 8 2/5 - 5 3/5 = 3 -1/5 = 2 + 5/5 - 1/5 = 2 4/5It is surprising that this is deemed a way that has never been donebefore by her 7th grade teacher (and his college friends), and alsoby her mother (a senior accounting technician). All of them shouldrecognize this as essentially equivalent to subtracting via borrowing.Even if, surprisingly, all of them had never seen such a techniqueemployed for fractions, they shouldnt be so naive to think that thisis a new discovery - especially given that the obvious web search onsubtracting fractions borrow yields over 700 matches. Are thereelementary school curricula which do not present such basic methods?Ive added k12.ed.math, who probably know more about such curricula. -Bill DubuqueSTUDENT INVENTS NEW MATH PROCESSby Mike Wallace, Aeronautical Systems Center Public Affairshttp://www.af.mil/stories/story.asp?storyID=123006043 Killie Rick found a new solution to subtraction problems involvingwhole numbers and fractions. She used the concept of negative numbersin a way that has never been done before, as far as her seventh-gradeteacher has been able to ascertain.The 12-year-old girl is the daughter of Terri Rick, a senioraccounting technician in Air Force Materiel Commands materielsystems group here.An example of a problem and Killies solution is: 8 2/5 - 5 3/5 = 3 -1/5 = 2 + 5/5 - 1/5 = 2 4/5By using negative numbers, a concept she began to get comfortablewith this year at Mary Help of Christians school in Fairborn, Ohio,Killie was able to simplify the process of subtracting fractions.Ive never seen anybody do this, said Colin McCabe, Killiesteacher. It simplifies it by taking out three steps (to find asolution). I went home and tried to find fault with it, but Icouldnt. I got online and did research, and I talked to friends of mine from college, and I cant find anybody whos seen this.Her process was not used in any of McCabes reference materials. He said he was so impressed that on Nov. 12 he presented her acertificate for outstanding achievement. The certificate was in recognition of her mathematical ingenuity in the discovery of a new method of solution to mixed number subtraction. McCabesaid he intends to teach what he calls Killies Way to studentsin his future classes.I think a lot of credit should go to the teacher, said Anne Steck,the schools principal. I know lots of math teachers who wouldvelooked at Killies work and just said it was wrong.I got this (math) problem, and I didnt remember what to do (tosolve it), Killie said. I thought (my solution) made sense, butI expected the teacher to say it was wrong.The use of negative numbers seemed reasonable to her, Killie said.Finding a new way to solve problems with fractions, having herteacher praise her work, and hearing her school principal call it aday of mathematical rejoicing; maybe it is not so surprising thatKillie said math is her favorite subject now.INSET: FAIRBORN, Ohio -- Killie Rick explains her new method ofsolving subtraction of fractions problems, as her teacher, ColinMcCabe, puts an example on the chalkboard. She uses negative numbers,which McCabe said he had never seen before. Killie is a seventh-gradestudent at Mary Help of Christians school here. Her mother is TerriRick, an employee at nearby Wright-Patterson Air Force Base. (U.S. Air Force photo by Spencer P. Lane)http://www.af.mil/media/photodb/photos/031118-F-0000S-008 .jpg-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: Interesting math news story, hard to say whether its overrated or not> 8 2/5 - 5 3/5 = 3 -1/5 = 2 + 5/5 - 1/5 = 2 4/5>> It is surprising that this is deemed a way that has never been done> before by her 7th grade teacher (and his college friends), and alsoThats interesting as...8 2/5 - 5 3/5 = 3 - 1/5because, for instance, 8 2/5 means 8 + 2/5 . So 3 - 1/5 can itself be afinal form...Obviously, this is just novel application...-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: Interesting math news story, hard to say whether its overrated or not> 8 2/5 - 5 3/5 = 3 -1/5 = 2 + 5/5 - 1/5 = 2 4/5>> It is surprising that this is deemed a way that has never been done> before by her 7th grade teacher (and his college friends), and also> > Thats interesting as...> > 8 2/5 - 5 3/5 = 3 - 1/5> > because, for instance, 8 2/5 means 8 + 2/5 . So 3 - 1/5 can itself be a> final form...> > Obviously, this is just novel application...Just a new way of teaching kids math, I think its good; but I wonderwhy if one has thought of it before. Personally thats often what I doin my head, along with 127 - 18 = 127 - 17 - 1 = 110 - 1 = 109 ... ifthats the way I happen to think of it at the time.bother to investigate, but jsut wanted to have something good to writeabout, so he did :-)Now, back to the mathematical pr0n, please? :-)-Grisha-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: Interesting math news story, hard to say whether its overrated or notI agree w/you, all the grownups in this story come across as cretins.> The story is at:> > http://www.af.mil/stories/story.asp?storyID=123006043> > Basically, a 12 year old girl figures out that> A - (B/C) = (A-1) + (C/C) - (B/C),> and uses it as an improved way to subtract fractions.> However, the artical comes off making it sound like she has invented> calculus or something. Obviously if a grown adult were to discover> the time.> On the other hand, the fact that she is 12 does lend her some credit. > Im not sure how much it lends, though. Certainly if she were 3 or> even 6, it would lend a lot more, but 12 is pushing it. When I was 12> I discovered that you could do those damned division problems where> youre required to give the remainder, using a calculator:> A/B = §oor(A/B) with a remainder of B*[A/B-§oor(A/B)]> (I used the QBasic int rather than §oor though)> When I showed that to my math teacher her reply was youre trying to> do algebra that you dont understand. She was an ignorant person to> one extreme, but the teacher of the girl in this story seems to be an> ignorant person to the opposite extreme: his claim that this girls> discovery is some kind of revolution does a lot to make him look like> an idiot. The fact that it got so far (the principal called it a day> of mathematical rejoicing and some ignorant USAF reporter put it on> the Air Force news) without anyone noticing that it is not really all> that big a deal, shines a very poor light on all parties involved.> Of course it would be cruel and heartless to express such sentiments> to the girl, but on the other hand, is it no less cruel or heartless> to decieve her so utterly? Suppose, hypothetically, that she had> independantly discovered the quadratic equation: certainly she would> deserve great praise and honors, but to tell her you are the first> person to ever solve this problem would be an outright black lie.> > It seems as though in the entire journey from the classroom to the> news, noone ever once thought of actually consulting a mathematician. > (I guess a lot of ignorant folk would think the teacher in the story> is story, hard to say whether its overrated or not> I agree w/you, all the grownups in this story come across as cretins.> I think thats a bit harsh. You forget that the typical elementary orjunior-high math teacher has a degree in education, rather thanmathematics. This teacher probably has his or her friends of similarbackgrounds. In addition, the education major is a haven for peoplewith little confidence in their mathematical abilities.All told, I would imagine that the technique was novel *to the teacher*,and certainly to the student. The amount of ingenuity or creativity thatthis modest step took should be recognized, and if the teacher wants toprint out a Certificate of Lah Di Doo Dah and award it to the student,or maybe the base commander could have a Lah Di Doo Dah parade, completewith brass band, key to the city, and a Blue Angels §y-over, it onlymakes them people who are easily impressed. Not cretins.When your child shows you a drawing she made, do you crumple it andsneer, Rembrandt cries bitter tears over your pitiful attempts at art,or do you make use of your refrigerator magnets?In short, which is the approach one *should* take over the efforts ofchildren: celebration, or ennui?Dale> >>The story is at:>>http://www.af.mil/stories/story.asp?storyID=123006043>>Basically, a 12 year old girl figures out that>>A - (B/C) = (A-1) + (C/C) - (B/C),>>and uses it as an improved way to subtract fractions.>>However, the artical comes off making it sound like she has invented>>calculus or something. Obviously if a grown adult were to discover>>the time. ... story deleted ...>>It seems as though in the entire journey from the classroom to the>>news, noone ever once thought of actually consulting a mathematician. >>(I guess a lot of ignorant folk would think the teacher in the story>>is a mathematician... say whether its overrated or notI generally try to be nonjudgemental about people who are not verymathematically bright, but there is something different about thisstory. IMO statements like, nobody has ever done this before in allof history are arrogant and make me want to make fun of the personwho said it. Especially when theyre so silly, as in this case. Canyou think of someone else who makes statemants like that on thisgroup? Lets not forget that this is a mathematics teacher, someonewho is supposed to be an expert relative to the kids theyre teaching.Maybe these people arent cretins, I dont know, but I think theyshould stay away from these statements, which to me make them seemlike idiots, which was all I said. And if my child came up w/somethinglike that Id be happy but I wouldnt exactly nominate them for theFields medal. And, anyway, isnt that the way everone subtractsfractions from whole numbers? Ive always done it that way.P.S. Its great to make people feel good about themselves, even whenthey have no reason to, but dont our public schools suck, and isntthat a problem, and isnt trying to make everyone feel like geniusesat least part of the reason?> I agree w/you, all the grownups in this story come across as cretins.> > > I think thats a bit harsh. You forget that the typical elementary or> junior-high math teacher has a degree in education, rather than> mathematics. This teacher probably has his or her friends of similar> backgrounds. In addition, the education major is a haven for people> with little confidence in their mathematical abilities.> > All told, I would imagine that the technique was novel *to the teacher*,> and certainly to the student. The amount of ingenuity or creativity that> this modest step took should be recognized, and if the teacher wants to> print out a Certificate of Lah Di Doo Dah and award it to the student,> or maybe the base commander could have a Lah Di Doo Dah parade, complete> with brass band, key to the city, and a Blue Angels §y-over, it only> makes them people who are easily impressed. Not cretins.> > When your child shows you a drawing she made, do you crumple it and> sneer, Rembrandt cries bitter tears over your pitiful attempts at art,> or do you make use of your refrigerator magnets?> > In short, which is the approach one *should* take over the efforts of> children: celebration, or hard to say whether its overrated or notX-DMCA-Notifications: http://www.giganews.com/info/dmca.html>I generally try to be nonjudgemental about people who are not very>mathematically bright, but there is something different about this>story. IMO statements like, nobody has ever done this before in all>of history are arrogant and make me want to make fun of the person>who said it. Especially when theyre so silly, as in this case. Can>you think of someone else who makes statemants like that on this>group? Lets not forget that this is a mathematics teacher, someone>who is supposed to be an expert relative to the kids theyre teaching.>Maybe these people arent cretins, I dont know, but I think they>should stay away from these statements, which to me make them seem>like idiots, which was all I said. And if my child came up w/something>like that Id be happy but I wouldnt exactly nominate them for the>Fields medal. And, anyway, isnt that the way everone subtracts>fractions from whole numbers? Ive always done it that way.>>P.S. Its great to make people feel good about themselves, even when>they have no reason to, but dont our public schools suck,Yes.> and isnt>that a problem,Yes.> and isnt trying to make everyone feel like geniuses>at least part of the reason?Yes, or so it seems to me.>> I agree w/you, all the grownups in this story come across as cretins.>> >> >> I think thats a bit harsh. You forget that the typical elementary or>> junior-high math teacher has a degree in education, rather than>> mathematics. This teacher probably has his or her friends of similar>> backgrounds. In addition, the education major is a haven for people>> with little confidence in their mathematical abilities.>> >> All told, I would imagine that the technique was novel *to the teacher*,>> and certainly to the student. The amount of ingenuity or creativity that>> this modest step took should be recognized, and if the teacher wants to>> print out a Certificate of Lah Di Doo Dah and award it to the student,>> or maybe the base commander could have a Lah Di Doo Dah parade, complete>> with brass band, key to the city, and a Blue Angels §y-over, it only>> makes them people who are easily impressed. Not cretins.>> >> When your child shows you a drawing she made, do you crumple it and>> sneer, Rembrandt cries bitter tears over your pitiful attempts at art,>> or do you make use of your refrigerator magnets?>> >> In short, which is the approach one *should* take over the efforts of>> children: celebration, or ennui?>> >> Dale>>David C. emailed replies to this message constitute permission for an E5 5E EC F3 04 26 4E BF 1A 92X-Tom-Swiftie: Turn that fan off, Tom said coldly> I think thats a bit harsh. You forget that the typical elementary or> junior-high math teacher has a degree in education, rather than> mathematics. This teacher probably has his or her friends of similar> backgrounds. In addition, the education major is a haven for people> with little confidence in their mathematical abilities.What you are doing is explaining perhaps why teacher may beincompetent.But story, hard to say whether its overrated or not>> I think thats a bit harsh. You forget that the typical elementary or> junior-high math teacher has a degree in education, rather than> mathematics. This teacher probably has his or her friends of similar> backgrounds. In addition, the education major is a haven for people> with little confidence in their mathematical abilities.>> What you are doing is explaining perhaps why teacher may be> incompetent.>> But that doesnt excuse it.>The real reason teachers are incompetent is that school systems pay lousyfor teaching, but great for babysitting. So what do you expect to get?I taught high school for a year, trying to sell math. I discovered Imnot a salesman, and the environment is really terrible. (Of course, Itaught at a science and engineering magnet school, so I dont have the fullmagnitude of the terrible environment.) The children believe that it istheir duty to resist any education that might be available around them.(Actually, many were closet learners, but they sure wouldnt let theirpeers know.) Worse than college freshmen.I gave up about $40,000 a year to do that, and discovered that I actuallyhad to pay to teach (meaning that if I had just stayed home and done nothingbut housework and cooking, I would have come out ahead).Will this story, hard to say whether its overrated or notW. Dale Hall grava .88 la saucisse et au marteau:> All told, I would imagine that the technique was novel *to the teacher*,> and certainly to the student. The amount of ingenuity or creativity that> this modest step took should be recognized, and if the teacher wants to> print out a Certificate of Lah Di Doo Dah and award it to the student,> or maybe the base commander could have a Lah Di Doo Dah parade, complete> with brass band, key to the city, and a Blue Angels §y-over, it only> makes them people who are easily impressed. Not cretins.I know someone who at age 12 or 13 discovered the formulae for (a+b)^nand sum(i=1 to n) i^s for any s by himself, though without being able toprove them. I dont think anyone ever had the idea to talk about it ina, even local, newspaper. But I dont think he suffered of this lack ofpublic recognition :)Tricks like the one were talknig about is VERY common and I see no needto emphasize it, but I may an ugly french coward fulfilled with fear andwithout any enthusiasm, who knows?-- NicolasPS: Xavier, if you read me, feel free to story, hard to say whether its overrated or notX-DMCA-Notifications: http://www.giganews.com/info/dmca.html>> I agree w/you, all the grownups in this story come across as cretins.>> >>I think thats a bit harsh. You forget that the typical elementary or>junior-high math teacher has a degree in education, rather than>mathematics. This teacher probably has his or her friends of similar>backgrounds. In addition, the education major is a haven for people>with little confidence in their mathematical abilities.>>All told, I would imagine that the technique was novel *to the teacher*,>and certainly to the student. The amount of ingenuity or creativity that>this modest step took should be recognized, and if the teacher wants to>print out a Certificate of Lah Di Doo Dah and award it to the student,>or maybe the base commander could have a Lah Di Doo Dah parade, complete>with brass band, key to the city, and a Blue Angels §y-over, it only>makes them people who are easily impressed. Not cretins.>>When your child shows you a drawing she made, do you crumple it and>sneer, Rembrandt cries bitter tears over your pitiful attempts at art,>or do you make use of your refrigerator magnets?>>In short, which is the approach one *should* take over the efforts of>children: celebration, or ennui?When _your_ child shows you a drawing she made, do you get out therefrigerator magnets or do you call the papers and have it declareda day of artistic rejoicing?>Dale>> >The story is at:>>http://www.af.mil/stories/story.asp?storyID=123006043>> Basically, a 12 year old girl figures out that>A - (B/C) = (A-1) + (C/C) - (B/C),>and uses it as an improved way to subtract fractions.>However, the artical comes off making it sound like she has invented>calculus or something. Obviously if a grown adult were to discover>the time.>> ... story deleted ...>>It seems as though in the entire journey from the classroom to the>news, noone ever once thought of actually consulting a mathematician. >(I guess a lot of ignorant folk would think the teacher in the story>is a mathematician... story, hard to say whether its overrated or not> I agree w/you, all the grownups in this story come across as cretins.>>I think thats a bit harsh. You forget that the typical elementary or>junior-high math teacher has a degree in education, rather than>mathematics. This teacher probably has his or her friends of similar>backgrounds. In addition, the education major is a haven for people>with little confidence in their mathematical abilities.>>All told, I would imagine that the technique was novel *to the teacher*,>and certainly to the student. The amount of ingenuity or creativity that>this modest step took should be recognized, and if the teacher wants to>print out a Certificate of Lah Di Doo Dah and award it to the student,>or maybe the base commander could have a Lah Di Doo Dah parade, complete>with brass band, key to the city, and a Blue Angels §y-over, it only>makes them people who are easily impressed. Not cretins.>>When your child shows you a drawing she made, do you crumple it and>sneer, Rembrandt cries bitter tears over your pitiful attempts at art,>or do you make use of your refrigerator magnets?>>In short, which is the approach one *should* take over the efforts of>children: celebration, or ennui?>> When _your_ child shows you a drawing she made, do you get out the> refrigerator magnets or do you call the papers and have it declared> a day of artistic rejoicing?You owe me a new keyboard. Ill forgo the mouthful of coffee.Jon and M(2,R) > The point is that M(2,R) can be thought of (being isomorphic to) a> 2-dimensional complex algebra with a basis given by {1,chi}> satisfying> > chi^2=1,> ichi + chi i=0.> an algebra is a vector space. its scalar multiplication cannotsatisfy the second of those identities. if your analogy is to benon-vacuous. the implication is that i is not a scalar, which leads toself-inconsistency.as chi looks like X, you want an algebra, whose elements are of theform,a + bX, where a,b are complex numbers. but bX should be the same asXb, which it isnt in your definition.this isnt to say you dont have something interesting to look at,just that it isnt an C and M(2,R)> > > The point is that M(2,R) can be thought of (being isomorphic to) a> 2-dimensional complex algebra with a basis given by {1,chi}> satisfying> > chi^2=1,> ichi + chi i=0.> > > an algebra is a vector space. its scalar multiplication cannot> satisfy the second of those identities. if your analogy is to be> non-vacuous. the implication is that i is not a scalar, which leads to> self-inconsistency.> > as chi looks like X, you want an algebra, whose elements are of the> form,> > a + bX, where a,b are complex numbers. but bX should be the same as> Xb, which it isnt in your definition. > this isnt to say you dont have something interesting to look at,> just that it isnt an algebra.Perhaps it should mean: scalars R, extra elements i, X [= chi],satisfying i^2 = relations between C and M(2,R)Wow, finally someone joined this discussion!Before entering into the details of each post, let me point out that(it *seems to me* that) different authors assume different meaningsfor the term algebra, let alone ring! (commutativity, unity, etc.)I dont know if anyone has ever adopted a definition so wide as tocapture the structure I need, but it is easy to understand which arethe requirements that need to be relaxed. The point is that M(2,R) can be thought of (being isomorphic to) a>> 2-dimensional complex algebra with a basis given by {1,chi}>> satisfying>> >> chi^2=1,>> ichi + chi i=0.>>an algebra is a vector space. its scalar multiplication cannot>satisfy the second of those identities. if your analogy is to be>non-vacuous. the implication is that i is not a scalar, which leads to>self-inconsistency.>>as chi looks like X, you want an algebra, whose elements are of the>form,FWIW in my handwritten notes I use a greek k, of the kind that inLaTeX reads varkappa.>a + bX, where a,b are complex numbers. but bX should be the same as>Xb, which it isnt in your definition.Good point!!Basically Im adjoining to C an element X that satisfies the abovecondition. Thus I obtain an enlarged ring that is (as usual) both aleft and a right module on its subring (identificable with) C.Of course, as you correctly pointed out, it cant be a vector spacebecause you do *not* have aX=Xa for all a in C. To be precise thenon-complex (or anti-complex?) part bX in a certain sense reallyaccounts for the non commutativity of the enlarged ring.In fact if you let A=a+bX, B=c+dX, then you have(AB-BA)/2i = Im(bd*)+ (Im(a)d-bIm(c))X.At this point I think it all boils down to a matter of terminology, ashinted above. But then I plainly dont know what to use instead ofalgebra: so, in the following Ill write pseudo-algebra unlesssomeone suggests a better/appropriate term.>> a + bX, where a,b are complex numbers. but bX should be the same as>> Xb, which it isnt in your definition.>> >> this isnt to say you dont have something interesting to look at,>> just that it isnt an algebra.>>Perhaps it should mean: scalars R, extra elements i, X [= chi],>satisfying i^2 = -1, X^2 = 1, iX + Xi = 0.Well, not really/not only: in fact I am doing that, thus obtaining a4-dimensional real algebra. If one wants to think of it in terms ofmatrices, then Im just using as a basis for M(2,R) the followingmatrices: [1, 0] [0,-1] [1, 0] [0, 1] [0, 1], [1, 0], [0,-1], [1, 0].BUT! What is meaningful and desirable to me is to *factorize* thisextension through the extension to the (2-dimensional, real) algebraof complex numbers *and* the extesion of the latter to a(2-dimensional, complex) pseudo-algebra obtained by adjoining thespecial element X.Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on between C and M(2,R)Ive written at least twice about this subject in the past, withoutreceiving any feedback. Id be glad to read any kind of comment!>>Exponentials, and logarithms of invertible elements, exist in any Banach >>algebra. Look up holomorphic functional calculus.>>I should qualify that. The holomorphic functional calculus exists in>complex Banach algebras, while the usual quaternions are an algebra>over the reals. Of course theres no trouble the OP, I would like to expand to someextent on the relationships between C (the complex *field*) and thethe *algebra* M(2,R) of 2x2 real matrices.Of course nothing of what Im saying has a sound mathematical meaning,but IMO my observations yield a very natural point of view in somerespects, e.g. when dealing with some particular problems. (HoweverIll give as a brief account as possible!)The point is that M(2,R) can be thought of (being isomorphic to) a2-dimensional complex algebra with a basis given by {1,chi}satisfying chi^2=1, ichi + chi i=0.On the other hand (the algebra isomorphic to) M(2,R) is a4-dimentional real algebra with a basis given by {1,i,chi,ichi}: inparticular these elements are not (the images of) the standard basisvectors of M(2,R).Note that in this sense the elements of M(2,R) are (numbers thatconstitute) another hypercomplex extension of C.Now, you can work abstractly with this extension of the ring ofcomplexes, and it is not important how you do intepret them. But ifyou want to have a direct expression of z+wchi (z,w in C) in terms ofsquare matrices, then a *possible choice* of i and chi for thetranslation is: i=[0 -1] chi=[1 0] [1 0], [0 -1].Its worth to notice that it is not a mere chance that the secondmatrix acts on a column vector as complex conjugation.Now, it is straightforward to realize that for A=z+wchi det(A)=|z|^2-|w|^2, A^{-1}=det(A)^{-1} (z*-wchi) if det(A)neq 0.(the latter identity works also if you abstractly *define* det(A) asabove). Interestingly the operatorial norm of A is ||A||=|z|+|w|.Now, as an example, lets find the solutions of the equation x^2=-1.Let x=a+bchi, a,b in C: the following two equations must besatisfied: a^2+|b|^2=-1, 2Re(a)b=0.If (i) b=0 then a^2=-1 => a=+i or a=-i; if (ii) bneq 0, then a=ik, kin R, |b|^2=k^2-1 => |k|>1. By allowing |k|>=1 one can express all thesolutions including those found at point (i) as x=ik+sqrt(k^2-1)e^{itheta}chi,period!If do the similar calculation for x^2=1, then you find that thesolutions found at the corresponding point (i) cannot be incorporatedin a general expression and one has x=1 or x=-1 or x=ih+sqrt(h^2+1)e^{iphi}chi, h in R.Another interesting exercise is to look for numbers/matrices I,X thatsatisfy the same identities as i,chi. (Since 1,-1 commute with everyelement of the algebra, then X must be chosen of the latter form!)Hope this was a TEASER!!Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Dixons character table algorithm: > Im working on implementing Dixons character table algorithm,: > more-or-less as a matter of personal interest at this point.: Your questions are rather difficult to follow: without knowing what text you are working from.10 (1967), 446-450). At the middle of p.448 (first paragraph ofsection 3) he discusses the steps I mention...It seems clear to me (though I could still be wrong), now, thatthe whole nullspace calculation is to be done mod p, in whichcase a second question arises -- how does one manage to find abasis for the nullspace of a matrix mod something?As I said originally, though, if you know of a text thats------------------------------------------------------- - --------- . . . Except when they dont, Because sometimes they wont. - Dr. Seuss--------------------------------------------------------- Help with Dixons character table algorithm>>: > Im working on implementing Dixons character table algorithm,>: > more-or-less as a matter of personal interest at this point.>>: Your questions are rather difficult to follow>: without knowing what text you are working from.>>10 (1967), 446-450). At the middle of p.448 (first paragraph of>section 3) he discusses the steps I mention...>>It seems clear to me (though I could still be wrong), now, that>the whole nullspace calculation is to be done mod p, in which>case a second question arises -- how does one manage to find a>basis for the nullspace of a matrix mod something?Finding a nullspace modulo a prime number is very easy, because you areworking over a finite field. It is a lot easier than working over therationals, say, because you do not get problems with large integers.If A is a matrix over any field, and you want a basis for the space of vectorsv with Av = 0, you just row-echelonize A, which does not change the nullspace,and then for each column of A that does not contain a pivot 1 entry, you caneasily write down a basis vector of the nullspace. For example, if theecelonized form of A is[ 1 2 0 1 ][ 0 0 1 3 ]Then the nullspace has basis [ 0 0 -3 1 ]^T, [-2 1 0 0 ]^T.Finding a nullspace modulo a non-prime or over the integers is slightly moredifficult, because you are no longer working over a field, but it can stillbe done.Derek Holt.>As I said originally, though, if you know of a text thats>>----------------------------------------------------- - -----------> . . . Except when they dont,> Because sometimes they wont. - Dr. Seuss>-------------------------------------------------------- Help with Dixons character table algorithm> Im working on implementing Dixons character table algorithm,> more-or-less as a matter of personal interest at this point.Your questions are rather difficult to followwithout knowing what text you are working from.-- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, in quarter-disc?Originator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow)Heres a simple-looking calculus problem that I found myself needing tosolve yesterday. Take a quarter of a unit disc and inscribe a rectanglein it, with one of the rectangles vertices on each of the straight sidesof the quarter-disc, equidistant from the center of the (original, full)disc. What is the maximum area that such a rectangle can have?Unless I messed up, the answer is sqrt(2) - 1. However, I arrivedat this answer via a rather clumsy and tedious calculation. Questions:1. Is there a slick solution? Perhaps not even needing calculus?2. Has anyone seen this problem published in a calculus text (or elsewhere)?-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues rectangle inscribed in quarter-disc?>Heres a simple-looking calculus problem that I found myself needing to>solve yesterday. Take a quarter of a unit disc and inscribe a rectangle>in it, with one of the rectangles vertices on each of the straight sides>of the quarter-disc, equidistant from the center of the (original, full)>disc. What is the maximum area that such a rectangle can have?>Unless I messed up, the answer is sqrt(2) - 1. However, I arrived>at this answer via a rather clumsy and tedious calculation. Questions:>1. Is there a slick solution? Perhaps not even needing calculus?Is this slick enough? Suppose one of the vertices is at (x,0), another at (0,x). A third must be at (x+y,y) on the circle, which means (x+y)^2 + y^2 = 1. The area isthe length of the cross-product of (-x,x,0) and (y,y,0), namely2 x y. So we want to maximize 2 x y subject to (x+y)^2 + y^2 = 1.Using Lagrange multipliers, we want to solve 2 y + 2 t (x+y) = 0 2 x + 2 t (x+2y) = 0 (x+y)^2 + y^2 = 1The first two equations, considered as linear equations in x and y, have nonzero solutions if and only if t is one of the two roots of [ t 1+t ]det([ 1+t 2t ]) = -1-2t+t^2For such t, we then have x+y = -y/t, so the third equation becomes y^2 (1/t^2+1) = 1; thus y = (+/-) t/sqrt(1+t^2) and x = (-/+) (1+t)/sqrt(1+t^2), and the area is 2 x y = - 2t(1+t)/(1+t^2) = -t (since 1+t^2 = 2+2t). Since this must be positive, t must be the negative root, namely 1-sqrt(2), and the maximal area is sqrt(2)-1.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Heres a simple-looking calculus problem that I found myself needing to> solve yesterday. Take a quarter of a unit disc and inscribe a rectangle> in it, with one of the rectangles vertices on each of the straight sides> of the quarter-disc, equidistant from the center of the (original, full)> disc. What is the maximum area that such a rectangle can have?> > Unless I messed up, the answer is sqrt(2) - 1. However, I arrived> at this answer via a rather clumsy and tedious calculation. Questions:> > 1. Is there a slick solution? Perhaps not even needing calculus?> 2. Has anyone seen this problem published in a calculus text (or elsewhere)?I havent seen it before, but here is a way not too hard. Half your rectangle sits in the 1/8 disk defined by x^2+y^2<1, 0 Heres a simple-looking calculus problem that I found myself needing to>> solve yesterday. Take a quarter of a unit disc and inscribe a rectangle>> in it, with one of the rectangles vertices on each of the straight sides>> of the quarter-disc, equidistant from the center of the (original, full)>> disc. What is the maximum area that such a rectangle can have?>> >> Unless I messed up, the answer is sqrt(2) - 1. However, I arrived>> at this answer via a rather clumsy and tedious calculation. Questions:>> >> 1. Is there a slick solution? Perhaps not even needing calculus?>> 2. Has anyone seen this problem published in a calculus text (or elsewhere)?>>I havent seen it before, but here is a way not too hard. Half your rectangle >sits in the 1/8 disk defined by x^2+y^2<1, 0(cost,sint) on the circle, and (sint,sint) on line y=x. So, your >rectangle has area: 2*(cost-sint)sint, which is 2*sint*cost-2*(sint)^2. >rectangle area is: A = sin(2t)+cos(2t)-1. dA/dt=2*cos(2t)-2*sin(2t)=0, >so tan(2t)=1. Thus 2t=Pi/4 and t=Pi/8. Plugging t=Pi/8 into the formula >for A gives A=sqrt(2)-1 as you found.>>--Jim BuddenhagenTo avoid using calculus at all, one can write A = sin(2t)+cos(2t)-1 = sqrt(2)sin(2t+pi/4)-1which obviously has a maximum of sqrt(2)-1 when sin(2t+pi/4) = 1.Rob Johnsontake out inscribed in quarter-disc?Originator: tchow@MARKOV.MIT.EDU.mit.edu (Timothy Chow)<>sits in the 1/8 disk defined by x^2+y^2<1, 0endpoints (cost,sint) on the circle, and (sint,sint) on line y=x. So, your <>rectangle has area: 2*(cost-sint)sint, which is 2*sint*cost-2*(sint)^2. [...]> elements?> Well, the prime factorization of 10! = 2^8 * 3^4 * 5^2 * 7^1>> You can do the rest.How? (Or is this just a joke? Is there a Subgroups of S_10mareg@mimosa.csv.warwick.ac.uk () of subgroups of the symmetric group S_10 on ten >>elements?>> According to Magma, there are 1593 conjugacy classes of that Sloanes sequence A000638 has the first number and the corresponding number for S_11. The second number is still missing from A005432, you could contribute your number (hint, OEIS?Anum=A000638http://www.research.att.com/projects/OEIS? is the number of subgroups of the symmetric group S_10 on ten >>elements?>>According to Magma, there are 1593 conjugacy classes of subgroups, and>29594446 subgroups altogether.>Can someone provide a reference?>>I dont know - I cant.>Can GAP do it?>>I expect so - I am just trying that as a check, but it is still thinking>about it.Yes, GAP can do it (but it takes a few hours) and gets the same answer asabove. Since the algorithms used by GAP and Magma are different, it seemsquite possible that the answer is correct. I hope it proves useful Optimizer in Excel? How do you find it? I tried looking for it that, but I am not used to>using Excel.>> Its a property of any system doing numerical computation. Matlab does> the same thing. So would a math library in FORTRAN or C++.>> I usually use Mathematica, or something.>> Systems like Mathematica and Maple escape round-off error by doing> calculations symbolically, as we do.>> I am just trying to>learn Excel so that I can add it to my resume.>> I recommend learning the optimizer. Its a powerful solver that does> integer programs, linear programs, and nonlinear optimization, and> many people Excel for math> Is the Optimizer in Excel? How do you find it? I tried looking for it with> the help function, but wasnt you can get it under Tools - Solver...If you dont see it there, you have to add it yourself. Its adequateas a solver of linear programs, but it has severe limitations.Proceed with tip! I just added it in. Now, I just have to figure outhow to use it.Lurch>> Is the Optimizer in Excel? How do you find it? I tried looking for itwith> the help function, but wasnt you can get it under Tools - Solver...> If you dont see it there, you have to add it yourself. Its adequate> as a solver of linear programs, but it has severe limitations.> Proceed I am just starting to play a little with MS-Excel for Math purposes. Does> anyone know of a way to calculate several values for a given function, i.e.> Cos(1, pi/2, pi/3, ...n) ? And, can a cell reference be given instead of> putting in the values? Such as, Cos(A1:A10). (I have also cross-posted to> =cos(a1:a10) as normal and pressCTRL+SHIFT+ENTER to enter it (see help index array topic Aboutarray formulas and how to enter them) you get what you want.Personally I find entering =cos(a1), selecting this entry and draggingit to extend it, is black holesJack,It is really interesting! Both the papers have direct relation to thecomplex structure of the Kerr source.The section 5 of the Kaiser paper describes thecomplex Kerr source ( up to specific notations used in many of my papers from1974).Yes, I realized that. I immediately thought of your presentation at Paris Vigier IVIn my opinion Kaisers interpretation of the source as pulsed beamslooks doubtful. It is a stationary rotating string in the zero point field(detailed description in gr-qc/0212048).In one paper Kaiser says no Zitterbewegung (did I spell that right?) andsays that normal ordering is automatic in his scheme. I am not sureif he means to say that there are no zero point energy §uctuations ever.That would not be compatible with Lamb shift for example and thedark energy.PS One possible conceptual puzzle in my own theory.I need to use Dirac eq gap 2mc^2 in globally §at false vacuum to get the BCS instability to have both Einsteins c-number gravity and zero point energy density w = -1 exotic vacuum unified dark energy/matter emerge in thevacuum phase transition.However, mc^2 ~ e^2|/zpf|^1/2Oh, its OK. I need to useMc^2 ~ e^2|/zpf(False Vacuum)|^1/2 ~ 10^16 Gev/zpf < 0andmc^2 ~ e^2|/zpf(True Vacuum)|^1/2 ~ 10^-3 MevThe beams of this type exist, buthave to correspond to more complicated solutions.Alexander----- Original Message === -----Subject: Re: Physical wavelets and their sources: Real quantum field theory and stringtheory in the wavelet transform generalization of the Fourier transformis important. Note how complex spacetime comes in. Possiblyhypercomplex non-commutative spacetime beyond that.http://www.arxiv.org/abs/math-ph/0303027Authors: Gerald KaiserJournalof Physics A: Mathematical and General, this http URLSubj-class: Mathematical Physics; Complex Variables; Analysis of PDEsFor the first time, complete source distributions for the emission andabsorption of acoustic and electromagnetic wavelets are defined andcomputed, both in spacetime and Fourier space. The biggest surprise isthegreat simplicity of the Fourier sources as compared to the ratherconvolutedspacetime expressions obtained from the original wavelets. Thissuggeststhat the associated pulsed-beam propagators may play a fundamentalrole inemission and absorption processes including focus or directivity. Italsoopens the way to constructing FFT-based algorithms for pulsed-beamanalysesof acoustic and electromagnetic waves.Electromagnetic Wavelets as Hertzian Pulsed Beams in Complex Spacetimehttp://www.arxiv.org/abs/gr-qc/0209031Authors: Gerald KaiserComments: 16 pages, 2 figures, Topics in Mathematical Physics, GeneralRelativity and Cosmology conference (in honor of Jerzy Plebanski) Subj-class: General Relativity and Quantum Cosmology; MathematicalPhysics;Complex VariablesElectromagnetic wavelets are a family of 3x3 matrix fields W_z(x)parameterized by complex spacetime points z=x+iy with y timelike. Theyaretranslates of a sl basic rm wavelet W(z) holomorphic in thefuture-oriented union T of the forward and backward tubes. Applied to acomplex polarization vector p (representing electric and magnetic dipolemoments), W(z) gives an anti-selfdual solution W(z)p of Maxwellsequationsderived from a selfdual Hertz potential Z(z)=-iS(z)p, where S is the slSynge function rm acting as a Whittaker-like scalar Hertz potential.Resolutions of unity exist giving representations of sourcelesselectromagnetic fields as superpositions of wavelets. With the choiceof abranch cut, S(z) splits into a difference of retarded and advanced slpulsed beams rm whose limits as yto 0 give the propagators of the waveequation. This yields a similar splitting of the wavelets and leads totheircomplete physical interpretation as EM pulsed beams absorbed andemitted bya sl disk source rm D(y) representing the branch cut. The choice of ydetermines the beams orientation, collimation and duration, givingbeams assharp and pulses as short as desired. The sources are computed asspacetimedistributions of electric and magnetic dipoles supported on D(y). Thewavelet representation of sourceless electromagnetic fields now splitsintorepresentations with advanced and retarded sources. Theserepresentationsare the electromagnetic counterpart of relativistic coherent-staterepresentations previously derived for massive Klein-Gordon and DiracNon-linear Vacuum Phenomena in Non-commutative QEDhttp://www.arxiv.org/abs/hep-th/0006209Authors: L. Alvarez-Gaume, J.L.F. BarbonComments: LaTeX, 23 ppReport-no: CERN-TH/2000-181Journal-ref: Int.J.Mod.Phys. A16 (2001) 1123-1146We show that the classic results of Schwinger on the exact propagationofcanbe readily extended to the case of non-commutative QED. It is shown thatnon-perturbative effects on constant backgrounds are the same as theircommutative counterparts, provided the on-shell gauge invariantdynamics isreferred to a non-perturbatively related space-time frame. For the caseofthe plane wave background, we find evidence of the effective extendednaturescattering. Besides the known `dipolar character of non-commutativeneutralextended, butthey MATHEMATICIANS READ WITH HALF A LIGHTBULB?[...deleted...]> However, your argument boils down to: ïyou cant do it, and the> results of trying to do so give results we call invalid, and thats> our proof.>> I still believe that those invalid forms hide a discernible underlying> logic that is consistent and goes deeper than weve looked at. As I> slowly learn math and more logic, Ill keep my mind and eyes open.Youre not alone. I believe Euler tried to formalize series like1 + 2 + 4 + 8 + 16 + ... = -1It has already been discussed that under conventional rules ofmathematics this is not allowed, but one should always generalize.How much mileage you get out of the MATHEMATICIANS READ WITH HALF A LIGHTBULB?Grisha---Excellent discussion, and I will reapproach the subject after Ive gotsome more logic and math under my belt. Im concerned about trying tolearn math when much of it may be super§uous if a deeper and morebasic understanding of math processes would eliminate many of thecomplexities of higher math.Very HALF A LIGHTBULB?> The problem is that thousands of people came and said You> mathematicians have gone blind by what youve studied. I, who didnt do> any mathematics in my whole life, I may bring a new look to what you> think is true. Why not? But, more or less, thats what everyone here> has done before knowing a bit of mathematics. I think everyone here> tried to define a proper division by zero, said oh my god, Ive got to> tell the world about my discoveries and finally understood that such a> new definition was leading to nothing interesting.> > But, anyway, there has always been end there the compliment for both me and humanity in general.Very HALF A LIGHTBULB?raydpratt grava .88 la saucisse et au marteau:> However, your argument boils down to: ïyou cant do it, and the> results of trying to do so give results we call invalid, and thats> our proof.The problem is that thousands of people came and said Youmathematicians have gone blind by what youve studied. I, who didnt doany mathematics in my whole life, I may bring a new look to what youthink is true. Why not? But, more or less, thats what everyone herehas done before knowing a bit of mathematics. I think everyone heretried to define a proper division by zero, said oh my god, Ive got totell the world about my discoveries and finally understood that such anew definition was leading to nothing interesting.But, anyway, there has always been end there will always be people likethat.-- Nicolas, thats my Re: Logic missing from Proofs from THE BOOKOriginator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow)>>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.>>The algebra/number theory/encoding is totally unnecessary.>>All that is needed is to have every formula/proposition/>theorem/proof have a positive integer assigned to it; one>way to do this is to use a direct ASCII (or other such)>representation.Hold on a moment...first of all, assigning a positive integer to syntacticobjects is what people often mean by encoding, so this doesnt show thatencoding is totally unnecessary.But more importantly, one needs more than just the assignment; one needsto represent the relevant number-theoretic functions (that correspond torules of inference in your sequent calculus, for example) in the languageof the system whose incompleteness youre proving. In particular, thesimplest encodings all assign integers that are exponentially large (inmagnitude, not in digit-length) in the length of the syntactic objects,so you need some way of representing exponentiation or something similar.Thats where you need to do some work. Its true that you dont necessarilyhave to use the Chinese remainder theorem, but you cant bypass all thiswork *totally*.By the way, I wouldnt describe this part of the proof as ugly. Itsquite beautiful that arbitrarily complex recursive functions can berepresented in such weak systems.-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Proofs from THE BOOK>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.>>The algebra/number theory/encoding is totally unnecessary.>>All that is needed is to have every formula/proposition/>>theorem/proof have a positive integer assigned to it; one>>way to do this is to use a direct ASCII (or other such)>>representation.> > Hold on a moment...first of all, assigning a positive integer to syntactic> objects is what people often mean by encoding, so this doesnt show that> encoding is totally unnecessary....> you cant bypass all this work *totally*.> > By the way, I wouldnt describe this part of the proof as ugly. Its> quite beautiful that arbitrarily complex recursive functions can be> represented in such weak systems.sure, the -fact- is wonderful but the -justification- using CRT or whatever else is kind of messy algebraically. I Re: Logic missing from Proofs from THE BOOK >topology is really just a fancy word for geometry, right?).> > Topology is NOT a fancy word for geometry.Ow. sorry. I was drawing a very, very general comparison there with telescoped words. What my intention was in the general scheme of things, if all of mathematics had to be pigeonholed into a few small areas say algebra, analysis, topology, then geometry would most likely fall into the last one. Yes, that is quite arguable (along with how I split After all, one could cavil that my original disgruntlement (at logic being left out) was misplaced given that mathematics -is- logic. (yes, that is a blatant |-|erc:>> Take 10 minutes to join the Astronomy R-I-N-G!!> > > well rot in hell then all of you, Ron Balke knows who I am> and youre all ded fukers> Wow. The window was open a whole 6 hours, 13 minutes, and 31 seconds.Its a pity you didnt announce this in your first post....:-)-- #191, ewill3@earthlink.netIts still legal to go minutes to join the Astronomy R-I-N-G!!> well rot in hell then all of you, Ron Balke knows who I am> and youre all ded fukers>Not sure of the posters location...But I wonder if he realizes that he just committed a felony 10 minutes to join the Astronomy R-I-N-G!!well rot in hell then all of you, Ron Balke knows who I amand youre all ded -1I(n)= S ----------------- dx 0 x^n oo x^(1/n) -1> I(n)= S ----------------- dx> 0 x^n -1>> n=3,4,5,.......And why do you think its nice?-- S ----------------- dx> 0 x^n -1>> n=3,4,5,.......> > > And why do you think its nice?Ugh. I suppose because the discontinuity at x = 1 is removable. Theresult certainly isnt pretty: (pi/n) (cot(pi/n) - cot((n+1)pi/n^2).And the result of the indefinite integration is hardly prettier. Alittle surprising, perhaps, that it can be written out explicitly.A better question is, why exclude n = 2? Thats the only value wherethe integral is (statistics)how to make date more like Laplacian distribution? ...> Jerry,> > Yes, we can ask but this will only happen about once every 49300 times.> > I was trying to illustrate two different points, the first was the case> where the number of die values doesnt divide evenly into the sample size.> And the other which is the more important is the case where we dont expect> each value (face value on a die) to show up in perfectly matched counts in> trial runs. I know that you know this, but I think Walala now gets the> point - I hope.> > ClayOf course I know it, and I hope I didnt offend. Im sure Walala knowsit too. I intended to point out that non-divisibility is both tangentialand trivial, but might give the wrong idea to someone truly ignorant ofthe matter.Jerry-- Engineering is the art of making what you want from things you (statistics)how to make date more like Laplacian distribution?walala> Why people think I want to lie upon seeing my question? Oh, its my> problem> that I did not clearly present the background...Are you quite sure you are describing your problem clearly? Reading betweenthe lines you have a poor non invertible integer DCT which you are trying tocompensate for.I am going to assume that is true for now ... my apologies if it isnt true.Statistics and deblocking are not the answer, your transform is the problem.Either use more precision to get closer to the true DCT transform, orcreate an invertible transform using lifting/ladder structures.> Here is the story: in deblocking of block DCT coded JPEG images, it was> known that the DCTed coefficients are Laplacian distributed... But now I> am> looking at low bit rate JPEG images, so there are someking of artifacts...> in order to reconstruct the original images... many algorithms have been> devised... one possibility is to make the image coefficients more Laplcian> like...This will do nothing, the idea of using the laplacian distribution withdeblocking works by using a reconstruction point for the quantizedcoefficient inside its quantization interval which will minimize theexpected distortion. In very simple terms, the coefficient is more likely tobe in the low end of the quantization interval, so we assign a value in thatregion to the dequantized coefficient.If you want to use this method to do deblocking you dont have to make theimage coefficients more laplacian, but you simple have to fit thedistribution of the coefficient (across the image) to a laplacian ... thenyou take the parameter for that laplacian and determine from it where in thequantization interval to reconstruct the coefficient.If you are trying to use this in some fashion to combat a poor transform youare barking up the wrong tree. You need to work on the transform, the paperI mentioned earlier has most of the relevant math :http://citeseer.nj.nec.com/kim98fixedpoint.htmlIf your goal is not to get that close to the DCT but simply get a DCT liketransform which can be implemented cheaply then you should it, someone on thecomp.dsp newsgroup posted a link to source code in the past I think).Without the invertible property the low precision integer math you want touse will create an unuseable transform, and no amount of deblocking orlaplacian (statistics)how to make date more like Laplacian data distribution should follow the shape> of Laplacian distribution... the data obtained from measurement is of course> a little off(not very symmtrical), how can I make the measured data more> Laplacian distribution like(make it at least a little more symmtrical)?> > Can anybody give me an example or detailed explanation? I am kind of afraid> of statistics... :=)> You could look at ïhistogram equalisation in an image processing textbook. That takes an image with a non-uniform histogram and produces a mapping to a new set of values that will have a ïsort ofuniform histogram. If you use the correct algorithm, its not limited touniform output -- the output histogram can be specified.However, the histogram will approximate the desired one only globally --there will be holes in it. I cannot imagine the technique being any useto Laplacian distribution?> Why people think I want to lie upon seeing my question? Oh, its my problem> that I did not clearly present the background...Oh, Im not calling you a liar. It does always helpthat you explain the source of your data. In thiscase, Ill define it as more like creative accounting. ;-)There are three kinds of lies: lies, damned lies,and statistics. Mark Twain, Autobiography. Ok, I realize you have a serious question. I donot have an explicit answer for you though.> Here is the story: in deblocking of block DCT coded JPEG images, it was> known that the DCTed coefficients are Laplacian distributed... But now I am> looking at low bit rate JPEG images, so there are someking of artifacts...> in order to reconstruct the original images... many algorithms have been> devised... one possibility is to make the image coefficients more Laplcian> like...> > So that came my question: how to make data more Laplican like...Its a little tough, since I dont know what thedeviation from Laplacian-ness is.If the empirical distribution of x does not havethe desired shape, one could transform x so thatits moments match the moments of a laplacian.Since there are many possible transformations,it is not at all obvious which one would beappropriate. (I honestly dont know.) You wouldneed to try out some different ones to see whichworks best. This would appear to involve anonlinear root-finding operation.Perhaps you do not appreciate why I cannot saymore. Ill give you an example. Suppose you hadtwo distinct cases. In the first, instead of aLaplacian distribution, the distribution ofcoefficients was normally distributed. In thesecond case, the coefficients turn out to be alleither zero or one, with probability 1/2. Youcan appreciate that each case would require adifferent class of transformation.Perhaps others will have better ideas.> (I am just a poor student, not lieing government agency, issurance company,> weapon dealer, lawyers, and politicians... so please help me!)Government agencies never lie. They do not need to.They just change the rules so that whatever theysay becomes the truth. ;-)HTH,John-- There are no questions ? about my real address.The best material model of a cat is another, orpreferably the same, cat.A. Rosenblueth, with minimum turning radiusI want to calculate shortest trajectory of a rocket between two pointsin 3D space. Directions in starting and ending points are given. Therocket has certain minimum turning radius.The trajectory has to consist of starting arc, straight line, andending arc. The line is tangent for both arcs. The first arc containsthe starting point and its tangent there is equal to the startingdirection. The second arc has similar properties for the ending pointand ending direction.When I tried to solve this problem I obtained a system of quadraticequations hard to solve. Therefore, I am looking for some relevantresources. Do in 3d with minimum turning radiusThis does not sound very realistic as the turning radius will depend on thespeed of the rocket and the thrust it is capable of delivering and perhapseven the amount of fuel it consumes if this is a significant proportion ofits mass.But if it is for a game, who cares.> I want to calculate shortest trajectory of a rocket between two points> in 3D space. Directions in starting and ending points are given. The> rocket has certain minimum turning radius.>> The trajectory has to consist of starting arc, straight line, and> ending arc. The line is tangent for both arcs. The first arc contains> the starting point and its tangent there is equal to the starting> direction. The second arc has similar properties for the ending point> and ending direction.>> When I tried to solve this problem I obtained a system of quadratic> equations hard to solve. Therefore, I am looking for some relevant> resources. Do you have any minimum turning radius> I want to calculate shortest trajectory of a rocket between two points> in 3D space. Directions in starting and ending points are given. The> rocket has certain minimum turning radius.> > The trajectory has to consist of starting arc, straight line, and> ending arc. The line is tangent for both arcs. The first arc contains> the starting point and its tangent there is equal to the starting> direction. The second arc has similar properties for the ending point> and ending direction.> > When I tried to solve this problem I obtained a system of quadratic> equations hard to solve. Therefore, I am looking for some relevant> resources. Do you have any suggestions?> > Mirek.This, assuming that arcs are segments of circles, appears to require certain starting and ending configurations be excluded.For example, if the ending point is the same as the starting point but with opposite direction required, it cannot be done with only an arc-line-arc sequence, but needs, at least, turning radiusIf an object enters this path from the bottom its direction isreversed in the point where the two arcs (270deg each) touches. *** **** * ** * * ******* * It consists of two arcs and one line. However, path made from threearcs looks definitely better.Mirek. > > I want to calculate shortest trajectory of a rocket between two points> in 3D space. Directions in starting and ending points are given. The> rocket has certain minimum turning radius.> > The trajectory has to consist of starting arc, straight line, and> ending arc. The line is tangent for both arcs. The first arc contains> the starting point and its tangent there is equal to the starting> direction. The second arc has similar properties for the ending point> and ending direction.> > When I tried to solve this problem I obtained a system of quadratic> equations hard to solve. Therefore, I am looking for some relevant> resources. Do you have any suggestions?> > Mirek.> > This, assuming that arcs are segments of circles, appears to require > certain starting and ending configurations be excluded.> > For example, if the ending point is the same as the starting point > but with opposite direction required, it cannot be done with only an > arc-line-arc sequence, but needs, at least, turning radius> equations hard to solve. Therefore, I am looking for some relevant> resources. Do you have any suggestions?You migth be able to extrapolate Quinlan and KhatibsElastic Bands to three dimensions: http://citeseer.nj.nec.com/quinlan93elastic.html-- study analysis?> 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.>> > Because youre an engineer !!http://www.princeton.edu/~rvdb offers an on-line book Real Analysisfor Engineers but the book offers no reasons for such study. Perhapsyou could get some explicit reasons from the Why should an engineer study analysis?> 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.>Because youre an engineer thought you may be interested on this linkhttp://www.thequantummachine.comand the Mars mounds research explained in the link at the right.Constructive -- martian mounds anomalyCesar Sirvent thought you may be interested on this link|| http://www.thequantummachine.com|| and the Mars mounds research explained in the link at the right.| Constructive cricism and comments welcome.As I see it, several of the labeled points are not placed exactly on the top ofa hill. If they had been, the picture would not show those nice triangles. Imafraid its a bit of wishful thinking by the artist, just like the Cydonianface that was just a §uke of light and shadows. I pity the ignorant fool whobelieves this mounds anomaly> As I see it, several of the labeled points are not placed exactly on the top of> a hill. If they had been, the picture would not show those nice triangles. Im> afraid its a bit of wishful thinking by the artist, just like the Cydonian> face that was just a §uke of light and shadows. I pity the ignorant fool who> believes this story.You are a slow thinker. The crater chains present in Mars, Moon andeven Earth does not display perfect alignments, not perfectinter-distances. The chances for the mounds if they fit perfectlywould be so low that not only they would be non-random, THEY WOULD BEARTIFICIAL most likely (Nature has never displayed === welcome -- martian mounds anomaly>Subject: Re: Comment welcome -- martian mounds anomaly>Message-id: >Cesar Sirvent thought you may be interested on this link>|>| http://www.thequantummachine.com>|>| and the Mars mounds research explained in the link at the right.>| Constructive cricism and comments welcome.>>As I see it, several of the labeled points are not placed exactly on the top>of>a hill. If they had been, the picture would not show those nice triangles.>Im>afraid its a bit of wishful thinking by the artist, just like the Cydonian>face that was just a §uke of light and shadows. I pity the ignorant fool who>believes this story.>>-- >Someonehttp://members.aol.com/mensanator666/cydonia/ welcome -- martian mounds anomaly> http://members.aol.com/mensanator666/cydonia/cydonia.htmWhat Comment welcome -- martian mounds anomaly> Cesar Sirvent thought you may be interested on this link> |> | http://www.thequantummachine.com> |> | and the Mars mounds research explained in the link at the right.> | Constructive cricism and comments welcome.> > As I see it, several of the labeled points are not placed exactly on the top of> a hill. If they had been, the picture would not show those nice triangles. Im> afraid its a bit of wishful thinking by the artist, just like the Cydonian> face that was just a §uke of light and shadows. I pity the ignorant fool who> believes this story.I think there was a planet between mars and jupiter and Comment welcome -- martian mounds anomaly> I think there was a planet between mars and jupiter and they blew it> up. but they left us a message.You are a slow thinker. Read the caution in the page. I ask if themounds are randomly distributed. Any other idea will be sent to thetrash, even if it is ironic by wondering if there is a simple way of solving this:x = r1 (mod 3) ; r1=1x = r2 (mod 5) ; r2=4x = r3 (mod 7) ; r3=5=> x = a + 3*5*7*n (n=0...oo)By inspection we get a = 19Suposing a = f(r1,r2,r3,3,5,7)How do I get ïf?TIA and please forgive my wondering if there is a simple way of solving this:> > x = r1 (mod 3) ; r1=1> x = r2 (mod 5) ; r2=4> x = r3 (mod 7) ; r3=5 => x = a + 3*5*7*n (n=0...oo)> > By inspection we get a = 19> > Suposing a = f(r1,r2,r3,3,5,7)> > How do I get ïf?> > TIA and please forgive my english!> > N.> > Goto http://mathworld.wolfram.com/ChineseRemainderTheorem.htmlor Google on Chinese Remainder Theorem Almost 15000 there is a simple way of solving this:>>x = r1 (mod 3) ; r1=1>x = r2 (mod 5) ; r2=4>x = r3 (mod 7) ; r3=5>>=> x = a + 3*5*7*n (n=0...oo)>>By inspection we get a = 19>>Suposing a = f(r1,r2,r3,3,5,7)>>How do I get ïf?>>TIA and please forgive my english!>>N.>There is a nice explanation of the chinese remainder theorem here:http://www.math.hawaii.edu/~lee/courses/Chinese.pdfFor a computer algorithm to solve the crt, seehttp://www.cacr.math.uwaterloo.ca/hac/index.html chapter 2 Professor at the University of Montana.>Hi all!>>I was wondering if there is a simple way of solving this:>>x = r1 (mod 3) ; r1=1>x = r2 (mod 5) ; r2=4>x = r3 (mod 7) ; r3=5>>=> x = a + 3*5*7*n (n=0...oo)It can be negative as well.Chinese Remainder Theorem gives you a formula. You can also solve itas follows:x = 1 (mod 3), so x = 1 +3a for some integer a.x = 4 (mod 5), so 1+3a = 4 (mod 5), so 3a = 3 (mod 5). So a=1+5b forsome integer b. That means thatx = 1+3a = 1+3(1+5b) = 4+15b.Since x= 5 (mod 7), 4+15b = 5 (mod 7), so 15b = 1 (mod 7), so b=1 (mod7). Therefore, b= 1 +7c, sox = 4 + 15(1+7c) = 19 + 105c for some integer c.>By inspection we get a = 19>>Suposing a = f(r1,r2,r3,3,5,7)>>How do I get ïf?The Chinese Remainder Theorem gives a formula, which requires you tosolve the congruencess*5*7 = 1 (mod 3)t*3*7 = 1 (mod 5)u*3*5 = 1 (mod 7).Once you found s, t, and u that satisfy these congruences, thenf(r1,r2,r3,3,5,7) = r1*s*5*7 + r2*t*3*7 + r3*u*3*5 + 3*5*7*n , n anarbitrary integer.In general, if m1,....,mk are pairwise relatively prime positiveintegers, and r1, r2, ..., rk are arbitrary integers, then thesolution tox = r1 (mod m1)x = r2 (mod m2) . . .x = rk (mod mk)may be found as follows: find integers s1,...,sk such thats1*m2*m3*...*mk = 1 (mod m1)m1*s2*m3*...*mk = 1 (mod m2)m1*m2*s3*...*mk = 1 (mod m3) . . .m1*m2*m3*...*sk = 1 (mod mk)And then letf(r1,...,rk,m1,....,mk) = r1*s1*m2*...*mk + ... + rk*m1*m2*...*sk +(m1*...*mk)nwith n an arbitrary what I accept as reality. --- Calvin (Calvin and a simple way of solving this:>>x = r1 (mod 3) ; r1=1>x = r2 (mod 5) ; r2=4>x = r3 (mod 7) ; r3=5>>=> x = a + 3*5*7*n (n=0...oo)>> It can be negative as well.>> Chinese Remainder Theorem gives you a formula. You can also solve it> as follows:>> x = 1 (mod 3), so x = 1 +3a for some integer a.>> x = 4 (mod 5), so 1+3a = 4 (mod 5), so 3a = 3 (mod 5). So a=1+5b for> some integer b. That means that>> x = 1+3a = 1+3(1+5b) = 4+15b.>> Since x= 5 (mod 7), 4+15b = 5 (mod 7), so 15b = 1 (mod 7), so b=1 (mod> 7). Therefore, b= 1 +7c, so>> x = 4 + 15(1+7c) = 19 + 105c for some integer c.>>By inspection we get a = 19>>Suposing a = f(r1,r2,r3,3,5,7)>>How do I get ïf?>> The Chinese Remainder Theorem gives a formula, which requires you to> solve the congruences>> s*5*7 = 1 (mod 3)> t*3*7 = 1 (mod 5)> u*3*5 = 1 (mod 7).>> Once you found s, t, and u that satisfy these congruences, then>> f(r1,r2,r3,3,5,7) = r1*s*5*7 + r2*t*3*7 + r3*u*3*5 + 3*5*7*n , n an> arbitrary integer.>> In general, if m1,....,mk are pairwise relatively prime positive> integers, and r1, r2, ..., rk are arbitrary integers, then the> solution to>> x = r1 (mod m1)> x = r2 (mod m2)> .> .> .> x = rk (mod mk)>> may be found as follows: find integers s1,...,sk such that>> s1*m2*m3*...*mk = 1 (mod m1)> m1*s2*m3*...*mk = 1 (mod m2)> m1*m2*s3*...*mk = 1 (mod m3)> .> .> .> m1*m2*m3*...*sk = 1 (mod mk)>> And then let>> f(r1,...,rk,m1,....,mk) = r1*s1*m2*...*mk + ... + rk*m1*m2*...*sk+(m1*...*mk)n> with n an arbitrary integer.>> I accept as reality.> --- Calvin (Calvin and Hobbes)> wondering if there is a simple way of solving this:>> x = r1 (mod 3) ; r1=1> x = r2 (mod 5) ; r2=4> x = r3 (mod 7) ; cauchy sequencesIf real numbers are thought of as equivalence classes of Cauchysequences, then what would be a formulation to say one real number isbigger than another? May we say something about the limits of theCauchy sequences, or what?Is this a practical way of defining real numbers, or something likeDedekind cuts (which I can see would admit an ordering of real numbersin a simple formulation) would be more practical and cauchy sequences> If real numbers are thought of as equivalence classes of Cauchy> sequences, then what would be a formulation to say one real number is> bigger than another? May we say something about the limits of the> Cauchy sequences, or what?> > Is this a practical way of defining real numbers, or something like> Dedekind cuts (which I can see would admit an ordering of real numbers> in a simple formulation) would be definition of r > s for the Cauchy constructed reals is:If f(n) is a Cauchy sequence of rationals representing (converging to) real number r, and g(n) is a Cauchy sequence of rationals representing real number s, then r > s if and only if there is some positive rational e such that f(n) > g(n) + e for almost all n (all but finitely many n). It may be shown, at some length, that this definition is independent of which representations are used, and that if neither r > s nor s > r, the difference, f(n) - g(n), must converge to zero, so the and cauchy sequences>If real numbers are thought of as equivalence classes of Cauchy>sequences, then what would be a formulation to say one real number is>bigger than another? May we say something about the limits of the>Cauchy sequences, or what?>>Is this a practical way of defining real numbers, or something like>Dedekind cuts (which I can see would admit an ordering of real numbers>in a simple formulation) would be Cauchy sequence of rationals is positive iff there exists a positive rational q such that eventually the terms of the sequence exceed q. Show that if two Cauchy sequences are equivalent and one is postive, then so is the other. (In fact, you can use the same q). Then a real number (class of Cauchy sequences) is positve iff each of its members (Cauchy sequences) are positive. a < b iff b-a is positive.In general, whether you use Cauchy sequences or Dedekind cuts is really immaterial, or perhaps a matter of taste. Personally, I prefer Cauchy sequences because I find the definition of multiplication with Dedekind cuts rather messy.-- Stephen J. Herschkorn cauchy sequences[...]> In general, whether you use Cauchy sequences or Dedekind cuts is really> immaterial, or perhaps a matter of taste. Personally, I prefer Cauchy> sequences because I find the definition of multiplication with Dedekind> cuts rather messy.But its not messy if you develop R+ from Q+ by Dedekind cuts, and thendevelop R from R+. Of course, many people prefer to develop R from Q.For them, I would suggest the trivial idea which I presented in sci.mathin Sept. 1998: Multiplying Dedekind cuts naturally If you look at it, Id be happy to hear any comments you might have. Forme at least, its the nicest way to multiply signed Dedekind cauchy sequences> Multiplying Dedekind cuts naturally> How does this compare to the method suggested by Conway in On sequences>> Multiplying Dedekind cuts naturally> How does this compare to the method suggested by Conway in On Numbers> and Games ?Looking at his definition of multiplication (p. 5 in the second ed.),I see no obvious relation between his method and mine. Of course, theremight be some nice relation which Re: Real numbers and cauchy thought of as equivalence classes of Cauchy> sequences, then what would be a formulation to say one real number is> bigger than another? May we say something about the limits of the> Cauchy sequences, or what?> > Is this a practical way of defining real numbers, or something like> Dedekind cuts (which I can see would admit an ordering of real numbers> in a simple formulation) would be more practical for these things?Equivalence classes of Cauchy sequences of RATIONAL numbers, presumably?Two equivalence classes are equal, [{x_n}] = [{y_n}], if and only i§im_n |x_n - y_n| = 0 --- by definition. (And the definition of limitis subtly different: for every RATIONAL eps > 0, there exists N suchthat n >= N implies |x_n - y_n| < eps.)If the two classes ARENT equal, then some subsequence of |x_n - y_n|stays bounded away from 0. Thus there exists a rational eps > 0 suchthat either x_n - y_n >= eps for infinitely many n, or y_n - x_n >= eps for infinitely many n. Using the fact that these are Cauchy sequences, therefore either x_n - y_n >= eps/2 for all sufficiently large n, or y_n - x_n >= eps/2 for all sufficiently large n. The dichotomy allows us to define [{x_n}] > [{y_n}] in the first response, i take the liberty to repostsomeone elses question as a different subject. Does anyonerecognize the calculation where the learner can clearly see why the> method is working. For example, the basic square root finding> method 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> --->> ArjoeDoes anyone recognize the method above to find square roots?[The method i know would look something like:225 -> (1)1100 [20] (5)125 0The figures (1) and (5) being the chance of response, i take the liberty to repost> someone elses question as a different subject. Does anyone> recognize such arithmetic methods where the learner can clearly see why the> method is working. For example, the basic square root finding> method 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> > Does anyone recognize the method above to find square roots?> [The method i know would look something like:> > 225 -> (1)> 1> 100 [20] (5)> 125> 0> > The figures (1) and (5) being the answer.> ]> > Many this:Compute: _____ 231.4Group the digits in pairs to the left and right of the decimal: __________ 2 31 . 40Find largest digit whose square doesnt exceed 1st group (1) and writeit above the 1st group: 1 __________ 2 31 . 40Subtract the square of the current root from 1st group and drop thenext group. 1 __________ 2 31 . 40 1 __ 1 31Multiply current root by 20, add the largest integer n such that(20*root + n)*n <= previous difference. This digit n is the next digitof the root, place it above next group. 1 5 . __________ 2 31 . 40 1 __20+5 1 31 *5 1 25Subtract previous product, drop the next group and proceed as in theprevious step. 1 5 . 2 __________ 2 31 . 40 1 __20+5 1 31 *5 1 25 ______20*15+2 6 40 *2 6 04Continue ... (The decimal in root is above decimal in radicand. But,ignore it while calculating and treat the root as an integer)Consider: Let r,n be decimal digits (This is the same n as above)Then, (10r+n)^2 = 100r^2 + 20rn + n^2 = 100r^2 + n(20r + n) <----- take the liberty to repost>>someone elses question as a different subject. Does anyone>>recognize the calculation methods where the learner can clearly see why the>method is working. For example, the basic square root finding>method 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>>Does anyone recognize the method above to find square roots?>>[The method i know would look something like:>>225 -> (1)>>1>>100 [20] (5)>>125>> 0>>The figures (1) and (5) being the answer.>>]>>Many same as this:> > Compute:> _____> 231.4> > Group the digits in pairs to the left and right of the decimal:> __________> 2 31 . 40> > Find largest digit whose square doesnt exceed 1st group (1) and write> it above the 1st group:> > 1> __________> 2 31 . 40> > Subtract the square of the current root from 1st group and drop the> next group.> > 1> __________> 2 31 . 40> 1> __> 1 31> > Multiply current root by 20, add the largest integer n such that> (20*root + n)*n <= previous difference. This digit n is the next digit> of the root, place it above next group. 1 5 .> __________> 2 31 . 40> 1> __> 20+5 1 31> *5 1 25> > Subtract previous product, drop the next group and proceed as in the> previous step.> > 1 5 . 2> __________> 2 31 . 40> 1> __> 20+5 1 31> *5 1 25> ______> 20*15+2 6 40> *2 6 04> > Continue ... (The decimal in root is above decimal in radicand. But,> ignore it while calculating and treat the root as an integer)> > Consider: Let r,n be decimal digits (This is the same n as above)> > Then, (10r+n)^2 = 100r^2 + 20rn + n^2> = 100r^2 + n(20r + n) <----- compare with above> > Hope this you. The explanation makes sense.>> ArjoeYes, but i dont recognize it in your calculation.Can you explain your 225 | 15 <----> 1> --> 25|125> 125> --->>Arjoe> I believe it is the same as this:>> Compute:> _____> 231.4>> Group the digits in pairs to the left and right of the decimal:> __________> 2 31 . 40>> Find largest digit whose square doesnt exceed 1st group (1) and write> it above the 1st group:>> 1> __________> 2 31 . 40>> Subtract the square of the current root from 1st group and drop the> next group.>> 1> __________> 2 31 . 40> 1> __> 1 31>> Multiply current root by 20, add the largest integer n such that> (20*root + n)*n <= previous difference. This digit n is the next digit> of the root, place it above next group.>> 1 5 .> __________> 2 31 . 40> 1> __> 20+5 1 31> *5 1 25>> Subtract previous product, drop the next group and proceed as in the> previous step.>> 1 5 . 2> __________> 2 31 . 40> 1> __> 20+5 1 31> *5 1 25> ______> 20*15+2 6 40> *2 6 04>> Continue ... (The decimal in root is above decimal in radicand. But,> ignore it while calculating and treat the root as an integer)>> Consider: Let r,n be decimal digits (This is the same n as above)>> Then, (10r+n)^2 = 100r^2 + 20rn + n^2> = 100r^2 + n(20r + n) <----- prime, so p(1)=2, p(2)=3, p(3)=5, etc. It is well-known that p(n)=O(n log n): that follows from Gauss prime number theorem. But the proof of that is difficult. Now, I only need to prove that p(n) grows at most polynomially, but the proof has to be very simple (say, not much more than one line). Any polynomial distribution of primes?> Now, I only need to prove> that p(n) grows at most polynomially, but the proof has to be very> simple (say, not much more than one line). Any ideas?You can use the fact that sum (1/p, p prime) diverges, which follows easilyfrom sum (1/n, n integer) diverges.-- primes?X-DMCA-Notifications: http://www.giganews.com/info/dmca.html>> Now, I only need to prove>> that p(n) grows at most polynomially, but the proof has to be very>> simple (say, not much more than one line). Any ideas?>You can use the fact that sum (1/p, p prime) diverges, which follows easily>from sum (1/n, n integer) diverges.There are plenty of sequences a(n) which are not of polynomial growth,for which the sum of 1/a(n) diverges.>-- polynomial distribution of primes?> There are plenty of sequences a(n) which are not of polynomial growth,> for which the sum of 1/a(n) diverges.Yes, but this one is monotonic.-- primes?X-DMCA-Notifications: http://www.giganews.com/info/dmca.html>> There are plenty of sequences a(n) which are not of polynomial growth,>> for which the sum of 1/a(n) diverges.>>Yes, but this one is monotonic.There are plenty of monotonic sequences a(n) which are not ofpolynomial growth, but for which the sum of 1/a(n) diverges.I cant give a simple formula for such a sequence, but I canexplain how to show one exists. This may take a few lines toexplain...Theres going to be an increasing sequence of positiveintegers N_j. If (1) a(N_j) = A_j then were going to have(2) a(N_j + k) = A_j + k (for all k with N_j + k < N_{j+1})and also(3) N_{j+1} = N_j + A_jand(4) A_j > 2^(N_j)and(5) A_{j+1} > 2*A_j.Supposing we can do all that then note that (2), (3) and(5) show that a(n) is monotonic, while (3) and (5) show thatthe sum of 1/a(n) diverges and (4) shows a(n) does nothave polynomial growth.Start with N_1 = 1, A_1 = 1. Now (3) determines N_2.Choose A_2 so (4) is satisfied (with j = 2) and (5)is satisfied (with j=1). Now A_2 determines N_3, by (3).Choose A_3 so (4) and (5) are still satisfied. Etc.QED.That was off the top of my head - there may be a fewsubscripts in the wrong place, whatever, but itsextremely clear to me that a simple construction likethat choose a bunch of consecutive integers to make thesum of 1/a(n) large, then choose another bunch ofconsecutive integers far enough out to contradictpolynomial growth, and long enough to make sum 1/a(n)large works. If you dont follow the constructionyou might try posting a proof that if sum 1/a(n)diverges and a(n) is monotonic then a(n) haspolynomial growth...>-- >Maxi>David === C. UllrichSubject: Re: Simple proof of polynomial distribution that there is a prime between n and n+sqrt(n), for every big enough n. (Or weaker: between n and n+n^(1-epsilon) for some Re: is tan(n)/n unbounded?Robert Israel> Robin Chapman> 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?>The poiunt is that tan(n) is big in absolute value iff n is close to>an odd multiple of pi/2, that is if pi/2 is approximately n/m>where 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 be>bounded away from zero. Its not obvious that there would be infintely>many cases with m odd, and to get a better result we need a better>Diophantine approximation result on pi/2. I dont know if there>is such a result.> Yes there is. See my posting on the Subject A limit problem with> explanation in sci.math.symbolic on 17 October 2001.Im not at all sure that the following will help, but ifn cos ngoes to zero on a subsequence, then(tan n)/n = (sin n)/(n cos n)gets arbitrarily big on those n. For if n cos n goes to 0 (on asubsequence), then so does cos n, and therefore sin n is bounded away fromzero.IMO at least, the problem looks tidier in terms of cos n than tan n, and itmight simplify the number-crunching experiments.Another question is what other cluster points (than 0) the sequence n cos nmight have. is what other cluster points (than 0) the sequence n cos n>might have. None?For n cos(n) ~= y when n is large we need n/m - pi/2 ~= (+/-) y/(nm) ~= (+/-) (2/pi) y/m^2 with odd m (the sign depending on m mod 4).If |y| < pi/4, this implies that n/m is a convergent of the continued fraction of pi/2. Now if a_n are the elements and p_n/q_n the convergents of this continued fraction, we have 1/(a_{n+1}+2) < q_n^2 |p_n/q_n - pi/2| < 1/a_{n+1}Thus if k is any positive integer that occurs infinitely many times as an element of the continued fraction of pi/2, there will be infinitely many n for which pi/(2k+4) < n |cos(n)| < pi/(2k).Of course we dont know any k for which its true, but its almostcertainly true for all positive integers k. And unless lim_{n -> infinity} a_n = infinity, it will be true for some k, sothere will be a finite cluster point.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of is tan(n)/n unbounded?> 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> minima in 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]}]]][...]> Is it proven that the extrema have to be on CF convergents numerators?> I know thts the best place to look for them, but there are a lot> of numbers between 65666... and 23724... which could be darned> close to k*Pi/2, but arent on the CF.I think that Erick has already answered that question nicely. Furthermore,toward the end of his response, he shows why, in describing my littleprogram above, I merely called it _fairly_ efficient. The efficiencycertainly could be improved a good bit.> Anyway, I ported the above to GP and ran it a bit further.> Perhaps youd like to independently verify these, and forward them to> Eric?Forward them yourself if you wish, but remember that Eric chose not to listthe larger numbers which I had already calculated. I doubt that he would beinterested in listing still larger ones.BTW, Phil, Im not really interested in number crunching per se.Of course, in cases when I think that number crunching might lead is tan(n)/n unbounded?>by a random variable or not. If so, whats the distribution, and what >kind of growth of the maxima can be expected? A parallel question is >what happens if we strictly limit ourselves to the contfrac convergents?It is well-known that if x is irrational and | x - p/q | < 1/2q^2, thenp/q is necessarily a continued fraction convergent for x. So it shouldbe impossible to find any record values of | tan(n)/n | where n does notcome from a convergent.>Ive pushed the extrema but due to the increased precision required >Im suffering from a bit of a slow-down.One thing that might help find locate values at the cost of some lossof precision is that you can predict the approximate value of tan(n)/njust by looking at the size of the subsequent continued fraction term.There exists a constant c such that if the corresponding numerator isodd (or is it denominator?), then |tan(n)/n| is approximated by c timesthe next c.f. term (provided said term is sufficiently large). The signof tan(n)/n naturally alternates with each term, and so the only checkyou need to perform is keeping track of parity to make sure the numerator(denominator?) is odd.Empirically from your table it looks like c is about 0.636, and Im toolazy to work out the exact value. If youre willing to sacrifice precisionyou dont need to do any high-precision arithmetic beyond the initialcontinued fraction; otherwise at the very least you should be able tovastly cut down on the number of tan() present using a kalman filter to predict the future motion of a ball> that has been thrown. I have at present implemented a simple model of the> situation with gravity and air resistance. However, I have only modelled air> resistance as directly proportional to speed....> I cant help with exactly what you ask, but are you aware of the exact analytic solution in this case? See for example D.E. Rutherford, Classical Mechanics, section 37 (pp.63-65 in the 2nd edition). He discusses resistance proportional to (speed)^n, then solves Ball Trajectory QuestionHi All,I am present using a kalman filter to predict the future motion of a ballthat has been thrown. I have at present implemented a simple model of thesituation with gravity and air resistance. However, I have only modelled airresistance as directly proportional to speed.1. I am tracking the ball (a cricket balls motion) over a distance of about2.5m and want to predict a further 3-4m into the future. Will modelling theair resistance as proportional to speed squared make much difference to theaccuracy of my prediction?2. If I do go for a model with air resistance as proportional to speedsquared, any suggestions how I can fit it to the kalman setup i.e.X(t+1) = A.X(t)where X(t+1) and X(t) are column vectors of x,x speed, y, y speed. and A isthe relating matrix between the current state and the state at t+1 of theaformentioned variables. Maybe I need to have a different things in mycolumn vectors? But they are the things I can calculate.Note:- I have tracked the third dimension, but I am just trying to get theidea of Geometric interpretation of Taylors seriesHi all,Is there any way to get a geometric idea of what is going on in the Geometric interpretation of Taylors seriesTry thisU have f(x) = Sum(1 Hi all,>> Is there any way to get a geometric idea of what is going on in the Taylor> series of a was a 2, I am sorry to havemistyped it. I will certainly have a more closer look at your postthis week-end, I just have too much work now.Anyway, thank you. I was really blocked by this === proof...> Subject: Big divisibility problem> > >prove that it can never be an integer...> >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 0...> What happened at 24 * ^(2p ?> > Assume monster is integer k. Then> 24 [ 24 * ^(2p+n) - 3^(n+1) (6 * 3^p + 4^p) ]> = k [27 * 3^(p+n) - 16 * 2^(2p+n)]> > 24 [ 24 * ^(2p+n) - 3^(p+n+3) - 3^(n+1) 2^2p ]> = k [3^(p+n+3) - 2^(2p+n+4)]> > as 24, 3^(p+n+3) - 2^(2p+n+4) coprime, 24 | k, so change k so that> > 24 * ^(2p+n) - 3^(p+n+3) - 3^(n+1) 2^2p> = k [3^(p+n+3) - 2^(2p+n+4)]> > Guessing the correction is> 24 * 2^(2p+n) - 3^(p+n+3) - 3^(n+1) 2^2p> = k [3^(p+n+3) - 2^(2p+n+4)]> > 3 * 2^(2p+n+3) - 3^(p+n+3) - 3^(n+1) 2^2p> = k [3^(p+n+3) - 2^(2p+n+4)]> > 3 [ 2^(2p+n+3) - 3^(p+n+2) - 3^n 2^2p]> = k [3^(p+n+3) - 2^(2p+n+4)]> > Again 3, 3^(p+n+3) - 2^(2p+n+4) coprime, 3 | k, so change k so that> > 2^(2p+n+3) - 3^(p+n+2) - 3^n 2^2p> = k [3^(p+n+3) - 2^(2p+n+4)]> > 2^(2p+n+3) = -k 2^(2p+n+4) (mod 3)> 1 = -2k; k = 1 (mod 3)> > -3^(p+n+2) = k 3^(p+n+3) (mod 2)> 1 = -1 = 3k = k (mod 2)> > k = 1 (mod 6)> > 2^(2p+n+3) - 3^(p+n+2) = 3^(p+n+3) - 2^(2p+n+4) (mod 6)> > 2^(2p+n+3) * (1 + 2) = 3^(p+n+2) * (3 + 1) (mod 6)> 3 * 2^(2p+n+3) = 4 * 3^(p+n+2) (mod 6)> 2^(2p+n+1) = 3^(p+n+1) (mod 6)> > 2^(2p+n+1) = 3^(p+n+1) + 6k for some new k> but this cannot be === for then 2 | 3 and 3 | 2> > ----Subject: Re: Big divisibility never be an integer... >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 0...What happened at 24 * ^(2p ?Assume monster is integer k. Then24 [ 24 * ^(2p+n) - 3^(n+1) (6 * 3^p + 4^p) ] = k [27 * 3^(p+n) - 16 * 2^(2p+n)]24 [ 24 * ^(2p+n) - 3^(p+n+3) - 3^(n+1) 2^2p ] = k [3^(p+n+3) - 2^(2p+n+4)]as 24, 3^(p+n+3) - 2^(2p+n+4) coprime, 24 | k, so change k so that24 * ^(2p+n) - 3^(p+n+3) - 3^(n+1) 2^2p = k [3^(p+n+3) - 2^(2p+n+4)]Guessing the correction is24 * 2^(2p+n) - 3^(p+n+3) - 3^(n+1) 2^2p = k [3^(p+n+3) - 2^(2p+n+4)]3 * 2^(2p+n+3) - 3^(p+n+3) - 3^(n+1) 2^2p = k [3^(p+n+3) - 2^(2p+n+4)]3 [ 2^(2p+n+3) - 3^(p+n+2) - 3^n 2^2p] = k [3^(p+n+3) - 2^(2p+n+4)]Again 3, 3^(p+n+3) - 2^(2p+n+4) coprime, 3 | k, so change k so that2^(2p+n+3) - 3^(p+n+2) - 3^n 2^2p = k [3^(p+n+3) - 2^(2p+n+4)]2^(2p+n+3) = -k 2^(2p+n+4) (mod 3)1 = -2k; k = 1 (mod 3)-3^(p+n+2) = k 3^(p+n+3) (mod 2)1 = -1 = 3k = k (mod 2)k = 1 (mod 6)2^(2p+n+3) - 3^(p+n+2) = 3^(p+n+3) - 2^(2p+n+4) (mod 6)2^(2p+n+3) * (1 + 2) = 3^(p+n+2) * (3 + 1) (mod 6)3 * 2^(2p+n+3) = 4 * 3^(p+n+2) (mod 6)2^(2p+n+1) = 3^(p+n+1) (mod 6)2^(2p+n+1) = 3^(p+n+1) + 6k for some new kbut this cannot be censorship, counting primes> A sociologist of science with whom I am cordial has taken up the topic> of who does what to whom and why, in NGs such as sci.math, sci.logic,> and sci.physics. This is an area that has not been explored in the> detail that it deserves to be, perhaps because even sociologists are> not entirely comfortable with what these NGs show about those, many of> them professors, who in§ict as much harm as possible on> seekers-after-knowledge such as yourself in these groups.Where JSHs interaction with this newsgroup is concerned, sociology is definitely more relevant than mathematics. But I think psychology and psychiatry would be even more apposite. JSH would undoubtedly make an interesting psychiatric case study, and the psychology of mathematicians who cannot resist engaging with him, despite years of evidence that he §outs (!) all the rules of mathematical argumentation, and has only a rudimentary level of mathematical Re: De facto censorship, counting primes> >[...]> > What Jesse said:> >A sociologist of science with whom I am cordial has taken up the topic>of who does what to whom and why, in NGs such as sci.math, sci.logic,>and sci.physics. > > Thats interesting.> > This is an area that has not been explored in the>detail that it deserves to be, perhaps because even sociologists are>not entirely comfortable with what these NGs show about those, many of>them professors, who in§ict as much harm as possible on>seekers-after-knowledge such as yourself in these groups.> > But suggesting that Harris is a seeker-after-knowledge is hilarious.> He has repeatedly said hes not interested in learning any math.> Hes the only person Ive ever seen state on sci.math that if what> he just said was wrong we shouldnt bother saying so because he> didnt want to know. Seeker after knowledge? Right.If you were a purveyor-of-knowledge rather than a frontmanfor logical orthodoxy--you would never have left uncorrectedthe blunders of yours that follow my sig.> People in power (or in positions of authority) do not like detractors.> They generally are not so bright either. That is, they> crave power in the first place because they are at least smart enough> to know that they cannot rely upon their sparse, actual abilities.> Did people understand Abels proof but sat quiet? Did they see> Galoiss ideas and decided to not have to address them?>> I would not be a bit surprised. People often do not like recognizing> others ... especially when the others are alive and can profit from> the recognition 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 toitself.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}] 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 the Last Digit is 5I posted a message about this some weeks ago, and its gotten lost in all the msgs since posted. I vaguely recalled some quick method for multiplying numbers like 35*45. This weekend, I was in a book store and happend upon Julius book on Rapid Math Tricks. I think I found what I distantly recalled. The method was for rapidly squaring two digit numbers with a 5 as the last digit. 35*35=1225. The trick is to take the first digit and add one to it, then multiply it by itself, 3*(3+1), then prepend this to 25, 12 .. 25 give 1225. That seems like what I was trying to recall-- Wayne T. Watson (121.015 Deg. W, 39.262 Deg. N, 2,701 feet, Nevada City, CA) If Im given six hours to cut down a tree, then I spend four hours sharpening my axe. -- Abraham Lincoln Web Page: sierra_mtnview -at- earthlink -dot- net Imaginarium Museum: Reality of response to my work> I have to mention here that I was giving you advice. So it was really> meta-advice. And how being *free* applies Im not sure, since I wouldnt> charge you for an admonition, as it would be unlikely that anyone would pay.> The tone of your reply threw me, I must admit.> > According to my newsreader I didnt top post. But Im using a microsoft> product and as such it might not be working as intended. If I am top> posting will someone email me and let method to reliably count primes may be> > substantial, perhaps many millions of $/yr. Such a method would> > enable someone to decide the probability of whether he he has tried> > all possible prime factors of a number for code breaking. Such a method combined with others could be worth a great deal of money.> >> > This doesnt really make sense to me. If you can generate all the > > possible> > prime factors, then you know how many there are. If you cant generate> > them, then certainly you cant use the fact that there are more factors to> > somehow generate more factors. Seriously, if you know that there must > > be at> > least 10^20 more primes to try, that still leaves you to find all these> > primes. If you know that there arent any more primes to try, then you > > know> > how many youve tried. (Im not an expert on this topic, so I may be > > way> > off.)>> If you know that there are 10^20 primes to try it will take you 10^11> seconds to try them all when you could try each prime in a nanosecond.> That is something like 3000 years. That would even get you moderate> 40 digit numbers out of reach.>> > There are very fast methods > > to> > count how many primes are less than a given number. There are VERY fast> > methods to determine the primality of even extremely large numbers, with> > some small chance of error.>> There are even VERY fast methods to determine the primality of even> extremely large numbers without a chance of error.> > > Just to clarify, as I understand it... knowing a number is composite does not> tell you the factors, to be commercially useful you have to find the factors, > otherwise> the ïdiscovery is mostly academic.But knowing that a large number does not have factors (is prime) may be quite valuable commercially.> > The main techinique is finding witnesses to compositeness. Over half of all > numbers> less than any composite are ïwitnesses to that composite, so to determine a > number has> factors just requires more and more random attempts until a witness is found, > unfortanetly> this only establishes a confidence of primality, you seem to be suggesting a > certain> technique.> > HercIf I understand things correctly, there are ways of showing that some large numbers are prime which do not require anything as slow as trying all primes up to its square root as potential factors.Search for witness at arctan(x/2) trig id question>Ive searched for simplifications to arctan(x/2) to no avail. Does>anyone know of one, or know that it does not exist?Well, arctan(x/2) = arctan(x) + arctan(x+2/x) - signum(x) pi/2, but Id hardly call that a simplification.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of arctan(x/2) trig id questionI may have misunderstood what you meant the first time. You are then askingfor a half angle formula identity for tangent or arctangent? One of the morehighly studied people may have information on it.G calipa@unity.ncsu.edu:>Ive searched for simplifications to arctan(x/2) to no avail. Does>anyone know of one, or know that it does not exist?>You must be missing some of the information. The tangent of some angle or arcis x/2; the value of this arc or angle is expressed if desired as ïthe arcwhich has the tangent of x/2 ï. Did you want this arctan(x/2) to be equatedto some expression or value, so as to be able to id question> Ive searched for simplifications to arctan(x/2) to no avail. Does> anyone know of one, or know that it does not exist?So, if y=arctan(x/2), then x = 2 tan y. So you want to write2 tan y = tan (??). I think there is no such nice GDDtIEZeoeN6aycevScZqs5gdLow9jWcayWS3WgxA4Zt5AJ2QbIg8oLet M be an n-dimensional C^infty-manifold.Does the set C^infty(R,M) have a canonical manifold-structure? In theinfinite-dimensional case, charts are homeomorphisms into some Banach spaceE, in this case I think we could take E=C^infty(R,R^n) (but already thefirst problem is: what is the correct norm?).Probably the answer is no. Is there at least a topology on C^infty(R,M)?Is the situation any different if we only take a (compact?) intervalC^infty(I,R) instead of C^infty(M,R)?-- Phyics is much too hard for physicists.reverse my forename for n-dimensional C^infty-manifold.>>Does the set C^infty(R,M) have a canonical manifold-structure? In the>infinite-dimensional case, charts are homeomorphisms into some Banach space>E, in this case I think we could take E=C^infty(R,R^n) (but already the>first problem is: what is the correct norm?).>Probably the answer is no. Is there at least a topology on C^infty(R,M)?>>Is the situation any different if we only take a (compact?) interval>C^infty(I,R) instead of C^infty(M,R)?I havent looked at this book in about 2 years, but I thinkhttp://www.ams.org/online_bks/surv53/ (available for free download) may have something relevant somewhere; its binomial coefficient. How can I computeC(n,k) mod 2? If its any easier, what Im really interested in isC(n-k,k) mod Lot-o-fun >Let C(n,k) stand for the binomial coefficient. How can I compute>C(n,k) mod 2? If its any easier, what Im really interested in is>C(n-k,k) mod 2.In http://www.whim.org/nebula/math/multinomialfactors.html it is shownthat for any prime p, that the number of factors of p that divide thebinomial coefficient C(n,k) is the number of carries generated when kis added to n-k in base p. Thus, the only way for C(n,k) to be 1 mod 2is that there be no carries generated by adding k to n-k in base 2.That is, the binary representations of k and n-k have no bits in common.The way to check this is to test k AND (n-k) == 0 using the bitwise ANDoperator.That is, C(n,k) = 1 mod 2 if and only if k AND (n-k) == 0; otherwise,C(n,k) = 0 mod 2.Examples:C(5,4): 4 AND 1 = 0, so C(5,4) = 1C(5,3): 3 AND 2 = 2, so C(5,3) = 0C(52,5): 5 AND 47 = 5, so C(52,5) = 0C(100,36): 36 AND 64 = 0 so C(100,36) = 1Rob Johnsontake out the trash before the binomial coefficient. How can I compute>C(n,k) mod 2? If its any easier, what Im really interested in is>C(n-k,k) mod 2.Write n and k in binary (or equivalently, as a sum of distinct powers of2). If the coefficient of any power of 2 is 0 in n and 1 in k, thenC(n,k) is even, otherwise C(n,k) is odd. So for C(9,2), 9 is 1001 inbinary and 2 is 0010. Since the second last digit of the binary expansionof 9 is 0 and the second last digit of the binary expansion for 2 is 1,then C(9,2) is even. For C(11,3), 11 is 1011 in binary and 3 is 0011. Since there is no power of 2 which has coefficient 0 in 11 and coefficient1 in 3, then C(11,3) is Math factorization, other sideNow Ive given enough time for quite a few of you to think about thefactorization I talked so much about, and given you other informationto allow you to form your opinions based on the data available to you.My assumption is that many of you have some internal opinion aboutyour own thinking ability: both your basic intelligence level, yourability to follow a rational argument, and your trust level forexperts.Now lets go from the other side by considering, yet again, afactorization:(5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where f_1(0) = f_2(0) = f_3(0) = 0.For some of you a quick way to serious mental confusion andconsternation is to simply try and answer the question of whether ornot f_1(x), f_2(x), and f_3(x) even exist.Ill leave those people stuck on that question, stuck, and moveforward.It turns out that what Im showing is a basic use of asymmetricalfactors of a polynomial. For that reason, its not possible toexplicitly display the functions.Many of you when faced with that impossibility simply disbelieved it,and so were easy prey for posters who simply either were incompetentor dedicated in just disagreeing with me, I guess to be disagreeable.But make no mistake, its outside of the ability of human beings atall, to give the fs explicitly, which is basically saying that thepolynomial is a non-monic primitive irreducible over Q.I want to see how many of you can think at a certain level. Or,unfortunately, given the situation, if ANY of you have the ability tothink at a certain level.If it helps, consider it a puzzle. James HarrisMy math discoveries, found for factorization, other sideNntp-Posting-Host: hera.cwi.nl > Now Ive given enough time for quite a few of you to think about the > factorization I talked so much about, and given you other information > to allow you to form your opinions based on the data available to you. > > My assumption is that many of you have some internal opinion about > your own thinking ability: both your basic intelligence level, your > ability to follow a rational argument, and your trust level for > experts. > > Now lets go from the other side by considering, yet again, a > factorization: > > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > where f_1(0) = f_2(0) = f_3(0) = 0. > > For some of you a quick way to serious mental confusion and > consternation is to simply try and answer the question of whether or > not f_1(x), f_2(x), and f_3(x) even exist.Yes, they exist. They are functions from the integers (for instance)to the algebraic numbers. (Note, they are *not* functions from theintegers to the algebraic integers.) Moreover, many such functionsexist and will fit the bill.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, factorization, other side> > Now Ive given enough time for quite a few of you to think about the> > factorization I talked so much about, and given you other information> > to allow you to form your opinions based on the data available to you.> > > My assumption is that many of you have some internal opinion about> > your own thinking ability: both your basic intelligence level, your> > ability to follow a rational argument, and your trust level for> > experts.> > > Now lets go from the other side by considering, yet again, a> > factorization:> > > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => > > 300125 x^3 - 18375 x^2 - 360 x + 22> > > where f_1(0) = f_2(0) = f_3(0) = 0.> > > For some of you a quick way to serious mental confusion and> > consternation is to simply try and answer the question of whether or> > not f_1(x), f_2(x), and f_3(x) even exist.> > Yes, they exist. They are functions from the integers (for instance)> to the algebraic numbers. (Note, they are *not* functions from the> integers to the algebraic integers.) Moreover, many such functions> exist and will fit the bill.>In particular, its not hard to show that f_1(1), f_2(1), and f_3(1)will not be algebraic other side> > Now Ive given enough time for quite a few of you to think about the> > factorization I talked so much about, and given you other information> > to allow you to form your opinions based on the data available to you.> > > > My assumption is that many of you have some internal opinion about> > your own thinking ability: both your basic intelligence level, your> > ability to follow a rational argument, and your trust level for> > experts.> > > > Now lets go from the other side by considering, yet again, a> > factorization:> > > > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => > > > 300125 x^3 - 18375 x^2 - 360 x + 22> > > > where f_1(0) = f_2(0) = f_3(0) = 0.> > > > For some of you a quick way to serious mental confusion and> > consternation is to simply try and answer the question of whether or> > not f_1(x), f_2(x), and f_3(x) even exist.> > Yes, they exist. They are functions from the integers (for instance)> to the algebraic numbers. (Note, they are *not* functions from the> integers to the algebraic integers.) Moreover, many such functions> exist and will fit the bill.There are actually an *infinite* number of functions that will work,which is a subtle point you missed in a reply on sci.math in adifferent thread--an indication of lack of competence.Now then, besides your say-so, what *mathematical reason* can you givefor them not being algebraic integer functions?Others should remember that my point is that Dik Winter isntmathematically competent enough to discuss these advanced topics as anexpert.He is merely a tag-along who hardly knows whats going on here, but Math factorization, other side>Now Ive given enough time for quite a few of you to think about the>factorization I talked so much about, and given you other information>to allow you to form your opinions based on the data available to you.>>My assumption is that many of you have some internal opinion about>your own thinking ability: both your basic intelligence level, your>ability to follow a rational argument, and your trust level for>experts.>>Now lets go from the other side by considering, yet again, a>factorization:>>(5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) =>> 300125 x^3 - 18375 x^2 - 360 x + 22>>where f_1(0) = f_2(0) = f_3(0) = 0.>>For some of you a quick way to serious mental confusion and>consternation is to simply try and answer the question of whether or>not f_1(x), f_2(x), and f_3(x) even exist.> Of course f_1(x), f_2(x), and f_3(x) exist. That is, given x, there are algebraic *numbers* f_1(x), f_2(x), and f_3(x)which satisfy your equation above. This follows from thefundamental theorem of algebra.>Ill leave those people stuck on that question, stuck, and move>forward.>>It turns out that what Im showing is a basic use of asymmetrical>factors of a polynomial. For that reason, its not possible to>explicitly display the functions.>>Many of you when faced with that impossibility simply disbelieved it,>and so were easy prey for posters who simply either were incompetent>or dedicated in just disagreeing with me, I guess to be disagreeable.>>But make no mistake, its outside of the ability of human beings at>all, to give the fs explicitly, which is basically saying that the>polynomial is a non-monic primitive irreducible over Q.> Well, no, not at all. You know that, for example, -1/f_1(x) isa root of the polynomial in a, 7*a^3 + 3*(-1 + 49*x)*a^2 - (2401*x^3 - 147*x^2 + 3*x)a third-degree polynomial with integer coefficients (for x aninteger), and there are formulas for the roots of third-degreepolynomials in terms of the coefficients. The formulas arevery messy to write down and in themselves they dont tell youmuch about what kind of numbers f_1(x), etc. actually are, butthey CAN be written down explicitly. Abel and Galois proved that for 5th degree or higher polynomials they cannot be written down, but here in the present post you are talking about a cubic. Irreducible does not mean you cannot give the roots explicitly; for example, x^2 + 3*x + 1 is irreducible (over the rationals), but it is easyto give the roots. >I want to see how many of you can think at a certain level. Or,>unfortunately, given the situation, if ANY of you have the ability to>think at a certain level.>>If it helps, consider it a puzzle. > This is such a bizarre post. Other than your silly false statement(its outside the ability of human beings, etc), which is equivalent to saying the roots of a cubic cannot be written down, you are not claiming anything at all !!! You are not saying, as you have in the past, that f_1(x) is an algebraic integer, or an object, or anything else. What are your loyal fans supposed to think? That the evil mathematician conspiracy has intimidated you to the point that you dare not make any assertion with any substance ? How can you say as you do above that people are disagreeing with you, when you are not claiming that anything is true or false??? Nora B. >>James Harris>>My math discoveries, found for profit>http:// side by considering, yet again, a> factorization:>> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) =>> 300125 x^3 - 18375 x^2 - 360 x + 22>> where f_1(0) = f_2(0) = f_3(0) = 0.>> For some of you a quick way to serious mental confusion and> consternation is to simply try and answer the question of whether or> not f_1(x), f_2(x), and f_3(x) even exist.Here are values for f_1(x), f_2(x) and f_3(x) which satisfy your equation:f_1 = -5.787021874819612193 xf_2 = -3.263829626960578769 xf_3 = +127.1187330391642012 xHence, they exist.> Ill leave those people stuck on that question, stuck, and move> forward.>> It turns out that what Im showing is a basic use of asymmetrical> factors of a polynomial. For that reason, its not possible to> explicitly display the functions.Excuse me, but you didnt show anything. You just posted an equation whichhas solutions which I provided above. If you want exact values for thef_1(x), f_2(x) and f_3(x), they are:f_1 = (-48 + (186589*2^(1/3))/(186589 + 539*I*Sqrt[62893])^(2/3) + (539*I*2^(1/3)*Sqrt[62893])/(186589 + 539*I*Sqrt[62893])^(2/3) - (2983*2^(2/3))/(186589 + 539*I*Sqrt[62893])^(1/3) - 2*(2*(186589 + 539*I*Sqrt[62893]))^(1/3) - Sqrt[16543717668*2^(2/3) + 201142942*I*2^(2/3)*Sqrt[62893] + 2226379948*(186589 + 539*I*Sqrt[62893])^(1/3) + 6431348*I*Sqrt[62893]*(186589 + 539*I*Sqrt[62893])^(1/3) - 53389734*2^(1/3)*(186589 + 539*I*Sqrt[62893])^(2/3) + 23864*(186589 + 539*I*Sqrt[62893])^(4/3) - 4*2^(2/3)*(186589 + 539*I*Sqrt[62893])^2]/ (186589 + 539*I*Sqrt[62893])^(2/3))/44 * xf_2 = (-4104958*2^(1/3) - 11858*I*2^(1/3)*Sqrt[62893] - 65626*2^(2/3)*(186589 + 539*I*Sqrt[62893])^(1/3) - 1056*(186589 + 539*I*Sqrt[62893])^(2/3) + 22*Sqrt[16543717668*2^(2/3) + 201142942*I*2^(2/3)*Sqrt[62893] + 2226379948*(186589 + 539*I*Sqrt[62893])^(1/3) + 6431348*I*Sqrt[62893]*(186589 + 539*I*Sqrt[62893])^(1/3) - 53389734*2^(1/3)*(186589 + 539*I*Sqrt[62893])^(2/3) + 23864*(186589 + 539*I*Sqrt[62893])^(4/3) - 4*2^(2/3)*(186589 + 539*I*Sqrt[62893])^2])/ (968*(186589 + 539*I*Sqrt[62893])^(2/3)) * xf_3 = -24 + (2983*2^(2/3))/(186589 + 539*I*Sqrt[62893])^(1/3) + (2*(186589 + 539*I*Sqrt[62893]))^(1/3) * x> Many of you when faced with that impossibility simply disbelieved it,> and so were easy prey for posters who simply either were incompetent> or dedicated in just disagreeing with me, I guess to be disagreeable.>> But make no mistake, its outside of the ability of human beings at> all, to give the fs explicitly,Hmmm. I just gave them -- explicitly. The solutions for f_1(x), f_2(x) andf_3(x) are valid for your equation and for all values of ïx.> which is basically saying that the> polynomial is a non-monic primitive irreducible over Q.???> I want to see how many of you can think at a certain level. Or,> unfortunately, given the situation, if ANY of you have the ability to> think at a certain level.Or, unfortunately, given the solutions above, if YOU have the ability tothink at all.> If it helps, consider it a puzzle.It might help you to see a professional therapist.> James Harris--It takes a village to raise an idiot.--Democracy: The triumph of popularity over factorization, other side> > Now Ive given enough time for quite a few of you to think about the> factorization I talked your errors. When you fixthem you dont have anything remaining.Just how inordinately refractorily stooopid are you, Harris? Are yousome kind of Black History advocate mindlessly incessantly spewingdingleberries about Hemet while any North African Arab accused ofbeing Black would rip your lungs out through your ears for theinsult? Are you so stooopid you cannot even do algebra consistently?http://www.crank.net/harris.html Its not every braying jackass that gets a whole page at crank.net-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The given enough time for quite a few of you to think about the> - every poster pointed out your errors. When you fix> them you dont have anything remaining.> > Just how inordinately refractorily stooopid are you, Harris? Are you> some kind of Black History advocate mindlessly incessantly spewing> dingleberries about Hemet while any North African Arab accused of> being Black would rip your lungs out through your ears for the> insult? Are you so stooopid you cannot even do algebra consistently?> > http://www.crank.net/harris.html> Its not every braying jackass that gets a whole page at crank.netWell, not surprisingly, Uncle Al failed the intelligence test.Heres more to help the rest of you along.Notice that you can go from(5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22to(5 g_1 x+ 1)(5 g_2 x + 1)(5 g_3 x + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where the gs *should* be algebraic integers, and, of course,f_1(x) = g_1 x, f_2(x) = g_2 x, and f_3(x) = g_3 x, so theyre justlinear functions!!!If you realized that simple step, then at least you know you may havean intellect at a certain level, if you did not, then you are alreadyout of your league.For those completely lost, but looking for help, notice that 300125 = 2401(125), and of course, 125 is just 5^3.Can you work out whats going on now? Do any of you think you cangive the gs?James HarrisMy math discoveries, found for factorization, other sideNntp-Posting-Host: hera.cwi.nl... > Notice that you can go from > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > > to (5 g_1 x+ 1)(5 g_2 x + 1)(5 g_3 x + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > > where the gs *should* be algebraic integers, and, of course,Again, why *should*?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; other sideIn sci.physics, Dik T. Winter:> ...> Notice that you can go from> (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> > to> > (5 g_1 x+ 1)(5 g_2 x + 1)(5 g_3 x + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> > where the gs *should* be algebraic integers, and, of course,> > Again, why *should*?If we substitute y = 1/x and multiply by y^3, we get(y + 5*g_1)*(y + 5*g_2)*(22*y + 5*g_3) = 22*y^3 - 360*y^2 - 18375*y + 300125If we substitute y = 5z, we get(5*z + 5*g_1)*(5*z + 5*g_2)*(110*z + 5*g_3) = 2750*z^3 - 9000*z^2 - 91875*z + 300125Now divide by 125; we get(z + g_1) * (z + g_2) * (22 * z + g_3) = 22*z^3 - 72*z^2 - 735*z + 2401We now have a minor problem, because of the asymmetry. If we divideagain by 22 we get(z + g_1) * (z + g_2) * (z + g_3/22) = z^3 - 36/11*z^2 - 735/22*z + 2401/22Im not sure what to conclude here but its clear that we cantautomatically state the g_i are algebraic integers, withoutmore manipulations/logic. Were this last equation to have allintegrer coefficients, I could conclude the g_i are in factalgebraic integers (and furthermore that g_3 has a factor of 22),but obviously thats not the case.Followups.-- #191, ewill3@earthlink.netIts still legal to go > Notice that you can go from> > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => > 300125 x^3 - 18375 x^2 - 360 x + 22> > > > to> > > > (5 g_1 x+ 1)(5 g_2 x + 1)(5 g_3 x + 22) => > 300125 x^3 - 18375 x^2 - 360 x + 22> > > > where the gs *should* be algebraic integers, and, of course,> > Again, why *should*?Well, g_3 clearly is an algebraic integer, but wouldnt you argue thatg_1 and g_2 are not?Now then, why would you so argue?For others, yes the intelligence test other side> ...> > Notice that you can go from> > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) => > 300125 x^3 - 18375 x^2 - 360 x + 22> > > > to> > > > (5 g_1 x+ 1)(5 g_2 x + 1)(5 g_3 x + 22) => > 300125 x^3 - 18375 x^2 - 360 x + 22> > > > where the gs *should* be algebraic integers, and, of course,> > Again, why *should*?> > Well, g_3 clearly is an algebraic integer, but wouldnt you argue that> g_1 and g_2 are not?> > Now then, why would you so argue?> > For others, yes the intelligence test continues.> > > James HarrisIn another post I showed that g1, g2 and 22*g3 are the roots of the polynomial equation 22*z^3 + 72*z^2 - 735*x - 2401 = 0. Details supplied on request.algebraic integers, and it easily follows that g3 is not one either.So what JSH seems to think is clearly true, is clearly not, and what JSH thinks should be again turns out not to be.It would seem that JSH did not pass his own side > ... > > Notice that you can go from > > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > > > to > > > > (5 g_1 x+ 1)(5 g_2 x + 1)(5 g_3 x + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > > > where the gs *should* be algebraic integers, and, of course, > > Again, why *should*? > > Well, g_3 clearly is an algebraic integer, but wouldnt you argue that > g_1 and g_2 are not? > > Now then, why would you so argue?Because I see no reason that g1 and g2 should be algebraic integers.Rather, they can not be algebraic integers. > > For others, yes the intelligence test continues.Ah, yes.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 Math factorization, other side>> > where the gs *should* be algebraic integers, and, of course,>> Again, why *should*?>Well, g_3 clearly is an algebraic integer, but wouldnt you argue that>g_1 and g_2 are not?>>Now then, why would you so argue?I think you are missing the point of his question. He asked why they*should be* algebraic integers, and you responded by questioning himabout his position on whether they *are* algebraic integers.If I understand you correctly, you say they arent algebraic integersbut should be. Regardless of whether they arent, what is your reasonfor saying that they should be?-- Richard-- Spam filter: to mail me from a .com/.net site, put my surname in the headers.FreeBSD where the gs *should* be algebraic integers, and, of course,> Again, why *should*?> >Well, g_3 clearly is an algebraic integer, but wouldnt you argue that>g_1 and g_2 are not?>>Now then, why would you so argue?> > I think you are missing the point of his question. He asked why they> *should be* algebraic integers, and you responded by questioning him> about his position on whether they *are* algebraic integers. If I understand you correctly, you say they arent algebraic integers> but should be. Regardless of whether they arent, what is your reason> for saying that they should be?> A good enough reason is that g1, g2 and 22*g3 can be shown to be roots of the polynomial equation 22*z^3 + 72*z^2 - 735*z - where the gs *should* be algebraic integers, and, of course,> Again, why *should*?> >Well, g_3 clearly is an algebraic integer, but wouldnt you argue that>g_1 and g_2 are not?>>Now then, why would you so argue?> > I think you are missing the point of his question. He asked why they> *should be* algebraic integers, and you responded by questioning him> about his position on whether they *are* algebraic integers. If I understand you correctly, you say they arent algebraic integers> but should be. Regardless of whether they arent, what is your reason> for saying that they should be?> > -- RichardSorry, but youre clearly out of your league here.Since Im making a point that several posters who have been dedicatedin replying to me arent competent to the task, I wont elaboratefurther with you.Unless youre someone who likes to disagree with me, then I can alsoinclude you as well.The math here involves *advanced* concepts, which is the factorization, other side>Sorry, but youre clearly out of your league here.>>Since Im making a point that several posters who have been dedicated>in replying to me arent competent to the task, I wont elaborate>further with you.>>Unless youre someone who likes to disagree with me, then I can also>include you as well.>>The math here involves *advanced* concepts, which is the point Im>driving home.>James HarrisGee, James, I guess anyone who disagrees with you, or fails to see howadvanced your mathematical thinking is, must be out of your league.Not a bad place to be, considering your incredible arrogance. You dont wantdiscussion, you want agreement.James, can you figure out what the following means?Alu tabaraka e foliki basoparta, eki laganato geri, eki hasto docui.No?Then I guess youre out of my league.Heh heh!-- Wolf Kirchmeir, Blind River ON CanadaNature does not deal in rewards or punishments, but only in consequences.(Robert Ingersoll)