mm-171 === Subject: Re: JSH: My use of my initials f his work ever turns out to have legitimacy then his heirs, assuming>he has any, can sue you as well Mr. Dudley.Really? I'm not very familiar with US law, but I'm surprised to hearthat. Do you know of any examples where such a thing has happened? ! === Subject: De facto censorship, counting primesSome of you were probably surprised to learn that I did indeed 'nd away to count prime numbers by integrating a partial differenceequation. Some of you probably STILL doubt that no one else inrecorded history has managed such a feat because you need to believein mathematicians.But my point is that mathematicians have gone rogue and act againstthe needs of society by de facto censorship of information that theydon't think makes them look good, like the information about mypartial difference equation.What do I mean by de facto censorship?Well, besides the active activity, like webpages labeling me a crank,there's the passive act of refusing to acknowledge the discoveryitself.After all, it's very compact, as here are the instructions, yet again:dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,sqrt(y-1))],S(x,1) = 0.And p(x, y) = §oor(x) - S(x, y) - 1, and you get S as the sum of dSfrom dS(x,2) to dS(x,y).That's it. That's the knowledge which mathematicians have purviewover, in terms of the expectation from society that importantinformation of a mathematical nature will be acknowledged bymathematicians.Note that it's a *discrete* function, so for you programmers thatmeans you need to use int's or long's or some discrete variable type. Also, if you wish to implement it, please sum from dS(x,2) up to andINCLUDING dS(x,y).Now if you're a programmer or have been taught as a programmer, didyou ever get an assignment to count prime numbers?Now then, think about kids currently in school who I doubt will seethe method I've just shown you, unless maybe they're out on Usenetreading my posts, because the mathematical establishment thinks it canignore my results.Have I contacted mathematicians? Yes.I've contacted mathematicians all over the world.But you see, what bene't do they see to their society by allowingthat someone NOT a mathematician found such a result?Worse, I have other math knowledge. But mathematicians can let theworld be convinced I'm just a crank, most of them passively justsitting by, and keeping quiet about my results--after all, that'squite effective, eh?Then they have de facto censorship because people BELIEVE theywouldn't do such a thing if my work were important!!!So you have a standstill with me pushing my research, and a fewmathematicians actively 'ghting its acceptance on Usenet, while mostjust do their best to ignore it.For instance, I contacted Georgia Tech and talked to a Professor ErnieCroot, giving him more information about my prime counting researchthan I've posted here. He replied back *once*, and seemed friendlyenough. I answered him and awaited further replies. After some weeksI sent a query to follow-up, and here to looking at it. I'll let you knowwhen I do.Best,ERnieOn Thu, checking to see if you still have any interest in my 'nd of a way to > count prime numbers by integrating a partial difference equation, as I > haven't heard from you since my last reply.If you've lost interest can you refer me back to the professor who sent me > to you because I'd like to check laziness is about deciding> ahead of time what you wish to believe,> and daring God to be different.> http://lostincomment.blogspot.com/Will I ever hear back from Professor Croot? Well, consider theevidence:I've given something new, a partial difference equation integrationfor counting prime numbers, a 'rst in recorded human history.Professor Croot has had some time to consider my work, but now begsoff, claiming not to have looked at it.It turns out that he's a 'rst year professor and I was referred tohim by another professor at Georgia Tech who *asked* him to look overmy work.I daresay that Professor Croot lied in his email.That's a professor at Georgia Tech.So I'm an *independent* researcher, which means that mathematicians atuniversities have a lot of power when it comes to acceptance of mywork, but may see little point in helping me.That leaves me Usenet, where there are mathematicians, like Ullrich, a tenured math professor at Oklahoma State University, toTELL people that my work is useless or wrong.Yet, I found a partial difference equation that you can integrate tocount prime numbers which NO ONE ELSE in recorded human history hasmanaged.I have other mathematical research, but as long as mathematiciansstick to their guns, who gets to hear it?Sure I can talk about it on Usenet, and watch as posters malign mywork, lie and generally act like asses, knowing that others will justsit, and wait, waiting for mathematicians in the mainstream to letthem know that it's important.To a large extent I now censor my *own* work in talking about it, as Ifocus on things that are hard for people to lie about, and hope forthe best.Right now, locked inside of me is information that could be lost tohumanity because I'm the genius maligned, trapped by a system thatlets mathematicians get away with hurting the society that feeds andclothes them, by de facto censorship.I know things, important things, that you may never know aboutnumbers, and mathematics.Mathematicians are no longer part of decent society, but are now roguehaving taken their own path into darkness. Don't believe me?Check my instructions for integrating that partial differenceequation.Check for yourself.My math discoveries, found for pro'thttp://mathforpro't.blogspot.com/ === Subject: Re: De facto censorship, counting primes :> I've given something new, a partial difference equation integration> for counting prime numbers,Partial difference equations are *not* integrated. Their solutions are foundusing the sum calculus. Integration refers to 'nding anti-derivatives.> a 'rst in recorded human history.I suppose it would be -- except that you haven't shown how to *integrate* anypartial difference equations.> Professor Croot has had some time to consider my work, but now begs> off, claiming not to have looked at it.It turns out that he's a 'rst year professor and I was referred to> him by another professor at Georgia Tech who *asked* him to look over> my work.I daresay that Professor Croot lied in his email.You would dare to say anything that af'rmed your own magni'cence.> That's a professor at Georgia Tech.So I'm an *independent* researcher, which means that mathematicians at> universities have a lot of power when it comes to acceptance of my> work, but may see little point in helping me.That leaves me Usenet, where there are mathematicians, like > Ullrich, a tenured math professor at Oklahoma State University, to> TELL people that my work is useless or wrong.Also there are non-mathematicians who tell people that your work is useless andwrong.> Yet, I found a partial difference equation that you can integrate to> count prime numbers which NO ONE ELSE in recorded human history has> managed.Partial difference equations are *not* integrated. Their solutions are foundusing the sum calculus.> I have other mathematical research, but as long as mathematicians> stick to their guns, who gets to hear it?You post your results repeatedly and incessantly. This thread is acounter-example to your claim. *Everyone* gets to hear it -- over and over andover and over...> Check my instructions for integrating that partial difference> equation.Partial difference equations are *not* integrated. Their solutions are foundusing the sum calculus.> Check for yourself.Did that. No integration found.> My math discoveries, found for pro't> http://mathforpro't.blogspot.com/--There are two things you must never attempt to prove: the unprovable -- and theobvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: plotting have to ask this in the tex newsgroup, but I think the math> people know more about this type of problem. I wanted to know if there is a> package I can use to plot elliptic curves in LaTeX, can one do it in gnuplot? I> may just be able to use plot command and square root of x^3+ax+b and then the> negative of it, then stick them together.. but that's messy, probably theres a> better way (I mean.. what if I have an algebraic curve de'ned by f(x,y)=0 and> I can't express y in terms of x very simply? is there no way to do it with> gnuplot? or maybe other free softwares, or latex packages).As a last resort, I might just travel to university just to use their> maple/mathematica to do these dirty work and copy the Jose Capcognuplot can do such plots and it can also output LaTeX code. Also, Postscript plotscan be imported using PSTricks. http://www.crbond.com === Subject: De'nition of IntegralI have some questions about the de'nition of the Riemann Stieltjesintegral of a real bounded function f over an interval [a,b], withrespect to a function g, also bounded on [a,b].According to one de'nition (like we see in Bartle's and Apostol'sbooks), the integral is related to the concept of tagged partition. Wesay the integral of f with respect to g is I if, for every eps>0,there's a partition Pe of [a,b], such that, if P is any re'nement ofPe, then |S(P,f,g) - I| 0, there's a delta>0 such that, for every partition withmesh originality. > I don't believe mathematical proofs are patentable. Sadly, software > algorithms *are* patentable in the United States, and so it is unclear > where to draw the distinction.Software is indeed patentable in the US (and in Japan I think). It is*not* patentable in the EU. Laws to allow such patenting have justsome months ago been rejected in the European parliament. So even *if*he had a patent on his software it would have no validity in Europe.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Reality of response to my workNntp-Posting-Host: apps.cwi.nl... > My experience with patents is that the provisional application which > but I advise him to contact a patent attorney to write it. This costs > me about $400 for this provisional application. The provisional > application protects him for 1 year till he 'les the complete > application. This also gives him time to shop the idea around to > companies. After a year he 'les the complete application. My > experience is that this could cost from $6000-10000 when using a > patent attorney. > I wish him luck as the world needs new ideas.That is a large amount of money to invest in something with which you*never* would recuperate your cost, *even if it was new and original*!If he had an astoundingly fast method to count primes, my estimateis that from the sales of the algorithm he would get at most $50-200.If he is lucky. The market for such algorithms is small to non-existing.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: A potentially rewarding challengeIn response to self...anandb@vsnl.com (Bhupinder Singh in base 2;...> For instance, the Goodstein sequence G(4) is given by:G(1, 4) = 1*(2^2) + 0*(2^1) + 0*(2^0) (base 2) > = 4> G(2, 4) = 1*(3^3) + 0*(3^1) + 0*(3^0) - 1 Because in hereditory representation 2 is 2^1, and thereforebumping the base turns it into 3^1.Sorry for any confusion.Phil-- Unpatched IE vulnerability: history.back method cachingDescription: cross-domain scripting, cookie/data/identity theft, command executionReference: http://safecenter.net/liudieyu/RefBack/ RefBack-Content.HTMExploit: http://www.safecenter.net/liudieyu/RefBack/RefBack-MyPage.HTM= === ==Subject: Re: De facto censorship, counting primes jstevh@msn.com () :Professor Croot has had some time to consider my work, but now begs> off, claiming not to have looked at it.> Clearly he's part of the conspiracy. === Subject: Re: Reality of response to my work> There is no question of the legitimacy of my method for counting prime> numbers by integrating a partial difference equation.Yes, there is. Partial difference equations are *not* solved by integration. They are solved using the sumcalculus. Integration is is used to 'nd anti-derivatives.> There is also no question that I'm the 'rst person in recorded> history to ever present such a method.Right now, no one has presented such a method. Your presentation did not involve integration of any kind, muchless of a partial difference equation.> There's also no question that partial difference equations are analogs> to partial differential equations.There are parallels, but not the ones you think.> So before I can make any money, I have to get past mathematicians, the> dark gatekeepers testing their ability to deny knowledge from the> general public.Your paranoia is showing. No one has *ever* stopped you from §inging your pseudo-mathematics all over theinternet. You are well-known to at least a segment of the readership as an uneducable crank.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: Reality of response to my workParallax :> My experience with patents is that the provisional application which> but I advise him to contact a patent attorney to write it. This costs> me about $400 for this provisional application. The provisional> application protects him for 1 year till he 'les the complete> application. This also gives him time to shop the idea around to> companies. After a year he 'les the complete application. My> experience is that this could cost from $6000-10000 when using a> patent attorney.> I wish him luck as the world needs new ideas.Your understanding of the patent requirements is incorrect. There are no provisional applicationcapabilities. There are disclosure documents which can be 'led to establish temporal priority, but theydo not permit the future applicant to publish or otherwise publicly disclose his work.Having said that, I think your suggestion that James apply for intellectual property protection is sound --especially if he retains a patent attorney.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: Reality of response to my work > expand the Cunningham tables (a list of factors of 2^n, etc.). Around > Bill, I think you meant /factors of large prime numbers/ ? > Large prime numbers don't have proper factors. > And Dik's name isn't Bill.O, I do not know. I do not have a birth certi'cate here at home. Butat least my passport says it is indeed not. Sheesh, a lucky escape.-- ; http://www.cwi.nl/~dik/ === Subject: Re: Newsgroup survey: Math and personality assessment> john_correy@yahoo.com What's up, dude? > My, oh my. You make a joke on my name, and in response I make one on your > The relevance is Jesse Hughis inability to distinguish between word > play and an insult--a distinction you were (apparently) able to make.Well, I agree with Jesse. You posted a wordplay that was intended tobe insulting (and childishly so), see the surrounding wording. I sawit as an insult. That I did not retaliate in kind was deliberate. Irarely insult people on Usenet. I think I have done that about two timesin 20 years of posting history.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Re: De'nition of IntegralHi Artur,I have found a link that might help, although I worry it is not advancedenough to answer your question.It is at:http://mathworld.wolfram.com/StieltjesIntegral.htmlI hope this helps!Kavon> Hi allI have some questions about the de'nition of the Riemann Stieltjes> integral of a real bounded function f over an interval [a,b], with> respect to a function g, also bounded on [a,b].According to one de'nition (like we see in Bartle's and Apostol's> books), the integral is related to the concept of tagged partition. We> say the integral of f with respect to g is I if, for every eps>0,> there's a partition Pe of [a,b], such that, if P is any re'nement of> Pe, then |S(P,f,g) - I| Riemann-Stieltjes sum of P, f and g over [a,b]. In this de'nition,> the sum is related to tag points chosen in each of the intervals of P,> and inequality (1) must be satis'ed regardless the choice of the tag> points. The only requirement about g is that it is bounded on [a,b].Other de'nition, like we 'nd in Rudin's book, involves the concepts> of upper and lower sums and there are no tag points. The function g,> however, is required to be monotonically increasing on [a,b]. If P is> a partition of [a,b], the Riemann-Stieltjes sum is de'ned by taking> the in'mum (lower sum) or the supremum (upper sum ) of f on each> interval of P. If the in'mum of the upper sums equals the supremum of> the lower sums, both taken over all possible partitions, themn this> common number is said to be the integral of f with respect to g. Isn't> the fact the g must be increasing a loss of generality when compared> to the 'rst de'nition? If we restrain our attention to the Riemann> integral, it doesn't matterm because then we always hace g(x) =x, but> in the general case it seems to be a loss of generality.There's a 3rd de'nition you can 'nd in a recent Bartle's book, which> relies on the concept of mesh (or norm ) of a partition (de'ned as> the lenght of the largest interval of P). It's similar to the 1st> de'nition, but instead of re'nements, Bartle says that, for every> eps>0, there's a delta>0 such that, for every partition with> mesh the usual eps-delta de'nition of limit of a function, but it renders> the proof of some theorems very cumbersome.Bartle still de'nes the Generalized Riemann Integral (seems this> concept is due to Henstock) by introducing a gauge function which is> simply a positive real function de'ned on [a,b]. And he says a> function is Lebesgue integrable if |f| is Generalized Riemann> integrable. But he doesn't de'ne Lebesgue integral. In the case of> generalized Riemann integral, there's no function g and f need not be> bounded.I think the 'rst and the second de'nition, if g is monotonically> incresing, are equivalent and lead to the same value for I, but I'm> not sure if the Henstock integral or the integral based on meshes are> equivalent to them.I'm a bit confused, could anyone please clarify such points?> Artur === Subject: Re: Big Number Game I'm working on a summer math assignment and it says to do problems9-27, mod 3. Does anyone know what mod 3 means?-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html =mod means module> I'm working on a summer math assignment and it says to do problems> 9-27, mod 3. Does anyone know what mod 3 means?>-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html => I'm working on a summer math assignment and it says to do problems> 9-27, mod 3. Does anyone know what mod 3 means?Probably that you are to do every third problem, i.e., 9, 12, 15, 18, 21,24, and 27.Rich-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html =>I'm working on a summer math assignment and it says to do problems>9-27, mod 3. Does anyone know what mod 3 means?Either you should select every third problem (9, 12, 15, 18, ..., 27), or you should give your answers modulo 3. If it's the latter, look up modular arithmetic in your textbook.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comFortunately, I live in the United States of America, where we aregradually coming to understand that nothing we do is ever ourfault, especially if it is really stupid. --Dave Barry-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html =Suppose we really did want to stir some computer programming into the math curriculum cauldron. How would we go about it?One point of contact is the boolean algebra segment that sometimes makes an appearance around the start of algebra. The focus is on truth tables and the semantics of AND, OR, NOT and IF/THEN. Truth values of 0 and 1 or F and T are assigned to p and q, and the various tables get worked out, based on all permutations of the inputs. A variety of syntax gets used,but the idea is the same:p q p & q p | q p->q------------------------------T T T T TT F F T FF T F T TF F F F T [cite: http://www.wikipedia.org/wiki/Truth_tables ]The segue here is to conditional branching in scripted algorithms. The §ow of execution is generally top to bottom, but, based on conditionals, we alter this §ow by means of expressions which evaluate to either true or false.These conditionals may consist of several expressions joined together by and and or statements.With kids (and adults too), one might use the metaphor of train tracks with switches (this metaphor serves a similar purpose to §ow charts, but has the advantage of being already familiar). When the train gets to a switch, it goes one way or the other, depending on a test.For example, in the program below, a slight modi'cation of one Gregor Lingi sent me recently, we have two Owhile' loops, one inside the other, each governed by a test: def primefactors(n): Return the prime fractors of n in ascending order factors = [] p=2 while p*p <= n: # outer loop while n % p == 0: # inner loop factors.append(p) n //= p p+=1 if n>1: # 'nal test factors.append(n) return factorsUsage: > primefactors(201) [3, 67] > primefactors(2011) [2011] > primefactors(3110) [2, 5, 311] > primefactors(10929331) [7, 233, 6701] p*p <= n evaluates to True or False, which in this language (Python) are simply synonyms for 1 and 0. As long as p squared is less than or equal to n, keep looping around in the indented portion of code below the Owhile' -- which portion includes an increment to p (i.e. p += 1) -- which is why the loop is eventually bound to end (provided n was positive -- we don't test for invalid inputs in this version (which we might do with another conditional)).n % p == 0, part of the above while loop, is likewise an expression equating the remainder (after division of n by p) to zero. As long as this expression evaluates to True (as long as p goes evenly into n, leaving no remainder), we append p to the list named Ofactors', and divide p out of n ( n //=p means: assign n the new value of n//p, where // just means Odivide to return an integer result').We also had an IF loop. If n is still standing after all those primes p have been divided out of it, then n must now contain the last remaining prime factor of interest.Another point of contact would be in the realm of data structures, one of the most important being the set. A set allows only one of any element i.e. all the elements must be unique. The prime factorization of a number may well include several copies of the same prime. But if we assign [2,2,3,5,7,7] to a set object in Python, we will end up with only the unique elemtns 2,3,5,7.This prohibition against duplication turns out to be a useful feature of sets in the case of a totient algorithm built around the prime factors of n.Given positive integer n, with prime factors p1, p2... pn, we know that every pth number <= n will divide n e.g. if 3 is a prime factor, then 3,6,9...n will be divisors of n -- which is 1/3rd of the numbers between 1 and n. So we diminish n by subtracting a 3rd of it, leaving a new value for n. We write: n = n - n/3 where O=' is not a declaration of equality, but an assignment operator.However, if 5 is likewise a prime factor of the original n, then 1/5th of the remaining numbers will be multiples of 5. This isnot immediately obvious, given 3*5,6*5... will have already been removed -- assuming large enough n -- but since n has been reduced by a third, the density of the remaining multiples of 5 will be unchanged. So again: n = n - n/5. Note that we use each prime factor only once, even if it is a factor more than once.By the time we've reduced n in this way, by all its unique primefactors, we're left with the number of totatives of n, i.e. thosepositives < n with no common factors with n. Earlier, we computedthis using a test involving the gcd. In the function below, we use the above reduction by prime factors algorithm. We use the Set object to ensure our prime factors are unique. from sets import Set def totient(n): Return the number of integers unique = Set(primefactors(n)) for p in unique: n -= n//p return nUsage: > totient(1000) 400 > totient(1001) 720n -= n//p is short-hand for n = n - n//p where // again ensures an integer result. If we want to make totient(0) return 1, we might accomplish this with a conditional.I think the above examples well demonstrate how fairly short programs, used interactively in shell mode, are able to reinforce basic numeracy concepts. Here the algorithms appear in a spelled out form -- we don't use any black boxes to hide critical conceptual steps. Once an algorithm is de'ned, it may be incorporated into another, as totient incorporates primefactors or gcd.I expect there would be a market for workbooks and other media using this approach. The idea is to introduce programming into mathematics, more than it is to introduce some mathematics into programming. It's an approach that has been tried (is being tried) in various ways. But I think the languages are getting easier and easier to use (we've come a long way since Pascal), and so the practicality of this approach is increasing as well.Kirby-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html =I saw this problem when Matt was in 6th grade, but it was couched as acensus question. I believe it was a Problem of the Week, so it mayhave be taken off the Math Forum PoW website!mattsmom-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html =Does anyone know of a similar problem to this one?The Chocolate Allergy Revelation - posted September 28, 1998A travelling salesperson shows up at the apartment of a math teacherto peddle her wares. During the course of her spiel, the salespersondiscovers that the teacher has three children and inquires as to theirages. The teacher answers, The sum of my children's ages is 13. The salesperson immediately declares, That's not enough informationto deduce your kids' ages! The teacher continues, Well, the product of their ages is the same asmy apartment number. After taking a quick check of the apartment number, the salespersonthinks for a minute but then sighs, That's STILL not enoughinformation for me to 'gure it out. So the teacher says, Oh, I forgot to mention, my eldest child isallergic to chocolate. At which point the salesperson exclaims, Aha, now I know all theirages! What are the ages of the children?-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html =>Does anyone know of a similar problem to this one?>Yes, there's one very like this in Sarah Flannery's best selling OIn Code'. She also gives the answer and the reasoning process behind it.Kirby-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html =Tux Math Scrabble v2.2 released July 27, 20003By Charlie Cosse, Asymptopia SoftwareTux Math Scrabble is a math version of the popular board game for ages4-40, which is highly entertaining as well as great educational value.The game challenges young people to construct compound equations andconsider multiple abstract possibilities. There are three skill-levelsfor practice from basic addition and subtraction through tomultiplication and division.http://www.asymptopia.com-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html =Michael asked for help on some homework:I will not DO you homework for you. (Don't take offense, but we'verecently had a few people boasting about not having to learn anymath because they con newgroup readers into doing their homework- and even exam questions - for them.) However here are somesuggestions:: 1. Suppose that (x_0,y_0)=(0,0) and for every positive interger n,: f(1/n,0)=1 and f(0.1/n)=-1. Does lim (x,y)-->(x_0,y_0) exist?: Prove your answer.This one SHOULD be really easy. Since you are studying multi-variantcalculus, you must have learned the concept of a LIMIT in a pre-requisitecourse. However it will be in your current text book as well. Youneed to do some serious reading.You may also have a problem with function concepts and notation.Can you partially visualize the graph of that function z=f(x,y)?(Only paritially because you are only told about a tiny bit ofthe function.): 2. Let g(x,y)= (sin(x-y))^2/(abs(x)+abs(y)), where abs(x)=absolute: value of x. Prove that lim (x.y)-->(0,0) g(x,y)=0 using either the: abs(sin(s+t))> Douglas Hofstadter once proposed a game called Hruska*, in which the>> object was to obtain the middle score. There was to be a series of>> games, in which the winner was not the one achieving the middle score>> most often, or least often, but (of course) the middlemost number of>> times.IIRC, he, or someone, analysed it in game theory terms, and it turned out>that the optimum strategy even when you allow in'nitely high numbers was>only to play numbers from 1 to 5, in a certain proportion.>That wasn't Hruska, but a two player game called IIRC underwhelm, wheretwo players choose a number, and the one with the smallest number gets thatnumber added to her score, unless the larger number is exactly one biggerthan the smaller, in wich case the one with the larger number gets the sumof the two numbers.A three person game can't have such a simple strategy where you just playa sequence of random numbers. The other player will just settle on a pairof consecutive numbers wich come up only rarely in your strategy and thenplay (n,n+1) and then (n+1, 1) both winning nearly half of the games.Every non-losing strategy depends on what the others do.-- Wim Benthem === Subject: Re: trans'nite charles ramsey :>>Take the in'nite series expansion for e and put it into the in'nite>>series expansion of e to the power of x and multiply the terms and you>>end up with a series of aleph 1 terms that sum to a 'nite result e to>>the power of e. If you repete this process for e to the power of e to>>the power of e do you end up with aleph 2 terms? You still only get aleph 0 terms. I think the OP is making a very common error: mistaking the set of>> all subsets of N with the set of all *'nite* subsets of N> Working in binary after the decimal point 'rst you have .000... and> its compliment .111... next comes those with one one .1 .01 .001 etc> and their compliment .0111... .00111... .000111... then those with> two ones .11 .101 .1001 etc .011 .0101 .01001 etc .0011 .00101 > .001001 etc etc and their compliments these can be arranged in two two> dimensional arrays all the way up to those with an in'nite number of> ones that can be arranged in two in'nite dimensional arrays. Which> real number did I miss? Each number in your 'rst list has a 'nite number of 1's, and eachnumber in your second list has a 'nite number of 0's. You never reach anumber such as 1/3 = (.01010101...)_2 or 2/3 = (.10101010...)_2, in whichboth the 0's and the 1's are in'nite.>So now we have a one to two correspondence> between my original construction and the reals between zero and one> therefore a series with a continium of nonzero terms sum to a 'nite> result. Mr fritz says this is in impossible. So does e^(e^e) have c> members or 2^c membersIt has aleph_0 members. -- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: De facto censorship, counting primesRight now, locked inside of me is information that could be lost to> humanity because I'm the genius maligned, trapped by a system that> lets mathematicians get away with hurting the society that feeds and> clothes them, by de facto censorship.>If you can't beat Oem ,j oin e'm. That is to say, get a PhD in math. For agenius like you that should be easy. Since you like math so much, why not doit full time?MB === <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h}YL`!h_XXr5Q>_nGsY2_If I remember correctly, Leonidas Alaoglu was born in Canada (Red Deer,Alberta). :> Aynur :>That being said: However, if you search Alaoglu Greek on Google, you don't>get any pages referring to Greeks aside from, possibly, Leonidas Alaoglu. >On the other hand, if you search Alaoglu Turkish on Google, you get>several pages referring to present-day Turks with that surname. Moreover,>the word ala also has a meaning in Turkish (i.e. very good, excellent).>Yet, clearly, Leonidas is a Greek name and not a Turkish name.>So, here's what I'm wondering: Is it possible that the mathematician >Leonidas Alaoglu had a Greek mother and a Turkish father?Unlikely -- marriages between Greeks (read Orthodox Christians) and Turks > (read Muslims) in the Ottoman Empire were very uncommon, moreover a Turkish> man marrying a Christian woman back then (if not today as well) would have > no incentive at all (to put it mildly) to give a non-Muslim/Turkish name to > his son ... while a possibility you left out (Greek father, Turkish mother) > would be even less likely for reasons I am asking you to guess :-)[Now look at me: Greek national, father's last name Turkish, father's 'rst > name Greek (Christos), father's parents both Greeks (Georgios, Antonia)...] baloglouAToswego.edu-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: trans'nite series currentresident@veloemail.com (charles ramsey) :> Working in binary after the decimal point 'rst you have .000... and> its compliment .111... next comes those with one one .1 .01 .001 etc> and their compliment .0111... .00111... .000111... then those with> two ones .11 .101 .1001 etc .011 .0101 .01001 etc .0011 .00101 > .001001 etc etc and their compliments these can be arranged in two two> dimensional arrays all the way up to those with an in'nite number of> ones that can be arranged in two in'nite dimensional arrays. Which> real number did I miss? So now we have a one to two correspondence> between my original construction and the reals between zero and one> therefore a series with a continium of nonzero terms sum to a 'nite> result. Mr fritz says this is in impossible. So does e^(e^e) have c> members or 2^c members.Your compliments seems to mean, based on your examples above, representations of reals betweeen 0 and 1 having two binary representations. There are only countably many reals with such dual binary representation (or dual representation in any larger integral base), as all such numbers are rational.I do not see from the above too brief description how you intend to count those reals whose binary representations contain both in'nitely many zeros and in'nitely many ones.Indeed, a trivial translation between base 2 and base 4 allows a direct application of Cantor's diagonal proof in base 4 that there is no surjection from the naturals to reals between 0 and 1.Each such real, between 0 and 1, has a base 4 representation a = sum[i=1..oo. a_i/4^1], with all a_i in {0,1,2,3}And any dual representations are all 0's or all 3's from some point onwards, so 1's and 2's do not involve the dual representation uncertainties. For any listingFf: N -> [0,1]: n -> f(n), construct x as follows:let the n'th digit of x equal 1 if the nth base four digit of f(n) is not 1 and let it be 2 otherwise: x_n = 1 if f(n)_n <> 1 x_n = 2 if f(n)_n = 1Then x is between 0 and 1 but is not equal to any of the f(n).Since no assumptions were made about f except that its domain is N and its codomain is [0,1], and there is shown to be some member of the codomain not in the image of f, such f cannot ever be surjections.In fact, similar explicit constructions can produce a least countably many members of [0,1] not in the image of any such f. === Subject: Re: Relativity is based on assumption. Androcles :> All theories, in physics and elsewhere, are based on assumptions.> No theory can bootstrap itself out of nothing.> So what?> OSo what' depends on the validity of the assumption. The PoR is an> assumption, and is intuitive as well. Making the assumption that the time it> takes for a signal to reach an object is the same as the time it take for> the signal to return, when in the meantime you've moved away or toward the> object, is a rather silly assumption that I will not accept. That's Oso> what'.> AndroclesThen ignore relativity. But unless you can come up with something that agrees with the experimental evidence better than relativity, everyone else will ignore you. === Subject: Re: De facto censorship, counting primes> Some of you were probably surprised to learn that I did indeed 'nd a> way to count prime numbers by integrating a partial difference> equation. Some of you probably STILL doubt that no one else in> recorded history has managed such a feat because you need to believe> in mathematicians.But my point is that mathematicians have gone rogue and act against> the needs of society by de facto censorship of information that they> don't think makes them look good, like the information about my> partial difference equation.What do I mean by de facto censorship?Well, besides the active activity, like webpages labeling me a crank,> there's the passive act of refusing to acknowledge the discovery> itself.After all, it's very compact, as here are the instructions, yet again:dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,> sqrt(y-1))],S(x,1) = 0.And p(x, y) = §oor(x) - S(x, y) - 1, and you get S as the sum of dS> from dS(x,2) to dS(x,y).That's it. That's the knowledge which mathematicians have purview> over, in terms of the expectation from society that important> information of a mathematical nature will be acknowledged by> mathematicians.Note that it's a *discrete* function, so for you programmers that> means you need to use int's or long's or some discrete variable type.Also, if you wish to implement it, please sum from dS(x,2) up to and> INCLUDING dS(x,y).Now if you're a programmer or have been taught as a programmer, did> you ever get an assignment to count prime numbers?Now then, think about kids currently in school who I doubt will see> the method I've just shown you, unless maybe they're out on Usenet> reading my posts, because the mathematical establishment thinks it can> ignore my results.Have I contacted mathematicians? Yes.I've contacted mathematicians all over the world.But you see, what bene't do they see to their society by allowing> that someone NOT a mathematician found such a result?Worse, I have other math knowledge. But mathematicians can let the> world be convinced I'm just a crank, most of them passively just> sitting by, and keeping quiet about my results--after all, that's> quite effective, eh?Then they have de facto censorship because people BELIEVE they> wouldn't do such a thing if my work were important!!!So you have a standstill with me pushing my research, and a few> mathematicians actively 'ghting its acceptance on Usenet, while most> just do their best to ignore it.For instance, I contacted Georgia Tech and talked to a Professor Ernie> Croot, giving him more information about my prime counting research> than I've posted here. He replied back *once*, and seemed friendly> enough. I answered him and awaited further replies. After some weeks> I sent a query to follow-up, and here is his reply:Did it ever occur to you that he may be too busy to reply? That doesn't I haven't gotten around to looking at it. I'll let you know> :> Professor Croot:> Just checking to see if you still have any interest in my 'nd of a wayto> count prime numbers by integrating a partial difference equation, as I> haven't heard from you since my last reply.> If you've lost interest can you refer me back to the professor who sentme> to you because I'd time what you wish to believe,> and daring God to be different.> http://lostincomment.blogspot.com/> Will I ever hear back from Professor Croot? Well, consider the> evidence:I've given something new, a partial difference equation integration> for counting prime numbers, a 'rst in recorded human history.Professor Croot has had some time to consider my work, but now begs> off, claiming not to have looked at it.It turns out that he's a 'rst year professor and I was referred to> him by another professor at Georgia Tech who *asked* him to look over> my work.I daresay that Professor Croot lied in his email.That's a professor at Georgia Tech.So I'm an *independent* researcher, which means that mathematicians at> universities have a lot of power when it comes to acceptance of my> work, but may see little point in helping me.That leaves me Usenet, where there are mathematicians, like > Ullrich, a tenured math professor at Oklahoma State University, to> TELL people that my work is useless or wrong.Yet, I found a partial difference equation that you can integrate to> count prime numbers which NO ONE ELSE in recorded human history has> managed.I have other mathematical research, but as long as mathematicians> stick to their guns, who gets to hear it?Sure I can talk about it on Usenet, and watch as posters malign my> work, lie and generally act like asses, knowing that others will just> sit, and wait, waiting for mathematicians in the mainstream to let> them know that it's important.To a large extent I now censor my *own* work in talking about it, as I> focus on things that are hard for people to lie about, and hope for> the best.Right now, locked inside of me is information that could be lost to> humanity because I'm the genius maligned, trapped by a system that> lets mathematicians get away with hurting the society that feeds and> clothes them, by de facto censorship.I know things, important things, that you may never know about> numbers, and mathematics.Mathematicians are no longer part of decent society, but are now rogue> having taken their own path into darkness. Don't believe me?Check my instructions for integrating that partial difference> equation.Check for yourself.>My math discoveries, found for pro't> http://mathforpro't.blogspot.com/ === Subject: Re: Advanced techniques, non-polynomial factorizationNntp-Posting-Host: apps.cwi.nlposted. I did comment on it already, but you did not answer. I thinkyou have not seen my response. So I will try again here. > Polynomials are well-known in science and mathematics, but while > 'nding roots of polynomials is typically the aim of the average > researcher, polynomials themselves can be used as powerful tools for > analyzing the roots of *other* polynomials.Oh, well, It appears you have modi'ed it a bit. Yes, polynomials arepowerful tools to analyse roots of other polynomials. Yup, thisparagraph was rewritten, as is the next. > The concepts are advanced, but can be approached by 'rst considering > a basic example. > The basic factorization to start is > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > 49(x^3 + 5x^2 + 3x + 1) > with the c's algebraic integers, notice that only two of the c's have > 7 as a factor.A comment I add this time and did not add in the last version. The c'scan *only* be algabraic integers because the constant factor of thepolynomial is 1. When that constant factor is 2, there is *no* suchdecomposition. > It might help to go the *other* way, and start with > (d_1 x + 1)(d_2 x + 1)( d_3 x + 1) = > x^3 + 5x^2 + 3x + 1 > and now multiply by 49.And again the same, newly added, comment: the d's are *only* algebraicintegers because the constant term of the cubic is 1. > In the 'rst example you're looking at a product and realizing that > from the distributive property a(b+c) = ab + ac, you know there's > *one* way it could be produced, which is to multiply something like > the second example by 49. > The distributive property is key here. Understanding it thoroughly, > is of prime importance.Hrm, an added paragraph. Does not add much information. > Now notice that you can abstract from here as you're looking at > *functions* of x, as introducing > f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, > you have > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). > Notice that dividing both sides by 49 gives > (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 > as long as you're in a ring where 7 is not a factor of 1.A repeat:This is independent of 7 being a factor of 1. Note however that itis *not* the only way to distribute 49 amongst the three factors onthe left hand side. This is the only way *only* if you require thatthe three factors on the left hand side are polynomials, if you havenot such an requirement it can be done differently. However, you canhave polynomials on the left hand side *only* because the roots of thepolynomial on the right hand side are units, i.e. divisors of 1. Itwill *not* work if that is not the case.An addition:If you do not have the requirement that the factors on the left handside are polynomials, the following factorisation is also possible(and gazillion others), and assuming x is integer: De'ne: w3(x) = gcd(f3(x) + 1, 7) { this can be 1 or some other divisor of 7. } w2(x) = 7 / w3(x).Now: [ (f1(x) + 7)/7 ][ (f2(x) + 7)/w2(x) ][ (f3(x) + 1)/w3(x) ]is a perfectly valid factorisation in the algebraic integer valuedfunctions on the integers of: x^3 + 5 x^2 + 3x + 1Note that the distributive property dictates: (a + b) * c = a * c + b * cin a particular ring. Not that (a + b) / c = a / c + b / cin a ring. That is, that when (a + b)/c is in a ring, there is norequirements that a/c and b/c are also in that ring. You keep onassuming that. > Which is consistent with what was found before, as only two of the > functions have the property that 7 is a factor.This is true *only* in some special cases... > Now I'll move on to a more complicated example. > Let > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so they are functions of x, and since one of the roots equals 3 at > x=0, I haveNote, that they are *not* polynomials. So the situation is differentfrom the one above. And indeed, in general *none* of the factors isdivisible by 7. Moreover, you can not even have the requirement thatwhen you divide by 49 the factors on the left must be polynomials,because you do not start with polynomials on the left. So there areother ways to distribute 49 amongst the three factors, however theway you distribute is a function of x.Remainder of repetition skipped.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ John Harrison :> For those who have the book, it is W. Rudin, Principles of> Mathematical Analysis, chapter 2 ( Basic Topology), problem 18:> [A set E is perfect if E is closed and every point of E is a limit> point of E] Is there a nonempty perfect set in R which contains no> rational number?> This one has really had me stumped. My 'rst guess was that the answer> was no, but I didn't get anywhere trying to prove it. I then tried to> construct such a set: since the set of rationals is countable, we can> write it as {r_1, r_2, r_3,...}. Let I_n be the open interval of> length 1/2^n centred at r_n, let I be the union of the I_n, and let X> be the complement of I in R. Since the total length of the intervals> is at most 1, X is nonempty. As the complement of an open set, X is> closed. But I can't show that X is perfect. So: Is the answer yes or> no? If yes, does my set work as an example? Any hints would be most> welcome.I think it depends on how you enumerate the rationals. For example,if r_1 = 0, r_2 = 3/4, ....., then the point 1/2 lies exactly on the boundaries of I_1 and I_2, and _may_ end up an isolated point in your X.I also believe that multiplying your 1/2^n by some irrational numberwould help, although I've not worked out the details. === Subject: Re: question about the riemann hypothesisBut is this the best bound?> Craig Feinstein :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 >= 2657> Then |pi(x)-li(x)| <= sqrt(x)log(x)/(8Pi)> Peter === Subject: Re: Relativity is based on assumption.> Androcles :> All theories, in physics and elsewhere, are based on assumptions.> No theory can bootstrap itself out of nothing.> So what?> OSo what' depends on the validity of the assumption. The PoR is an> assumption, and is intuitive as well. Making the assumption that thetime it> takes for a signal to reach an object is the same as the time it takefor> the signal to return, when in the meantime you've moved away or towardthe> object, is a rather silly assumption that I will not accept. That's Oso> what'.> Androcles>Then ignore relativity.But unless you can come up with something that agrees with the> experimental evidence better than relativity, everyone else will> ignore you.What experimental evidence? Moving clocks running slow? They don't. The GPSclocks run fast. MMX? It's pretty obvious to anyone with half a brain thatthe result is what you would expect if the speed of light were sourcedependent. I would say that agreed with the experimental evidence muchbetter than relativity.Androcles === Subject: Re: Squares that end with four integers can have squares that end with four identical digits?I believe that all integers that end in two zeroes have squares that endin four identical digits.If x^2 is an integer that ends in 4 identical digits aaaa thenx^2 is congruent to aaaa modulo 10^4. In this case the generalizedLegendre symbol L(aaaa/10^4) will be 1 and if aaaa is not a square modulo10^4 then it will be -1. Maple's procedure numtheory[quadres] (c,d)implements L(c/d) and the following calculation shows that a squarecan end in 4 identical digits only when they are 0's:> for i from 0 to 9 do> a:=i+i*10+i*10^2+i*10^3:> print(a,numtheory[quadres](a,10^4));> od: 0, 1 1111, -1 2222, -1 3333, -1 4444, -1 5555, -1 6666, -1 7777, -1 8888, -1 9999, -1 === Subject: Re: A potentially rewarding challenge Goodstein sequence, G(m), > only terminates if m < 4.Others in the thread have shown your claim is false; however, for more detail about the structure of these sequences, you might want to see the text 'le athttp://r.s.home.mindspring.com/GoodsteinSequenceswhich shows, incidentally, that the Goodstein sequence that starts at 4 has (two) maximum terms equal to N = (j+1)*2^j-1, where j=3*2^27-1, and terminates at 0 on the (2N)th term.--r.e.s. === Subject: Re: Leonidas AlaogluG. A. Edgar :>If I remember correctly, Leonidas Alaoglu was born in Canada (Red Deer,>Alberta).Perhaps, I do not know much about his life. But my guess is that his parents were Greek immigrants from Asia Minor (Anatolia/Turkey) -- whereGreek refers *not* to Ocitizenship' but to language and/or religion... baloglouAToswego.edu John Harrison :> For those who have the book, it is W. Rudin, Principles of> Mathematical Analysis, chapter 2 ( Basic Topology), problem 18:> [A set E is perfect if E is closed and every point of E is a limit> point of E] Is there a nonempty perfect set in R which contains no> rational number?> Look up the Cantor-Bendixson Theorem. === Subject: 2nd CFV: misc.metric-systemSupersedes: <1067921070.23664@isc.org>Archive-Name: news.announce.newgroups iD8DBQE/tuReXMotZRinPKkRAi2LAJ9+ 8hmR21RIWOsTBRhkRMlAfm5SkQCff0Ob JMmThuHtRd9MxytWJT9ivNc= =kTW0 LAST CALL FOR VOTES (of 2) unmoderated group misc.metric-system[ Note: This document is multiposted in 3 copies because too many large service providers implement crossposting limits and would otherwise drop it. - n.a.n moderation team ]Newsgroups line:misc.metric-system The International System of Units.Votes must be received by neutral third party. Questions aboutthe proposed group should be directed to the proponent.Proponent: Markus Kuhn Votetaker: Bill Aten RATIONALE: misc.metric-systemUnits of measurement and related standards affect many aspects of ourdaily lives. The global standardization of a single consistentInternational System of Units was a major breakthrough for humancivilization and signi'cantly simpli'ed communication, learning,work and trade all over the planet.The introduction of the metric system still faces delays in someareas. Notable examples are consumer communication and traf'cregulations in the United States and United Kingdom, as well as partsof the aeronautical and typographic industry. It is therefore nosurprise that discussions about the metric system §are up regularlyin many different newsgroups. In particular the slow progress withmetrication in the United States promises to fuel such debates formany years to come.A dedicated newsgroup will focus expertise and will provide a mediumfor professionals and hobbyists to 'nd advice and suggestions onmetric product standards and conventions. None of the newsgroups inwhich metric-system issues §are up frequently is particularly suitedfor this topic by charter and readership. The popularity of theexisting US and UK Metric Associations' mailing lists demonstratesthat there is a signi'cant number of people interested in the topic.Considering the important role that units of measurement play ineveryone's life, this promises to become a quite lively newsgroup.The name of the proposed group has been the subject of some debate.The present proposal is motivated by these considerations: - Although discussions about the metric system focus much on its slow progress in a small number of countries, the topic is inherently international in nature and discussions tend to bene't very signi'cantly from world-wide participation. Therefore, placing the group under us.* or uk.* would be inappropriate. - The metric system affects many 'elds, including consumer communication and road traf'c. Discussions about the metric system range from basic science and applied engineering considerations to economic, social, psychological, legal, public policy and media aspects. This excludes sci.* and leaves misc.* as the most appropriate hierarchy. - The metric system is the only system of units used in almost every region and 'eld of application. Imperial and U.S. Customary units are usually discussed in relation to the metric system, which is within the scope of the proposed group. Other unit systems have very limited applications and are better discussed in specialized science or history groups. The proposed group is far more likely to have specialized children rather than equivalent siblings, which speaks against an entire misc.measurement.* hierarchy and justi'es a place directly under misc. - The term metric system remains the most well known and most easily recognized English language term for what is more formally called the International System of Units (SI). This speaks against group names such as *.si or *.metric.The proposed charter has equally been the subject of some debate. Thepresent proposal is motivated by these considerations: - It refers equally to both the of'cial modern name International System of Units (SI) and the colloquial English term metric system. This is in the interest of rapid recognition by both readers and search-engine users. Any further distinction between these terms is deliberately left to explanatory periodic postings. - It is broad enough to cover discussions about different historic variants of the metric system (e.g., CGS, MKS, various European customary units) as well as contemporary units that compete with the SI (e.g., inch, pound, Fahrenheit, calorie). - It is narrow enough to exclude topics that are not related to metric units (e.g., the history and rede'nition of calendars). - It covers product standards and conventions that are not part of any of'cial de'nition of the metric system, but that are closely related to metric units (e.g., metric clothing sizes, paper formats, engineering components, traf'c regulations). These can be considerably more complex topics than the metric system itself, leading to discussions of particular interest to consumers and practitioners. - It leaves room for the possible later creation of a separate misc.metric-system.advocacy group for those with a particular interest in political activities related to metrication. - It was written with the expectation that the topic is unlikely to attract large non-plain-text postings, commercial advertising or unsocial behaviour in any particular way, leaving these issues to common sense and USENET etiquette. - It is brief.CHARTER: misc.metric-systemThis newsgroup is for discussion about the International System ofUnits (SI) or metric system, including its use in scienti'c,technical, and consumer applications, its history and de'nition, andits adoption in 'elds and regions where other units of measurementare still prevalent (metrication). 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When in doubt, ask thevotetaker.DISTRIBUTION:The only of'cial sources for copies of this CFV are the locations listedbelow, the UVV web site at http://www.uvv.org/, and the votetaker's e-mailCFV server which can be reached at .This CFV has been posted to the following newsgroups: news.announce.newgroupsnews.groupscomp.std.internatmisc.transpo rt.roadsci.engrsci.mathsci.physicssoc.culture.canadasoc.culture .europesoc.culture.usaPointers directing readers to this CFV will be posted in these newsgroupsand mailing lists:comp.std.miscde.comp.standardsrec.backcountryMailing list name: U.S. Metric AssociationSubmission address: usma@colostate.eduMailing list name: UK Metric AssociationSubmission address: metric@smartgroups.commisc.metric-system - Ack Bounce List---Because the Vote Ack email bounced, the email addresses and associatedballots listed below are considered invalid and will not be counted in the'nal result. Ballots must be submitted from a valid and verifyable emailaddress in order to be processed. The individuals listed below will needto revote from a valid email address prior to the poll closing in order tocorrect this problem.---chemgurl77.trash [at] hotmail.com chemm.collado [at] aaron.ls.'.upm.es Manuel Collado-- Bill Aten, UVV === Subject: Re: Indian 'rsts in Pifer) :INVENTIONS & DISCOVERIES> INVENTION OF NUMERALS> Numerals are found in the inscriptions of Ashoka The Great in the 3rd> Century BC. This knowledge traveled from there to Europe and West. In> Arab countries even now numerals are known as HINDSE: from India. La> time, It is India that gave us the ingenious method of expressing all> numbers by means of ten symbols - Prof. O.M. Mathew in Bhavan's> JournalINVENTION OF ZERO> Brahmagupta was the 'rst mathematician to treat ZERO (0) as a number> and showed its mathematical operationsINVENTION OF ARITHMETIC> Arithmetic was discovered by Indians in about 2nd Century BC.> Bhaskaracharya's book Lilavathi is regarded as the 'rst book on> modern arithmetic. The Arabs learnt and adopted it from India and> spreaded it to Europe. In 499 AD Aryabhatta 'nished his work> Aryabhatt, giving rules of Arithmetic (Encyclopedia Britannica)INVENTION OF ALGEBRA> In Western Europe the knowledge of Algebra was borrowed, not from> Greece but from Arabs, who acquired this from India. Algebra is the> only Arabic name for Bijaganitha. Aryabhatta was one of the 'rst to> use Algebra (Encyclopedia Britannica)INVENTION OF GEOMETRY AND TRIGNOMETRY> The brick work of Harappa and Mohenjodaro excavations show that people> of ancient India (2500 BC) possessed knowledge of Geometry. Aryabhatta> formulated the rules for 'nding the area of a ?triangle', which led> to the origin of Trignometry.DISCOVERY OF ASTRONOMY> The knowledge of the motion of heavenly bodies was discovered by> Aryabhatta (499 AD), Latadeva (505 AD) and Brahmagupta (628 AD), for> calculating the timing of eclipses. In Surya Sidhanta' Latadeva,> talked about the earth's axis and called it SUMERU. That the earth is> a sphere and it rotates on its own axis, was known to Varahamihira> and other Indian astronomers much before Copernicus published this> theory. (Jewish Encyclopedia)INVENTION OF CALENDAR MAKING> Discovery of measurement of time and discovery of nomenclature of> days, month and years and invention of calendar making was made in> India. In his book ?Surya Sidhanta' Latadeva (505 AD) divided the year> into 12 months. Seven planets of the solar system effect the earth's> atmosphere and their names were added to the seven days of the week,> which was accepted all over the world.DISCOVERY OF THEORY OF GRAVITATION> In his book ?Sidhanta Shiromani' Bhaskaracharya mentions about force> of attraction resembling gravity, discovered centuries later by> Newton. (Jewish Encyclopedia)INVENTION OF IRON PRODUCTS IN 3000 BC> The word AYAS occurs in the four Vedas which denotes iron. Ashoka> pillar at Mehrauli, New Delhi and another iron pillar in Karnataka> stand proof of India's metallurgical heritage (A study published in> the magazine ?The Current Science').INVENTION OF COPPER, BRONZE AND ZINC> The copper and bronze artifacts dates back to Indus Valley> Civilization (2500 BC). According to treatise RASARATNAKAR, zinc was> made in around 50 BC at Zawar in Rajasthan (India).INVENTION OF CHEMICALS PROCESSES, DYES AND CHEMICAL COLORS> Chemistry known as RASAYAN SHASTRA was invented in India. Elphinstone> to prepare the sulphate of copper, zinc and iron and carbonates of> lead and iron. RASAVIDYA or Indian alchemy made its appearance around> 5th Century AD (National Science Centre, New Delhi)At http://ancienthistory.about.com/gi/dynamic/offsite.htm?site= http%3A%2F%2Fmembers.aol.com%2Fbbyars1%2F'rst.html is the title The First Mathematicians.At http://www.stormloader.com/ajy/zero.html is the title Zero. === or emailed replies to this message constitute permission for an E5 5E EC F3 04 26 4E BF 1A 92X-Zippy-Says: Toes, knees, NIPPLES. Toes, knees, nipples, KNUCKLES... Nipples, dimples, knuckles, NICKLES, wrinkles, pimples!! I don't like FRANK SINATRA or his CHILDREN.Joona I Palaste Europe and Asia is being replaced by US things> nowadays. Reverse occurrences are rare to the point of non-existence.Ah well, the US is a replacement of American things (remember theamerinds?) by European things.The UK thing is a replacement of the English thing with a Frenchthing. The English were a replacement of a British thing with anGerman thing.The French are the replacement of a Gallic thing with a German thing.And so it goes.Thomas === Subject: Re: billion (was: Big Number message constitute permission for an emailed EC F3 04 26 4E BF 1A 92X-Zippy-Says: There's enough money here to buy 5000 cans of Noodle-Roni!Richard Heath'eld dictionary, area can certainly mean that, but it has > numerous other meanings as well, including part of a building. If he does > not wish to be misinterpreted, perhaps he should make his meaning clearer.He means: if you don't know the usage of an area, the best guess isto assume that.... And he's right...You will be right more often (by vastly more) if you make hisassumption than the other assumption. === Subject: Re: Implementing the partial difference this message constitute permission for an emailed EC F3 04 26 4E BF 1A 92 If elected, Zippy pledges to each and every American a 55-year-old houseboy...Why don't you just post the program? === Subject: Re: Squares that end with four John Waterford :> Which integers can have squares that end with four identical digits?>[snip Jyrki's solution]10/10 for part 1 Jyrki, now generalise to other bases :-)>Here are some questions for other bases:For base 3: I 'nd that the square of [1, 1, 1, 0, 2, 1, 2, 2, 1, 2, 1]_3is [2, 0, 1, 1, 2, 2, 0, 2, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,1, 1]_3This square ends in 12 ones. Is there a limit to the number of 1's a squarecan end in for base b = 3.For base 5: I 'nd:The square of [1, 3, 4, 4, 0, 3, 3, 1]_5is [3, 1, 0, 1, 4, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1]_5and the square of [3, 2, 4, 3, 1, 2, 1, 2]_5is [2, 2, 4, 1, 2, 3, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4]_5Can the number of 1's and 4's be arbitrary for base 5?--Edwin Clark === Subject: Re: trans'nite seriesTreme: C&C,DWSIn ramsey) said:>Take the in'nite series expansion for e and put it into the in'nite>series expansion of e to the power of x and multiply the terms and>you end up with a series of aleph 1 terms No, you end up with a series of Aleph_0 terms.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Subject: Re: compact de'nition....Treme: C&C,DWSIn , said:>which of de'niton is right??Both. One de'nes a compact space, the other defrines a compact subsetof a topological space. A subset of a topological space is compact if't is a compact space when given the subspace topology. It's a matterof taste which you start with.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Subject: Re: Mathematical IntegrityTreme: C&C,DWSIn said:>Sometimes a student would post an innocent question but he or she>would get scold for posting such a stupid question and would be>called a moron. What does that have to do with integrity? >It seems like these dese days, some mathematicians would use>profanity against others mathematicians.What does that have to do with integrity?>Where is the purity and integrity in mathematics?Neither purity nor integrity has anything to do with civility or withthe ability to suffer fools gladly.>Anyhow, I still love math by heart and I would like to give my>respect to many great math fanatics in this forum.You could best show your respect by not referring to them as fanatics.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Subject: Re: Axiom of Foundation (absymally stupid question) Treme: C&C,DWSIn Elaine Jackson said:>The whole problem is just that you're misquoting the axiom.No he is not.>You say: Every nonempty B contains a y >with (B intersect y) = empty.Not only him. Everybody who uses GBN or ZF says so as well.>I say: Every nonempty B contains a y for which >there is no z with z in y and z in B.In what context do you say it?>My axiomWhat set theory is your axiom a part of? What are the other axioms?>but it allows for the possibility that there exist>citrus fruits that are not sets.Then it's not part of the same set theory, it is your responsibilityto give the complete set of axioms that you are using.>Citrus fruits that have no elements, but aren't the>empty set, are technically called individuals. No. There are sets theories that have individuals without having toabandon extentionality. Again, if you wish to be taken seriously youwill need to state what set theory it is that you are using.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Subject: Re: De'nition of IntegralArtur> I have some questions about the de'nition of the Riemann Stieltjes> integral of a real bounded function f over an interval [a,b], with> respect to a function g, also bounded on [a,b].According to one de'nition (like we see in Bartle's and Apostol's> books), the integral is related to the concept of tagged partition. We> say the integral of f with respect to g is I if, for every eps>0,> there's a partition Pe of [a,b], such that, if P is any re'nement of> Pe, then |S(P,f,g) - I| Riemann-Stieltjes sum of P, f and g over [a,b]. In this de'nition,> the sum is related to tag points chosen in each of the intervals of P,> and inequality (1) must be satis'ed regardless the choice of the tag> points. The only requirement about g is that it is bounded on [a,b].This is equivalent to the next de'nition _if_ g is of bounded variation,because any such g is the sum two functions, one monotonic up and the othermonotonic down.> Other de'nition, like we 'nd in Rudin's book, involves the concepts> of upper and lower sums and there are no tag points. The function g,> however, is required to be monotonically increasing on [a,b]. If P is> a partition of [a,b], the Riemann-Stieltjes sum is de'ned by taking> the in'mum (lower sum) or the supremum (upper sum ) of f on each> interval of P. If the in'mum of the upper sums equals the supremum of> the lower sums, both taken over all possible partitions, themn this> common number is said to be the integral of f with respect to g. Isn't> the fact the g must be increasing a loss of generality when compared> to the 'rst de'nition? If we restrain our attention to the Riemann> integral, it doesn't matterm because then we always hace g(x) =x, but> in the general case it seems to be a loss of generality.For more on this (plus some of the Henstock business, by another name) Isuggest Avner Friedman, _Foundations of Modern Analysis_, in the exercisesfor section 2.11.LH Earlier today I noticed that it is very easy to derive Ceva's theorem from Menelaus' theorem: with the Cevians AD, BE, CF of triangle ABC meeting at G, apply Menelaus' theorem to triangles ABD and ACD and multiply the outcomes;to be precise, (AF/FB)*(BC/CD)*(DG/GA) = -1 and (AG/GD)*(DB/BC)*(CE/EA) = -1lead to (AF/FB)*(BD/DC)*(CE/EA) = +1 via DB/CD = BD/DC, DG/GD = AG/GA = -1, and, *of course*, (-1)*(-1) = +1 :-)The emphasis on the multiplication above alludes to the eccentic observationthat, just as +1 cannot generate -1, Ceva's theorem does not seem to be capable of proving Menelaus' theorem: in view of the well established andcelebrated duality, I 'nd this a bit surprising; moreover, and in view ofthe simplicity of the novel (?!) derivation above, one has to wonder why the two theorems are separated by no less than 16 centuries... Comments, anyone?[I would like to dedicate this post to the memory of my father ChristosBaloglou (1919-2002), who passed away a year ago this week, one and half years after publishing Scattered Drops of Geometry (in Greek).] baloglouAToswego.edu === Subject: Re: Argument with professor> A cross product is a special case of the Clifford wedge product. The> wedge product of two vectors is called a bi-vector. It just so> happens that in R^3 bi-vectors are dual to vectors, so we can> interpret the three dimensional bi-vector as corresponding to a> special type of vector (usually referred to as an axial vector). This> correspondence breaks down in higher dimensions, since the ranks of> dual objects must add up to the number of dimensions. Hence, in R^4,> bi-vectors are dual to themselves and cannot be interpreted as vectors> at all. In even higher dimensions, the situation becomes all the more> hopeless.Did you insult me sometime recently? I can't remember.However, even if it was you, all is forgiven for that nice precis. === Subject: Re: Question on numerical integration (Simpson)Carlos Moreno :>Is the Simpson's method optimal with respect to the>number of calculations? (i.e., does it yield the>lowest error for a 'xed number of calculations?)No.>I know that there may not exact and absolute answer>to this -- but in general, for non-pathological,>non-super-obscure situations, is it true?No. There is no such thing as an optimal method for a 'xednumber of calculations, unless you severely restrict the set of possible functions to be integrated. Each method will give 0 error for some functions, not for others.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: JSH: My use of the|Mathematical Association of America also did a lot of work, 'nding|all sorts of precedents where a person was called a scab, a traitor,|and other nasty things in print and the courts let the authors get|away with it.MAA lawyers-- what a concept! :-)Was the issue of whether it was factually correct that he wasa crank ever raised in court?Keith Ramsay === Subject: Please help with generating function...HelloCould you please help me on the following as I simply do not understandhow to attack the problem?Use generating functions to 'nd the number of ways to select 14 ballsfrom a Jar holding 100 red balls, 100 blue balls and 100 green balls sothat no fewer than 3 and no more than 10 blue balls are selected. Assumethat the order in which the balls are drawn does not matterPLEASE HELP explain how to do this as I have 3 more to do also....muchappreciated...TIA === Subject: Re: Reality of response to my work jstevh@msn.com () :> So before I can make any money, I have to get past mathematicians, the> dark gatekeepers testing their ability to deny knowledge from the> general public.> If JSH really wants to make money, he should try some 'eld where there is more of it than in mathematics. === Subject: Brouwer's 'xed point theorem variationIf A is a retract of B^2 (the disk in R^2), then every continuous map f :A ---> A has a 'xed point.How can I prove this, knowing Brouwer's 'xed point theorem for B^2?Mike John Harrison :>For those who have the book, it is W. Rudin, Principles of>Mathematical Analysis, chapter 2 ( Basic Topology), problem 18:>[A set E is perfect if E is closed and every point of E is a limit>point of E] Is there a nonempty perfect set in R which contains no>rational number?>This one has really had me stumped. My 'rst guess was that the answer>was no, but I didn't get anywhere trying to prove it. I then tried to>construct such a set: since the set of rationals is countable, we can>write it as {r_1, r_2, r_3,...}. Let I_n be the open interval of>length 1/2^n centred at r_n, let I be the union of the I_n, and let X>be the complement of I in R. Since the total length of the intervals>is at most 1, X is nonempty. As the complement of an open set, X is>closed. But I can't show that X is perfect.Your example might not be perfect. However, it can be 'xed.Instead of 1/2^n, make the length be irrational (so all the intervals haveirrational endpoints), and small enough so that either I_n is contained ineither the union of the previous I_k's or its closure is in the interiorof the complement of that union.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: What constitutes an implicit use of a function? (Was: a confusing group theory conundrum) said:>Let P(x) denote the logic statement x < x^2>Let Q(x) denote the logic statement x < f(x)>P and Q are unequal iff there is some real x such that P(x) != Q(x)>Else, they are equalNo. They can be different propositional functions even if their valuesare provably the same.>It seems to me upon inspection (though I keep an open mind to>correctionNo. You haven't kept an open mind.>And since they are equal, if one has a given property, the other>must as well. sniz and SNIZ refer to the same person, so they must be the same.Since the 'rst is lower case, the second must be lower case as well.Do you see the fallacy in that?>This is all quite confusing,To you. Others see the distinctions quite clearly.>know that I am just a naive person who humbly desires>to understand these things.Were that the ask you would ask more and assert less.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Subject: Re: Prime conjecture - puzzleTapio Hurme :>The arithmetic mean of two consecutive primes minus the geometric mean of>the same two consecutive primes equals less than 1/2.>What's the name GM(x,y) = (x+y)/2 - sqrt(xy) = (sqrt(x)-sqrt(y))^2/2so you're saying sqrt(p_(n+1)) - sqrt(p_n) < 1, which is Andrica's Conjecture.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 John Harrison :> For those who have the book, it is W. Rudin, Principles of> Mathematical Analysis, chapter 2 ( Basic Topology), problem 18:> [A set E is perfect if E is closed and every point of E is a limit> point of E] Is there a nonempty perfect set in R which contains no> rational number?This one has really had me stumped. My 'rst guess was that the answer> was no, but I didn't get anywhere trying to prove it. I then tried to> construct such a set: since the set of rationals is countable, we can> write it as {r_1, r_2, r_3,...}. Let I_n be the open interval of> length 1/2^n centred at r_n, let I be the union of the I_n, and let X> be the complement of I in R. Since the total length of the intervals> is at most 1, X is nonempty. As the complement of an open set, X is> closed. But I can't show that X is perfect. So: Is the answer yes or> no? If yes, does my set work as an example? Any hints would be most> welcome.Try showing (1) X is not countable (2) the isolated points of X constitute at most a countable set (3) X {isolated points of X} is closed === Subject: Re: Squares that end with four identical digitsEdwin Clark :>Here are some questions for other bases:>For base 3: I 'nd that the square of> [1, 1, 1, 0, 2, 1, 2, 2, 1, 2, 1]_3>is> [2, 0, 1, 1, 2, 2, 0, 2, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,>1, 1]_3>This square ends in 12 ones. Is there a limit to the number of 1's a square>can end in for base b = 3.No.Suppose 2 x^2 + 1 == 0 mod 3^n (so x^2 ends in n 1's in base 3) but not mod 3^(n+1), where n > 1. Of course x is not divisible by 3. Then2 (x+3^n)^2 - 1 = 2 x^2 - 1 + 3^n x mod 3^(n+1) and2 (x-3^n)^2 - 1 = 2 x^2 - 1 - 3^n x mod 3^(n+1) so one of these == 0 mod 3^(n+1). Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of === British Columbia Vancouver, BC, Canada V6T 1Z2Subject: Re: Please check my boolean simpli'cation Maelstrom :> in the following boolean equation simpli'cation:R = (C1'+C0'+A'+B')(C 1'+A+B)(C1'+C0+A+B)(C1+C0'+A[Ca pitalOTilde])(C1+A+B')>The best and my 2nd O'nal' version, has only four disjunctions.I doubt there's anything shorter.R = (C1'+C0'+A'+B')( C1'+A+B)(C1'+C0+A+B)(C1+C0'+A[C apitalOTilde])(C1+A+B') = (C1'+C0'+A'+B')( C1'+A+B)(C1+C0'+A')(C1+A+B[Capi talOTilde])(C1'+A+B)(C1+A+B') = A + (C1'+B)(C1+B') = A + C1'B' + BC1(C1'+C0'+A'+B[CapitalOTilde ])(C1+C0'+A') = C0' + A' + (C1' + B)C1 = C0' + A' + B'C1R = (C1'+C0'+A'+B')( C1'+A+B)(C1'+C0+A+B)(C1+C0'+A[C apitalOTilde])(C1+A+B') = (C1'+C0'+A'+B')( C1'+A+B)(C1+C0'+A')(C1+A+B[Capi talOTilde]) = (A' + C0' + B'C1)(A + C1'B' + BC1) = A'C1'B' + A'BC1 + C0'A + C0'C1'B' + C0'BC1 + B'C1A = A'C1'B' + A'BC1 + A'C0'C1'B' + A'C0'BC1 + AC0'C1'B' + AC0'BC1 + C0'A + B'C1A = A'C1'B' + A'BC1 + C0'A + B'C1A = A'(B'C1' + BC1) + A(C0' + B'C1)---- === Subject: Re: Please help with generating Hello> Could you please help me on the following as I simply do not understand> how to attack the problem?Use generating functions to 'nd the number of ways to select 14 balls> from a Jar holding 100 red balls, 100 blue balls and 100 green balls so> that no fewer than 3 and no more than 10 blue balls are selected. Assume> that the order in which the balls are drawn does not matterPLEASE HELP explain how to do this as I have 3 more to do also....much> appreciated...TIA You want the coef'cient of x^14 in (x^3 + x^4 + ... + x^10)*(1 + x + x^2 + ...)^2 = x^3*(1 + x + x^2 + ... + x^7)*(1 + x + x^2 + ... )^2.So, you want the coef'cient of x^11 in (1 + x + x^2 + ... + x^7)*(1 + x + x^2 + ... )^2 =[(1 - x^8)/(1 - x)]*[1/((1-x)^2)] =(1 - x^8)/((1 - x)^3) = (1 - x^8)*(1 - x)^(-3).Now, use the generalized binomial theorem and note there aren't toomany ways to get x^11.-- Post in haste; repent at leisure. Paul SperryColumbia, SC (USA) === Subject: Re: Factorial/Exponential Identity, In'nityFish, or cut bait.Solve problems.Borel says almost all sequences are normal and have equal zero and onedensities, combinatorics say almost no sequences do, I think half do.How is reconciled Borel and asymptotic combinatorics?Sets of numbers contain only === sets.RossSubject: Re: Bredon corrections. teaching myself homotopy theory> from GB's Geometry and Topology book, and noticed a shed load of typos. Does> anyone know if they've all been catalogued and whether I could access a list> of them somewhere?>What you think the typos? What's the 4 or 5 simplest? === Subject: Re: Largest number ever written down?> C.Stevens scribbled the following:> Joona I Palaste -->> If you allow 'nite numbers constructed from a series of mathematical>> formulae, I think Graham's number takes the cake, by far.> FYI,Harvey Friedman's lower bound for n(4) in his block subsequence theorem is> much larger than Graham's number- it involves the Ackermann numbers A(n,n).> let A(n) equal A(n,n)> A(A(A........(A(1)..) where there are A(187,196) A'sAren't Ackermann's functions something like A(1,1)=1+1, A(2,2)=2*2,> A(3,3)=3^3, A(4,4)=(((4^4)^4)^4)^4, and so on? In that case A(187, 196)> must be pretty huge.> I read that webpage about writing big numbers on a piece of paper. If> I am allowed to refer to Ackermann's function only by name, I think I> can write a 'nite natural number that is much larger than Harvey> Friedman's number.Actually,Friedman uses a streamlined Ackermann hierarchy,to base 2,whereA(1,n) is doubling, A(2,n) is 2^n, A(3,n) is 2^^n,etc.In the expression above,he uses the unary form--A(n) = A(n,n).So the startingpoint, A(1) = A(1,1) = 2.And, there are A(187,196) = A(187196,187196) A's in total, which dwarfs theGraham number.How you would Ackermann down that many times,I haven't a clue. === Subject: Re: PSWFsOn Thu, point me toward a good method for evaluating the prolate> spheroidal wave functions?There's a book Computation of Special Functions, by Shan-jie Zhangand Jianming Jin. The programs themselves are available on theweb, in Fortran-77. Here:http://iris-lee3.ece.uiuc.edu/~jjin/routines/ routines.html-- Andrew === Subject: Re: billion (was: Big Number GameThomas Bushnell, BSG :> You will be right more often (by vastly more) if you make his> assumption than the other assumption.[Slartibartfast] gave a hollow laugh. OWhat does it matter? Science has achieved some wonderful things of course, but I'd far rather be happy than right any day.'I am content with my great big billions, and I will continue to use them when appropriate (which is hardly ever, of course!). If others are content with milliscopic billions, then that's entirely up to them. :-)-- Richard Heath'eld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === Subject: Re: ?Any good, feedback...I seem to have many choices;so I'll check out the various things you suggested and see how they go.Gene === Subject: Re: De'nition of IntegralFirst, when distinguishing different de'nitions,you should leave out the Stieltjes part -- thatis sort of an extra which can be added on later,if desired.I will refer to your de'nitions as: (in orderof appearance):(1) tagged partition(2) inf/sup(3) mesh limit(4) gauged partitionThere are three popular de'nitions of the integralin semi-common use today. They are the Riemann,Lebesgue, and Henstock (sometimes called Henstock-Kurzweil). Lebesgue is more powerful than Riemann,and Henstock is more powerful than Lebesgue. In myopinion, the Riemann de'nition is far out-dated andshould be left behind. The Lebesgue de'nition isvery beautiful, in that it gives a very nice Fund.Thm of Calc, and it basically can be generalizedto 'nd the integral of any function from ameasure space to a Banach space. Lebesgue is alsothe default among many mathematicians.Henstock, as far as I know, is not so easy togeneralize. So how is it more powerful? Take theanalogy of in'nite series. We have absoluteconvergence (series which are well-behaved), andconditional convergence (not as well-behaved).Go from countable sums to continuum-sized sums:Lebesgue captures the idea of absolute convergencehere, but not conditional convergence. Henstockgets both. Why can't Henstock generalize as easily?Because for conditional convergence, you need toinclude the idea of order within the sum (sincea conditionally convergent sum depends on theorder of the summands), and a measure space missesthe idea of order.Finally, to your de'nitions. I must admit thatI am not 100% sure about (1), but it looks very muchlike Henstock to me. (4) is de'nitely Henstock.And (2) and (3) are both Riemann. Where is Lebesgue?I don't know -- I 'nd it curious that you are missingit. Maybe you just haven't learned measure theory yet?You should -- fun stuff!Hope that helps, Tyler === Subject: The Gauge IntegralI'm looking for some resources to learn a bit more about the gauge integral (Alternately called the Generalised Riemann Integral and the Henstock-Kurzweil integral or probably a couple others).It seems like an extremely neat idea (although for most purposes I'd probably prefer lebesgue integration anyway, as it's so much more general) and worth learning more about. Also I'm in the process of (slowly) writing an analysis textbook, and it seems a fairly natural extension of the way I'm introducing Riemann integration, so I thought it might be worth pursuing to see if it's a good idea to include it.At the moment I know the de'nition, and a couple basic facts about it - not much more than I learned from http://www.math.vanderbilt.edu/~schectex/ccc/gauge/ .An online treatment would of course be preferable as, well, I wouldn't have to pay money for it. :) But if you can reccomend some good books on the subject that would be great as well.Now lets see if this gets through... my college newsserver has been === Subject: Re: Axiom of Foundation (absymally stupid question)Shmuel (Seymour J.) Metz PM, Elaine Jackson said:||>The whole problem is just that you're misquoting the axiom.||No he is not.||>You say: Every nonempty B contains a y |>with (B intersect y) = empty.||Not only him. Everybody who uses GBN or ZF says so as well.In a careful presentation, it would be usual to write it out in termsof epsilon and =, and not de'ned terms like intersect which technicallyare not part of the language of ZFC.I did a web search, and I see that on Mathworld the axiom is given as aform of the axiom scheme of epsilon-induction, which again does not involvereferring directly to intersections.|>I say: Every nonempty B contains a y for which |>there is no z with z in y and z in B.||In what context do you say it?The point here is that speaking of intersections presupposes that theobjects in question are sets, whereas set theories with urelementstypically consider epsilon to be a relationship on the whole domain,including both urelements and sets. So saying there doesn't exist a zsuch that z in y and z in B allows for the possibility that y or Bis an urelement (like a piece of fruit). Thus the same statement ofthe foundation axiom would suf'ce for a set theory with urelementsof this kind.|>My axiom||What set theory is your axiom a part of? What are the other axioms?ZFU would be a familiar example of this kind of set theory.|>but it allows for the possibility that there exist|>citrus fruits that are not sets.||Then it's not part of the same set theory, it is your responsibility|to give the complete set of axioms that you are using.Seymour Metz is being overly formal again.|>Citrus fruits that have no elements, but aren't the|>empty set, are technically called individuals. ||No.But they are.|There are sets theories that have individuals without having to|abandon extentionality.She didn't say this was the only sense in which the term was used. True,another way to model urelements/atoms/individuals is to have them bearthe epsilon relation to themselves. But that requires either abandoningor adjusting the foundation axiom.|Again, if you wish to be taken seriously you|will need to state what set theory it is that you are using.The context was adequate.Keith Ramsay === Subject: Re: what spin 1/2 is in physical reality Re: === Robertson versus Philips screwdriver Re:Subject: Re: what spin 1/2 is in physical reality Re: Robertsonversus Archimedes Plutonium NOdtgEMAIL whole entire Universe is just one big atom where dots ofthe electron-dot-cloud are galaxies sci.physics, sci.math, Sefton :> http://rapfast.petcom.com/~john/Be.GIF> Start a point orbitting a globe at a 'xed radius> from the north to the south pole.> At the same time take this circular orbit and> start it precessing at twice the rate at which> the point is travelling within the orbit.> The point will make one complete trip around the globe> in the east-west or west-east direction (whichever> way you precessed it) by the time it gets> to the south pole. Then it will make another> trip around, cutting all the lines of longitude,> on its way back up to the north pole.> Place another point exactly opposite the> 'rst. It follows exactly the same path BUT> it is THE ONLY OTHER POINT which follows that> pathway. Divide the ring into four, then eight, then> sixteen points. You will 'nd that sixteen points> completes the galaxy pattern:Not sure why a nice pattern should have any meaning for physicsitself.http://rapfast.petcom.com/~john/ galaxypattern.gif> Then consider this; growth of rings of 16> can be directly related to the Periodic Table:> http://rapfast.petcom.com/~john/periodicpattern.GIFTHIS is what spin * (Alt0189) is in physical reality.> JohnI have seen the Periodic Table placed into a circular pattern andthereis reallyno connection with a geometric 'gure that has 16 points. One can justas easily seek out a nice pattern with triangles or trapezoids where anumber of 16 comes into play and then claim that it relates to thePeriodic Table.No, John, I do not buy your above. But your above does con'rm mysuspicions in the way that you need 2 prime movers. You needorbitingand you needprecession and that ties and 'ts with my need for 2 forces asrepresented by2 rectangles inside a circumscribing ellipse (or the 3rd dimensionalequivalents).Both of us need 2 entities, better to call them forces.Yours John is purely a fanciful geometry pattern.Mine however, is more than geometry but tells of the Prime Mover ofspininside or interior where at least 2 forces are diametrical to oneanother. Inside an electron which has spin 1/2 there are the twoforcesof Gravity and Antigravity and they can be visualized as a Philipsheadof a cross where one force is one line of the cross and the otherforceis the other line of the Philips cross.And since forces are in motion the Philips head embedded in eachelectronis spinning. Not that there exists the screwdriver to cause theelectronto turn and spin. But that the interior forces of Gravity andAntigravity themselves generate the spin motion.would havespin 3/2 and then you would have a different Philips head patternbecause you have more than 2 diametrically opposing internal forces.Sorry to say John, yours seems to be no more than a cute mathematicalpattern and nothing of any Physics. At least nothing I can salvage atthe moment.Archimedes Plutonium, a_plutonium@hotmail.comwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies === Subject: Re: The Gauge Integral R. MacIver :> I'm looking for some resources to learn a bit more about the gauge > integral (Alternately called the Generalised Riemann Integral and the > Henstock-Kurzweil integral or probably a couple others).It seems like an extremely neat idea (although for most purposes I'd > probably prefer lebesgue integration anyway, as it's so much more > general) and worth learning more about. Also I'm in the process of > (slowly) writing an analysis textbook, and it seems a fairly natural > extension of the way I'm introducing Riemann integration, so I thought > it might be worth pursuing to see if it's a good idea to include it.At the moment I know the de'nition, and a couple basic facts about it - > not much more than I learned from > http://www.math.vanderbilt.edu/~schectex/ccc/gauge/ .An online treatment would of course be preferable as, well, I wouldn't > have to pay money for it. :) But if you can reccomend some good books on > the subject that would be great as well.I suppose there must be online lecture notes abot this, but I don't knowthem. Try this book: Lee Peng Yee & Rudolf Vyborny Integral: An Easy Approach After Kurzweil and Henstock Santos === Subject: Re: Brouwer's 'xed point theorem :> If A is a retract of B^2 (the disk in R^2), then every continuous map f :> A ---> A has a 'xed point.How can I prove this, knowing Brouwer's 'xed point theorem for B^2?Let r be a retraction of B^2 onto A. Then f r (the composition) mapsB^2 into A (which is a subset of B^2), therefore has a 'xed-point x. But x is in the range of f, hence in A, hence r(x) = x, hence x =f(r(x)) = f(x), QED.This is one of the simplest and most fundamental tricks in 'xed-pointtheory...--Ron Bruck === Subject: Analysis question (primitives)Hi all,Let a and b be two real numbers with a < b and consider a function ffrom [a,b] into the reals which has the intermediate value propertyand is Riemann-integrable. Let F(x) be the integral of f between aand x.Question: Must F be a primitive of f?Best === Subject: Re: The Gauge IntegralJos.8e Carlos Santos :> R. MacIver :>> I'm looking for some resources to learn a bit more about the gauge >> integral (Alternately called the Generalised Riemann Integral and the >> Henstock-Kurzweil integral or probably a couple others).>> It seems like an extremely neat idea (although for most purposes I'd >> probably prefer lebesgue integration anyway, as it's so much more >> general) and worth learning more about. Also I'm in the process of >> (slowly) writing an analysis textbook, and it seems a fairly natural >> extension of the way I'm introducing Riemann integration, so I thought >> it might be worth pursuing to see if it's a good idea to include it.>> At the moment I know the de'nition, and a couple basic facts about it >> - not much more than I learned from >> http://www.math.vanderbilt.edu/~schectex/ccc/gauge/ .>> An online treatment would of course be preferable as, well, I wouldn't >> have to pay money for it. :) But if you can reccomend some good books >> on the subject that would be great as well.> I suppose there must be online lecture notes abot this, but I don't know> them. Try this book: Lee Peng Yee & Rudolf Vyborny Integral: An Easy Approach After Kurzweil and Santos> Cambridge University Press? Woo hoo! Student discount. :)I thought I'd seen a book by CUP about it, and I was there the other day looking for it without succes, but I was looking under Gauge integration rather than Kurzweil and Henstock. Glad to know I'm not going completely insane. I shall go take a look the obvious way) === Subject: Re: Geometrical proof of the have suggested is true, I do not seehow that is related to the maximization problem I am thinking about...The intuition can explain how can one decompose H as V D V^T with V orthonormal, but I cannot see how it can be related to the maximizationproblem I am thinking about.Mind elaborating? :-)Larry Hammick :> Hiu Chung Law>> Hi all,>> It is well-known that, for a symmetric matrix H of size n by n,>> the following optimization problem>> Maximize 1/2 x' H x + 1/2 y' H y>> subject to ||x|| = ||y|| = 1, x' 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 x' 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 the> largest eigenvalue is duplicated, then the two summands in> 1/2 x' H x + 1/2 y' H y> are the same, so the geometric interpretation is similar to that in the> 'rst case.> I'm not a good arm-waver :)> LH === Subject: Re: billion (was: Big Number Game :>Thomas Bushnell, BSG :> You will be right more often (by vastly more) if you make his>> assumption than the other assumption.[Slartibartfast] gave a hollow laugh. OWhat does it matter? Science has >achieved some wonderful things of course, but I'd far rather be happy than >right any day.'Ha! You didn't transcribe the (important) next two lines! === Subject: Re: importance of Riemann hypothesistylerneylon@yahoo.com (Tyler Neylon) in|one post, so what I'd really like to see are|subject, or any good books which discuss some of the|fruitful implications the hypothesis has.It tells us facts about the distribution of primes. One usualway to prove the number of primes ::> 18. Let G be the group of all real 2-by-2 matrices:> (a b):> (c d), with ad - bc non-zero, under matrix multiplication, and let N:> be the subgroup consisting of those elements of G with ad - bc = 1.:> Prove that N contains G', the commutator subgroup of G.:> No. 18 is OK, I think. Any matrix of the form A B A^(-1) B^(-1) has:> determinant 1, and thus so do all products of elements of this form.:> Thus G' is contained in N.::> 19. In problem 18 show, in fact, that N = G'.: I assume, that your matrices have their coef'cients in some commutative: 'eld k. If k* are the nonzero elements of k (with multiplication),: you can verify that: (a b): (c d) --> ad-bc: does indeed give a surjective homomorphism det: G ---> k*.: This will exhibit N = ker(det) and G/N = k* commutative. >All true, but this only shows that N contains G'; the hard part is >the other direction (problem 19, show that N = G'), which is not >coming to me right now.Apply theorem: Let G be a group with commutator subgroup G'.(a) The subgroup G' is normal in G, and the factor group G/G' is abelian.(b) If N is any normal subgroup of G, then the factor group G/N is abelian if and only if G' is contained in N.-- G metabelian?Now from N = G' is it possible to show N is Abelian?In general, when G is invertible nxn matrices with *,then N = G' as shown above.Is it still true that N is Abelian?That is do rigid rotations commute?It seems so. How's it shown and does N = G' help?---- === Subject: Re: Need help on algebraic geometryAndersbrasil :>>Can you give a page reference? It is page 41 in the last version of those course notes (page 35 for those> using the old version). If you can help me I would be very grateful.Gamma(D(h_i),O_V) is the ring of regular functions on D(h_i).This is a subring of the 'eld of rational functions on V(Milne denotes this 'eld as k(V)). Milne is asserting that,as subrings of k(V), the intersection Gamma(D(h_i),O_V)is Gamma(U,O_V), the ring of regular functions on U.Naturally each regular function on U is regular on D(h_i)so that Gamma(U,O_V) is a subring of, and so a subring of their intersection. On the other hand if f in k(V) lies inall Gamma(D(h_i),O_V), then it is a regular function on U dueto the sheaf property of O_V. (A function regular on an opencover of U is regular on U).-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Argument with professorIgor :> A cross product is a special case of the Clifford wedge product. Really this is an exterior algebra construction, not a Cliffordconstruction.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: billion (was: Big Number Game Discussion, :> at least a county, rather than an individual dwelling.According to my dictionary, area can certainly mean that, but it has > numerous other meanings as well, including part of a building. If he does > not wish to be misinterpreted, perhaps he should make his meaning clearer.If you're going to willfully misinterpret commonplace uses of area,he can hardly hope to include your usage of words generally, includingbillion, can he?-- Jesse HughesLOL. How arrogant you are. Now when you realize that I DID proveGoldbach's conjecture and that I proved Fermat's Last Theorem as well,how are you going to feel then? -- === Subject: Question on generation of large prime numbersI have looking at a few web pages dealing with the largest knowncalculated primes and a great deal of computational time is takinginto searching for these numbers and verifying they are primes. I haveseen the Euclid's proof of in'nitude primes and it occurs that methat super-large prime numbers can be calculated using the following:p1=2 < p2=3A large prime number, p, can be generated usingp = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbersexceeding currently known largest primes can be generated ratherquickly. For example, we calculate the 'rst million primes, multiplythem and just add one.If this is true, then calculating super-large primes should be acomputational trivial task, shouldn't it? Then why the big fuss overcalculating large primes?I suspect that I am missing fundamental here. Can super large primesreally be calculated using the trivial method outlined above? Hopethat someone can enlighten me here. === Subject: Question on generation of large prime numbersI have looking at a few web pages dealing with the largest knowncalculated primes and a great deal of computational time is takinginto searching for these numbers and verifying they are primes. I haveseen the Euclid's proof of in'nitude primes and it occurs that methat super-large prime numbers can be calculated using the following:p1=2 < p2=3A large prime number, p, can be generated usingp = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbersexceeding currently known largest primes can be generated ratherquickly. For example, we calculate the 'rst million primes, multiplythem and just add one.If this is true, then calculating super-large primes should be acomputational trivial task, shouldn't it? Then why the big fuss overcalculating large primes?I suspect that I am missing fundamental here. Can super large primesreally be calculated using the trivial method outlined above? Hopethat someone can === Subject: Re: plotting elliptic curves for LaTeX>gnuplot can do such plots and it can also output LaTeX code. Also, Postscript plots>can be imported using PSTricks.>could you give me an example code? given a polynomial f(x,y) .. I want to plot the points that solve Capco === Subject: Re: Limit at in'nityThis is just to thank you is possible to de'ne the limit of a> function f as x tends to +oo, where f is de'ned on R^+, except on an> in'nite countable number of points of R^+ (for instance, except on> the set of integers).or for instance, something like gamma(-x) ...I think you can safely use the standard limit de'nitions:> For all P>0, there is a Q such that for all x of dom(f):> x > Q ==> f(x) < -P> For all e>0, there is a Q such that for all x of dom(f):> x > Q ==> |f(x)-b| < e> For all P>0, there is a Q such that for all x of dom(f):> x > Q ==> f(x) > PI think these de'nitions work 'ne for the functions you have in mind.> Of course the gamma(-x) would have no limit for x --> +ooDirk Vdm === Subject: Re: Question on generation of large prime numbers> it occurs that me> that super-large prime numbers can be calculated using the following:p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = .... Alas, it usually doesn't work. 2*3*5*7*11*13+1 = 30031 = 59*509: not prime; See http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha102. htm http://www.research.att.com/projects/OEIS?Anum=A014545 Subtracting one at the end is also a good try, and usually fails to produce a prime.--Don Reble djr@nk.ca === Subject: Re: 01:51:47 -0800, bluebear@pokernetwork.com (BlueBear) :[...]p1=2 < p2=3> A large prime number, p, can be generated using> p = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5> p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7> p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbers> exceeding currently known largest primes can be generated rather> quickly. For example, we calculate the 'rst million primes, multiply> them and just add one.If this is true, then calculating super-large primes should be a> computational trivial task, shouldn't it? Then why the big fuss over> calculating large primes?There are at least three problems in relying on this constructionto produce primes:Firstly in the main real-life application of large prime numbers,namely cryptography, the primes must be hard (in practice impossible)to guess, and for this they must be taken from a large/random Opool'of possibilities. Relying on a simple recursive formula would meanyour decrypter would 'nd it trivial to run through all practicalpossibilities.That brings us to the next problem - Values produced by the abovemethod very soon become impractically large, although you couldgeneralize it by spreading the factors between two factors, i.e.p_{n+1} = p_1.p_3.p_4...p_m + p_2.p_5...p_n. This also increasesthe pool of values, which somewhat ameliorates the 'rst problem.But I've saved the worst snag until last - Values obtained by yourconstruction, and the two-product (jr@adslate.com)Eternity is a long time, especially towards the end. Woody Allen === Subject: Re: Prime conjecture - puzzleRight! Maybe too easy. :-)Tapio> Tapio Hurme :>The arithmetic mean of two consecutive primes minus the geometric mean of>the same two consecutive primes equals less than 1/2.>What's the name of the original - sqrt(xy) = (sqrt(x)-sqrt(y))^2/2> so you're saying sqrt(p_(n+1)) - sqrt(p_n) < 1, which is Andrica's> Conjecture.Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2> === Subject: Semi-normal numbers Was: if x has normally distributed digits, Myerson :>The set of non-normal numbers has Lebesgue measure zero. >If you pick a real uniformly at random from the interval [0, 1] >then with probability 1 you have picked a normal number. >This leads many to take the position that if some number pops up >somewhere & there's no good reason to think it's not normal - some >number like pi, or like the reciprocal of the number you get when >you replace every other digit in the decimal expansion of pi with >a zero - then it's safe to assume that the number is normal.I started thinking about this and the same result could possibly beextended to tell something about the reciprocals of normal numbers.Please inform if I've made a logical error with the de'nition ofalmost all.---De'nition:A normal number whose reciprocal is non-normal is called a seminormalnumber. A number whose reciprocal is normal is called an inverse-normal number.Theorem: Almost all real numbers are inverse-normal.Proof:Assume two subsets of R called R_1 and R_2 so that for each r inR{0}, r is either in R_1 or R_2 and its reciprocal 1/r is in theother one. This divides the reals to two subsets and creates thebijection:T: R_1->R_2, T(r) = 1/rbetween them. Now, assume the set of all seminormal numbers in R_1called S_1. We call the image of S_1:U_1 = {T(s) | s in S_1}By de'nition the set U_1 contains only non-normal numbers. Sincealmost all reals are normal, we know that almost all numbers in R_2don't belong in U_1.Since there is a bijection between S_1 and U_1, we know that almostall numbers in R_1 don't belong in S_1 either, so we have proved thatalmost all numbers in R_1 aren't semi-normal. We now use the sameargument to obtain a similar result for R_2.Since almost all numbers in R_1 union R_2 = R{0} aren't non-normaleither, the only option is that almost all numbers in R are in factinverse-normal. === Subject: Re: My research, publication announcement> john_correy@yahoo.com (John) agenda>--constraints of scienti'c dogma and orthodoxy>--by de'nition, *peer* reviewers are usually limited to the same>paradigm and philosophical traditionsFalse. Counter-example: me.I'm sure you're in a class quite by yourself. === Subject: Re: Question on generation of large prime numbers bluebear@pokernetwork.com (BlueBear) :> I have looking at a few web pages dealing with the largest known> calculated primes and a great deal of computational time is taking> into searching for these numbers and verifying they are primes. I have> seen the Euclid's proof of in'nitude primes and it occurs that me> that super-large prime numbers can be calculated using the following:p1=2 < p2=3> A large prime number, p, can be generated using> p = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5> p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7> p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbers> exceeding currently known largest primes can be generated rather> quickly. For example, we calculate the 'rst million primes, multiply> them and just add one.> If this is true, then calculating super-large primes should be a> computational trivial task, shouldn't it? Then why the big fuss over> calculating large primes?I suspect that I am missing fundamental here. Can super large primes> really be calculated using the trivial method outlined above? Hope> that someone can === Subject: Re: Please check my boolean simpli'cation - NEW VERSION!> Dirk Van draw truth tables and calculate prime implicants,> feel free to download from> http://users.pandora.be/vdmoortel/dirk/Boole/boole.html> (carful: you'll have to replace C0 with C and C1 with D)This a new version (1.1) of the program.It had a little bug. Please replace.Dirk Vdm === Subject: Re: Please check my boolean following boolean equation simpli'cation:> R = (C1'+C0'+A'+B')( C1'+A+B)(C1'+C0+A+B)(C1+C0'+A[C apitalOTilde])(C1+A+B')> The best and my 2nd O'nal' version, has only four disjunctions.> I doubt there's anything shorter.R = (C1'+C0'+A'+B')( C1'+A+B)(C1'+C0+A+B)(C1+C0'+A[C apitalOTilde])(C1+A+B')> = (C1'+C0'+A'+B')( C1'+A+B)(C1+C0'+A')(C1+A+B[Capi talOTilde])(C1'+A+B)(C1+A+B') = A + (C1'+B)(C1+B')> = A + C1'B' + BC1> (C1'+C0'+A'+B')( C1+C0'+A') = C0' + A' + (C1' + B)C1> = C0' + A' + B'C1R = (C1'+C0'+A'+B')( C1'+A+B)(C1'+C0+A+B)(C1+C0'+A[C apitalOTilde])(C1+A+B')> = (C1'+C0'+A'+B')( C1'+A+B)(C1+C0'+A')(C1+A+B[Capi talOTilde])> = (A' + C0' + B'C1)(A + C1'B' + BC1)> = A'C1'B' + A'BC1 + C0'A + C0'C1'B' + C0'BC1 + B'C1A= A'C1'B' + A'BC1 + A'C0'C1'B' + A'C0'BC1> + AC0'C1'B' + AC0'BC1 + C0'A + B'C1A> = A'C1'B' + A'BC1 + C0'A + B'C1A> = A'(B'C1' + BC1) + A(C0' + B'C1)Neat.Verifying this, it helped me spot a little bug in my program.Version 1.1 put in in === FPU emulator and as expected, itcontained a signi'cant number of constants. A fair amount of these are very obvious, but a quite a few are complete mysteriesto me. For all but three I 'gured out the (likely) relationship between the successive terms, but as for what they are al used? I hope someone in this group can tell me or point me to a sitethat has information about mathematical series. The constants are:-0.005208333333333333423683514 -1/192+0.000048828124999993646575269 1/20480 *320/3-0.000000544956752147722229412 -1/1835008 *448/5+0.000000006622737896066696605 1/150994944 *576/7-0.000000000084664710789036923 -1/11811160064 *704/9+0.000000000001118158950846881 1/big *832/11-0.000000000000014377930269982 -1/bigger *960/13+0.12451171875+0.498751787292275571965858085+ 0.000262287674813145674657804+0.249050582915605093375140816+ 0.498101165831210186750281632+0.002034561246815280125422448+ 0.249977256935346610686307297+0.482158790777797965568110885+ 0.798637441128639956279457873+0.250000000000000000162630325 1/4+0.041666666666666666576316485 1/24 /6+0.005208333333333335212617098 1/192 /8+0.000520833333333347420733561 1/1920 /10+0.000043402777777744446993657 1/23040 /12+0.000003100198412569815306638 1/322560 /14+0.000000193762401041715562333 1/5160960 /16+0.000000010764578285932073498 1/92897280 /18+0.000000000538227910807861920 1/1857945600 /20+0.000000000024464188541715309 1/408748032000 /22+0.000000000001021270795795864 1/9809952768000 /24+0.000000000000039701152521748 1/255058771968000 /26+0.020833333333333334345255360 1/48 *48/1+0.000781250000000021185310450 1/1280 *80/3+0.000034877232141512120363463 1/28672 *112/5+0.000001695421027976551104493 1/589824 *144/7+0.000000086697509034958444851 1/11534336 *176/9+0.000000004585603636012795109 1/218103808 *208/11+0.000000000246930795093098370 1/4026531840 *240/13+0.000000000015363636849275580 1/73014444032 === Prinsprino@bigfoot.comSubject: Re: what spin 1/2 is in physical reality Re: Robertson versus Philips screwdriver Re: Electron spin of 1/2 is just a myth invented to fudge over bad physics!!!!!It is magnetic §ux that is quantised. The hydrogen atom in ground statecontains one quantum §uxoid most of which tightly wraps the electronsorbit. The electrons behaviour is coverned by the laws of induction. In thenext orbital, two §uxoids 'll the orbit. Either they both tightlly wrapthe orbit, or one tightly wraps it wit the other free to link with other§uxoids. So in an experiment to measure the magentic moment, you loose atleast one unit of magnetic moment.Seehttp://users.powernet.co.uk/bearsoft/ncqt.pdfAndhttp ://users.powernet.co.uk/bearsoft/ClasAtom.htmwhich will Harveybruce@bearsoft.co.ukThe Alternative Physics === Sitehttp://users.powernet.co.uk/bearsoft> Subject:> Re: what spin 1/2 is in physical reality Re: Robertson> versus Archimedes Plutonium whole entire Universe is just one big atom where dots of> the electron-dot-cloud are galaxies> sci.physics, sci.math, Sefton :> http://rapfast.petcom.com/~john/Be.GIF> Start a point orbitting a globe at a 'xed radius> from the north to the south pole.> At the same time take this circular orbit and> start it precessing at twice the rate at which> the point is travelling within the orbit.> The point will make one complete trip around the globe> in the east-west or west-east direction (whichever> way you precessed it) by the time it gets> to the south pole. Then it will make another> trip around, cutting all the lines of longitude,> on its way back up to the north pole.> Place another point exactly opposite the> 'rst. It follows exactly the same path BUT> it is THE ONLY OTHER POINT which follows that> pathway. Divide the ring into four, then eight, then> sixteen points. You will 'nd that sixteen points> completes the galaxy pattern:Not sure why a nice pattern should have any meaning for physics> itself.> http://rapfast.petcom.com/~john/galaxypattern.gif> Then consider this; growth of rings of 16> can be directly related to the Periodic Table:> http://rapfast.petcom.com/~john/periodicpattern.GIF> THIS is what spin * (Alt0189) is in physical reality.> JohnI have seen the Periodic Table placed into a circular pattern and> there> is really> no connection with a geometric 'gure that has 16 points. One can just> as easily seek out a nice pattern with triangles or trapezoids where a> number of 16 comes into play and then claim that it relates to the> Periodic Table.No, John, I do not buy your above. But your above does con'rm my> suspicions in the way that you need 2 prime movers. You need> orbiting> and you need> precession and that ties and 'ts with my need for 2 forces as> represented by> 2 rectangles inside a circumscribing ellipse (or the 3rd dimensional> equivalents).Both of us need 2 entities, better to call them forces.Yours John is purely a fanciful geometry pattern.Mine however, is more than geometry but tells of the Prime Mover of> spin> inside or interior where at least 2 forces are diametrical to one> another. Inside an electron which has spin 1/2 there are the two> forces> of Gravity and Antigravity and they can be visualized as a Philips> head> of a cross where one force is one line of the cross and the other> force> is the other line of the Philips cross.And since forces are in motion the Philips head embedded in each> electron> is spinning. Not that there exists the screwdriver to cause the> electron> to turn and spin. But that the interior forces of Gravity and> Antigravity themselves generate the spin motion.would have> spin 3/2 and then you would have a different Philips head pattern> because you have more than 2 diametrically opposing internal forces.Sorry to say John, yours seems to be no more than a cute mathematical> pattern and nothing of any Physics. At least nothing I can salvage at> the moment.Archimedes Plutonium, a_plutonium@hotmail.com> whole entire Universe is just one big atom where dots> of the === electron-dot-cloud are galaxiesSubject: Re: Question on numerical integration (Simpson) In practice, it is a trade off between the number of divisions and andnumber of bits of accuracy of your computer.If you keep repeatig, but each time doubling the number of divisions, youwill see the answers home BruceBruce Harveybruce@bearsoft.co.ukThe Alternative Physics Sitehttp://users.powernet.co.uk/bearsoft> Carlos Moreno :>Is the Simpson's method optimal with respect to the>number of calculations? (i.e., does it yield the>lowest error for a 'xed number of calculations?)No.>I know that there may not exact and absolute answer>to this -- but in general, for non-pathological,>non-super-obscure situations, is it true?No. There is no such thing as an optimal method for a 'xed> number of calculations, unless you severely restrict the set> of possible functions to be integrated. Each method will give> 0 error for some functions, not for others.Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2> === Subject: probability .........let~ rate 0.002 of decayed tooth patients are a cancer of mouth patient.Until 'rst cancer of mouth patient have discovered in any dentist'sconsultation room, we diagnose.'nd the expected value whether we treat and you must diagnose the couple ofpatients.--um......i think.....answer is reciprocal of 1/500it is right?? === Subject: probability 2......we cuts a two place into a wire at random.let length of wire is 1We got the wire of a three piece.when we made a triangle from three wire piece,'nd that probability of possibility.---------very dif'cult......um....help....me === please.......Subject: Re: 've vectors You are confusing vectors with the algebra of real numbers!!!!!Vectors are mathematical representations of real physical entities. Aproblem is meaningless unless you de'ne the nature of the Alternative Physics Sitehttp://users.powernet.co.uk/bearsoft> Hallo,Ich have given 5 vectors and have to 'nd a> base for V = L(v1, v2, v3, v4, v5). Then I must express the 'fth> vector through this base.I think the general approach is to try every possible linear> combination:1.) a*v1+b*v2+c*v3+d*v4 = v5> or> 2.) a*v1+b*v2+c*v3+d*v5 = v4> or> 3.) a*v1+b*v2+c*v5+d*v4 = v3> or> 4.) a*v1+b*v5+c*v3+d*v4 = v2> or> 5.) a*v5+b*v2+c*v3+d*v4 = v1The vectors are:> ^^^^^^^^^^^^^^^^^^> v1 = (1,-2,0,3); v2 = (2,-5,-3,6);> v3 = (0,1,3,0); v4 = (2,-1,4,-7)> v5 = (5,-8,1,2)I got the following results:1.) a + 2c = 1 and b - c = 1 and d = 1> 2.) a + 2b = -3 and c - b = 1 and d = 1> 3.) a - d = 2 and b - d = -1 and c + d = 0> 4.) a - 3d = 2 and b + d = 0 and c + d = -1> 5.) a + d = 0 and 2b - 3d - 1 = 0 and 2c - d - 1 = 0I have expected to get simple numerical values for a,b,c> and d (like: a = 2,...) but not you!Karl.> === Subject: Dream problem :-)This night my dream was crucially based on thefact that sec(2x)=(sec(x)+sec(4x))/2.Uhm, I get the nagging feeling that in that stateI'm just as lousy as a mathematician as awake :-)Do me a favor and compute the x for which thatequation is accidentally right before someonegets hurt...-- Hauke Reddmann <:-EX8 Private email:fc3a501@math.uni-hamburg.deFor our chemistry anything else === Subject: Re: Question on generation of large prime numbersBlueBear :> I have looking at a few web pages dealing with the largest known> calculated primes and a great deal of computational time is taking> into searching for these numbers and verifying they are primes. I have> seen the Euclid's proof of in'nitude primes and it occurs that me> that super-large prime numbers can be calculated using the following:Others have already given you an example of why your idea won't §y, but I don't want you to get the idea that Euclid is broken. His proof that there is no largest prime (and therefore an in'nite number of primes) is as follows:If there is a largest prime, P, then the number of primes is 'nite. We can therefore form a new number, N, which is one greater than the product of all primes. Therefore, N > P (and, indeed, it would be very much greater than P).When divided by any prime in our list, it leaves remainder 1. Therefore, /either/ it is prime /or/ it is the product of two or more primes, at least one of which is greater than P.In /either/ case, P has been shown not to be the largest prime number.Don't forget the second case! :-)-- Richard Heath'eld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === Subject: Re: billion (was: Big Number GameDgates :> :>>Thomas Bushnell, BSG :> You will be right more often (by vastly more) if you make his> assumption than the other assumption.>>[Slartibartfast] gave a hollow laugh. OWhat does it matter? Science has>>achieved some wonderful things of course, but I'd far rather be happy than>>right any day.'Ha! You didn't transcribe the (important) next two lines! :-oThat's right. I didn't. :-)-- Richard Heath'eld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === Subject: Re: Reality of response to my workI think this is morally wrong. Don't tell James to waste his money.Mocking the town idiot is bad enough, but actually convincing him to jumpdown a well is much worse.And James, If you want to make money in math you should try one of the Clayproblems. Their worth a million dollars each if you prove one. It might bethe only way to make alot of money in math.http://www.claymath.org/Millennium_Prize_Problems/Justin Van Winkle> Parallax :> My experience with patents is that the provisional application which> but I advise him to contact a patent attorney to write it. This costs> me about $400 for this provisional application. The provisional> application protects him for 1 year till he 'les the complete> application. This also gives him time to shop the idea around to> companies. After a year he 'les the complete application. My> experience is that this could cost from $6000-10000 when using a> patent attorney.> I wish him luck as the world needs new ideas.Your understanding of the patent requirements is incorrect. There are noprovisional application> capabilities. There are disclosure documents which can be 'led toestablish temporal priority, but they> do not permit the future applicant to publish or otherwise publiclydisclose his work.Having said that, I think your suggestion that James apply forintellectual property protection is sound --> especially if he retains a patent attorney.--> There are two things you must never attempt to prove: the unprovable -- and the obvious.> --> Democracy: The triumph of popularity over === principle.> --> http://www.crbond.comSubject: Re: conjugate subgroups > Why can't a conjugate of a subgroup be a proper subset of it?See BruceBruce Harveybruce@bearsoft.co.ukThe Alternative Physics Sitehttp://users.powernet.co.uk/bearsoft === Subject: Re: Question on generation of large prime looking at a few web pages dealing with the largest known> calculated primes and a great deal of computational time is taking> into searching for these numbers and verifying they are primes. I have> seen the Euclid's proof of in'nitude primes and it occurs that me> that super-large prime numbers can be calculated using the following:p1=2 < p2=3> A large prime number, p, can be generated using> p = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5> p = (p1*p2*p3) + 1 = 31Look at the page on Primorial Primes at http://primepages.org/> p1=2 < p2=3 < p3=5 < p4=7> p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbers> exceeding currently known largest primes can be generated rather> quickly. For example, we calculate the 'rst million primes, multiply> them and just add one.In the same way that 2*3*5*7*11*13+1 = 30031 = 59*509 is prime?> If this is true, then calculating super-large primes should be a> computational trivial task, shouldn't it? Then why the big fuss over> calculating large primes?I suspect that I am missing fundamental here. Can super large primes> really be calculated using the trivial method outlined above? Hope> that someone can at:http://primepages.org/Phil-- Unpatched IE vulnerability: Notepad popupsDescription: Opening popup windows without scriptingReference: http://computerbytesman.com/security/notepadpopups.htmFollowup : http://msgs.securepoint.com/cgi-bin/get/bugtraq0308/55.html === 01:00:37 -0800, anandb@vsnl.com (Bhupinder Singh Anand)> :For any given natural number m, the Goodstein sequence, G(m), only>terminates if m < 4.G(4) terminates, so the argument is invalid. I don't have a formal> proof, but just look at the following and it should be obvious (unless> I made a mistake somewhere):Bravo, indeed - I could not spot any mistake in the reasoning. If Ihad bothered to apply myself to the tedious drudgery of speci'callyworking out the steps you did - instead of attempting to visualisethem generally - I would have, of counts. Firstly for helping me spot the §aw (itis not as obvious as I thought it might be); secondly, since I didn't- fortunately - hold back my argument, for helping me bring intobetter focus the issue that triggered my investigation:As I see it, your reasoning that we can see that G(4) terminatesshould, surely, be capable of expression as a formal proof sequence inPA. Now, assuming the standard interpretation of Goodstein's argumentis valid, then, for any given natural number m, we should also,reasonably, be able to see, similarly, that G(m) must terminate -necessarily in 0. Thus, for any given m, we should also be able tomirror this vision, without appeal to trans'nite induction, as aformal proof sequence in PA.In other words, assuming that Goodstein's theorem cannot be formallyproved in PA, there should, reasonably, be some constructivemeta-proof - in Goedel's sense, i.e., one that does not appeal totrans'nite concepts - to the effect that, for any given naturalnumber m, there is some proof sequence in PA that implies that [G(m)]terminates === Subject: Re: conjugate subgroupsBruce Harvey :>> Why can't a conjugate of a subgroup be a proper subset of it?> See http://mathworld.wolfram.com/ConjugateSubgroup.htmlSurely the question doesn't deserve an URL?-- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: De facto censorship, counting primes :Some of you were probably surprised to learn that I did indeed 'nd a> way to count prime numbers by integrating a partial difference> equation. Some of you probably STILL doubt that no one else in> recorded history has managed such a feat because you need to believe> in mathematicians.But my point is that mathematicians have gone rogue and act against> the needs of society by de facto censorship of information that they> don't think makes them look good, like the information about my> partial difference equation.What do I mean by de facto censorship?Well, besides the active activity, like webpages labeling me a crank,> there's the passive act of refusing to acknowledge the discovery> itself.No, censorship is when you are stopped from expressing your views. Whatkind of delusions of grandeur do you have to suffer from to think thatpeople are and invalid to be removed if you're e-mailing me. === Subject: Re: Question on generation of large prime numbersRichard :I have looking at a few web pages dealing with the largest known> calculated primes and a great deal of computational time is taking> into searching for these numbers and verifying they are primes. I have> seen the Euclid's proof of in'nitude primes and it occurs that me> that super-large prime numbers can be calculated using the following: Others have already given you an example of why your idea won't §y, but I > don't want you to get the idea that Euclid is broken. His proof that there > is no largest prime (and therefore an in'nite number of primes) is as > follows:If there is a largest prime, P, then the number of primes is 'nite. We can > therefore form a new number, N, which is one greater than the product of > all primes. Therefore, N > P (and, indeed, it would be very much greater > than P).When divided by any prime in our list, it leaves remainder 1. Therefore, > /either/ it is prime /or/ it is the product of two or more primes, at least > one of which is greater than P.In /either/ case, P has been shown not to be the largest prime number.Nope. Nowhere have you assumed that the list of primes known is complete up to P. You've simply proved that the 'nite list ofknown primes is not the list of all primes.If you use Euclid's proof as constructive, then the sets you generate (assuming you have a factorisation oracle) are: {2}{2,3}{2,3,7}{2,3,7,43}{2,3,7,13,43,139}{ 2,3,7,13,43,139,3263443}and thence you 'nd the new primes {547,607,1033,31051}, all of which are less than the currently largest known prime 3263443.> Don't forget the second case! :-)Don't forget that if you're pretending that you don't know what the primes are, you can't suddenly pull otherwise well-known features of the prime numbers out of a hat.Phil-- Unpatched IE vulnerability: ADODB.Stream local 'le writingDescription: Planting arbitrary 'les on the local 'le systemExploit: http://ip3e83566f.speed.planet.nl/eeye.html (but unrelated to the EEye exploit) === Subject: Re: Squares that end with four identical digitsEdwin Clark :> If x^2 is an integer that ends in 4 identical digits aaaa then> x^2 is congruent to aaaa modulo 10^4. In this case the generalized> Legendre symbol L(aaaa/10^4) will be 1 and if aaaa is not a square modulo> 10^4 then it will be -1. That's not quite accurate.L(a/n) = 1 if a is a square mod n,but the converse is not true in general,eg L(2/15) = L(2/3) L(2/5) = -1.-1 = 1.-- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === vectors and have to 'nd a > base for V = L(v1, v2, v3, v4, v5). Then I must express the 'fth> vector through this base.Look up Gram-Schmidt orthogonalization.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Real torsion (0,1) matricesIf A is a real nXn (0,1) matrix such that A^m = I (the identitymatrix) === Subject: Re: De facto censorship, counting primes > To a large extent I now censor my *own* work in talking about it, as I> focus on things that are hard for people to lie about, and hope for> the best.>And we appreciate that you do that! Now if you could only put the primecounting differebce equation, polynomial factorization and FLT back into thebox too, that would be superb!> Right now, locked inside of me is information that could be lost to> humanity because I'm the genius maligned, trapped by a system that> lets mathematicians get away with hurting the society that feeds and> clothes them, by de facto censorship.>Could you please keep all of those locked up?When you die, then maybe, just maybe your INCREDIBLE genius will be sharedwith the world and all will know just how tortured you were by theconspirators!> I know things, important things, that you may never know about> numbers, and mathematics.>Do you know that you actually have more than nine of these symptoms? Thatalso deals with numbers, including 9 which is the perfect square 3^2!***Diagnostic criteria for 301.81 Narcissistic Personality Disorder (cautionarystatement)A pervasive pattern of grandiosity (in fantasy or behavior), need foradmiration, and lack of empathy, beginning by early adulthood and present ina variety of contexts, as indicated by 've (or more) of the following:(1) has a grandiose sense of self-importance (e.g., exaggerates achievementsand talents, expects to be recognized as superior without commensurateachievements)(2) is preoccupied with fantasies of unlimited success, power, brilliance,beauty, or ideal love(3) believes that he or she is special and unique and can only beunderstood by, or should associate with, other special or high-status people(or institutions)(4) requires excessive admiration(5) has a sense of entitlement, i.e., unreasonable expectations ofespecially favorable treatment or automatic compliance with his or herexpectations(6) is interpersonally exploitative, i.e., takes advantage of others toachieve his or her own ends(7) lacks empathy: is unwilling to recognize or identify with the feelingsand needs of others(8) is often envious of others or believes that others are envious of him orher(9) shows arrogant, haughty behaviors or attitudesReprinted with permission from the Diagnostic and Statistical Manual ofMental Disorders, fourth Edition. Copyright 1994 American PsychiatricAssociation*** === Subject: Re: Dream problem :-) > This night my dream was crucially based on the> fact that sec(2x)=(sec(x)+sec(4x))/2.> Uhm, I get the nagging feeling that in that state> I'm just as lousy as a mathematician as awake :-)> Do me a favor and compute the x for which that> equation is accidentally right before someone> gets hurt...>Perhaps you can plot the left function and the right function and see foryourself before you go off and hurt someone? === explain to me what's the difference between basis transformation and a similarity transformation.I'm really George Baloglou> Earlier today I noticed that it is very easy to derive Ceva's theorem from> Menelaus' theorem: with the Cevians AD, BE, CF of triangle ABC meeting atG,> apply Menelaus' theorem to triangles ABD and ACD and multiply theoutcomes;> to be precise, (AF/FB)*(BC/CD)*(DG/GA) = -1 and (AG/GD)*(DB/BC)*(CE/EA)= -1> lead to (AF/FB)*(BD/DC)*(CE/EA) = +1 via DB/CD = BD/DC, DG/GD = AG/GA= -1,> and, *of course*, (-1)*(-1) = +1 :-)The emphasis on the multiplication above alludes to the eccenticobservation> that, just as +1 cannot generate -1, ...Nice. Eccentric, but I don't mind :)> ... Ceva's theorem does not seem to be capable of proving Menelaus'stheorem...Not exactly. This page goes into it:http://www.cut-the-knot.org/Generalization/Menelaus.shtmlIt seems odd that the ancient theorem in the pair (Ceva/Menelaus) is the onewhich involves negative numbers.The above website, which is high-school level in the main, doesn't emphasizethat Ceva and Menelaus belong properly to af'ne geometry, meaning that wedon't do any comparison of distances in two independant directions.Ceva and Menelaus can be thought of as statements about determinants inhomogeneous coordinates (a 19th Century invention). In that vein, Routh gavea formula for the area of the triangle formed by any three cevians,concurrent or not. The story is in Coxeter's celebrated book _Introductionto Geometry_.Thx for this interesting note, GB.LH === Subject: Re: plotting elliptic curves for LaTeXJose Capco :>gnuplot can do such plots and it can also output LaTeX code. Also, Postscript> plots>can be imported using PSTricks.>could you give me an example code? given a polynomial f(x,y) .. I want to plot> the points that solve f(x,y)=0 ... say documentation, demos and example 'les -- including plottingimplicit functions. I could not presume to do justice to it in a limited emailresponse. Do a search.--There are two things you must never attempt to prove: the unprovable -- and theobvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: The Gauge Integral> I'm looking for some resources to learn a bit more about the gauge > integral (Alternately called the Generalised Riemann Integral and the > Henstock-Kurzweil integral or probably a couple others).It seems like an extremely neat idea (although for most purposes I'd > probably prefer lebesgue integration anyway, as it's so much more > general) and worth learning more about. Also I'm in the process of > (slowly) writing an analysis textbook, and it seems a fairly natural > extension of the way I'm introducing Riemann integration, so I thought > it might be worth pursuing to see if it's a good idea to include it.At the moment I know the de'nition, and a couple basic facts about it - > not much more than I learned from > http://www.math.vanderbilt.edu/~schectex/ccc/gauge/ .An online treatment would of course be preferable as, well, I wouldn't > have to pay money for it. :) But if you can reccomend some good books on > the subject that would be great as well.Now lets see if this gets through... my college newsserver has Sherbert's Introduction to Real Analysis is agood choice. I guess that book is not that expensive.Artur === Subject: Re: Bredon corrections.> I made the (possibly controversial) step of teaching myself homotopytheory> from GB's Geometry and Topology book, and noticed a shed load of typos.Does> anyone know if they've all been catalogued and whether I could access alist> of them somewhere?> What you think the typos? What's the 4 or 5 simplest?What difference does that make? Is there a minimum dif'culty level beforeprinted typo lists are considered?I don't remember off-hand so I'll have to wait until I get to my of'cetomorrow. === Subject: Re: 've vectors SfgBt2ZBwe2hx5866T8vas-VrYwb+0Kkc0WTtm0MaQluH3NRSfzr44> On given 5 vectors and have to 'nd a> base for V = L(v1, v2, v3, v4, v5). Then I must express the 'fth> vector through this the explanations I have found on it so far are tocomplicatedfor me. Can you give me at least one really short example for theapplication of this method? Please!Best understand how this method could behelpfulfor me. (:-{ ] === Subject: Re: Reality of response to my workJustin Van Winkle :> I think this is morally wrong.Since you top posted, I have no idea exactly *what* was said that you regard asmorally wrong.> Don't tell James to waste his money.I only recommended that he get the advice of a patent attorney if he wanted tosecure intellectual property rights. I assume you regard your *free* adviceabout the matter as superior to the advice of a professional. If so, you may beguilty of practicing law without a license.--There are two things you must never attempt to prove: the unprovable -- and theobvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: De facto censorship, counting primesX-DMCA-Noti'cations: -0800, jstevh@msn.com () :>Some of you were probably surprised to learn that I did indeed 'nd a>way to count prime numbers by integrating a partial difference>equation. Some of you probably STILL doubt that no one else in>recorded history has managed such a feat because you need to believe>in mathematicians.But my point is that mathematicians have gone rogue and act against>the needs of society by de facto censorship of information that they>don't think makes them look good, like the information about my>partial difference equation.What do I mean by de facto censorship?Well, besides the active activity, like webpages labeling me a crank,>there's the passive act of refusing to acknowledge the discovery>itself.Um, those acts are not censorship. The only person around here I'veever seen attempt any actual censorship is _you_, complaining to myemployer because you wanted him to get me to shut up.>After all, it's very compact, as here are the instructions, yet again:[Yet another repetition of the supposedly censored algorithm snipped]Now then, think about kids currently in school who I doubt will see>the method I've just shown you, unless maybe they're out on Usenet>reading my posts, because the mathematical establishment thinks it can>ignore my results.Have I contacted mathematicians? Yes.I've contacted mathematicians all over the world.And not _one_ of them has the decency to acknowledge the _obvious_importance of your work. Does that really seem plausible, thatthere could be such _universal_ agreement on a lie? Universalagreement on the truth seems much more plausible to me.>But you see, what bene't do they see to their society by allowing>that someone NOT a mathematician found such a result?Worse, I have other math knowledge. But mathematicians can let the>world be convinced I'm just a crank, most of them passively just>sitting by, and keeping quiet about my results--after all, that's>quite effective, eh?Then they have de facto censorship because people BELIEVE they>wouldn't do such a thing if my work were important!!!Do people really believe that? Huh.>So you have a standstill with me pushing my research, and a few>mathematicians actively 'ghting its acceptance on Usenet, while most>just do their best to ignore it.For instance, I contacted Georgia Tech and talked to a Professor Ernie>Croot, Why? You're not going to stop until you've contacted everymathematician on the planet?>giving him more information about my prime counting research>than I've posted here. He replied back *once*, and seemed friendly>enough. I answered him and awaited further replies. After some weeks>I sent a query to follow-up, and here is his looking at it. I'll let you know>when I do.Best,ERnieOn Thu, to see if you still have any interest in my 'nd of a way to >> count prime numbers by integrating a partial difference equation, as I >> haven't heard from you since my last reply. If you've lost interest can you refer me back to the professor who sent me >> to you because I'd like to check ahead of time what you wish to believe,>> and daring God to be different.>> http://lostincomment.blogspot.com/>Will I ever hear back from Professor Croot? Well, consider the>evidence:I've given something new, a partial difference equation integration>for counting prime numbers, a 'rst in recorded human history.Professor Croot has had some time to consider my work, but now begs>off, claiming not to have looked at it.It turns out that he's a 'rst year professor and I was referred to>him by another professor at Georgia Tech who *asked* him to look over>my work.I daresay that Professor Croot lied in his email.Possibly he intends to look at it. Possibly he's trying to bepolite: Suppose for the sake of argument that he's recognizedyour crackpottery nature somehow. Would you really be happierif he came out and said so?>That's a professor at Georgia Tech.So I'm an *independent* researcher, which means that mathematicians at>universities have a lot of power when it comes to acceptance of my>work, Actually they don't, because the mathematical world does not decidethese things the way you think they do. If Barry Mazur and AndrewWiles published a paper saying your proof of FLT was correct itmight make an impression at 'rst, but _very_ soon people wouldstart saying no it was nonsense and wonder what they were smoking.>but may see little point in helping me.That leaves me Usenet, where there are mathematicians, like >Ullrich, a tenured math professor at Oklahoma State University, to>TELL people that my work is useless or wrong.Yet, I found a partial difference equation that you can integrate to>count prime numbers which NO ONE ELSE in recorded human history has>managed.I have other mathematical research, but as long as mathematicians>stick to their guns, who gets to hear it?Sure I can talk about it on Usenet, and watch as posters malign my>work, lie and generally act like asses, Like asses and also like ing dog.>knowing that others will just>sit, and wait, waiting for mathematicians in the mainstream to let>them know that it's important.To a large extent I now censor my *own* work in talking about it, as I>focus on things that are hard for people to lie about, and hope for>the best.Right now, locked inside of me is information that could be lost to>humanity because I'm the genius maligned, trapped by a system that>lets mathematicians get away with hurting the society that feeds and>clothes them, by de facto censorship.I know things, important things, that you may never know about>numbers, and mathematics.You know things that you've never told us about? That's hard tobelieve.>Mathematicians are no longer part of decent society, but are now rogue>having taken their own path into darkness. Don't believe me?Check my instructions for integrating that partial difference>equation.Check for yourself.Check _what_? Nobody denies that it gives the right answer,the only dispute is about importance and originality. Howis checking going to settle that?>My math discoveries, found for pro't>http://mathforpro't.blogspot.com/ C. Ullrich === Subject: Re: Analysis question (primitives)X-DMCA-Noti'cations: 07:53:14 +0000, Jos.8e Carlos Santos :>Hi all,Let a and b be two real numbers with a < b and consider a function f>from [a,b] into the reals which has the intermediate value property>and is Riemann-integrable. Let F(x) be the integral of f between a>and x.Question: Must F be a primitive of f?No. Say a < 0 < b. Let f(x) = sin(1/x) for x <> 0, f(0) = 0.Let g(x) = sin(1/x) for x <> 0, g(0) = 1. Then both f and gare Riemann integrable and have the intermediate value property.But they have the same inde'nite integral, and Santos C. Ullrich === Subject: Re: Question on generation of large prime numbers[Disclaimer: Phil Carmody is a damn sight better at mathematics than I am. But I think I got a case. :-) ]Phil Carmody :> Richard Heath'eld have looking at a few web pages dealing with the largest known>> calculated primes and a great deal of computational time is taking>> into searching for these numbers and verifying they are primes. I have>> seen the Euclid's proof of in'nitude primes and it occurs that me>> that super-large prime numbers can be calculated using the following: > Others have already given you an example of why your idea won't §y, but>> I don't want you to get the idea that Euclid is broken. His proof that>> there is no largest prime (and therefore an in'nite number of primes) is>> as follows: If there is a largest prime, P, then the number of primes is 'nite. We>> can therefore form a new number, N, which is one greater than the product>> of all primes. Therefore, N > P (and, indeed, it would be very much>> greater than P). When divided by any prime in our list, it leaves remainder 1. Therefore,>> /either/ it is prime /or/ it is the product of two or more primes, at>> least one of which is greater than P.>> In /either/ case, P has been shown not to be the largest prime number.Nope. Nowhere have you assumed that the list of primes known is> complete up to P.On the contrary, I formed N by multiplying all primes and then adding 1. To multiply all primes together, I must 'rst have a complete list of primes. If the list is not complete, then I cannot perform the multiplication. the primes are, you can't suddenly pull otherwise well-known> features of the prime numbers out of a hat.It's not necessary to know all the primes in order to construct the argument. It is suf'cient to know that it is /in theory/ possible to know them, e.g. by sieving. It is only necessary to know all the primes if you wish to perform the multiplication in real life.-- Richard Heath'eld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === Subject: Integral of a function over [a,b]HelloSuppose g:[a,b]->R is continuous on [a,b] and let f[a,b]->R be de'nedin such a way that, at every irrational x in [a,b], f is continuousand f(x) = g(x). Then, is it true that the Riemann integral of f over[a,b] equals the integral of g? Since the set of discontinuities of fon [a,b] is at most countable and, therefore, has measure 0, itfollows f is certainly integrable over [a,b], but I couldn't come to aconclusion if its integral needs to equal the integral of g or if itdepends on how we de'ne f for the rationals of [a,b].I know that if f is Thomae's function, then the answer is yes and theinegralas vanish, but in this case we have f(x) =0 for everyirrational, which seems to imply a loss of generality when compared tothe function f de'ned above.In case the answer is yes, does it remain yes if we replace theassumption that g is continuous by the weaker one that g is onlyRiemann integrable over [a,b]?Now, suppose in the questions above we replace the set Q cap [a,b], ofthe rationals in [a,b], by an uncountable set E with measure zero?Then, is there anything very much for any help or suggestions.Amanda === Subject: Re: Squares that end with four identical digits> Edwin Clark :> If x^2 is an integer that ends in 4 identical digits aaaa then> x^2 is congruent to aaaa modulo 10^4. In this case the generalized> Legendre symbol L(aaaa/10^4) will be 1 and if aaaa is not a squaremodulo> 10^4 then it will be -1.That's not quite accurate.> L(a/n) = 1 if a is a square mod n,> but the converse is not true in general,> eg L(2/15) = L(2/3) L(2/5) = -1.-1 = 1.>I think you are confusing the jacobi function with the generalized Legendresymbol (called quadres in Maple):> jacobi(2,15); 1> quadres(2,15); -1Here are the descriptions from Maple the help for these two functions aswell as for the Legendre symbol.The function quadres will compute a generalized Legendre symbol L(a/b) of aand b, which is de'ned to be 1 if a is a quadratic residue (mod b) and -1if a is a quadratic non-residue (mod b) . The number a is a quadraticresidue of b if it has a square root (mod b); i.e., an integer c exists suchthat c^2 is congruent to a (mod b).The function jacobi will compute the Jacobi symbol J(a/b) of a and b. If thefactorization of b is p1^k1 * ... * ps^ks, then jacobi(a, b) = legendre(a,p1)^k1 * ... * legendre(a, ps)^ks , where legendre(a,p) is the Legendresymbol of a and p.The legendre(a, p) function computes the Legendre symbol L(a/p) of a and p(a prime), which is de'ned to be 1 if a is a quadratic residue (mod p), -1if a is a quadratic non-residue (mod p), and 0 if a is congruent to 0 (modp). The number a is a quadratic residue of p if it is not a multiple of pand has a square root (mod p), that is, there is an integer c such that c^2is congruent to a (mod p). The number a is a quadratic non-residue of p i't is not a multiple of p and does not have a square root (mod p).Edwin === Subject: Re: Question on generation of large prime numbersIn rec.puzzles Richard Heath'eld :> It's not necessary to know all the primes in order to construct the > argument. It is suf'cient to know that it is /in theory/ possible to know > them, e.g. by sieving. It is only necessary to know all the primes if you > wish to perform the multiplication in real life.So, since you can't sieve all of them there is a (practical) largest prime! === Subject: Re: :>>Hi all,>>Let a and b be two real numbers with a < b and consider a function f>>from [a,b] into the reals which has the intermediate value property>>and is Riemann-integrable. Let F(x) be the integral of f between a>>and x.>>Question: Must F be a primitive of f?> No. Say a < 0 < b. Let f(x) = sin(1/x) for x <> 0, f(0) = 0.> Let g(x) = sin(1/x) for x <> 0, g(0) = 1. Then both f and g> are Riemann integrable and have the intermediate value property.> But they have the same inde'nite integral, and a function has> at most one Santos === Subject: Re: Question on numerical integration (Simpson)Robert Israel :>>Is the Simpson's method optimal with respect to the>>number of calculations? (i.e., does it yield the>>lowest error for a 'xed number of calculations?)No.>>I know that there may not exact and absolute answer>>to this -- but in general, for non-pathological,>>non-super-obscure situations, is it true?No. There is no such thing as an optimal method for a 'xed> number of calculations, unless you severely restrict the set > of possible functions to be integrated. Each method will give > 0 error for some functions, not for others.Yes, this was clear to me (as stated in the aboveparagraph you quoted).I was hoping to 'nd out if in general, or in acertain class of functions, or say, in a certainapproximate percentage of the cases it might betrue that Simpson's rule would beat higher orderpolynomials in terms of accuracy per number ofcalculations.I assume that even that is hard to 'nd out?I guess I'll do some more reading on this, andmaybe some experimentation (what the hell, itwill be === Subject: Re: Squares that end with four identical digitsEdwin Clark :>> If x^2 is an integer that ends in 4 identical digits aaaa then>> x^2 is congruent to aaaa modulo 10^4. In this case the generalized>> Legendre symbol L(aaaa/10^4) will be 1 and if aaaa is not a square> modulo>> 10^4 then it will be -1.>> That's not quite accurate.>> L(a/n) = 1 if a is a square mod n,>> but the converse is not true in general,>> eg L(2/15) = L(2/3) L(2/5) = -1.-1 = 1.>I think you are confusing the jacobi function with the generalized> Legendre symbol (called quadres in Maple):I was indeed.I never saw the term generalised Legendre symbol used in this sense.though I have seen the term applied to the Jacobi symbol, eg by Hecke.Do you have a reference for this usage?I just looked in a few number theory books I have,and none of them used the term.It seems to me both misleading and useless --useless because there is actually no way of calculating itas there is with the Jacobi symbol,except by computing the Legendre symbol for each factor p(probably using the Jacobi symbol!).-- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Reality of response to my work> Justin Van Winkle :> I think this is morally wrong.Since you top posted, I have no idea exactly *what* was said that youregard as> morally wrong.> Don't tell James to waste his money.I only recommended that he get the advice of a patent attorney if hewanted to> secure intellectual property rights. I assume you regard your *free*advice> about the matter as superior to the advice of a professional. If so, youmay be> guilty of practicing law without a license.Not if it's free. === Subject: Re: Integral of a function over [a,b]Amanda :> Hello> Suppose g:[a,b]->R is continuous on [a,b] and let f[a,b]->R be de'ned> in such a way that, at every irrational x in [a,b], f is continuous> and f(x) = g(x). Then, is it true that the Riemann integral of f over> [a,b] equals the integral of g? Since the set of discontinuities of f> on [a,b] is at most countable and, therefore, has measure 0, it> follows f is certainly integrable over [a,b], but I couldn't come to a> conclusion if its integral needs to equal the integral of g or if it> depends on how we de'ne f for the rationals of [a,b].In fact, your argument to prove that f is integrable over [a,b] is notcorrect. Indeed, in order that a function de'ned on [a,b] is integrable(in the sense of Riemann), the set of discontinuities must have measure0 *and* f must be bounded. However, if f is supposed to be bounded, yourargument is correct.On the other hand, the integrals of f and g must be equal (stillassuming that f is bounded, of course), since the function f - g isintegrable and takes the value 0 at every irrational form [a,b].> In case the answer is yes, does it remain yes if we replace the> assumption that g is continuous by the weaker one that g is only> Riemann integrable over [a,b]?Sure. I did not use the assumption g is continuous. > Now, suppose in the questions above we replace the set Q cap [a,b], of > the rationals in [a,b], by an uncountable set E with measure zero? > Then, is there anything interesting we can af'rm about the integral > of f?I don't think so.Best === Subject: Permutation: how to permutation:(3 5 1 2 4 6)After detaching into cycles we have:(1 3)(2 5 4)(6)The same with transpositions:(1 2 3 4) = (1 4)(1 3)(1 2)My question is how to translate these algorithms into anycomputer-understandable code. It's easy to transform it on a sheet of paper,but implementation seems a StobiskiRepublic of Poland === Subject: Re: Dream problem :-)In 07:15:10 -0800<1068995840.212611@news-1.nethere.net>:>> This night my dream was crucially based on the>> fact that sec(2x)=(sec(x)+sec(4x))/2.>> Uhm, I get the nagging feeling that in that state>> I'm just as lousy as a mathematician as awake :-)>> Do me a favor and compute the x for which that>> equation is accidentally right before someone>> gets hurt... Perhaps you can plot the left function and the right function and see for> yourself before you go off and hurt someone?> There are also multiple solutions -- although one of themshould jump right out and bite the OP (were the value tohave any teeth, that is -- and this particular numeralobviously hasn't hatched any yet :-) ).Consider that 1 = (1 + 1)/2. Now when does sec(2x) = 1?This hint should make solving the problem a dream... :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: Question on generation of large prime numbersIn sci.math, Christian 11:14:35 svr.pol.co.uk>:> bluebear@pokernetwork.com (BlueBear) :>> I have looking at a few web pages dealing with the largest known>> calculated primes and a great deal of computational time is taking>> into searching for these numbers and verifying they are primes. I have>> seen the Euclid's proof of in'nitude primes and it occurs that me>> that super-large prime numbers can be calculated using the following: p1=2 < p2=3>> A large prime number, p, can be generated using>> p = (p1 * p2) + 1 = 7 p1=2 < p2=3 < p3=5>> p = (p1*p2*p3) + 1 = 31 p1=2 < p2=3 < p3=5 < p4=7>> p = (p1*p2*p3*p4) + 1 = 211 p1=2 < p2=3 < p3=5 ..... < pn=...>> p = (p1*p2*p3*p4*....*pn) + 1 = .... It seems to me that using the above method, super large prime numbers>> exceeding currently known largest primes can be generated rather>> quickly. For example, we calculate the 'rst million primes, multiply>> them and just add one.>> If this is true, then calculating super-large primes should be a>> computational trivial task, shouldn't it? Then why the big fuss over>> calculating large primes? I suspect that I am missing fundamental here. Can super large primes>> really be calculated using the trivial method outlined above? Hope>> prime?30031 = 2*3*5*7*11*13 + 1 = 59 * 509Regardless of whether the number is prime or not, one always getsa prime not in the original set, though.As it is, RSA's algorithm requires generation of largeprimes and, in order to crack it, the factorizationof a product of two large primes (the product, not theprimes, is given as part of the public key). Large primegeneration isn't all that dif'cult unless one gets stuckin a desert (pick a random odd number N of the requisitesize; if it's prime, stop; otherwise add 2 to N and loop).A known desert is just after a factorial; it's obviousthat none of N! + 2, N! + 3, ..., N! + N can be prime.Or one can take the lcm of the numbers 1 through N, forthe same effect.Another interesting generator is x^2 + x + 41. The 'rsttime this fails is x = 40, as one can verify using asimple program.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: Which series do these constants come from?Robert AH Prins contained a signi'cant number of constants. A fair amount of> these are very obvious, but a quite a few are complete mysteries> to me. For all but three I 'gured out the (likely) relationship> between the successive terms, but as for what they are al used?> I hope someone in this group can tell me or point me to a site> that has information about mathematical series.I don't have anything speci'c but maybe search on inverse symboliccalculator?> The constants are:... [snipped, see OP]There is this pattern:48 = 2^4 *31280 = 2^8 *528672 = 2^12 *7589824 = 2^16 *911534336 = 2^20 *11218103808 = 2^24 *134026531840 = 2^28 *1573014444032 = 2^32 *17No doubt the software is calculating a logarithm, because these numbers arethe inverses of the coef'cients of x^n in the power series for ln(1+4x).Similarly (with alternating signs)192 = 2^6 *320480 = 2^12 *51835008 = 2^18 *7150994944 = 2^24 *911811160064 = 2^36 *11big = 2^42 *13 (I presume)bigger = 2^48 *15and we are looking at a power series for ln(1+6x).4 = 2 *2!24 = 2^2 *3!192 = 2^3 *4!1920 = 2^4 * 5!23040 = 2^5 *6!322560 = 2^6 *7!etc, and it's calculating an exponential.The isolated constants in the list aren't so easy :)Larry === Subject: Re: diff EQ on strings, check out the math type=text/plain; boundary=----=_NextPart_000_001A_01C3AC3C.A7D70E10------------ differential equation for strings starting from Stokes> Theorem to show that energy is conserved along a world-sheet. These diff> eq's involve connection coef'cients. And I'm not really sure what it all> means yet. I would appreciate it if some who are more skilled in the art> would take a look at this and comment.The math can be found link in original post.What does it mean that ? Does this mean that the surface is a geodesic?And what does it mean that ? Does this mean that there is no force in thedirection of time? If F is still an arbitrary 'eld, then does this meanthat the string must be traveling in a frame where the time component iszero? === Subject: Re: Squares that end with four identical digits> Edwin Clark :>> I think you are confusing the jacobi function with the generalized> Legendre symbol (called quadres in Maple):I was indeed.> I never saw the term generalised Legendre symbol used in this sense.> though I have seen the term applied to the Jacobi symbol, eg by Hecke.Do you have a reference for this usage?Not except for Maple. I was afraid someone might ask about this. I alsotried unsuccessfully to 'nd it in my number theory texts.> I just looked in a few number theory books I have,> and none of them used the term.It seems to me both misleading and useless --> useless because there is actually no way of calculating it> as there is with the Jacobi symbol,> except by computing the Legendre symbol for each factor p> (probably using the Jacobi symbol!).>Well, it was useful for this problem. :-) Here's the Maple code for quadres:> showstat(numtheory:-quadres);numtheory:-quadres := proc(a, p)local fac, ig, n, pf, s; 1 if nargs <> 2 then 2 error invalid arguments end if; 3 n := modp(a,p); 4 if type(n,integer) and issqr(n) then 5 return 1 end if; 6 if not (type(n,integer) and type(p,integer)) then 7 return Oprocname(args)' end if; 8 if p < 6 then 9 return -1 end if; 10 if isprime(p) then 11 return mods((Opower')(n,1/2*p-1/2),p) end if; 12 ig := igcd(n,p); 13 if ig <> 1 and igcd(denom(n/ig^2),p/ig) = 1 then 14 return numtheory:-quadres(n/ig^2,p/ig) end if; 15 pf := ifactor(p); 16 if type(pf,`^`) then 17 return legendre_pow(n,pf) end if; 18 for fac in pf do 19 if type(fac,`^`) then 20 s := legendre_pow(n,fac) else 21 s := numtheory:-quadres(n,op(fac)) end if; 22 if s = -1 then 23 return -1 end if end do; 24 1end proc---Edwin === Subject: Re: Problem from Herstein 4>Marc Olschok :> :> 18. Let G be the group of all real 2-by-2 matrices> :> (a b)> :> (c d), with ad - bc non-zero, under matrix multiplication, and let N> :> be the subgroup consisting of those elements of G with ad - bc = 1.> :> Prove that N contains G', the commutator subgroup of G.> :> No. 18 is OK, I think. Any matrix of the form A B A^(-1) B^(-1) has> :> determinant 1, and thus so do all products of elements of this form.> :> Thus G' is contained in N.> :> 19. In problem 18 show, in fact, that N = G'.> : I assume, that your matrices have their coef'cients in some commutative> : 'eld k. If k* are the nonzero elements of k (with multiplication),> : you can verify that> : (a b)> : (c d) --> ad-bc> : does indeed give a surjective homomorphism det: G ---> k*.> : This will exhibit N = ker(det) and G/N = k* commutative.>All true, but this only shows that N contains G'; the hard part is>the other direction (problem 19, show that N = G'), which is not>coming to me right now.Apply theorem: Let G be a group with commutator subgroup G'.> (a) The subgroup G' is normal in G, and the factor group G/G' is abelian.> (b) If N is any normal subgroup of G, then the factor group G/N is> abelian if and only if G' is contained in N.I don't think I follow you here. We already know that G' is containedin N. The problem is the reverse inclusion, namely that N is containedin G'.-- G metabelian?> Now from N = G' is it possible to show N is Abelian?> In general, when G is invertible nxn matrices with *,> then N = G' as shown above.Is it still true that N is Abelian?> That is do rigid rotations commute?> It seems so. How's it shown and does N = G' help?----Again, I think you've lost me here. Certainly N is not abelian in theoriginal question. For example, the matrices(3 1)(8 3)and(2 5)(1 3) do not commute.John Harrison === Subject: Accumulation pts. in R and C ?Hi all,I just started Complex variables and applications by Churchill, et al. Iam §ummoxed by something already. In Churchill, they say that anaccumulation point is...a point z0 ... of a set S if each neighborhood if z0 contains at leastone point of S distinct from z0. Why is it different from R? To wit,every neighborhood of x0 has an in'nite number of points of, say, E.TIA,Lurch === Subject: Re: Accumulation pts. in R and C ?Charlie Johnson> I just started Complex variables and applications by Churchill, et al.I> am §ummoxed by something already. In Churchill, they say that an> accumulation point is> ...a point z0 ... of a set S if each neighborhood if z0 contains at least> one point of S distinct from z0. Why is it different from R? To wit,> every neighborhood of x0 has an in'nite number of points of, say, E.It is equivalent.It is always equivalent in a metric space.-- Maxi === Subject: Re: 've 22:38:38 +0100, Karl Pech :> Hallo,> Ich have given 5 vectors and have to 'nd a> base for V = L(v1, v2, v3, v4, v5). Then I must express the 'fth> vector through this base.> Look up explanations I have found on it so far are to> complicated> for me. Can you give me at least one really short example for Take a look at-- Paul SperryColumbia, SC (USA) === Subject: Re: cuts a two place into a wire at random.>let length of wire is 1>We got the wire of a three piece.>when we made a triangle from three wire piece,>'nd that probability of possibility.I believe what you're saying is that the three pieces must be able toform a genuine triangle. WLOG, assume that the pieces are cut in increasing order of size, so thelength of the pieces are a < b < c. Then c = Sqrt[1-a^2-b^2], whichalso must be larger than Sqrt[a^2+b^2].Proceed in this fashion.Doug === Subject: Relational algebra helpIm having trouble understanding some questions. Some help would beappreciated. I have read the book, but dont follow it.Problem 1F={B->A,A->C}1. the non-trivial dependencies are: B->A, A->C, B->Cis that right? I was wondering if it was a trick question.2. Find a non-empty instance of R that satis'ed every FD in F, botnot A->Bany idea what that question MEANS. Does it want a real example? Imconfused.3. Find an instance of R that satis'es every FD in F, bot not A-> B. again does this mean an example? Im confused. === Subject: Re: probability 2......> we cuts a two place into a wire at random.let length of wire is 1We got the wire of a three piece.when we made a triangle from three wire piece,'nd that probability of possibility.---------very dif'cult......um....help....me please.......Let say that lenghts of pieces are A,B,C. To form triangle pieces mustsatisfy:(A+B)>C(A+C)>B(B+C)>AThese conditions have another geometric interpretation: A B C0----------X-------------Y----------1Let points X,Y be random numbers (let say that they are places where we cutoriginal wire to make three pieces)Then we may restate conditions as:(x+y-x)>1-y <=> 2y>1 <=> y>0.5(x+1-y)>y-x <=>2x-2y+1>0 <=> that 0<=x<=1, 0<=y<=1.Draw these in x-y plane - the area of part of square ( 0<=x<=1, 0<=y<=1) inwhich all conditions are satis'ed represents your probability.Goran === Subject: Re: Squares that end with four identical digits> Edwin Clark :>Here are some questions for other bases:>For base 3: I 'nd that the square of> [1, 1, 1, 0, 2, 1, 2, 2, 1, 2, 1]_3>is> [2, 0, 1, 1, 2, 2, 0, 2, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1,1,>1, 1]_3>This square ends in 12 ones. Is there a limit to the number of 1's asquare>can end in for base b = 3.No.Suppose 2 x^2 + 1 == 0 mod 3^n (so x^2 ends in n 1's in base 3) but> not mod 3^(n+1), where n > 1. Of course x is not divisible by 3. Then> 2 (x+3^n)^2 - 1 = 2 x^2 - 1 + 3^n x mod 3^(n+1) and> 2 (x-3^n)^2 - 1 = 2 x^2 - 1 - 3^n x mod 3^(n+1)> so one of these == 0 mod 3^(n+1).>After a little effort in deciphering your proof I agree. Very nice!But shouldn't the 3rd and 4th lines of your proof be> 2 (x+3^n)^2 + 1 = 2 x^2 + 1 + 4*3^n x mod 3^(n+1) and> 2 (x-3^n)^2 + 1 = 2 x^2 + 1 - 4*3^n x mod 3^(n+1) === Subject: Re: probability 2......> we cuts a two place into a wire at random.let length of wire is 1We got the wire of a three piece.when we made a triangle from three wire piece,'nd that probability of possibility.---------very dif'cult......um....help....me please.......Put the wire on a number line with left end-point at 0 and right end-point at 1. Let c and d be the points at which the wire iscut such that 0 < c < d < 1. Now use the fact that in a trianglethe sum of the lengths of any two sides must be greater than thethird. For example, we must have c + (d - c) > 1 - d. Whenyou get the appropriate inequalities between c and d, plot themin the Cartesian plane with horizontal axes labelled c and d.Then 'nd the area of the solution region. This will be yourprobability. === Subject: chess bishops - copy of ALL headers otherwise we will be unable to process your complaint properly.I understand that it is possible to express the number of ways ofhaving k bishops on an n*n board such that they are not attacking eachother as a combinatorial formula and i have been trying to derive theformula without any luck, can anyone help me out ? === Subject: Re: Permutation: how to detach cycles/transpositionsPawe Stobi.96ski a .8ecrit> Let's have a permutation:(3 5 1 2 4 6)After detaching into cycles we have:(1 3)(2 5 4)(6)I don't understand... Your cycle is already decomposed into a product ofdisjoint cycles! (1 3)(2 5 4)(6) is not equal to (3 5 1 2 4 6).-- Maxi === Subject: Re: chess bishops - combinationsparmito :I understand that it is possible to express the number of ways of> having k bishops on an n*n board such that they are not attacking each> other Why would bishops attack one another? Oh--over homosexuality perhaps?> as a combinatorial formula and i have been trying to derive the> formula without any luck, can anyone help me out e-mailing me. === Subject: Re: Permutation: how to detach cycles/transpositions> I don't understand... Your cycle is already decomposed into a product of> disjoint cycles! (1 3)(2 5 4)(6) is not equal to (3 5 1 2 4 6).>By (3 5 1 2 4 6) he meant p(1)=3, p(2)=5, etc... in which case thedecomposition holds.--@+ sur le forum fran.8dais !! === Subject: Re: Relativity is based on assumption. <8pr7rvojo5acmso735p31bo58d7g0rq43v@4ax.com> <4C0tb.4282$_g6.527@news-binary.blueyonder.co.uk>Treme: C&C,DWSIn <4C0tb.4282$_g6.527@news-binary.blueyonder.co.uk>, === said:>Subject: Relativity is based on assumption.Duh! So is any scienti'c theory. But it has nothing to do withMathematics. Please take it to sci.physics.relativity.crackpots.Oh, BTW, welcome to my 'lters.*PLONK*-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Subject: Re: Squares that end with four identical digits> Edwin Clark :>Here are some questions for other bases:>For base 3: I 'nd that the square of> [1, 1, 1, 0, 2, 1, 2, 2, 1, 2, 1]_3>is> [2, 0, 1, 1, 2, 2, 0, 2, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,>1, 1]_3>This square ends in 12 ones. Is there a limit to the number of 1's a square>can end in for base b = 3.No.Suppose 2 x^2 + 1 == 0 mod 3^n (so x^2 ends in n 1's in base 3) but > not mod 3^(n+1), where n > 1. Of course x is not divisible by 3. Then> 2 (x+3^n)^2 - 1 = 2 x^2 - 1 + 3^n x mod 3^(n+1) and> 2 (x-3^n)^2 - 1 = 2 x^2 - 1 - 3^n x mod 3^(n+1) > so one of these == 0 mod 3^(n+1). More generally, for any odd b, if (b-1) x^2 + d == 0 mod b^n but not modb^(n+1), where gcd(d,b) = 1, then gcd(x,b) = 1 and (b-1) (x + y b^n)^2 - d == (b-1) x^2 - d + 2 (b-1) b^n x y mod b^(n+1)which is 0 for the appropriate value of y mod b. So if d is a quadratic residue mod b with gcd(d,b)=1 there are squares ending in arbitrarily many d's in base b. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Need advice on letters of recommendation> Rudolph :>>mmanch01@my-deja.com (maky ludicrous mechanism in>> place in academia - it's a shame. institutional admissions exams>> would be far superior. Letters of recommendation may well be ludicrous. I have no reason>> to believe, and much reason to disbelieve, that institutional>> admissions exams would be any better at deciding who would be>> a good graduate student.>perhaps you do not believe that a solid undergraduate>preparation is key component in graduate studies success.>is that it?I suppose I believe that, for a value of key that may be> lower than your value of key. What I'm sure I believe is that> (1) institutional admissions exams probably aren't particularly > good at measuring that component in graduate studies success> (since I don't have much faith in exams), and thatwell, your faith regarding institutional exams has to be the weakestdefense you have for your belief. perhaps if you made a statement, andjusti'ed it, regarding subjectivity, perhaps you would have betterluck.> (2) as Bart Goddard has pointed out, letters of recommendation do> have at least some chance of indicating that other (for me, much more> key) component in graduate studies success, namely, the > ability to *do mathematics*, while (3) institutional admissions> exams surely couldn't do anything at all towards indicating that> other component.the ability to solve mathematical problems, say in an admissions exam,is without a doubt, a much better indicator of ability to domathematics. in fact, the ability to solve mathematics problems is theonly accurate indicator of ability in mathematics. do you understandwhere i am coming from?>> Go into detail on how you'd make such an exam, if you don't mind.>i do not need to go into any detail to ascertain that a>resounding reason for the lack of institutional exams in the>united states is correlated with the economic and logistics>required to put such systems in place.You certainly do not need to go into any detail to do that> on *my* behalf, because I didn't express (and don't have) any> interest in your opinion on that question, which I didn't raise.> The question I raised was, How would you make such an exam (with> some reasonable hope that it would do what you want it to do)?you propose the question, you seek the answer. not from me though.> Lee Rudolph === Subject: Re: Question on generation of large prime numbersRichard Heath'eld Carmody is a damn sight better at mathematics than I am. > But I think I got a case. :-) ]I hope you've got a hot-shot lawyer...Your premise is:>> If there is a largest prime, P, and that's all, as can be deduced from the then and therefores hereafter.>> then the number of primes is 'nite. We>> can therefore form a new number, N, which is one greater than the product>> of all primes. Therefore, N > P (and, indeed, it would be very much>> greater than P). When divided by any prime in our list, it leaves remainder 1. Therefore,>> /either/ it is prime /or/ it is the product of two or more primes, at>> least one of which is greater than P.> In /either/ case, P has been shown not to be the largest prime number.Nope. Nowhere have you assumed that the list of primes known is> complete up to P.On the contrary, I formed N by multiplying all primes Erm, says who? You've not de'ned all in this context. Nowhere do you say anything about having a contiguous set of primes from 2 to P. The only thing you've said is that your set of primes has a maximumelement P. Full stop. There were no further premises -- look above if you don't believe me.Anyway, your maximum prime P is a red herring and completely unnecessary - why did you even introduce it in the 'rst place?> and then adding 1. To > multiply all primes together, I must 'rst have a complete list of primes.You must have a list of primes. What does Ocomplete' mean to you in this context? If you wish to multiply together all primes between 2 and P then you must know /a priori/ all the primes between 2 and P. How do you know that you know all those primes? If you wish to use that as an assumption your argument you must state so. You didn't.But why would you want to?Let the complete list of primes be {2}.The multiplication/addition yields 3. The factorisation oracle tells you that that's prime. Try again - let the complete list of primes be {2,3}The multiplication/addition yeilds 7. The factorisation oracle tells you that that's prime.Try again - let the complete list of primes be {2,3,7}Do you permit me to do the above?Are you going to claim that I must insert 5 into that set?If so, then why? Would Euclid insist that I inserted 5? What would bethe justi'cation?If not, then your argument seems to be too transient and 'ckle to be able to be countered rationally.> If the list is not complete, then I cannot perform the multiplication.Erm??? If the assumed 'nite set of primes is {2,3,7}, then the multiplication is trivial, and yields the new prime 43. Can you not multiply 2, 3, and 7 together? Completeness is not a clearly de'ned in this context. At this point we _don't know_ that 5 is prime. Unless your book of axioms says oh, you may assume 5 is prime whenever you like, but _my_ book of axioms differs from yours in that regard.The same for _any_ 'nite set of primes. Euclid's proof simply requires that one assume there's a 'nite set of primes, it does _not_ ask one to assume that this set has any other properties. Why do you wish to give it extra unnecesary properties? In particular, properties that rely on an external or prior prime generation?As I said before, and gave an example which you for some reasonsnipped, if you use the Euclid method as a constructive prime generator you very quickly end up generating primes smaller than the previously known maximum. (On the assumption that this generator is the only generator you have.)> Don't forget that if you're pretending that you don't know what> the primes are, you can't suddenly pull otherwise well-known> features of the prime numbers out of a hat.It's not necessary to know all the primes in order to construct the > argument. It is suf'cient to know that it is /in theory/ possible to know > them, e.g. by sieving. It is only necessary to know all the primes if you > wish to perform the multiplication in real life.That makes it sound like you've missed the entire point of Euclid's proof.Euclid's proof shows that you never need to to _any multiplications at all_ in real life.As an _existance_ proof, Euclid's algorithm is non-constructive.If you're trying to use it as a _constructive_ algorithm, why the heck do you want/need to invoke a second constructive algorithm (the one that you used to generate all primes up to P)?Phil-- Unpatched IE vulnerability: 'le-protocol proxyDescription: cross-domain scripting, cookie/data/identity theft, command executionReference: http://safecenter.net/liudieyu/WsOpenFileJPU/ WsOpenFileJPU-Content.HTMExploit: http://safecenter.net/liudieyu/WsOpenFileJPU/ WsOpenFileJPU-MyPage.HTM === Subject: Re: Real torsion (0,1) matricesBill :> If A is a real nXn (0,1) matrix such that A^m = I (the identity> matrix) for some m, must A be a permutation matrix ?Yes. AB = I where B = A^{m-1} has nonnegative integer entries.are nonnegative integer linear combinations of the rows of A --this is only possible in trivial ways.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Mathematical IntegrityJ.) Metz purity and integrity in mathematics?||Neither purity nor integrity has anything to do with civility or with|the ability to suffer fools gladly.They are separate issues, but they are not unrelated. I 'nd thatthe people I've gotten acquainted with, who have a reputation fornot suffering fools gladly or who are proud of not doing so,nearly all can be seen erring on the side of prematurely concludingthat some idea or person was to be rejected. Incivility conduces tohostility, and hostility clouds the mind. On a crude level, it ispossible to be accurate about people who annoy one, but I 'nd ithard, and I think most people do too, to be very observant in sucha frame of mind. One feels a much greater temptation to slack offon efforts at describing these fools accurately, which is a lapsein integrity.For example, one mathematician I met had considered whetherit would work well to use an alternative treatment of a certainissue in functional analysis. He looked for anyone who mighthave addressed the issue, and found that of all the places helooked, it was solely in a textbook by a second mathematicianthat a reason was given why we should do it the usual way. Butthe 'rst mathematician had a counterargument, though, whichhe wanted to present to the second mathematician. The secondone (who gave me an impression of being usually conscientiousabout addressing mistakes in his work and so on) had thisdon't suffer fools gladly attitude, and when approached by the'rst, stormed angrily away declaring he had no time to discusssuch a thing. I realize that I'm expecting the reader to believe mewhen I indicate that this was an overreaction, although it's hardto demonstrate. It wasn't a matter of the second mathematicianbeing interrupted during something important and the like. Andthe discussion was not simply postponed to a more convenienttime.It is sometimes said that this kind of stern response is needed tokeep at bay the large volumes of nonsense that we are all exposedto. I think the value of such hostility is much overrated, however.Plenty of people are civil and relatively nonjudgemental withoutletting very much of their time be wasted on pointless pursuits.My opinion is that society in the U.S. is generally erring on theside of glibness, and that there are many involved in politics whoare keen to encourage us to err still farther in that direction.Rather than reasoning things out, we get discussions in whichpeople express shock and horror that the plain, obvious answeris not being accepted immediately. I see pundits trying to get menot only to agree with them, but to agree that it's so painfullyobvious that I needn't bother considering the other point of view,at least not seriously. This goes beyond incivility, but it's the kindof problem that incivility encourages.Keith Ramsay === Subject: Re: Question on generation of large prime :|> Richard Heath'eld number of primes is 'nite. We|>> can therefore form a new number, N, which is one greater than the product|>> of all primes. Therefore, N > P (and, indeed, it would be very much|>> greater than P).|>> |>> When divided by any prime in our list, it leaves remainder 1. Therefore,|>> /either/ it is prime /or/ it is the product of two or more primes, at|>> least one of which is greater than P.|>>|>> In /either/ case, P has been shown not to be the largest prime number.|> |> Nope.Yes it has.|> Nowhere have you assumed that the list of primes known is|> complete up to P.He doesn't need to assume that all the primes up to P are known for theargument to work. He just assumes (for the sake of proof by contradiction)that P is the largest prime.|On the contrary, I formed N by multiplying all primes and then adding 1. To |multiply all primes together, I must 'rst have a complete list of primes. |If the list is not complete, then I cannot perform the multiplication.There are two versions of the proof. One proceeds like you just did,starting with a (false) assumption that there is a largest prime. Theother just starts by considering an arbitrary 'nite set of primes. If Iremember correctly, the latter is more like Euclid's own proof. The crux isthat the N you get by adding one to the product is always divisible by aprime not in the set. It's perhaps worth noting that you're not reallyusing the assumption that P is the largest prime, then, until at the endwhen you contradict it.proofs as if it was a proof by contradiction, by starting it with anassumption that the conclusion of the proposition was false, and deducinga contradiction in each case where the result was shown true. This is astylistic weakness to avoid.In this case starting with the assumption that P is the largest prime doesnot seem so bad; at least it seems more popular. It doesn't really play afunctional role, though, and it's perhaps worth doing it the other way tomake it more obvious that we're seeing as well a method for 'nding a primeoutside of any 'nite set of primes.Keith Ramsay === Subject: Re: chess bishops - combinationsparmito :> I understand that it is possible to express the number of ways of> having k bishops on an n*n board such that they are not attacking each> other as a combinatorial formula and i have been trying to derive the> formula without any luck, can anyone help me out ?Is this right?.. Each bishop has a coordinate (r,c) (row, column). For allthe bishops, the r's are different to each other, and the c's are differentto each other.So for k bishops the total amount of possible coordinates is how manydifferent ways you can take k r's out of 8, times how many different waysyou can take k c's out of 8. And that should be enough for you.. If it'sright..I think so.-- Quaternion === Subject: Re: 've have given 5 vectors and have to 'nd a>> base for V = L(v1, v2, v3, v4, v5). Then I must express the 'fth>> vector through Dave. But the explanations I have found on it so far are to> complicated> for me. Can you give me at least one really short example for the> application of this method? Please!> Best understand how this method could be> helpful> for me. (:-{ ]You are given 've vectors in R^4. The Gram-Schmidt algorithm allows youto 'nd a minimal set of vectors that spans the same subspace as thegiven vectors. If your vectors happen to span the entire space, thenGram-Schmidt gives you a basis for that space.As a bonus, the vectors you get from Gram-Schmidt are orthonormal, whichmakes it very easy to decompose any given vector as a linear combinationof the basis vectors.The important thing is not to get bogged down in computation until youunderstand what the method is doing.The 'rst vector is very easy. If the vector v1 is nonzero, then {v1} isa linearly independent set, and therefore all you need to do isnormalize. Take u1 = v1 / ||v1||. That is, divide the 'rst vector byits Euclidean norm, and call that u1, which is a unit vector pointing inthe same direction as v1.Next consider your second vector, v2, and its projection along u1, whichis (v2 . u1) u1. If v2 happens to be a scalar multiple of u1, this dotproduct will just give you v2. If it's not a scalar multiple, then thedot product gives you the component of v2 in the direction of u1. Youneed to subtract this from v2 in order to get the orthogonal complement,which is w2 = v2 - (v2 . u1) u1and then normalize, obtaining u2 = w2 / ||w2||.Next consider v3 and its components in the directions of u1 and u2.Since we now have two vectors in our orthonormal set, the orthogonalcomplement is w3 = v3 - (v3 . u1) u1 - (v3 . u2) u2.If w3 is zero, it means v3 is already in the subspace spanned by u1 andu2, and therefore you can ignore it and move on to the next vector. Ifw3 is nonzero, then you normalize it and add it to the basis you arebuilding: u3 = w3 / ||w3||and so on.If it turns out that v5 is a linear combination of the other vectors thatyou started with, then it will reduce to a linear combination in terms ofyour orthonormal basis, and you can easily compute its components. Thenjust substitute what each of the u_i's is in terms of the v_j's, andyou're done.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Mathematical IntegrityVery well said. I agree.Have a tolerable existence. Eli.KRamsay :> |Neither purity nor integrity has anything to do with civility or with> |the ability to suffer fools gladly.They are separate issues, but they are not unrelated. I 'nd that> the people I've gotten acquainted with, who have a reputation for> not suffering fools gladly or who are proud of not doing so,> nearly all can be seen erring on the side of prematurely concluding> that some idea or person was to be rejected. Incivility conduces to> hostility, and hostility clouds the mind. On a crude level, it is> possible to be accurate about people who annoy one, but I 'nd it> hard, and I think most people do too, to be very observant in such> a frame of mind. One feels a much greater temptation to slack off> on efforts at describing these fools accurately, which is a lapse> in integrity.For example, one mathematician I met had considered whether> it would work well to use an alternative treatment of a certain> issue in functional analysis. He looked for anyone who might> have addressed the issue, and found that of all the places he> looked, it was solely in a textbook by a second mathematician> that a reason was given why we should do it the usual way. But> the 'rst mathematician had a counterargument, though, which> he wanted to present to the second mathematician. The second> one (who gave me an impression of being usually conscientious> about addressing mistakes in his work and so on) had this> don't suffer fools gladly attitude, and when approached by the> 'rst, stormed angrily away declaring he had no time to discuss> such a thing. I realize that I'm expecting the reader to believe me> when I indicate that this was an overreaction, although it's hard> to demonstrate. It wasn't a matter of the second mathematician> being interrupted during something important and the like. And> the discussion was not simply postponed to a more convenient> time.It is sometimes said that this kind of stern response is needed to> keep at bay the large volumes of nonsense that we are all exposed> to. I think the value of such hostility is much overrated, however.> Plenty of people are civil and relatively nonjudgemental without> letting very much of their time be wasted on pointless pursuits.My opinion is that society in the U.S. is generally erring on the> side of glibness, and that there are many involved in politics who> are keen to encourage us to err still farther in that direction.> Rather than reasoning things out, we get discussions in which> people express shock and horror that the plain, obvious answer> is not being accepted immediately. I see pundits trying to get me> not only to agree with them, but to agree that it's so painfully> obvious that I needn't bother considering the other point of view,> at least not seriously. This goes beyond incivility, but it's the kind> of problem that incivility encourages.Keith Ramsay === Subject: Re: Semi-normal numbers Was: if x has normally distributed digits, does 1/x?Toni Lassila :> [...]De'nition:A normal number whose reciprocal is non-normal is called a seminormal>number. A number whose reciprocal is normal is called an inverse->normal number.Theorem: Almost all real numbers are inverse-normal.Proof: [omitted]>The reciprocal is a measure automorphism on the nonzero reals. Given that almost all reals are normal, doesn't the conclusion follow immediately? One might need to invoke absolute continuity of measures.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Question on generation of large prime numbersRichard largest prime, P, then the number of primes is 'nite. We can >> therefore form a new number, N, which is one greater than the product of >> all primes. Therefore, N > P (and, indeed, it would be very much greater >> than P). When divided by any prime in our list, it leaves remainder 1. Therefore, >> /either/ it is prime /or/ it is the product of two or more primes, at least >> one of which is greater than P.>> In /either/ case, P has been shown not to be the largest prime that the list of primes known is > complete up to P. You've simply proved that the 'nite list of> known primes is not the list of all primes.Well, Richard and Phil are both wrong here. The proof could be expressedin terms of a 'nite number of known primes, as Phil seems to have assumedit was, but that's not the way Richard expressed it -- he spoke explicitlyof the product of *all* primes.But Richard's conclusion that P has been shown not to be the largestprime number is also wrong; what has been shown is that the assumptionof a 'nite number of primes is wrong. The only reason to assume thatat least one of the additional primes discovered must be larger thanP is that we *thought* P was the largest prime, before we discovered them.-- Mark Brader | Sir, your composure baf§es me. A single counterexampleToronto | refutes a conjecture as effectively as ten... Hands up!msb@vex.net | You have to surrender. -- Imre Lakatos === Subject: Re: Reality of response to my work> Justin Van Winkle :> I think this is morally wrong.Since you top posted, I have no idea exactly *what* was said that youregard as> morally wrong.> Don't tell James to waste his money.I only recommended that he get the advice of a patent attorney if hewanted to> secure intellectual property rights. I assume you regard your *free*advice> about the matter as superior to the advice of a professional. If so, youmay be> guilty of practicing law without a license.Is it illegal to practice law without a clue?[snip!]Justin Van Winkle === Subject: Re: probability 2......norrisdt@rintintin.colorado.edu (Doug increasing order of size, so the>length of the pieces are a < b < c. Then c = Sqrt[1-a^2-b^2], which>also must be larger than Sqrt[a^2+b^2].Yeah, I messed the details on this one, as I've found by playing aroundwith it some more. It's always a treat to post upon waking. :-PDoug === Subject: Re: Relativity is based on assumption.> Androcles :> All theories, in physics and elsewhere, are based on assumptions.> No theory can bootstrap itself out of nothing.> So what?> OSo what' depends on the validity of the assumption. The PoR is an> assumption, and is intuitive as well. Making the assumption that the> time it> takes for a signal to reach an object is the same as the time it take> for> the signal to return, when in the meantime you've moved away or toward> the> object, is a rather silly assumption that I will not accept. That's Oso> what'.> Androcles> Then ignore relativity.> But unless you can come up with something that agrees with the> experimental evidence better than relativity, everyone else will> ignore you.> What experimental evidence? Moving clocks running slow? They don't. The GPS> clocks run fast. MMX? It's pretty obvious to anyone with half a brain that> the result is what you would expect if the speed of light were source> dependent. I would say that agreed with the experimental evidence much> better than relativity.> AndroclesAndrocles1. Are you saying that the speed of light is source dependent? 2. What is your de'nition or explanation of Osource dependency'?Peter Riedt === Subject: Andrei Linde speak last night at UC Berkeley as my updated version ofhttp://qedcorp.com/APS/EmergentGravity.doc shows. Also a rather large crowd showed up to hearmy talk right in the middle of the talk before mine.Linde explained that he does not believe most of the current textbook presentations of in§ationary cosmology.The essence of his talk is in my 2 equations II.9 and II.10 in the new version of http://qedcorp.com/APS/EmergentGravity.docThe key is the friction term in II.9 which is a linearization of my II.8 near the bottom of a local minimum in the effective potential of the vacuum coherence 'eld in the FRW large scale limit assuming homogeneity.The friction term that is the root cause of chaotic in§ation with slow descent at large order parameter followed by oscillatory reheating to make the post-in§ationary Big Bang is quite simply the effect of the connection 'eld for parallel transport in the second application of the covariant time derivative to the scalar 'eld. Of course Linde has no micro-dynamics for the emergence of the scalar 'eld as I do nor does he derive Einstein's gravity together with the uni'ed exotic vacuum dark energy/matter 'eld in a two-way bootstrap of the never-ending spontaneous self-organizing parallel universes splitting off into baby universes in the sense of Andrei Sakharov's metric elasticity a special case of P.W. Anderson's More is different.Steve Carlip gave a interesting talk on topology change in WKB approx to quantum gravity with a possibleexplanation of why space is homogeneous on large scale consistent with in§ation. === Subject: Product of matrix elements?Given two nx1 column matrices (vectors) X=(x1 x2 ... xn)^T and Z=(z1 z2 ... zn)^T, all elements real numbers (T=transpose):1) Is there a simple matrix operation to create the nx1 matrix(x1*z1 x2*z2 ... xn*zn)^T from X and Z? I am referring to a mathematical operation like inner product, rather than an algorithm or computer program.2) Is there a way using matrix operations to produce the column vector (exp(x1) exp(x2) ... exp(xn))^T (exp=exponential function) from X?3) Related to (2), is there a way using matrix operations to produce the nxn diagonal matrix D(x1 0 0 ... 0)(0 x2 0 ... 0)( ... )(0 0 0 ... xn)from X? And conversely, given D as above, to write X in terms of D using matrix e-mail) === Subject: Proving if an inverse function to a matrix trouble understanding how to apply the inverse functiontheorem to matrix transformations. The textbook that I am using hasexamples of using the inverse function theorem for ordinary R(n)->R(m)functions but not for matrix transformations such as S(X)=X^3, where X isin Mat(3,3) for example. Here is the problem that I need to solve and mysolution. I would greatly appreciate if someone could go over my reasoningand point out any §aws that I have and explain to me how to solveproblems of 0 0 ]Note that A^3 = I (identity matrix). Is there a cont. differentiablefunction g such that g(I)=A and (g(A))^3=A in the neighborhood of I? My Solution:Let f: X -> X^3 (X is a matrix)then f(g(x)) = [g(x)]^3 = xif X = If(g(I)) = INow use the chain rule for the derivative of f(g(I)):[D(fog)(X)] = [Df(g(X))][Dg(X)] (by de'nition)Let X = I:[D(fog)(I)] = [Df(g(I))][Dg(I)] = [Df(A)][Dg(I)] (since g(I)=A)Finding [Df(A)] is a bit tricky. I used the general de'nition of thederivative to show that[Df(A)]H = 3A^2H + 3AH^2since lim {(1/H) * (f(A+H) - f(A) - (3A^2H + 3AH^2))} = 0. Now I let H = Iso that [Df(A)]I = 3A^2 + 3AI then went on to show that the determinant of this matrix (using A fromthe problem) = 54 (is nonzero).g(X), asked for in the problem exists.However, I am a bit confused myself as to whether this reasoning iscorrect and whether I am using the inverse function theorem correctly. Iwould greatly appreciate if someone could comment on my solution andexplain to me how to do these types of problems. === Subject: Re: Axiom of Foundation (absymally stupid question)Since we're discussing the axiom of foundation (in the textbooks I've seen, it's called the axiom of regularity), does anyone know what the intuitive justi'cation for this axiom is? I mean, all the other axioms seem pretty natural to me, even the infamous axiom of choice. But where in world did they come up with the axiom of regularity?Have a tolerable existence. Eli === Subject: Re: Permutation: how to detach cycles/transpositions napisa w wiadomoci> I don't understand... Your cycle is already decomposed into a product of> disjoint cycles! (1 3)(2 5 4)(6) is not equal to (3 5 1 2 4 6).> By (3 5 1 2 4 6) he meant p(1)=3, p(2)=5, etc... in which case the> decomposition holds.Right, sorry for lack of any description to my question.-- Pawe StobiskiRepublic of Poland === Subject: Vector Calculus problemOk, I'm trying to work on my homework and am stuck on 4.9 #10 of VectorAnalysis by Davis. The question states By means of Stokes' theorem, 'nd S F*dR around theellipse x^2+y^2=1, z=y, where F=xi+(x+y)j+(x+y+z)k.I got the curl of F and that equalled i-j+k but I'm not really sure how to dothe rest of the problem. Any help would be appreciated. I've wasted a lot oftime and gotten almost nowhere. === Subject: Re: Vector Calculus problemI forgot to mention that the answer is in the back of the book:+-2pi depending upon the direction of integration.I just can't 'gure out how to get that. === Subject: and I am having trouble with the de'nition of such a group. What exactly are the open sets?Any help would be === Subject: ADE Groups, Cosmology and M-M Sirag :Jack & Jonathan,The *Roots of Consciousness* by Jeffrey Mishlove on the web does NOT includemy 40 page appendix paper, Consciousness: a Hyperspace View. Jeffreywanted to include it but there was some dif'culty with the complexity ofthe 'gures -- if I remember correctly. Of course, it had never been in pdfform, since the web (& home computers) were in a very primitive state in1993. This paper has the most detailed account of what I now call ADEXtheory -- the application of the A-D-E Coxeter graphs to mathematics,physics, and other 'elds such as consciousness theory. This appendix paperwas written in 1989, and published in 1993. A short summary and update ofthis paper (with more ADEX examples) was titled, A Mathematical Strategyfor a Theory of Consciousness. It was written in 1994 and published in thebook, *Toward a Science of Consciousness: the First Tucson Discussions andDebates* (edited by Stuart R. Hameroff, Alfred W. Kaszniak, and Alwyn C.Scott), MIT Press, 1996.People have told me that my papers are hard to read without much moremathematical knowledge than they possess. Jeffrey tells me that Russianscientists have less trouble with the math since their mathematical trainingis more advanced than that of American psychologists and biologists.Actually much of the mathematics of ADEX theory is of very recent vintage,but the key area of mathematics is quite old -- group theory (both 'nitegroups and Lie groups and relationships between them). Our understanding ofthese relationships depends on both algebra and geometry -- especiallyhyperspace geometry -- including differential geometry and algebraicgeometry.Physicists who study general relativity know some differential geometry, butthey have not studied the more recently developed algebraic geometry -- andvery few physicist (or mathematicians) have seen the great utility of theA-D-E Coxeter graphs which have been used to classify more than 20mathematical objects. The advantage of having these classi'cations, is thatthe A-D-E graphs provide the relationships between all these mathematicalobjects. I will mention a few of these objects -- the study and applicationof which I call ADEX theory: 1. Finite re§ection groups (Coxeter groups also called Weyl groups) 2. Hyperspace polytopes and thus crystallographic lattices 3. Coxeter arrangements (mirrors in re§ection space) 4. Lie algebras and Lie groups (& also Kac-Moody Lie algebras) 5. Thom-Arnold catastrophe bundles (useful for Jack's version of Bohm) [BTW: Thom claimed (1975) that it models the mind-body relationship][Yes, this is the aspect I want to §esh out with you.] 6. McKay groups ('nite subgroups of SU(2) -- unit length quaternions) 7. Gravitational Instantons (closely related to Penrose twistors) 8. 2-d Conformal 'eld theories (which live on hyperspace strings)[Jack: It is interesting that O(2) macro-quantum order parameter in ordinary space has string defects e.g. Hagen Kleinert also books on soft condensed matter physics and cosmic strings. Then introduce extra space dimensions including fermi dimensions for supersymmetry to get higher dim branes from the macro-quantum order parameters perhaps with higher O(N) internal symmetry, hyper-complex matrix order parameters over hyper-complex generalized space-time manifolds.My model in http://qedcorp.com/APS/EmergentGravity.doc is only the low energy tail of that. BTW new version shows how to go from my BIT FROM IT Landau-Ginburg eq to Andrei Linde's speci'c equations for chaotic in§ation as he presented them Fri nite friction term from my equations a year ago not realizing their crucial role in Linde's theory of the continuous creation of the parallel universes.] 9. Error-correcting codes (related to Jack's IT <--> BIT idea)That too. The mind 'eld must have error correcting codes built into it.] 10. Quantizing lattices (analog to digital transforms)As the motto for Plato's academy said, Let no one enter here withoutgeometry. Today, this geometry must include hyperspace geometry -- whichdates back to the 19th century.BTW: *Roots of Consciousness* is mentioned at the end of John McKay's veryshort paper, A Rapid Introduction to ADE Theory. The URL for this paperis: http://math.ucr.edu/home/baez/ADE.htmlThis is on John Baez's very extensive website, and from the above URL youcan access 4 much longer (tutorial) papers by John Baez on the ADE relatedmathematics.Nuff said!Saul-Paul----------,Jeffrey === Mishlove Subject: Re: Roots of Consciousnesssent to Saul-Paul and MishloveAll the old stuff is really obsolete IMHOit's pre-science both Leibniz and Young.It's not asking the right questions not using the right concepts.It's mainly metaphor and not useful given today's advances. === Subject: Re: How do you prove that the circumference of a circle is proportional to it's radius? panamars@otenet.gr (Eur Ing Panagiotis Stefanides) modern area formula (pi)(r^2) or (pi)(d^2)/4 ....Ken ,> Is this really a modern formula for circle's area?.... What I meant was modern _notation_ for the formula. I'm sorry ifthat wasn't clear. Ken Pledger. === Subject: Playing with Fibonacci, (n^2 - n + 2)^2 / 4 - n )was quite interesting, as soon as n is very great (if n is small,the property is still true, but not easy to see).Take n=2^67; then, you can notice that starting from the 4-th term in the expansion, you will 'nd blocks starting with a great value, and a few very small terms just after. What is interesting is that a formula f(n,i) can be built that returns rationals equal to the rational built from the i-th block. Here is an example: for n=2^67 The 38-th block is [22801,14,1,1,5,15,1,1,3,3,1,2,1,1,1,18,...] (length is 33). Considering this block is the continued fraction of a rational, we have the 38-th rational, being: 238...709 / 'bonacci(38+2)^2Now, I found the formula that gives rational equal to the successiveblocks: §oor( (n-epsilon) / ('bo(i+1)*'bo(i+2)) ) [Times] ( ((-1)^i [Times] 'bo(i+1)[Times](n-'bo(i+2)-1)[Times][CapitalTho rn]bo(i+2)-1) mod ('bo(i+2)^2)) / ('bo(i+2)^2)Now, just put epsilon=0, and you will see the formula is very good.BUT... The integer part is sometimes greater (diff=1) than the realvalue; the rest of the block is 'ne.Thus, I think that an epsilon value should be put in the formula.It looks like epsilon depends on both n and i.It means that §oor( n / ('bo(i+1)*'bo(i+2)) )is almost the right value for the successive big values in theinitial continued fraction expansion, but not quite exact.Could someone 'x the formula ?PS - may my formula be simpli'ed (for instance, is it possible to remove the (-1)^i ?)Cordially, === Subject: Re: Problems book toolshed37@yahoo.com (Nobody) :> Is there a book similar to Berkeley Problems in Mathematics, but> geared more towards the undergraduate, and perhaps more 'nding> stuff rather than proving stuff? I would hope the problems would be> of the same relative dif'culty to an undergraduate as the BPM book> would be to a graduate. You may 'nd some of what you want in D.K. Faddeev & I.S. Sominskii(trans. J.L. Brenner), Problems in Higher Algebra, Freeman, 1965. It'sold-fashioned, but has a lot of good problems. Ken Pledger. === Subject: Re: Question on generation of large prime numbersRichard Heath'eld and Phil Carmody write:>> If there is a largest prime, P, and that's all, as can be deduced from the then and therefores hereafter.>> then the number of primes is 'nite. We>> can therefore form a new number, N, which is one greater than the product>> of all primes. ...> Nope. Nowhere have you assumed that the list of primes known is> complete up to P.On the contrary, I formed N by multiplying all primes Erm, says who? You've not de'ned all in this context. Nowhere do you > say anything about having a contiguous set of primes from 2 to P. Phil, stop blathering about de'ning Oall'. You misread or mis-interpreted Richard's wording, and you've been caught.> The only thing you've said is that your set of primes has a maximum> element P. Full stop. There were no further premises -- look above if > you don't believe me.We're dealing with positive integers. If there is a largest one in aset of them, then that set is 'nite, as Richard said, and if it's'nite, then a product of all its members exists. It is not necessaryto specify an algorithm to construct the set (although in this case itwould be simple to do so), nor to de'ne Oall' in this context.-- Mark Brader, Toronto | Professor, I think I have a counterexample.msb@vex.net | That's all right; I have two proofs. === Subject: Re: Squares that end with four identical digitsIn the code for the Maple procedure quadres that I previously posted therewas a procedure legendre_pow that is not normally accessible. However,Robert Israel and Alec Mihailovs showed me two ways to get it. I haveattached the method by Alec to the end of this message. --Edwin> Here's the Maple code for quadres:> showstat(numtheory:-quadres);numtheory:-quadres := proc(a, p)> local fac, ig, n, pf, s;> 1 if nargs <> 2 then> 2 error invalid arguments> end if;> 3 n := modp(a,p);> 4 if type(n,integer) and issqr(n) then> 5 return 1> end if;> 6 if not (type(n,integer) and type(p,integer)) then> 7 return Oprocname(args)'> end if;> 8 if p < 6 then> 9 return -1> end if;> 10 if isprime(p) then> 11 return mods((Opower')(n,1/2*p-1/2),p)> end if;> 12 ig := igcd(n,p);> 13 if ig <> 1 and igcd(denom(n/ig^2),p/ig) = 1 then> 14 return numtheory:-quadres(n/ig^2,p/ig)> end if;> 15 pf := ifactor(p);> 16 if type(pf,`^`) then> 17 return legendre_pow(n,pf)> end if;> 18 for fac in pf do> 19 if type(fac,`^`) then> 20 s := legendre_pow(n,fac)> else> 21 s := numtheory:-quadres(n,op(fac))> end if;> 22 if s = -1 then> 23 return -1> end if> end do;> 24 1> end proc>Here's the code for legendre_pow following Alec's method:kernelopts(opaquemodules=false):showstat(numtheory:- legendre_pow);numtheory:-legendre_pow := proc(nn, pow)local b, n, r; 1 r := op(2,pow); 2 b := op(op(1,pow)); 3 n := modp(nn,b^r); 4 if n <> 0 then 5 while igcd(n,b) <> 1 do 6 if not type(n/b^2,integer) then 7 return -1 else 8 n := n/b^2 end if end do end if; 9 if has({0, 1, 4, 9, 16},n) then 10 return 1 end if; 11 if b = 2 then 12 if modp(n,2^min(3,r)) = 1 then 13 return 1 else 14 return -1 end if end if; 15 quadres(n,b)end proc === Subject: Re: Axiom of Foundation (absymally stupid question)Eli :>Since we're discussing the axiom of foundation (in the textbooks I've seen, >it's called the axiom of regularity), does anyone know what the intuitive >justi'cation for this axiom is? I mean, all the other axioms seem pretty >natural to me, even the infamous axiom of choice. But where in world did >they come up with the axiom of regularity?>The Axiom of Foundation is not really necessary - exactly 99.93% (:-)) of mathematics can be done without it. In fact, it even limits our universe of discourse when we accept it. However, it is there to make things nice. Without it, it is consistent to have pathologies such as a set which is an element of itself, two distinct sets each of which is a member of the other, and a sequence x of distinct sets such that x_(n+1) is an element of x_n.The axiom also allows an ice-cream cone construction of the class of sets. With the axiom, there is a sequence R through the ordinals such that a < b implies R(a) is a subset of R(b) and the universe of sets is the well-founded sets, viz., theunion of all these sets. (In particular, R(0) = 0 and R(a+1) = P(R(a)).) This structure allows certain proofs in set theory; for example, the well-founded sets provide a model for ZF.I get most of this from Kunen.One more question: If we didn't mind invoking the axiom of foundation, wouldn't {a, {a,b}} suf'ce as a de'nition of the ordered pair (a,b)? Seems to me the answer is yes.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Inversion? jon@pidham.vispa.com (Jonathon) :> When I did A level maths at school, about 30+ years ago, in my> analytical geometry course I was taught a subject called Oinversion'.> .... > I have never come across this again, and the subject seems to heve> disappeared from text books.... You'll 'nd it in Chapter 5 of the recent text-book by Brannan,Esplen & Gray, Geometry, Cambridge U.P., 1998. They show how inversionis connected with Moebius transformations of the complex plane.> .... Does anyone know how/when this concept came about?.... I'd like to know more about that, too. Can anyone help? Ken Pledger. === Subject: Bottom line on prime counting issueSome of you may have noticed frenetic activity from posters trying toconvince you that there's nothing sinister about mathematicians doingtheir best to downply my 'nd of a way to count prime numbers byintegrating a partial difference equation, but what's the bottom line?Does what I found work or not?It does. End of story, so mathematicians should acknowledge it. I't's not important they can just put it in some math text somewhere,or in some journal and drive on.No big deal.But they're 'ghting to totally ignore it. Translation: Sinisterattempt by academic types to hide something really important.Otherwise, why go to so much effort to 'ght me, when a simple way toshut me up on the issue is just record it somewhere? And it is aFIRST in human history, so use your common sense.The loser academic world is 'ghting me over something that works. End of story.These posters trying to convince you otherwise are just insulting yourbasic intelligence.My math discoveries, found for pro'thttp://mathforpro't.blogspot.com/ === Subject: Re: topological groups> I'm reading about topological groups and I am having trouble with the> de'nition of such a group. What exactly are the open sets?>Hey,You can't talk about the open sets since a group is said topological if't is continuous ( (x,y)-> x+y and x->x^(-1)are continuous); sometimes it isalso required to be Hausdorff. But considering a group (G,+) you can putdifferent topologies on G so that G is a topological group (provided that Gis continuous (and sometimes, Hausdorff)). We're not de'nining a specialtopology here.--Julien Santini,France. === Subject: Re: topological groups>> I'm reading about topological groups and I am having trouble with the>> de'nition of such a group. What exactly are the open sets? Hey,You can't talk about the open sets since a group is said topological> iff it is continuous ( (x,y)-> x+y and x->x^(-1)are continuous);> sometimes it is also required to be Hausdorff. But considering a group> (G,+) you can put different topologies on G so that G is a topological> group (provided that G is continuous (and sometimes, Hausdorff)).> We're not de'nining a special topology here.--> Julien === Subject: Re: Bottom line on prime counting issue :Some of you may have noticed frenetic activity from posters trying to> convince you that there's nothing sinister about mathematicians doing> their best to downply my 'nd of a way to count prime numbers by> integrating a partial difference equation, but what's the bottom line?Does what I found work or not?>Now that is an interesting question, isn't it! pssst, hey Harris... your stuff doesn't work.... but don't tell anybody... === Subject: Re: 've vectors Karl Pech :> have given 5 vectors and have to 'nd a> base for V = L(v1, v2, v3, v4, v5). Then I must express the 'fth> vector through this base.> The vectors are:> ^^^^^^^^^^^^^^^^^^> v1 = (1,-2,0,3); v2 = (2,-5,-3,6);> v3 = (0,1,3,0); v4 = (2,-1,4,-7)> v5 = Dave. But the explanations I have found on it so far are to> complicated> for me. Can you give me at least one really short example for the> application of this method? Please!Best understand how this method could be> helpful> for me. (:-{ ]> Using Gram-Schmidt here is rather like using a sledge hammer to crack eggs. It is way too much work for the result desired.A much better approach is to try to determine whether each vector, in (v1,v2,v3,v4,v5) is a linear combination of the previous ones.This can be done pretty much by eyeball analysis. One 'nds that v1 is trivially independent,v2 is not dependent on (v1),v3 is depentent on (v1,v2), v3 = 2*v1 - 1*v2v4 is not dependent on (v1,v2) so not on (v1,v2,v3) either, andv5 is dependent on (v1,v2,v3,v4), v5 = v1 + v2 + v4Thus (v1,v2,v4) is a basis, v1, v2 and v4 can trivially be expressed as linear conbinations of basis vectors and v3 and v5 can be expressed less trivially as linear combinations of these basis vectors. === Subject: Re: Question on generation of rec.puzzles Richard Heath'eld :>> It's not necessary to know all the primes in order to construct the>> argument. It is suf'cient to know that it is /in theory/ possible to>> know them, e.g. by sieving. It is only necessary to know all the primes>> if you wish to perform the multiplication in real life.So, since you can't sieve all of them there is a (practical) largest> prime!Great! Please share...-- Richard Heath'eld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === Subject: Re: Axiom of Foundation (absymally stupid question)eli :|Since we're discussing the axiom of foundation (in the textbooks I've seen, |it's called the axiom of regularity), does anyone know what the intuitive |justi'cation for this axiom is? I mean, all the other axioms seem pretty |natural to me, even the infamous axiom of choice. But where in world did |they come up with the axiom of regularity?i think the answer is that (in the presence of the usual other axioms)it's equivalent to the statement every set belongs to some level ofthe cumulative hierarchy, which is a powerful statement because itallows you to prove theorems applying to all the sets in the universe,by trans'nite induction with respect to the level of the cumulativehierarchy to which they belong (the levels of the hierarchy beingindexed by ordinal numbers).the cumulative hierarchy starts out with the empty set as the bottomlevel, then the power set of the empty set as the next level, and soforth on upwards. at limit ordinals you take the union of the levelsbelow.speaking as a non-specialist in set theory, what most bugged me aboutthe axiom of foundation when i was 'rst trying to learn axiomatic settheory (after already learning 'rst-order logic) was that on the onehand it sounds like it's trying to prevent weird loops and weirdin'nite regresses of membership chains, while on the other hand themeans by which it's trying to prevent it (namely the expressive powerof axioms of 'rst-order logic) is notoriously ineffectual at actuallypreventing the existence of models with in'nite regresses. but inretrospect i guess the idea is supposed to be that that shouldn'treally bother me any more than the similar fact that 'rst-order peanologic is trying to prevent the same kind of in'nite regresses, andessentially fails to do so, for essentially the same reason. inboth cases you still end up with a powerful means of proving theoremswithin the theory by some kind of induction even though some of themodels of the theory contain in'nite regresses which naively seemto violate the spirit of the inductive principles.-- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Question on generation of large prime numbersPhil Carmody :> Richard Heath'eld Carmody is a damn sight better at mathematics than I>> [am.>> But I think I got a case. :-) ]I hope you've got a hot-shot lawyer...He might end up as just a shot lawyer at this rate...Your premise is:> If there is a largest prime, P,and that's all, as can be deduced from the then and therefores> hereafter.Right. And the point of the premise (and the argument) is to take us to reductio ad absurdum - i.e. to show that there is no largest prime.>> Nope. Nowhere have you assumed that the list of primes known is>> complete up to P.>> On the contrary, I formed N by multiplying all primesErm, says who? You've not de'ned all in this context.Haven't I? Okay, guilty, your honour. all means every prime between 2 and P, including 2 and P.> Anyway, your maximum prime P is a red herring and completely> unnecessary - why did you even introduce it in the 'rst place?I was merely explaining to the OP that Euclid's proof of the in'nitude of primes is no use in attempting to construct a prime.>> and then adding 1. To>> multiply all primes together, I must 'rst have a complete list of>> primes.You must have a list of primes. What does Ocomplete' mean to you in> this context? If you wish to multiply together all primes between 2> and P then you must know /a priori/ all the primes between 2 and P.Well, I can get that list by starting at 2, using your rather handy factorisation oracle, and iterating up to and including P. As I said before, and gave an example which you for some reason> snipped,Brevity, my dear chap. Mere brevity.> if you use the Euclid method as a constructive prime> generator you very quickly end up generating primes smaller than> the previously known maximum. (On the assumption that this> generator is the only generator you have.)I wasn't trying to construct primes. I was trying to demonstrate why you can't construct them in that way.-- Richard Heath'eld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === Subject: Re: Question on generation of large prime numbersMark Brader :> Well, Richard and Phil are both wrong here. The proof could be expressed> in terms of a 'nite number of known primes, as Phil seems to have assumed> it was, but that's not the way Richard expressed it -- he spoke explicitly> of the product of *all* primes.Yes.> But Richard's conclusion that P has been shown not to be the largest> prime number is also wrong; what has been shown is that the assumption> of a 'nite number of primes is wrong. The only reason to assume that> at least one of the additional primes discovered must be larger than> P is that we *thought* P was the largest prime, before we discovered them.Hmmm. I seem to have taken a step for granted. Okay, let me make it more explicit. We have shown that the number we thought was the largest prime cannot be the largest prime, so there must be a bigger one. So now we 'nd out what that new largest prime is, and then iterate. We can do this in'nitely often (if we have the time to spare).Is that a bit closer to the mark?-- Richard Heath'eld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === Subject: Re: question Feinstein schrieb> Peter Luschny > Assume RH, assume x >= 2657> Then |pi(x)-li(x)| <= sqrt(x)log(x)/(8Pi)> But is this the best bound?Well, the best bound I know. If you 'nda better bound, please inform us! ;-)> But is this the best bound?IMHO there is no reason known which excludesthe possibility of the existence of N and Bsuch that the following is valid: Assume x Peter === Subject: Re: Question on generation of large prime numbers bluebear@pokernetwork.com (BlueBear) :> I have looking at a few web pages dealing with the largest known> calculated primes and a great deal of computational time is taking> into searching for these numbers and verifying they are primes. I have> seen the Euclid's proof of in'nitude primes and it occurs that me> that super-large prime numbers can be calculated using the following:p1=2 < p2=3> A large prime number, p, can be generated using> p = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5> p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7> p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbers> exceeding currently known largest primes can be generated rather> quickly. For example, we calculate the 'rst million primes, multiply> them and just add one.> If this is true, then calculating super-large primes should be a> computational trivial task, shouldn't it? Then why the big fuss over> calculating large primes?I suspect that I am missing fundamental here. Can super large primes> really be calculated using the trivial method outlined above? Hope> that someone can 59*509 === Subject: Re: Implementing the partial difference integration> Why don't you just post the program?Ok. Here's a straight-forward Java implementation. Nothing fancy. It just does the job. ___JSH----------------------public class DefPrimeCounter { /** Creates a new instance of Main */ public DefPrimeCounter() { } /** * @param args the command line arguments */ public static void main(String[] args) { int x = Integer.parseInt(args[0]); DefPrimeCounter Counter = new DefPrimeCounter(); System.out.println(Counter.p(x, (int)Math.sqrt(x))); } public int S(int x, int y){ if (y==1) return 0; int max = y+1; int sum=0, j; for (j=2; j :It's not necessary to know all the primes in order to construct the > argument. It is suf'cient to know that it is /in theory/ possible to know > them, e.g. by sieving. It is only necessary to know all the primes if you > wish to perform the multiplication in real life.So, since you can't sieve all of them there is a (practical) largest prime!But the (practical) largest prime is time dependant -- as time passes and better computers and programs come into existence, the largest know prime keeps getting larger. With, as yet, no end in sight. === Subject: Re: Semdirect product of categories?michele dondi :|jdolan@math-rs-n03.math.ucr.edu (James Dolan) :||>michele dondi :|>|>|Now it occurs to me that another simple generalization step can bring|>|the concept to a categorical formulation: let A,B be categories and|>|let (A,A) (the monoid of functors A->A) be regarded as a category|>|itself. Then choose a functor sigma:B->(A,A) and de'ne|>||>|(a2,b2)(a1,b1):=(a2sigma(b2)a1,b2b1)|>||>|for all a2,b2,a1,b1 s.t. Dom(a2)=sigma(b2)Cod(a1), Dom(b2)=Cod(b1).|>|>actually i'm not sure i understand your notation well enough to see|>whether what you're describing actually works, but i'm a bit skeptical|>about it because you don't seem to be giving the most straightforward|>generalization of the semi-direct product construction to a context|>involving arbitrary categories.||Well, I didn't know the construction mentioned hereafter, and in this|respect my one is indeed extremely naive. But I don't see how it|could not work, and what is not clear:||sigma is a functor B->(A,A), so sigma(b) is a functor A->A for all|morphisms b. Its action on objects is obviously trivial. Note that|sigma(e)=1_A (the identical functor) for all identities e of B.||The notation (a2,b2)(a1,b1):=(a2sigma(b2)a1,b2b1) is a|straightforward generalization of that used for groups/monoids. The|only difference being that the obvious compatibility relation for the|composition of the given morphisms is required:||a1:X->Y, a2:sigma(b2)Y->Z,|b1:S->T, b2:T->U.||Veri'cation that both the associative and the identity properties|hold is merely a mechanical task.||>the most straightforward such generalization is some version of the|>homotopy colimit construction. you start with a category c and a|>functor (or in some versions a pseudo-functor) f from c to the|>category of categories. the homotopy colimit of f is the category|>where an object is a pair (x,y) with x an object in c and y an object|>in f(x), and a morphism from (x,y) to (z,w) is a pair (m,n) with|>m:x->z in c and n:f(m)(y)->w in f(z), with composition of morphisms|>de'ned in a semi-obvious way.|>|>semi-direct product of monoids is then the special case where c has a|>unique object x and f(x) has a unique object y.||Then if I'm not mistaken (or misunderstanding), my construction is|the special case when c is arbitrary and f(x) has a unique object A|(in the notation above). In other words it is the same construction|with the difference that the codomain of f is restricted to a|subcategory of the category of categories, namely the category of|functors A->A.certainly the special case where [for any object x in c, f(x) has aunique object] can be considered, but i'm still confused about whetherwhat you tried to describe is really the same thing. if [for anyobject x in c, f(x) has a unique object], then what we're dealing withis something like a functor from c to (some version of) the categoryof monoids, but i didn't see anything in your description that reallysounded like that. but it's not implausible to me that you might betrying to describe the same thing in different language, becausedespite what you say i honestly 'nd your notation and terminologyconfusing, and it makes me wonder whether you might have made what icall a level slip somewhere- getting concepts on the level ofobjects mixed up with concepts on the level of morphisms, or somethinglike that.-- [e-mail address jdolan@math.ucr.edu] === Subject: Re: I NEED HELP BADLY (sorry, 13:14:34 +0100, Paul B. Andersen > :>> Of course you are funny.> If electric and magnetic 'elds acted instantaneously, light would travel at> in'nite speed.>I note with interest that your statement is so ambiguously put>>that it may be right as well as wrong.>But I take for granted that your statement is meant to be>>relevant to my claim, which was:>> as it enters a static electric 'eld.>So assuming that acting instantly is to be understood>>in this sense, an unambiguous version of your statement>>becomes:>> travel at in'nite speed.>Was this what you meant to say, Henry?>>If not, what DID you mean to say, and what is the relevancy>>of what you meant to say to your action time of the static>>accelerating 'eld in an accelerator?>Paul, 'nding the acrobatic show a little monotonous> Remember the old vacuum tube triodes.> Acccording to you, a signal on the grid would be instantly felt at the anode.>So if:> as it enters a static electric 'eld.>it follows:>A signal on the grid would be instantly felt at the anode>Explain why, please.>> If that's not instantaneous communication, what is?>Henry, you know of course that you are babbling nonsense.>There simply isn't possible to be as stupid as you pretend.>Paul>Let's get this straight.You say that the force on a charge due to an electric 'eld acts> instantaneously. Correct?Why do you ask what I am saying, when what I amsaying is quoted right above?I am saying: as it enters a static electric 'eld.> Let us consider a charged sphere somewhere in the universe. It exerts a force> on every other charge. If we can arrange for it to lose that charge somehow,> you are claiming that all those forces disappear INSTANTLY.And why the hell do you think I would claim something as stupid as that?You MUST know this is stupid nonsense, Henry.Why the hell do you pretend that it is possible to interpret mystatement above to mean that the electric 'eld will act instantlyelectric 'eld?This is actually too bloody stupid even for you, Henry.Why do you pretend to be such a moron?> I will continue when I receive your answer (if one is forthcoming).Answer what?You are desperated to evade the point, are you?Lets take this from the beginning.Of course there are really no static electric'elds in an accelerator, and I never said it was.that is a resonance cavity with a very powerful EM-'eld.The typical frequency is in the order of 100MHz (or higher).where the E-'eld is longitudinal.The crucial point is that the phase of this resonator isadjusted such that the E-'eld peaks (in the same direction)enters the accelerating stretch.That is the point. Why is that so hard to get?So what I am claiming is still:it enters the (quasi) static electric 'eld. And the forceit speed.Can you please explain how you can use this forinstant communication, Henry?Of course you cannot.The only decent thing you can do is to admit thatthis claim is wrong.But you won't do that, will you?You will rather keep insisting that when I say thatthe RF-cavity in an accelerator, then I really have statedthat a force acts on an electron in the Andromeda galaxyin the accelerator.Won't you?Paul === Subject: Re: A potentially rewarding challenge[...] for more detail about the structure of these> sequences, you might want to see the text 'le at > http://r.s.home.mindspring.com/GoodsteinSequences > [...]r.e.s. analysis of Goodsteinsequences is, indeed, illuminating, and the graphical analysis isfascinating. I shall certainly study === Subject: Re: Accumulation pts. in R and C ?How is it equivalent? If one has the usual metric in R and C, thenaccording to the books I am reading, they are different. C requiring onlyone point and R requiring in'nite amount of points. (The metric in C isabs(z-z0) < e and the metric in R being the usual: abs(x - x0) < e.)Lurch> Charlie Johnson> I just started Complex variables and applications by Churchill, et al.> I> am §ummoxed by something already. In Churchill, they say that an> accumulation point is> ...a point z0 ... of a set S if each neighborhood if z0 contains atleast> one point of S distinct from z0. Why is it different from R? To wit,> every neighborhood of x0 has an in'nite number of points of, say, E.It is equivalent.> It is always equivalent in a metric space.-- > Maxi === Subject: Re: I NEED HELP BADLY 22:55:41 -0000, Androcles :> get this straight.>> You say that the force on a charge due to an electric 'eld acts>> instantaneously. Correct?>> Let us consider a charged sphere somewhere in the universe. It exerts a>force>> on every other charge. If we can arrange for it to lose that charge>somehow,>> you are claiming that all those forces disappear INSTANTLY.>> I will continue when I receive your answer (if one is forthcoming).> Interesting question, Henry. I've used it myself, on the discussion of>gravity propagation. If a star were to convert *all* it mass to radiation in>one super-supernova, how long would it take for us to detect its loss of>gravity? I don't mean watch its planet suddenly §y away, that would be a>distant observation and would reach us at c. I mean detect the gravity loss>right here. This would be the biggest gravity pulse (negative going) that we>could possibly detect. But in order to detect the loss, where would have to>be aware of it in the 'rst place. As it turns out, the nearest star to us>(other than the sun) is 3.9 light-years away, and that is just too far to>detect it's gravity directly. So I fail to understand why anyone would>attempt an experiment to search for gravity waves, other than to give>themselves some funding on a futile attempt. There's a lot of money to be>made out of relativity, and very few people are altruistic.>Androcles>Exactly.There is one problem that I'm sure Paul will pounce on. That is, how to makecharge suddenly disappear.Here is another interesting question. The OMass' of an object is made up of a major proportion, the total mass of allenergy.Presumably, the two portions play an equal role in Newton's gravitation Law. If there is no distinction between the two, is it possible that all MATTER isnothing but some kind of manifestation of various O'elds'?Also, since mass is lost/gained in a Nuclear explosion, would we not expectmeasurable gravity waves to be produced. Even though the amount of matter isrelatively small, the sum total of all the potential energy associated withthat matter is quite large. (If the nuclear bomb is considered to be the centre of the universe, what isthe energy required to raise all the other matter in the universe to itspresent distance from that point. Might it just turn out to be mc^2?). Henri Wilson. See the Stupidity of Relativity.www.users.bigpond.com/hewn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych)Expires: 28 daysOn 14 object being accelerated by a idealistic jet of water or a> continuous Ostream of elastic ping pong balls'. What is its subsequent velocity> pattern?>> [snippage] > M.dv/dt=m(Vo-v) or: >>[more snippage]>>Henri, you've got major problems with this. It assumes>>Newtonian momentum transfer, and so implies a Newtonian>>equation for kinetic energy. And there's a real problem>>most assuredly not 1/2 mv^2. So your model has gross>>disagreement with observation.>>Socks What an idiot!>> Naturally, if one includes assumptions that a theory is correct in an attempt>> to prove it wrong, that attempt is likely to fail.Yes you are, and you did, and it did. You assumed Newton, and>you got Newton, and those are clearly not similar to what is>obverved in accelerators.>> A moving charge surrounds itself with a volume of Oreverse 'eld' because the>> main 'eld takes time to operate. The energy associated with this Oback 'eld'>> is equivalent to a mass increase.And their speeds are also not observed to go above c.It is very hard to accelerate anything to c let alone beyond it.The experiments are probably incapable of measuring superluminal speeds anyway.And you have not provided any theory of E&M that allows any such>thing as a reverse 'eld. Nor why there should be any kind of>speed limit involved. Nor why it should follow any such thing>as the kinetic energy formula observed in accelerators. Nor have>you provided a relation between energy and mass if you don't>accept relativity.>Socksradiation from an acceleraed charge!'elds associated with a moving charge!The OBack EMF' concept.I would be most amazed if a moving charge DID NOT alter the 'eld arounditself, wouldn't you?Henri Wilson. See the Stupidity of Relativity.www.users.bigpond.com/hewn/index.htm === Subject: Re: Accumulation pts. in R and C ?> How is it equivalent? If one has the usual metric in R and C, then> according to the books I am reading, they are different. C requiring only> one point and R requiring in'nite amount of points. (The metric in C is> abs(z-z0) < e and the metric in R being the usual: abs(x - x0) < e.)If every neighbourhood of z0 contains one point of E distict from z0, thenby considering the open balls centered in z0 of radius 1/n you see thatthere is in'nitely many (otherwise there would be none in some of thesedisks for n suf'ciently large).-- Maxi === Subject: Re: Bottom line on prime counting issueJames I believe that the people responsible for covering up the Roswellincident are the same people responsible for suppressing your worldchanging ideas about algebraic integers, and your prime counting function.I would be careful about what you say of the government. You may end upspeaking with Mulder and Scully.Lurch> Some of you may have noticed frenetic activity from posters trying to> convince you that there's nothing sinister about mathematicians doing> their best to downply my 'nd of a way to count prime numbers by> integrating a partial difference equation, but what's the bottom line?Does what I found work or not?It does. End of story, so mathematicians should acknowledge it. If> it's not important they can just put it in some math text somewhere,> or in some journal and drive on.No big deal.But they're 'ghting to totally ignore it. Translation: Sinister> attempt by academic types to hide something really important.Otherwise, why go to so much effort to 'ght me, when a simple way to> shut me up on the issue is just record it somewhere? And it is a> FIRST in human history, so use your common sense.The loser academic world is 'ghting me over something that works.End of story.These posters trying to convince you otherwise are just insulting your> basic intelligence.>My math discoveries, found for pro't> http://mathforpro't.blogspot.com/ === Subject: Re: I 12:55:09 -0000, Androcles> :> Remember the old vacuum tube triodes.> Acccording to you, a signal on the grid would be instantly felt at the>anode.> If that's not instantaneous communication, what is?> Henri Wilson.>I think Paul's acrobatics are no match for yours, H. His plea of monotony is>his way of giving up. Paul wants a static 'eld at the grid. That's ok,>until the electron reaches the grid, then that static charge has to reverse>sign, and is no longer static. Leaving aside the slew rate of the ampli'er>driving the grid (which in any real experiment we could not do), if the grid>reversed from positive (attracting the electron) to negative as it passed>through (propelling the electron on), it would still take a 'nite amount of>time for the 'eld beyond the grid where the electron is heading to change,>because c is 'nite. Hence at the instant the electron reaches the grid,>even if the grid potential is zero, there is a negative 'eld ahead of the>electron to slow it down. Hence the electron cannot attain c, by this>method. However, a positron coming down the other way would have a closing>velocity with the electron that was greater than c.Precisely. Paul Andsernon doesn't know what he's talking about.I note with interest that you think Androcles' incoherent babbleshows that I don't know what I am talking about. :-)Did you actually read it, Henry?in accelerators, isn't it? :-)Hatch or Anrocles - doesn't matter.They are 100% correct as long as they disgree with relativity.Right? :-)Paul, always amused when Henry agrees to nonsense === 22:33:36 -0500, Matt Timmermans :> Can anyone point me toward a good method for evaluating the prolate> spheroidal wave functions?There's a book Computation of Special Functions, by Shan-jie Zhang> and Jianming Jin. The programs themselves are available on the> web, in Fortran-77. Here:> http://iris-lee3.ece.uiuc.edu/~jjin/routines/routines.html-- > Andrew> === Subject: Re: Question on numerical integration (Simpson)Carlos Moreno :>Robert Israel :>Is the Simpson's method optimal with respect to the>number of calculations? (i.e., does it yield the>lowest error for a 'xed number of calculations?) >> No. >I know that there may not exact and absolute answer>to this -- but in general, for non-pathological,>non-super-obscure situations, is it true?>> No. There is no such thing as an optimal method for a 'xed>> number of calculations, unless you severely restrict the set >> of possible functions to be integrated. Each method will give >> 0 error for some functions, not for others.>Yes, this was clear to me (as stated in the above>paragraph you quoted).>I was hoping to 'nd out if in general, or in a>certain class of functions, or say, in a certain>approximate percentage of the cases it might be>true that Simpson's rule would beat higher order>polynomials in terms of accuracy per number of>calculations.No.Consider, e.g., polynomials f(x) = sum_{j=0}^5 a_j x^j of degree 5, integrated on [0,1]. If you allow 5 function evaluations,Simpson's rule gives you 1/12*f(0)+1/3*f(1/4)+1/6*f(1/2)+1/3*f(3/4)+1/12*f(1)which has an error of a_4/1920 + a_5/768. With the same 5 functionevaluations, the Newton-Cotes rule of order 4 (Bode's rule) gives you7/90*f(0)+16/45*f(1/4)+2/15*f(1/2)+16/45*f(3/4)+7/90*f(1) which has error 0. This Bode's formula will have a smaller error thanSimpson's on int_0^1 x^n dx for any integer n > 3. So there'sno way Simpson's rule can be considered to be optimal.On the other hand, if you allow 5 function evaluations anywhere in the interval, 5-point Gaussian quadrature gives you a formula thatis exact for polynomials of degree up to 9:sum_{j=1}^5 w_j f(x_j) wherew_1 = 161/900-13/1800*70^(1/2), x_1 = -1/42*(245+14*70^(1/2))^(1/2)+1/2w_2 = 161/900+13/1800*70^(1/2), x_2 = -1/42*(245-14*70^(1/2))^(1/2)+1/2w_3 = 64/225, x_3 = 1/2w_4 = 161/900+13/1800*70^(1/2), x_4 = 1/42*(245-14*70^(1/2))^(1/2)+1/2w_5 = 161/900-13/1800*70^(1/2), x_5 = 1/42*(245+14*70^(1/2))^(1/2)+1/2In general, n-point Gaussian quadrature will be exact on polynomialsof degree up to 2n-1, and it's easy to see there can't be an n-point formula that is exact on all polynomials of degree 2n. In this sense,Gaussian quadrature is the optimal method with a 'xed number of function evaluations.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: 've vectors Tl6b12ZpQe696epJPxXSTE84fYPri-2i8CL3mNkB0RpA71M7HIbaQ7> Thus (v1,v2,v4) is a basis, v1, v2 and v4 can trivially be expressed > as linear conbinations of basis vectors and v3 and v5 can be > expressed less trivially as linear combinations of these basis > vectors.Hi Virgil,But I'm working in a R^4 room. So how can a 4D-room have a3D-basis? Or did I misunderstand === Subject: Re: JSH: My use of my initials Adjunct Assistant Professor at the University of Montana.KRamsay :>(Underwood Dudley) America also did a lot of work, 'nding>|all sorts of precedents where a person was called a scab, a traitor,>|and other nasty things in print and the courts let the authors get>|away with it.MAA lawyers-- what a concept! :-)Was the issue of whether it was factually correct that he was>a crank ever raised in court?Prof. Dudley will obviously be in a better position to answer than Ican, but judging from the opinion in the 7th Circuit, I would guessthat the answer is no. The case was dismissed on a question of law,that the label crank, as de'ned and used by Prof. Dudley in hisbook, cannot be held to be difamatory (beginning of the 3rd paragraphand 4th paragraph of the opinion:http://www.law.emory.edu/7circuit/jan96/95-2282.html) Matters of fact are for the jury, matters of law are for the judge. Ifthe term is held incapable of being defamatory (as crank was here),then there is no question of fact of whether it was correct or not. Itis immaterial whether the person 'ts or not the description, sincethe term cannot cause defamation. The question of whether the personwas factually deserving of the word would be a question of fact, andthus left to a jury... So it was never addressed at all. In fact, theopinion states that ->if<- the label was incorrectly applied, thenthere was a ready remedy available for the other person: This is especially clear where, as in this case, the word is used in a work of scholarship. As we emphasized in the Underwager case, judges are not well equipped to resolve academic controversies, of which a controversy over Cantor's diagonal process is a daunting illustration, and scholars have their own remedies for unfair criticisms of their work--the publication of a rebuttal. Unlike the ordinary citizen, a scholar generally has ready access to the same media by which he is allegedly defamed. If Dudley's criticisms of Dilworth are unsound, Dilworth should be able to publish a rebuttal in the same journal in which he published the publish a rebuttal since its own reputation was impugned by Dudley's charges.(paragraph 7 of the opinion). == ==It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu === Subject: Problem calculating limits - Need Help!I'm having dif'culties solving these two limits.I mustn't use L'hospital rule:a) lim(x*(2^(1/x))-x) where x increases to in'nite.b) lim(cosh(x)-1)/(x^2) where x approaches 0. === Subject: Re: Axiom of Foundation (absymally stupid message constitute permission for an emailed simple nonproblems.Stephen J. Herschkorn we didn't mind invoking the axiom of> foundation, wouldn't {a, {a,b}} suf'ce as a de'nition of the> ordered pair (a,b)? Seems to me the answer is yes.Sure, but what makes {a, {a,b}} better than {{a}, {a,b}}? The key isto prove that (a,b) = (c,d) => a=c & b=d. With your proposedde'nition, the proof is a royal pain, and the resulting thingiesaren't really any simpler.Thomas === Subject: Re: A potentially rewarding structure of these> sequences, you might want to see the text 'le at > http://r.s.home.mindspring.com/GoodsteinSequences > [...]r.e.s.> analysis of Goodstein> sequences is, indeed, illuminating, and the graphical analysis is> fascinating. I shall certainly study them carefully.One correction to my post: The graph of the Goodstein sequence is piecewise linear, having a plateau consisting of (N+1)/2 terms-- not 2, of course. Rather, it's the 2nd to last, and 2nd longest linear segment (relative to the number of terms). All the terms in this plateau are equal to N = (j+1)*2^j-1, j=3*2^27-1.--r.e.s. === Subject: Re: Reality of response to my workI have to mention here that I was giving you advice. So it was reallymeta-advice. And how being *free* applies I'm not sure, since I wouldn'tcharge you for an admonition, as it would be unlikely that anyone would pay.The tone of your reply threw me, I must admit.According to my newsreader I didn't top post. But I'm using a microsoftproduct and as such it might not be working as intended. If I am topposting will someone email me and let me know?Justin Van Winkle> Justin Van Winkle :> I think this is morally wrong.Since you top posted, I have no idea exactly *what* was said that youregard as> morally wrong.> Don't tell James to waste his money.I only recommended that he get the advice of a patent attorney if hewanted to> secure intellectual property rights. I assume you regard your *free*advice> about the matter as superior to the advice of a professional. If so, youmay be> guilty of practicing law without a license.--> There are two things you must never attempt to prove: the unprovable -- and the> obvious.> --> Democracy: The triumph of popularity over principle.> --> http://www.crbond.com === Subject: Re: JSH: My use of my initials> :>If his work ever turns out to have legitimacy then his heirs, assuming>he has any, can sue you as well Mr. Dudley.Really? I'm not very familiar with US law, but I'm surprised to hear> that. Do you know of any examples where such a thing has happened?-- RichardAll his heirs would have to do is show harm, that is, a tort.If his results were legitimate, then proper recognition would havebeen a bene't, both to him, and I'd think to his family.Like would you rather have it known that a relative was a crank or adiscoverer? === Subject: Can someone help with these with some numbercrunchingpower help solve these following diophantine equations (there arein'nitely many solutions but I am looking for the one with thesmallest A) (and of course, nonnegative integers only please!):A^2 = 729B + 364A^2 = 2187B + 1093A^2 = 6561B + 3280A^2 = 19683B + 9841The general pattern isA^2 = (3^k)B + SUM( i = 0 to k-1 )3^i k > 5But help with even just 1 of the equations would be much === Subject: Re: Problem calculating limits - Need Help!Roy escribi.97 en el two limits.> I mustn't use L'hospital rule:> a) lim(x*(2^(1/x))-x) where x increases to in'nite.> b) lim(cosh(x)-1)/(x^2) where x approaches 0.Can you use series developments?-- Best (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Problem calculating limits - Need Help!> I'm having dif'culties solving these two limits.> I mustn't use L'hospital rule:The only alternative which springs to mind is to expand in series validfor the limit under consideration.> a) lim(x*(2^(1/x))-x) where x increases to in'nite.Write y = 1/x; take limit y -> 0:lim (2^y - 1)/y = lim (exp(y log2) - 1) / y> b) lim(cosh(x)-1)/(x^2) where x approaches 0.>This is an easier exercise in Taylor series expansion.-- P.A.C. want to live in a peaceful, free world.And we will 'nd these people and we will bring them to justice. - George W. Bush === Subject: Re: Bottom line on prime counting issue> Some of you may have noticed frenetic activity from posters trying to> convince you that there's nothing sinister about mathematicians doing> their best to downply my 'nd of a way to count prime numbers by> integrating a partial difference equation, but what's the bottom line?Does what I found work or not?He's talking about this === one:Subject: Partial difference equation, counting primesFor newbies: www.crank.net/harris.htmlFor harrisologists: mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=705 === Subject: Re: Why Derivative is Inverse to Integral; geometric that you didn't forget this discussion, only don'tunderstand, why did you take just this place of it. Someone likestrapezoid, someone other - picket fence, this is of no importancefor integral. Of course, if you integrate numerically (even with avery small step), there will be the difference, and the more step themore difference. Numerical integration generally is better inparabolic curves, and even in curves of higher order. When you pass toin'nitesimals, all these differences disappear, because ALWAYS only'rst term of expansion remains - just as in differentiating. The factthat expansion in Fourier series in irreversible, especially afteroperations with this series - this also is a well-known dif'cultywhich one can solve only 'nding some class of analytical functionswhich would generalise all existing functions and would continuouslytransform from one analytic function into another with the parametersvariation. I'm afraid, this task in general case is some alike seekingthe philosophic stone. But in particular problems it is possible. Idon't remember, have I told it in that discussion, but we dealt withsuch functions when studied the models of resistant elastic lines. Theshape of one such function you can see in Fig. 2, page 22 of our paperSome features of vibrations in homogeneous 1d resistant elasticlumped line,http://angel're.lycos.com/la3/selftrans/v2_1/resist22/ resist22.htmlThis function is described by the algebraic expression (16) in page 20of that paper,http://angel're.lycos.com/la3/selftrans/v2_1/resist20/ resist20.htmlAs you can see, the function varies with varied r from the asymptoticline to a line with a bend, retaining always analytical and algebraic!Besides, in the limiting case r = 0, this curve is described by asystem of two functions - linear and asymptotically descending, with abend at critical regime (see (25) in the page 22)! With it theanalytical function approaches to a broken curve WITH ANY ACCURACY.Anyway, lest to build a separate plot in the diagram in Fig. 2, webuilt the limiting curve with the use of analytical function. Yes,this is a complicated function even in the complex form. Itscomplicacy grows even more in passing to real variables. Itsintegration is out of question. But we needn't to integrate, as thisis just the solution of problem. If we have another elastic line,there will be simply another solution. ;-) Have you ever thought inthis direction? ;-)I can show you a function that changes the number of resonance peakswith changing parameter, and many other. This means, transformingfunctions exist, just as exist the functions that combine severalanalytical functions. This is just what differs our solutions fromconventional techniques. We yield these complicated functions in theend of solving, not trying to pluck the solution out of integral or torape the Green function. This is why we have not problems which make atrouble to others. Though we have other problems, indeed. ;-)However this pragmatism is not utilitarianism. Our solutions areexact, analytic, not replicable by other techniques, they can 'nework also in combination with numerical techniques, providing theaccuracy and effectiveness of these last. But the main, our solutionsare in full agreement with experiments. This is why I cannotunderstand, what for have you raised from archive that thought fromour dialogue with Archimedes. Could you explain of course, relation between the geometricrepresentation of second derivative and acceleration in physics. ;-)> On 02 Jun Karavashkin :> just this Archimedes Plutonium meant when said of a slope of one of>> rectangle sides. To simplify a trapezium to a rectangle is simpler>> than to resume the lost information of the slope. So in most cases we>> are unable to integrate. The same, we can easily expand a function>> into Fourier series, but to reconstruct the original by the Fourier>> series - it's practically unsolvable problem. Or, if Trapezoid or trapezium much>better than picketfence.>Can you think of a term for a rectangle that has an end-sides of just a>mere point rather than a true rectangle whose four sides are more than>just a point?>In the derivative there are no strange objects but in the integral there>is this>strange object of a rectangle whose end-sides are mere one point.>So in differentiation, there appears to be no real strange objects for a>set of trapezoids is normal geometric objects but in integration we have>this strange set of objects of rectangles whose end sides are one point.>And thus, I see geometrically the inverse relationship between derivative>and integral as that between trapezoids to one-point-sided rectangles. When>differentiating one makes trapezoids and when one integrates they collapse>the trapezoids into these strange rectangles.>Sergey, can you comment on the issue that derivative has normal geometric>objects of trapezoids (what I call picketfences) yet the integral relies>upon these>strange abnormal geometric objects of a collapsed rectangle whose end-sides>consist of one mere point.>i think trapezoids are stupid!!!! >So can we say that the essence of the Inverse relation between>derivative and>integral is the geometric idea that one is to expand strange rectangles into>trapezoids and the other is to collapse trapezoids into one-point end sided>rectangles.>P.S. I do not know if I should bother with a geometric explanation of the>2nd derivative that is acceleration in physics.>Archimedes Plutonium, a_plutonium@hotmail.com>whole entire Universe is just one big atom where dots>of the electron-dot-cloud are galaxies === Subject: square root of -1 mod pis there a way to compute sqrt(-1) mod p easily (for a large prime p,say 7-digits)? i know that -1 has a square mod p by the law ofquadratic reciprocity, but that doesnt help constructing an x suchthat x^2+1=0 (mod p). i also know that the problem is equivalent to'nd numbers of order 4 (mod p), since order of sqrt(-1) is 4, but yetagain that seems like a dif'cult problem.i am trying to do this without checking every number === Subject: Re: De facto censorship, counting primes :Some of you were probably surprised to learn that I did indeed 'nd a> way to count prime numbers by integrating a partial difference> equation.What you found was somebody else's work that you could notcompetently plagiarize.http://www.crank.net/harris.html It's not every braying jackass that gets a whole page at crank.netThe NSA is all hot and bothered about prime numbers for crackingtrapdoor encryptions. Of course, idiot , if you screw theNSA you disappear into indetreminate offshore Federal detention whereeven a bleeding heart ACLU Liberal lawyer can't 'nd you. Go ahead,make their day.-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! === Subject: Re: square root of -1 mod p Adjunct Assistant Professor at the University of Montana.billy d. :>is there a way to compute sqrt(-1) mod p easily (for a large prime p,>say 7-digits)? i know that -1 has a square mod p by the law of>quadratic reciprocity,You mean, you know that -1 has a square mod p when p is congruent to 1mod 4; this does not use quadratic reciprocity per se (which is abouttwo odd primes), but rather the Legendre symbol and similar results.> but that doesnt help constructing an x such>that x^2+1=0 (mod p). i also know that the problem is equivalent to>'nd numbers of order 4 (mod p), since order of sqrt(-1) is 4, but yet>again that seems like a dif'cult problem.i am trying to do this without checking every number between 1 and p.Usually, one proves that x^2 = -1 (mod p) has a solution if and onlyif p=2 or p=1 (mod 4) by ->constructing<- a solution. Namely, take(p-1)/2, and compute its factorial mod p. That's a solution, aconsequence of Wilson's Theorem. =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: question about the riemann hypothesisKate Ebneter >> But I was wondering something about the Riemann Hypothesis: What if>> it were proven to be false? I know that it's an important resultfor>> number theory, but what would the consequences be, if any, if it>> turned out to be false?>Rainer Rosenthal : >Hello Kate,>in de.sci.mathematik, there is a thread of that kind, where I>learned the following astonishing fact:>The following 18 numbers >1 2 4 6 10 18 22 30 42 58 70 78 102 130 190 210 330 462>are the only known ones not of the form > pq + qr + rp with p,q,r > 0Astonishing indeed!What is also interesting about these 18 integers is, add 1 to eachinteger and you have 18 primes.Also an interesting fact is that 210,330 and 462, the last 3 integersin this list, where each are @ a different level in the Collatz treeand each have this same property. Mainly (n-1) == 2(mod)3. Also allthere subsequent higher level integers have this property thus nobranching occurs in higher levels above these integers.Could there be a possible relation to the Collatz Conjecture?Dan >According to Borwein and Choi, there are are no others and>there are certainly no more than 19 of them.>BUT: if the Riemann Hypothesis fails, there could be another>number not of this form. And it must be > 10^11 then.>(1) Sequence A025052 in the njas archive> http://www.research.att.com/~njas/sequences/>(2) Posting by Hermann Kremer in de.sci.mathematik (german)> from 20.06.2001 00:15>related sequence A006093. ( See the thread pq+p+q === Subject: Re: Need advice on letters of recommendation>> The question I raised was, How would you make such an exam (with>> some reasonable hope that it would do what you want it to do)?you propose the question, you seek the answer. not from me though.> There is a GRE Subject test in Mathematics which covers thea general knowledge one would expect a math major to have mastered at a pretty good school. I downloaded the practiceexam a few months ago and worked it. It is, of course, multiple choice, but one has to evaluate contour integral,say which of the following is not a basis for a vector space,do arithmetic in a dihedral group, and so on. No proofs,but there are a couple questions from every normal coursein the undergraduate sequence: topology, number theory,algebra, analysis, probability, statistics, etc. And Calculus,just to check.Some graduate school require the test. The reason I downloadedthe test was that I was designing a new math major, and wantedto go through the questions and how many of the them a potentialgraduate from my new program would have a chance of answering.The practice test was good for pointing out holes in my program. (Namely, complex analysis and topology.) Not thatI have to kowtow to the test designers, but the test designersare trying to 'nd what's typical for an American math major,and so I thought the test would be at least somewhat normative.Anyway, if one is curious, the practice test is downloadablefrom the GRE website.Bart === Subject: Re: Question on generation of large prime a step for granted. Okay, let me make it more > explicit. We have shown that the number we thought was the largest prime > cannot be the largest prime,Yes, but not directly. You have shown that the assumption that there werebe the largest prime.> so there must be a bigger one.No. If you had proved *only* that the number P is not in fact the largestprime, this would *not* prove that a larger prime exists. It would alsobe possible that P is not prime after all, or that *no* largest primeexists.Now in fact we know that primes exist, and we have proved that there aremany, and since we're dealing with integers, from *this* it follows thatthere exists a prime larger than any given number P.In other words, you've taken a valid proof and made it invalid by falselyinserting, as an intermediate step, a fact that you actually only knowto be true because of the 'nal conclusion.-- Mark Brader, Toronto Logic is logic. That's all I say.msb@vex.net -- Oliver Wendell Holmes === Subject: Re: Question on generation of large prime numbersPhil Carmody :> Nope. Nowhere have you assumed that> the list of primes known is > complete up to P.Well, he said, ``form a new number, N, which is one greater than the product of_all_ primes [emphasis added].'' It seems to me that this would necessarily bea complete list.I am not sure what you're getting at. If your point is that RichardHeath'eld's proof is not the same as Euclid's, that's one thing. But hisargument seems sound to me. Filling in a little, but I think really rephrasingRichard's argument: If there's a largest prime P, then there are 'nitely many primes. If there are 'nitely many primes, we can take the product of all primes, and add 1 (call this N). We know 2 is a prime and 3 is a prime, and there are others, so N > P. No prime less than or equal to P (we took the product of all primes, where P was assumed to be the largest prime) divides N (they all leave a remainder of 1). So, any prime that divides N must be greater than P. There must be a prime that divides N. This contradicts P being the largest prime.Euclid's proof, if I recall correctly, is just given a 'nite list of primes,we can show there must be another prime. So that is different from Richard'sproof. But that doesn't mean there's anything wrong with Richard's (apart fromthe claim that it's Euclid's).Your {2, 3, 7} is not ``all primes,'' as Richard speci'ed. If we think 7 isthe largest prime, then the list is {2, 3, 5, 7} and any prime that divides2.3.5.7+1 is bigger than 7. He's not required to produce the list. We knowanything less than P is a prime or not a prime; to get a list of all primes, wetake everything less than P that is a prime.John Robertson === Subject: Re: probability 2......thank...you....very much....i think......if we only use 0-x-y-1in this case, probability is 1/8but, if we use 0-x-y-1, 0-y-x-1in this case, probability is 1/4which of case is right??advice....please....sir~ === Subject: Re: Need advice on letters of recommendationmmanch01@my-deja.com (maky Goddard has pointed out, letters of recommendation do>> have at least some chance of indicating that other (for me, much more>> key) component in graduate studies success, namely, the >> ability to *do mathematics*, while (3) institutional admissions>> exams surely couldn't do anything at all towards indicating that>> other component.the ability to solve mathematical problems, say in an admissions exam,>is without a doubt, a much better indicator of ability to do>mathematics. Scarcely without a doubt! The conditions of an exam, admissionsexam or otherwise, are entirely different from the conditions under which (real, practicing) mathematicians (of my broadpersonal acquaintance--which, I admit, does not include you,who may for all I know work along entirely different lines)actually do mathematics. And such problems as must,necessarily, be posed on an exam (admissions exam orotherwise), are just that--problems (with known solutions)to solve, not new mathematics to do (and possibly fail at).>in fact, the ability to solve mathematics problems is the>only accurate indicator of ability in mathematics. I would trust the seriously considered opinion of almost anyof my professional acquaintances, expressed in a letter ofrecommendation, that a third party had--or had not--the ability to do mathematics, far more than I would trust anadmissions exam. I wouldn't trust a stranger's opinionas much, but I'd still trust it more than an exam. >do you understand>where i am coming from?Either (if I am to be charitable) from somewhere far from whereI am, or (otherwise) from a position of profound ignorance ofwhat it means to do mathematics.Lee Rudolph === it dif'cult to 'nd the homology group ofS^n x S^m. Can anyone give me some ideas? I am thinking of using the MVsequence (exact):... -->H_n(A and B)-->H_n(A) + H_n(B)--> H_n(A union B)-->H_(n-1) (A andB) --> ... === Subject: Re: Accumulation pts. in R and C ?I don't quite get it yet, but I will work on metric in R and C, then> according to the books I am reading, they are different. C requiringonly> one point and R requiring in'nite amount of points. (The metric in Cis> abs(z-z0) < e and the metric in R being the usual: abs(x - x0) < e.)>If every neighbourhood of z0 contains one point of E distict from z0, then> by considering the open balls centered in z0 of radius 1/n you see that> there is in'nitely many (otherwise there would be none in some of these> disks for n suf'ciently large).-- > Maxi === Subject: Re: chess bishops - copy of ALL headers otherwise we will be unable to process your complaint properly.>parmito :> I understand that it is possible to express the number of ways of>> having k bishops on an n*n board such that they are not attacking each>> other as a combinatorial formula and i have been trying to derive the>> formula without any luck, can anyone help me out ?Is this right?.. Each bishop has a coordinate (r,c) (row, column). For all>the bishops, the r's are different to each other, and the c's are different>to each other.>So for k bishops the total amount of possible coordinates is how many>different ways you can take k r's out of 8, times how many different ways>you can take k c's out of 8. And that should be enough for you.. If it's>right..I think so.Thats a little too simplistic, as you can see fromhttp://acm.uva.es/p/v102/10237.html with 6 bishops on an 8*8 boardthere are 5599888 ways and with 5 bishops on a 30*30 board there are3127859642656 ways. === Subject: Re: Bottom line on prime counting issue :Some of you may have noticed frenetic activity from posters trying to> convince you that there's nothing sinister about > My math discoveries, found for pro'thttp://www.crank.net/harris.html It's 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! === Subject: Re: JSH: All the dumb crap in journalsJames Dolan :>if you work on creating>algorithms and writing computer programs then you can get concrete>answers about whether you've done things correctly simply by running a>computer program.>Excellent advice! Few professionals will 'nd an amateur's proof interesting. On the other hand, today's amateur can perform carefully crafted computationsthat are sometimes of interest to the professional. It's one of the reasons why I believe the 21st century is a great time to be === Subject: Re: square root of -1 mod p magidin@math.berkeley.edu (Arturo Magidin) :> billy d. :>is there a way to compute sqrt(-1) mod p easily (for a large prime p,>say 7-digits)? > Usually, one proves that x^2 = -1 (mod p) has a solution if and only> if p=2 or p=1 (mod 4) by ->constructing<- a solution. Namely, take> (p-1)/2, and compute its factorial mod p. That's a solution, a> consequence of Wilson's Theorem.Yes, but if you actually want to compute a solution, 'nding (p - 1) / 2 factorial mod p is a really bad way to do it, at any rate for the 7-digit p that OP has. A couple of algorithms are given in Crandall & Pomerance, Prime Numbers, Section 2.3.2-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Question on generation of Hmmm. I seem to have taken a step for granted. Okay, let me make it more>> explicit. We have shown that the number we thought was the largest prime>> cannot be the largest prime,Yes, but not directly. You have shown that the assumption that there were> be the largest prime.>> so there must be a bigger one.No. If you had proved *only* that the number P is not in fact the largest> prime, this would *not* prove that a larger prime exists. It would also> be possible that P is not prime after all, or that *no* largest prime> exists.But I de'ned P to be the largest prime. Therefore, it /is/ prime, and therefore /either/ no largest prime exists /or/ there is a prime larger than P, which is what I was obviously failing to say.> Now in fact we know that primes exist, and we have proved that there are> many, and since we're dealing with integers, from *this* it follows that> there exists a prime larger than any given number P.In other words, you've taken a valid proof and made it invalid by falsely> inserting, as an intermediate step, a fact that you actually only know> to be true because of the 'nal conclusion. A mathematician cut out to be I was not.-- Richard Heath'eld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === Subject: Re: Bottom line on prime counting issueX-DMCA-Noti'cations: -0800, jstevh@msn.com () :>Some of you may have noticed frenetic activity from posters trying to>convince you that there's nothing sinister about mathematicians doing>their best to downply my 'nd of a way to count prime numbers by>integrating a partial difference equation, but what's the bottom line?Does what I found work or not?It does. End of story, so mathematicians should acknowledge it. If>it's not important they can just put it in some math text somewhere,>or in some journal and drive on.No big deal.But they're 'ghting to totally ignore it. Translation: Sinister>attempt by academic types to hide something really important.Fascinating. If people were raving about how important it wasof course that would prove it was important. In fact people areignoring it, and curiously that also proves it's important.Do you really think anyone's buying this?>Otherwise, why go to so much effort to 'ght me, when a simple way to>shut me up on the issue is just record it somewhere? Huh? First, nobody's going to any trouble - what looks to you likepeople going to a lot of trouble is just people having a bit ofgood-natured fun with the village idiot. And second, all yourstuff _is_ recorded, right there on Google. (You're going to 'ndthat fact embarassing if you ever sober up...)> And it is a>FIRST in human history, so use your common sense.The loser academic world is 'ghting me over something that works. End of story.These posters trying to convince you otherwise are just insulting your>basic intelligence.Uh, right. Our common sense tells us that when people on sci.math,journal editors, mathematicians you harass with email, mathematiciansyou harass in person _all_ 'nd your work of no interest the onlypossible explanation is that they all realize it's tremendouslyimportant, and every single one of them is quick enough to realizehe needs to lie about his opinion before once saying oh my godthat's incredible even once.That's not common sense as we know it, Jim.>My math discoveries, found for pro't>http://mathforpro't.blogspot.com/ C. Ullrich === Subject: Re: Can someone help with these diophantine equations? snizpilbor@yahoo.com (Sniz Pilbor) :> This is not homework. Could anyone with some numbercrunching> power help solve these following diophantine equations (there are> in'nitely many solutions but I am looking for the one with the> smallest A) (and of course, nonnegative integers only please!):A^2 = 729B + 364A = 254 (you can work out the B value)> A^2 = 2187B + 1093254> A^2 = 6561B + 32801933> A^2 = 19683B + 98418494> The general pattern is> A^2 = (3^k)B + SUM( i = 0 to k-1 )3^i k > 5You're after the 3-adic expansion of sqrt(-1/2). I wouldn't expect much of a pattern in the answers.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Can someone help with these diophantine equations? > But help with even just 1 of the equations would be much appreciatedA^2 = B*c + d----------------------254^2 = 29*2187 + 1093254^2 = 88*729 + 364475^2 = 309*729 + 3641933^2 = 569*6561 + 3280983^2 = 1325*729 + 3641933^2 = 1708*2187 + 10931204^2 = 1988*729 + 3642441^2 = 2724*2187 + 10934628^2 = 3264*6561 + 32808494^2 = 3665*19683 + 98411712^2 = 4020*729 + 3641933^2 = 5125*729 + 36411189^2 = 6360*19683 + 98414120^2 = 7761*2187 + 10932441^2 = 8173*729 + 3642662^2 = 9720*729 + 3644628^2 = 9793*2187 + 10938494^2 = 10996*6561 + 32803170^2 = 13784*729 + 3643391^2 = 15773*729 + 3646307^2 = 18188*2187 + 109311189^2 = 19081*6561 + 32803899^2 = 20853*729 + 3646815^2 = 21236*2187 + 10934120^2 = 23284*729 + 3644628^2 = 29380*729 + 3644849^2 = 32253*729 + 3648494^2 = 32989*2187 + 109315055^2 = 34545*6561 + 32809002^2 = 37053*2187 + 10935357^2 = 39365*729 + 36428177^2 = 40336*19683 + 98415578^2 = 42680*729 + 36417750^2 = 48020*6561 + 328030872^2 = 48421*19683 + 98416086^2 = 50808*729 + 36410681^2 = 52164*2187 + 10936307^2 = 54565*729 + 36411189^2 = 57244*2187 + 10936815^2 = 63709*729 + 3647036^2 = 67908*729 + 36421616^2 = 71216*6561 + 328012868^2 = 75713*2187 + 10937544^2 = 78068*729 + 36413376^2 = 81809*2187 + 10937765^2 = 82709*729 + 36424311^2 = 90081*6561 + 32808273^2 = 93885*729 + 3648494^2 = 98968*729 + 36415055^2 = 103636*2187 + 109315563^2 = 110748*2187 + 10939002^2 = 111160*729 + 36447860^2 = 116373*19683 + 98419223^2 = 116685*729 + 36428177^2 = 121009*6561 + 328050555^2 = 129848*19683 + 98419731^2 = 129893*729 + 3649952^2 = 135860*729 + 36417242^2 = 135933*2187 + 109317750^2 = 144061*2187 + 109330872^2 = 145264*6561 + 328010460^2 = 150084*729 + 36410681^2 = 156493*729 + 36411189^2 = 171733*729 + 36419429^2 = 172604*2187 + 109311410^2 = 178584*729 + 36419937^2 = 181748*2187 + 109334738^2 = 183924*6561 + 328011918^2 = 194840*729 + 36412139^2 = 202133*729 + 36437433^2 = 213569*6561 + 328021616^2 = 213649*2187 + 109312647^2 = 219405*729 + 36422124^2 = 223809*2187 + 109312868^2 = 227140*729 + 36467543^2 = 231776*19683 + 984113376^2 = 245428*729 + 36470238^2 = 250641*19683 + 984113597^2 = 253605*729 + 36423803^2 = 259068*2187 + 109341299^2 = 259961*6561 + 328024311^2 = 270244*2187 + 109314105^2 = 272909*729 + 36414326^2 = 281528*729 + 36443994^2 = 294996*6561 + 328014834^2 = 301848*729 + 36425990^2 = 308861*2187 + 109315055^2 = 310909*729 + 36426498^2 = 321053*2187 + 109315563^2 = 332245*729 + 36415784^2 = 341748*729 + 36447860^2 = 349120*6561 + 328028177^2 = 363028*2187 + 109316292^2 = 364100*729 + 36416513^2 = 374045*729 + 36428685^2 = 376236*2187 + 109387226^2 = 386545*19683 + 984150555^2 = 389545*6561 + 328017021^2 = 397413*729 + 36417242^2 = 407800*729 + 36489921^2 = 410800*19683 + 984130364^2 = 421569*2187 + 109317750^2 = 432184*729 + 36430872^2 = 435793*2187 + 109317971^2 = 443013*729 + 36454421^2 = 451401*6561 + 328018479^2 = 468413*729 + 36418700^2 = 479684*729 + 36432551^2 = 484484*2187 + 109357116^2 = 497216*6561 + 328033059^2 = 499724*2187 + 109319208^2 = 506100*729 + 36419429^2 = 517813*729 + 36419937^2 = 545245*729 + 36434738^2 = 551773*2187 + 109320158^2 = 557400*729 + 36460982^2 = 566804*6561 + 328035246^2 = 568029*2187 + 1093106909^2 = 580680*19683 + 984120666^2 = 585848*729 + 36420887^2 = 598445*729 + 364109604^2 = 610325*19683 + 984163677^2 = 618009*6561 + 328036925^2 = 623436*2187 + 109321395^2 = 627909*729 + 36437433^2 = 640708*2187 + 109321616^2 = 640948*729 + 36422124^2 = 671428*729 + 36422345^2 = 684909*729 + 36467543^2 = 695329*6561 + 328039112^2 = 699473*2187 + 109322853^2 = 716405*729 + 36439620^2 = 717761*2187 + 109323074^2 = 730328*729 + 36470238^2 = 751924*6561 + 328023582^2 = 762840*729 + 36423803^2 = 777205*729 + 36441299^2 = 779884*2187 + 109341807^2 = 799188*2187 + 109324311^2 = 810733*729 + 364126592^2 = 814181*19683 + 984124532^2 = 825540*729 + 36474104^2 = 836976*6561 + 3280129287^2 = 849216*19683 + 984125040^2 = 860084*729 + 36443486^2 = 864669*2187 + 109325261^2 = 875333*729 + 36443994^2 = 884989*2187 + 109376799^2 = 898961*6561 + 328025769^2 = 910893*729 + 36425990^2 = 926584*729 + 36445673^2 = 953828*2187 + 109326498^2 = 963160*729 + 36446181^2 = 975164*2187 + 109326719^2 = 979293*729 + 36480665^2 = 991745*6561 + 3280for 1 <= B <= 1,000,000 courtesy of GP-Pari and about 4 minutes time. === Subject: Re: Can someone help with these diophantine equations?Originator: twomack@chiark.greenend.org.uk ([193.201.200.170])Sniz Pilbor anyone with some numbercrunching>power help solve these following diophantine equations (there are>in'nitely many solutions but I am looking for the one with the>smallest A) (and of course, nonnegative integers only please!):A^2 = 729B + 364A^2 = 2187B + 1093A^2 = 6561B + 3280A^2 = 19683B + 9841OK, you're looking at successive approximations to the 3-adic square root of -1/2So if you ask magmap := pAdicRing(3, 100);Sqrt(p!(-1/2))you get 14918756739905062250418133874814682552290432142and if you consider that large number mod 3^n you get the values of A that workat each step.For example, 475^2 = 729 * 309 + 365 1933^2 = 2187 * 1708 + 1093You just need to look at X, X+3^n and X+2*3^n, modulo 3^(n+1), to see which digitto put on the beginning at each stage.I'm afraid I don't have a good reference for the p-adic numbers; they turn upincidentally in most number-theory books, but you're expected to absorb theminstantaneously.Tom === Subject: Re: chess bishops - combinations Quaternion :> parmito :I understand that it is possible to express the number of ways of> having k bishops on an n*n board such that they are not attacking each> other as a combinatorial formula and i have been trying to derive the> formula without any luck, can anyone help me out ?Is this right?.. Each bishop has a coordinate (r,c) (row, column). For all> the bishops, the r's are different to each other, and the c's are different> to each other.No, bishops attack on diagonals, not rows & columns. OP; have you tried starting small? n = 2, n = 3, see whether there are any patterns? Or try searching Sloane's On-Line Encyclopedia of Integer Sequences for the word bishop?-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Can anyone recommend a good lie algebra book?Looking for a good === Subject: Re: Accumulation pts. in R and C ?They are equivilent. Let me see if I can explain it on the internet. Ifthis helps let me knowyou can just think of R for simplicity, although I don't think my argumentwill use this fact.let's say we want to check if x0 is an accumulation point of some set S.I want to show that:Every open neighborhood contains at least one point of S distinct from x0is equivelent toEvery open neighborhood contains in'nitely many points of SSuppose every open neighborhood contains at least one point of S distinctfrom x0. Fix an open neighborhood. (just pick any one) Pick one of thepoints distinct from x0, call it x1. (again, pick one). Now, clearly thereexists an open neighborhood of x0 that doesn't contain x1. For example, if|x0 - x1| = a, then take the a/2 neighborhood of x0 (this is just sayingthat x1 is some distance away from x0 since it is a different point, so ifwe take a neighborhood that only goes out half as far, x1 is not in thatneighborhood).Now, this smaller neighborhood must contain a point of S. call it x2. nowtake an even smaller neighborhood that doesn't contain x2. this evensmaller neighborhood doesn't contain x1 or x2, but it does contain somepoint of S distinct from x0. So we can keep repeating this forever. Inother words, for any n, however large n is, there is some x_n in S.I hope that made sense. To recap, if you keep taking smaller neighborhoods,you keep 'nding points. So if every neighborhood contains one point, everyneighborhood must contain in'nitely many points, since each neighborhoodalso contains every smaller neighborhood.This might not be very rigorous, but hopefully it will get you started.Obviously, if every neighborhood contains in'nitely many points, itcontains at least one point.Justin Van WinkleSuppose----- Original === Message ----- Subject: Accumulation pts. in R and C ?> Hi all,I just started Complex variables and applications by Churchill, et al.I> am §ummoxed by something already. In Churchill, they say that an> accumulation point is> ...a point z0 ... of a set S if each neighborhood if z0 contains at least> one point of S distinct from z0. Why is it different from R? To wit,> every neighborhood of x0 has an in'nite number of points of, say, E.TIA,Lurch === Subject: The Cipher of Genesisre: http://qedcorp.com/APS/EmergentGravity.dochttp://qedcorp.com/ APS/StarGate1.movCommentary 1The 'ber bundle as an idea has 4 parts.1. A structure symmetry group G.2. The total hyperspace H or, in some applications Wheeler's BIT.3. The projection map P.4. The base space M or, in some applications. Wheeler's IT.The hyperspace H consists of 'bers f(x) that areeither copies of or representations of the symmetrygroup G.The projection map P collapses a 'ber f(x) in the hyperspace H toa point x in the base space M.All of these objects are continuum differential manifoldsdepending on the continuum of real numbers which itsassociated issues of Cantor's in'nity of in'nities ofCabalistic Aleph's in an ascending Jacob's Ladder.This is not a discrete combinatoric mathematics althoughsuch a skeletal structure is associated with it as inHerman Weyl's Theory of Groups and Quantum Mechanicsand as in Saul-Paul Sirag's presentation of V.I. Arnold'sA-D-E mathematics of everything.The base space is covered by an atlas of local coordinate patcheswith all important overlap transition functions sewing thepatches together like a quilt.M is space-time in local micro-quantum 'eld theory of pointThe extra-dimensions of hyperspace formthe Calabi-Yau space of vibrations of thesuperstring beyond space-time.The connection on the total hyperspace H is the potentialof a local gauge force.Examples of connections is the 4 potential Au(x) inMaxwell's electromagnetism with G as U(1).There are similar connections for the Yang-Mills weak forcewith G = SU(2) and the strong force with G = SU(3).Classical general relativity, as distinct from local micro-quantum'eld theory, has the torsion-free symmetric three-index non-tensorLevi-Civita connection with G as the Diff(4) group.The latter comes from locally gauging the 4 parameter translation subgroup(generated by the 4-momentum Pu of globally §at special relativity )of the 15 parameter conformal group of Roger Penrose's massless twistors.Bottom -> Up: Given base space M and symmetry group G construct thehyperspace H as a quilt patchwork.Top -> Down: Given hyperspace H and symmetry group G construct thebase space M as the non-overlapping partition of hyperspace into G-orbitscalled the quotient space of H mod G in the principal bundle.Micro-quantum source renormalizable local 'elds of spin 1/2 lepto-quarks are associated vector bundles.Micro-quantum force renormalizable local 'elds of spin 1 gauge force bosons (electro-weak and strong) arefrom the principal bundle.There is no renormalizable quantum gravity in this precise sense.This is because classical Einstein gravity is a More is different (P.W. Anderson)emergent collective effect as in Andrei Sakharov's metric elasticity of aninstability in the globally §at false vacuum of the interacting lepto-quark source/electroweak-strong force.Einstein's gravity + uni'ed exotic vacuum dark energy/matter with Andrei Linde's chaotic in§ationary cosmology are the result of the continual phase transitions from globally §at false high entropy micro-quantum vacua to locally curved macro-quantum low entropy metastable vacua.to be continued: === Subject: Re: conjugate subgroups Timothy Murphy :> Bruce Harvey :> Why can't a conjugate of a subgroup be a proper subset of it?See http://mathworld.wolfram.com/ConjugateSubgroup.htmlSurely the question doesn't deserve an URL?It doesn't have one - at any rate, if it does, that's not it. That URL just de'nes conjugacy & links to some other stuff, it doesn't address the original question.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: square root of -1 mod pbilly d. :>is there a way to compute sqrt(-1) mod p easily (for a large prime p,>say 7-digits)? i know that -1 has a square mod p by the law of>quadratic reciprocity, but that doesnt help constructing an x such>that x^2+1=0 (mod p). i also know that the problem is equivalent to>'nd numbers of order 4 (mod p), since order of sqrt(-1) is 4, but yet>again that seems like a dif'cult problem.It's not dif'cult at all (unless you demand a deterministic algorithm).For any b, b^[(p-1)/4] has order at most 4 (mod p), so if you pick brandomly there is a 50% chance that it will have order exactly 4 (theonly other possibilities are order 1 and order 2 which are +1 and -1respectively). -- Erick === Subject: Re: plotting elliptic curves for LaTeX > could you give me an example code? given a polynomial f(x,y) .. I want to plot > the points that solve f(x,y)=0 ... say f(x,y)=y^2+y-x^3+x I tried to get splot to work with f(x,y) = y^2 + y - x^3 + x, but it refused, soI completed the square for y^2, and created two real functions of y, calledf(x) and g(x) depending upon the +/- value of the square root functionHere's the gnuplot codegnuplot> f(x)=-0.5+sqrt(x*x*x-x+0.25)gnuplot> g(x)=-0.5-sqrt(x*x*x-x+0.25)gnuplot> h(x)=0gnuplot> set xrange [-2:3]gnuplot> plot f(x),g(x),h(x)crude, but a start.and yes, you can set up output for LaTex:gnuplot> set terminal latexTerminal type set to Olatex'Options are O(document speci'c font)'gnuplot> set output my elliptic curvegnuplot> replotto get back to your terminalgnuplot> set terminal X11 (or whatever)gnuplot> set output === Subject: Re: Can someone help with these diophantine equations? snizpilbor@yahoo.com (Sniz Pilbor) :> This is not homework. Could anyone with some numbercrunching> power help solve these following diophantine equations (there are> in'nitely many solutions but I am looking for the one with the> smallest A) (and of course, nonnegative integers only please!):A^2 = 729B + 364A^2 = 2187B + 1093A^2 = 6561B + 3280A^2 = 19683B + 9841The general pattern is> A^2 = (3^k)B + SUM( i = 0 to k-1 )3^i k > 5See also Robert Israel's posts in the thread, Squares that end with four identical digits-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Integral of a function over [a,b]> Amanda :Hello> Suppose g:[a,b]->R is continuous on [a,b] and let f[a,b]->R be de'ned> in such a way that, at every irrational x in [a,b], f is continuous> and f(x) = g(x). Then, is it true that the Riemann integral of f over> [a,b] equals the integral of g? Since the set of discontinuities of f> on [a,b] is at most countable and, therefore, has measure 0, it> follows f is certainly integrable over [a,b], but I couldn't come to a> conclusion if its integral needs to equal the integral of g or if it> depends on how we de'ne f for the rationals of [a,b].In fact, your argument to prove that f is integrable over [a,b] is not> correct. Indeed, in order that a function de'ned on [a,b] is integrable> (in the sense of Riemann), the set of discontinuities must have measure> 0 *and* f must be bounded. However, if f is supposed to be bounded, your> argument is correct.Yes, sure. I forgot to say f was bounded. On the other hand, the integrals of f and g must be equal (still> assuming that f is bounded, of course), since the function f - g is> integrable and takes the value 0 youAmanda === Subject: Re: square root of -1 mod p tungsteneer3@yahoo.com (billy d.) asked:> is there a way to compute sqrt(-1) mod p> easily (for a large prime p,> say 7-digits)You can use the continued fractionexpansion of sqrt(p) to solve x^2 - py^2 = -1and then x = sqrt(-1) (mod p). For anef'cient way to do this, see Solving x^2 - Dy^2 = +-1in Solving the generalized Pell equationat http://hometown.aol.com/jpr2718Other than that, Henri Cohen, A Course in Computational Algebraic NumberTheory, Section 1.5 might help.John Robertson === Subject: Re: Vector Calculus === problem>Subject: Vector Calculus problemOk, I'm trying to work on my homework and am stuck on 4.9 #10 of Vector>Analysis by Davis.The question states By means of Stokes' theorem, 'nd S F*dR around the>ellipse x^2+y^2=1, z=y, where F=xi+(x+y)j+(x+y+z)k.z=y ? what?I got the curl of F and that equalled i-j+k but I'm not really sure how to do>the rest of the problem. Any help would be appreciated. I've wasted a lot of>time and gotten almost nowhere.>write down an expresion for the (vector) dA for your circle...I think that you should use (vector) dA = (scalar) dA kthen curl F dot dA is just (scalar) dAand a circle of radius one has area 2 Piadam === Subject: Re: Vector Calculus problem>>Ok, I'm trying to work on my homework and am stuck on 4.9 #10 of Vector>>Analysis by Davis.>>The question states By means of Stokes' theorem, 'nd S F*dR around the>>ellipse x^2+y^2=1, z=y, where F=xi+(x+y)j+(x+y+z)k.z=y ? what?>I got the curl of F and that equalled i-j+k but I'm not really sure how to do>>the rest of the problem. Any help would be appreciated. I've wasted a lot of>>time and gotten almost nowhere.>>write down an expresion for the (vector) dA for your circle...I think that you should use (vector) dA = (scalar) dA kthen curl F dot dA is just (scalar) dAand a circle of radius one has area 2 Piadamuhhh. duh no a circle of radius one has area Pi.hmmm, maybe the problem here is that I'm notsure what the z=y condition means...sorryadam === Subject: Re: Relativity is based on assumption.No way, Henri. I've been debating with him for years, and he's just one of> those relativists that refuse to see what is right under their nose. It's a> psychological hang-up that most people have. The don't want to admit they've> been made a fool of by Einstein, they have to appear in the eyes of others> as the one who has all the answers, but that's provided the question suits> them, of course. When it doesn't, they ignore it.> When Einstein 1/2[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v)) ] = tau(x',0,0,t+x'/(c-v))> ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/> Where does he get that 1/2 from?I send you a letter on the 10th. I get a reply from you on the> 16th. So I assume you received the letter and replied to it on> the 13th:> 1/2( 10 + 16) = 13Right... but perhaps still a little too far out. > I will try to improve the analogy, who knows, it may just make the> difference:Without mentioning it, Dirk also calculated that the total return time> was 16-10 =6 days.> Therefore, if we don't know anything about the speed of either> trajectory, we have no other option but to assume - or declare by> convention - that the one way delivery time is 1/2 * 6 = 3 days.You may have noticed that at that time Einstein was careful to declare> the total average return speed to be constant, in full agreement with> measurements and LET. No assumption there!> The later SRT assumption that the one-way speed is truly c in one's> own frame of reference, can not be proved if the Lorentz equations are> correct. ;-)Haraldxein: How simple. Bravo. When was the last OWLS-TWLS discussion? Was it decided by vote or physics? === Subject: Re: 've vectors Karl Pech :> Thus (v1,v2,v4) is a basis, v1, v2 and v4 can trivially be expressed > as linear conbinations of basis vectors and v3 and v5 can be > expressed less trivially as linear combinations of these basis > vectors.Hi Virgil,But I'm working in a R^4 room. So how can a 4D-room have a> 3D-basis? Or did I misunderstand of the entire space, rather than a basis of the space spanned by the 5 vectors given? If so, that meaning was not clear to me. And such a basis is impossible from just the 5 vectors given.If you did want a basis of the entire space, what is wrong withb1 = (1,0,0,0), b2 = (0,1,0,0), b3 = (0,0,1,0), b4 = (0,0,0,1)as such a basis? === Subject: Re: I NEED HELP BADLY (sorry, maths not psych)>> as it enters a static electric 'eld.Let us consider a charged sphere somewhere in the universe. It exerts a force>on every other charge. If we can arrange for it to lose that charge somehow,>you are claiming that all those forces disappear INSTANTLY.Parse the bloody sentence in quotes, Henry. It doesn't say that. Isn'tEnglish your 'rst language? - Randy, grabbing his popcorn and going back to watch theentertainment === Subject: Re: Question on generation of large prime numbersmsb@vex.net (Mark Brader) is a largest prime, P, and that's all, as can be deduced from the then and therefores hereafter.>> then the number of primes is 'nite. We>> can therefore form a new number, N, which is one greater than the product>> of all primes. ...Nope. Nowhere have you assumed that the list of primes known is> complete up to P.On the contrary, I formed N by multiplying all primes Erm, says who? You've not de'ned all in this context. Nowhere do you > say anything about having a contiguous set of primes from 2 to P. Phil, stop blathering about de'ning Oall'. You misread or mis-> interpreted Richard's wording, and you've been caught.I did not. He introuced concepts which were unnecessary, and I claim harmful. > The only thing you've said is that your set of primes has a maximum> element P. Full stop. There were no further premises -- look above if > you don't believe me.We're dealing with positive integers. If there is a largest one in a> set of them, then that set is 'nite, as Richard said, and if it's> 'nite, then a product of all its members exists. Since when has that been at issue? I've even said as much myself. You must have misread.> It is not necessary> to specify an algorithm to construct the set (although in this case it> would be simple to do so), nor to de'ne Oall' in this context.Nor do I say it is. You must have misread.Phil-- Unpatched IE vulnerability: Click hijackingDescription: Pointing IE mouse events at non-IE/system windowsReference: http://safecenter.net/liudieyu/HijackClick/ HijackClick-Content.HTMExploit: http://safecenter.net/liudieyu/HijackClick/HijackClick2- MyPage.HTM === Subject: Re: Vector Calculus problem>Subject: Vector Calculus problemOk, I'm trying to work on my homework and am stuck on 4.9 #10 of Vector>Analysis by Davis.The question states By means of Stokes' theorem, 'nd S F*dR around the>ellipse x^2+y^2=1, z=y, where F=xi+(x+y)j+(x+y+z)k.I got the curl of F and that equalled i-j+k but I'm not really sure how to do>the rest of the problem. Any help would be appreciated. I've wasted a lot of>time and gotten almost nowhere.ahh, z=y means the ellipse is tilted in the z-y plane and theprojection of it on the x-y plane is x^2+y^2=1so the area vector of the ellipse is pointing in the(0,-1,1) direction so that you get two contributionsfrom curlF dot dA I.e. integrate (-j+k)dot(-j+k)= 2 over the circle to get 2 PiAlso, Check that the back of book is right by doing it explicitly:x=costy=sintz=sintt==0..2PidR=i(-sint)+j(cost)+k( cost)F=i(cost)+j(cost+sint)+k(cost+2sint)FdotdR=(-costsint)+( cos^2t+costsint)+cos^2t + 2costsintintegrate from 0 to 2pi = 0 + Pi + 0 + Pi + 0 = 2 Piadam === Subject: Re: Bottom line on prime counting issueAnd that are my dogs!Mulder and Scully are two lovely dogs, Amstaffs of course. They love Harrisas I do. I and them would like Harris to be honest about his work andcommit to failure. Or else!!!!What I really want is that Harris fully gave his work a scrutiny whichincluded all the corrections from the bistanders and helping hands. Then hecan conclude and put forward his proof.One last question to Harris:What is the difference between algebraic numbers, algebraic integers,numbers, integers and complex numbers?Karl-Olav Nyberg === Subject: Re: Bottom line on prime counting issue :> Some of you may have noticed frenetic activity from posters trying to> convince you that there's nothing sinister about mathematicians doing> their best to downply my 'nd of a way to count prime numbers by> integrating a partial difference equation, but what's the bottom line?Your work does *not* involve the integration of a partial differenceequation. Integration is used to 'nd anti-derivatives. Differenceequations are solved using the sum calculus. Nowhere in the exposition ofyour 'nd do you ever integrate any equation whatsoever, much less apartial difference equation.--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: The Cipher of GenesisJack Sarfatti :Nothing, as usual, and was prolix about it.Society national meeting in Denver next year? Uncle Al knows a fellowwho was in your audience last year. He said you had everybody rollingon the §oor clutching their tummies - some from laughter, otherspuking, and the remainder collecting lint for tinder for igniting yourfaggots.--Uncle Alhttp://www.mazepath.com/uncleal/qz.pdfhttp:// www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: Implementing the partial difference 13:02:41 -0800, :>> Why don't you just post the program?Ok. Here's a straight-forward Java implementation. Nothing fancy. > It just does the job. ___JSHI re this code in C.It seems to work, at least for a few input values:# primes < 100 = 25# primes < 1000 = 169# primes < 1000000 = 78498C code follows:#include #include #include #de'ne STATS 0#de'ne CACHE 1#if CACHE == 1#de'ne I_CACHE_SIZE 1000#de'ne XDIVI_CACHE_SIZE 10000long term_cache[I_CACHE_SIZE];char term_found[I_CACHE_SIZE] = {0};long xtra_cache[XDIVI_CACHE_SIZE][I_CACHE_SIZE];char xtra_found[XDIVI_CACHE_SIZE][I_CACHE_SIZE] = {{0}};#if STATS == 1long i_max = 0;long xdivi_max = 0;long i_out_of_range = 0;long xdivi_out_of_range = 0;#endifstatic long S (long x, long y){ long i; long sum; long prev_term; long this_term; if (y < 2) return 0; sum = this_term = 0; for (i = 2; i < y+1; ++i) { long xdivi; long xtra_term; prev_term = this_term; xdivi = x/i;#if STATS == 1 if (i > i_max) i_max = i; if (xdivi > xdivi_max) xdivi_max = xdivi; if (i >= I_CACHE_SIZE) ++xtra_i_out_of_range; if (xdivi >= XDIVI_CACHE_SIZE) ++xdivi_out_of_range;#endif if (xdivi < XDIVI_CACHE_SIZE && i < I_CACHE_SIZE) { if (xtra_found[xdivi][i] == 0) { xtra_cache[xdivi][i] = (xdivi-1) - S(xdivi,i-1); xtra_found[xdivi][i] = 1; } xtra_term = xtra_cache[xdivi][i]; } else { xtra_term = (xdivi-1) - S(xdivi,i-1); } if (i < I_CACHE_SIZE) { if (term_found[i] == 0) { term_cache[i] = (i-1) - S(i,sqrt(i)); term_found[i] = 1; } this_term = term_cache[i]; } else { this_term = (i-1) - S(i,sqrt(i)); } sum += (xtra_term - prev_term) * (this_term - prev_term); } return sum;} #else /* CACHE != 1 *//* no caching */static long S (long x, long y){ long i; long sum; long prev_term; long this_term; if (y < 2) return 0; sum = this_term = 0; for (i = 2; i < y+1; ++i) { long xdivi; long xtra_term; prev_term = this_term; xdivi = x/i; xtra_term = (xdivi-1) - S(xdivi,i-1); this_term = (i-1) - S(i,sqrt(i)); sum += (xtra_term - prev_term) * (this_term - prev_term); } return sum;} #endif /* CACHE == 1 */int main (int argc, char * argv[]){ long n,r; if (argc != 2) { printf(usage: %s 0>n, argv[0]); exit(EXIT_FAILURE); } n = strtol(argv[1], NULL, 10); if (n <= 0) { printf(usage: %s 0>n, argv[0]); exit(EXIT_FAILURE); } r = (n-1) - S(n,sqrt(n)); printf(n%ldnn,r);#if CACHE == 1#if STATS == 1 printf(i_max:%ld xdivi_max:%ldnn, i_max, xdivi_max); printf(i_out_of_range:%ld xdivi_out_of_range:%ldn, i_out_of_range, xdivi_out_of_range);#endif#endif return EXIT_SUCCESS;} === Subject: Re: Bottom line on prime counting issuec. bond :| :||> Some of you may have noticed frenetic activity from posters trying to|> convince you that there's nothing sinister about mathematicians doing|> their best to downply my 'nd of a way to count prime numbers by|> integrating a partial difference equation, but what's the bottom line?||Your work does *not* involve the integration of a partial difference|equation. Integration is used to 'nd anti-derivatives. Difference|equations are solved using the sum calculus. Nowhere in the exposition of|your 'nd do you ever integrate any equation whatsoever, much less a|partial difference equation.you're lucky herman rubin isn't reading these threads.-- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Implementing the partial difference integration :> Why don't you just post the program?Ok. Here's a straight-forward Java implementation. Nothing fancy.> It just does the job. ___JSH----------------------> public class DefPrimeCounter { /** Creates a new instance of Main */> public DefPrimeCounter() {> } /**> * @param args the command line arguments> */> public static void main(String[] args) { int x = Integer.parseInt(args[0]); DefPrimeCounter Counter = new DefPrimeCounter(); System.out.println(Counter.p(x, (int)Math.sqrt(x)));> } public int S(int x, int y){ if (y==1) return 0; int max = y+1;> int sum=0, j; for (j=2; j (int)Math.sqrt(j)) - p(j-1, (int)Math.sqrt(j-1))); }> return sum; } public int p(int x, int y){ return x - S(x,y) - 1; }}Notice also that this program does not perform any integration. It simply forms a 'nite sum.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: I NEED HELP -0000, Androcles> :>> On get this straight.> You say that the force on a charge due to an electric 'eld acts>> instantaneously. Correct?> Let us consider a charged sphere somewhere in the universe. It exerts a>force>> on every other charge. If we can arrange for it to lose that charge>somehow,>> you are claiming that all those forces disappear INSTANTLY.> I will continue when I receive your answer (if one is forthcoming).> Interesting question, Henry. I've used it myself, on the discussion of>gravity propagation. If a star were to convert *all* it mass to radiationin>one super-supernova, how long would it take for us to detect its loss of>gravity? I don't mean watch its planet suddenly §y away, that would be a>distant observation and would reach us at c. I mean detect the gravityloss>right here. This would be the biggest gravity pulse (negative going) thatwe>could possibly detect. But in order to detect the loss, where would haveto>be aware of it in the 'rst place. As it turns out, the nearest star tous>(other than the sun) is 3.9 light-years away, and that is just too far to>detect it's gravity directly. So I fail to understand why anyone would>attempt an experiment to search for gravity waves, other than to give>themselves some funding on a futile attempt. There's a lot of money to be>made out of relativity, and very few people are altruistic.>Androcles>Exactly.There is one problem that I'm sure Paul will pounce on. That is, how tomake> charge suddenly disappear.Here is another interesting question.> The OMass' of an object is made up of a major proportion, the total massof all> energy.Presumably, the two portions play an equal role in Newton's gravitationLaw.> If there is no distinction between the two, is it possible that all MATTERis> nothing but some kind of manifestation of various O'elds'?Since matter is mostly empty space anyway, why can't I pass through a brickwall?And of course the answer is that I am repelled by the electrons. So what areelectrons?> Also, since mass is lost/gained in a Nuclear explosion, would we notexpect> measurable gravity waves to be produced. Even though the amount of matteris> relatively small, the sum total of all the potential energy associatedwith> that matter is quite large.As I understand it, a mountain is just detectable. An nuclear device issmall enough to carry on a plane. Therefore you would need a *Very*sensitive device to measure the gravity of the weapon, and that would implyproximity. If you detonate the device, I think you are likely to destroy theinstrument before it could react :)(If the nuclear bomb is considered to be the centre of the universe, whatis> the energy required to raise all the other matter in the universe to its> present distance from that point. Might it just turn out to be mc^2?).Henri Wilson.> See the Stupidity of Relativity.> www.users.bigpond.com/hewn/index.htm === Subject: Re: square root of -1 mod pJpr2718 :> tungsteneer3@yahoo.com (billy d.) asked:> is there a way to compute sqrt(-1) mod p>> easily (for a large prime p,>> say 7-digits)You can use the continued fraction>expansion of sqrt(p) to solve x^2 - py^2 = -1and then x = sqrt(-1) (mod p). For an>ef'cient way to do this, seeAre you sure this is wise? For p of the range indicated by the OP, oneexpects to 'nd continued fractions with periods up to the thousands.Even though this is not overwhelmingly large in terms of space or timecomplexity, I'm not sure I see the point of doing it this way. -- Erick === Subject: Re: Question on generation of large prime numbers> I have looking at a few web pages dealing with the largest known> calculated primes and a great deal of computational time is taking> into searching for these numbers and verifying they are primes. I have> seen the Euclid's proof of in'nitude primes and it occurs that me> that super-large prime numbers can be calculated using the following:p1=2 < p2=3> A large prime number, p, can be generated using> p = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5> p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7> p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbers> exceeding currently known largest primes can be generated rather> quickly. For example, we calculate the 'rst million primes, multiply> them and just add one.It seems that you haven't really understood Euclid's proof then. Thenew number constructed does not need to be prime; the point is that itis itself a product of primes and all of its factors must be newprimes. 2*3*5*7*11*13+1=59*509, as was mentioned earlier on thisthread. Also, 2*3*5*7-1=11*19. In either case, 59 and 509 are not in{2,3,5,7,11,13} and 11 and 19 are not in {2,3,5,7}. So long as yourset of primes is 'nite, you can extend it. Therefore the full set,which is not extensible by de'nition, must be in'nite.> If this is true, then calculating super-large primes should be a> computational trivial task, shouldn't it? Then why the big fuss over> calculating large primes?But it's not trivial. See my examples, which are very standard andwell-known.I suspect that I am missing fundamental here. Can super large primes> really be calculated using the trivial method outlined above? Hope> that someone can === Subject: Re: Bottom line on prime counting issueIn sci.physics, Sam Wormley> Some of you may have noticed frenetic activity from posters trying to>> convince you that there's nothing sinister about mathematicians doing>> their best to downply my 'nd of a way to count prime numbers by>> integrating a partial difference equation, but what's the bottom line? Does what I found work or not?>> Now that is an interesting question, isn't it! > pssst, hey Harris... your stuff doesn't work.... but don't tell anybody...Actually, his stuff worked more or less 'ne ... but I can'tsay it was the fastest.http://home.earthlink.net/~ewill3/math/primecounters/ index.htmlwas a somewhat tongue-in-cheek contest I sponsored 2 months backthat produced a few bizarre results and some interestingalgorithms. (However, Christian Bau has a better one anyway,although he didn't submit that particular one for my contest.Perhaps it was because my contest was unworthy thereof. :-) )James' wasn't the best, especially after a round ofmemoization which I for one did not foresee. In fact,yours truly beat him with a rather simple Legendrephi entrythat was more than twice as fast. However, an even simplersieving entry (sieve6bool, which did take advantage of aprime being of one of the forms 6k+1 or 6k-1, after thetwo entries 2 and 3) was 10.5 times as fast, once I gotit working -- and the best entry (edgar3), apart from acouple of what were essentially large table lookups byyours truly, was submitted by a hitherto unknown posterby E-mail.James did win a consolation prize for elegance. Of courseprize here is a bit of a misnomer. :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: Bottom line on prime counting issue> One last question to Harris:What is the difference between algebraic numbers, algebraic integers,> numbers, integers and complex numbers?Hint: Use Harris's theorem, Q=R.LH === Subject: Re: Question on generation of large prime numbersmsb@vex.net Euclid, primes is 'nite. We can >> therefore form a new number, N, which is one greater than the product of >> all primes. Therefore, N > P (and, indeed, it would be very much greater >> than P). When divided by any prime in our list, it leaves remainder 1. Therefore, >> /either/ it is prime /or/ it is the product of two or more primes, at least >> one of which is greater than P.> In /either/ case, P has been shown not to Nowhere have you assumed that the list of primes known is > complete up to P. You've simply proved that the 'nite list of> known primes is not the list of all primes.Well, Richard and Phil are both wrong here. Richard: Phil: Nope.Both wrong? !a AND !!a?> The proof could be expressed> in terms of a 'nite number of known primes, as Phil seems to have assumed> it was, but that's not the way Richard expressed it -- he spoke explicitly> of the product of *all* primes.If you view it in terms of sets, subsets of Z, my view is perfectly standard one, it's what Ribenboim calls Euclid's proof, and as such is applicable to the premise that Richard was assuming. Richard was the one who was introducingmore assumptions. All I need is a 'nite set of primes. A maximum prime P, and the fact that we're in Z, gets me that. Euclid's proof just follows immediately. Let all primes be all primes in that set, and the connectionis made. This is why I was blathering about what all meant. > But Richard's conclusion that P has been shown not to be the largest> prime number is also wrong; what has been shown is that the assumption> of a 'nite number of primes is wrong.Yup, which is why I said Nope..> The only reason to assume that> at least one of the additional primes discovered must be larger than> P is that we *thought* P was the largest prime, before we discovered them.Euclid permits me to assume {2,3,7,13,43,139,3263443} is the 'nite set of primes, with maximum prime P=3263443. In what way does {2,3,7,13,43,139,3263443}, P=3263443 violate the premise Let P be the largest prime? In which case, surely Richard, as he originally stated hisargument should permit it too. However, _none_ of the primes that Euclid's construction discovers is larger than P. If the premise is satis'ed, but the conclusion isn't, then there's a syllogistic error somewhere.If Richard has simply added let all primes <=P be known to his premiseI wouldn't have jumped on it that way. However, if he had, then it would _not_ have been Euclid's proof (and he claimed what followed was Euclid's proof so I would have jumped on that instead), and wouldn't (on its own) have been a proof of the in'nitude of primes (as you've added another assumption, so the proof by contradiction only disproves the _conjunction_of the assumptions, not either one individually).Ockham has some wise words for moments like this.Phil-- Unpatched IE vulnerability: Web Archive buffer over§owDescription: Possible automated code execution.Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0303/107.html== === =Subject: Re: Indian 'rsts in Maths <8DWsb.193431$HS4.1622428@attbi_s01> <3fb4d62b.18805594@news.clara.net>Treme: C&C,DWSIn === habshi@anony.com (habshi) said:>Subject: Indian 'rsts in MathsROTF,LMAO! You list a bunch of tripe that has nothing to do withMathematics, But you forget G. H. Hardy's greatest discovery,Ramanujuan! Plus, of course, you misrepresented some and you neglectedChinese and Greek work much earlier than some things you list.-- Shmuel (Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail will be subject to legal action. I reservethe right to publicly post or ridicule any abusive E-mail.Reply to domain Patriot dot net user shmuel+news to contact me. Donot reply to === Subject: Re: square root of -1 mod pbilly d. :|>is there a way to compute sqrt(-1) mod p easily (for a large prime p,|>say 7-digits)? i know that -1 has a square mod p by the law of|>quadratic reciprocity, but that doesnt help constructing an x such|>that x^2+1=0 (mod p). i also know that the problem is equivalent to|>'nd numbers of order 4 (mod p), since order of sqrt(-1) is 4, but yet|>again that seems like a dif'cult problem.In Maple 9, for example:> msolve(x^2=-1, 1299709); {x = 329008}, {x = 970701}takes very little time, even on my rather slow computer.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Product of matrix elements?> Given two nx1 column matrices (vectors) X=(x1 x2 ... xn)^T and Z=(z1 z2 > ... zn)^T, all elements real numbers (T=transpose):1) Is there a simple matrix operation to create the nx1 matrix> (x1*z1 x2*z2 ... xn*zn)^T from X and Z? I am referring to a mathematical > operation like inner product, rather than an algorithm or computer program.let Em be the square matrix (eij) such that emm = 1 and eij = 0otherwise.consider Em.Z.X, m = 1, 2, 3, ..., n> 2) Is there a way using matrix operations to produce the column vector > (exp(x1) exp(x2) ... exp(xn))^T (exp=exponential function) from X?assuming exp(A) = Sum A^n/(n!) over n = 0, 1, 2, ..., in'nityyou can obtain the jth entry of the sought vector:exp(A)ej,where A is the diagonal matrix aii = xi, aij = 0 for i not equal to j,and ej is the jth elementary vector.> 3) Related to (2), is there a way using matrix operations to produce the > nxn diagonal matrix D(x1 0 0 ... 0)> (0 x2 0 ... 0)> ( ... )> (0 0 0 ... xn)from X? And conversely, given D as above, to write X in terms of D using > matrix operations?consider X.ejform a linear combo.> e-mail) === Subject: Re: Question on generation of large prime Nowhere have you assumed that> the list of primes known is > complete up to P.Well, he said, ``form a new number, N, which is one greater than the product of> _all_ primes [emphasis added].'' It seems to me that this would necessarily be> a complete list.Let all known primes be {2,3,7,13,43,139,3263443}, the largest prime being 3263443. > I am not sure what you're getting at. If your point is that Richard> Heath'eld's proof is not the same as Euclid's, that's one thing. But his> argument seems sound to me. Filling in a little, but I think really rephrasing> Richard's argument: If there's a largest prime P, then there are 'nitely> many primes. If there are 'nitely many primes, we can take the product> of all primes, and add 1 (call this N). We know 2 is a prime and 3 is a prime, and there are> others, so N > P.Yup to here. > No prime less than or equal to P (we took the product> of all primes, where P was assumed to be the largest> prime) divides N (they all leave a remainder of 1).i.e. you want to add the assumption that you know all primes up to P. > So, any prime that divides N must be greater than P. There must be a prime that divides N. This contradicts P being the largest prime.Or it contradicts the assumption that you know all the primes up to P.If you use proof by contradiction with two assumptions, then all you know is that the conjunction of your assumptions is false.> Euclid's proof, if I recall correctly, is just given a 'nite list of primes,> we can show there must be another prime. So that is different from Richard's> proof. Quoth Richard: His proof [...] is as follows:Call me a pedant, if you like, as that's what I am. I see it, I call it.> But that doesn't mean there's anything wrong with Richard's (apart from> the claim that it's Euclid's).Your {2, 3, 7} is not ``all primes,'' as Richard speci'ed. This is why I asked Richard to de'ne all primes. He gave no de'nition.Euclid's gave one, which permits {2, 3, 7}. Which is why I don't view myinsistance that we all know what all primes means to be blathering.Richard's insistance that all primes up to P be known and there be no primesgreater than P is cannot work directly, as is, to prove that there's a prime greater than P, as the contradiction doesn't tell you whether there be no primes greater than P is false or all primes up to P be known is known.Adding extra assumptions is almost always a bad thing to do when performinga proof by contradiction.Phil-- Unpatched IE vulnerability: WMP local 'le bounceDescription: Switching security zone, arbitrary command execution, automatic email-borne command executionReference: http://www.ntbugtraq.com/default.asp?pid=36&sid=1&A2=ind0307&L =ntbugtraq&F=P&S=&P=6783Exploit: http://www.malware.com/once.again!.html === Subject: Re: Reality of response to my work> Justin Van Winkle :> I think this is morally wrong.> Since you top posted, I have no idea exactly *what* was said that you> regard as> morally wrong.> Don't tell James to waste his money.> I only recommended that he get the advice of a patent attorney if he> wanted to> secure intellectual property rights. I assume you regard your *free*> advice> about the matter as superior to the advice of a professional. If so, you> may be> guilty of practicing law without a license.Is it illegal to practice law without a clue?[snip!]Justin Van WinkleRegarding patents, a previous poster is correct, a provisionalapplication establishes who was 'rst only. However, I think that itis possible in the US to publish up to a year prior to 'ling a patentapplication. After 'ling the provisional application, the personshould obtain a simple non-disclosure agreement from anyone hediscusses it with. The original poster should contact an attorney forgood advice.Now, the market for a method to reliably count primes may besubstantial, perhaps many millions of $/yr. Such a method wouldenable someone to decide the probability of whether he he has triedall possible prime factors of a number for code breaking. Such amethod combined with others could be worth a great deal of money. === Subject: Re: The Cipher of GenesisMr. Sarfatti SirHow are you..how do you feel?dave === Subject: Re: Bottom line on prime counting issue> :Some of you may have noticed frenetic activity from posters trying to> convince you that there's nothing sinister about mathematicians doing> their best to downply my 'nd of a way to count prime numbers by> integrating a partial difference equation, but what's the bottom line?Does what I found work or not?> Now that is an interesting question, isn't it! > pssst, hey Harris... your stuff doesn't work.... but don't tell anybody...And Sam Wormley, surprise, surprise, is making a false statement as it*does* work, but I'm not terribly surprised by his immature behavior.After all, I found this partial difference equation, which integratesover a certain range to give a count of prime numbers!!!Here's the equation:dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) -p(y-1,sqrt(y-1))],Here are the instructions for the integration:S(x,1) = 0.And p(x, y) = §oor(x) - S(x, y) - 1, and you get S as the sum of dSfrom dS(x,2) to dS(x,y).Now for someone like Sam Wormley it probably doesn't seem fair thatI'm here posting on Usenet stealing thunder from everyone else, buthey, blame the mathematicians.The bottom line on my work is that it DOES work.Now then, do any of you know *how* it works?Maybe some poster will reply to this post to explain it to you, buthow do you trust them?And what about the story--my story--of the discovery?What was I thinking? What motivated me to look in this area? What'sthe story?If mathematicians hadn't decided to break faith with you and the restof the world, probably there'd be a book, some popular work,explaining the story.But how can you get that story if mathematicians are playing theiracademic games?Bottom line: What I have works.So what if I sell my story and get rich. That's how capitalism works. These mathematicians are worse than communists, as how do you explaintheir behavior?I *am* the American Dream, 'ghting for what should be mine, having toget past weak-minded academics who are 'ghting to block my success. But I shall prevail!!!I'm sure some hate that I'm in it for the money. But why screw overthe world, why screw *you* over by blocking the story of my success?What's their motivation?My math discoveries, found for pro'thttp://mathforpro't.blogspot.com/ Larry Hammick :>George Baloglou> ... Ceva's theorem does not seem to be capable of proving Menelaus's>theorem...>Not exactly. This page goes into it:>http://www.cut-the-knot.org/Generalization/ Menelaus.shtmlIndeed! I had done my web search before posting, by the way, and I had beenat the sister page http://www.cut-the-knot.org/Generalization/ceva.shtml, where a variety of proofs did not include the one I posted yesterday; butI see now that there is a remark toward the bottom of the page, following anumber of applications, that does allude to the equivalence between the two theorems...Thx for this correction. baloglouAToswego.edu === Subject: Re: Integral of a function over [a,b] sca18@hotmail.com (Amanda) :> On the other hand, the integrals of f and g must be equal (still> assuming that f is bounded, of course), since the function f - g is> integrable and takes the value 0 at every irrational form [a,b].> How can we prove this?ffffThm: If f is Riemann integrable on [a,b] and = 0 a.e., then int_[a,b] f(x) dx = 0.Proof sketch: Let eps > 0. Then there is an open set U in [a,b] such that m([a,b] U) = 0 and such that |f| < eps everywhere on U. Now [a,b] U is compact, and so can be covered with 'nitely many disjoint open intervals, the sum of whose lengths is < eps. The endpoints of these intervals form a partition P of [a,b] such that the upper Riemann sum of |f| < eps*(b-a) + eps*M (or something like that). === Subject: Re: Question on generation of large prime numbersThere appears to be some incomprehensible logic in this thread.No-one is claiming that Euclid's method can generate all primes. Butit appears that someone IS saying that Richard is claiming this. He'snot.Let us all agree that if we are ever to multiply all primes up to P,we need a way of generating all of them, and knowing that they reallyare all prime, and that there are none omitted. I'm sure even Euclidhimself would agree with this - so would any rules-lawyer.So let's start. We'll keep it simple with digits we can all manage,and work our way up.A. Euclid says there is no largest prime, because for any givenprime P, a larger one Q can be found. (Well, not exactly that, but Idon't speak Greek, and certainly not the Greek of Euclid's time.)B. It's not hard to see that once Q has been found it can be treatedas though it were P, and its own Q found, and the new Q can again betreated as though it were P, and a further Q found, and so on. So ifthe 'rst Q is found, we know there will be an in'nite chain ofgreater and greater prime numbers.C. Euclid then de'nes a number N, which is one more than theproduct of all primes up to and including P. Note that Euclid neverrequres the value of N to be known.D. He states that either N (which is larger than P) is prime, orthat it has at least one prime factor M which is larger than P. Notethat Euclid never requires the value of M to be known.E. His reasoning is that N can not be divisible by any prime numberup to and including P, as they all give a remainder of 1.F. If N is composite, its prime factors are greater than P. Therecould be just one prime factor M, in which case N is a square number. More likely, there are two or more prime factors. Any one of thesefactors will do for M, and thus for Q. Because M is larger than P, Qis larger than P.G. If N is not composite, it's prime (as it's greater than 1), so Nwill do for Q. Because N is larger than P, Q is larger than P.H. Because neither the value of N nor M need be known, the value ofQ need not be known, but we do know that is is prime and that it islarger than P.If Euclid's method is true, is should work for ANY prime number - evensmall ones. So let us start at P = 5. Note, we could have startedanywhere. 5 is small enough that calculations are easy, without beingtrivial.Using a sieve of Eratosthenes, we know the primes up to and including5 are {2,3,5}. The number N Euclid's method calculates is 31. Usinga seive, we know 31 is prime, so Euclid is correct.Let's work with the next highest prime after 5, i.e. P = 7. N = 211, which is prime.Next, P = 11. N = 2311, which is prime.Next, P = 13. N = 30031, which is NOT prime as it is 59 * 509. ButEuclid claims that either N is prime OR it has at least one primefactor M greater than P. Either 59 or 509 will do for M. They areboth prime, and both greater than P (13).Next, P = 17. N = 223092871, which is NOT prime as it is 19 * 97 *277. M could be any of the three factors.So lets's check (adding in the smaller primes): P N N is prime? M exits? Q? Q > P? 2 3 true false Q = N = 3 true 3 7 true false Q = N = 7 true 5 31 true false Q = N = 31 true 7 211 true false Q = N = 211 true 11 2311 true false Q = N = 2311 true 13 30031 false true (59) Q = M = 59 true 17 510511 false true (19) Q = M = 19 trueSo what happens when we reach the last KNOWN value of P where we alsoKNOW all previous values. For any higher values of P (such as a knownmersenne prime) we will not be able to calculate N.Well, it doesn't matter, as Euclid never required us to know allprimes smaller than P. It is suf'cient to know than N is not evenlydivisible by any prime smaller than P or by P itself, simply becauseof the way N is calculated (see paragraphs E and C).We can then follow through paragraphs F, G and H to know that Euclidwas correct.The red herring that Phil threw in about how do we get 5 into thelist of prime numbers is nonsense. No-one ever claimed the only wayto get the list of prime numbers was to generate N from each P bysubstituting the previous N as the current P.Phil has his P - (N => P) - (N => P) - (N => P) chain {2,3,7,31}. Sure that chain doesn't include 5. Nor does it include 11, 13, 17,19, 23 or 29. Nor 37, 41, 43 and a whole lot more. And Phil seems tothink that 5 therefore needs special dispensation to be allowed as aprime number.But Phil: What justi'cation do you have for starting at 2? Why is2 a prime number, and 5 not?ObPuzzle: The de'nition of a prime number is a counting number thathas exactly two factors - 1 and itself (which is why 1 is consideredas not being prime [which is odd, given the derivation of the wordprime] - neither is it composite, as those numbers have a factorother than 1 or itself). If we glibly change the de'nition fromcounting number to integer, how many prime numbers are there? What are they (if any)? === Subject: Re: Vedic Mathematics --- Myth and Reality> :>> Steven G. Johnson :>> Equivalently, M*N is the same as M*N mod (M + N - 1).> Sorry, this should be multiplication of M digits with N digits, base b,>> is equivalent to multiplication modulo b^(M + N - 1), i.e. M+N-1 digits.>Oh well, so FFT or not, looks like multiplying M by N by any method means>MN multiplications! Well, when these multiplications are hardwired (as in>human memory for single digits) the computational issues (On*n) becomes>really irrelevant, for they all are done in no time at. Like, the video>extraction for radar data processing is done by NAND gates - its all done in>real time!You are in error. The number of multiplications required for> multiplying two numbers with the FFT method is O(n*log(n) where n is> the larger of the two numbers; it is not m*n.Fine, just multiply 12345 by 67809 using FFT with less than 25multiplications. Do it here.> Richard Harter, cri@tiac.net> http://home.tiac.net/~cri, http://www.varinoma.com> We have people from every planet on the earth in this State.> -- California Governor Gray Davis === Subject: Re: Relativity is based on assumption.> Androcles :> All theories, in physics and elsewhere, are based on assumptions.> No theory can bootstrap itself out of nothing.> So what?> OSo what' depends on the validity of the assumption. The PoR is an> assumption, and is intuitive as well. Making the assumption that the> time it> takes for a signal to reach an object is the same as the time ittake> for> the signal to return, when in the meantime you've moved away ortoward> the> object, is a rather silly assumption that I will not accept. That's'so> what'.> Androcles> Then ignore relativity.> But unless you can come up with something that agrees with the> experimental evidence better than relativity, everyone else will> ignore you.> What experimental evidence? Moving clocks running slow? They don't. TheGPS> clocks run fast. MMX? It's pretty obvious to anyone with half a brainthat> the result is what you would expect if the speed of light were source> dependent. I would say that agreed with the experimental evidence much> better than relativity.> AndroclesAndrocles> 1. Are you saying that the speed of light is source dependent?Yes.> 2. What is your de'nition or explanation of Osource dependency'?Peter RiedtJust throw a ball forward from your car window as you move.It should be obvious, and no different from throwing a photon from a star.The velocity of light in interstellar space is c, with respect to the star.If the star moves, then it still c with respect to the star. To an observerthe velocity is c+v, where v is the velocity of the star.Now create a mathematical model of a star in elliptical orbit accordingto Kepler's laws, determine when the light will arrive, calculate thevariation in brightness, calculate a spectrum, and see if it matchesempirical data. If it does, then Einstein's second postulate hasno foundation. A match to empirical data has been found.There are three models to consider.1) Newton-Galileo - photons and PoR, SoL is source dependent.2) Maxwell-Lorentz - wave and PoR, SoL is aether dependent3) Einstein - photon or wave, SoL is observer dependent.I don't know of any sensible fourth choice.MMX thows out 2), but cannot differentiate between 1) and 3).Therefore 1) should be given due consideration, someone has to do it,and that someone is me.Androcleshttp://www.androc1es.pwp.blueyonder.co.uk/ index.html === Subject: Re: Was Lucy the split image - Jabriol is a creationist.Psycho-hypocrite Antonio once again resorts to the same ad-hom he sowhiningly accuses others of! How the mighty have fallen!.Look at this hilarious bull:> oh by the way since you are in an informative mood.. tell them how you> have been caught lying so much.. that nobody belives you anymore..Nobody ia a bigger liar than you, as I have repeatedly demonstrated. BTW, you have now proven evolution by evolving into a weasel.Budikka === Subject: Re: Reality of response to my work> Now, the market for a method to reliably count primes may be> substantial, perhaps many millions of $/yr. Such a method would> enable someone to decide the probability of whether he he has tried> all possible prime factors of a number for code breaking. Such a> method combined with others could be worth a great deal of money.This doesn't really make sense to me. If you can generate all the possibleprime factors, then you know how many there are. If you can't generatethem, then certainly you can't use the fact that there are more factors tosomehow generate more factors. Seriously, if you know that there must be atleast 10^20 more primes to try, that still leaves you to 'nd all theseprimes. If you know that there aren't any more primes to try, then you knowhow many you've tried. (I'm not an expert on this topic, so I may be wayoff.)Now, if you had a function that returned how many primes are less than agiven number that would be something. However, if you must iterate throughevery number less than that given number, your method isn't really thatuseful (unless it is much faster than any current method). If you had analgorithm to factor a number, that would be useful. If that algorithminvolves testing every prime less than sqrt(n) to test the primality of n,then the algorithm is useless (and trivial).In the end, however, you may still have a valid point. But that point restson the assumption that this is the best method to do such work. (or it iscomperable to good methods) There are already pretty good estimates of howmany primes are less than a given number. There are very fast methods tocount how many primes are less than a given number. There are VERY fastmethods to determine the primality of even extremely large numbers, withsome small chance of error. === Subject: Re: Need advice on letters of Rudolph :> ...>> (2) as Bart Goddard has pointed out, letters of recommendation do>> have at least some chance of indicating that other (for me, much more>> key) component in graduate studies success, namely, the >> ability to *do mathematics*, while (3) institutional admissions>> exams surely couldn't do anything at all towards indicating that>> other component.>the ability to solve mathematical problems, say in an admissions exam,>is without a doubt, a much better indicator of ability to do>mathematics. Scarcely without a doubt! The conditions of an exam, admissions> exam or otherwise, are entirely different from the conditions > under which (real, practicing) mathematicians (of my broad> personal acquaintance--which, I admit, does not include you,> who may for all I know work along entirely different lines)> actually do mathematics. And such problems as must,> necessarily, be posed on an exam (admissions exam or> otherwise), are just that--problems (with known solutions)> to solve, not new mathematics to do (and possibly fail at).if you were wishing to establish that the conditions between takingexams and being a mathematician are different, congratulations, youhave done a superv job! i wonder though, how does that damage mystatement?>in fact, the ability to solve mathematics problems is the>only accurate indicator of ability in mathematics. I would trust the seriously considered opinion of almost any> of my professional acquaintances, expressed in a letter of> recommendation, that a third party had--or had not--the > ability to do mathematics, far more than I would trust an> admissions exam. I wouldn't trust a stranger's opinion> as much, but I'd still trust it more than an exam. i reckon that is a problem you and many others share.>do you understand where i am coming from?Either (if I am to be charitable) from somewhere far from where> I am, or (otherwise) from a position of profound ignorance of> what it means to do mathematics.perhaps now you will reject the notion that a mathematician is de'nedby his ability to solve math problems?no, i am not referring to ones with known solutions...> Lee Rudolph Mike Kent :> John Harrison :>>For those who have the book, it is W. Rudin, Principles of>>Mathematical Analysis, chapter 2 ( Basic Topology), problem 18:>>[A set E is perfect if E is closed and every point of E is a limit>>point of E] Is there a nonempty perfect set in R which contains no>>rational number?>>This one has really had me stumped. My 'rst guess was that the answer>>was no, but I didn't get anywhere trying to prove it. I then tried to>>construct such a set: since the set of rationals is countable, we can>>write it as {r_1, r_2, r_3,...}. Let I_n be the open interval of>>length 1/2^n centred at r_n, let I be the union of the I_n, and let X>>be the complement of I in R. Since the total length of the intervals>>is at most 1, X is nonempty. As the complement of an open set, X is>>closed. But I can't show that X is perfect. So: Is the answer yes or>>no? If yes, does my set work as an example? Any hints would be most>>welcome.> Try showing (1) X is not countable> (2) the isolated points of X constitute at most a> countable set> (3) X {isolated points of X} is closedI don't think this will work. Try considering points of X such thatevery neighborhood contains an uncountable in'nity of points of Xinstead. === Subject: Re: Can someone help with these diophantine equations? Thomas Womack :> Sniz Pilbor anyone with some numbercrunching>power help solve these following diophantine equations (there are>in'nitely many solutions but I am looking for the one with the>smallest A) (and of course, nonnegative integers only please!):>A^2 = 729B + 364>A^2 = 2187B + 1093>A^2 = 6561B + 3280>A^2 = 19683B + 9841OK, you're looking at successive approximations to the 3-adic square root of > -1/2So if you ask magmap := pAdicRing(3, 100);> Sqrt(p!(-1/2))you get 14918756739905062250418133874814682552290432142That's one of the square roots - its negative is the other> and if you consider that large number mod 3^n you get the values of A that > work at each step.But you might not get the smallest A, which OP asked for, as that might sometimes come from the given square root and sometimes from the negative.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: === question about the riemann hypothesis>Subject: Re: question about the riemann hypothesis>Message-id: >> But I was wondering something about the Riemann Hypothesis: What if> it were proven to be false? I know that it's an important result>for> number theory, but what would the consequences be, if any, if it> turned out to be false?>Rainer Rosenthal : >Hello Kate,>in de.sci.mathematik, there is a thread of that kind, where I>>learned the following astonishing fact:>The following 18 numbers >1 2 4 6 10 18 22 30 42 58 70 78 102 130 190 210 330 462>are the only known ones not of the form > pq + qr + rp with p,q,r > 0Astonishing indeed!What is also interesting about these 18 integers is, add 1 to each>integer and you have 18 primes.Also an interesting fact is that 210,330 and 462, the last 3 integers>in this list, where each are @ a different level in the Collatz tree>and each have this same property. Mainly (n-1) == 2(mod)3. Isn't that also true for 6, 18, 30, 42, 78 and 102? If (n-1) == 2(mod3), thenn==0(mod3), so the numbers you selected are just those divisible by 3.So what's the story with 1, 2, 4, 10, 22, 58, 70, 130 and 190? Don't youhave to explain why some numbers on the list have that property while others don't?>Also all>there subsequent higher level integers have this property thus no>branching occurs in higher levels above these integers.That's true on any number ==0(mod3). What distinguishes this small group from the in'nite lsit of other numbers with that property?Could there be a possible relation to the Collatz Conjecture?What relation? That some of them are divisible by 3? I think you need more than that. Does that particular list of numbers show up in theCollatz Conjecture somwhere? As an example, the list of knowninteger crossover points is {1, -1, -5, -17}. And sure enough, one numberon the list matches. But that's just coincidence. If _every_ number matched,then we should start looking for a relationship.Dan >According to Borwein and Choi, there are are no others and>>there are certainly no more than 19 of them.>>BUT: if the Riemann Hypothesis fails, there could be another>>number not of this form. And it must be > 10^11 then.>(1) Sequence A025052 in the njas archive>> http://www.research.att.com/~njas/sequences/>(2) Posting by Hermann Kremer in de.sci.mathematik (german)>> from 20.06.2001 00:15>related sequence A006093. ( See the thread pq+p+q === Subject: Re: Problem calculating limits - Need Help! drbr@bezeqint.net (Roy) :> I'm having dif'culties solving these two limits.> I mustn't use L'hospital rule:> a) lim(x*(2^(1/x))-x) where x increases to in'nite.We want lim_{x->oo} [2^(1/x)) - 1]/(1/x). As x -> oo, 1/x -> 0. So this is nothing but the derivative of 2^x at x = 0.> b) lim(cosh(x)-1)/(x^2) where x approaches 0.Use Taylor series as others have suggested, or if you're feeling adventurous, show that for any C^2 function f in a neighborhood of 0 with f'(0) = 0, (f(x) - f(0))/x^2 -> f''(0)/2 as x -> 0. === Subject: Re: Vedic Mathematics --- Myth and Reality (a working out)> adda1234@bigpond.com (Arindam at all!> It's 4 things to combine, no matter how it's done.> Oh, it matters a lot how it's done. Because it is done differently, it> is a different method. Because it is done more easily and quickly> both in the mind and on paper, it is Vedic?That's what I too would like to know. I think it is, for I suspect myfather knew this method. If a number of Indian grandparents from allover India jump up saying that they knew this method, that it wastaught to them traditionally, then it would be Vedic (and not a fraudderived from say Tractenberg or someone and put in a book and passedoff as Vedic) unless one showed that they were all lying.> 2. is it really such a big deal?Oh yes. The idea is that it should be taught in primary schools,replacing the current multi-line method. If that is indeed done allover the world, then certainly is is a big deal.> Ad 1: The expression crosswise vertical doesn't say anything about the order of addition.That's the beauty of it.> Ad 2: Math (at least western math) is much more than arithmetic. It sure is a cool thing> to do 345 x 167 in the mind. Especially for school children. But it's not an advance> in the 'eld of mathematics. It's simply arithmetic.And isn't arithmetic really wonderful? Especially when you know whatit is all about? Vedic arithmetic shows you that. If implemented, itwill be an advance in arithmetic, and arithmetic was taught to me inschool as a part of mathematics. We started with arithmetic, then wehad algebra, then geometry, then trigonometry, then co-ordinategeometry and 'nally calculas.Look, let us do the following sum using Vedic arithmetic:(1234*5678 + 5659873 - 9987654)*345 just like that. Well, we get( (5)(16)(34)(60)(61)(52)(32) + 5659873 - 9987654)*345or ((10)(22)(39)(69)(69)(59)(35) - 9987654)*345 or ((1)(13)(31)(62)(63)(54)(31))*345or (3)(43)(150)(379)(632)(724)(624)(394)(155)or 925010495.I'm not sure if I have worked it out correctly, for I was only tryingto show the Vedic method, how beautifully and elegantly it works, thatis, without bothering about all the carries from one operation toanother, as done in current methods taught in school. Concentratingon the place values, and not the carries, makes life lot easier forthe working-out. If the beauty of all this does not strike you oranyone, I have nothing further to say.And still I don't know the one-line division and square rootextraction method. As far as I can see, that is also unknown toanyone around. But these methods have been published in the book onVedic mathematics, that is hailed as a fraud.Arithmetic is the basis of all computation. And computation is soimportant, in business, 'nance, engineering, etc. Faster computationwill naturally lead to greater ef'ciencies.> And for the purpose of math education, learning people calculation Otricks'> is generally considered poor education of math.We are talking about teaching the fundamentals of arithmetic tochildren using better methods. After 44 years of dealing withmathematics, I am of the 'rm conviction that mathematics is an art,and like all arts, it is a bag of tricks. I am con'dent that peoplewill learn to love maths, once they get over their dread ofarithmetic, taught the usual way. Anyway, the best judges for thiswill not be experienced mathematicians, but primary school teachersand their charges, oce the green signal has been given by Jurjus === Subject: Re: Need advice on letters of recommendation Bart Goddard :> 1. When a student (or anyone) asks for a letter, it is > implicit that they are asking Will you write me a _good_> recommendation. Although I do have one example to the contrary. Many years ago, an undergraduate asked me to write her a letter of recommendation for mathematics graduate school. I told her that I wouldn't be able to say very much beyond she completed most of the homework assignments and she passed the course, but she still wanted me to write the letter, which I did. The story turned out to be that she was an overseas student whose government had been paying her expenses. The catch was that she had to apply to graduate school. Didn't have to go to grad school, but had to apply. She didn't want to go, and knew as well as I did that it wasn't a good idea for her to go, so what she actually wanted was a letter that would get her rejected! Unfortunately for her, one of the places she applied to was so desperate for students that it accepted her despite the lukewarm letters of recommendation she presented. I regret that I don't know how things turned out in the end.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: SymbMath.com: web-based computer algebra system > SymbMath For Java is web-based symbolic math and computeralgebra> system,> which runs in any computer with Java. You can play it online.> www.SymbMath.com === Subject: Re: Data analysis software It plots and analyses any x-y data for peak location, peak height,> peak> width, semi-derivative, derivative, integral, semi-integral,> convolution,> deconvolution, curve 'tting, and separating overlapped peaks> and> background.> www.chemSoftware.com === Subject: Re: Reality of response to my workThis doesn't really make sense to me. If you can generate all thepossible> prime factors, then you know how many there are. If you can't generate> them, then certainly you can't use the fact that there are more factors to> somehow generate more factors. Seriously, if you know that there must beat> least 10^20 more primes to try, that still leaves you to 'nd all these> primes. If you know that there aren't any more primes to try, then youknow> how many you've tried. (I'm not an expert on this topic, so I may be way> off.)There is a simple proof that there can be no largest prime. Even my kidsunderstand it (10, 11 and 13). === Subject: Re: online math handbook > web-based symbolic calculator, math handbook and computer algebrasystem.> www.mathHandbook.com === Subject: Re: square root of -1 mod pHenri Cohen's book A Course in Computational Algebraic NumberTheory(Graduate Texts in Mathematics, Vol 138), Springer-Verlag has anice explanation of ef'cient ways to compute square roots modulo p ingeneral, although as one poster pointed out, there are may well bequicker ways to compute the square root of -1.JoeS> is there a way to compute sqrt(-1) mod p easily (for a large prime p,> say 7-digits)? i know that -1 has a square mod p by the law of> quadratic reciprocity, but that doesnt help constructing an x such> that x^2+1=0 (mod p). i also know that the problem is equivalent to> 'nd numbers of order 4 (mod p), since order of sqrt(-1) is 4, but yet> again that seems like a dif'cult problem.i am trying to do this without checking every number === Subject: Re: Bottom line on prime counting issueNntp-Posting-Host: apps.cwi.nl... > After all, I found this partial difference equation, which integrates > over a certain range to give a count of prime numbers!!! > Here's the equation: > dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - > p(y-1,sqrt(y-1))],As written this does not look like a partial difference equation to me.perhaps you intend: S(x, y) - S(x, y-1) = the expression on the right,or: S(x, y+1) - S(x, y) = the expression on the right.It is not clear which of the two you intend here. > Here are the instructions for the integration: > S(x,1) = 0.Instruction? Looks more like a boundary condition. > And p(x, y) = §oor(x) - S(x, y) - 1, and you get S as the sum of dS > from dS(x,2) to dS(x,y).There must be something wrong. You now instruct how to get S. But Ithought in a partial difference equation you had to get S by solvingthe equation. So how can you now give explicit instructions on howto calculate S?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Relativity is based on assumption.> I provided proof for my claim. But of course you *conveniently*> snipped the link. That proof consists of fairly straightforward> reasoning.> LOL!!> I provided the proof that Einstein makes assumption, AFTER you denied it,> and called me a liar.You did no such thing.You are a disgusting lying pig.> Then YOU snipped my response, windbag.I did no such thing.You are a disgusting lying pig.> Not only that,> but you gave no proof at all.I did.You are a disgusting lying pig.> Assertion isn't proof, even if you think it is. I never claimed it were.You, however, have provided nothing but assertion for the subject line.> The subject line remains true, whether you like it or not, Ostrich.The subject line has been solidly disproven.You are a disgusting lying pig. Zbigniew Fiedorowicz :> Mike Kent :>> John Harrison :>>For those who have the book, it is W. Rudin, Principles of>Mathematical Analysis, chapter 2 ( Basic Topology), problem 18:>[A set E is perfect if E is closed and every point of E is a limit>point of E] Is there a nonempty perfect set in R which contains no>rational number?> ... [ contructs open set containing rationals and with measure < 1, takex X as its complement ] ... >> Try showing (1) X is not countable>> (2) the isolated points of X constitute at most a>> countable set>> (3) X {isolated points of X} is closedI don't think this will work. Try considering points of X such that> every neighborhood contains an uncountable in'nity of points of X> instead.By Cantor-Bendixson, (2) must be true. (3) seems clear since X, being closed, contains all its accumulation points, and these are also exactly the accumulation points of X{isolated points}.Is there a problem with (1) or C-B too much to take as known? === Subject: Re: Algebra questionArnold> Can someone explain to me what's the difference between> basis transformation and a similarity transformation.No doubt basis transformation refers to a square matrix which describes achange of basis. For any two given basesx={x1,...,xn}y={y1,...,yn}the change of basis from x to y is described by a unique such matrix M.Perforce M has a nonzero determinant and is invertible.The only meaning of similarity transformation that I know of is this onefrom geometry: a mapping f of a Euclidean space into itself such that[f(x),f(y)]=k[x,y] for any x,y and some real constant k>0, where [ , ]refers to the distance between two points. If this doesn't seem to makesense, maybe you could tell us more about the context in which you cameacross the term.LH === Subject: Re: Reality of response to my workNntp-Posting-Host: apps.cwi.nl...You do not know the 'eld, yes? > Now, the market for a method to reliably count primes may be > substantial, perhaps many millions of $/yr.You are off by a factor of many millions. > Such a method would > enable someone to decide the probability of whether he he has tried > all possible prime factors of a number for code breaking. Such a > method combined with others could be worth a great deal of money.Such a method would be the most stupid to be done. Trying all possiblefactors of a number for code breaking is about the slowest possiblemethod you can 'nd, even if each possible factor is found and triedeach femtosecond.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Reality of response to my workNntp-Posting-Host: apps.cwi.nl > Now, the market for a method to reliably count primes may be > substantial, perhaps many millions of $/yr. Such a method would > enable someone to decide the probability of whether he he has tried > all possible prime factors of a number for code breaking. Such a > method combined with others could be worth a great deal of money. > This doesn't really make sense to me. If you can generate all the possible > prime factors, then you know how many there are. If you can't generate > them, then certainly you can't use the fact that there are more factors to > somehow generate more factors. Seriously, if you know that there must be at > least 10^20 more primes to try, that still leaves you to 'nd all these > primes. If you know that there aren't any more primes to try, then you know > how many you've tried. (I'm not an expert on this topic, so I may be way > off.)If you know that there are 10^20 primes to try it will take you 10^11seconds to try them all when you could try each prime in a nanosecond.That is something like 3000 years. That would even get you moderate40 digit numbers out of reach. > There are very fast methods to > count how many primes are less than a given number. There are VERY fast > methods to determine the primality of even extremely large numbers, with > some small chance of error.There are even VERY fast methods to determine the primality of evenextremely large numbers without a chance of error.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: DO MATHEMATICIANS READ WITH HALF A LIGHTBULB?I'm trying to get the deadwood off my reading shelf, and I just'nished BITCH, by Elizabeth Wurtzel, which turned out to be worthreading for her powerful, insightful, and soul-baring epilogue, buttoo much of the rest of the book was just too much, over and over,about O.J. Simpson. Worse, Ms. Wurtzel is a cinema-holic whodismisses the woes of real-life victims who don't play their partswell, as if we should all have just the right life scripts for ourtragedies and have the necessary acting abilities to earn her respectand sympathy, or at least a tepid encore?And, one of the next space-hoggers on my shelf is INFINITY AND THEMIND: THE SCIENCE AND PHILOSOPHY OF THE INFINITE, by Rudy Rucker.Rudy ticked me off yesterday -- I had hoped to get a good start on'nishing the rest of the book, but then he stopped writing his barelycomprehensible Math-eze and added some stinking number questions,including the following:Prove that 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a)So I did:(1 - a) [1 + (a + a^2 + a^3 . . .)] = (1 - a) (1 / (1 - a)) {multiplying both sides by (1 - a)}(1 - a) + [(1 - a) (a + a^2 + a^3 . . .)] = 1 {interim result}(1 - a) + [(a + a^2 + a^3 . . .) - (a^2 + a^3 + a^4 . . .)] = 1{interim result}(1 - a) + a = 1 {interim result}1 = 1 {'nal proof of equality}This proves that the original equation is true no matter what value ofa we use, but Rudy argues that the original formula makes no sensewhen we substitute certain values for a.However, before we discuss it, let's give a step-by-step de'nition ofdivision so that everybody can follow along:A / B asks as to count how many times we can subtract all of B fromA to the limit of A, plus count any fractional units of B that we cansubtract from A, and to state the total as our answer.For example, 9 / 5 asks us to count how many times we can subtract 5from 9 (once), plus how many fractional units of 5 we can subtractfrom the remainder of 4 (and our total answer is 1 & 4/5ths).Now, in the above formula:1+ (a + a^2 + a^3 . . .) = 1 / (1 - a), Rudy notes that if we use 2 for a, then the formula becomes:1 + (2 + 4 + 8 . . .) = 1 / (1 - 2), or1 + (2 + 4 + 8 . . .) = 1 / -1The brain stops working when we see an ever-increasing series on theleft side and a simple negative fraction on the right side, and weconclude that the two expressions are not equal. However, if we forcethat supposedly simple negative fraction through the true processrequired by division, as explained above, we will see how complex thatsimple fraction really is, and that it exactly equals the expressionon the left side:1 / -1 says to state how many times we can subtract -1 from 1 tothe limit of 1 plus state the fractional units.The 'rst time that we subtract -1 from 1 we get 2 and move away fromour limit of 1 instead of towards it (which makes sense as a reverseprocess from a division process using numbers with the same sign), andwe immediately have a whole count of 1 subtraction:1 + (. . .)After subtracting -1 from 1, the 1 now has a remainder of 2 afterbeing subtracted from, and thus our division limit is now 2 instead of1, and the fractional units of -1 that can subtracted from the limitof 2 becomes 2 / -1, or -2. As such, the Ofractional' units to beadded to our count comes from the next division of 2/ -2., which saysto subtract -2 from 2, which equals 4 (to the limit of 2), and to addthe interim answer to our series:1 + (2 + . . .)And so on as above:1 + (2 + 4 + 8 . . .) = 1 / -1The rote learners among us will immediately object that 1 / -1 = -1,and thus that -1 cannot possibly equal 1 + (2 + 4 + 8 . . .). Theargument ignores the fact that transforming 1 / -1 into -1 is analgebraic process (-1 = [-1] * 1 = [-1] * 1 / 1 = etc.), and is not agiven fact, and any such transformation should be performed on bothsides of the equation to be truly and logically valid in all cases,regardless that it is irrelevant in most cases. Thus, whenever thedivision process itself is crucial to the meaning of an equation, itcannot simply be dispensed with or collapsed on one side of theequation alone.Thus, the above idea is not contrary to algebra, but rather states apoint that is not being taught. It's a distinction, not acontradiction.Rudy gives a more interesting but nonetheless complying example of asubstitution for a where the substitution is -1:If a = -1 in: 1 + (a + a^2 + a^3 + a^4 + a^5 . . .) = 1 / (1 - a),then all the even powers of a will equal 1 and all the odd powers ofa will equal -1:1 + (-1 + 1 - 1 + 1 - 1 . . .) = 1 / (1 - [-1]) {interim}1 + (-1 + 1 - 1 + 1 - 1 . . .) = 1 / 2 {'nal}What Rudy notes is that at any point in the in'nite series on theleft, the total can be either 0 or 1, and thus that it can be arguedthat the evidence shows that the sum is ultimatlely either 0 or 1. Rudy then 'nds it amusing that the right side of the equation seemsto average this mind bender by declaring the sum to be the average of0 and 1, which is 1/2.No such metaphysics are needed to understand the formula. Although 1/2 is in its simplest form, it is still in the form of a requireddivision process that continues ad in'nitum. Although I haven'tworked out the exact mechanics of the math process on the right sidethat produces exactly the sequence on the left side, I imagine that itis not much more dif'cult than a 2-cycle engine the produces +1 atthe start and alternately procuces -1 every other cycle. (Perhapssomething like the following: 1 / 2 = zero 2's subtractable (leaving +1 untouched) + 1 / 2, and 1 - 2 = -1 (forced past the limit of 1),and -1 - (-2) = 3, and 3 / 2 = +1 with a remainder of 1 / 2, adin'nitum.)1 + (-1 + 1 - 1 + 1 - 1 . . .) = 1 / 2 {ad in'nitum}Rudy argues that the left side of the above equation represents aThompson Lamp where it can be argued that the 'nal state of thelamp is on, or off, however one chooses to argue the case. Theargument is spurious since it is plainly stated that the light, if itwere such, is being turned on and off at speci'c points in themathematical process, ad in'nitum.I'll 'nish reading the book, and I might even look at all theproblems, and perhaps I'll be enriched by it, but it really makes mewonder how much logic mathematicians really study.Very Respectfully,Ray === Subject: [JSH] Super-Crank Troll: Re: Bottom line on prime counting issue What is the difference between algebraic numbers, algebraic integers,> numbers, integers and complex numbers?I believe you might have to wade through JSHs Object Oriented Mathematicsand the brilliance that that has yet to shine on all of us losers before youcan even ask the self-proclaimed highest ranking number theorist in theworld.In fact, Mr. Harris ego is growing at a super-exponential rate and soon noone in sci/math will be able to contain his NPD! === Subject: Re: DO MATHEMATICIANS READ grava .88 la saucisse et au marteau:[SNIP]I hope it's a bad, ugly troll. Otherwise, it's a bad, uglymathematician.-- Nicolas, who wonders why there is always such people posting on Usenetnewsgroups. === Subject: Re: Question on generation of large prime numbersPhil Carmody :> Let all known primes be > {2,3,7,13,43,139,3263443},> the largest prime being> 3263443.If you change Richard's proof, you can make it wrong. Richard did not say,``take all _known_ primes'', or ``take all primes found by process X.'' Hesaid, ``take _all_ primes.''Your list above is missing a few primes less than 3263443. If you multiply_all_ primes up to 3263443 and add 1, the result will not be divisible by anyprime less than or equal to 3263443.that there are a 'nite number of primes. Not a 'nite number of _known_in principle, take the product of _all_ primes and add 1, and label that N. Then there has to be a prime dividing N. That prime is necessarily larger thanP.I agree that if you take something less than _all_ primes and apply the aboveprocess, that N could be divisible by something less than P. So what?Also, we all understand that this is not Euclid's proof. Again, so what?>> No prime less than or equal to P>> divides N.> i.e. you want to add the assumption> that you know all primes up to P.I don't see this as adding an assumption. If there's a largest prime, thereare 'nitley many primes, and in principle they can be listed. There is noaddtional assumption after that of a largest prime.I did not use the word ``blathering.''John Robertson === Subject: Re: Question on generation of large prime numbersIn rec.puzzles Richard Heath'eld :> Great! Please share...10^80 - 3? === Subject: Fiber Bundle Physics/Consciousness 2Commentary 2Given coordinate patch C(x) in the base space M in a neighborhood of point x and 'ber f(x)form the local Cartesian product C(x)f(x) with ordered pair X = (x,fo).Take the union C(x)f(x)/C(x')f(x')/... of all such local products.There are redundant ordered pairs X because the coordinate patches C(x) and C(x') as sets overlapwith non-vanishing intersection C(x)/C(x')=/= Empty Set.Identify the redundant multiple images of the same actual point of the base space M usingthe symmetry group G as an equivalence relation. That is, two ordered pairs X and X' areidenti'ed or equivalent if x = x' < C(x)/C(x') and if fo' = gfo where g < G to form disjointequivalence classes {f(x)} that are the distinct points of the 'ber in hyperspace H.This is all local at a 'xed base point x like in an internal gauge force symmetry.g is also called a transition function.The hyperspace H is the factor space of the union C(x)f(x)/C(x')f(x')/ ... mod G.The projection map P:(x,{fo}) -> xWhen M is the curved space-time of Einstein's gravity theory in addition to the G equivalencein the extra space dimensions of the 'ber, x'(E) = Diff(4)x(E) at 'xed event Eto make disjoint equivalence classes {x(E)} mod Diff4(E).One can imagine a hybrid where the 'ber is a discrete space of strings of c-bits.One can also imagine a 'ber of strings of qubits.1 qubit is a parallel in'nity of c-bits.i.e.|qubit> = |1 c-bit><1c-bit|qubit> + |0 c-bit><0 c-bit|qubit>Where there is a continuous in'nity of different c-bit basesor orthonormal frames each corresponding, for example,the the angular orientation of an inhomogeneous 'eldmagnet in a Stern-Gerlach 'lter for spin qubitsin the DARPA spintronics project or like the billion billionSingle Electron Transistors inside the human brain at thesub-microtubular protein dimer hydrophobic cage level formingthe hardware interface with external world whose software is our stream of inner consciousness.Each possible orientation is a primitive parallel quantum universe.The quantum computer computes in all possibleorientations simultaneously like a continuousin'nity of classical Turing machines in adistributed network working on the same problem - or so the folklore goes.to be continued.On Sunday, November 16, as an idea has 4 parts.1. A structure symmetry group G.2. The total hyperspace H or, in some applications Wheeler's BIT.3. The projection map P.4. The base space M or, in some applications. Wheeler's IT.The hyperspace H consists of 'bers f(x) that areeither copies of or representations of the symmetrygroup G.The projection map P collapses a 'ber f(x) in the hyperspace H toa point x in the base space M.All of these objects are continuum differential manifoldsdepending on the continuum of real numbers which itsassociated issues of Cantor's in'nity of in'nities ofCabalistic Aleph's in an ascending Jacob's Ladder.This is not a discrete combinatoric mathematics althoughsuch a skeletal structure is associated with it as inHerman Weyl's Theory of Groups and Quantum Mechanicsand as in Saul-Paul Sirag's presentation of V.I. Arnold'sA-D-E mathematics of everything.The base space is covered by an atlas of local coordinate patcheswith all important overlap transition functions sewing thepatches together like a quilt.M is space-time in local micro-quantum 'eld theory of pointThe extra-dimensions of hyperspace formthe Calabi-Yau space of vibrations of thesuperstring beyond space-time.The connection on the total hyperspace H is the potentialof a local gauge force.Examples of connections is the 4 potential Au(x) inMaxwell's electromagnetism with G as U(1).There are similar connections for the Yang-Mills weak forcewith G = SU(2) and the strong force with G = SU(3).Classical general relativity, as distinct from local micro-quantum'eld theory, has the torsion-free symmetric three-index non-tensorLevi-Civita connection with G as the Diff(4) group.The latter comes from locally gauging the 4 parameter translation subgroup(generated by the 4-momentum Pu of globally §at special relativity )of the 15 parameter conformal group of Roger Penrose's massless twistors.Bottom -> Up: Given base space M and symmetry group G construct thehyperspace H as a quilt patchwork.Top -> Down: Given hyperspace H and symmetry group G construct thebase space M as the non-overlapping partition of hyperspace into G-orbitscalled the quotient space of H mod G in the principal bundle.Micro-quantum source renormalizable local 'elds of spin 1/2 lepto-quarks are associated vector bundles.Micro-quantum force renormalizable local 'elds of spin 1 gauge force bosons (electro-weak and strong) arefrom the principal bundle.There is no renormalizable quantum gravity in this precise sense.This is because classical Einstein gravity is a More is different (P.W. Anderson)emergent collective effect as in Andrei Sakharov's metric elasticity of aninstability in the globally §at false vacuum of the interacting lepto-quark source/electroweak-strong force.Einstein's gravity + uni'ed exotic vacuum dark energy/matter with Andrei Linde's chaotic in§ationary cosmology are the result of the continual phase transitions from globally §at false high entropy micro-quantum vacua to locally curved macro-quantum low entropy metastable vacua.to be continued: === Subject: Re: Accumulation pts. in R it.Lurch> They are equivilent. Let me see if I can explain it on the internet. If> this helps let me knowyou can just think of R for simplicity, although I don't think my argument> will use this fact.let's say we want to check if x0 is an accumulation point of some set S.I want to show that:> Every open neighborhood contains at least one point of S distinct from x0> is equivelent to> Every open neighborhood contains in'nitely many points of SSuppose every open neighborhood contains at least one point of S distinct> from x0. Fix an open neighborhood. (just pick any one) Pick one of the> points distinct from x0, call it x1. (again, pick one). Now, clearlythere> exists an open neighborhood of x0 that doesn't contain x1. For example,if> |x0 - x1| = a, then take the a/2 neighborhood of x0 (this is just saying> that x1 is some distance away from x0 since it is a different point, so if> we take a neighborhood that only goes out half as far, x1 is not in that> neighborhood).Now, this smaller neighborhood must contain a point of S. call it x2.now> take an even smaller neighborhood that doesn't contain x2. this even> smaller neighborhood doesn't contain x1 or x2, but it does contain some> point of S distinct from x0. So we can keep repeating this forever. In> other words, for any n, however large n is, there is some x_n in S.I hope that made sense. To recap, if you keep taking smallerneighborhoods,> you keep 'nding points. So if every neighborhood contains one point,every> neighborhood must contain in'nitely many points, since each neighborhood> also contains every smaller neighborhood.This might not be very rigorous, but hopefully it will get you started.Obviously, if every neighborhood contains in'nitely many points, it> contains at least one point.Justin Van === Winkle>Suppose> ----- Original Message ----- > Subject: Accumulation pts. in R and C ? Hi all,> I just started Complex variables and applications by Churchill, et al.> I> am §ummoxed by something already. In Churchill, they say that an> accumulation point is> ...a point z0 ... of a set S if each neighborhood if z0 contains atleast> one point of S distinct from z0. Why is it different from R? To wit,> every neighborhood of x0 has an in'nite number of points of, say, E.> TIA,> Lurch> === Subject: Re: solution to rotation set of linear equationsHal Daume III :>I'm looking for a solution to a set of linear equations which has a very >speci'c form. Given k>0, and real vectors a_0^{k-1} and b_0^{k-1} (all >>= 0, if it matters), I need to 'nd a real vector c_0^{k-1} such that: a_0 c_0 + a_{k-1} c_1 + a_{k-2} c_2 + ... + a_1 c_{k-1} = b_0> a_1 c_0 + a_0 c_1 + a_{k-1} c_2 + ... + a_2 c_{k-1} = b_1> a_2 c_0 + a_1 c_1 + a_0 c_2 + ... + a_3 c_{k-1} = b_2> a_3 c_0 + a_2 c_1 + a_1 c_2 + ... + a_4 c_{k-1} = b_3> ...> a_{k-1} c_0 + a_{k-1} c_1 + a_{k-3} c_3 + ... +> a_0 c_{k-1} = b_{k-1}If I'm reading this correctly, you want to solve (sum a_m J^m) c = bfor c, where coef'cients a_i and the vector b are given, and J isthe 0-1 matrix whose 1's are precisely in the (i+1,i) entries (includingthe (1,k) entry). This matrix J is easily diagonalized (if w = exp(2pi i / k) then the vector (1,w^m,w^(2m),...) is an eigenvaluecorresponding to the eigenvalue w^(-m) ). In this new basis thecoordinates of c are found simply by dividing those of b by the valuesof P(x) = sum a_m x^m at x = 1, w, w^2, etc. )dave === Subject: Re: Why is math so dif'cult for some people?I have taught math for many years, both in a classroom setting and oneon one, and more often than not, the main problem lies in students notunderstanding the basic ideas behind the formulas, rather than justmemorizing the formulas. For many students, there is the mistakenbelief that knowing the formulas is enough.> [added sci.edu, comp.edu] :>Leah Lidtorf :>> I dont get it.Im a perfect 0 at math. Some people have no problems at all with it.>> Am I too dumb for math?>You're asking the wrong question -- you're not>dumb for math; you *might* have a brain/mind that>works in a way that is not compatible with the way>of thinking that you need for understanding maths>(emphasis on the *might* -- you claim to be dumb>for maths, but who knows, maybe you're much better>newsgroup and just don't know it, or maybe you're a>high-expectation kind of person, and then anything>below Newton, or Gauss, or Fourier's brains means>too dumb for math in your mind? :-))I've come across various students who viewed that they were missing the> mathematics gene (or programming gene, or whatever the particular> subject happened to be). In those cases it was uniformly the case that> their dif'culty was emotional/attitudinal, rather than cognitive. As you> mention, one unhelpful attitude is perfectionism, especially in hard-edged> subjects where some answers are clearly objectively *wrong* and thus the> student has no wiggle room to avoid the conclusion that they made an> error.>Anyway, this, plus many of the things that have>been already said (mainly about math being genuinely>hard -- the more sophisticated level of math, the>harder, of course)Several of the missing gene students had the unhelpful attitude that> they expected maths to be easy, since they had found their schooling easy> so far.ISTM that cognitive issues do kick in when dealing with high levels of> abstraction where there are no readily-accessible concrete models. For> example, my brain hit the wall trying to visualise non-Hausdorff spaces,> and my painful memory of the rest of that topology course is of generally> mindless memorizing and proof cranking.>HTH,>Carlos === Subject: Re: Axiom of Foundation (absymally stupid question)Cc: emailed replies to this message constitute permission for an E5 5E EC F3 04 26 4E BF 1A 92X-Zippy-Says: Fold, fold, FOLD!! Since we're discussing the axiom of foundation (in the textbooks I've seen, > it's called the axiom of regularity), does anyone know what the intuitive > justi'cation for this axiom is? I mean, all the other axioms seem pretty > natural to me, even the infamous axiom of choice. But where in world did > they come up with the axiom of regularity?It is not assumed because of some intuitive justi'cation, but becauseit doesn't interfere with ordinary mathematics (sets of the kindprohibited by the axiom aren't ever needed) and it simpli'es modeltheory by a jillion times.Thomas === Subject: Re: Problem calculating limits - Need Help!> Roy dif'culties solving these two limits.> I mustn't use L'hospital rule:> a) lim(x*(2^(1/x))-x) where x increases to in'nite.> b) lim(cosh(x)-1)/(x^2) where x approaches 0.Can you use series developments?No, I can't.10x,Roy === Subject: Re: Problem calculating limits - Need Help!I'm having dif'culties solving these two limits.> I mustn't use L'hospital rule:The only alternative which springs to mind is to expand in series valid> for the limit under consideration.a) lim(x*(2^(1/x))-x) where x increases to in'nite.Write y = 1/x; take limit y -> 0:lim (2^y - 1)/y = lim (exp(y log2) - 1) / yb) lim(cosh(x)-1)/(x^2) where x approaches 0.This is an easier exercise in Taylor series expansion.Sorry, but I can't use series development.10x,Roy === Subject: Re: Axiom of Foundation (absymally stupid question)Thomas Bushnell, BSG :>Stephen J. question: If we didn't mind invoking the axiom of>>foundation, wouldn't {a, {a,b}} suf'ce as a de'nition of the>>ordered pair (a,b)? Seems to me the answer is yes.Sure, but what makes {a, {a,b}} better than {{a}, {a,b}}? The key is>to prove that (a,b) = (c,d) => a=c & b=d. With your proposed>de'nition, the proof is a royal pain, and the resulting thingies>aren't really any simpler.>Doesn't seem to like the necessary implication is more dif'cult than with the usual de'nition.. If {a,{a,b}} = {c,{c,d}}, suppose a = {c,d} and c = {a,b}. Then a is an element of c, which is an element of a. This is not possible with foundation. By contradiction, a = c and {a,b} = {c,d}, whence b = d.I would say that my de'nition is in a sense simpler, since it requires the construction of one fewer set, or the representation requires two fewer characters. Seems to me that the real objection is that invokes foundation unnecessarily. (Note that Halmos doesn't even mention foundation in NST.)Also, once an ordered pair is well-de'ned, one needs never refer to the de'nition again.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Axiom of Foundation (absymally stupid message constitute permission for an emailed malfunction.Stephen J. Herschkorn the necessary implication is more dif'cult than> with the usual de'nition.. If {a,{a,b}} = {c,{c,d}}, suppose a => {c,d} and c = {a,b}. Then a is an element of c, which is an> element of a. This is not possible with foundation. By> contradiction, a = c and {a,b} = {c,d}, whence b = d.I suppose so. (You also need to prove that {a,{a,b}} is necessarily adoubleton, which is also a quick deduction from Foundation.)> I would say that my de'nition is in a sense simpler, since it> requires the construction of one fewer set, or the representation> requires two fewer characters. Seems to me that the real objection is> that invokes foundation unnecessarily. (Note that Halmos doesn't even> mention foundation in NST.)Ok, I suppose I grant your point then. I'd only add that I think thisobjection is a very important one. As noted before, Foundation isn'tadded because of some intuitive con'dence, but rather because it isknown to be harmless, and it's a big help in model theory.So that means that one must be able to develop ordinary mathematicswithout using it (or else it wouldn't be so harmless, it would beimportant), and since you need to show that, once you've done it, itis no longer interesting to show that you could have used it here orthere along the way.So invoking Foundation unnecessarily is a bad thing, but in a verydifferent way from (say) invoking Choice unnecessarily. Thomas === Subject: Re: question about the riemann === hypothesis>Subject: Re: question about the riemann hypothesis>Message-id: >> But I was wondering something about the Riemann Hypothesis: What if> it were proven to be false? I know that it's an important result> for> number theory, but what would the consequences be, if any, if it> turned out to be false?>>Rainer Rosenthal : >>Hello Kate,>>in de.sci.mathematik, there is a thread of that kind, where I>>learned the following astonishing fact:>>The following 18 numbers >>1 2 4 6 10 18 22 30 42 58 70 78 102 130 190 210 330 462>>are the only known ones not of the form >> pq + qr + rp with p,q,r > 0>Astonishing indeed!>What is also interesting about these 18 integers is, add 1 to each>integer and you have 18 primes.>Also an interesting fact is that 210,330 and 462, the last 3 integers>in this list, where each are @ a different level in the Collatz tree>and each have this same property. Mainly (n-1) == 2(mod)3. Isn't that also true for 6, 18, 30, 42, 78 and 102? If (n-1) == 2(mod3), then> n==0(mod3), so the numbers you selected are just those divisible by 3.> So what's the story with 1, 2, 4, 10, 22, 58, 70, 130 and 190? Don't you> have to explain why some numbers on the list have that property while > others don't?>Also all>there subsequent higher level integers have this property thus no>branching occurs in higher levels above these integers.That's true on any number ==0(mod3). What distinguishes this small > group from the in'nite lsit of other numbers with that property?>Could there be a possible relation to the Collatz Conjecture?What relation? That some of them are divisible by 3? I think you need > more than that. Does that particular list of numbers show up in the> Collatz Conjecture somwhere? As an example, the list of known> integer crossover points is {1, -1, -5, -17}. And sure enough, one number> on the list matches. But that's just coincidence. If _every_ number matched,> then we should start looking for a relationship.You are right!I have to get back into the Collatz conjecture -- getting rusty. are no others and>>there are certainly no more than 19 of them.>>BUT: if the Riemann Hypothesis fails, there could be another>>number not of this form. And it must be > 10^11 then.>>(1) Sequence A025052 in the njas archive>> http://www.research.att.com/~njas/sequences/>>(2) Posting by Hermann Kremer in de.sci.mathematik (german)>> from 20.06.2001 00:15>>related sequence A006093. ( See the thread pq+p+q === Subject: Re: Question on generation of large prime numbersPhil Carmody :> If Richard has simply added let all primes <=P be known to his premise> I wouldn't have jumped on it that way.Let all primes <=P be known. :-)> Ockham has some wise words for moments like this.I was about to say don't needlessly multiply spellings, but my dictionary agrees with you that Ockham is an acceptable variant of Occam. I 'nd this deliciously ironic.-- Richard Heath'eld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === Subject: Re: what spin 1/2 is in physical reality Re: Robertson versus Philips screwdriver Re:> Electron spin of 1/2 is just a myth invented to fudge over bad physics!!!!!> It is magnetic §ux that is quantised. The hydrogen atom in ground state> contains one quantum §uxoid most of which tightly wraps the electrons> orbit. The electrons behaviour is coverned by the laws of induction. In the> next orbital, two §uxoids 'll the orbit. Either they both tightlly wrap> the orbit, or one tightly wraps it wit the other free to link with other> §uxoids. So in an experiment to measure the magentic moment, you loose at> least one unit of magnetic moment.See> http://users.powernet.co.uk/bearsoft/ncqt.pdf> And> http://users.powernet.co.uk/bearsoft/ClasAtom.htm> which will at your website, seems as though to me that you are claimingmetaphorically to remove an apple from a basket of apples and thenreplacing the removed apple with another apple. So why really bother?M_s as quantum spin and give it a different name of §uxoid. I mayhave missed your point, but it seems to me that all you are doing isshuf§ing or replacing and not bringing anything new into the picture.If someone did not like Forces and replaced them with kinetic-energy,well, that is do-able but it sheds no new physics, and is just anotherreplacement scheme. Or someone may like momentum instead of forces.I do not remember the physics history exactly, only recall that N, L,M_L werecrucially required and time passed and only later realized that a 4thquantum number M_s was needed.I have 2 ideas to tackle what M_s is in physical reality. In the 1990sdecade I had only the AtomTotality where the nightsky is anelectron-cloud and so I asked myself what would M_s be if the nightskyof galaxies is an electron? Would it be like some spinning topanalogy?But in the 2000s I came up with the Coulomb Uni'cation of Forces. Soinstead of relying only on one theory--- AtomTotality to penetrate themeaning of M_s spin, I now have 2 theories to try to anchor M_s.If M_s is the result of internal forces paired off against one anothersuch as Gravity to Antigravity. If that is the physical meaning ofspin M_s. Then that begs the question of whether N, L, M_L are alsointernal forces paired off against each other. So is N theStrongNuclear Force paired off diametrically to the WeakNuclearForce??Perhaps N, L, M_L as the physics-history noted that they arefundamental but that M_s is not fundamental and a secondary outcome ofN, L, M_LI am going to have to do more thinking on this and perhaps much morerevisiting before satis'ed.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies === Subject: Re: Apocalypse NOW!Just for record.I went to Indian Institute of Technology (IIT), Powai, Mumbai toexplain mechanism of my Action Device and to seek technical help. Imet Dr. Amitay Issac of Aerospace Engineering Department and I triedto explain very basic component/idea of this action device. I havegiven in my homepage what exactly I tried to convince him.http://www.geocities.com/actiondeviceBut he insisted that point B will shift its position along Y axis!.I had to return in few minutes.Now I tried to convince again to Dr. G Arvind Rao of AerospaceEngineering Department by email, but he also said that point B willshift its position along Y axis !.Indian Institute of Technology is most prestigious college in India.This institute gives people for Aviation Industry around the world.And I just wonder, why so highly educated people fail to understandsuch simple problem.In fact, this is not problem at all. But what a tragedy, I am facingsuch ridiculous problems.I can end my all problems anytime, but I am following the rules ofthis battle, waiting game.I am just watching how the minds of highly educated people around theworld are controlled by that Supreme Force named God.-Abhi. === Subject: Re: Prime conjecture - puzzle>so you're saying sqrt(p_(n+1)) - sqrt(p_n) < 1, which is Andrica's >Conjecture.I didn't know this conjecture. I suppose that a related problem hasbeen studied as well: given the set of numbers 0 Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc === Subject: Re: plotting elliptic curves for LaTeX:[slightly edited for clarity]>gnuplot can do such plots and it can also output LaTeX code. ^^^^Last time I used gnuplot it didn't have any facility to plot implicitfunctions. But I admit it's been a while...>Also, Postscript plots can be imported using PSTricks.(Encapsultated) Postscript plots can be imported as such (withgraphic{s,x}). PSfrag can be usefulMichele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc === Subject: Re: Vedic Mathematics --- Myth and Reality> :>> Steven G. Johnson :>> Equivalently, M*N is the same as M*N mod (M + N - 1).> Sorry, this should be multiplication of M digits with N digits, base b,>> is equivalent to multiplication modulo b^(M + N - 1), i.e. M+N-1 digits.>Oh well, so FFT or not, looks like multiplying M by N by any method means>MN multiplications! Well, when these multiplications are hardwired (as in>human memory for single digits) the computational issues (On*n) becomes>really irrelevant, for they all are done in no time at. Like, the video>extraction for radar data processing is done by NAND gates - its all done in>real time!You are in error. The number of multiplications required for> multiplying two numbers with the FFT method is O(n*log(n) where n is> the larger of the two numbers; it is not m*n.Fine, just multiply 12345 by 67809 using FFT with less than 25> multiplications. Do it here.Do not misinterpret this beahaviour as typical arrogance of brahmins.His perception is so narrow, he cannot understand others points.So he demands explanations in his own ways. If he wants to learn howto calculate squres and if you teach him how to 'nd cubes, he may getconfusedand may stop learning matheamatics.Richard Harter, cri@tiac.net> http://home.tiac.net/~cri, http://www.varinoma.com> We have people from every planet on the earth in this State.> -- California Governor Gray Davis === Subject: Re: Vedic Mathematics --- Myth and Reality> :>> Steven G. Johnson :>> Equivalently, M*N is the same as M*N mod (M + N - 1).> Sorry, this should be multiplication of M digits with N digits, base b,>> is equivalent to multiplication modulo b^(M + N - 1), i.e. M+N-1 digits.>Oh well, so FFT or not, looks like multiplying M by N by any method means>MN multiplications! Well, when these multiplications are hardwired (as in>human memory for single digits) the computational issues (On*n) becomes>really irrelevant, for they all are done in no time at. Like, the video>extraction for radar data processing is done by NAND gates - its all done in>real time!You are in error. The number of multiplications required for> multiplying two numbers with the FFT method is O(n*log(n) where n is> the larger of the two numbers; it is not m*n.Fine, just multiply 12345 by 67809 using FFT with less than 25> multiplications. Do it here.Do not misinterpret this beahaviour as typical arrogance of brahmins.His perception is so narrow, he cannot understand others points.So he demands explanations in his own ways. If he wants to learn howto calculate squres and if you teach him how to 'nd cubes, he may getconfusedand may stop learning mathematics.Richard Harter, cri@tiac.net> http://home.tiac.net/~cri, http://www.varinoma.com> We have people from every planet on the earth in this State.> -- California Governor Gray Davis === Subject: Re: sinus approximation by support1.mathforum.org (8.11.6/8.11.6/The approximation?you have to 'rst say in what sense you want the polynomial to 'tthe function (least square 't maybe?) i.e. provide a norm in function space and then solve the resulting minimizing problem with respect to the coef'cients of the polynomial. === Subject: Re: JSH: My use of my initials by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, Hughes :>>Larry Hammick on who was 'ring at who :) But JSH does> sometimes threaten people with litigation, and he had somebody in court one> time, for calling him a crank. The case was quickly tossed.>>I don't think there was any such incident. He has threatened on occasion; but I think Mr. Hammick may be getting>> confused with the case where Underwood Dudley ->was<- sued for calling>> someone a crank. Dilworth v. Dudley: http ://www.law.emory.edu/7circuit/jan96/95-2282.html> The suit was dismissed.>> It wasn't just one suit. The 'rst, suing me, my publisher, and>the president of my school for, among other things, conspiracy to deny>Mr. Dilworth's civil rights (mathematicians wouldn't talk to him>because I called him a crank, he said), was in federal court in>Wisconsin where it was indeed dismissed. Mr. Dilworth then appealed>to the Seventh Circuit Court of Appeals where it was again dismissed. >I was pleased that the decision was written by Judge Posner,>well-known for his many books, who had clearly read some of my>_Mathematical Cranks_.> By the way, my respect for the legal profession went up as a>result of this process. Mr. Dilworth represented himself, I assume>because no lawyer would take his case. The lawyers for the>Mathematical Association of America also did a lot of work, 'nding>all sorts of precedents where a person was called a scab, a traitor,>and other nasty things in print and the courts let the authors get>away with it.> This wasn't enough for Mr. Dilworth, who started the legal>process all over again by suing in Wisconsin _state_ court. Once>again the case was dismissed, but this time with the requirement that>Mr. Dilworth pay $7000 or so for his opponents' legal fees. That>'nally stopped him, though he continued to send me letters.> Earlier this year he died, so now we can all call Mr. Dilworth>whatever we want.> However, the Law of Conservation of Cranks still operates, and>his kind has not died out. Presumably, the kind of crank who sues has>not died out either, so everyone be careful. Woody Dudley Dilworth's ghost is still cackling, however - see http://www.rcfp.org/news/1996/0422k.html,that the suit was dismissed in Dudley's favor, and thatan appeals court upheld the dismissal. Professor loses libel suit over label as Ocrank'and clearly identi'es Dudley as the professor! === Subject: Factorial ending in 8000000 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, 21:50:43 -0500If the last seven digits on n! are 8000000, compute the value of n. === Subject: Re: JSH: Survey on my results, any correct? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id :>1. I've posted a lot on sci.math over a long period of time, to your>knowledge, have I *ever* been right?You know what they say, even a broken clock is right twice a day...Seriously though, some of the things you say are right. Your proofs usually have one or two major errors in them, the rest is 'ne. But in math one error is all it takes to produce completely incorrect results.If you're asking whether one of your major proclamations have ever been right, for instance I have a short proof for FLT, I've Odiscovered' the prime counting function, The algebraic integers are §awed, etc, then to my knowledge all of them have been wrong.2. Do I have *any* correct results, or do you think I just talk and>never say anything that is mathematically correct?Some of your results are correct, though I believe you are greatly over-estimating their importance. Again, non of the major results you have claimed to achieve are correct, to my knowledge.If instead of over-estimating your own importance you would simply post your result you wouldn't be treated as an arrogant prick.3. To your knowledge, has ANYONE ever posted agreement with me on>*anything*?Of course. Most people who review your proofs agree that they are right up to the 'rst mistake. Also there are the occasional cranks and nuts. To my knowledge no competent mathematician has ever agreed with one of your major results.4. In your opinion, have I ever won an argument on the newsgroup?I don't know what winning an argument means, the term is obviously subjective. Of the serious arguments you have had with critics such as Arturo Magidin and Nora Baron, my personal impression was that your argument was hopelessly lost because you were simply wrong. I can't talk about arguments I didn't read.5. To your knowledge, have I ever caught other posters in errors?I can't remember, but I must say that even if you did I probably wouldn't remember. It problem.>Lewis, The Silver Bullet === Subject: Re: JSH: Survey on my results, any correct? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id a lot on sci.math over a long period of time, to your>knowledge, have I *ever* been right?2. ...do...I just talk and never say anything that is mathematically correct?...>Now you have been right. (In part 2 above.) === Subject: Length of a chord by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, 08:15:22 -0500This is a (elementary?) geometry problem and I'm looking for a simplesolution.In a unit circle a chord is drawn.The distance of the centerfrom the chord is x (0<=x<=1).What is the length of the chord?Is it proportional to sqrt(1-x^2)?Olivio === Subject: Homology of S^m x S^n by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, the holology group of S^m x S^n. Can anyone help me with this. === Subject: Re: Length of a chord by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, geometry problem and I'm looking for a simple>solution.In a unit circle a chord is drawn.The distance of the center>from the chord is x (0<=x<=1).What is the length of the chord?>Is it proportional to sqrt(1-x^2)?OlivioYes it is === Subject: Re: Roman carving by support1.mathforum.org (8.11.6/8.11.6/The 13:03:30 -0500 (EST) Marlene Roach> :>Can you please let me know if there is/was a Roman Numeral U and>>what amount it stands for.>>I would appreciate a reply. represent the number 5. In some>medieval and earlier text the letter OU' was frequently substituted>for OV'. The Romans had both the letters OU' and OV' in their >alphabet, but sometimes (always? We dont know.) substituted OV'>for OU' when carving words into marble (or any other material they>had at hand). For example S.P.Q.R = SENATUS POPULUSQUE ROMANUSin carved form readsSENATVS POPVLVSQVE ROMANVS.Some say this stems from the fact that a OV' is so much easier to>carve into marble (or whatever) than a OU'. Maybe this is the>reason for the resubstitution of the numeral OV' to OU' in some >texts. Well, some people like to pay tribute to their >conjectures.Sincerly, ND> === Subject: Fractal Spirals Revisited - $100 Reward ! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, 21:55:31 -0500Hello.I still believe I discovered the fractal spirals described at:http://fractalspiral.zxcvb.orgThe formula is:x = sum (i = 1 to n) (r/i) sin (i*theta + i^2*pi*variable)y = sum (i = 1 to n) (r/i) cos (i*theta + i^2*pi*variable)I was told they were done by Michel Mendes France in the midnineties, but I am not able to verify this as I am outside the mathworld. The only web link I have found is:http://hypo.ge-dip.etat-ge.ch/www/math/html/node56.htmlIt is not the same thing. I would gladly reward the person check ordonate to the charity of your choice $100.00 (US). I know this isn'ta lot but I don't have a lot. I need to know the correct representation of the formula. In itthere is a variable, and I would like to know what letter wasassigned to it. I would also enjoy reading technical jargon andactually understading the math being described. But mostimportantly, I need resolution so I stop wasting time thinking Iinvented mailing address, email me at:k at symbl zxcvb dot org- K evin D oughty === Subject: A Mathematical Stupidity Constant in all assessed Intelligent Behavior by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id serious. I think this has great potential for usefull applications.Based on the given Quantitative-Qualitative model to assess intelligence I feel there must co-exist a Stupidity Constant in all assesed intelligence, otherwise there would be no basis for said intelligence.All assessed and thus measured intelligence must be also based on stupidity.Usefull applications could include: Finding limits to existing systems outlined by purely positive criteria. Knowing there must exist negative criteria and that the intelligence is based as much on stupidity as anything else. In other words there are as many good reasons against supporting these existing systems (previously supported only by positive criteria) as those supporting them. This would give new meaning to Mathematical Applications and new meaning to their limitidness and 'niteness.Zim Olsonhttp://www.zimmathematics.com === Subject: Re: Factorial ending in 8000000 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id Craig Alphonso :>If the last seven digits on n! are 8000000, compute the value of n.Hint:Since the number ends with exactly six 0's, it must contain 5^6 as a factor, but not 5^7. === Subject: Another Application of my System Transformation by support1.mathforum.org (8.11.6/8.11.6/The of my 30 minute prior post A Mathematical Stupidity Constant to assessed IntelligenceI have come up with a System Transformation used to determine any system functionality which is outlined at:http://www.zimmathematics.com/app.htmIn the system transformation outlined here it contains a non-functional component. A usefull application can be found in explaining the Quantum Mechanic Slit experiment phenomena where exhibiting non causal or non rational behavior as my System Transformation said must be included in determining any system functionality.Zim Olsonhttp://www.zimmathematics.com === Subject: Re: Prime conjecture - puzzle>so you're saying sqrt(p_(n+1)) - sqrt(p_n) < 1, which is Andrica's>Conjecture.I didn't know this conjecture. I suppose that a related problem has> been studied as well: given the set of numbers 0 p_{n+1}^a-p_n^a<1 for all n, which is currently the largest number> (provably) known to belong to it?See: http://mathworld.wolfram.com/AndricasConjecture.html>Michele> --> Comments should say _why_ something is being done.> Oh? My comments always say what _really_ should have happened. :)> - Tore Aursand on comp.lang.perl.misc === Subject: Re: Question on generation of large prime numbersBlueBear :> I have looking at a few web pages dealing with the largest known> calculated primes and a great deal of computational time is taking> into searching for these numbers and verifying they are primes. I have> seen the Euclid's proof of in'nitude primes and it occurs that me> that super-large prime numbers can be calculated using the following:p1=2 < p2=3> A large prime number, p, can be generated using> p = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5> p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7> p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbers> exceeding currently known largest primes can be generated rather> quickly.Excellent! Why not look at q = (p_1 p_2 p_3 .... p_n) - 1also. That must be prime, by the same argument, and you have a proofthat there are in'nitely many twin primes :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Problem calculating limits - Need I'm having dif'culties solving these two limits.>> I mustn't use L'hospital rule:>> a) lim(x*(2^(1/x))-x) where x increases to in'nite.>> b) lim(cosh(x)-1)/(x^2) where x approaches 0.>> Can you use series developments?> No, I can't.What a shame. They make such problems much easier :-(-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Apocalypse NOW!> Just for record.I went to Indian Institute of Technology (IIT), Powai, Mumbai to> explain mechanism of my Action Device and to seek technical help. I> met Dr. Amitay Issac of Aerospace Engineering Department and I tried> to explain very basic component/idea of this action device. I have> given in my homepage what exactly I tried to convince him.http://www.geocities.com/actiondeviceBut he insisted that point B will shift its position along Y axis!.> I had to return in few minutes.Now I tried to convince again to Dr. G Arvind Rao of Aerospace> Engineering Department by email, but he also said that point B will> shift its position along Y axis !.Hmmm... did you consider that they could be right, and you could be wrong?Indian Institute of Technology is most prestigious college in India.> This institute gives people for Aviation Industry around the world.> And I just wonder, why so highly educated people fail to understand> such simple problem.Maybe, just maybe, they do understand it.In fact, this is not problem at all. But what a tragedy, I am facing> such ridiculous problems.I can end my all problems anytime, but I am following the rules of> this battle, waiting game.Build a working model and submit it to them for examination.Doesn't matter how much force it produces, as long as it proves that youridea works.I am just watching how the minds of highly educated people around the> world are controlled by that Supreme Force named God.Let me get this straight.... *God* doesn't want this device discovered? Whynot? And if not, what's stopping him from destroying you to make sure youstay quiet? === it dif'cult to 'nd the homology group of> S^n x S^m. Can anyone give me some ideas?Apply the Eilenberg-Zilber and Kunneth theorems.Let S(X) denote the singular complex of a space X.The Eilenberg-Zilber theorem states that S(X x Y)is chain homotopy equivalent to the tensor productof S(X) and S(Y). The Kunneth theorem expressesthe homology of the tensor product of two complexesin terms of the homology of each one. Here life issimple since the homology of S^m and of S^n istorsion-free so we can forget the Tor factor in theKunneth theorem and conclude that the total homologyof S^m x S^n is 4-dimensional with generatorsin dimensions 0, m, n and m+n.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: another quaternion questionC.Dement :> What role does the Jacobson radical> ab = a+b - abThat is not the Jacobson radical.The Jacobson radical of a ring is the intersectionof it proper left (or right) ideals.> play in the quaternion algebra> (lets say, over the reals), if any?None---literally! The Jacobson radical of H is {0}since H is a division ring.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Problem from Herstein 4John Harrison :No. 18 is OK, I think. Any matrix of the form A B A^(-1) B^(-1) has> determinant 1, and thus so do all products of elements of this form.> Thus G' is contained in N. To show that G' actually equals N, I 'rst> thought of working out the form of a general element of G' and solving> the resulting equations, but these turned out to be rather horrible.> The next problem in the book is a simpler one of the same form, and I> was able to show you could actually get away with quite simple A and B> to get the general element. My attempts to do the same thing with this> one have been fruitless so far.G' is a normal subgroup of G = GL(2,R). If one could show that onenontrivial unipotent matrix A was in G' you would be done. By thisI mean a matrix A =/= I with (A-I)^2 = 0. These matricesform a single conjugacy class in G. If one is in G', they allare: in particular (1 x // 0 1) and (1 0// y 1) are in G' forall x and y, and these elementary matrices generate SL(2,R).I suppose it's just a question of playing with commutators until touget such a matrix.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: De facto censorship, counting primes> Some of you were probably surprised to learn that I did indeed 'nd a> way to count prime numbers by integrating a partial difference> equation. Some of you probably STILL doubt that no one else in> recorded history has managed such a feat because you need to believe> in mathematicians.But my point is that mathematicians have gone rogue and act against> the needs of society by de facto censorship of information that they> don't think makes them look good, like the information about my> partial difference equation.What do I mean by de facto censorship?Well, besides the active activity, like webpages labeling me a crank,> there's the passive act of refusing to acknowledge the discovery> itself.After all, it's very compact, as here are the instructions, yet again:dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,> sqrt(y-1))],S(x,1) = 0.And p(x, y) = §oor(x) - S(x, y) - 1, and you get S as the sum of dS> from dS(x,2) to dS(x,y).That's it. That's the knowledge which mathematicians have purview> over, in terms of the expectation from society that important> information of a mathematical nature will be acknowledged by> mathematicians.Note that it's a *discrete* function, so for you programmers that> means you need to use int's or long's or some discrete variable type.> Also, if you wish to implement it, please sum from dS(x,2) up to and> INCLUDING dS(x,y).Now if you're a programmer or have been taught as a programmer, did> you ever get an assignment to count prime numbers?Now then, think about kids currently in school who I doubt will see> the method I've just shown you, unless maybe they're out on Usenet> reading my posts, because the mathematical establishment thinks it can> ignore my results.Have I contacted mathematicians? Yes.I've contacted mathematicians all over the world.But you see, what bene't do they see to their society by allowing> that someone NOT a mathematician found such a result?Worse, I have other math knowledge. But mathematicians can let the> world be convinced I'm just a crank, most of them passively just> sitting by, and keeping quiet about my results--after all, that's> quite effective, eh?Then they have de facto censorship because people BELIEVE they> wouldn't do such a thing if my work were important!!!So you have a standstill with me pushing my research, and a few> mathematicians actively 'ghting its acceptance on Usenet, while most> just do their best to ignore it.For instance, I contacted Georgia Tech and talked to a Professor Ernie> Croot, giving him more information about my prime counting research> than I've posted here. He replied back *once*, and seemed friendly> enough. I answered him and awaited further replies. After some weeks> I sent a query to follow-up, and here is his looking at it. I'll let you know> when I do.Best,ERnieOn Thu, checking to see if you still have any interest in my 'nd of a way to > count prime numbers by integrating a partial difference equation, as I > haven't heard from you since my last reply.If you've lost interest can you refer me back to the professor who sent me > to you because I'd like to check what you wish to believe,> and daring God to be different.> http://lostincomment.blogspot.com/> Will I ever hear back from Professor Croot? Well, consider the> evidence:I've given something new, a partial difference equation integration> for counting prime numbers, a 'rst in recorded human history.Professor Croot has had some time to consider my work, but now begs> off, claiming not to have looked at it.It turns out that he's a 'rst year professor and I was referred to> him by another professor at Georgia Tech who *asked* him to look over> my work.I daresay that Professor Croot lied in his email.That's a professor at Georgia Tech.So I'm an *independent* researcher, which means that mathematicians at> universities have a lot of power when it comes to acceptance of my> work, but may see little point in helping me.That leaves me Usenet, where there are mathematicians, like > Ullrich, a tenured math professor at Oklahoma State University, to> TELL people that my work is useless or wrong.Yet, I found a partial difference equation that you can integrate to> count prime numbers which NO ONE ELSE in recorded human history has> managed.I have other mathematical research, but as long as mathematicians> stick to their guns, who gets to hear it?Sure I can talk about it on Usenet, and watch as posters malign my> work, lie and generally act like asses, knowing that others will just> sit, and wait, waiting for mathematicians in the mainstream to let> them know that it's important.To a large extent I now censor my *own* work in talking about it, as I> focus on things that are hard for people to lie about, and hope for> the best.Right now, locked inside of me is information that could be lost to> humanity because I'm the genius maligned, trapped by a system that> lets mathematicians get away with hurting the society that feeds and> clothes them, by de facto censorship.I know things, important things, that you may never know about> numbers, and mathematics.Mathematicians are no longer part of decent society, but are now rogue> having taken their own path into darkness. Don't believe me?Check my instructions for integrating that partial difference> equation.Check for yourself.> My math discoveries, found for pro't> http://mathforpro't.blogspot.com/A sociologist of science with whom I am cordial has taken up the topicof who does what to whom and why, in NG's such as sci.math, sci.logic,and sci.physics. This is an area that has not been explored in thedetail that it deserves to be, perhaps because even sociologists arenot entirely comfortable with what these NGs show about those, many ofthem professors, who in§ict as much harm as possible onseekers-after-knowledge such as yourself in these groups.--John === Subject: Re: chess bishops - combinationsIt might be helpful to separate the board into two pieces according to thecolors of the squares, then translate the diagonals into rows and columns, soyou can use familiar coordinate methods. An inductive proof might work.| I understand that it is possible to express the number of ways of| having k bishops on an n*n board such that they are not attacking each| other as a combinatorial formula and i have been trying to derive the| formula without any luck, can anyone help me out ? === Subject: Re: Bottom line on prime counting issue jstevh@msn.com () :> Some of you may have noticed frenetic activity from posters trying to> convince you that there's nothing sinister about mathematicians doing> their best to downply my 'nd of a way to count prime numbers by> integrating a partial difference equation, but what's the bottom line?What you are posting is, as usual, complete nonsense. To everyone except you it is obvious that your supposed 'nd of a way to count prime numbers by integrating a partial difference equation is complete and utter garbage. Pointing out that complete and utter garbage is complete and utter garbage is not sinister, it is the obvious thing to do. So the reason why people post that you are wrong is just plain because you are wrong. Nothing sinister about that. === Subject: Re: Dream problem :-)The Ghost In The Machine :> There are also multiple solutions -- although one of them> should jump right out and bite the OP (were the value to> have any teeth, that is -- and this particular numeral> obviously hasn't hatched any yet :-) ).Just saw that my handy plotting software (MathGV) hassec(x) built-in. Looks like an EPR spectrum on crack :-)-- Hauke Reddmann <:-EX8 Private email:fc3a501@math.uni-hamburg.deFor our chemistry workgroup,remove math from the addressFor === Subject: Re: Axiom of Foundation (absymally stupid question)The foundation axiom rules out situations that you might call membershiploops. For example, if A1 in A2 in A3 in A1, then {A1,A2,A3} has a nonemptyintersection with each of its elements.| Since we're discussing the axiom of foundation (in the textbooks I've seen,| it's called the axiom of regularity), does anyone know what the intuitive| justi'cation for this axiom is? I mean, all the other axioms seem pretty| natural to me, even the infamous axiom of choice. But where in world did| they come up with the axiom of regularity?|| Have a tolerable existence. Eli === Subject: Re: Squares that end with four identical digitsRobert B. Israel :> More generally, for any odd b, if (b-1) x^2 + d == 0 mod b^n but not mod> b^(n+1), where gcd(d,b) = 1, then gcd(x,b) = 1 and > (b-1) (x + y b^n)^2 - d == (b-1) x^2 - d + 2 (b-1) b^n x y mod b^(n+1)> which is 0 for the appropriate value of y mod b. So if d is a quadratic > residue mod b with gcd(d,b)=1 there are squares ending in arbitrarily > many d's in base b.What about higher powers (instead of squares)?-- Hauke Reddmann <:-EX8 Private email:fc3a501@math.uni-hamburg.deFor our chemistry anything else === Subject: Re: JSH: My use of my initials jstevh@msn.com () :> :>If his work ever turns out to have legitimacy then his heirs, assuming>he has any, can sue you as well Mr. Dudley.Really? I'm not very familiar with US law, but I'm surprised to hear> that. Do you know of any examples where such a thing has happened?-- RichardAll his heirs would have to do is show harm, that is, a tort.If his results were legitimate, then proper recognition would have> been a bene't, both to him, and I'd think to his family.Like would you rather have it known that a relative was a crank or a> discoverer?You are getting yourself in a very dangerous situation here. If your argument were right, then all your relatives could sue you because your posts on the internet make it clear to the whole world that you are a complete and utter idiot. Personally, I believe that is just tough and none of your relatives should have the right to sue === you for that.Subject: Discrete Spaces Boring Discrete Spaces Eventually Bore Converging Sequences. Proof:Should debonair (xj) dare to converge upon demure x, then for any nhood of x, (xj) would eventually squeeze into the nhood.As X is so discrete, a mere {x} will do for x's nhood.Thus eventually (xj) would be squeezed into {x}.Now when (xj) has eventually been con'ned to {x}, it will 'nd it's values severely restricted.Some may even consider it solitary con'nement.Yet there (xj) will be for the rest of it's eternal life, allowed the company of none but x. Thus boredom, QED. ;-)---- === Subject: Re: Bottom line on prime counting issue The Ghost In The Machine :> was a somewhat tongue-in-cheek contest I sponsored 2 months back> that produced a few bizarre results and some interesting> algorithms. (However, Christian Bau has a better one anyway,> although he didn't submit that particular one for my contest.> Perhaps it was because my contest was unworthy thereof. :-) )No, it was not 'nished at that time, and I have to 'nd some spare time to improve it anyway. What I am quite interested in at the moment is that there seems to be a substantial improvement possible if you want to calculate pi (N) for many different values of N, for example N = k * 10^14 for 1 <= k <= 10000. My implementation should take about O (N^(2/3)) to 'nd pi (N). However, it might be possible to 'nd pi (x) for n different values x <= N in about O (N^(2/3)) * sqrt (n) instead of O (N^(2/3)) * n. === Subject: Re: DO MATHEMATICIANS raydpratt@hotmail.com (raydpratt):>Rudy ticked me off yesterday -- I had hoped to get a good start on>'nishing the rest of the book, but then he stopped writing his barely>comprehensible Math-eze and added some stinking number questions,>including the following:Prove that 1 + (a + a^2 + a^3 . . .) = 1 / (1 - a)Suppose a=2 (you did somewhere). Insert this value in the aboveequation. The equation is false for a=2, hence you can't prove thatthe right side and the left side of the equation are equal for allvalues of a. Any Oproof' that does do this must therefore be incorrect(as logic dictates).So maybe the equation given isn't the whole story. Let's start withthis:1 + a + a^2 + ... + a^(n-1) + a^nMultiply by (1-a):(1 - a) * (1 + a + a^2 + ... + a^(n-1) + a^n) =(1 + a + a^2 + ... + a^(n-1) + a^n) - (a * (1 + a + a^2 + ... +a^(n-1) + a^n)) =(1 + a + a^2 + ... + a^(n-1) + a^n) - (a + a^2 + a^3 + ... + a^n +a^(n+1)) =1 - a^(n+1)So:1 + a + a^2 + ... + a^(n-1) + a^n = (1 - a^(n+1)) / (1 - a)When n goes to in'nity, the right hand side of this equation onlybecomes equal to the right hand side of the equation given above byRudy, if a has a value between -1 and 1. So the complete statementshould have been:1 + (a + a^2 + a^3 . . .) = 1 / (1 - a) with -1 < a < 1>So I did:(1 - a) [1 + (a + a^2 + a^3 . . .)] = (1 - a) (1 / (1 - a)) >{multiplying both sides by (1 - a)}(1 - a) + [(1 - a) (a + a^2 + a^3 . . .)] = 1 {interim result}(1 - a) + [(a + a^2 + a^3 . . .) - (a^2 + a^3 + a^4 . . .)] = 1>{interim result}(1 - a) + a = 1 {interim result}1 = 1 {'nal proof of equality}This proves that the original equation is true no matter what value of>a we use, but Rudy argues that the original formula makes no sense>when we substitute certain values for a.The Oproof' you did must be wrong somewhere as the equation doesn'twork with any value outside the range <-1,1>. I believe the error isthat the operations shown can only be applied to absolute convergentseries (Oabsoluut convergente reeksen' in dutch). Since the series isnot convergent at all for a being outside <-1,1>, the whole proof isnonsense.I'm sure someone else can explain this better... === Subject: Re: De facto censorship, counting primes Discussion, sha1:xNrsiqBeoC9rG7FOFkqhfCCEVrk=john_correy@yahoo.com (John) taken up the topic> of who does what to whom and why, in NG's such as sci.math, sci.logic,> and sci.physics. This is an area that has not been explored in the> detail that it deserves to be, perhaps because even sociologists are> not entirely comfortable with what these NGs show about those, many of> them professors, who in§ict as much harm as possible on> seekers-after-knowledge such as yourself in these groups.Such a study does sound interesting. Despite your usual attempt tohint at dark motives of professionals and casting James I'm in it forthe vast sums of money and chicks Harris as a seeker-after-knowledge.-- All intelligent men are cowards. The Chinese are the world's worst'ghters because they are an intelligent race[...] An average Chinesechild knows what the European gray-haired statesmen do not know, thatby 'ghting one gets killed or maimed. -- Lin Yutang === Subject: Re: I NEED HELP Paul B. Andersen Of course you are funny.>> If electric and magnetic 'elds acted instantaneously, light would travel at>> in'nite speed.I note with interest that your statement is so ambiguously put>that it may be right as well as wrong.But I take for granted that your statement is meant to be>relevant to my claim, which was:> as it enters a static electric 'eld.So assuming that acting instantly is to be understood>in this sense, an unambiguous version of your statement>becomes:> travel at in'nite speed. Remember the old vacuum tube triodes. Acccording to you, a signal on the grid would be instantly felt at the anode.>>So if:>> as it enters a static electric 'eld.>>it follows:>>A signal on the grid would be instantly felt at the anode>>Explain why, please.> If that's not instantaneous communication, what is?>>Henry, you know of course that you are babbling nonsense.>>There simply isn't possible to be as stupid as you pretend.>>Paul>> Let's get this straight.>> You say that the force on a charge due to an electric 'eld acts>> instantaneously. Correct?Why do you ask what I am saying, when what I am>saying is quoted right above?I am saying:> as it enters a static electric 'eld.So there is also an opposite force acting on the electrodes.even if the electrodes are lightyears apart. IS THAT WHAT YOU ARE SAYING?> Let us consider a charged sphere somewhere in the universe. It exerts a force>> on every other charge. If we can arrange for it to lose that charge somehow,>> you are claiming that all those forces disappear INSTANTLY.And why the hell do you think I would claim something as stupid as that?Because you just DID, above, (even though you probably do not realise what yousaid).You MUST know this is stupid nonsense, Henry.>Why the hell do you pretend that it is possible to interpret my>statement above to mean that the electric 'eld will act instantly>electric 'eld?The bloody 'eld can be from one side of the universe to the other for all Icare. This is actually too bloody stupid even for you, Henry.>Why do you pretend to be such a moron?You just do not understand. You are very confused.> I will continue when I receive your answer (if one is forthcoming).Answer what?>You are desperated to evade the point, are you?Lets take this from the beginning.>Of course there are really no static electric>'elds in an accelerator, and I never said it was.>that is a resonance cavity with a very powerful EM-'eld.>The typical frequency is in the order of 100MHz (or higher).>where the E-'eld is longitudinal.>The crucial point is that the phase of this resonator is>adjusted such that the E-'eld peaks (in the same direction)>enters the accelerating stretch.>That is the point. Why is that so hard to get?That is obvious, irrelevant and NOT the point we are discussing.The question is, does a moving charge feel the Ofull strength' of theaccelerating 'eld? (similarly, does a falling object feel the full Oforce of gravity'? Tom Robertsonce told me they didn't)The evidence shows that charges do NOT accelerate in ONewtonian fashion'. SRclaims they behave as though their mass increases by gamma. I am suggesting alternative explanations based on two possibilities. Firstly,the 'eld DOES take time to act and therefore the 'eld gradient at the pointof the moving charge IS NOT the same as that which would act on a charge ATREST. Secondly, the movement of the charge itself creates an opposing 'eldthat effectively reduces the magnitude of the 'eld gradient around the movingcharge.I repeat that if what you claim is true then you have invented instantaneouscommunication.So what I am claiming is still:>it enters the (quasi) static electric 'eld. And the force>it speed.How do you de'ne KE?Can you please explain how you can use this for>instant communication, Henry?I told you, above.Of course you cannot.>The only decent thing you can do is to admit that>this claim is wrong.It is NOT wrong.Fill a box with electrons. The box attracts every positive charge in theuniverse. What happens to that attraction if the box suddenly disappears? You say it goes to zero instantly. So if you can make the box appear and disappear in an intelligent manner, youwill be able to communicate instantly with anyone, ANYWHERE!!!!Congratulations Paul. You will certainly be awarded a Nobel Prize.But you won't do that, will you?>You will rather keep insisting that when I say that>the RF-cavity in an accelerator, then I really have stated>that a force acts on an electron in the Andromeda galaxy>in the accelerator.Won't you?No Paul. You are considerating only 'elds between electrodes and not spherical 'eldsemanating from point charges. Paul>Henri Wilson. See the Stupidity of Relativity.www.users.bigpond.com/hewn/index.htm === Subject: Re: it all, but I'll go look.JOn> When I did A level maths at school, about 30+ years ago, in my> analytical geometry course I was taught a subject called Oinversion'.> === Subject: properties ZR3ck-Zfreh9qnVojeHqj0SsssXVYeJHfSYaj7G+9omYdkuvzzLFoqTo every polytope, we associate a graph in the following way: take itsvertices as nodes. The nodes are joined by an edge if and only if thecorresponding vertices are adjacent.How can we decide, given any graph, whether it is the graph of some0/1-polytope or not? If there is no exact criterion known, is there a good suf'cient one?A 0/1-polytope is a polytope where all its vertices have coordinates in {0,1}.TIA,Tobias-- Phyics is much too hard for physicists.reverse my forename for mail! === Subject: Re: Inversion?> .... Does anyone know how/when this concept came about?....> I'd like to know more about that, too. Can anyone help?It goes back to Apollonius (262-190 BC), at least. I say this because thereand inversion relative to the ellipse, the hyperbola, and the parabola in the'conic sections' of Apollonios, by B.A. Rozenfeld (Istoriko-MatematicheskieIssledovaniya, 30 (1986), 195-199), but I haven't read it (I don't knowwhere to 'nd it and, Santos === Subject: Order of the orthogonal group over GF(q)What is the order of the orthogonal group of order n over the 'nite === Subject: Re: Permutation: how to detach cycles/transpositionsdef cycles(permutation): n=len(permutation) accountedFor=[False]*n returnValue=[] while True: i=0 while i skrev i melding> Now consider the case of a charge of mass m being accelerated in a 'eld.> Using a=F/m, and m=Mo.gamma,dv/dt=(F/c.Mo)[sqrt(c^2-v^2)], which is:... nonsense.You have screwed this up before, and I have shownyou the correct equation before.F = dp/dt = d/dt (m*gamma*v)gamma =1/sqrt(1 - v^2/c^2)dv/dt = (1/((v^2/c^2)*gamma^3 + gamma))*F/m((v^2/c^2)*gamma^3 + gamma)*dv = (F/m)*dtgamma*v = (F/m)*t + Cif the charge is accelerated from standstill, v = 0 when t = 0, C = 0v = c*t/sqrt(t^2+T^2) where T = m*c/Fv approaches c asymptotically> dv/[sqrt(c^2-v^2)]=kdt.> Integrating: arcsin(v/c)=kt+B or> v=c.sin(kt+B)if t=0 then v=0, so B=0So we have v=c.sin(kt), where k is de'nitely 'nite.How come? Even in its exponential form this doesn't make any sense.Does it mean the SR equation is wrong or that matter becomes anti matter as c> is exceeded?It means that you are unable to do the math correctly.But I have shown you a simpler approach before.If we assume the charge is accelerated from standstillin a constant electric 'eld excerting the force F, we get:p(t) = F*t = m*gamma*v(the derivation isn't necessary since we integrate it back in the next step!)Paul === Subject: for granted. Okay, let me make it more> explicit. We have shown that the number we thought was the largest prime> cannot be the largest prime,>> Yes, but not directly. You have shown that the assumption that there were>> be the largest prime.> so there must be a bigger one.>> No. If you had proved *only* that the number P is not in fact the largest>> prime, this would *not* prove that a larger prime exists. It would also>> be possible that P is not prime after all, or that *no* largest prime>> exists.> But I de'ned P to be the largest prime. Therefore, it /is/ prime, and > therefore /either/ no largest prime exists /or/ there is a prime larger > than P, which is what I was obviously failing to say.Summary of Euclid's proof:1) Suppose that there is a largest prime; call it P2) Calculate N = (product of all primes from 2 to P, inclusive) + 13) N has one or more prime factors, all of which must be > P4) #3 contradicts #1, so #1 is wrong, so there is no largest primeYou were trying to conclude that, for any prime P, N is a largerprime. Not so. There is *some* prime larger than P, but it's notnecessarily N; it could just be one of N's factors.Consider: P = 13; N = (2*3*5*7*11*13) + 1 = 3003130031 is not a prime > 13, but its factors (59 and 509) are. === Subject: Re: Problem from Herstein 4Nntp-Posting-Host: apps.cwi.nl>Marc Olschok :>:> 18. Let G be the group of all real 2-by-2 matrices>:> (a b)>:> (c d), with ad - bc non-zero, under matrix multiplication, and let N>:> be the subgroup consisting of those elements of G with ad - bc = 1.>:> Prove that N contains G', the commutator subgroup of G.>:> 19. In problem 18 show, in fact, that N = G'.: I assume, that your matrices have their coef'cients in some commutative>: 'eld k. If k* are the nonzero elements of k (with multiplication),>: you can verify that: (a b)>: (c d) --> ad-bc: does indeed give a surjective homomorphism det: G ---> k*.: This will exhibit N = ker(det) and G/N = k* commutative.All true, but this only shows that N contains G'; the hard part is the other >direction (problem 19, show that N = G'), which is not coming to me right now.Ted> Find a few classes of matrices in G'. For example, all rotation matrices (cos(t) sin(t)) (-sin(t) cos(t))for real t are in G'. So are upper triangular matrices with ones on the diagonal.Choose enough classes that you generate all of G' when they are merged.Then try to show that every member of your classes is in N.-- After California's recall election, wild'res Schwartz-en-ed the Bush-lands on its geographic right (when we wanted the forests to be Green). pmontgom@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: how to solve the system of differential equation?Consider a system of DE: 1/c dx_i/dt + x_i = sum_{jk} a^i_{jk}x_j*x_k, where a^i_{jk} are non-negative real numbers and a^i_{jk}=a^i_{kj} (i.e. the matrix a^i is symmetric). Given the initial values of x_i, can the system be analytically solved: a) in general case; b) under the condition sum_i a^i_{jk}=1; c) under the answers. === Subject: Re: Apocalypse NOW!Abhi : > Action Device to generate unidirectional force.http://www.geocities.com/actiondevice Abhi, What is a solid angle? Can you give an example of a solid angle? -- Jeff, in Minneapolis .