mm-1250 the pole and polar depicted in :http://mathworld.wolfram.com/Inversion.htmlThe circle is as much ïstraightÍ in Hyperbolic geometry as thestraight line is in the Euclidean.The mappings are geodesics in theirrespective geometries, i.e., are in some interpreted way theshortest.> > I have never come across this again, and the subject seems to have> disappeared from text books. It seemed a bit of an odd topic to me at> the time. Does anyone know how/when this concept came about? Did it> ever have any practical application?>Hyperbolic ( & Elliptic) geometries were born out of accomodating deviations possible in EuclidÍs Fifth Parallel postulate that wasacccepted for more than 20 centuries. A young Johanes Bolyai,K.F.Gauss and Lobachevsky were poineers.In Non-Eucliden geometry text books Inversion is currently andactively available. It is the basis of Hyperbolic geometry appearingin PoincareÍs unit disk model. It is an important component ofbilinear transformation in complex variable conformal mappings wherecircles are mapped on as circles. In the application of 3 Dimensionalsterographic projection maps the north pole is the center o'nversion, the globe and plane through south pole now corrospond tocircle getting inverted as a straight line, through an imagined bowlmirror of inversion. In Inversion, angles are conserved, the senseis interchanged. Log spirals become Loxodromes.It is exciting to look slowly at all these , so simple yet deep,together, as you appreciate their inter- relationships better thatway. Hope this would rekindle your interest in this exciting subject. When I did A level maths at school, about 30+ years ago, in my> analytical geometry course I was taught a subject called ïinversionÍ.> > As I recall, one de'ned a point O (centre of inversion?) and used> this to transform a curve. Any point P on the curve was transformed> to PÍ, such that O, P, & PÍ were \ on a straight line and OP*PPÍ = k^2> (k = radius of inversion?).> > The one result I can remember is that a circle can be transformed into> a straight line, and vice versa.> > I have never come across this again, and the subject seems to heve> disappeared from text books. It seemed a bit of an odd topic to me at> the time. Does anyone know how/when this concept came about? Did it> ever have any practical application?Readhttp://mathworld.wolfram.com/Inversion.htmlIt has a long bibliography.Jose Carlos Santos =I need a general technique to solve linear inequation systems. Forexample:|3* x1 - x2 - 6| <= 6|x1 - x2| <= 3x1, x2 >= 0In fact, IÍm interested in the minimum values for x1 and x2 thatsatisfy the previous conditions. Any ideas about how to do it? I need a general technique to solve linear inequation systems. For>example:>|3* x1 - x2 - 6| <= 6>|x1 - x2| <= 3>x1, x2 >= 0Simplex method. Lots of good references available; to pick one, IÍd recommend David LuenbergerÍs Linear and Nonlinear Programming if you can get a hold of it.Doug =May be by solving a linear programming problem such as:min x1 + x23* x1 - x2 - 6 >= -63* x1 - x2 - 6 <= 6x1 - x2 >= -3x1 - x2 <= 3x1, x2 >= 0 http://www.gams.comerwin@gams.com, http://www.gams.com/~erwin-- --> I need a general technique to solve linear inequation systems. For> example:> |3* x1 - x2 - 6| <= 6> |x1 - x2| <= 3> x1, x2 >= 0> In fact, IÍm interested in the minimum values for x1 and x2 that> satisfy the previous conditions. Any ideas about how to do it? I need a general technique to solve linear inequation systems. For> example:>If |a| < b, thereÍs two cases 0 <= a or a <= 0giving two cases |a| = a or |a| = -a 0 < a < b or -a < 0 < bwhich gives -b < a < bThus to convert your system> |3* x1 - x2 - 6| <= 6> |x1 - x2| <= 3> x1, x2 >= 0>-6 <= 3x1 - x2 - 6 <= 6-3 <= x1 - x2 <= 30 <= x1, x20 <= 3x1 - x2 <= 120 <= x1 - x2 + 3 <= 60 <= x1, x2x2 <= 3x1 <= x2 + 12x2 <= x1 + 3 <= x2 + 60 <= x1, x2x1 <= x2/3 + 4x2 <= x1 + 3x1 <= x1/3 + 1 + 42x1/3 <= 5x1 <= 15/2x2 <= x1 + 3 <= 21/2> In fact, IÍm interested in the minimum values for x1 and x2 that> satisfy the previous conditions. Any ideas about how to do it?>15/2, 21/2 be the max and 0,0 the min. =Well just foing this in mathematica gave me this<< Algebra`InequalitySolve`InequalitySolve[{Abs[3*x1 - x2 - 6] <= 6, Abs[x1 - x2] <= 3, x1 >= 0, x2 >= 0}, {x1, x2}](x1 == 0 && x2 == 0 ||0 < x1 <= 3/2 && 0 <= x2 <= 3 x1 || 3/2 < x1 <= 3 && 0 <= x2 <= 3 + x1 || 3 < x1 <= 9/2 && (-3) + x1 <= x2 <= 3 + x1 || 9/2 < x1 < 15/2 && (-12) + 3 x1 <= x2 <= 3 + x1 || x1 == 15/2 && x2 == 21/2)So from that your min vlaues are x1=x2=0 and i have a feeling that the maxvalues are x1=15/2 and x2=21/2If you had to do it by hand you would end up doing some linear program(operations research) where you replace Abs[3*x1 - x2 - 6] <= 6 by takingthe case that x1 is pos or neg and that x2 is either pos or neg then use thesimplex method to 'nd the solution i am not to sure about that part as ihave never came accross that type of problem b4.hope this helped I need a general technique to solve linear inequation systems. For> example: |3* x1 - x2 - 6| <= 6> |x1 - x2| <= 3> x1, x2 >= 0 In fact, IÍm interested in the minimum values for x1 and x2 that> satisfy the previous conditions. Any ideas about how to do it?> Well just foing this in mathematica gave me this> << Algebra`InequalitySolve`>Blach, no math just computhink.> InequalitySolve[{Abs[3*x1 - x2 - 6] <= 6, Abs[x1 - x2] <= 3, x1 >= 0,> x2 >= 0}, {x1, x2}] (x1 == 0 && x2 == 0 || 0 < x1 <= 3/2 && 0 <= x2 <= 3 x1 ||> 3/2 < x1 <= 3 && 0 <= x2 <= 3 + x1 ||> 3 < x1 <= 9/2 && (-3) + x1 <= x2 <= 3 + x1 ||> 9/2 < x1 < 15/2 && (-12) + 3 x1 <= x2 <= 3 + x1 ||> x1 == 15/2 && x2 == 21/2) So from that your min vlaues are x1=x2=0 and i have a feeling that the max> values are x1=15/2 and x2=21/2 If you had to do it by hand you would end up doing some linear program> (operations research) where you replace Abs[3*x1 - x2 - 6] <= 6 by taking> the case that x1 is pos or neg and that x2 is either pos or neg then use the> simplex method to 'nd the solution i am not to sure about that part as i> have never came accross that type of problem b4. hope this helped> I need a general technique to solve linear inequation systems. For> example: |3* x1 - x2 - 6| <= 6> |x1 - x2| <= 3> x1, x2 >= 0 In fact, IÍm interested in the minimum values for x1 and x2 that> satisfy the previous conditions. Any ideas about how to do it? =IÍm having dif'culty proving that ||v||_p <= ||v||_2 for p>=2, here vis a vector in R^n. IÍm having dif'culty proving that ||v||_p <= ||v||_2 for p>=2, here v> is a vector in R^n.Prove it when ||v||_p = 1 and then remember these are norms. IÍm having dif'culty proving that ||v||_p <= ||v||_2 for p>=2, here v>is a vector in R^n.David C. Ullrich =Some of you might have noticed that I took a break from talking aboutthe key proof with which I can show an over one hundred year old errorto once again bring up my 'nd of a way to count prime numbers byintegrating a partial difference equation.Here you can see the truth, clearly presented, and how posters reactto it.My point is that these posters have an agenda, theyÍre not logical,and they will simply attack my work no matter how correct it is.They are apparently *very* angry people, and if you look over my postson that partial difference equation, you should be hard-pressed to'nd a good reason for their responses.Maybe you think IÍm using hyperbole in my posts by saying that IÍm theonly one in recorded human history to 'nd a way to count primenumbers by integrating a partial difference equation.But that is a fact. ItÍs not hyperbole to simply state a fact.However, letÍs imagine that youÍre a jealous and angry person, who'nds it INFURIATES you that thereÍs this person daring to post abouttheir discovery, which makes you look bad, you think.Now then, if you believe that you can smear this person, andnegatively impact by posting something against them, why not?My fear is that some of you are attributing traits to posters becauseyou believe that mathematicians as a group *always* have certaintraits of intelligence and rationality.But are not prime numbers important? ShouldnÍt a method for countingthem that is a 'rst in recorded human history be of some interest? Why should people get away with attacking a discoverer who is tellingthe truth?Why do *you* let them?James HarrisMy math discoveries, found for pro'thttp://mathforpro't.blogspot.com/ Some of you might have noticed that I took a break from talking about> the key proof with which I can show an over one hundred year old errorYou must realise by now that people fall into these two groups:a) Those that donÍt understand the maths and donÍt care. They see you whining. They see shades of Ed Conrad, Jack Sarfatti, George Hammond etc. They might poke you with a stick to see what noises you make but otherwise they donÍt care.b) Those that do understand the maths and still donÍt care. They know youÍre wrong, theyÍve told you are wrong, \ they have demonstrated you are wrong (countless times) and at this late stage have taken to ignoring or laughing at you. They believe that you donÍt have the good grace, wit, or intelligence to tone down the posturing, or concede that you may actually be wrong.Now continue your appeal in the court of public opinion for as long as you like but really no one cares now and nor will they ever. You could a hundred messages a day and still no one would care - perhaps with the exception of your ISP abuse admin and the KoTM judging panel.Did I mention that no one cares? Some of you might have noticed that I took a break from talking about> the key proof with which I can show an over one hundred year old error> to once again bring up my 'nd of a way to count prime numbers by> integrating a partial difference equation.> By took a break James means not replying to repeatedquestions from a number of posters. Rest assured thatwhen James returns from this break he will notstart by answering these questions.James has speci'cally said he was not going totalk about his prime counting function.To bring people up to speed: James claims he has found a new way to enumerate Pi(x) (the number of primes <= x). Everyone else claims JamesÍ method is a simple variation on LegendreÍs method. James claims that his method is the only method that leads to a partial differential equation for Pi(x) Everyone else claims that James takes his difference equation and magically turns it into a (generalized) partial differential equation. It has been pointed out (with examples) that in general there need be no relationship between the solutions to the difference equation and the differential equation. James claims to have defeated his critics. No one else can 'nd any evidence of this. - William Hughes Maybe you think IÍm using hyperbole in my posts by saying that IÍm the> only one in recorded human history to 'nd a way to count prime> numbers by integrating a partial difference equation.So what. Your approach seems to be mathematically equivalent to an olderone. ItÍs not a big deal at all to come up with new ways of deriving oldresults. I may be the only one to have come up with a way of derivingFibonacciÍs numbers from the Gaussian factorisation of a Poissonoperator, but itÍs not a big deal. When that happened I said Cool,showed it to a buddy, and erased the blackboard again.V.-- homepage: cs utk edu tilde lastname > Some of you might have noticed that I took a break from talking about> the key proof with which I can show an over one hundred year old error> to once again bring up my 'nd of a way to count prime numbers by> integrating a partial difference equation.> Crank Information http://www.crank.net/harris.html http://www.crank.net/usenet.html 3Ausers.pandora.be possible rational tesponse to your work:http://www.crank.net/harris.html ItÍs 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 Net! =... > But are not prime numbers important?Interesting, barely important. > ShouldnÍt a method for counting > them that is a 'rst in recorded human history be of some interest? Well, if it was a 'rst, perhaps, yes. It has been shown that yourmethod is not so very original. > Why should people get away with attacking a discoverer who is telling > the truth?We do not know whether you are talking the truth. You have a partialdifference equation and create from that a partial differentialequation and claim they lead to the same results. It is to you to*prove* that that is so, because it is not necessarily true. Exampleshave been given that show that it is indeed not true in a number ofcases.Moreover, you state it is easy to program, you have said earlier thatyou are a programmer, and now you ask others to provide numericalresults. Why do you not do the programming yourself?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ...> > But are not prime numbers important?> > Interesting, barely important.> > > ShouldnÍt a method for counting> > them that is a 'rst in recorded human history be of some interest? > > Well, if it was a 'rst, perhaps, yes. It has been shown that your> method is not so very original.> > > Why should people get away with attacking a discoverer who is telling> > the truth?> > We do not know whether you are talking the truth. You have a partial> difference equation and create from that a partial differential> equation and claim they lead to the same results. It is to you to> *prove* that that is so, because it is not necessarily true. Examples> have been given that show that it is indeed not true in a number of> cases.> > Moreover, you state it is easy to program, you have said earlier that> you are a programmer, and now you ask others to provide numerical> results. Why do you not do the programming yourself?If this guy really has found a way to count prime numbers or to easilydetermine if a number is prime or even easily sieve out numbers thatare either prime or not prime, then it is a major discovery. If it isoriginal, he should get credit regardless of his ego or the egos ofanybody else here. If it is real, he should look into intellectualproperty aspects of the work. The market will determine itsusefulness, not the egos of posters here who can only call names. = > ... > > But are not prime numbers important? > > Interesting, barely important.... > If this guy really has found a way to count prime numbers or to easily > determine if a number is prime or even easily sieve out numbers that > are either prime or not prime, then it is a major discovery.He has indeed found a way to count prime numbers. Counting prime numbersis not very important, nor very interesting. It only serves as a methodto check how good an approximation function will do. But that only forsmall number in reasonable time.He has *not* found a method to easily determine whether a number is primeor not (nor dmm-123 == Undeniable is silly. Old Man has provided a general closed form> analytic solution, t = t(r), and Undeniable complains because the> inverse, r = r(t) is transcendental, yet, if the situation were reversed,> wherein r = r(t) is analytic, while t = t(r) is transcendental, no> complaint would be heard. Petty personal preferences for a solution> of the form r = r(t) aside, the motion is completely described by> t = t(r) via a straight-forward, non-iterative process.Old Man makes silly statement. Old M3 or so, the program would take 8 minuteson a Cray-1 \ to determine whether a 300 digit number was prime. Currentdesktops are an order of magnitude faster for this kind of program.50 MHz vs. 2 GHz?) > If it is > original, he should get credit regardless of his ego or the egos of > anybody else here.He has received all the credit for the correctness of his work. Notfor the originality, as it appears to be a modi'cation of a methodhundreds of years old. But when he shifts from difference equationto differential equation, a lot is left out. Even the proof that itcan be done with the same result. > If it is real, he should look into intellectual > property aspects of the work. The market will determine its > usefulness, not the egos of posters here who can only call names.The market is not interested in prime counting functions. The marketis barely interested in primality proving algorithms. That is alla niche market where most participators have not much cash at theirhands. I do not think that Henri Cohen or Arjen Lenstra (the C and Lin APR-CL) have gotten rich through their method. I even doubt thatthey have gotten rich at all, although both have contributed quite abit more to mathematics. When we had implemented APR-CL the 'rsttime it was thought that the program could be marketed (and APR-CLwas a huge improvement over previous methods). I think the pricewas about US$ 50 for the program, including shipping and help. Iwas even sent to our legal councellor to write the contract. Iwould be surprised if even 5 copies have ever been sold. But I knowwe sold at least one (sucker ;-)). If you want to get rich, doingmathematics is not the way.-(*) On the history of APR-CL, as I have seen it. Around 1981 HenriCohen came to the CWI (where Arjen Lenstra did work) with ideas forimprovement of APR. Henri and Arjen worked out the method and I didthe programming to make it a feasable operation on a CDC Cyber; usingstandard Fortran would be much too slow. This program was used toexpand the Cunningham tables (a list of factors of 2^n, etc.). Around1982 the computer centre we used wanted to buy a supercomputer, eithera Cray-1 or a CDC Cyber 205 and it became my task to set tests on them.Amongst other test programs, I modi'ed the APR-CL program to run atspeed on a Cray-1. Arjen and I have next been a week in London to runthe program on the Cray-1 of the University of London Computer centre,and that greatly expanded the Cunningham tables. (I could remain onlyone week for this and some more tests. Two weeks after my return tothe Netherlands my daughter was born.)-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ If this guy really has found a way to count prime numbers or to easily> determine if a number is prime or even easily sieve out numbers that> are either prime or not prime, then it is a major discovery. If it is> original, he should get credit regardless of his ego or the egos of> anybody else here. If it is real, he should look into intellectual> property aspects of the work. The market will determine its> usefulness, not the egos of posters here who can only call names.JSH has what he claims is a method of counting exactly the number of or easier that extant methods, and is said to look remarkably like a long know method.Its advantages, if any, have not yet become apparent, at least to anyone except its author. ...> > But are not prime numbers important? Interesting, barely important. > ShouldnÍt a method for counting> > them that is a 'rst in recorded human history be of some interest? Well, if it was a 'rst, perhaps, yes. It has been shown that your> method is not so very original. > Why should people get away with attacking a discoverer who is telling> > the truth? We do not know whether you are talking the truth. You have a partial> difference equation and create from that a partial differential> equation and claim they lead to the same results. It is to you to> *prove* that that is so, because it is not necessarily true. Examples> have been given that show that it is indeed not true in a number of> cases. Moreover, you state it is easy to program, you have said earlier that> you are a programmer, and now you ask others to provide numerical> results. Why do you not do the programming yourself? If this guy really has found a way to count prime numbers or to easily> determine if a number is prime or even easily sieve out numbers that> are either prime or not prime, then it is a major discovery. If it is> original, he should get credit regardless of his ego or the egos of> anybody else here. If it is real, he should look into intellectual> property aspects of the work. The market will determine its> usefulness, not the egos of posters here who can only call his method. Thegeneral consensus here is that it is1) not original2) slower than current established methods to to the same thing If this guy really has found a way to count prime numbers or to easily> determine if a number is prime or even easily sieve out numbers that> are either prime or not prime, then it is a major discovery. There is some dispute as to whether or not he found a *new* way, buteven supposing that he has, itÍs not so clear that it would be a majordiscovery. He corresponded to a professor with some interest in thearea, and the professor replied that about 20% of his graduatestudents come up with similar results.I personally claim no special knowledge about what counts as a majordiscovery in this 'eld.> If it is original, he should get credit regardless of his ego or the> egos of anybody else here. If it is real, he should look into> intellectual property aspects of the work. The market will> determine its usefulness, not the egos of posters here who can only> call names.Intellectual property? The market? How exactly do you expect hisproof to be used in the market?Are you *sure* you understand what heÍs claimed to have shown?Whether or not a brand new characterization of the pi function countsas a major mathematical discovery, it is not likely to lead topractical applications -- especially since this new characterizationhas no evident computational advantages.-- Sale or rental of this disc is ILLEGAL. If you have rented orpurchased this disc, please call the MPAA at 1-800-NO-COPYS. -- The MPAA begins a new anti-piracy program, found on a DVD purchased in China Some of you might have noticed that I took a break from talking about>the key proof with which I can show an over one hundred year old error>to once again bring up my 'nd of a way to count prime numbers by>integrating a partial difference equation.Here you can see the truth, clearly presented, and how posters react>to it.My point is that these posters have an agenda, theyÍre not logical,>and they will simply attack my work no matter how correct it is.They are apparently *very* angry people, and if you look over my posts>on that partial difference equation, you should be hard-pressed to>'nd a good reason for their responses.Maybe you think IÍm using hyperbole in my posts by saying that IÍm the>only one in recorded human history to 'nd a way to count prime>numbers by integrating a partial difference equation.But that is a fact. ItÍs not hyperbole to simply state a fact.Except that itÍs not a fact.>However, letÍs imagine that youÍre a jealous and angry \ person, who>'nds it INFURIATES you that thereÍs this person daring to post about>their discovery, which makes you look bad, you think._You_ can imagine what you want about all this. The rest ofus are more interested in the truth of the matter.>Now then, if you believe that you can smear this person, and>negatively impact by posting something against them, why not?My fear is that some of you are attributing traits to posters because>you believe that mathematicians as a group *always* have certain>traits of intelligence and rationality.Could be that some people are basing their opinions on silliness likethat. But many people are instead basing their opinions on lookingat whoÍs right and whoÍs wrong.And I _bet_ that many other people base their opinions on the factthat you do things like announce youÍve decided to use marketingtactics - someone wouldnÍt have to know anything about the mathto decide that if you were right youÍd use logical argument instead.>But are not prime numbers important? ShouldnÍt a method for counting>them that is a 'rst in recorded human history be of some interest? >Why should people get away with attacking a discoverer who is telling>the truth?Why do *you* let them?>James HarrisMy math discoveries, found for pro't>http://mathforpro't.blogspot.com/David C. Ullrich =James Harris> Some of you might have noticed that I took a break from talking about> the key proof with which I can show an over one hundred year old error> to once again bring up my 'nd of a way to count prime numbers by> integrating a partial difference equation.etc. Crossposted to sci.logic, to boot.www.crank.net/harris.html =I was just doing some pleasure reading, and maybe this is too advanced forme, but could someone explain why this is true?If a subgroup H of a topological group G is open, then it is also closed.The reason the notes give is that the cosets gH are all open (why is thistrue?) ; and so H, as the complement of the union of all other cosets, isclosed.For example, the subgroup R+ as a subset of R* (where R* is R {0} withmultiplication and the metric d(x,y) = |x-y| and R+ = {x in R such that x >0}). But isnÍt R+ = (0,in'niti) an open set in the standard topology? Howis it also closed???WhatÍs going on here?Mike I was just doing some pleasure reading, and maybe this is too advanced for>me, but could someone explain why this is true?If a subgroup H of a topological group G is open, then it is also closed.>The reason the notes give is that the cosets gH are all open (why is this>true?) Because the group operations are continuous by hypothesis. Hence for a'xed a, the map x |-> ax is continuous. Since this map has a continuous inverse (given by x |-> (a^(-1))x) it is a homeomorphism.>; and so H, as the complement of the union of all other cosets, is>closed.For example, the subgroup R+ as a subset of R* (where R* is R {0} with>multiplication and the metric d(x,y) = |x-y| and R+ = {x in R such that x 0}). But isnÍt R+ = (0,in'niti) an open set in the standard topology? Yes.> How>is it also closed???Well, what do _you_ think the closure of R+ is? ThatÍs the closurein R*, not the closure in R.>WhatÍs going on here?>Mike>David C. Ullrich I was just doing some pleasure reading, and maybe this is too advanced for> me, but could someone explain why this is true?> > If a subgroup H of a topological group G is open, then it is also closed.> The reason the notes give is that the cosets gH are all open (why is this> true?) ; and so H, as the complement of the union of all other cosets, is> closed.> Well, by the de'nition of a topologic group the map h -> gh is continuous. Thus the map h -> g^-1 h is also continuous (as g^-1 is in the group). But if f is a continuous function, and U an open set then the inverse image f^-1 (U) is open (by de'nition).So g(H) = (g^-1) ^-1 (H) is open, as g^-1 is continuous and H is open. Thus the image of an open set under the function g is open, but the image of H under this function is just a coset gH. So gH is open. Thus the union of all cosets not equal to H is open, and because the cosets partition the group this must be equal to H complement. Hence H is closed.> For example, the subgroup R+ as a subset of R* (where R* is R {0} with> multiplication and the metric d(x,y) = |x-y| and R+ = {x in R such that x 0}). But isnÍt R+ = (0,in'niti) an open set in the standard topology? How> is it also closed???It is open, but its also closed as youÍre not quite using the standard topology - youÍre using the subset topology. 0 has been removed from R, disconnecting the set. Thus (0, in'nity) is indeed closed, as itÍs complement is (-in'nity, 0) which is open.Hope this helps,David > I was just doing some pleasure reading, and maybe this is too advanced for> me, but could someone explain why this is true?> > If a subgroup H of a topological group G is open, then it is also closed.> The reason the notes give is that the cosets gH are all open (why is this> true?) ;Multiplication by an element is a homeomorphism, so it takes open setsto open sets. and so H, as the complement of the union of all other cosets, is> closed.> > For example, the subgroup R+ as a subset of R* (where R* is R {0} with> multiplication and the metric d(x,y) = |x-y| and R+ = {x in R such that x 0}). But isnÍt R+ = (0,in'niti) an open set in the standard topology? How> is it also closed???The complement of R+ in the group R^* is R-=(-in'nity,0). Surely youagree that this is an open set, so its complement is closed.Disconnected spaces have non-trivial sets that are simultaneously open and closed.Jyrki Lahtonen I got to thinking about that big numbers thread. HereÍs one quite big> number.> > Let f: N^2->N be f(i, 0)=i, f(i, j)=f(i, j-1)! if j>0.> My number is:> f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, 9))))))))))).> There are 11 nested iterations of f.> > How big is this number? Is it smaller or bigger than GrahamÍs number?> It ends in a rather long strip of zeroes but I havenÍt the foggiest> what it starts with.> Hmm. I tried to post this to the last thread, but the post got eaten for some reason and I forgot to save a copy. HereÍs a brief description of ïbigÍ numbers which can be obtained in a computable way. ItÍs mostly taken from an informal talk given by Professor Gowers here in Cambridge last year. (Very undergraduate level, but still neat).Ok. Consider a starting function f_0 (n) = n + 1De'ne inductively f_m(n) by f_(m+1) (n) = f_0 ( f_m (n) )As you can see, this is a bit similar to what youÍre doing (just with a smaller starting function).So f_m(n) = n+m. Very exciting, and not a very fast growing function.But now we extend this by a process of diagonalisation to associate a function with the 'rst in'nite ordinal, w. De'ne f_w(n) = f_n(n) = 2n.Still not very impressive.Analogously to our previous de'nition we de'ne f_(w + (m+1) ) (n) = f_w ( f_(w+m) (n) ).So f_{w+1} (n) = 2^nf_{w+2} (n) = 2^2^...^2^n with m 2s.(where x^y^z = x^(y^z) )So these functions are already getting extremely big.But we can diagonalise again, to get f_(w^2)(n) = f_(w+n)(n). Repeating this we can de'ne f_(w^n), diagonalise again to get f_(w^w), etc.ItÍs kindof hard to make our choice of the function for each ordinal precise, and we canÍt de'ne functions like this for every countable ordinal (because given this way of construction weÍre only going to obtain countably many ordinals, and there are uncountably many countable ordinals), but we donÍt really need to.If we take something like f_(e) (2) where e = w^w^w^... then this number is going to be so unimaginably much bigger than yours that itÍs not even funny. IÍd be willing to bet that f_(w^w)(2) was also stupidly larger than it. And we can keep going on like this.Even when weÍre limiting ourself to computable functions, this sortof shows why itÍs kindof silly to ask about ïreally big numbersÍ - itÍs incredibly easy to construct numbers so collosally huge that one can form no sort of intuition about them at all, or really 'nd a practical use for them. ItÍs very easy to write down arbitrarily big numbers using the above notation. TheyÍre not quite as large as the busy beaver numbers, but theyÍre much easier to get a handle on.Not entirely relevant to your OP, but interesting nonetheless. :)David I got to thinking about that big numbers thread. HereÍs one quite big> number.> > Let f: N^2->N be f(i, 0)=i, f(i, j)=f(i, j-1)! if j>0.> My number is:> f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, 9))))))))))).> There are 11 nested iterations of f.> > How big is this number? Is it smaller or bigger than GrahamÍs number?> It ends in a rather long strip of zeroes but I havenÍt the foggiest> what it starts with.WhatÍs the biggest number that can be made with ïthreeÍ nines? I got to thinking about that big numbers thread. HereÍs one quite big> number.> > Let f: N^2->N be f(i, 0)=i, f(i, j)=f(i, j-1)! if j>0.> My number is:> f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, 9))))))))))).> There are 11 nested iterations of f.> > How big is this number? Is it smaller or bigger than GrahamÍs number?> It ends in a rather long strip of zeroes but I havenÍt the foggiest> what it starts with.Your number is certainly pretty large, but I believe GrahamÍs numberis unimaginably larger. I shall call your number J for the purposes ofthis post.We de'ne a function T : N -> N (the ïtower taming functionÍ). It isthe monotonically increasing step-function which changes value ontowers of 9s, so that T takes the value k on a tower of k 9s. You canthink of T as a sort of super-logarithm. We now perform some verycrude estimates on the function f de'ned above:f(9,1) = 9! < 9^9f(9,2) = f(9,1)! < (9^9)! < (9^9)^(9^9) < 9^9^9^9f(9,n) < a tower of 2^n 9sThus T(f(9,n)) <= 2^n, and hence (making the extremely crude estimatethat T increases more slowly than log_2) T^2(f(9,n)) < n.Thus T^2(f(f(9,n))) < f(9,n), and so T^4(f(f(9,n)) < T^2(f(9,n)) < n.Thus T^22(J) < 9, so T^23(J) <= 1. So J is ïtamedÍ by 23 applicationsof the taming function T.The page at mathworld.com says that GrahamÍs number is @(3,3,64),where @ is the function de'ned recursively by:@(m, n, 1) = m^n@(m, n, r) = @(m, @(m, @(...@(m, @(m, m, r-1))..., r-1), r-1), r-1)(n-1 @s)Firstly observe that @(m, n, 2) is a tower of n ms.@(3,3,3) = @(3, @(3, 3, 2), 2) = @(3, 3^3^3, 2) = tower of 3s 3^27 high@(3,3,4) = @(3, @(3, 3, 3), 3) = @(3, (tower of 3s 3^27 high), 3) = @(3, @(3, @(3, @(3, ..., @(3, 3, 2), 2), 2), ..., 2)))) (the number of terms in this expression is the huge towerof 3s)If we try to work out what this expression means, language andintuition fail us. Working from the inside outwards:@(3,3,2) = 3^3^3@(3, @(3, 3, 2), 2) = tower of 3s 3^27 high@(3, @3, @(3, 3, 2), 2), 2) = tower of 3s (tower of 3s 3^27 high) highand so on for (tower of 3s 3^27 high) times. At this point, it becomesclear that the taming function T barely makes a dent, even if we applyit lots of times in succession. This effect only gets worse as theparameter r increases. We have only got as far as @(3,3,4), andGrahamÍs number is @(3,3,64).Best wishes,-- Simon Nickerson =Some physicist will be daft enough to produce a theory that this is thenumber of parallel universes!!!!!!Wait and see!!!!-- Bruce Harveybruce@bearsoft.co.ukThe Alternative Physics Sitehttp://users.powernet.co.uk/bearsoft> I got to thinking about that big numbers thread. HereÍs one quite big> number. Let f: N^2->N be f(i, 0)=i, f(i, j)=f(i, j-1)! if j>0.> My number is:> f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, f(9, 9))))))))))).> There are 11 nested iterations of f. How big is this number? Is it smaller or bigger than GrahamÍs number?> It ends in a rather long strip of zeroes but I havenÍt the foggiest> what it starts with. -- > /-- Joona Palaste (palaste@cc.helsinki.') - Finland --> -- http://www.helsinki.'/~palaste rules! --/> Show me a good mouser and IÍll show you a cat with bad breath.> - Gar'eld > Some physicist will be daft enough to produce a theory that this is the> number of parallel universes!!!!!!> > Wait and see!!!!> -- No! The number of parallel universes would be much larger! =Bruce Harvey scribbled the following:> Some physicist will be daft enough to produce a theory that this is the> number of parallel universes!!!!!!> Wait and see!!!!I am fairly sure my number is bigger than e to the power of the-- /-- Joona Palaste (palaste@cc.helsinki.') - Finland - http://www.helsinki.'/~palaste rules! --/Bad things only happen to scoundrels. - Moominmamma = ..................>No particular reason, I just thought it would be interesting to see if>there was some noticeable pattern in the probabilities. I suppose as>n tends to in'nity the probability of an arbitrary n-th degree>polynomial having n real roots tends to 0, but it would still require>proof. Anyway it basically boils down to just being curious.There are quite a few papers on the distribution of the number of real roots of random polynomials. I believethat some of them are by Arnold.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University =INVENTIONS & DISCOVERIES INVENTION OF NUMERALSNumerals are found in the inscriptions of Ashoka The Great in the 3rdCentury BC. This knowledge traveled from there to Europe and West. InArab countries even now numerals are known as HINDSE: from India. Latime, It is India that gave us the ingenious method of expressing allnumbers by means of ten symbols - Prof. O.M. Mathew in BhavanÍsJournalINVENTION OF ZEROBrahmagupta was the 'rst mathematician to treat ZERO (0) as a numberand showed its mathematical operationsINVENTION OF ARITHMETICArithmetic was discovered by Indians in about 2nd Century BC.BhaskaracharyaÍs book Lilavathi is regarded as the 'rst book onmodern arithmetic. The Arabs learnt and adopted it from India andspreaded it to Europe. In 499 AD Aryabhatta 'nished his workAryabhatt, giving rules of Arithmetic (Encyclopedia Britannica)INVENTION OF ALGEBRAIn Western Europe the knowledge of Algebra was borrowed, not fromGreece but from Arabs, who acquired this from India. Algebra is theonly Arabic name for Bijaganitha. Aryabhatta was one of the 'rst touse Algebra (Encyclopedia Britannica)INVENTION OF GEOMETRY AND TRIGNOMETRYThe brick work of Harappa and Mohenjodaro excavations show that peopleof ancient India (2500 BC) possessed knowledge of Geometry. Aryabhattaformulated the rules for 'nding the area of a triangleÍ, which ledto the origin of Trignometry.DISCOVERY OF ASTRONOMYThe knowledge of the motion of heavenly bodies was discovered byAryabhatta (499 AD), Latadeva (505 AD) and Brahmagupta (628 AD), forcalculating the timing of eclipses. In Surya SidhantaÍ Latadeva,talked about the earthÍs axis and called it SUMERU. That the earth isa sphere and it rotates on its own axis, was known to Varahamihiraand other Indian astronomers much before Copernicus published thistheory. (Jewish Encyclopedia)INVENTION OF CALENDAR MAKINGDiscovery of measurement of time and discovery of nomenclature ofdays, month and years and invention of calendar making was made inIndia. In his book Surya SidhantaÍ Latadeva (505 AD) divided the yearinto 12 months. Seven planets of the solar system effect the earthÍsatmosphere and their names were added to the seven days of the week,which was accepted all over the world.DISCOVERY OF THEORY OF GRAVITATIONIn his book Sidhanta ShiromaniÍ Bhaskaracharya mentions about forceof attraction resembling gravity, discovered centuries later byNewton. (Jewish Encyclopedia)INVENTION OF IRON PRODUCTS IN 3000 BCThe word AYAS occurs in the four Vedas which denotes iron. Ashokapillar at Mehrauli, New Delhi and another iron pillar in Karnatakastand proof of IndiaÍs metallurgical heritage (A study published inthe magazine The Current ScienceÍ).INVENTION OF COPPER, BRONZE AND ZINCThe copper and bronze artifacts dates back to Indus ValleyCivilization (2500 BC). According to treatise RASARATNAKAR, zinc wasmade in around 50 BC at Zawar in Rajasthan (India).INVENTION OF CHEMICALS PROCESSES, DYES AND CHEMICAL COLORSChemistry known as RASAYAN SHASTRA was invented in India. Elphinstoneto prepare the sulphate of copper, zinc and iron and carbonates o§ead and iron. RASAVIDYA or Indian alchemy made its appearance around5th Century AD (National Science Centre, New Delhi) INVENTIONS & DISCOVERIES> > INVENTION OF NUMERALS> INVENTION OF ZERO> INVENTION OF ARITHMETIC> INVENTION OF ALGEBRA> INVENTION OF GEOMETRY AND TRIGNOMETRY> DISCOVERY OF ASTRONOMY> INVENTION OF CALENDAR MAKING> DISCOVERY OF THEORY OF GRAVITATION> INVENTION OF IRON PRODUCTS IN 3000 BC> INVENTION OF COPPER, BRONZE AND ZINC> INVENTION OF CHEMICALS PROCESSES, DYES AND CHEMICAL COLORSGoodness Gracious Me!(Middle of 2nd paragraph, just after Mr ïCooperÍ.)Phil-- Unpatched IE vulnerability: document.domain parent DNS resolverDescription: Improper duality check leading to 'rewall breach Published: July 29 2002Reference: http://online.securityfocus.com/archive/1/284908/2002-07-27/ 2002-08-02/0 > INVENTIONS & DISCOVERIES>> >> INVENTION OF NUMERALS>> INVENTION OF ZERO>> INVENTION OF ARITHMETIC>> INVENTION OF ALGEBRA>> INVENTION OF GEOMETRY AND TRIGNOMETRY DISCOVERY OF ASTRONOMY>> INVENTION OF CALENDAR MAKING>> DISCOVERY OF THEORY OF GRAVITATION>> INVENTION OF IRON PRODUCTS IN 3000 BC INVENTION OF COPPER, BRONZE AND>> ZINC>> INVENTION OF CHEMICALS PROCESSES, DYES AND CHEMICAL COLORS All that pathetic nationalism does not change the FACT that India is avery minor player in modern science (leaving aside Ramanujan, Bose, Ramanand the like, to be expected on a statistical basis from such a humongouspopulation) a place that people who can leave, do leave, a focus apoverty, disease and misery. Instead of proclaiming how bloody good andimportant India is, why not try and get hundreds of millions of Indian outtheir miserable ignorance and poverty? Eldon Moritz grava .88 la saucisse et au marteau:> > Your problem is similar, but it adds a lot of conjecture. Conjecture> changes the question; sometime it changes the answer.> > Consider our question, Two coins were §ipped and at least one is a> tail. What are the chances for two tails?> > Ok, IÍll tell you my answer (I donÍt even recall who was arguing for> which answer, so the other one, feel free to yell):> > I guess the answer is 1/2> > The one we saw has a chance 1/2 to be the forst coin and 1/2 to be the> second. Then> > P(TT|one is a tail) = P(we saw the 'rst)P(TT|the 'rst is tail) + P(we> saw the second)P(TT|the second is tail) = 1/2 * 1/2 + 1/2 * 1/2 = 1/2> > One other way to see it is that there are four possible double-throws:> > H H> T H> H T> T T> > Assuming we saw a tail, the 'rst one must be removed. It remains:> > T H> H T> T T> > But those three throws have not the same probability to have been seen.> P(it was the 'rst throw|we see a tail) = P(it was the 'rst throw and we> see a tail)/P(we see a tail) = (1/6)/(4/6) = 1/4> > The 1/6 comes that there are 6 possible coins to be seen and only one> tail in the 'rst throw (by throw, I mean line, not column). The 4/6> comes from the fact that there are 4 tails over the 6 coins.> > The same probability holds for the second throw.> > For the third one, we have:> > P(it was the second throw|we see a tail) = P(it was the third throw and> we see a tail)/P(we see a tail) = (2/6)/(4/6) = 1/2> > Therefore, given that the 'rst coin we see is a tail, the probability> that the two coins are tails is 1/2 .I agree. Notice that if we see one and only one, donÍt know which one,then the probability for two of the same is 1/2.Also:The popular answer for, Two coins were §ipped and at least one is ahead. What are the chances for two heads?, is 1/3.Notice that to set up a coin §ip sequence which will give the answer1/3, 1)heads must be selected prior to the §ip, 2)the coins must bere§ipped on TT.Notice that to §ip around TT, both coins must be investigated.The statement at least one is a head gives no evidence that thewriter of the statement, either 1)selected heads prior to the §ip,or, 2)investigated more than one coin.Therefore to answer our question 1/3 is in error.Also notice that Doctor Ullrich and his cronies donÍt wish to hearabout it.Eldon Moritz =Were you able to recollect other questions from the GRE Math test? Ifso, can you please share them with me? Do you have any other studymaterial that might be able to assist me when I take the GRE Mathtest? If so, how can I get a copy of them?Looking forward to hearing from you at your earliest convenience.> [GRE questions]>1. Hom(Z/2Z x Z/2Z, S_3) is isomorphic to what?>2. Let f(x) = x^2 + Bx + C, let B and C be independent random variables>uniformly distributed on [0,1]. What is the probability that the roots> of>f(x) are distinct and real? After 10 seconds of thought: 2:1 odds against. > That seems to be incorrect. Explanation? We look for the probability that> B^2/4 > C...We can immediately> put an upper bound on the probability at 1/4 since if B = 1, then C must be> less than or equal to 1/4...> > > I 'rst learned to count at the poker table, at the age of 1 or so, going> A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. :) =I am not a mathematician by trade, but I was talking to a mathsprofessor and he absolutely refused to acknowledge the concept of across product for two vectors that are not 3 dimensional.For me,Let A be vector in N spaceLet B be vector in N spaceA x B = CNow, I can prove that C has the propertyC.A / |C||A| = C.B / |C||B|and that C exist and not be a 3-vector.So, one could de'ne the cross product between two n-vectors as ann-vector with the propertyC.A / |C||A| = C.B / |C||B|where C = A x BBut, am I wrong? Or is the professor wrong? JS =I am not a mathematician by trade, but I was talking to a maths>professor and he absolutely refused to acknowledge the concept of a>cross product for two vectors that are not 3 dimensional.>IÍd try three vectors in R^4 with: CROSS(A,B,C) = det([i,,j,k,l;a1,a2,a3,a4;b1,b2,b3,b4;c1,c2,c3,c4]). Here i=(1,0,0,0), ect. and A=(a1,a2,a3,a4), ect.Probably wonÍt be an exact generalization, but you never know! Rich Burge I am not a mathematician by trade, but I was talking to a maths>professor and he absolutely refused to acknowledge the concept of a>cross product for two vectors that are not 3 dimensional.The key properties of the cross product AxB = -BxA; Ax(BxC) + Bx(CxA) + Cx(AxB) = 0identify this as the product operation of a Lie Algebra, sincethatÍs how Lie Algebras are de'ned.The classi'cation of all 'nite dimensional Lie Algebras is alreadywell-known, and is standard material in math and probably evenphysics curricula at the graduate level. So, itÍs not a questionthat needs to be surmised.Also, if C = AxB>Now, I can prove that C has the propertyC.A / |C||A| = C.B / |C||B|both of these would be 0, since (AxB).A = 0. YouÍre not goingto get a lot from requiring that 0 = 0, nor from requiring that(AxB).A = 0. >I am not a mathematician by trade, but I was talking to a maths>>professor and he absolutely refused to acknowledge the concept of a>>cross product for two vectors that are not 3 dimensional.The key properties of the cross product> AxB = -BxA; Ax(BxC) + Bx(CxA) + Cx(AxB) = 0>identify this as the product operation of a Lie Algebra, since>thatÍs how Lie Algebras are de'ned.The classi'cation of all 'nite dimensional Lie Algebras is already>well-known, and is standard material in math and probably even>physics curricula at the graduate level. So, itÍs not a question>that needs to be surmised.Also, if C = AxB>Now, I can prove that C has the property>>C.A / |C||A| = C.B / |C||B|both of these would be 0, since (AxB).A = 0. YouÍre not going>to get a lot from requiring that 0 = 0, nor from requiring that>(AxB).A = 0.Well, heÍs going to get zero. Which is what he got:-)Mati Meron | When you argue with a fool,meron@cars.uchicago.edu | chances are he is doing just the same I am not a mathematician by trade, I.e. you are admitting a lack of education. It is good to admit such athing.> but I was talking to a maths> professor and he absolutely refused to acknowledge the concept of a> cross product for two vectors that are not 3 dimensional.Such a concept is in fact unde'ned.> > For me,> > Let A be vector in N space> Let B be vector in N space> > A x B = C> De'ned how? Anybody can write down an equation like the above, but itwill forever remain meaningless until you tell us what C is actuallysupposed to be.In a three-dimensional, oriented, euclidean space (thatÍs the threeconditions right there) the vector product w = u x v of twonon-collinear vectors u and v is usually de'ned such thata) w is orthogonal to u and vb) (u,v,w) are positively oriented andc) ||w|| = ||u|| . ||v|| . sin( arc(u,v) )(a) is already not uniquely de'nable for spaces larger than threedimensions, i.e. there are always at least two linearly independentvectors w1, w2, ... that satisfy it.Your attempt stops right there.> Now, I can prove that C has the property> > C.A / |C||A| = C.B / |C||B|> > and that C exist and not be a 3-vector.No, you cannot prove any such thing. Because you never actuallyde'ned what you mean with the letter C. Which reders it ameaningless symbol.> So, one could de'ne [...]One can de'ne a lot of things. One cannot, however, de'ne anoperation in general R^n that behaves anywhere near similar to thecross product in 3 dimensions. Thus it would be at least confusing(and really dishonest) to call anything de'ned in a general R^n across product.> > But, am I wrong? Or is the professor wrong? > You are wrong.But that is not a helpful answer. It is a correct answer, though, tothe question you asked. What you probably WANTED to ask was somethingalong the lines of Why did the professor think that IÍm wrong?.ThatÍs a much better question as it yields answers from which you canlearn something.Ask yourself: if you can come up with something as utterly trivial asthis and someone a whole lot smarter than you whoÍs answered similarquestions from students for years disagrees -- doesnÍt it strike youas a lot more plausible to ask obviously IÍm missing something here,could someone 'll me in what that is? or some such? I am not a mathematician by trade, but I was talking to a maths> professor and he absolutely refused to acknowledge the concept of a> cross product for two vectors that are not 3 dimensional.> > For me,> > Let A be vector in N space> Let B be vector in N space> > A x B = C> > Now, I can prove that C has the property> > C.A / |C||A| = C.B / |C||B|> > and that C exist and not be a 3-vector.> > So, one could de'ne the cross product between two n-vectors as an> n-vector with the property> > C.A / |C||A| = C.B / |C||B|> > where C = A x B> > But, am I wrong? Or is the professor wrong? > > JSSure you can generalize the cross product. You need to go into aGrassmann or Clifford (geometric) algebra though. For example, see thepapers at the websitehttp://ModelingNTS.la.asu.eduPatrick I am not a mathematician by trade, but I was talking to a maths> professor and he absolutely refused to acknowledge the concept of a> cross product for two vectors that are not 3 dimensional.> > For me,> > Let A be vector in N space> Let B be vector in N space> > A x B = C> > Now, I can prove that C has the property> > C.A / |C||A| = C.B / |C||B|> > and that C exist and not be a 3-vector.> > So, one could de'ne the cross product between two n-vectors as an> n-vector with the property> > C.A / |C||A| = C.B / |C||B|> > where C = A x B> > But, am I wrong? Or is the professor wrong? > > JSYou havenÍt explained what A x B is. For 3-vectors A x B is uniquelydetermined by A and B. How is your n-dimensional crossproduct de'nedexplicitly? So, one could de'ne the cross product between two n-vectors as an>n-vector with the property>C.A / |C||A| = C.B / |C||B|>where C = A x B unique, then youÍve proven your professor wrong.Doug I am not a mathematician by trade, but I was talking to a maths> professor and he absolutely refused to acknowledge the concept of a> cross product for two vectors that are not 3 dimensional.> > For me,> > Let A be vector in N space> Let B be vector in N space> > A x B = C> > Now, I can prove that C has the property> > C.A / |C||A| = C.B / |C||B|> > and that C exist and not be a 3-vector.> > So, one could de'ne the cross product between two n-vectors as an> n-vector with the property> > C.A / |C||A| = C.B / |C||B|> > where C = A x B> > But, am I wrong? Or is the professor wrong? > The problem with your de'nition is that C is not unique.For dimensions > 3, there are in'nitely many such vectorsC. The cross-product in R^3 has the properties that A X B is orthogonal to A and B, with direction determinedby the right-hand rule, and it has magnitude |A X B| = |A|*|B|*sin(theta),where theta is the angle between A and B. This magnitude happensto be the area of the parallelogram formed by A, B, and A + B. One could extend this de'nition to, e.g., R^4, but as a functionof 3 vectors. De'ne X(A, B, C) as a vector orthogonal to A, B, and C and having magnitude equal to the volume of the parallelopiped spanned by A, B, C, and A + B + C. This is still not de'ned uniquely;in general there are two vectors of a given magnitude orthogonal to all of A, B, and C (one is the negative of the other). To de'nea unique direction for X(A, B,C), one needs to de'ne some kind oforientation of A, B, and C, and I am not sure how that should be done. Nora B. > JS I am not a mathematician by trade, but I was talking to a maths> professor and he absolutely refused to acknowledge the concept of a> cross product for two vectors that are not 3 dimensional.> > For me,> > Let A be vector in N space> Let B be vector in N space> > A x B = C> > Now, I can prove that C has the property> > C.A / |C||A| = C.B / |C||B|> > and that C exist and not be a 3-vector.> > So, one could de'ne the cross product between two n-vectors as an> n-vector with the property> > C.A / |C||A| = C.B / |C||B|> > where C = A x B> > But, am I wrong? Or is the professor wrong? You are, I am afraid. First of the the cross product of two vectors is not a vector, it is an anti-symmetric tensor that happens to have 3 components. It turns out that the number of ways of selecting 2 things from a set of 3 is 3. If we were to form an anti-symmetric object from two n-th order objects by choosing pairs in indexes i,j to produce a term containing:x_i.y_j - x_j.y_1you would need nC2 terms where nC2 is the number of ways of selecting 2 things from n. If n > 3 nC2 is not equal to n. Whereas 3C2 = 3 (what a lucky break!). That is why we can pretend the cross product is a vector of order 3.Bob Kolker> > I am not a mathematician by trade, but I was talking to a maths> professor and he absolutely refused to acknowledge the concept of a> cross product for two vectors that are not 3 dimensional.It depends what properties you want your cross product to have. Forthe usual properties you are restricted to dimensions 1, 3 and 7.> > For me,> > Let A be vector in N space> Let B be vector in N space> > A x B = C> > Now, I can prove that C has the property> > C.A / |C||A| = C.B / |C||B|> > and that C exist and not be a 3-vector.> > So, one could de'ne the cross product between two n-vectors as an> n-vector with the property> > C.A / |C||A| = C.B / |C||B|> > where C = A x B> > But, am I wrong? Or is the professor wrong?YouÍre not actually _saying_ anything to be wrong about; but you cantell the professor that dimensions 1, 3 and 7 have cross products. References supplies if you want them.> > JS-- G.C. = I am not a mathematician by trade, but I was talking to a maths> professor and he absolutely refused to acknowledge the concept of a> cross product for two vectors that are not 3 dimensional. It depends what properties you want your cross product to have. For> the usual properties you are restricted to dimensions 1, 3 and 7.> For me, Let A be vector in N space> Let B be vector in N space A x B = C Now, I can prove that C has the property C.A / |C||A| = C.B / |C||B| and that C exist and not be a 3-vector. So, one could de'ne the cross product between two n-vectors as an> n-vector with the property C.A / |C||A| = C.B / |C||B| where C = A x B But, am I wrong? Or is the professor wrong? YouÍre not actually _saying_ anything to be wrong about; but you can> tell the professor that dimensions 1, 3 and 7 have cross products.> References supplies if you want them.>Does that also work for 11, 13, 17 dimensions? IÍm sure it should work for11 dimensions. I am not a mathematician by trade, but I was talking to a maths> professor and he absolutely refused to acknowledge the concept of a> cross product for two vectors that are not 3 dimensional.> > For me,> > Let A be vector in N space> Let B be vector in N space> > A x B = C> > Now, I can prove that C has the property> > C.A / |C||A| = C.B / |C||B|> > and that C exist and not be a 3-vector.We already have this concept. We call it vectororthogonal to A and B. There are in'nitely many vectorsC with the property that C.A / |C||A| = C.B / |C||B| = 0But thatÍs not whatÍs generally meant by cross product.Even in 3-space, there are in'nitely many such vectors.But only one is A x B.> So, one could de'ne the cross product between two n-vectors as an> n-vector with the property> > C.A / |C||A| = C.B / |C||B|> > where C = A x B> > But, am I wrong? Or is the professor wrong? YouÍve focused on one property thatÍs easy to generate.You havenÍt stated a general way to calculate thisthing, nor have you shown a way to do so that givesthe other properties associated with cross products,such as anti-symmetry.However, IÍve heard (in this newsgroup) that a crossproduct can be de'ned for 7 dimensions. - Randy I am not a mathematician by trade, but I was talking to a maths> professor and he absolutely refused to acknowledge the concept of a> cross product for two vectors that are not 3 dimensional.> > For me,> > Let A be vector in N space> Let B be vector in N space> > A x B = C> > Now, I can prove that C has the property> > C.A / |C||A| = C.B / |C||B|> > and that C exist and not be a 3-vector.> > We already have this concept. We call it vector> orthogonal to A and B. There are in'nitely many vectors> C with the property that > C.A / |C||A| = C.B / |C||B| = 0> > But thatÍs not whatÍs generally \ meant by cross product.> > Even in 3-space, there are in'nitely many such vectors.> But only one is A x B.> > So, one could de'ne the cross product between two n-vectors as an> n-vector with the property> > C.A / |C||A| = C.B / |C||B|> > where C = A x B> > But, am I wrong? Or is the professor wrong? > > YouÍve focused on one property thatÍs easy to generate.> > You havenÍt stated a general way to calculate this> thing, nor have you shown a way to do so that gives> the other properties associated with cross products,> such as anti-symmetry.> > However, IÍve heard (in this newsgroup) that a cross> product can be de'ned for 7 dimensions.What is it that makes something a cross product? Is it suf'cient tobe a Lie product (anti-symmetric and Jacobi identity)?-- However, IÍve heard (in this newsgroup) that a cross> product can be de'ned for 7 dimensions.Nope. The combinatorics do not work out.Bob Kolker However, IÍve heard (in this newsgroup) that a cross> product can be de'ned for 7 dimensions.> ThatÍs interesting, how does that work??-- Tobias Fritz, Student of Mathematics and PhysicsUniversity of Heidelberg, Germany > ....> > However, IÍve heard (in this newsgroup) that a cross> product can be de'ned for 7 dimensions.Or 1!-- G.C. I am not a mathematician by trade, but I was talking to a maths>professor and he absolutely refused to acknowledge the concept of a>cross product for two vectors that are not 3 dimensional.For me,Let A be vector in N space>Let B be vector in N spaceA x B = CNow, I can prove that C has the propertyC.A / |C||A| = C.B / |C||B|and that C exist and not be a 3-vector.So, one could de'ne the cross product between two n-vectors as an>n-vector with the propertyC.A / |C||A| = C.B / |C||B|where C = A x BBut, am I wrong? Or is the professor wrong? JSIn more than three dimensions, there is no single mutually perpendicular axis for C to represent. To see why, think of a line in a plane. Through a given point, thereÍs only one perpendicular line that is still in the plane. Add another dimension and you can draw an in'nite number of perpendicular lines through that point because they can be at any angle relative to the original plane. So in three dimensions draw two perpendicular lines. Now add another mutually perpendicular line, but also add another dimension. If you have Cartesian axes w, x, y, and z, draw a line in the x direction and a line in the y direction. A mutually perpendicular line could be in the z direction, in the w direction, or at any angle into w relative to the z axis.But concepts like torque still make sense in higher dimensions. In an arbitrary number of dimensions except 0 or 1, but including 2, there is a unique plane of rotation. That would be the plane that both A and B are in. If they had unit vectors u_A and u_B, the plane of rotation can be given u_A^u_B, the wedge product. DonÍt even try to reduce it to a single axis of rotation, thatÍs something you can only do in three dimensions.The wedge has i^i=0, i^j=-j^i. And a dual space can be de'ned such that in three dimensions i^j=k, but in four dimensions it would be something like i^j=k^l, you canÍt reduce it to a single direction. The dual is basically all the left-over dimensions you havenÍt used yet, and a sign convention. You donÍt have to worry about any of that when you use Cartesian coordinates in three dimensions.-- Let us learn to dream, gentlemen, then perhaps we shall 'nd the truth... But let us beware of publishing our dreams before they have been put to the proof by the waking understanding. -- Friedrich August Kekul.8e I am not a mathematician by trade, but I was talking to a maths> professor and he absolutely refused to acknowledge the concept of a> cross product for two vectors that are not 3 dimensional.> > For me,> > Let A be vector in N space> Let B be vector in N space> > A x B = CCome again? What do you mean by A x B?> Now, I can prove that C has the propertyWhat is C?> C.A / |C||A| = C.B / |C||B|What is C? > and that C exist and not be a 3-vector.What is C? > So, one could de'ne the cross product between two n-vectors as an> n-vector with the property> > C.A / |C||A| = C.B / |C||B|> > where C = A x BWhat is A x B? > But, am I wrong? Or is the professor wrong?It is not a question between right and wrong, but betweento explain what you meant by it.Unless you do so there is no content to your witterings.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) > I am not a mathematician by trade, but I was talking to a maths> professor and he absolutely refused to acknowledge the concept of a> cross product for two vectors that are not 3 dimensional.He is correct. > For me,> > Let A be vector in N space> Let B be vector in N space> > A x B = C> > Now, I can prove that C has the property> > C.A / |C||A| = C.B / |C||B|> > and that C exist and not be a 3-vector.> > So, one could de'ne the cross product between two n-vectors as an> n-vector with the property> > C.A / |C||A| = C.B / |C||B|> > where C = A x B> > But, am I wrong? Or is the professor wrong?You need better math under your belt. If in more than 3-D, what isthe unique direction of the resultant crossproduct vector? ThereisnÍt any. Right-handed and left-handed coordinate systemsinterconvert without re§ection in 4-D, do they not?-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! I am not a mathematician by trade, but I was talking to a maths> professor and he absolutely refused to acknowledge the concept of a> cross product for two vectors that are not 3 dimensional.Probably because the term cross product is the name given to an operationon vectors in 3-space. See - http://mathworld.wolfram.com/CrossProduct.htmlWhat youÍre probably thinking of is the wedge product. See -http://mathworld.wolfram.com/WedgeProduct.html For me, Let A be vector in N space> Let B be vector in N space A x B = CPlease de'ned the cross product of two vectors in spaces higher than 3.> Now, I can prove that C has the property C.A / |C||A| = C.B / |C||B| and that C exist and not be a 3-vector.Please de'ne the ïxÍ operator for spaces of higher dimension than 3 'rst I am not a mathematician by trade, but I was talking to a maths> professor and he absolutely refused to acknowledge the concept of a> cross product for two vectors that are not 3 dimensional.> > For me,> > Let A be vector in N space> Let B be vector in N space> > A x B = C> > Now, I can prove that C has the property> > C.A / |C||A| = C.B / |C||B|> > and that C exist and not be a 3-vector.> > So, one could de'ne the cross product between two n-vectors as an> n-vector with the property> > C.A / |C||A| = C.B / |C||B|> > where C = A x B> > But, am I wrong? Or is the professor wrong? > > JSWhy do you call your vector A x B ? In 4 or more dimensions, there areMANY MANY vectors C with the properties you mention, not just the twovectors that you get in 3 dimensions (and we use a right-hand rule tochoose one of the two). Can somebody help me recall the details of an old maths joke? It is one of> those situations where a computer scientist, a physicist and a mathematician> (or some such combination) have to solve a certain problem, and the> mathematician starts by doing something crazy in order to reduce to a> problem that is already solved.The de'nitive collection:http://www.math.utah.edu/~cherk/mathjokes.html - Randy =|Can somebody help me recall the details of an old maths joke? It is one of|those situations where a computer scientist, a physicist and a mathematician|(or some such combination) have to solve a certain problem, and the|mathematician starts by doing something crazy in order to reduce to a|problem that is already solved.A mathematician, a physicist and an engineer are separatelyshown rooms, each with a gas stove, matches, a sink, and a pot,and are asked to boil some water. All three 'll the pot from thesink, light the stove, and put the pot on the stove.Then each is shown a similar room, but with the pot already'lled with water and on the stove. The physicist and engineereach light the stove. The mathematician pours out the waterto reduce the problem to a previously solved one.Keith Ramsay =It is well-known that, for a symmetric matrix H of size n by n, the following optimization problemMaximize 1/2 xÍ H x + 1/2 yÍ H y subject to ||x|| = ||y|| = 1, xÍ y = 0can be solved by the eigenvectors corresponding to the two largesteigenvalues of H. Here, both x and y are n by 1 vectors.I am just wondering if there is any geometrical proof of this?Or, does this property contain any geometrical intuition?P.S. I can be reached by lawhiu at cse dot msu dot edu. = It is well-known that, for a symmetric matrix H of size n by n,> the following optimization problem Maximize 1/2 xÍ H x + 1/2 yÍ H y> subject to ||x|| = ||y|| = 1, xÍ y = 0 \ can be solved by the eigenvectors corresponding to the two largest> eigenvalues of H. Here, both x and y are n by 1 vectors. I am just wondering if there is any geometrical proof of this?> Or, does this property contain any geometrical intuition?Well, for any two eigenvectors x,y with distinct eigenvalues, we have xÍ y =0. So, if all the eigenvalues are distinct, we can separate x and y(improvised jargon) and the geometric interpretation is pretty clear. If thelargest eigenvalue is duplicated, then the two summands in1/2 xÍ H x + 1/2 yÍ H yare the \ same, so the geometric interpretation is similar to that in the'rst case.IÍm not a good arm-waver :)LHX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft = at 01:53 PM, Tobias Fritz said:>Is this equivalent to the following: two>tangent vectors (at maybe different points on the manifold) are>identical in one chart if and only if they are identical in all>charts (with the appropriate range).>Because there are no speci'c charts mentioned, I assume that this >is the case for every atlas equivalent to the atlas consisting of >the two stereographic projection charts.The term parallelizable has nothing to do with speci'c charts,although you could certainly formulate it in terms of an atlas withthe proper properties. The normal way to de'ne it is as the existenceof a §at connection.What do you mean by identical? Tangents at two different points aredifferent.>But this is not true; even for S^1, it is easy to show that, given>any non-constant differentiable curve on S^1, tangent vectors in one>chart may be parallelWhat do you mean by parallel? How can you de'ne the word in ameaningful fashion without stipulating a curvature-free connection?-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to =Watch Neo play ping-pong! Ping-pong MatrixSee this short video of Neo playing ping pong:http://go.jitbot.com/ping-pong-matrixMitch = I asked this question in math.research, but no one has responded.Maybe someone here knows the answer:The Riemann Hypothesis says that pi(n)=Li(n)+O(sqrt(n)log n).If it were true, is it known if it would follow that O(sqrt(n) log n)is the best bound for pi(n)-Li(n)? Certainly O(n^{.5-delta}) fordelta>0 would not be an upper bound, but might there be something inbetween?I read somewhere that sqrt(n) log n/(8pi) is an exact bound, nothingbetter, if the RH is true. Is this true?Craig I read somewhere that sqrt(n) log n/(8pi) is an exact bound, nothing> better, if the RH is true. Is this true?Assume RH, assume x >= 2657Then |pi(x)-li(x)| <= sqrt(x)log(x)/(8Pi)This is a result of Shoenfeld 1976, Math. Comp. 30. =Craig Feinstein> I asked this question in math.research, but no one has responded.> Maybe someone here knows the answer: The Riemann Hypothesis says that pi(n)=Li(n)+O(sqrt(n)log n). If it were true, is it known if it would follow that O(sqrt(n) log n)> is the best bound for pi(n)-Li(n)? Certainly O(n^{.5-delta}) for> delta>0 would not be an upper bound, but might there be something in> between?The discrepancy pi(n)-Li(n) is known to be at least O(Li(sqrt(n)) log loglog n). This rules out the bound O(n^k) for any k<0.5. There is a bit aboutthis in the Clay wanted poster; go tohttp://www.claymath.org/Millennium_Prize_Problems/Riemann_ Hypothesis/and click on Of'cial Problem Description.pdf.> I read somewhere that sqrt(n) log n/(8pi) is an exact bound, nothing> better, if the RH is true. Is this true?Dunno, but I doubt it.LH In many analysis-type texts IÍve found statements like since the series>converges uniformly, we can apply term-by-term differentiation and uniform>convergence allows us to differentiate term-by-term, but IÍve never>actually seen a proof or explanation of WHY you need uniform convergence toYou need uniform convergence of the series of derivatives.Uniform convergence of the series of functions need notimply convergence of derivatives at all.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University In many analysis-type texts IÍve found statements like since the series>converges uniformly, we can apply term-by-term differentiation and uniform>convergence allows us to differentiate term-by-term, but IÍve never>actually seen a proof or explanation of WHY you need uniform convergence toAs others have noted, in general uniform convergence of a seriessum_n u_n(x) is not suf'cient to be able to differentiate term by term.Perhaps the analysis was complex analysis? If sum_n u_n(z) isa sequence of analytic functions converging uniformly on compact subsetsof some domain in C, then you can differentiate term-by-term.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 In many analysis-type texts IÍve found statements like since the series>converges uniformly, we can apply term-by-term differentiation and uniform>convergence allows us to differentiate term-by-term, but IÍve never>actually seen a proof or explanation of WHY you need uniform convergence to> > First we need to clarify exactly what result weÍre talking about! Your> post _sounds_ like youÍre saying youÍve read this:> > (1) If \ the sum f = sum(f_n) converges uniformly then fÍ = sum(f_nÍ).> > I hope youÍve never read that, that statement is simply false.> > WhatÍs true is this:> > (2) If f = sum(f_n) and if sum(f_nÍ) converges uniformly then > fÍ = sum(f_nÍ).> > \ Yes, uniform convergence is needed in (2). Some people have pointed> out that uniform convergence is not a necessary condition; itÍs> not - thatÍs not what it means to say the condition is needed,> what I mean by that is that if we simply omit the condition> (2) becomes false. That is, (3) is false:> > > (3) If f = sum(f_n) and if sum(f_nÍ) converges (pointwise) then > fÍ = sum(f_nÍ).> > A simple counterexample to \ (3): Let I_n be the ooen interval> (1/(n+1), 1/n). Choose a differentiable function f_n such that> f(n)x = 0 for all x not in I_n, but such that f_n = 1 at the> center of I_n. Then the two series f(x) = sum(f_n(x)) and > sum(f_nÍ(x)) both converge for _all_ x, but f is not even> continuous at 0, hence not differentiable at 0.> > > David C. UllrichThe police and so forth only exist insofar as they can demonstratetheir authority. They say theyÍre here to preserve order, but in facttheyÍd go absolutely mad if all the criminals of the world went onstrike for only a month. TheyÍd be on their knees waiting for acrime. ThatÍs the only existence they have.William S. Burroughs (American writer) Guardian, 1966huffy =...> > The police and so forth only exist insofar as they can demonstrate> their authority. They say theyÍre here to preserve order, but in fact> theyÍd go absolutely mad if all the criminals of the world went on> strike for only a month. TheyÍd be on their knees waiting for a> crime. ThatÍs the only existence they have.> > William S. Burroughs (American writer)> Guardian, 1966> > huffyIÍm beginning to think youÍre a one-trick pony.--Ron Bruck > In many analysis-type texts IÍve found statements like since the series> converges uniformly, we can apply term-by-term differentiation and uniform> convergence allows us to differentiate term-by-term, but IÍve never> actually seen a proof or explanation of WHY you need uniform convergence toI would throw away a book which makes the 'rst statement. ItÍs theseries of DERIVATIVES which you need to converge uniformly (and theseries of original functions needs to converge only at a single point).The second form is true, but vague. Again, the uniform convergence ofWHAT?Finally, as others have pointed out, the uniform convergence of sum f_nÍ(x) (and the convergence of sum f_n(x) at one point) is aSUFFICIENT condition, but not necessary.Another very useful situation is when the f_nÍ are all non-negative. The monotone convergence theorem then allows us to integrate the seriesterm-by-term on an interval [a,x], and if the f_n are all absolutelycontinuous (and sum_n f_n(a) converges) this leads to the uniformconvergence of sum_n f_n(x).--Ron Bruck > > In many analysis-type texts IÍve found statements like since the series> converges uniformly, we can apply term-by-term differentiation and uniform> convergence allows us to differentiate term-by-term, but IÍve never> actually seen a proof or explanation of WHY you need uniform convergence to> > I would throw away a book which makes the 'rst statement. ItÍs the> series of DERIVATIVES which you need to converge uniformly (and the> series of original functions needs to converge only at a single point).> > The second form is true, but vague. Again, the uniform convergence of> WHAT?> > Finally, as others have pointed out, the uniform convergence of > sum f_nÍ(x) (and the convergence of sum f_n(x) at one point) is a> SUFFICIENT condition, but not necessary.> > Another very useful situation is when the f_nÍ are all non-negative. > The monotone convergence theorem then allows us to integrate the series> term-by-term on an interval [a,x], and if the f_n are all absolutely> continuous (and sum_n f_n(a) converges) this leads to the uniform> convergence of sum_n f_n(x).button before completing it...About DiniÍs theorem, I asked if the compactness of K is reallyessential - and IÍm almost sure it is - because IÍm working on asequence of functions (they represent a cost) that sati'es almostall of Dini`s theorem requirements , except that the set K is abounded but open subset of R^n. In addition, I know K is convex. Isthere any interesting conclusion that comes from these facts? Atleast, I can assure the convergence is uniform on every closed subsetof K.The second theorem says that if (f_n) is a sequence of monotonic realvalued functions (in my case, all of them are increasing) de'ned onan interval [a,b] that converges pointwise to a continuous function f,then the convergence is uniform, right? (IÍve never seen the proof ofthis theorem, but I think I could prove it if f is not constant - notsure if it remais true if f is constant. My proof was based on theuniform continuity of f on {a,b]). Anyway, in my case f is notconstant because I know none of the f_n is. But unfortunately, all thef_n have positive jumps on [a,b] and the set of discontinuities of thef_ns are, in general, distinct. I don`t have the slightest reason tosuppose f is continuous on [a,b] and, actually, IÍm almost quite sureit is not. Anyway, I know that, for every n, the set ofdiscontinuities of f_n on [a,b] is 'nite, which implies so is the setof discontinuities of f (on [a,b]). This allows me to break up [a,b]into a 'nite number, say m, of closed subintervals where theconvergence is uniform, right? And since the number of suchintervals is 'nite, for every eps>0 I can 'nd m natural numbers thatsatisfy the uniform convegence condition on each of the msubintervals. Taking the minimum of such m numbers, I get uniformconvergence of f_n to f, right?Sorry for so many questions, answer thwem if you have time. Any helpis welcome.Artur binary tree where I am the begin,my parents areforeard,and so on to 1000 years in past.Where all my grandparentswhich lived then,are in the end(top) of tree.If it is binary tree,then(if I say there where 40 generations in that1000 yers) then I had 2^40 ancestors.Of course there was I lot of common ancestors for my father andmather.And If I say that every fourth level of generations have mademarrage(fourth level of familyship) then I should remove every4.generation from number 40. It means I had 2^(40-40/4)=2^30ancestors.However this is an aproximation.THE FINAL QUESTION is:What is minimal number N such that some population of N people(noneare in RELATIONSHIP-means blood/familyship of grade 6 or less.),also that these N people can leave 1000 years so that all of they sonsand descendants always make marrage whith some which is not with himin RELATIONSHIP)Some my friend 'nd N=32,but he missed to tell me the prove.So how to make marrages through 1000 years.Help! =meeting at Lawrence Berkeley Labhttp://pdg.lbl.gov/aapt-aps/Jack,Your recollection is absolutely correct. A Levi-Civita connection maybe represented as the sum of Af'ne connection (noncovariant entity) anda covariant tensor of nonmetricity (which, by the way, in my view,describes gravitational 'eld preserving energy-momentum, as describedAHow do you mean nonmetricity.?I use metricity in Hagen KleinertÍs senseguv^;v = 0i.e. Diff(4) divergence vanishes.I rewrite EinsteinÍs zero torsion 1915 geometrodynamic classical local 'eld equatonGuv = -(8piG/c^4)TuvasGuv = -alphaÍTuvalphaÍ = (string tension)^-1 = Witten parameterIn'nite string tension means no gravity because space-time geometry is too stiff to bend.The local stress-energy density tensor of pure geometry is then triviallyTuv(Geometry) = (alphaÍ)^-1GuvEinsteinÍs 'eld equation is then simply the balanceTuv(Geometry) + Tuv(Ordinary Mass-Energy) = 0Adding random micro-quantum zero point energy density from all quantum 'elds of spin 1/2 lepto-quarks and spin 1 gauge force bosons gives additional termtuv(zpf) = (alphaÍ)-1/zpfguv/zpf > 0 is exotic vacuum dark energy with w = -1 negative pressure./zpf < 0 is exotic vacuum dark matter with w = -1 positive pressureDark matter detectors will never click except by false positives in my theory.EinsteinÍs equation is thenTuv(Geometry) + Tuv(Ordinary Mass-Energy) + tuv(zpf) = 0In the 1915 theory with /zpf = 0Tuv(Geometry)^;v = 0from the Bianchi identities.However these identities FAIL IMHO when /zpf =/= 0 and is variableand if there are torsion 'elds.In my theory (with WittenÍs h = c =1 convention)/zpf = (alphaÍ)^-1[(alphaÍ)^3/2[|MACRO-QUANTUM VACUUM \ COHERENCE|^2 - 1]guv = Minkowski metric + Kleinert World Crystal Lattice Strain TensorMake the Levi-Civita connection from guv in the usual way.World Crystal Lattice Distortion Field = du(x) = alphaÍ(Goldstone Phase of MACRO-QUANTUM VACUUM COHERENCE),uStrain Tensor = du(x),v + dv(x),uDiff(4) Landau-Ginzburg eq for VACUUM COHERENCE in a two-way feedback loop between IT World Crystal Lattice Distortion Field andBIT VACUUM COHERENCE in sense of BohmÍs interpretation of IT(hidden variable) + BIT(Pilot Wave of Active Information)Torsion 'elds meandu(x),v - dv(x),u =/= 0The ordinary 1915 GR theory with the Bianchi identities decouples the geometry stress-energy density currents from the ordinary matter-radiation stress-energy density currents i.e.Tuv(traditional mass-energy sources)^;v = 0This is no longer true in the extended theory.There are profound technological applications from this conjecture if true and I bet it is.bet is that we can and will bottle dark energy.I think we are so confused that we should keep an open mind to tinkering with gravity, said Dr. Michael Turner, a cosmologist at the University of Chicago.previouslyQ to Ed Witten: How can the cosmological constant be so close to zero but not zero?Ed WittenÍs answer: I really donÍt know. ItÍs very perplexing that astronomical observations seem to show that there is a cosmological constant. ItÍs de'nitely the most troublesome, for my interests, de'nitely the most troublesome, observation in physics in my lifetime. In my career that is.My answer to the same question is at http://qedcorp.com/APS/StarGate1.movand http://qedcorp.com/APS/EmergentGravity.doc or same with .pdf extension for AcrobatGeneral relativity predicted the bending of light, the expansion of the universe and black holes, and has served as the foundation for modern cosmology, but theorists have never presumed that it would be the last word on gravity.For one thing, it is mathematically incompatible with the quantum laws very small distances or very high energies corresponding to the 'rst moments after the Big Bang, where space and time become discontinuous, general relativity has to be merged with quantum theory, a project that has engrossed the present generation of physicists.But recently some experts have been wondering out loud if it is time to rewrite EinsteinÍs version of the law as it applies to the other end of the length scale, to very long distances. The motivation comes from the predominance of what is sometimes called the dark sector in the universe.According to what has recently become a highly celebrated standard model, ordinary atoms make up only 5 percent of the stuff of the cosmos. Some kind of mysterious dark matter, perhaps consisting of while the rest a whopping 70 percent consists of something even more mysterious, known as dark energy.Obviously a theory that leaves 95 percent of the universe unexplained is less than a complete triumph.Neither dark energy nor dark matter has been observed or detected directly. Each has been inferred from its gravitational effects on the tiny fraction of stuff we can see. As a result, some scientists have suggested that what astronomers have discovered in the last 20 years is their own ignorance of gravity.In particular, the discovery, 've years ago, that the expansion of the universe is apparently accelerating, under the in§uence of that dark energy, has occasioned a re-evaluation of the old certainties.The simplest explanation for dark energy is something called the cosmological constant, 'rst invented by Einstein, a cosmic repulsion caused by the energy residing in empty space. But attempts to calculate this energy have resulted in numbers 1060 bigger than what astronomers have measured so large that the universe would have blown apart before atoms or galaxies could have formed causing theorists to throw up their hands.I think we are so confused that we should keep an open mind to tinkering with gravity, said Dr. Michael Turner, a cosmologist at the University of Chicago.As a result of all this, physics literature has become peppered with suggestions of ways to change gravity. This fall, given a choice of explanations for dark energy during cosmology workshop at the Kavli Institute for Theoretical Physics in Santa Barbara, Calif., 20 of the 44 participants voted for some variation of Einstein was wrong.Some of these proposals take their cue from the science-'ction-sounding string theory, the putative theory of everything, which holds out the possibility that our universe might be a 4-dimensional membrane (or brane) in an 11-dimensional space.nature in string theory would be stuck to the brane, like the nap on a rug. But the strings responsible for transmitting gravity would be able to drift away or leak into the meta-space surrounding the brane as they traveled along it from distant objects, according to a theory set forth in 2000 by Dr. Gia Dvali, Dr. Gregory Gabadadze and Dr. Massimo Porrati of New York University. The effect, they say, would be to make distant galaxies appear as if they were accelerating as they moved away from us.Also in a stringy vein is Cardassian expansion, named after the villainous race on Star Trek, and dreamed up by Dr. Katherine Freese and Dr. Matthew Lewis of the University of Michigan. According to their theory, the universe accelerates as a result of other branes tugging on our own. One can get an accelerating universe without having any dark energy, Dr. Freese said.Other theorists are going back and modifying general relativity possible equations that would carry out his ideas. But more complicated equations might be necessary. That was the approach taken by Dr. Turner and his colleagues, Dr. Sean Carroll and Dr. Vikram Duvvuri of Chicago, and Dr. Mark Trodden of Syracuse. The result was a universe that would speed up as it got bigger and emptier.That might sound crazy, Dr. Turner said, but not any crazier than the idea 80 years ago that the universe would be expanding.The model raises as many questions as it answers, but it and others like it are still worth pursuing, Dr. Carroll said.Something funny is going on when the universe gets to be 10 billion years old, he said, and none of our current ideas is standing up and declaring itself to be the right answer, so we have to be bold. > Integrate[Cos[x]*Log[-Cos[x] + 1 + Sqrt[D^2 + 2 + 2*Cos[x]]],{x,0,Pi}]>> >> I responded in the latter newsgroup but basically quit simplifying when>> it became clear the answer would involve elliptic integrals.>Is there something readable about that ïgeometricÍ view>(not restricted to modular forms) in a book? I liked it.I have to confess that everything I know is in my notes from a course Itook from Sid Webster years ago. But he gave some references, and I wenttoday to look at them. George SpringerÍs Introduction to Riemann Surfaces seems to match this thread pretty well; the 'rst chapter alone makesfor a nice survey of the ideas (and the pictures are very nice!). You canget something of the §avour of an older generationÍs perspective withAppell and GoursatÍs Theorie des fonctions algebriques et de leurs integrales (1895). The material goes back much further: it was Abel who'rst looked systematically at the functions de'ned by these integrals,although many special cases date back at least to Euler. (See below, andlook in indices of complex-variables books for AbelÍs Theorem.)I canÍt resist illustrating a bit more of AbelÍs perspective, since itties in with a question from another thread:The World Wide Wade asked,>> Ok, the real log function can be de'ned as the de'nite integral>> of 1/x. The real exponential can then be de'ned as its inverse. sin>> and cos can be de'ned as the parameterization of the unit circle>> in cartesian coordinates. So how exactly do you de'ne sin and cos this way?AbelÍs answer would have been that in all these cases we are de'ninga function to be the inverse of a relationship given by integrals.We have a (complex) curve de'ned by a (polynomial) equation F(z,w)=0,and we use path integrals G(p) = int_a^p dz/w to de'ne a (typically multi-valued) function G on the curve. We then constructa (single-valued) function H as the inverse of G. (In practice wecompose H with the coordinate function z to get an ordinary functionfrom the complex line to itself.)Example: F(z,w) = z - w. The curve is just the complex line and theintegral is (say) G(p) = int_1^p dz/z = log(z(p)). The value of the integral depends on the path chosen from 1 to p, or more precisely onthe winding number of the path around the origin. So G is actuallya function from this Riemann surface to the quotient space C / (2 pi Z).Then the inverse H is a function from this quotient space back tothe Riemann surface or, to put it into a common setting, itÍs a singly-periodic function from C to itself, namely H(z) = exp(z).Example: F(z,w) = z^2+w^2-1 . This time we have G(p) = int_{(0,1)}^p dz/w =int_0^p dz/sqrt(1-z^2) (where sqrt is the branch of the square rootfunction chosen so that sqrt(1 - z(p)^2) = w(p) ). This integral isusually called arcsin(z), and as before itÍs multivalued, but wecan de'ne sin(z) to be its inverse.Example: F(z,w) = z^4+w^2-1. Now the integral G(p) = int_{(0,1)}^p dz/wtakes on even more values, all of them differing by integer linear combinations of the integrals of dz/w over the two closed curves inthe Riemann surface X, based at (0,1), which describe a basis for the homology group H^1( X, Z ). That is, G is really a map X --> C / Lambdawhere Lambda is the lattice (i.e. the subgroup of the additive group of complex numbers) spanned by those two integrals over closed loops.Its inverse is then a doubly-periodic function on the complex line.(This G happens to be MapleÍs EllipticF(z(p),I) .)You can extend this example to other elliptic curves w^2=cubic orquartic polynomial in z. Topologically, these curves are surfaces ofgenus 1, meaning they look like the quotient C/Lambda, but thisconstruction gives them not only the topological structure of a torusbut the complex-analytic structure of a torus. Size matters, as they say,and this procedure with the integrals allows one to compute the actualperiods of the torus, that is, the size of the loops around it.Abel found a general pattern of addition laws of the form int_0^p1 omega + int_0^p2 omega = int_0^p3 omega, .......... (*)that is G(p1) + G(p2) = G(p3) in the notation above, where the relationship between p3 and p1 and p2 depends on the differential1-form omega. Inverting this relationship, we obtain an additionlaw expressing H( z1 + z2 ) in terms of H(z1) and H(z2). In our 'rst example, this is exp(z1+z2)=exp(z1) exp(z2).In the second example, itÍs sin(z1+z2) = sin(z1) sqrt(1-sin(z2)^2) + sin(z2) sqrt(1-sin(z1)^2)In the third example the formula is due to Euler: H(z1+z2) = ( H(z1) sqrt(1-H(z2)^4) + H(z2) sqrt(1-H(z1)^4) )/ (1 + H(z1)^2 H(z2)^2 )which of course is much less known but can be veri'ed numerically(if you can numerically compute the inverse of EllipticF, that is!).I confess I donÍt know how to derive this exact formula from 'rstprinciples, but it does check out, that is, d(p3)/d(p1) computedhere agrees with the value obtained from differentiating (*) w.r.t. p1 .dave =Are you trading the markets? Are you an engineer, scientist, computerprogrammer or mathematician moonlighting as a trader? 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Click on http://www.traderiskmanagement.comor http://www.forextrm.comand reduce your stress in trading down to your sleeping level.Remember: There is no other site on or off the web that has charts andanalysis like Trade Risk Management! =|>> There is also a conjecture of Erdos which I think states that|>> the sum diverges iff Set contains arbitrarily long arithmetic|>> progressions.||That should be , not iff . Obviously we could have arbitrarily|long arithmetic progressions and still have convergence, e.g. if|Set = { 10^k + m , 0 < m < k }. I have a notes saying Erdos offered|$3000 for a proof of . You can still collect.If someone gets it, itÍs probably the hardest $3000 theyÍll ever earn.I think itÍs kind of remarkable that it hasnÍt been proven false.