mm-6 === >Decker noted an error in my original post, so here's one with>corrections.I've been using an example put forward by Rick Decker, a professor at>Hamilton College, for a few days, but now I've seen need to add to it.Consider,7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7(xy - y^2)(5) + 7^2 y^2so(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) >where the a's are roots of >a^2 - (x - y)a + 7(x^2 + xy).Now letting x=0, gives a(a + y^2) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, I let>Of course you mean a(a + y) = 0 here.Yeah, I missed changing that one.>a_2(x,y) = b_2(x,y) - y, so(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2) Which shows that dividing both sides by 7 must result in(5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2>Which is just ?e, as long as you realize that a_1(x, y)/7>will in general not be an algebraic integer. Try x = y = 1 > Yeah, that's my point!!!Mathematically, y as an independent variable isn't affected by the> value of x, which shoots down claims of people like Dik Winter that> the coef?ient IS in fact dependent on the value of x.>or, if you don't want x and y to be equal, x = 2, y = 1.>In the latter case you'll have the a's equal to (1 + sqrt(-167))/2, ' (1 - sqrt(-167))/2and it's not hard to show that neither of these are>divisible in the alg. ints by 7 (or, in fact, by sqrt(7)). > I'm getting a sense of a disconnect here Decker.Do you still believe that somehow there's a non-unit algebraic integer> factor of 7 shared by *both* of those roots? > That's correct. It's not a matter of belief, ' though, in the usual> sense of that word. I know it to be true. As I said in another> thread, the two roots are associates, so are indistinguishable> from the point of divisibility. > If so, how do you explain the coef?ients of the independent y's in(5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2? > Simple, you've just obfuscated the more obvious factorization (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 25x^2 + 30xy + 2y^2As I said earlier, your insistence on the so-called constant> terms is just confusing the issue.That factorization is misleading as it appears that both termsindependent of x have a factor of 7. Having x mixed in and thecomplexity of the a_1(x,y) and a_2(x,y) functions prevents just*looking* at the expression to get answers.It *has* to be adjusted as notice that it appears your termsindependent of x multiply to give 7y(7y) = 49y^2, but on the rightside you have 7(2y^2) as the term independent of x, so *something* hasto be adjusted, right?Lily it's possible to clear things up by setting x=0 to clear itout.Doing so shows that one of the a's is NOT zero when x=0, but in factequals -y when x=0, so there's something hidden in one of them, whichis revealed by setting x=0.So Decker your statement is the opposite of the truth.Letting a_1(0,y) = 0, means that a_2(0,y) = -y, so it makes sense tointroduce b_2(x,y), wherea_2(x,y) = b_2(x,y) - y, which gives(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2).Now I can understand that result might trouble you Decker, and itmight make you feel VERY unsettled, but that factorization is aclearer view of the picture, as both a_1(0,y) = 0, and b_2(0,y) = 0,and notice now the terms independent of x match up on both sides:7y(2y) from the left side matching with 7(2y^2) from the right. >If we let z = (1 + sqrt(-167))/(2sqrt(7))for instance, then we see that z satis?s a non-monic,>primitive, irreducible polynomial with integer coef?ients: 7z^4 + 166z^2 + 1008 = 0 > As I corrected elsewhere, that should be 7z^4 + 83z^2 + 252 = 0>so z isn't an algebraic integer and thus (1 + ' sqrt(-167))/2>isn't divisible by sqrt(7) (thus not divisible by 7 either). > Actually operator ambiguity keeps you from determining *which* of the> roots you have there, as sqrt() has two valid solutions, and it> doesn't matter if mathematicians by convention choose the positive> solution, as what if they change their minds tomorrow? > It doesn't matter which root you use.> The result will be the same.That's obstinance to the factorization(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)and the independence between x and y.Ok Decker, let's say that somehow you're right though, then you ' haveto lose independence between variables, as then somehow, x and y musthave some kind of hidden dependency, but they don't have anydependency as they're independent variables, so your position requiresa collapse of logic in mathematics.That is, Decker, your position requires the failure of mathematicalconsistency.It requires the loss of logic in mathematics.James Harris === =and the independence between x and y.Ok Decker, let's say that somehow you're right ' though, then you have> to lose independence between variables, as then somehow, x and y must> have some kind of hidden dependency, but they don't have any> dependency as ' they're independent variables, so your position requires> a collapse of logic in mathematics.> AS your method was incorrect for a speci? value of y, why on Earthshould it have more chance of working for an arbitrary y?> That is, Decker, your position requires the failure of mathematical> consistency.It requires the loss of logic in mathematics.> James Harris === I'd suggest you stop looking at the so-called constant> terms. They're doing you far more harm than good.> Why would you so suggest?The problem is that you take something like p P(x) = g1(x)g2(x)and conclude that you can determine how p dividesinto g1 and g2 by looking at the constant terms ofg1 and g2 (which you de?e to be g1(0) and g2(0)).For polynomial g1 and g2 this works. However, if g1 and g2are not polynomials, then the way that p divides g1(0)g2(0)does not tell us anything about the way p divides g1(x)g2(x)for x not equal to 0 (see below for a trivial example).We are dealing with non-polynomial factors. Thus looking atthe constant term cannot help, it can only hurt. - William HughesLet P(x) = 1; g1(x) = 2 for rational x and 1 for irrational x;g2(x) = 1 for rational x and 2 for irrational x. Then 2 P(x) = g1(x)g2(x) is true for all x. 2 divides g1(0) and does not divide g2(0).But for any irrational number y, 2 does not divide g1(y) and 2 doesdivide g2(y). === I'd suggest you stop looking at the so-called constant> terms. They're doing you far more harm than good.> Why would you so suggest?> The problem is that you take something like p P(x) = g1(x)g2(x)and conclude that you can determine how p divides> into g1 and g2 by looking at the constant terms of> g1 and g2 (which you de?e to be g1(0) and g2(0)).> For polynomial g1 and g2 this works. However, if g1 and g2Well, I found that posters like yourself could get away withnonsensical statements like that so I added a variable to take awaythe phrase constant terms.I also added that variable to a *quadratic* example given by RickDecker, a professor at Hamilton College.Here's the result:(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of a^2 - (x - y)a + 7(x^2 + xy).Notice that x and y are *independent* variables, as I haven't relatedthem to each other in any way.Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you havea_2(x,y) = b_2(x,y) - y, so(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). Now for readers wondering about it that expression isn't evenquestioned by posters, who are pushing the result that *both*(5a_1(x,y) + 7y) and (5b_2(x,y) + 2y) share non-unit factors in commonwith 7, where those factors vary as a function of x.However, their position requires that the terms *independent* of x,which on the left are 7y and 2y, and on the right 7(2y^2) do in facthave some kind of dependency, to force different factors in commonwith 7 as x varies.Notice that these people don't have the option of now claiming an xand y dependency as there's no mechanism for w_1(x,y) w_2(x,y) = 7,where the w's are factors of those terms that vary with x, and y, asnecessarily then the degree of the y exponent in 7y and 2y wouldchange once you divided by 7, so I think trying to claim a w_1(x,y)and w_2(x,y) is too wacky for even Dik Winter to try!(Maybe Nora Baron might though!!!)Remember that my position merely requires that you accept what you'vealready been taught in mathematics when it comes to *independent*variables.People trying to ?d some dependency between the terms independent ofx, like 7y and 2y, and x, are simply not talking logically.They're attacking the algebra that you should know and love.James Harris === I'd suggest you stop looking at the so-called constant> terms. They're doing you far more harm than good.> Why would you so suggest?> The problem is that you take something like p P(x) = g1(x)g2(x)and conclude that you can determine how p divides> into g1 and g2 by looking at the constant terms of> g1 and g2 (which you de?e to be g1(0) and g2(0)).> For polynomial g1 and g2 this works. However, if g1 and g2Well, I found that posters like yourself could get away with> nonsensical statements like that so I added a variable to take away> the phrase constant terms.I also added that variable to a *quadratic* example given by Rick> Decker, a professor at Hamilton College.Here's the result:(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of a^2 - (x - y)a + 7(x^2 + xy).Notice that x and y are *independent* variables, as I haven't related> them to each other in any way.Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you havea_2(x,y) = b_2(x,y) - y, so(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). Now for readers wondering about it that expression isn't even> questioned by posters, who are pushing the result that *both*> (5a_1(x,y) + 7y) and (5b_2(x,y) + 2y) share non-unit factors in common> with 7, where those factors vary as a function of x.However, their position requires that the terms *independent* of x,> which on the left are 7y and 2y, and on the right 7(2y^2) do in fact> have some kind of dependency, to force different factors in common> with 7 as x varies.Notice that these people don't have the option of now claiming an x> and y dependency as there's no mechanism for w_1(x,y) w_2(x,y) = 7,> where the w's are factors of those terms that vary with x, and y, as> necessarily then the degree of the y exponent in 7y and 2y would> change once you divided by 7, so I think trying to claim a w_1(x,y)> and w_2(x,y) is too wacky for even Dik Winter to try!(Maybe Nora Baron might though!!!)> De?itely! Here is the correct approach: First, note that 7(25x^2 + 30xy + 2y^2) = 7((x^2 + xy)*5^2 + (x - y)*y*5 + 7*y^2. Assume a factorization of the form (a_1(x)*5 + 7y)*(a_2(x)+ 5 + 7y) = 7((x^2 + xy)*5^2 + (x - y)*y*5 + 7*y^2). Thinking of this as a factorization in 5, -7*y/a_1 and -7*y/a_2 are roots of 7*((x^2 + xy)*s^2 + (x - y)y*s + 7*y^2 = 0, where s is the variable. This implies that a_1 and a_2 satisfy a^2 - (x - y)*a + 7*(x^2 + x*y) = 0. This of course means that a_1 * a_2 = 7*(x^2 + x*y). Now assume (as is more often the case than not) that x^2 + x*y is coprime to 7. De?e w1 = GCD(a_1, 7) and w2 = GCD(a_2, 7). This implies that a_1/w1, 7/w1, a_2/s2, and 7/w2 are all algebraic integers [by the de?ition of GCD in the ring of algebraic integers]. This in turn implies that ((a_1/w1)*5 + (7/w1)*y)*((a_2/w2)*5 + (7/w2)*y) is a factorization of (x^2 + xy)*5^2 + (x - y)*y*5 + 7*y^2 in which all the coef?ients are algebraic integers. Now: how might you know that neither w1 nor w2 are units? If w1 or w2 were a unit, then one may assume that the other one is equal to 7. If w1 = 7, one readily ?ds that a_1 cannot be an algebraic integer. One therefore concludes that w1 <> 7 and similarly w2 <> 7. Thus also neither w1 nor w2 can be units. Note that all of a1, a2, w1, and w2, are functions of both x and y. === Complicating this thing by adding in the variable y really makes no difference. The key fact is, YOUR factorization, in which you choose w1 = 7 and w2 = 1, does not work. [A singular exception occurs when x = 0.] The de?ition of of w1 and w2 in terms of GCDs as given above DOES work, whenever x^2 + x*y is coprime to 7, to give a factorization of your original polynomial in which all the coef?ients are algebraic integers. This was what you were aiming for in the ?st place. Contrary to what you have said repeatedly, it is not impossible. Dik shows that with a little more complication, you can even dispense with the requirement that x^2 + x*y is coprime to 7. Nora B. > Remember that my position merely requires that you accept what you've> already been taught in mathematics when it comes to *independent*> variables.People trying to ?d some dependency between the terms independent of> x, like 7y and 2y, and x, are simply not talking logically.They're attacking the algebra that you should know and love.> James Harris === =... > (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). > Now for readers wondering about it that expression isn't even > questioned by posters, who are pushing the result that *both* > (5a_1(x,y) + 7y) and (5b_2(x,y) + 2y) share non-unit factors in common > with 7, where those factors vary as a function of x. > However, their position requires that the terms *independent* of x, > which on the left are 7y and 2y, and on the right 7(2y^2) do in fact > have some kind of dependency, to force different factors in common > with 7 as x varies.You are *again* misstating the position. The divisor of(5 a_1(x,y) + 7y) that is in common with 7 is only a divisor ofthe complete expression. The 7y is independent of x. > Notice that these people don't have the option of now claiming an x > and y dependency as there's no mechanism for w_1(x,y) w_2(x,y) = 7, > where the w's are factors of those terms that vary with x, and y, as > necessarily then the degree of the y exponent in 7y and 2y would > change once you divided by 7, so I think trying to claim a w_1(x,y) > and w_2(x,y) is too wacky for even Dik Winter to try!Oh, come on, this is simply too simple... v1(x, y) = gcd(5 a_1(x, y) + 7y, 7) v2(x, y) = gcd(5 b_2(x, y) + 2y, 7) k(x, y) = v1(x, y) * v2(x, y) ; is 7 or a multiple of 7 (%) g(x, y) = k(x, y) / 7 ; the excess h1(x, y) = gcd(v1(x, y), g(x, y)) ; split in h1 * h2 where h1 is a h2(x, y) = g(x, y) / h1(x, y) ; factor of v1 and h2 of v2 w1(x, y) = v1(x, y) / h1(x, y) w2(x, y) = v2(x, y) / h2(x, y)You may verify that for all algebraic integers x and y, the abovefunctions are algebraic integer functions, that w1(x, y) * w2(x, y) = 7and that w1(x, y) is a factor of (5 a_1(x, y) + 7y) and w2(x, y) a factorof (5 b_2(x, y) + 2y) in the algebraic integers.(% Note that for any three algebraic integers p, q and r, gcd(pq, r)divides gcd(p, r)*gcd(q, r).)Now I wonder what you mean with the change of exponent... > People trying to ?d some dependency between the terms independent of > x, like 7y and 2y, and x, are simply not talking logically.There is no such dependency. Where do you see it?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === ...> (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). > > Now for readers wondering about it that expression isn't even> questioned by posters, who are pushing the result that *both*> (5a_1(x,y) + 7y) and (5b_2(x,y) + 2y) share non-unit factors in common> with 7, where those factors vary as a function of x.> However, their position requires that the terms *independent* of x,> which on the left are 7y and 2y, and on the right 7(2y^2) do in fact> have some kind of dependency, to force different factors in common> with 7 as x varies.You are *again* misstating the position. The divisor of> (5 a_1(x,y) + 7y) that is in common with 7 is only a divisor of> the complete expression. The 7y is independent of x.Any factor in common with 7 has to show up in the coef?ient of theterm independent of x, and it does as 7 is that factor, in anappropriately inclusive ring, while the ring of algebraic integers byvirtue of its de?ition excludes *any* algebraic integer factor incommon with 7 in this case.So I'm not misstating anything. If as you seem intent on claimingthere is some other factor than 7, dependent on x, or x and y, then ithas to show up as a factor of 7y. > Notice that these people don't have the option of now claiming an x> > and y dependency as there's no mechanism for w_1(x,y) w_2(x,y) = 7,> where the w's are factors of those terms that vary with x, and y, as> necessarily then the degree of the y exponent in 7y and 2y would> change once you divided by 7, so I think trying to claim a w_1(x,y)> and w_2(x,y) is too wacky for even Dik Winter to try!Oh, come on, this is simply too simple...> v1(x, y) = gcd(5 a_1(x, y) + 7y, 7)> v2(x, y) = gcd(5 b_2(x, y) + 2y, 7)> k(x, y) = v1(x, y) * v2(x, y) ; is 7 or a multiple of 7 (%)> g(x, y) = k(x, y) / 7 ; the excess> h1(x, y) = gcd(v1(x, y), g(x, y)) ; split in h1 * h2 where h1 is a> h2(x, y) = g(x, y) / h1(x, y) ; factor of v1 and h2 of v2> w1(x, y) = v1(x, y) / h1(x, y)> w2(x, y) = v2(x, y) / h2(x, y)> You may verify that for all algebraic integers x and y, the above> functions are algebraic integer functions, that w1(x, y) * w2(x, y) = 7> and that w1(x, y) is a factor of (5 a_1(x, y) + 7y) and w2(x, y) a factor> of (5 b_2(x, y) + 2y) in the algebraic integers.(% Note that for any three algebraic integers p, q and r, gcd(pq, r)> divides gcd(p, r)*gcd(q, r).)Now I wonder what you mean with the change of exponent...Dividing off 7 from both sides, and check at x=0, as then you shouldnotice that the terms independent of x are now y, and 2y, so 7 wasdivided off from 7y.When there were constants you were able to ?he notion thatsomehow, you could divide a constant as a function, but now you'verevealed just how little algebra you really understand by claiming youcan have w1(x,y), w2(x,y), and w3(x,y), as how do you suppose they candivide 7 from 7y without changing the exponent of y, if they arefunctions of x and y?I'm actually rather surprised that you'd betray ' such ignorance ofbasic algebra, but then, what choice do you have?You've been ?hting algebra all along relying on social forces toconvince people.Maybe you ?ure that readers on the sci.math newsgroup just don'tcare how bizarre and erroneous the mathematics you present is.You're in too deep now to worry about the truth, eh?James Harris === =[snip latest attempt to weasel out of the checkmate]You can run, James Harris, but you cannot hide!--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.comX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === said:>Admitting either the set of all sets or the set of everything allows>you to select a subset from either that corresponds to the>self-contradictory Russell Set (the set of all sets that are not>elements of themselves).No. There are set theories in which there is a universal set butRussell's Paradox does not exist.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === Admitting either the set of all sets or the set of everything allows>you to select a subset from either that corresponds to the>self-contradictory Russell Set (the set of all sets that are not>elements of themselves). No. There are set theories in which there is a universal set but> Russell's Paradox does not exist.>Only by doing things like arbitrarily banning self-membership of sets orpostulating some sort of in?ite hierarchy of sets. To me, it all seems sounnecessary and so inelegant. By simply treating set membership like anyother logical predicate of two variables, you can prove that the so-calledRussell set cannot exist (my website). The job of any set theory then ismake sure that you cannot prove that it does exist. One way -- and I don'tknow if it is really anything new -- is to have a simpli?d set theory thatdoes not postulate the existence of any particular set or sets. This is theapproach I have taken in my proof writing program. It seems to work andneatly avoid problems like RP that have plagued naive set theory. I would bevery interested to see if you can derive any contradictions from it.DanVisit DC Proof Online at http://www.dcproof.com -- FREE download === Admitting either the set of all sets or the set of everything allows>you to select a subset from either that corresponds to the>self-contradictory Russell Set (the set of all sets that are not>elements of themselves). No. There are set theories in which there is a universal set but> Russell's Paradox does not exist.> Only by doing things like arbitrarily banning self-membership of sets or> postulating some sort of in?ite hierarchy of sets. To me, it all seemsso> unnecessary and so inelegant. By simply treating set membership like any> other logical predicate of two variables, you can prove that the so-called> Russell set cannot exist (my website). The job of any set theory then is> make sure that you cannot prove that it does exist. One way -- and I don't> know if it is really anything new -- is to have a simpli?d set theorythat> does not postulate the existence of any particular set or sets. This isthe> approach I have taken in my proof writing program. It seems to work and> neatly avoid problems like RP that have plagued naive set theory. I wouldbe> very interested to see if you can derive any contradictions from it. Dan>Your system can not have the set of all natural numbers, N.The Axiom of In?ity is needed to guarantee that N exists.N can not be proven to exist using the other axioms of set theory.I ?d it interesting that the AoI doesn't actually say N is in?ite.One can derive this from the other axioms.Russell- 2 many 2 count === =in> In <5uqKb.7862$G1.38799@tor-nn1.netcom.ca>, on 01/05/2004> at 11:05 PM, Dan Christensen said:> >Admitting either the set of all sets or the set of everything allows>you to select a subset from either that corresponds to the>self-contradictory Russell Set (the set of all sets that are not>elements of themselves).> No. There are set theories in which there is a universal set but> Russell's Paradox does not exist.Only by doing things like arbitrarily banning self-membership of sets or> postulating some sort of in?ite hierarchy of sets. To me, it all seems> so> unnecessary and so inelegant. By simply treating set membership like any> other logical predicate of two variables, you can prove that theso-called> Russell set cannot exist (my website). The job of any set theory then is> make sure that you cannot prove that it does exist. One way -- and Idon't> know if it is really anything new -- is to have a simpli?d set theory> that> does not postulate the existence of any particular set or sets. This is> the> approach I have taken in my proof writing program. It seems to work and> neatly avoid problems like RP that have plagued naive set theory. Iwould> be> very interested to see if you can derive any contradictions from it. Dan> Your system can not have the set of all natural numbers, N.It's true that number theory is not built into my simpli?d set theory, theway it is in ZF. To develop number theory, I start with a premisecorresponding to Peano's Axioms. (Click on the Numbers menu.)> The Axiom of In?ity is needed to guarantee that N exists.[snip]You mean Peano's axiom that every natural number has a successor? See above.At any rate, there is no guarantee that anything actually existences in myset theory. Nevertheless, I think you should still be able to develop alarge part if not all of mathematical theory using my program.DanVisit DC Proof Online at http://www.dcproof.com -- FREE download (newrelease today)X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === =In <20040107022133.00545.00003418@mb-m02.aol.com>, on 01/07/2004 at 07:21 AM, kramsay@aol.com (KRamsay) said:>New Foundations is de?itely Quine's famous set theory.Which doesn't ' say whether it was his ?st.>What you're describing sounds like NFU, New Foundations modi?d to>permit urelements. The strati?ation is stricter, however; the>variables in P have to be ranked so that a is a member of b occurs>only for b of rank exactly one more than the rank of a. The one that I'm talking about didn't have the exactly one morerestriction.>Also in the presentations I've seen, the urelements are elements >which don't have members,The one that I'm talking about satis?d extensionality.>I would guess this might make a big difference in NFU, since the >formula for has no members can be strati?d, and has only >itself as a member can't.Well, obviously Quine had reasons for the change. I'm wondering-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org <3ffa32dd$8$fuzhry+tra$mr2ice@news.patriot.net> <8765fp8z1v.fsf@becket.becket.net>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === What Cantor (and others today) mean by the in?ite is not what the>ancient or medieval meaning was. Was there ever an ancient or medieval meaning. They may have slund theword around, but I'm not aware that they ever actually de?ed it.>Alternatively, that the old category of ?ite>had to come apart into two: what we now call ?ite, and>trans?ite.Only if the old category actually was de?ed and actually did includetrans?ite cardinals.>So the presentation in which the ancients medievals had confused>notions of the in?ite, which Cantor and others straightened out,>is really rather backwards. I understand the claim, but unless you can present a de?ition ofwhat the ancients meant and show that they really did mena that, I?d it implausible.>If anything, later developments in set>theory vindicated the convictions of Aristotle and others that a>completed in?ite cannot exist as at once. Omitting circular de?itions, did he ever de?e what eithercompleted in?ite or as at once meant? Aristotle had a lot o?eas that did not stand close scrutiny.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === What Cantor (and others today) mean by the in?ite is not what the>ancient or medieval meaning was. Was there ever an ancient or medieval meaning. They may have slund the> word around, but I'm not aware that they ever actually de?ed it.Of course they did; it means a thing without limit; in somecontexts, more exactly, without any limit in some regard.>Alternatively, that the old category of ?ite>had to come apart into two: what we now call ?ite, and>trans?ite.Only if the old category actually was de?ed and actually did include> trans?ite cardinals.Finite means whatever can be limited. Now they certainly took ?ite to include the same things weinclude, and they were entirely wrong in failing to see that betweentheir ?ite and their in?ite was a realm which we call thetrans?ite. In other words, that the lack of one kind o?tion on extent does not imply the lack of any limitationwhatsoever in that same direction.>So the presentation in which the ancients medievals had confused>notions of the in?ite, which Cantor and others straightened out,>is really rather backwards. I understand the claim, but unless you can present a de?ition of> what the ancients meant and show that they really did mena that, I> ?d it implausible.Well, it's my stock and trade, yes. I have a de?ition: the in?iteis whatever is not limited (and something could be in?ite in one wayor other); ?ite is whatever can be limited. Indeed, this is simplywhat the Latin words mean, plainly and directly.>If anything, later developments in set>theory vindicated the convictions of Aristotle and others that a>completed in?ite cannot exist as at once. Omitting circular de?itions, did he ever de?e what either> completed in?ite or as at once meant? Aristotle had a lot of> ideas that did not stand close scrutiny.A very easy thing to say; a very hard thing to prove. But yes, hedid. Thomas === So the presentation in which the ancients medievals had confused>notions of the in?ite, which Cantor and others straightened out,>is really rather backwards. I understand the claim, but unless you can present a de?ition of> what the ancients meant and show that they really did mena that, I> ?d it implausible.Well, it's my stock and trade, yes. I have a de?ition: the in?ite> is whatever is not limited (and something could be in?ite in one way> or other); ?ite is whatever can be limited. Indeed, this is simply> what the Latin words mean, plainly and directly.I should be more precise. In some ways, of course they were confused.But what Cantor straightened out, and what we now know as thetrans?ite, was generally *not* what they meant by the in?ite.Indeed, you have medieval philosophers who are concerned todistinguish very carefully between what is quodammodo in?itus(in?ite in a certain way), that is, in?ite in extent, or in?itein time, or in?ite in wisdom, or whatever; and penitus in?itus(entirely in?ite), which would be in?ite in every way, and referonly to God.They will then disagree with Aristotle about whether something can bea completed in?ite, if God is in question, and assert that therecertainly can be something penitus in?itus: God. As for things which are quodammodo in?itus, they will takeAristotle's line, and say there can be nothing which is a completein?ite, that is, a thing which is in?ite in extent say, but therecan be a process which proceeds without end. Now they allow for in?ite processes to continue into the future, andso there can be in?itus temporalis, at least of a future-directedsort. They are led to reject a backwards-in?ite duration of time,but normally for strictly theological reasons. Thomas Aquinas wasadamant that there can be no strictly philosophical objection to theeternity of the world, though he adhered to the theological convictionthat the world did not always exist.So we know fairly precisely what kinds of completed in?itesweren't supposed to be possible: the ones that are supposed to existall at once, at a single time. Indeed, we ?d exactly thatformulation sometimes. The reason for this conviction seems to have been really one of a morematerialistic vein that people suspect: that a completed in?ite hasto be built--made--somehow, and so since there are an in?ite numberof steps in building a completed in?ite, you just can't have ever?ished the job, though potential in?ity is quite possible.(Potential from potentia, meaning power or ability: the pointbeing that the power to produce yet more elements of the sequence isin?ite.)Now there are obviously problems: most notably, that if time can besubdivided inde?itely, or processes can be arbitrarily fast, then wecertainly can complete in?ite tasks in ?ite times. But consideredas physics, we have no reason to think that processes can proceedarbitrarily fast; indeed these days we have excellent reasons to saythey cannot.Mathematics in antiquity and the middle ages (and even surprisinglylate) was thought in fairly concrete terms; consider Euclid's rulethat you can extend a line inde?itely. A line, for Euclid, is notan in?ite thing; it's a *?ite* thing, but you can extend it asmuch as you like--with it always remaining ?ite. This is par forthe course throughout.So the notion of an in?ite line means, for an ancient or a medieval,a ?ite line that you have in fact extended to in?ity. And thisthey rejected, for physical reasons if not more as well.What about really big integers? Once nominalism won, a really biginteger is just a kind of name for a really big collection of realphysical things; I can make up whatever names I like, but there is nocollection of all the number names--since there are not, in fact,that many words. (For the early nominalists, there aren't that manywords because a word is an actual spoken thing; for the laternominalists, there aren't that many because a word only exists once ithas been given a denotation by some rational agent.)So in a very important sense, the question can't even get going if youwant to ask about the set of all integers. Thomas === with a little more work i will have x+y de?ed in terms of> multiplication > If you really can do x+1 in terms of multiplication only then it's> trivial to do x+y since (x+1)(y+1)=xy+(x+y)+1(x+y) = [(x+1)(y+1)-1]-xyI can reduce the term in square brackets to muls but I have gonenowhere since i still need to add -xy at the end.Also my solution for x+1 is a delta function so it still requires morework I suspect.JS === =We are working on software that will speak mathematics embedded in a webpage. Our main goal is to make math in web pages accessible to visuallyimpaired readers for which we've recently been awarded an NSF grant. Afterdemonstrating an early version of this software to teachers that happened tohave normal sight, they claimed that they could see an additional use inteaching normally sighted students how math is spoken. This would of coursebe most useful in situations where a teacher is not present. Does anyoneknow of any work done on this subject within the educational community? Arethere any opinions on the worth of this sort of thing? Any comments arewelcome.Paul ToppingDesign Science, Inc.www.dessci.comX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i07GT6r02702; === See some f(x):> x^n - 2 n^u a b x - a^v - b^v = 0>also such polynomial of x where coef?ients are expressed>stationary with some parameters a;b and ?ed values 2;n;u;v> using input a^v + b^v = x C .................(*)>primary we'll have x^(n-1) - 2 n^u ab - C = 0> { 2 n^u ab = x^(n-1) - C }^v> 2^nv a^v b^v = [x^(n-1) - C]^v> now inputting from (*) a^v = x C - b^v> 2^nv (x C - b^v) b^v = [x^(n-1) - C]^v ................(**)> and so on we'll have F(x) as a function of (n-1)v degree>of x as changing value.> Now coef?ients of F(x) are expressed with a;C parameters>and stationary ?ed values of 2;n;u;v> Could we use 1-st derivative for to judge correct>completation of such F(x) with some real x value ?> Also once n is some odd number, so at least ?st derivative>should equal 0.*** sorry for fault: ?st derivative of this F(x) should be: 2^nv C b^v = v[x^(n-1) - C]^(v-1) (n-1) x^(n-2) v(n-1) [x^(n-1) - C]^(v-1) x^(n-2) - 2^nv C b^v = 0 And again, could be used Eisenstein criterion once everycoef?ient except of the oldest one has C factor ? (from these C shuold be at least as C = product of (ci)^2 ) Ro*** _ > 2^nv C = v [(n-1) x^(n-2) - C]^(v-1) > v[(n-1) x^(n-2) - C]^(v-1) - 2^nv C = 0> Could we use now Eisenstein criterion for to express C>as at least C = product of (ci)^2 ??> Ro > X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i07HoGD09222; === =A point I made is that if you have a constant times a polynomial, like49(x+1)(x+2)(x+3) = (7x + 7)(7x + 14)(x+3)you can set x=0, to see how factors of 49 distributed, which is> trivial here as you can *see* how it distributed easily enough.So the short answer is that my position is that the distributive> property *always* holds, while these posters are trying to argue that> there are ways to break the distributive property: a(b+c) = ab + ac. > When I've given examples like49(x+1)(x+2)(x+3) = (7x + 7)(7x + 14)(x+3)in the past, they have claimed that it matters that those are> polynomials versus other expressions I use that are not polynomials,> but the reality is that they're STILL attacking the distributive> property, which is a bizarre math position to take.> Actually, no one is attacking the distributive property.Your argument goes as follows. You start with (a1(x) + p)(a2(x) + q) = pP(x)with a1(0)=a2(0)=0 and p and q coprime.You then claim (i) p clearly divides (a1(x) + p)(a2(x) + q). At x=0> p must divide (a1(x) + p), therefore p must divide> (a1(x) + p) for all x (ii) if p divides (a1(x) + p) then by the reverse distributive law> p must divide a1(x) ****** I think, that according to Your example for x=0 a1(x)=0 and a2(x)=0 and once P(x)=P1(x)+P where also P1(x)=0 so: (0+p)*(0+q)=p[0+ P]; p*q = p*P ; q=P what is some trivial state, but for other values of x there are still much more questions: 1)JHS and You claim only a1(x)+p is potentially divided by p but it can happen also to a2(x)+ q 2)in some other therms for some real values of x it can be a1(x)+p and a2(x)+q divided by sqrt of p What JHS is doing in the last time, I think can be similar to some tries to input, that factorisations always going such way: some polynomes with absoluthe therm divided by p should divide by p too. Dear Good, there is so much ways of factorisations in every FLT polynomians compositions, that I've ?d it some years ago not so easy to control: some parameters could give some reason not to do so but changing them into some other You can see some completation of Your questionable division. Some more complex metods are required, until we could have some l: See my universal input and decomposition of the primary FLT equations: sci.math topic from 23 december 03: FLT:advanced developments with some ideal parameters:TAB/tabunp Simply I am coming to secoundary FLT equations (You can see twoplotted equations from n=5 and so on You can chose only one of them) of (n-1)n degree. Now more comfortable is to use ?st derivativein the point of our hipotetic solution, then trying some true factorisations ... Ro*** * >And then note that (i) and (ii) lead to the conclusion a1(x) is divisible>by p for all x.Your critics reply: No, (i) is wrong so you are wrong.You reply: If you think that I am wrong then you must think that> (ii) is wrong. You are attacking the distributive> property.But no one is saying that (ii) is wrong, What people are saying is>that (i) is wrong. This is where polynomials come in. If a1(x) and a2(x) >are polynomials then (i) is correct. The examples you give>to support your reasoning all use polynomials for a1(x) and a2(x).>However, if a1(x) and a2(x) are not polynomials, then (i) is not>correct (explicit examples have been given). -William HughesX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i07HoGt09226; === =I think we have to center the tread, because at this moment it looks like the chats in the auditorium before the concert.POINTSPoint 1:The surjection f: N -> R is not possible because the de?ition of naturals and reals do not allow it.Point 2:Cantor clearly stated that we can say that an in?ite set K is countable if and only if it is possible to establish a one-to-one correspondence between K and N, i.e. f: N <-> K.Point 3:As the one-to-one correspondence N <-> R is not possible (Point 1), the criterion settled by Cantor (Point 2) is not valid to resolve whether R is countable or not. In other words, we cannot use this criterion with R. (See note below)Point 4: If a proof (any) comes to the conclusion that R is not countable because it is not possible to accomplish the criterion established in Point 2, then the proof is useless because that criterion is not valid for that purpose (Point 3). Note that the proof could be correct and the conclusion too.DEFINITIONUseless proof: A correct proof that proves nothing.QUESTIONDoes somebody know any other criteria (different from the one established in Point 2) to resolve whether R is countable or not? If you know one of them, I would like to know it please (but ?st be sure that it doesnt involve implicitly the criterion of Point 2).NOTE: Someone could interpret that if N <-> R is not possible, then R is uncountable. This conclusion is true, but in this case, it will imply that R is uncountable because we havent got a suitable tool to count its elements, since the properties of N and R are incompatible. It will never imply that R has more elements than N.Nicolas de la Foz === =On Wed, 7 Jan 2004 17:50:17 +0000 (UTC), nico80@jazzfree.com (Niclas>I think we have to center the tread, because at this moment it looks >like the chats in the auditorium before the concert.>POINTS>Point 1:>The surjection f: N -> R is not possible because the de?ition of >naturals and reals do not allow it.This needs proof. Although it can easily be proved, so in somesense it's true. In exactly the same way it is true to say thatsqrt(2) is irrational because the de?tion of the rationalsdo not allow it to be rational. So if Point 1 makes the standardproof of the uncountability of R invalid it also makes the standardproof of the irrationality of sqrt(2) invalid. (And it makes _any_proof by contradiction invalid.)>Point 2:>Cantor clearly stated that we can say that an in?ite set K is >countable if and only if it is possible to establish a one-to-one >correspondence between K and N, i.e. f: N <-> K.Point 3:>As the one-to-one correspondence N <-> R is not possible (Point 1), >the criterion settled by Cantor (Point 2) is not valid to resolve >whether R is countable or not. In other words, we cannot use this >criterion with R. (See note below)Huh? This makes just as much sense as saying because sqrt(2)rational is not possible, it is not valid to resolve whether sqrt(2)is rational by using the de?ition of the word ?rational'.You really are making no sense - these rules you state aboutwhat's valid are simply not correct, they're just ' things you made up.>Point 4: >If a proof (any) comes to the conclusion that R is not countable >because it is not possible to accomplish the criterion established in >Point 2, then the proof is useless because that criterion is not >valid for that purpose (Point 3). Note that the proof could be >correct and the conclusion too.DEFINITION>Useless proof: A correct proof that proves nothing.QUESTION>Does somebody know any other criteria (different from the one >established in Point 2) to resolve whether R is countable or not? If >you know one of them, I would like to know it please (but ?st be >sure that it doesnt involve implicitly the criterion of Point 2).NOTE: >Someone could interpret that if N <-> R is not possible, then R is >uncountable. This conclusion is true, but in this case, it will imply >that R is uncountable because we havent got a suitable tool to count >its elements, since the properties of N and R are incompatible. It >will never imply that R has more elements than N.Nicolas de la FozDavid C. Ullrich === I think we have to center the tread, because at this moment it looks > like the chats in the auditorium before the concert.> POINTS> Point 1:> The surjection f: N -> R is not possible because the de?ition of > naturals and reals do not allow it.Point 1 is what some may call begging the question. In what sense dothe de?itions not allow it? In the sense that, using thede?itions, one can show there is no such surjection. One way to dothat is the diagonal argument.> Point 2:> Cantor clearly stated that we can say that an in?ite set K is > countable if and only if it is possible to establish a one-to-one > correspondence between K and N, i.e. f: N <-> K. Point 3:> As the one-to-one correspondence N <-> R is not possible (Point 1), > the criterion settled by Cantor (Point 2) is not valid to resolve > whether R is countable or not. In other words, we cannot use this > criterion with R. (See note below)Huh? Because there is no surjection N <-> R, we cannot tell whether Ris countable or not?> Point 4: > If a proof (any) comes to the conclusion that R is not countable > because it is not possible to accomplish the criterion established in > Point 2, then the proof is useless because that criterion is not > valid for that purpose (Point 3). Note that the proof could be > correct and the conclusion too.Remarkable.> QUESTION> Does somebody know any other criteria (different from the one > established in Point 2) to resolve whether R is countable or not? Of course not. The criterion in Point 2 is the de?ition ofcountable. There is no other relevant meaning of the term.> NOTE: > Someone could interpret that if N <-> R is not possible, then R is > uncountable. This conclusion is true, but in this case, it will imply > that R is uncountable because we haven't got a suitable tool to count > its elements, since the properties of N and R are incompatible. It > will never imply that R has more elements than N.Remarkable.-- Jesse F. HughesNot all features that are found on the Security tab are designed tohelp make your documents and ?es more secure. -- Microsoft clari?s the Of?e security features. === I think we have to center the tread, because at this moment it looks > like the chats in the auditorium before the concert.> POINTS> Point 1:> The surjection f: N -> R is not possible because the de?ition of > naturals and reals do not allow it.This statement reqires proof, as it is not self evident.Point 2:> Cantor clearly stated that we can say that an in?ite set K is > countable if and only if it is possible to establish a one-to-one > correspondence between K and N, i.e. f: N <-> K.More precisely, a set K is countable if there is a surjective map from N to K and countably in?ite (or some say denumerable) if there is a bijective map.Point 3:> As the one-to-one correspondence N <-> R is not possible (Point 1), > the criterion settled by Cantor (Point 2) is not valid to resolve > whether R is countable or not. In other words, we cannot use this > criterion with R. (See note below)You assert it to be impossible, but you provide no proof. In mathematics, no assertion need be accepted without proof. Since you provide no proof, we may assume that you do not know whether the reals are countable.Point 4: > If a proof (any) comes to the conclusion that R is not countable > because it is not possible to accomplish the criterion established in > Point 2, then the proof is useless because that criterion is not > valid for that purpose (Point 3). Note that the proof could be > correct and the conclusion too.>There were, at last count, well over 100 proofs of the Pythagorean theorem. Which ones are invalid? > DEFINITION> Useless proof: A correct proof that proves nothing.This is an empty de?ition, since it can never be instantiated. A correct proof, by its own de?ition, proves something. You are de?ing the memebers of the empty set.This Futz is nearly as wacky as JSH. === I think we have to center the tread, because at this moment it looks >like the chats in the auditorium before the concert.>POINTS>Point 1:>The surjection f: N -> R is not possible because the de?ition of >naturals and reals do not allow it.So what is it about the de?itions of the naturals and the reals thatallows you to con?ently state that such a surjection is not possible?Obviously we cannot use the de?itions of the naturals and the rationals to prove that there is no surjection f : N -> Q, since we know that thereis such a surjection. This means that you cannot use the argument thatrN is a proper subset of R (N is a proper subset of Q and there exists a surjection f : N -> Q).>Point 2:>Cantor clearly stated that we can say that an in?ite set K is >countable if and only if it is possible to establish a one-to-one >correspondence between K and N, i.e. f: N <-> K.>Point 3:>As the one-to-one correspondence N <-> R is not possible (Point 1), Your claim is to the effect that it is trivial to demonstrate thatsuch a one-to-one correspondence does not exist. Let's see yourtrivial proof.>the criterion settled by Cantor (Point 2) is not valid to resolve >whether R is countable or not.Point 1 means that R is not countable, so it does provide an answer,except that you have not deemed it necessary to provide the trivialproof that makes the non-existence of a surjection trivially obvious to you, but not to others. Point 2 is valid. It is the DEFINITIONof countability of a set. If you can show that R does not satisfy Cantor's criterion for countability as espoused in Point 2, then Ris not countable. What is so dif?ult about this point?> In other words, we cannot use this >criterion with R. (See note below)Point 1 implies that R is uncountable. What more do you want?>Point 4: >If a proof (any) comes to the conclusion that R is not countable >because it is not possible to accomplish the criterion established in >Point 2,In other words, we conclude that R is uncountable because it fails the criterion for countability.> then the proof is useless because that criterion is not >valid for that purpose (Point 3).Yes, it is. It is the DEFINITION of countability of a set. It laysdown the one and only criterion for countability. R fails that criterion, so R does not satisfy the de?ition for countability,so R is not countable.> Note that the proof could be >correct and the conclusion too.>DEFINITION>Useless proof: A correct proof that proves nothing.You have not demonstrated why we should accept that there is a trivial proof of the nonexistence of a surjection f : N -> R.>QUESTION>Does somebody know any other criteria (different from the one >established in Point 2) to resolve whether R is countable or not?It is Cantor's de?ition for countability. This means that anyin?ite set which satis?s the criterion is countable, and anyin?ite set which does not satisfy the criterion is uncountable.> If >you know one of them, I would like to know it please (but ?st be >sure that it doesn't involve implicitly the criterion of Point 2).>NOTE: >Someone could interpret that if N <-> R is not possible, then R is >uncountable. This conclusion is true, but in this case, it will imply >that R is uncountable because we haven't got a suitable tool to count >its elements, since the properties of N and R are incompatible. It >will never imply that R has more elements than N.We know that R has more elements than N since there exists an injectionf : N -> R, but there is no surjection f : N -> R. That is ALL thatyou need.David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i07He2G08473; === =Occasionally it is useful to be reminded of the Characteristics of a Pseudoscientist. I have a great discovery! I have discovered it! It is mine! It overturns all of current mathematical thought! How silly of you to not have noticed it before! I am just an amateur with no formal training, but it doesn't matter! Look at my discovery! Let me explain my discovery! You must look at it with an open mind! Do not be prejudiced! What? There is NO error in my discovery! You hate me! You don't understand! Please respond to my major points! I haven't made an error. I am right an all of you mathematicians are wrong! You are closed minded! My discovery is mine! It is right! Sound familiar?>I've been using an example put forward by Rick Decker, a professor at>Hamilton College, for a few days, but now I've seen need to add to it.Consider,7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7(xy - y^2)(5) + 7^2 y^2so(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) >where the a's are roots of >a^2 - (xy - y^2)a + 7(x^2 + xy).Now letting x=0, gives a(a + y^2) = 0, so a = 0, or -y^2, and letting a_2(0,y) = -y^2, I leta_2(x,y) = b_2(x,y) - y^2, so(5a_1(x,y) + 7y)(5b_2(x,y) - 5y^2 + 7y) = 7(25x^2 + 30xy + 2y^2) what's interesting here is that the 7 is seen visibly to have been>multiplied by the ?st factor but not the second because of the>-5y^2.My hope is that *someone* of you can trust algebra and look at the>result with an open mind.Notice that when x=y, some of the terms with y as a coef?ient are>clipped out.Previous discussions were basically looking over the special case>where y=1.Want more advanced polynomial factorization?Then check out my blog archives:James Harris === Occasionally it is useful to be reminded of the Characteristics of a Pseudoscientist. I have a great discovery! I have discovered it! It is mine!> It overturns all of current mathematical thought! How silly of you to not have noticed it before!> I am just an amateur with no formal training, but it doesn't matter! Look at my discovery!> Let me explain my discovery! You must look at it with an open mind! Do not be prejudiced!> Sound familiar?James Harris. (OK, it's not strictly an *auto*biography, but it soundsso much like him that it almost could pass as something he's written.)-- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, SilverlockX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i07IEh811354; === 
> Andrew McMillan claimed that the
solution did not use the law of>> cosines, then outlined
how the solution was obtained. The method>> outlined is
one of the standard ways of deriving the law of>> cosines.
To see this just do the suggested calculations with sides>>
a, b and c.I used the below formula which is derived from
Pythagoras' therom:x = (a^2 + b^2 - c^2) / 2awhere for any
given triangle, x is the distance along line a from
the>corner ab to where the line from corner bc intersects
line a at the>perpendicular.
X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i07HoGG09218; === = Logic (mathematical or not) tell us that in a > proof it is not possible to conclude something reasonable and > trustable from a false premise.Logic also tells us that if an unreasonable and untrustable conclusion>follows from a certain premise, that premise was false. Which is the>basis of reducto ad absurdum. >-- Yes, but even the reductio ad absurdum needs a viable premise. The proof by reductio ad absurdum can be accepted as a valid proof technique when nobody knows in advance if the initial premise is false or true. But, when you already know that the initial premise is false, then the *absurdum* is to use this technique, since the result is not reliable.Nicolas de la Foz>Daniel W. Johnson>panoptes@iquest.net>http:// members.iquest.net/~panoptes/ proof by reductio ad absurdum can be accepted as a valid proof > technique when nobody knows in advance if the initial premise is > false or true. But, when you already know that the initial premise is > false, then the *absurdum* is to use this technique, since the result > is not reliable.This is, of course, utter nonsense.-- No feeling sympathy for mathematicians who start marching with signslike ?Will work for food' in the future... I will not show mercygoing forward. I was trained as a soldier in the United States Armyafter all... We play to win. --James Harris, feel his wrath! === = Logic (mathematical or not) tell us that in a > proof it is not possible to conclude something reasonable and > trustable from a false premise.Logic also tells us that if an unreasonable and untrustable conclusion>follows from a certain premise, that premise was false. Which is the>basis of reducto ad absurdum. >-- Yes, but even the reductio ad absurdum needs a viable premise. The proof by reductio ad absurdum can be accepted as a valid proof technique when nobody knows in advance if the initial premise is false or true. But, when you already know that the initial premise is false, then the *absurdum* is to use this technique, since the result is not reliable. If you do not have of a proof (in advance) that something is false, then you do not know it to be false, at least in any mathematical sense. What you may believe without proof is irrelevant.How do you know that the reals are uncountable? What is your proof? Unless _you_ have a proof, you do not know it. And if you do not know it, then there nothing outre in assuming it in order to see what consequences may develop.However, Cantor did not use reductio by absurdum in that proof, so the issue is moot. === Logic (mathematical or not) tell us that in a > proof it is not possible to conclude something reasonable and > trustable from a false premise.No, it doesn't. If I square both sides of the false statement 5 = -5, then I get the true statement 25 = 25. I was thereforeable to conclude a true consequence of a false premise.>Logic also tells us that if an unreasonable and untrustable conclusion>follows from a certain premise, that premise was false. Which is the>basis of reducto ad absurdum. >-- >Yes, but even the reductio ad absurdum needs a viable premise. The >proof by reductio ad absurdum can be accepted as a valid proof >technique when nobody knows in advance if the initial premise is >false or true. But, when you already know that the initial premise is >false, then the *absurdum* is to use this technique, since the result >is not reliable.This is pure nonsense. I take a premise P and conclude a contradiction.You are stating that this proves that P is false if P is initiallyviable, but it does not provide a proof that P is false if P is notviable. So you expect that P is false if it is not viable, but youassert that reductio ad absurdum ?ds it more dif?ult to prove thatP is false if it is easier to believe beforehand that P is false.Such an attitude is unaccountable.How do you cope with consequences of the Axiom of Choice which also look unviable, like the theorem of Banach and Tarski? The statement of the theorem looks about as unviable as you can get, butit still has a proof.David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. === Yes, but even the reductio ad absurdum needs a viable premise. The > proof by reductio ad absurdum can be accepted as a valid proof > technique when nobody knows in advance if the initial premise is > false or true. But, when you already know that the initial premise is > false, then the *absurdum* is to use this technique, since the result > is not reliable.So a reductio ad absurdum proof can be valid, but only if you have never laideyes on it before? Perhaps it should be a requirement that all reductio adabsurdum proofs should be written in disappearing ink.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i07Kmfc23146; === =We can construct analytic function f(z) from given its real or imaginarypart using Milne Thomson method. But we do not have proof of this method.is there any proof? or why this method work technically?Please give your views.msk === We can construct analytic function f(z) from given its real or imaginary> part using Milne Thomson method. But we do not have proof of this method.is there any proof? or why this method work technically?> Please give your views.mskI don't know what the Milne Thomson method is, but I do know theCauchy-Riemann equations. Suppose f is an analytic function from C toC. Write f(x+iy)=u(x,y) + iv(x,y) where x and y are real and u and vare real-valued functions of x and y.Then the Cauchy-Riemann equations tell us that du/dx = dv/dy and du/dy = -dv/dx. Here d should be the partial differentiation symbol,but I don't have it in plain text. By integrating the ?st equationwith respect to y, you can determine v up to a function of x. Byintegrating the second equation with respect to x, you can determine vup to a function of x. This will determine v up to a constant. This isthe best you can hope for, since you can add a purely imaginaryconstant to one f which works and get another one.If you want to determine u given v, use the same ideas but interchangethings as necessary.---- David === Write f(x+iy)=u(x,y) + iv(x,y) where x and y are real and u and v> are real-valued functions of x and y.Either u(x,y) or iv(x,y) is given. Try: let y=0 and replace x by z=(x,y)=x+iy and calculate.> I don't know what the Milne Thomson method isMe too, i hope msk is so polite to tell us.Have funHeroBy the way: i=(0,1) and i-is-no-longer-imaginaryX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i07LVX326660; === =Let p(z) be polynomial of degree n such that |p(z)|<= M for all z in the unit disc i.e. |z|<1Prove that |p(z)|<= M|z|^n for all |z|>1I cannot prove this inequality but I don't think it's ' too hard andI expect that someone will give me hint or solution.Thanx! === Let p(z) be polynomial of degree n such that |p(z)|<= M for all z in the unit disc i.e. |z|<1Prove that |p(z)|<= M|z|^n for all |z|>1X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i07KmfA23138; === =[snip]>Want more advanced polynomial factorization?[snip]*Incapable*, remember ?The mathematicians are *incapable*, aren't they ?*incapable*.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i07MefJ32053; === Let p(z) be polynomial of degree n such that |p(z)|<= M for all z in the unit disc i.e. |z|<1Prove that |p(z)|<= M|z|^n for all |z|>1>I cannot prove this inequality but I don't think ' it's too hard and>I expect that someone will give me hint or solution.Thanx! Why not start this homework problem by writing out the terms of p(z) , eh?philX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08045D06623; === =[/snip]>Posters in attacking algebra here relied on various tricks, mostly>social from what I've seen, where they have banded together as a group>to support each other, and simply kept repeating assertions that I>shot down with the algebra.Here I think that adding that variable should allow most of you to see>how they did it. My hope is that some of you actually care about>algebra, and feel more than a little perturbed that such people would>act for so long, with such persistence, to attack it.These people have been attacking the base of mathematics itself.Mathematics isn't about social truth, but about what is true. If>you can't handle a particular mathematical result because it makes you>feel bad, then maybe you need to go to some other ?ld, like>politics, or fashion!!!>James HarrisI feel like I've reas thius numerous times, but with only a different clsoing paragraph.Couldn't be, though... === =As a part of the automated decision support system for an empire-basedstrategy game, I would like to determine the ?ancial condition of aplayer. It should be a simple evaluation, stating how better or worse thanthe others the player is (this value is later fuzzi?d).The value should be 0 if the evaluated player has the same amount of moneyas each of the other players, non-zero otherwise (negative when he's poorer,positive when richer). The values should differ for [1, 0, 0, 0] and [5, 4,4, 4] cases (?st player is being evaluated), i.e. not only a difference,but also a proportion should count.What equation would you suggest?--BB === As a part of the automated decision support system for an empire-based> strategy game, I would like to determine the ?ancial condition of a> player. It should be a simple evaluation, stating how better or worse than> the others the player is (this value is later fuzzi?d).> The value should be 0 if the evaluated player has the same amount of money> as each of the other players, non-zero otherwise (negative when he'spoorer,> positive when richer). The values should differ for [1, 0, 0, 0] and [5,4,> 4, 4] cases (?st player is being evaluated), i.e. not only a difference,> but also a proportion should count.> What equation would you suggest?>P1 has 100 and P2 has 50. That's (100 - 50) / 50 = 1orP1 has 50 and P2 has 100. That's (50 - 100) / 50 = -1 .That's (P1 - P2) / Sv where Sv = P1 if P1 <= P2 or Sv = P2 if P2 < P1 . === As a part of the automated decision support system for an> empire-based strategy game, I would like to determine the> ?ancial condition of a player. It should be a simple evaluation,> stating how better or worse than the others the player is (this> value is later fuzzi?d). The value should be 0 if the evaluated> player has the same amount of money as each of the other players,> non-zero otherwise (negative when he's poorer, positive when> richer). The values should differ for [1, 0, 0, 0] and [5, 4, 4, 4]> cases (?st player is being evaluated), i.e. not only a difference,> but also a proportion should count. What equation would you suggest?> P1 has 100 and P2 has 50. That's (100 - 50) / 50 = 1> or> P1 has 50 and P2 has 100. That's (50 - 100) / 50 = -1 .> That's (P1 - P2) / Sv where Sv = P1 if P1 <= P2 or Sv = P2 if P2 < P1 .This is quite nice for 2 players, but my game has 3-5 ones. I think it wouldbe appropriate to take both average difference and biggest neg/posdifferences into account, e.g.P1: 2, P2: 1, P3: 1, P4: 4 (we are P1)Average difference: 2 - (1 + 1 + 4) / 3 = 0 (we are like the whole userpopulation)Biggest negative difference: 2 - 4 = -2 (we are poorer)Biggest positive difference: 2 - 1 = 1 (we are richer)I know it could be seen as a tri?rouble ;)), but I'd like to go downthe best path.--BB === With the new axioms, it is easy to construct a bijection from the> universal set to its power set. How?In a set theory with a universal set U and suf?ient power to reasonabout it, P(U)=U. So, the identity function is a suitable bijection.If P(U)=U is controversial (given a set theory with a universal setand no urelements), I would be interested in hearing so. I realizethat some universal set theories preclude analysis of P(U)=U on thegrounds of strati?ation, but if a theory allows one to reason aboutP(U)=U then surely it must either be true or undecidable.Note that Cantor's construction showing that for non-empty x,|P(x)|>|x|, fails on such large sets as U: the set one constructs torefute the possibility of a bijection involves negation and the law ofexcluded middle, so the set membership relation of the refutation setmay be partial, and that partiality precludes the contradiction. === = [.snip.]>(in case mathforum's post did not show up on your reader)Don't much mind breaking this to you, but the reason I did not reply> to you before is simple: I have absolutely no interest in engaging in> an exchange with you.just do not forget that the ball is in your court, should you decideto change your mind about engageing in an exchange with me. OK?> You have absolutely no interest in discussing anything, all you want> is someone to agree with you.how did you ?ure that? when i challenged your fairy conceptionsabout american employment practices?> Even how you phrase questions shows that you have already made up your> mindreally? can you provide one such question?> (and as usual, in an absurd way and frozen it in an absurd, ignorant,> and stupid position).let me get this straight.you disagree with my reports that math degrees in general are treatedwith mediocrity, and then charge that my position is somehow absurdand ignorant. have you looked in the mirror lately?> So, I am not interested in discussing adjuncts with you,if you change your mind, you know where to ?d me.> and I am not interested in your pathetic attempts to back off from> your insult,what insult? am i in some way responsible for your current academictitle? or are you under the impression that i actually implied thatyou are indeed incompetent? here is the 411, i DID NOT.> or your apologies, with their quotation marks, that are nothing but> further insults to my intelligence. what should be insulting to your intelligence is your apparentdishonesty with your own self...> -- > === == === = It's not denial. I'm just very selective ' about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> === == === ==Arturo Magidin> magidin@math.berkeley.eduwhat insult? am i in some way responsible for your current academic> title? or are you under the impression that i actually implied that> you are indeed incompetent? here is the 411, i DID NOT.Well, it certainly looked to me like that was *exactly* the implicationyou were making.-- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === > Well what the did [you] may well be right on this one... mean,> if it didn't mean that your comments increased the likelihood that the> original [im]poster was right? if i had written you could be correct... > I see no practical difference.in that case, i have a question for you - how often do you getinvolved in pointless discussions?> geez, i would hate to assume that you are having problems with> existential quanti?rs. is that it?> Well, I think I know a thing or two about logic, but please enlighten> me about what error I made regarding existentials.> well, ?st tell me how much logic you know. can you handle venn> diagrammes?I have more than suf?ient background in logic, I am sure. Honest. evidence?I have a piece of paper saying that I have a PhD in Logic,> Computability and Methodology.well, well. i guess you might have well proven that it is possible tohave a PhD in logic while at the same time being very careless whenexamining a colloquial statement.how good are you in your ?ld?would i be able to ?d publications under your name?> Did you mean only this?> what else could i have meant?> do you have further objections?> (1) Some adjuncts are incompetent.> (2) Arturo is an adjunct.> -------> Therefore, it is possible that Arturo is incompetent.I just want to be clear: The argument above is the argument you know> claim to have advocated? what do you think i said? no baseless assumptions.If I was certain, I wouldn't have asked.judgeing from your initial reaction, seems that you jumped the gun abit too soon.> Why won't you answer the question?i answered your question long ago. go up the thread.> nor advocated the above argument.here is what i meant:arturo magidin is ADJUNCT assistant professor. ADJUNCTS are thepariahsin the academic caste system. so whomever jstevh@yahoo.com is, it maywell be right on this one...let me know what you think i said. OK? === =nor advocated the above argument.here is what i meant:arturo magidin is ADJUNCT assistant professor. ADJUNCTS are the> pariahs> in the academic caste system. so whomever jstevh@yahoo.com is, it may> well be right on this one...let me know what you think i said. OK?I know you addressed this question to Jesse, but anyway, here's what*I* think:Your statement was made right after you quoted jstevh@yahoo.com sayingthis:> I maintain that Magidin is indeed either incompetent as a> mathematician, or more likely a liar, or both.So what I think you said was, essentially, Because Arturo Magidin isan adjunct professor, that means jstevh@yahoo.com may well be rightthat Arturo is incompetent or a liar. So if you weren't insultingProf. Magidin by saying that being an adjunct professor implies thathe's incompetent, then what *did* you mean? I can't see any otherplausible interpretation.-- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock <188f56bf.0312291220.6dbfebe6@posting.google.com> <87fzf3ku82.fsf@phiwumbda.org> <188f56bf.0312301023.5bde1c59@posting.google.com> <87isjxvg7q.fsf@phiwumbda.org> <188f56bf.0312310753.7950837@posting.google.com> <8765fwy57l.fsf@phiwumbda.org> <188f56bf.0401021033.36ff26f@posting.google.com> <87znd53oh3.fsf@phiwumbda.org> <188f56bf.0401051615.62a77c73@posting.google.com> <87ekudegc8.fsf@phiwumbda.org> <188f56bf.0401071625.6e98557c@posting.google.com> === > Well what the did [you] may well be right on this one... mean,> if it didn't mean that your comments increased the likelihood that the> original [im]poster was right? if i had written you could be correct... > I see no practical difference. in that case, i have a question for you - how often do you get> involved in pointless discussions?At least as often as I talk to you. > geez, i would hate to assume that you are having problems with> existential quanti?rs. is that it?> Well, I think I know a thing or two about logic, but please enlighten> me about what error I made regarding existentials.> well, ?st tell me how much logic you know. can you handle venn> diagrammes?I have more than suf?ient background in logic, I am sure. Honest. evidence?I have a piece of paper saying that I have a PhD in Logic,> Computability and Methodology. well, well. i guess you might have well proven that it is possible to> have a PhD in logic while at the same time being very careless when> examining a colloquial statement. how good are you in your ?ld?> would i be able to ?d publications under your name?Who cares? The question was whether I would understand your tutelageon the existential quanti?r. Now that I hope I am up to your highstandards, please enlighten me.> Did you mean only this?> > what else could i have meant?> do you have further objections?> (1) Some adjuncts are incompetent.> (2) Arturo is an adjunct.> -------> Therefore, it is possible that Arturo is incompetent.I just want to be clear: The argument above is the argument you know> claim to have advocated? what do you think i said? no baseless assumptions.If I was certain, I wouldn't have asked. judgeing from your initial reaction, seems that you jumped the gun a> bit too soon.Maybe. Why don't you explain what you meant then, jackass. All theselame attempts at sophistry and Socratic/Freudian techniques (What doyou think I meant? How would that make you feel?) are painful anddull. Why won't you answer the question? i answered your question long ago. go up the thread.No, you didn't. You have never stated what your comments weresupposed to mean. > nor advocated the above argument. here is what i meant: arturo magidin is ADJUNCT assistant professor. ADJUNCTS are the> pariahs> in the academic caste system. so whomever jstevh@yahoo.com is, it may> well be right on this one... let me know what you think i said. OK?That doesn't tell one a goddamn thing what you meant, since it givesno further clue of the relevance between Arturo's position and thecomment that the poster is likely right.You're a complete moron. If you want to insult Arturo, then have theballs to admit it. I will drop out of this conversation. -- Sure, [my Usenet presence is] like Shaq playing against you in yourbackyard, but that has its perks, as I ?d ways to have my fun *and*I can send messages to certain people in the United States Governmentwithout concern that the rest of you understand them. -- James HarrisX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i080vd210108; === =Occasionally it is a good idea to be reminded of the Characteristics of a Pseudoscientist. (I posted something similar on another thread.) I have a new discovery! It is mine! I will tell you my discovery! My discovery overturns all of mathematics as you know it! How silly of you not to have discovered my discovery before! I am just an amateur with no formal math training, but I still discovered my discovery even though the best math minds have overlooked it for hundreds of years! Here is my math discovery! Remember, it is mine! Remember to read it with an open mind! Don't be prejudiced! What? You dare to argue with my discovery! You are prejudiced! I am not wrong! I am a genius and this is my discovery! You need to repent! I am not wrong! This discovery is mine! Sound familiar?>For a while I've been talking about some rather basic mathematics,>which has been continually attacked or questioned by some rather>persistent posters. While I've tried to explain to readers on this>newsgroup and others, how they were attacking and questioning algebra>itself, I've found myself watching as they used social tricks and lies>rather successfully to fool apparently, all of you.However, by adding another variable to an example given by Rick>Decker, a professor at Hamilton College, I can ?ally explain to you>how these people have been assaulting algebra.Consider(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of >a^2 - (x - y)a + 7(x^2 + xy).Now letting x=0, gives a(a + y^2) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you havea_2(x,y) = b_2(x,y) - y, so(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). Now then, x and y are INDEPENDENT of each other, so my setting x=0 was>just a way to ?d out what the coef?ients of the the terms with y>that do NOT have x as a factor were. It's a neat technique used to>clear out the effects of one particular variable to focus on another.That revealed that in one factor the term independent of x, is 7y,>while in the other it's 2y, which reveals immediately which term was>multiplied by 7.Notice that arguments claiming that somehow 7 divides both sides>dependent on the value of x, are simply bizarre, irrational, and>CANNOT BE TRUE!!!Claiming that 7 divides as some kind of factor of x to try and>maintain that both (5a_1(x,y) + 7y) and (5b_2(x,y) + 2y) can have some>non-unit factor in common with 7, against the reality shown above that>one must be coprime to 7, is a bizarre math position contrary to>mathematics! It also just doesn't make even common sense.Posters in attacking algebra here relied on various tricks, mostly>social from what I've seen, where they have banded together as a group>to support each other, and simply kept repeating assertions that I>shot down with the algebra.Here I think that adding that variable should allow most of you to see>how they did it. My hope is that some of you actually care about>algebra, and feel more than a little perturbed that such people would>act for so long, with such persistence, to attack it.These people have been attacking the base of mathematics itself.Mathematics isn't about social truth, but about what is true. If>you can't handle a particular mathematical result because it makes you>feel bad, then maybe you need to go to some other ?ld, like>politics, or fashion!!!>James Harris === =Zero is not needed as a placeholder in the Alternate Number System.Also in this system in base 1, the digits are one's as required and issimply a tally system.In base (b) the digits are 1 to b which is more logical than didgits 0 to(b-1) in our existing system.http://my.tbaytel.net/forslund/ans.html>*>Base 8 uses the numbers 0,1,2,3,4,5,6,7>Base 7 uses the numbers 0,1,2,3,4,5,6>Base 6 uses the numbers 0,1,2,3,4,5> ...>Base 3 uses the numbers 0,1,2>Base 2 uses the numbers 0,1>Base 1 uses the numbers 0 Interesting point, but I don't think ' it's correct. Unary is special, or> maybe it is zero that is special. Zero was introduced into Arabic notation as a placeholder, to indicate> the signi?ance of particular digits. One hundred and one was written CI by the Romans, and could be written> as 1 x 10^2 + 1 x 10^0. The zero in 101 doesn't add anything to the> value, it just moves the hundreds digit one step to the left so we know> what it means by its place alone. In unary notation, position has no signi?ance so the zero is> unnecessary. But, you ask, why does binary use 1 and unary also? I guess unary IS a> little bit special. It would be useless to put it formally, because> there are still anomalies, but note that in decimal, the property of the> largest digit (9) is to be the largest single digit before the ?st> two-digit number, and in unary whatever digit we choose serves this> purpose. Zero is wrong, as I showed above. 1 is as good as any, because> differentiation between digits is not needed. Maybe we should really> write unary 3 as xxx to indicate this. x is a digit, but we don't need> separate symbols like 1, 2 and 3 etc. Gerry Quinn> -- > http://bindweed.com> Screensavers, Games, Kaleidoscopes> Download free trial versions> === Zero is not needed as a placeholder in the Alternate Number System.> Also in this system in base 1, the digits are one's as required and is> simply a tally system.I'm afraid your paper is hopeless when it comes to discussing ancientmathematical documents. Positional notation simply didn't exist atall, either in the modern way or in your alternative system. At thevery least, scholarship involves doing the hard work of tracking thedetails down, and not guessing about what you might ?d if youlooked. You claim interestingly that you can easily extend your system tonegative and fractional values.Extension to negative values is trivial, but I'm interested in exactlywhat you propose for fractional values.I think the most worrisome problem with your proposal is that theinteger zero must be represented by the empty string. While that hassome conceptual elegance, it is a huge practical liability.Thomas === =Oh, one ?al criticism. A strong advantage of the usual positionalsystem is that it maps conveniently to our language. Of course, the Roman system maps even more closely to language, andthe ancient Greek one even more so. But those are non-positional, andsuch a pain to do arithmetic with, that we don't consider them.In other words, we have existing words for numbers, and the usualpositional system connects very easily with them.You give the number 10203 as an example. In English, this is tenthousand, two hundred and three. Isn't that convenient? In yoursystem the same number is A1A3, and requires some thinking to turnit into ten thousand, two hundred and three.You might simply reply that we should change the way we speak. Butthat is pretty tough, and not just a matter of changing positionalnotation. Setting out to change the way people speak is nearlyimpossible to succeed in, and the language here drives the notation,not the other way round. (The language existed thousands of yearsbefore the notation...)If we did it, then 10203 would now be not ten thousand, two hundredand three; it would be written A1A3, and we would read it tenthousand, one hundred and tenty-three. [I have of course inventedtenty as the successor to ninety.]And, there is a much more serious problem if you want to change theway we speak. We have a bunch of convenient number words for powersof ten. We can quickly estimate a thing by giving poor-man'sscienti? notation: about three billion, or about 6 quintillion.In fact, we do this all the time, and simply drop the word about,because we rely on implicit speci?ations of precision.So take that one, three billion. Today we write it 3,000,000,000,unless you're English, and then you write it 3,000,000,000,000.In your system we would write 2,AAA,AAA,AAA. If we continue topronounce it three billion, then we have the problem that it's apain to convert on the ? the digital to the spokenrepresentation. But if we change the pronunciation, then we have tosay: two billion, ten hundred and tenty-ten million, ten hundred andtenty-ten thousand, ten hundred and tenty-ten.And the poor English, who would now write 2,AAA,AAA,AAA,AAA have tosay two billions, ten hundred and tenty-ten thousand millions, tenhundred and tenty-ten millions, ten hundred and tenty-ten thousand,ten hundred and tenty-ten.Thomas === So take that one, three billion. Today we write it 3,000,000,000,> unless you're English, and then you write it 3,000,000,000,000.In your system we would write 2,AAA,AAA,AAA. If we continue to> pronounce it three billion, then we have the problem that it's a> pain to convert on the ?m the digital to the spoken> representation. But if we change the pronunciation, then we have to> say: two billion, ten hundred and tenty-ten million, ten hundred and> tenty-ten thousand, ten hundred and tenty-ten.ITYM 2,999,999,99A and two billion, nine hundred and ninety-nine,nine hundred and ninety-nine thousand, nine hundred and ninety-ten.> And the poor English, who would now write 2,AAA,AAA,AAA,AAA have to> say two billions, ten hundred and tenty-ten thousand millions, ten> hundred and tenty-ten millions, ten hundred and tenty-ten thousand,> ten hundred and tenty-ten.Similarly. === =So take that one, three billion. Today we write it 3,000,000,000,> unless you're English, and then you write it 3,000,000,000,000.In your system we would write 2,AAA,AAA,AAA. If we continue to> pronounce it three billion, then we have the problem that it's a> pain to convert on the ?m the digital to the spoken> representation. But if we change the pronunciation, then we have to> say: two billion, ten hundred and tenty-ten million, ten hundred and> tenty-ten thousand, ten hundred and tenty-ten.ITYM 2,999,999,99A and two billion, nine hundred and ninety-nine,> nine hundred and ninety-nine thousand, nine hundred and ninety-ten. === Zero is not needed as a placeholder in the Alternate Number System.> Also in this system in base 1, the digits are one's as required and is> simply a tally system.I am a little confused by your claim in your paper that the integers0 are concrete concepts and zero is abstract. In what way iszero more abstract than two?For example, I have two laptops, so maybe that's concrete; but I alsohave zero elephants. Why is the one more concrete than the other?Thomas === =I look at the _digit_ zero as behaving differently from the other digits. Isee zero as abstract in the same way that the concept of in?ity isabstract. In contrast the integers 1, 2, 3, ... are concrete ideas that ican see. Zero i see as the absene of something just as in?ity representsendlessness. The important point i want to make is simply that a positionalnumber system does not need the digit zero. I leave the discussion of zeroand in?ity to the philosophers and don't want to dwell on that.bob Zero is not needed as a placeholder in the Alternate Number System.> Also in this system in base 1, the digits are one's as required and is> simply a tally system. I am a little confused by your claim in your paper that the integers> 0 are concrete concepts and zero is abstract. In what way is> zero more abstract than two? For example, I have two laptops, so maybe that's concrete; but I also> have zero elephants. Why is the one more concrete than the other? Thomas === I look at the _digit_ zero as behaving differently from the other digits. I> see zero as abstract in the same way that the concept of in?ity is> abstract. In contrast the integers 1, 2, 3, ... are concrete ideas that i> can see. I can see two laptops, but I can't see the number two. I can also seezero elephants.Your paper is much stronger if you leave out the specious claim aboutabstract and concrete; if you don't want to discuss them, that's?e, but then you shouldn't ' raise a rather controversial topic, or atleast, not in such a controversial way.If you should try to get your paper published in a respectedpeer-reviewed journal, you have to worry about such things.Thomas === = peer-reviewed journal, you have to worry about such things. ThomasThe paper IS published in a peer reviewed journal - The Southwest journal ofPure and Applied Mathematics.unaware of mine) in the American Mathematical Monthly i believe.Bob === peer-reviewed journal, you have to worry about such things. Thomas> The paper IS published in a peer reviewed journal - The Southwest journal> of Pure and Applied Mathematics.> unaware of mine) in the American Mathematical Monthly i believe.Essentially the same system as Mr Forslund's[Southwest J. Pure Appl. Math., vol. 1, 27--29 (1995)]appeared inJames E. Foster,Mathematics Magazine, vol. 21, 39--41 (1947).The only difference I could discern between the Foster and Forslundsystems is that for the ten digit, Foster uses T while Forslund uses A.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === peer-reviewed journal, you have to worry about such things.> The paper IS published in a peer reviewed journal - The Southwest journal of> Pure and Applied Mathematics.I said a respected peer-reviewed journal. Sorry, but the one youcite just ain't it.Thomas === I have determined that %77 of all ghosts are spotted within a 5 mile>radius of UFO sightings, and that %82 of all UFO sightings are>accompanied by a general increase in the intensity and overall>magnitude of ghost related hauntings.Therefore, I conclude, via statistics, that ghosts and UFO's aresomehow>related, and that ghosts are in all probability using UFO's for>transportation purposes.Will someone in the sci.math or sci.math.stat please stand up and>tell me why this is ? and no I am not kidding. I need an>independent opinion.You are confused as to what you are studying. The correct conclusion>is that CLAIMED SIGHTINGS of UFOs correlate with CLAIMED SIGHTINGS of>ghosts. You are not studying UFOs or ghosts, but claims concerning>them.Len.> So then Chemistry does not concern real atoms, but the claimedobservances> of atoms ?? Where does reality ? in ?? Music does not concern real notes, but merely the claim that someoneheard> music ? I would like to know if you can do valid science uopn things like UFO'sand> ghosts, things which cannot be proven to exist in the ?st place. I am not confused on this. Address the question. Well, for one thing Len has relieved my mind about getting> AIDS from unprotected anal sex, since I now realize that all> this time the researchers have been identifying, not actual> infections transmitted by butting, but CLAIMS of infection.> Now I can get rid of all those annoying, uncomfortable condoms> and enjoy sex for a change.>I turned to the mathematicians with a poorly worded example, composed inhaste.Predictably, they provided a very good critique as would be expected, andcon?med a few things which were more or less obvious to begin with. === The point, being, that you cannot measure the velocity of Superman in> ? because you cannot collect real data on mythical or nonexistent> objects/people. Nonsense. Of course you can. Superman's ?velocity has> been described, in various scripts, as approaching the speed> of light. Therefore, we can establish the following inequality: V(superman) > .98ic where c is the commonly accepted symbol for the speed of light,> and i is the mathematical symbol for sqrt(-1). Superman's speed IS imaginary, isn't it? Indeed : ' )Equally as imaginary as any psychological truths. === = > The critique from Klaus Nagel in that thread is: ...> (Klaus Nagel gives the square as a counterexample here.> And he is mentioning some others, too.)Entschuldigung, it was an error, a bad one even for a square. It cameabout due an assumption that disallows rotation of the ?ure...taking rotation into the Rechnung improves it. === =1. Abstract: In this paper, Bernoulli's effect is used to interpret the magnetic force. See my posting:http://groups.google.com/groups?selm= 9f181401.0312250209.55762999%40posting.google.comAll comments are welcome.Ka-In Yenyenkain@yahoo.com.tw === =: And God said Let there be numbers And there *were* numbers. Odd and even created he them, He said to them be fruitful and multiply And he commanded them to keep the laws of induction.--------------------------------------------------- --------------------------- MATH: the discovery, clari?ation and rigorous study of precise relationships in number, pattern, and structure.--------------------------------------------------- --------------------------- The chief difference between mathematics and physics is that in mathematics we have much more direct contact with reality. (Hardy)------------------------------------------------------ ------------------------ === I don't really see how to do this. Do you know that every computable> number has an in?ite number of digits? If not, how do you know that> f^{-1}(j) (i.e., the j-th computable number) has at least j digits, so> that you can de?e a new number whose j-th digit is different from the> j-th digit of f^{-1}(j)?I don't see that this is necessary for a proper diaganolization argument (ingeneral, not just here). If the j-th number has less than j digits, then itis suf?ient to pick 1 for the j-th digit of the constructed number. Itwon't match the j-th number, and that is all that is necessary.The problem with the original argument is that diaganolization constructsa number with an in?ite number of digits. That's ?e when you are usingit to product a real number, but not ?e when trying to product acomputable number, at least for most reasonable de?itions of computable.Cantor's diaganol argument is one of those proofs that is very beautiful.It is also one of those proofs that seems a lot less subtle than it reallyis, so it is easy to misapply it. It also seems to be a big sticking pointfor mathematical crackpots. I sometimes wonder if it would be better inmath courses to put it off, and establish the uncountability of the realssome other way, and give diaganolization later.The uncountability of the reals is pretty easy to do with elementaryarguments, without diaganolization:1. Show that there is no one-to-one corresponce between a set and the set o?s subsets. This is doable on the ?st day of set theory class.2. Apply #1 to the set of positive integers, showing that the set of subsetsof positive integers is uncountable.3. Exhibit a one-to-one correspondence between the set of subsets ofpositive integers and the reals. Here is such a one-to-one corresponce: For non-negative real r, take the continued fraction for r, r = [a0, a1, a2, ... ] The corresponding subset of positive integers is { 2+a0, 3+a0+a1, 4+a0+a1+a2, ... } For negative real r, take the continued fraction for |r|, |r| = [a0, a1, a2, ...] The corresponding subset of positive integers is { 1, 2+a0, 3+a0+a1, 4+a0+a1+a2, ... }-- --Tim Smith === =The ?discovery' of dark energy - invisible stuff that makes up the bulkof the Universe - was heralded as a breakthrough in 1998. New researchsuggests that it's just not there. Marcus Chown investigates07 January 2004 427IMHO this claim that dark matter is not there will not hold up. The dark energy is there. Look at Nature Jan 1 2004 p. 45 by Boughn and Crittenden that directly contradicts the conclusion below, see also all converges nicely to the FRW cosmological parameters of Einstein's General Relativity.Omega(Repulsive Dark Energy) ~ 0.7Omega (Attractive Stuff) ~ 0.3The ordinary matter in the Attractive Stuff is only ~ 0.04 which is same order as the error bars on these numbers.Not only that I have a simple explanation of both dark energy and dark matter as exotic vacua zero point energy based on battle-tested mundane physics!Not only that, the technological potential of manipulating the exotic vacua in metric engineering for fast space and also time travel is compelling. Herman Bondi saw some of this coming 45 years ago when he was Chief Scientist of the British Ministry of Defence. I heard him lecture on this at Cornell back then.It's invisible, it permeates every pore of space and it is remorselesslydriving the galaxies apart. The discovery of dark energy nearly sixyears ago was a bombshell dropped on the world of cosmology - and votedinternational team of astronomers is controversially claiming that itcould all be a big mistake. The dark energy could be a huge cosmicmirage, says Subir Sarkar of the University of Oxford. It may notexist at all.My bet is that they are wrong.Dark energy burst on to the world stage in 1998 when a team ofscientists led by physicist Saul Perlmutter at the University ofCalifornia at Berkeley announced that they had discovered something oddabout Type Ia supernova, a kind of exploding star thought to detonatewith a standard luminosity. It seemed that supernovae very far away werefainter than they should be, given their distance from the Earth.Everything made sense if, while the light from the supernovae had beentravelling across space to us, the expansion of the Universe hadactually speeded up, pushing the supernova farther away than expectedand making them appear fainter in telescopes.The problem was that the galaxies - the building blocks of the Universe- were known to be ?apart from each other like pieces of cosmicshrapnel in the aftermath of a titanic explosion known as the Big Bang.The only force acting on them should be their mutual gravity, which bypulling them together should be braking the expansion of the Universe,not speeding it up. Remarkably, the theory went, empty space must be?led with some kind of weird anti-gravity stuff - dark energy - whichwas counteracting gravity and driving the galaxies apart.It's this claim that is now in dispute. A team led by Alain Blanchard ofLaboratoire d'Astrophysique de l'Observatoire ' Midi-Pyrenees in Toulousehas used the European Space Agency's XMM satellite to measure the X-raysemitted by ultra-hot gas in clusters of galaxies. The galaxy clustersare at very large distances from the Earth and consequently we see themas they were very far back in time.The key thing that the observations have allowed is to be able to countthe number of clusters with a particular X-ray luminosity, so that acluster seen far back in time can be compared with a more recent clusterwith the same luminosity. This is crucial, says Sarkar, who isBlanchard's collaborator. It allows a like-for-like comparison of thenumber of clusters of a given luminosity back then and in today'sUniverse.What Blanchard's group has found is ' that nowadays there are far moregalaxy clusters in a given volume of space than there were in the past.This is at odds with what is expected in a universe with dark energy,says Sarkar.Because dark energy is a property of space, as space expands, the amountof dark energy grows and, with it, its anti-gravity effect.Consequently, the gravity that is sing in matter to make new galaxyclusters is eventually overwhelmed by dark energy, preventing any moregalaxy clusters from being born. If the Universe contains dark energy,therefore, the number of galaxy clusters in a given volume should havelevelled off at some time, not continued increasing until the presenttime, says Sarkar.Astronomers such as Cambridge's Sir Martin Rees think the behaviour ofX-ray emitting gas in galaxy clusters is highly complex and that ourlack of understanding of the complexities may be leading Blanchard andhis colleagues to an erroneous conclusion. Taken at face value, however,the XMM results imply that there is no dark energy.The results are compatible with us living in a high-density universe inwhich the total matter - visible stars and galaxies plus invisible, ordark, matter - adds up to close to the so-called critical density.This is a value strongly favoured by theorists because it is required byin?, the current best theory of what happened in the Universe's?st split-second of existence.If ' Blanchard's people are right, the question is: how did astronomersmanage to miss all the extra matter in their surveys of the Universe?Sarkar points out that astronomers cannot see dark matter directly butmust infer its existence from the motion of visible material such asstars and galaxies. They could easily have underestimated the ratio ofdark matter to ordinary matter, he says. And this could have causedthem to underestimate the total amount of dark matter.NO! What these people do not realize is that dark matter and dark energy ARE BOTH THE SAME VIRTUAL STUFF!They are both zero point energy that is of positive pressure in the former and negative pressure in the latter!All this data shows are dynamical changes in the con?uration of my /zpf ?ld!1. /zpf = (Quantum of Area)^-1[(Quantum of Volume)|Vacuum Coherence|^2 - 1]Using Baez's et-al loop quantum gravity ' terms.Witten's et-al string term is also hereWitten's alpha' = ' (Quantum of Area)/hc = (String Tension)^-1Einstein's gravity ?ld of curved space-time is simply2. guv = Einstein's metric of Special Relativity + (Quantum of Area)(Phase of Vacuum Coherence)(,u,v)Susskind's Hologram is already in my equation 2.whizzing around above the vacuum. This is a wrong idea that leads to the confusion here in the thinking of Sarkar and Blanchard. I am sure they are wrong because they do not have the right basic picture of what this stuff is.Clearly, if there is no dark energy, then the original supernovaobservations which indicated the existence of dark energy were wrong.Sarkar thinks so. The basic assumption made by Perlmutter and hiscolleagues is that supernovae in the ancient Universe were exactly thesame as supernovae today. This may be wrong. Certainly, two Indianphysicists have recently drawn attention to a peculiarity of thesupernova analysis. Perlmutter and his colleagues use a correction tothe brightnesses of supernovae that are extremely far away and oneswhich are relatively nearby. They then deduce that the Universe'sexpansion has speeded up as a result of dark energy. According to RoyChoudhury of the International School for Advanced Studies in Triesteand Thanu Padmanabhan of the University of Pune, if the two categoriesof supernovae are analysed separately, the two sets of data canindividually be explained without any need for dark energy.Not surprisingly, the scientists who made the original supernovaobservations that led to the announcement of dark energy stand by theirresult. According to Alexei Filippenko of the University of Californiaat Berkeley, leader of a second team at Berkeley beside Perlmutter's,the supernova observations are really pretty convincing now.Bizarrely, even if the supernova observations are wrong, it may notmatter. According to Rees, this is because the dark energy result is nownot dependent on a single observation but on a large, interlocked set ofobservations. One of the most important is an observation made last yearby Nasa's Wilkinson Microwave Anisotropy Probe (WMAP).The probe observed the cosmic background radiation, the relic heat ofthe Big Bang ?eball. Tiny variations in the temperature of thisafterglow across the sky turned out to be compatible with the Universehaving the critical density, with 30 per cent of its mass in the form ofmatter - both visible and dark - and 70 per cent in the form of darkenergy. In other words, they con?med the existence of dark energy.Sarkar and his colleagues, however, point out that the analysis of theprobe's data was based on a number of cosmic assumptions. If any wereslightly wrong, the Big Bang radiation might not be compatible with darkenergy after all.Sarkar and his colleagues, who include Michael Rowan-Robinson ofLondon's Imperial College and Marian Douspis of the University ofOxford, need the lumpiness of the matter in the Big Bang, which led tothe temperature variations seen by the probe, to be slightly differentfrom the standard assumption.They also require that 12 per cent of the mass of the Universe be in thethat neutrinos have mass, says Sarkar. The only thing not known iswhether they have suf?ient mass for our purposes.Critics such as David Spergel of Princeton, who has worked on the WMAPdata, point out that a lot of things need to be wrong for observationsof the Universe to make sense without dark energy. It is better to beslightly wrong about a number of things than incorporate a parameterwhich is 10 followed by 123 zeroes bigger than theory predicts!counters Sarkar.This is a reference to the fact that our best theory of physics -quantum theory - predicts a value for the energy density of the darkenergy which is 10-followed-by-123-zeroes times bigger than observed.This spectacular discrepancy has been described by Nobel laureate StevenWeinberg of the University of Texas as the worst failure of anorder-of-magnitude estimate in the history of science.Certainly, if the dark energy goes away, physicists will sleep easier intheir beds. Nature, however, is under no obligation to respect thesensibilities of physicists or anyone else. Ultimately, it will befurther observations of the Universe that determine whether dark energyis real or one of the biggest mirages in the history of science.Marcus Chown is the author of ?The Universe Next Door', published byHeadline (7.99)-- Many are stubborn in pursuit of the path they have chosen, few inpursuit of the goal. - Friedrich Nietzschehttp://annevolve.sourceforge.net is what I'm into nowadays. === =In sci.math, G. A. Edgaron Wed, 07 Jan 2004 11:15:26 -0500> ||x+y|| +||x-y|| = 2||x||+2||y||there should be some squares in this, of course.> Why? Did he say it was for a box kite?*evil grin*-- #191, ewill3@earthlink.net -- someone had to say it...It's still legal to go .sigless. === What if all that we know about X is that it is a real>normed space and:>for all x, y in X, for all t in reals:>||x|| = ||y|| -> ||x + t(x + y)|| = ||y + t(x + y)|| >Informally, the structure sais If x and y are of the same norm, >then x and y are the same distance away from each point on the line>passing through 0 - orthogonal to the segment passing through >x and y. >It is fairly easy to show that the kite equation above holds>for R^n. But I have not found it easy to show either of the two >general real normed space X. Am I missing something easy?it will be possible to have eigenvalues exp((+/-) i theta)where theta is not a rational multiple of pi. If w = a + i b(with a and b in R^2, corresponding to vectors in V) is an eigenvector of the matrix for eigenvalue exp(-i theta), we have T a = cos(theta) a + sin(theta) b and T b = - sin(theta) a + cos(theta) b.Moreover, T^n a = cos(n theta) a + sin(n theta) b, and so ||cos(n theta) a + sin(n theta) b|| = ||a||. But the vectors[cos(n theta), sin(n theta)] are dense in the unit circle, so weactually have ||cos(t) a + sin(t) b|| = ||a|| for all t. And thisimplies that the mapping [x,y] -> x a + y b is an isometry of R^2onto V. In particular we have the parallelogram law in V. But since V was an arbitrary two-dimensional subspace of X, you ?dthat the parallelogram law holds in X, which means X can be madeinto an inner product space.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i07Jnfb18522; === Note: Perhaps I could just as well substitue real with complex,>below.>If X is a real normed space, we know it has>at least a bit of structure if the parallelogram equation:>||x+y|| +||x-y|| = 2||x||+2||y|| holds; i.e. if it is>possible to de?e a scalar product and speak of orthogonal>vectors, etc.Here is another type of structure...>What if all that we know about X is that it is a real>normed space and:for all x, y in X, for all t in reals:>||x|| = ||y|| -> ||x + t(x + y)|| = ||y + t(x + y)|| Informally, the structure sais If x and y are of the same norm, >then x and y are the same distance away from each point on the line>passing through 0 - orthogonal to the segment passing through >x and y. It is fairly easy to show that the kite equation above holds>for R^n. But I have not found it easy to show either of the two >general real normed space X. Am I missing something easy?c = {(t_n): t_n in R, (t_n) converges }, || ||_c = lim n->infty ||tn||. === c = {(t_n): t_n in R, (t_n) converges }, >|| ||_c = lim n->infty ||tn||. This is not a norm: there are too many vectors with ||t||_c = 0.And if you mod out those vectors, you just get R, which isRobert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === =This may sound simple, but it's not my ?ld of expertise. I wonder ifanyone can help me create a unique number from two others.Let's say I have two columns of numbers:col_A col_B 1 4 2 2 3 3 4 1... but I know I won't get true duplicates such as:col_A col_B 1 4 1 4How can I create a unique number from the two that will not beproduced by another pair?The best I've come up with so far is adding a prime to the secondcolumn, but I'm not sure if this produces unique results every time:a * (b+3)Am I just working to get to the point where it's highly unlikely I'llget the same ' result, or is this possible?thanks for any suggestions === Let's say I have two columns of numbers:col_A col_B> 1 4> 2 2> 3 3> 4 1... but I know I won't get true duplicates such as:> col_A col_B> 1 4> 1 4How can I create a unique number from the two that will not be> produced by another pair?If you know that both numbers are positive integers, thenthe following formula will work.f(A,B) = (A+B-1)*(A+B-2)/2 + BIn fact, this is a bijection between ordered pairs of positiveintegers and positive integers. === =On 7 Jan 2004 21:01:21 -0800, webmaster@elken.com (Chumpmeister)>How can I create a unique number from the two that will not be>produced by another pair?The classic method for this is through use of the chinese remaindertheorem. Here's an example using three numbers.w = a * x + b * y + c * z where x, y, and z are the numbers to beencoded and a, b, and c are produced as follows:Let p, q, and r be three mutually prime numbers.Let m = p * q * r i * inv( m / p % p , p ) where inv(x,y) represents themodular inverse of x modulo y j = inv( m / q % q , q ) k = inv( m / r % r , r ) a = i * m / p b = j * m / q c = k * m / rif w = a * x + b * y + c * zthen x = w % p y = w % q z = w % rhttp://home.rochester.rr.com/jbxroads/interests/sci.crypt/old _trunk/CHINREMT.TXTprovides a more extensive example, using four numbers.>This may sound simple, but it's not my ?ld of expertise.Find the required modular inverses by using the modular inverse appletat: http://www.ryerson.ca/~dgrimshahttp://www.ryerson.ca/~ dgrimsha/courses/cps530/applets/gcd/inversepowerapplet.htmlis the speci? page for the applet (the URL for which may not surviveposting)John Bailey === =This may sound simple, but it's not my ?ld of expertise. I wonder if> anyone can help me create a unique number from two others.Let's say I have two columns of numbers:col_A col_B> 1 4> 2 2> 3 3> 4 1... but I know I won't get true duplicates such as:> col_A col_B> 1 4> 1 4How can I create a unique number from the two that will not be> produced by another pair?The best I've come up with so far is adding a prime to the second> column, but I'm not sure if this produces unique results every time:a * (b+3)Am I just working to get to the point where it's highly unlikely I'll> get the same result, or is this possible?thanks for any suggestionsIf you know the maximum possible values of all a,b, and a>=0, b>=0 thenselect c > max(col_A,col_B). d=a+c*b is then unique.For c=2^n you can avoid the multiplication and use logical + shiftoperations to build the binary image d.Hugo Pfoertner === This may sound simple, but it's not my ?ld of expertise. I wonder if> anyone can help me create a unique number from two others. Let's say I have two columns of numbers: col_A col_B> 1 4> 2 2> 3 3> 4 1 ... but I know I won't get true duplicates such as:> col_A col_B> 1 4> 1 4 How can I create a unique number from the two that will not be> produced by another pair?One way is the following:if you have two integers i and j you can associate them with (2^i)*(3^j)(That is 2 to the i-th power times 3 to the j-th power.Another way is as follows:If your two integers are between 0 and m then let b = m+1 and associate withi and j the integer i + j*b.For example:1 4 -> (2^1)*(3^4)= 1624 1 ->(2^4)*(3^1)= 48Using the second method if the largest number you will be using is, say m =10, the b=m+1 = 11 and we have, for example:1 4 ->1 + 4*11 = 454 1-> 4 + 1*11= 15The ?st method is justi?d by unique factorization of primes (2 and 3 areprimes).The second method is thinking of the pair i j as being the base brepresentation of a number which in decimal notation is given by i + j*b--Edwin Clark === =Woops.. I'm sorry that I neglected to mention I was trying to arrive at an Integer> This may sound simple, but it's not my ?ld of expertise. I wonder if> anyone can help me create a unique number from two others. Let's say I have two columns of numbers: col_A col_B> 1 4> 2 2> 3 3> 4 1 ... but I know I won't get true duplicates such as:> col_A col_B> 1 4> 1 4 How can I create a unique number from the two that will not be> produced by another pair?One way is the following:if you have two integers i and j you can associate them with (2^i)*(3^j)> (That is 2 to the i-th power times 3 to the j-th power.Another way is as follows:If your two integers are between 0 and m then let b = m+1 and associate with> i and j the integer i + j*b.For example:1 4 -> (2^1)*(3^4)= 162> 4 1 ->(2^4)*(3^1)= 48Using the second method if the largest number you will be using is, say m 10, the b=m+1 = 11 and we have, for example:1 4 ->1 + 4*11 = 45> 4 1-> 4 + 1*11= 15The ?st method is justi?d by unique factorization of primes (2 and 3 are> primes).The second method is thinking of the pair i j as being the base b> representation of a number which in decimal notation is given by i + j*b> --Edwin Clark === =thanks very much for the guidance... I'm experimenting with thisalgorithm straight away> This may sound simple, but it's not my ?ld of expertise. I wonder if> anyone can help me create a unique number from two others. Let's say I have two columns of numbers: col_A col_B> 1 4> 2 2> 3 3> 4 1 ... but I know I won't get true duplicates such as:> col_A col_B> 1 4> 1 4 How can I create a unique number from the two that will not be> produced by another pair?One way is the following:if you have two integers i and j you can associate them with (2^i)*(3^j)> (That is 2 to the i-th power times 3 to the j-th power.Another way is as follows:If your two integers are between 0 and m then let b = m+1 and associate with> i and j the integer i + j*b.For example:1 4 -> (2^1)*(3^4)= 162> 4 1 ->(2^4)*(3^1)= 48Using the second method if the largest number you will be using is, say m 10, the b=m+1 = 11 and we have, for example:1 4 ->1 + 4*11 = 45> 4 1-> 4 + 1*11= 15The ?st method is justi?d by unique factorization of primes (2 and 3 are> primes).The second method is thinking of the pair i j as being the base b> representation of a number which in decimal notation is given by i + j*b> --Edwin Clark === The ?st method is justi?d by unique factorization of primes (2 and 3are> primes).Of course, I meant unique factorization of integers as a product ofprimes.--Edwin === =Dear all,I am facing an algorithm design problem for an assignment problem:I am trying to factor a matrix D into the form of two matrices(A@A)U+(B@B)V... where @ is the Kronecker product. A, B are square. U, V arediagonal... I have write D into D1+D2, so now the task is to assign (A@A)Uto D1 and (B@B)V to D2...For example, for 2x2 matrices A=[a1 a2; a3 a4], B=[b1 b2; b3 b4],U=[u1 0 0 0; 0 u2 0 0; 0 0 u3 0; 0 0 0 u4], V=[v1 0 0 0; 0 v2 0 0; 0 0 v3 0;0 0 0 v4],hence,(A@A)U=[ a1^2*u1, a1*a2*u2, a1*a2*u3, a2^2*u4][ a1*a3*u1, a1*a4*u2, a3*a2*u3, a2*a4*u4][ a1*a3*u1, a3*a2*u2, a1*a4*u3, a2*a4*u4][ a3^2*u1, a3*a4*u2, a3*a4*u3, a4^2*u4](B@B)V=[ b1^2*v1, b1*b2*v2, b1*b2*v3, b2^2*v4][ b1*b3*v1, b1*b4*v2, b3*b2*v3, b2*b4*v4][ b1*b3*v1, b3*b2*v2, b1*b4*v3, b2*b4*v4][ b3^2*v1, b3*b4*v2, b3*b4*v3, b4^2*v4]So I just need to match elements in (A@A)U with those elements in D1, andelements in (B@B)V with those elements in D2.For example:a1^2*u1=32=4*4*2b1^2*v1=-4=2*2*(-1)a1*a3*u1=16=4*2* 2b1*b3*v1=2=-1*2*(-1)...(totally there are 28 different equations..., and 4x4=16 unknown variables)So I can try to assign a1=4, b1=2, a3=4, b1=-1, u1=2, v1=-1, etc.The problem is that the assignments are interdependent, one assignment inone equation may be con?g with assignment to the same variable inother equation...This is especially the case when the size of matrices becomes larger...sometimes there are no exact assignment that can be made; I need to seekbest such appromixation as my assignment...Best approximation means that the mean square error of the left hand side ofthe equation and the right hand side should be minimized:(A@A)U+(B@B)V = D1+D2=DI feel such assignment problem is very interesting and I guess it is verygeneral and it should have been researched previously...Can anybody give me some ideas? or some pointers?-Walala === = >Dear all, >I am facing an algorithm design problem for an assignment problem: >I am trying to factor a matrix D into the form of two matrices >(A@A)U+(B@B)V... where @ is the Kronecker product. A, B are square. U, V are >diagonal... I have write D into D1+D2, so now the task is to assign (A@A)U >to D1 and (B@B)V to D2... >For example, for 2x2 matrices A=[a1 a2; a3 a4], B=[b1 b2; b3 b4], >U=[u1 0 0 0; 0 u2 0 0; 0 0 u3 0; 0 0 0 u4], V=[v1 0 0 0; 0 v2 0 0; 0 0 v3 0; >0 0 0 v4], >hence, > >(A@A)U= >[ a1^2*u1, a1*a2*u2, a1*a2*u3, a2^2*u4] >[ a1*a3*u1, a1*a4*u2, a3*a2*u3, a2*a4*u4] >[ a1*a3*u1, a3*a2*u2, a1*a4*u3, a2*a4*u4] >[ a3^2*u1, a3*a4*u2, a3*a4*u3, a4^2*u4] >(B@B)V= >[ b1^2*v1, b1*b2*v2, b1*b2*v3, b2^2*v4] >[ b1*b3*v1, b1*b4*v2, b3*b2*v3, b2*b4*v4] >[ b1*b3*v1, b3*b2*v2, b1*b4*v3, b2*b4*v4] >[ b3^2*v1, b3*b4*v2, b3*b4*v3, b4^2*v4] >So I just need to match elements in (A@A)U with those elements in D1, and >elements in (B@B)V with those elements in D2. >For example: >a1^2*u1=32=4*4*2 >b1^2*v1=-4=2*2*(-1) >a1*a3*u1=16=4*2*2 >b1*b3*v1=2=-1*2*(-1) >... >(totally there are 28 different equations..., and 4x4=16 unknown variables) >So I can try to assign a1=4, b1=2, a3=4, b1=-1, u1=2, v1=-1, etc. >The problem is that the assignments are interdependent, one assignment in >one equation may be con?g with assignment to the same variable in >other equation... > This is especially the case when the size of matrices becomes larger... >sometimes there are no exact assignment that can be made; I need to seek >best such appromixation as my assignment... >Best approximation means that the mean square error of the left hand side of >the equation and the right hand side should be minimized: >(A@A)U+(B@B)V = D1+D2=D >I feel such assignment problem is very interesting and I guess it is very >general and it should have been researched previously... >a reformulation of your previous post, giving up the Kronecker structure in D.as written down here, it makes no sense at all since the additive decomposition of D is arbitrary as is the additive decomoposition of the left hand side,hence without any further restrictions it boils down to given D Frobeniusnorm squared of ( (A*A)V-D) to be minimized with respect to A and V over the reals, a nonlinear least squares problem for which lot of codes exist. again, by arbitrariness of scaling , bounds on V have to be introduced, say ||V||=1, i.e. V(i,i) in {-1,0,1} which again makes it mixed integer nonlinear. hthpeter === a reformulation of your previous post, giving up the Kronecker structurein D.> as written down here, it makes no sense at all since the additivedecomposition> of D is arbitrary as is the additive decomoposition of the left hand side,> hence without any further restrictions it boils down to> given D> Frobeniusnorm squared of ( (A*A)V-D) to be minimized with respect to Aand V over the reals, a nonlinear least squares problem for which lot of> codes exist. again, by arbitrariness of scaling , bounds on V have to be> introduced, say ||V||=1, i.e. V(i,i) in {-1,0,1}> which again makes it mixed integer nonlinear.> hth> peterDear Peter,is just that was too dif?ult so I relaxed a little bit...Please see the example:a1^2*u1=32=4*4*2b1^2*v1=-4=2*2*(-1)a1*a3*u1=16=4*2* 2b1*b3*v1=2=-1*2*(-1)...(totally there are 28 different equations..., and 4x4=16 unknown variables)So I can try to assign a1=4, b1=2, a3=4, b1=-1, u1=2, v1=-1, etc.Now the problem turned into an optimized assignment problem:how to assign those variables to the factors of 4*4*2, 2*2*(-1), 4*2*2,since there are 28 equations and 16 unknowns, there might be some con?n assignment somewhere, now we need to ?d the best approximation of suchassignment...I hope that some search algorithm can be devised... any more ideas?-Walala === Consider a liquid-?led sphere, surrounded by some sort of solid membrane>in space holding the liquid together, with no gravity. Assume the inner>(liquid) sphere has radius r_1, say, and the outer annulus has a radius r_2,>so that the total radius of the sphere is r_1 + r_2. As we allow r_1 to>increase, the mass increases proportionally to r_1^3 and the surface area>proportionally to r_1^2. But in order to keep the structural integrity (ie>so the bubble doesn't just break apart), the membrane thickness r_2 needs to>increase proportionally to r_1^a, where a is some power. My problem is, I'm>not sure what a should be. I'm ' guessing 2 or 3 but I'm useless at mechanics.In order to keep from boiling off, the liquid must be under some pressureP (independent of the volume of liquid) which presumably must be supplied by tension in the membrane. In order for a membrane material of constant tensile strength (force per unit area) to withstand this tension, I get that the thickness must be proportional to the radius. Think of it thisway: consider a small circular patch of membrane of radius d. The pressure produces a force proportional to d^2 on this patch. A tangentialforce F at the boundary has component approximately proportional to Fd/rin the direction of that presssure force, so each unit length of the boundary of the patch has to provide a force proportional to r. Sincethe allowed force per unit area is constant, the thickness must be proportional to r.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === =In sci.math, Troncon Wed, 7 Jan 2004 16:35:48 -0000:> Dear allI've got a mechanics-type problem at the moment - ' don't worry, it really> isn't for homework.Consider a liquid-?led sphere, surrounded by some sort of solid membrane> in space holding the liquid together, with no gravity. Assume the inner> (liquid) sphere has radius r_1, say, and the outer annulus has a radius r_2,> so that the total radius of the sphere is r_1 + r_2. As we allow r_1 to> increase, the mass increases proportionally to r_1^3 and the surface area> proportionally to r_1^2. But in order to keep the structural integrity (ie> so the bubble doesn't just break apart), the membrane thickness r_2 needs to> increase proportionally to r_1^a, where a is some power. My problem is, I'm> not sure what a should be. I'm guessing 2 or 3 but I'm useless ' at mechanics.Anyway, if anyone could help, it would be very much appreciated.Tronc> If one assumes that the membrane must always be a constant thickness(r_2 - r_1 = t) then one can do the following.The liquid volume of course will be equal to 4/3 * pi * r_1^3.The volume of ? membrane will be 4/3 * pi * (r_1 + t)^3.The difference gives one4/3 * pi * (3 * r_1^2 * t + 3 * r_1 * t^2 + t^3)which would be the amount of material one would need toconstruct that shell.For t << r_1, one can simply use the spherical shell approximation4*pi*r_1^2*twhich falls into one's lap anyway from the above difference,by assuming t^2 and t^3 are approximately equal to 0.A similar analysis can be done if one assumes the membranemust be a constant *ratio* (r_2 / r_1 = t); I'll leave thedetails to the interested reader.Other analyses are possible, depending on how one wants to set upthe problem. As a rough rule of thumb the weight of the fuel tankis about 3% of the weight of the fuel, IIRC. The difference inthis case will depend on the materials used for the tank and forthe fuel.HTH-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Knowing how to use the calculator is more important than all the bells and> whistles. You could easily get by with just the basic trigonometric stuff;> the rest is really just showoff. My calculator does not have e.g. symbolic> manipulation, and I've never missed it. Even my calculator (HP 32S) has some> bits that I've never touched.The real key to success is learning how to exploit your calculator to its> maximum potential.Just my two cents...-Michael.If your going to have a lot of stat classes the ti- 83 is the way togo. The 89 is nice butthere are things that I wont even use... theti-83 is supported a lot in textbooks. the choice is yours. i mean the89 has the stats stuff too but its really hard to get into while the83 is more simplistic. === The math course I'm taking right now recommends buying the TI-83 ($100),>but for $40 more I could get the TI-89 ($140). I'm assuming that I will>be using the calculator for future higher-level math courses. Not necessarily a good assumption. Typical high-school and some universitymath classes in the US involve lots of symbolic computation, graphing,and number crunching, but only up through roughly the end of calculus;if you're already near that stage, you may never use a calculator againin your higher-level math courses. If you can ?d a calculator which,can, for example, prove the set of rational numbers to be countable,do let me know; it could be very useful.>Is there>anything that the TI-83 can do that the TI-89 can't? Yes: it can be used on some exams where the TI-89 is _calculatora non grata_.Your instructors may very well prohibit the ?89 on the grounds that theywant to test _you_, not your machine. Why don't you ask the people in your school? Since the answers candepend on a lot of local variables, they may be the only ones in aposition to answer your question.davePS -- The TI machines do a good job at what they claim to do. But you'llneed to learn to use them! Study and save the manual. Also, remember toengrave your name on them somewhere; they tend to develop legs... === =Dave Rusin scribbled the following:>The math course I'm taking right now recommends buying the TI-83 ($100),>but for $40 more I could get the TI-89 ($140). I'm assuming that I will>be using the calculator for future higher-level math courses. > Not necessarily a good assumption. Typical high-school and some university> math classes in the US involve lots of symbolic computation, graphing,> and number crunching, but only up through roughly the end of calculus;> if you're already near that stage, you may never use a calculator again> in your higher-level math courses. If you can ?d a calculator which,> can, for example, prove the set of rational numbers to be countable,> do let me know; it could be very useful.Don't hold your breath. Last I checked, none of the calculators on themarket supported the concept of in?ite sets at all, countable oruncountable.-- /-- Joona Palaste (palaste@cc.helsinki.? ------------- Finland ---------- http://www.helsinki.?~palaste --------------------- rules! --------/We sorcerers don't like to eat our words, so to say. - Sparrowhawk === The math course I'm taking right now recommends buying the TI-83 ($100),>but for $40 more I could get the TI-89 ($140). I'm assuming that I will>be using the calculator for future higher-level math courses. > Not necessarily a good assumption. Typical high-school and some university> math classes in the US involve lots of symbolic computation, graphing,> and number crunching, but only up through roughly the end of calculus;> if you're already near that stage, you may never use a calculator again> in your higher-level math courses. If you can ?d a calculator which,> can, for example, prove the set of rational numbers to be countable,> do let me know; it could be very useful.>Is there>anything that the TI-83 can do that the TI-89 can't? > Yes: it can be used on some exams where the TI-89 is _calculatora non grata_.> Your instructors may very well prohibit the ?89 on the grounds that they> want to test _you_, not your machine. Why don't you ask the people in your school? Since the answers can> depend on a lot of local variables, they may be the only ones in a> position to answer your question.davePS -- The TI machines do a good job at what they claim to do. But you'll> need to learn to use them! Study and save the manual. Also, remember to> engrave your name on them somewhere; they tend to develop legs...didn't occur to me.) === I'm looking for a little advice on ?ding a Real Analysis text that> includes at least brief discussions between the de?itions that> motivates why such-and-such a de?ition is being used[snip]I think the only solution is to get all the standard reference books(Rudin, Royden, Zygmund & Wheeden, Apostol, etc) and see what theyeach say about whatever topic you are interested in. I recall seeinggood, but by that time I had caught on to the point where I felt Ididn't need it (probably a mistake!). Probably a lot of these abstractconcepts only make sense after working with them for a (long) while.Consider the Dominated Convergence Thm and friends Fatou's Lemma andMonotone Convergence Thm. I have the idea that the whole reason for people wanting the Lebesgue integral is to be able to interchange limitoperations and integration, which you can't prove you can do oftenenough with Riemann. I ' don't think any book I read stated thatas a motivation though.--Jeff === =You could perhaps try ?Analysis in Euclidean Space' by KennethHoffman. I dont know if it would be to your liking, but that was theonly book i was able to read after looking at all those real analysistexts.once you are comfortable, i guess Rudin comes handy. > I'm looking for a little advice on ?ding a Real Analysis text that> includes at least brief discussions between the de?itions that> motivates why such-and-such a de?ition is being used> [snip]I think the only solution is to get all the standard reference books> (Rudin, Royden, Zygmund & Wheeden, Apostol, etc) and see what they> each say about whatever topic you are interested in. I recall seeing> good, but by that time I had caught on to the point where I felt I> didn't need it (probably a mistake!). Probably a lot of these abstract> concepts only make sense after working with them for a (long) while.Consider the Dominated Convergence Thm and friends Fatou's Lemma and> Monotone Convergence Thm. I have the idea that the whole reason for > people wanting the Lebesgue integral is to be able to interchange limit> operations and integration, which you can't prove you can do often> enough with Riemann. I don't think any book I read stated that> as a motivation though.--Jeff === =PS One can also write for the loop quantum gravity concepts(Quantum of Volume) = (Quantum of Area)^3/2Quantum of Area ~ Bekenstein BITas in Wheeler's IT FROM BIT.Witten's alpha' = (Quantum of Area)/hcThis is ' how string theory relates to loop quantum gravity.The Quantum of Area is a stringy edge with discrete links in the spin network./zpf = (Quantum of Area)^-1[(Quantum of Area)^3/2|Vacuum Coherence|^2 - 1]All this can be written in terms of Witten's alpha' with Vacuum Coherence as the keymetric engineering control parameter.My equations are consistent with G. Volovik's The Universe in a Helium Dropand with G. Chapline's general idea although my dynamical vacuum instability mechanism is much simpler.Critics such as David Spergel of Princeton, who has worked on the WMAPdata, point out that a lot of things need to be wrong for observationsof the Universe to make sense without dark energy. It is better to beslightly wrong about a number of things than incorporate a parameterwhich is 10 followed by 123 zeroes bigger than theory predicts!counters Sarkar.This is a reference to the fact that our best theory of physics -quantum theory - predicts a value for the energy density of the darkenergy which is 10-followed-by-123-zeroes times bigger than observed.This spectacular discrepancy has been described by Nobel laureate StevenWeinberg of the University of Texas as the worst failure of anorder-of-magnitude estimate in the history of science.I have solved this problem with the notion of Vacuum Coherence missing from the above.Chapline has also realized this. Volovik has also a similar explanation using an analogywith condensed matter physics. Susskind also invokes supersymmetry to get / = 0, butthen must explain why the supersymmetry is broken in the actual Universe we are in.He invokes WAP. Einstein's / is a limiting case of the local /zpf ?ld on the large scale./ = 0 is only for a Universe in Antony Valentini's sub-quantal heat death with signal locality.Signal nonlocality in our sub-quantal non-equilibrium Universe impacts on Susskind's ideas for information loss/recovery in theevaporating event horizon of a black hole. === =but he has not answered that.I conclude that my conjecture was right.Franz === =machine,> but he has not answered that.> I conclude that my conjecture was right.I boobed, as Dave Rusin pointed out to me in a private letter.write-only machine.As he still has not answered, I still conclude that my conjecture was right.Franz === =but he has not answered that.> I conclude that my conjecture was right.FranzI don't see you advancing any ideas, Sir.Perhaps some conjectures on Physics from yourquarter rather than dissing those whocontinue to try to ?ure things out?I remember reading one interesting ideafrom you several (8?) years ago but itwas miniscule in scope. Come on! You stillgot it in you to wonder and there's moredata all the time. The more we know theless we know we know. Step up to theplate!! Is anything jelling in that worthycranium or is it all ossifying? :-)John === =My teacher assigns me some challenge problems to solve. But it seemsto be too hard for me. Can someone give me some ideas. Thanx !This is one of them :> 1)The function f(x) has f ?'(x) > 0 for all x >0 and the graph is asymptotic to y = ax + b as x. > Show that f(x) - ax - b has negative derivative for all x > 0,and that f(x) > ax + b for all x >0.[/quote:b29fe8cde7]--------------------------------------- ------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **---------------------------------------------------------- http://www.usenet.com === =http://www.cnn.com/2004/US/West/01/ 07/math.mystery.ap/index.html === =Michael Varney tipped off:|> http://www.cnn.com/2004/US/West/01/07/math.mystery.ap/ index.htmlThe basics is that we can show the surface of a 2-sphere (surface of aball) is simply connected but a 2-torus is not (surface of a doughnut ortire tube). Poincare conjectured it for a 3-sphere (the set of pointsin four dimensional space at unit distance from the origin).More detail is here:http://www.claymath.org/Millennium_Prize_Problems/ Poincare_Conjecture/-- cu,Brucedrift wave turbulence: http://www.rzg.mpg.de/~bds/ === Michael Varney tipped off:|> http://www.cnn.com/2004/US/West/01/07/math.mystery.ap/ index.htmlThe basics is that we can show the surface of a 2-sphere (surface of a> ball) is simply connected but a 2-torus is not (surface of a doughnut or> tire tube). Poincare conjectured it for a 3-sphere (the set of points> in four dimensional space at unit distance from the origin).> It sounds like you're saying that the conjecture is to prove a 3-sphereis simply connected; that's easy. The conjecture is that the 3-sphereis the *only* simply-connected closed 3-manifold. === =|> |> Michael Varney tipped off:|> |> |> http://www.cnn.com/2004/US/West/01/07/math.mystery.ap/ index.html|> |> The basics is that we can show the surface of a 2-sphere (surface of a|> ball) is simply connected but a 2-torus is not (surface of a doughnut or|> tire tube). Poincare conjectured it for a 3-sphere (the set of points|> in four dimensional space at unit distance from the origin).|> |> It sounds like you're saying that the conjecture is to prove a 3-sphere|> is simply connected; that's easy. The conjecture is that the 3-sphere|> is the *only* simply-connected closed 3-manifold.I got confused by the layman's explanation on the claymath.org link (theclaymath.org after I posted. They talk about setting up an imaginaryrubber band on the surface and then asking if it can be deformedsmoothly onto a point. They did give me the impression this was aquestion on a 3-sphere. So maybe they should re-write theirintroduction to make that clearer. The PDF did make it absolutelyclear: is every simply connected 3-domain homologous to a 3-sphere.|> Whoa! Are you telling us that the surface of a torus can be decomposed |> into two open sets (with respect to the local surface topology of the |> torus). Could you show us how to do that?I didn't say that. Maybe I confused you as much as that layman's linkdid me *g*-- cu,Brucedrift wave turbulence: http://www.rzg.mpg.de/~bds/ === =| |> The basics is that we can show the surface of a 2-sphere (surface of a> |> ball) is simply connected but a 2-torus is not (surface of a doughnut or> |> tire tube). Poincare conjectured it for a 3-sphere (the set of points> |> in four dimensional space at unit distance from the origin).> ||> It sounds like you're saying that the conjecture is to prove a 3-sphere> |> is simply connected; that's easy. The conjecture is that the 3-sphere> |> is the *only* simply-connected closed 3-manifold.I got confused by the layman's explanation on the claymath.org link (the> claymath.org after I posted. They talk about setting up an imaginary> rubber band on the surface and then asking if it can be deformed> smoothly onto a point. They did give me the impression this was a> question on a 3-sphere. So maybe they should re-write their> introduction to make that clearer. Milnor), and it is a little sloppy:If we stretch a rubber band around the surface of an apple, then wecan shrink it down to a point by moving it slowly, without tearing itand without allowing it to leave the surface. On the other hand, if weimagine that the same rubber band has somehow been stretched in theappropriate direction around a doughnut, then there is no way ofshrinking it to a point without breaking either the rubber band or thedoughnut. We say the surface of the apple is simply connected, butthat the surface of the doughnut is not. Poincar, almost a hundredyears ago, knew that a two dimensional sphere is essentiallycharacterized by this property of simple connectivity, and asked thecorresponding question for the three dimensional sphere (the set ofpoints in four dimensional space at unit distance from the origin).This question turned out to be extraordinarily dif?ult, andmathematicians have been struggling with it ever since.It's not wrong, but I can see how it's not clear what thecorresponding question is. I think it's very easy for a layman tomiss the phrase essentially characterized in the statement about howsimple-connectivity basically determines the surface is a 2-sphere. And then the bit about the corresponding question for the threedimensional sphere follows up very quickly on that. So to make things clearer, I would replace essentially characterizedby essentially determined and actually *state* what thecorresponding question is. Something like ...asked the analogousquestion for the three dimensional sphere: is the three dimensionalsphere determined by the property of simple connectivity?of sending in corrections and suggestions. I guess I could post on theClay Math forum.> The PDF did make it absolutely> clear: is every simply connected 3-domain homologous to a 3-sphere.> Instead of homologous you want homeomorphic. There are in factmany homology 3-spheres, 3-manifolds that have the same homology as a3-sphere. One usually doesn't say homologous even in this contexthowever. One usually says have the same homology or something tothat effect. === Michael Varney tipped off:|> http://www.cnn.com/2004/US/West/01/07/math.mystery.ap/ index.htmlThe basics is that we can show the surface of a 2-sphere (surface of a> ball) is simply connected but a 2-torus is not (surface of a doughnut or> tire tube). Poincare conjectured it for a 3-sphere (the set of points> in four dimensional space at unit distance from the origin).Whoa! Are you telling us that the surface of a torus can be decomposed into two open sets (with respect to the local surface topology of the torus). Could you show us how to do that?I also point out that the surface of the torus is acrwise connect. What is different about a torus is that an arc going around the hole cannot be homotopically contracted to a point. The same is not true of a sphere.Bob Kolker === =Michael Varney tipped off:|> http://www.cnn.com/2004/US/West/01/07/math.mystery.ap/ index.htmlThe basics is that we can show the surface of a 2-sphere (surface of a> ball) is simply connected but a 2-torus is not (surface of a doughnut or> tire tube). Poincare conjectured it for a 3-sphere (the set of points> in four dimensional space at unit distance from the origin).Whoa! Are you telling us that the surface of a torus can be decomposed > into two open sets (with respect to the local surface topology of the > torus). Could you show us how to do that?The torus is connected. But not simply connected. These are differentnotions.I also point out that the surface of the torus is acrwise connect. What > is different about a torus is that an arc going around the hole cannot > be homotopically contracted to a point. The same is not true of a sphere.Bob Kolker> === =In Marshall Hall's Theory of Groups, the following conjecture ismentioned as unresolved:Let G be a ?ite group, and let n be a divisor of |G|. If there areexactly n solutions to x^n=1 in G, then these solutions form asubgroup of G.Is it still unresolved? If not, when and by whom was it resolved?---- David === In Marshall Hall's Theory of Groups, the following conjecture is>mentioned as unresolved:Let G be a ?ite group, and let n be a divisor of |G|. If there are>exactly n solutions to x^n=1 in G, then these solutions form a>subgroup of G.Is it still unresolved? If not, when and by whom was it resolved?I don't know. The most recent reference to this problem that I know o? the case when n and |G|/n are coprime, the problem has been reduced toG simple, and so should be solvable using the classi?ation of ?ite simplegroups. Whether it ever was solved, even in that special case, I do notknow. Please post again if you ?d out! Derek Holt.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i086HfQ31717; === =We are working on software that will speak mathematics embedded in a web>page. Our main goal is to make math in web pages accessible to visually>impaired readers for which we've recently been awarded an NSF grant. After>demonstrating an early version of this software to teachers that happened to>have normal sight, they claimed that they could see an additional use in>teaching normally sighted students how math is spoken. This would of course>be most useful in situations where a teacher is not present. Does anyone>know of any work done on this subject within the educational community? Are>there any opinions on the worth of this sort of thing? Any comments are>welcome.Paul Topping>Design Science, Inc.>www.dessci.com>Been there, done that, got the tee shirt.You need to Google a bit === =| I am more interested right now in ?ding an explicit |computation of w1 = gcd((1 + sqrt(-167))/2, 7), i.e., in ?ding|a monic polynomial with constant term 7^k of which w1 is a root.Ok. :-)It appears that Q(sqrt(-167)) has a class group of order 11.Let r stand for (1+sqrt(-167))/2 (which is an algebraic integer,in spite of the 2 in the denominator).r^11 = (-592764018-86559857*r) = (44555-222*r) (-12882-2017*r)and7^11 = (44555-222*r)(44555+222*r).Also (-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11, which isrelatively prime to 7^11.The factor 44555-222*r = 44444 - 111 sqrt(-167) and its conjugate44444 + 111 sqrt(-167) are roots of x^2-88888*x+7^11=0. One GCD ofr with 7 is then (44444-111 sqrt(-167))^(1/11), which is a root ofx^22 - 88888 * x^11 + 7^11 = 0.Keith Ramsay === | I am more interested right now in ?ding an explicit > |computation of w1 = gcd((1 + sqrt(-167))/2, 7), i.e., in ?ding> |a monic polynomial with constant term 7^k of which w1 is a root.Ok. :-)It appears that Q(sqrt(-167)) has a class group of order 11.> Let r stand for (1+sqrt(-167))/2 (which is an algebraic integer,> in spite of the 2 in the denominator).r^11 = (-592764018-86559857*r)> = (44555-222*r) (-12882-2017*r)and7^11 = (44555-222*r)(44555+222*r).Also (-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11, which is> relatively prime to 7^11.The factor 44555-222*r = 44444 - 111 sqrt(-167) and its conjugate> 44444 + 111 sqrt(-167) are roots of x^2-88888*x+7^11=0. One GCD of> r with 7 is then (44444-111 sqrt(-167))^(1/11), which is a root of> x^22 - 88888 * x^11 + 7^11 = 0.Keith Ramsayand could not have carried out the computation myself, though Iknow enough theory to know it can be done. I was guessing theanswer might have been of lower degree. Do you make use ofsome computing package for this? Harris of course has claimed that one of (1 + sqrt(-167))/2 or(1 - sqrt(-167))/2 is coprime to 7. Your computation leads to anexplicit proof that this is not so. Nora B. === | I am more interested right now in ?ding an explicit > |computation of w1 = gcd((1 + sqrt(-167))/2, 7), i.e., in ?ding> |a monic polynomial with constant term 7^k of which w1 is a root.Ok. :-)It appears that Q(sqrt(-167)) has a class group of order 11.> Let r stand for (1+sqrt(-167))/2 (which is an algebraic integer,> in spite of the 2 in the denominator).r^11 = (-592764018-86559857*r)> = (44555-222*r) (-12882-2017*r)and7^11 = (44555-222*r)(44555+222*r).Also (-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11, which is> relatively prime to 7^11.The factor 44555-222*r = 44444 - 111 sqrt(-167) and its conjugate> 44444 + 111 sqrt(-167) are roots of x^2-88888*x+7^11=0. One GCD of> r with 7 is then (44444-111 sqrt(-167))^(1/11), which is a root of> x^22 - 88888 * x^11 + 7^11 = 0.> Nice work, Keith. As I guessed, it's certainly not simple (thoughall those repeated digits are pretty nice).Rick === =|It is not 5.00, instead 9.901475.....e-001 = 0.9901475.....|Does that ?ld have some automatic computations?The inverse symbolic calculator at www.cecm.sfu.casuggests the possibility it is sqrt(50/51). Does theoriginal poster have any idea what it's supposed to becomputing? Try checking to see whether the rest ofthe given digits match sqrt(50/51) too.Keith Ramsay === =|It is not 5.00, instead 9.901475.....e-001 = 0.9901475.....> |Does that ?ld have some automatic computations?The inverse symbolic calculator at www.cecm.sfu.ca> suggests the possibility it is sqrt(50/51). Does the> original poster have any idea what it's supposed to be> computing? Try checking to see whether the rest of> the given digits match sqrt(50/51) too.Keith Ramsaya cad program, it wasn't meant to do anything else but accept it, whati fear is happening is that it is ignoring me and writing it as 1.00but with it's own wierd way of corrupting 1 to a number near1.00........ that sqrt 50/51 thing is quite clever, thanks forthat. anyway i know that the thing doesn't do what i want it to,so i'll give up on it now.thanks for the advice allkieran === =First of all, thank you for your answer.> It is just saying that it is a property (by construction) of the> product bundle that smooth cross sections are preserved. It is a> special case of the embedding of C^{infty}(M_{1}, xi^{1})otimes> C^{infty}(M_{2}, xi^{2}) into C^{infty}(M_{1}x M_{2}, xi^{1}otimes> xi^{2}) using the de?ition of the product bundle.> KarinOk, the fact is thati am not able to relate the standard result:*Let X be a topological space (or a differentiable manyfold)and U = ( U_i )_{i in I} an open covering of X.forall i in I, consider the trivial vector bundle U_{i} x R^{p}and suppose that for each(i,j) in I^{2} such that U_{i} cap U_{j} not equal emptysetis assigned a continuous (or differentiable if it makes sense) applicationf_{ij}:U_{i} cap U_{j} --> GL(R^{p}).In the disjoint union of such trivial bundles, let's glue a point(x,v_{i}) in U_{i} x R^{p} with (y,v_{ j}) in U_{ j} x R^{p}if and only if x = y in U_{i}cap U_{j} and v_{j} = f_{ij}(x)[v_{i}].If, forall (i,j,k) in I^{3} and x in U_{i}cap U_{j}cap U_{k}we have f_{ij}(x) o f_{jk}(x) = f_{ik}(x) = the quotient for such arelation de?es a vector bundle over X.*with the fact that the bundle structure of xi^{1}otimes ... otimes xi^{k}is ------characterized------- by the property that f^{1}otimes ... otimesf^{k}}in C^{infty}(M, xi^{1} otimes ... otimes xi^{k}) whereverf ^ {i} in C^{infty}(M_{i}, xi^{i}).May you help me?Tern === =Dear all,Suppose I have a collection X, consisting of different sets,which are subset of some other set U.The sets in X satisfy this property: for all subsets Y of X, the intersection of all sets in Y is again an element of XDoes this property have some common name?Johan-- web: http://losderover.be:7832/jabber: johan@losderover.be (jabber onder Windows? zie http://exodus.sourceforge.net/) === ....> Suppose I have a collection X, consisting of different sets,> which are subset of some other set U.The sets in X satisfy this property: for all subsets Y of X,> the intersection of all sets in Y is again an element of X> Does this property have some common name?.... A closure system (Moore, 1910). Have you still got access to last month's sci.math thread called generator? On the 22nd December I posted there a message about closure systems and the corresponding closure operations. Let me know if you have trouble ?ding it, and I can send you a copy. Ken Pledger. === =* Johan Vervloet> Dear all,Suppose I have a collection X, consisting of different sets,> which are subset of some other set U.The sets in X satisfy this property: for all subsets Y of X,> the intersection of all sets in Y is again an element of X> Does this property have some common name?What comes to mind is that X is _closed on intersection_.-- Jon Haugsand === * Johan Vervloet> Suppose I have a collection X, consisting of different sets,> which are subset of some other set U. The sets in X satisfy this property:> for all subsets Y of X,> the intersection of all sets in Y is again an element of X Does this property have some common name?> What comes to mind is that X is _closed on intersection_.>To be more speci?, X is closed for arbitrary intersections.For example, the closed (different sense of closed) subsets of atopologically space are closed under arbitrary intersections while theopen subsets are closed under (?ite) intersection. === I have the following question:>Does the most slowly diverging seriesexist?>Let an and bn (n:integer) be series of real numbers.>If lim an and bn --> in?ite (n-->in?ite) and lim an/bn --> 0>(n-->in?ite)>then let's call an diverges slower than bn and designate it with an<Is there any lower limit of these series?>I think such lower limit doesn't exist intuitively but is it right?Let's answer ' your given subject ?st.No, there is no most slowly diverging function.Let f(n) be a diverging function.>f(n) - 1 is also a diverging function which is less than f(x).But not less in the sense de?ed.>Oh, so you want a function where more slowly means where the derivatives >aren't equal.How about f(n)/2 ?Still not in the sense de?ed. Did you read the question?Mea culpa. My apologies. No, I did not catch that the OP was de?ingg n0, g(n) < f(n).So all the stuff I said about sums is irrelevant (to the OPs request),but that about log and log* is suf?ient for the OPs de?ition and to show that there is no smallest diverging function (er.. not show but give examples of increasingly smaller ones n a way that can never end).Mitch Harris === =On Thu, 08 Jan 2004 11:05:29 +0100, Mitch Harris>I have the following question:>Does the most slowly diverging seriesexist?>Let an and bn (n:integer) be series of real numbers.>If lim an and bn --> in?ite (n-->in?ite) and lim an/bn --> 0>(n-->in?ite)>then let's call an diverges slower than bn and designate it with an<Is there any lower limit of these series?>I think such lower limit doesn't exist intuitively but is it right?>Let's answer your given subject ?st.>No, there is no most slowly diverging function.>Let f(n) be a diverging function.>f(n) - 1 is also a diverging function which is less than f(x).But not less in the sense de?ed.>Oh, so you want a function where more slowly means where the derivatives >aren't equal.>How about f(n)/2 ?Still not in the sense de?ed. Did you read the question?Mea culpa. My apologies. No, I did not catch that the OP was de?ing>gmaking from a few key words just to be pedantic: exists n0, for all n >n0, g(n) < f(n).So all the stuff I said about sums is irrelevant (to the OPs request),>but that about log and log* is suf?ient for the OPs de?ition and to >show that there is no smallest diverging function I don't see how it does that.>(er.. not show but >give examples of increasingly smaller ones n a way that can never end).Yes, it does that. But an ordered set can contain an in?ite decreasing sequence and also a smallest element.Mitch HarrisDavid C. Ullrich === On Thu, 08 Jan 2004 11:05:29 +0100, Mitch Harris So all the stuff I said about sums is irrelevant (to the OPs request),>but that about log and log* is suf?ient for the OPs de?ition and to >show that there is no smallest diverging function I don't see how it does that.>(er.. not show but >give examples of increasingly smaller ones n a way that can never end).Yes, it does that. But an ordered set can contain an in?ite > decreasing sequence and also a smallest element.Ah, yes. I see now the problem. We need to show that there is no bottom.I think the OPs solution does that: assume you have a bottom, call it f,but show you can construct a function g that diverges but is less than f. But maybe the details about log are missing.Convolving the ideas of divergence and order, let diverging functionsbe those in omega(1), i.e. f such that lim n->oo 1/f(n) = 0. so thenyou'd have to show that, no matter what f is, as long as it isdivergent, then lim (log f)/f = 0. The classic CS solution is to ' invokel'Hopital's rule (classic CS means only consider nice functions thatdon't mess up the assumptions). so lim (log f)/f = lim f'/(f ' f') =lim 1/f = 0.Mitch Harris === =On Thu, 08 Jan 2004 15:15:52 +0100, Mitch Harris> On Thu, 08 Jan 2004 11:05:29 +0100, Mitch Harris>So all the stuff I said about sums is irrelevant (to the OPs request),>but that about log and log* is suf?ient for the OPs de?ition and to >show that there is no smallest diverging function I don't see how it does that.> >(er.. not show but >give examples of increasingly smaller ones n a way that can never end).Yes, it does that. But an ordered set can contain an in?ite > decreasing sequence and also a smallest element.Ah, yes. I see now the problem. We need to show that there is no bottom.I think the OPs solution does that: assume you have a bottom, call it f,>but show you can construct a function g that diverges but is less than >f. But maybe the details about log are missing.The missing details are very simple. There _is_ a subtle point: Itcould happen that f(nn) <= 0 for some n even though lim f(n) =+in?ity. So instead of talking about log(f(n)) if we're careful wesay let g be such that g(n) = log(f(n)) for all n greater thansome N. Below I'll not worry about this and just say log(f(n)).>Convolving the ideas of divergence and order, let diverging functions>be those in omega(1), i.e. f such that lim n->oo 1/f(n) = 0. so then>you'd have to show that, no matter what f is, as long as it is>divergent, then lim (log f)/f = 0. I really think that diverging is the wrong word for this, but it'sthe word that's used by the OP, ' so let's call it that. Yes, youhave to show that lim log(f)/f = 0.>The classic CS solution is to invoke>l'Hopital's ' rule (classic CS means only consider nice functions that>don't mess up the assumptions). so lim (log f)/f = lim f'/(f ' f') lim 1/f = 0.This is not a great way to do it, because it _doesn't_ apply to alldivergent sequences f(n). Also because it's trivial to give anargument that _does_ apply: the fact that lim log(x)/x = 0 (asx -> +in?ity) shows immediately that lim log(f(n))/f(n) = 0for any f(n) which -> +in?ity.If you want the actual detailed proof: Let eps > 0. The factthat log(x)/x -> 0 as x -> in?ity shows that there existsA with log(x)/x < eps for all x > A. The fact that f(n) -> in?ityshows that there exists N such that f(n) > A for all n > N.Hence log(f(n))/f(n) < eps for all n > N, qed.That's all the missing details - nothing to it, really.>Mitch HarrisDavid C. Ullrich === On Thu, 08 Jan 2004 15:15:52 +0100, Mitch HarrisThe classic CS solution is to invoke>l'Hopital's rule (classic CS means only ' consider nice functions that>don't mess up the assumptions). so lim (log f)/f = lim f'/(f f') lim 1/f = 0.This is not a ' great way to do it, because it _doesn't_ apply to all>divergent sequences f(n). Understood.>Also because it's trivial to give an>argument that _does_ apply: the fact that lim log(x)/x = 0 (as>x -> +in?ity) shows immediately that lim log(f(n))/f(n) = 0>for any f(n) which -> +in?ity.If you want the actual detailed proof: Let eps > 0. The fact>that log(x)/x -> 0 as x -> in?ity shows that there exists>A with log(x)/x < eps for all x > A. The fact that f(n) -> in?ity>shows that there exists N such that f(n) > A for all n > N.>Hence log(f(n))/f(n) < eps for all n > N, qed.That's all the missing details - nothing to it, really.Yes, that's good. (and lim log(x)/x = 0 by l'Hopital if there were any doubts!)Mitch Harris === =Sorry for using inappropriate terminology.I noticed that non-existence of the most slowly diverging sequence could beeasily demonstrated using reduction to absurdityThat is:Assume there be the sequence pn which most slowly diverges.Then we make log(pn).lim (log(pn)/pn) ----->zero (n--->in?ite)So log(pn) is a slower sequence than pn.This is contradictory to the previous assumption.So there is not the most slowly diverging sequence.But I have another quetion about the following value related to the abovesequence while I thought about the above.Let fn(x) be fn(x) = log(log(log(....log(x))..) n times log nesting.What is the value of the following suppose an be a diverging sequence?lim fn(an) =? (n--->in?ite)Is it 0?or another value?Does it depend on the sequence an?G.C. I have the following question:> Does the most slowly diverging seriesexist?> Let an and bn (n:integer) be series of real numbers.> If lim an and bn --> in?ite (n-->in?ite) and lim an/bn --> 0> (n-->in?ite)> then let's call an diverges slower than bn and designate it withan< Is there any lower limit of these series?> I think such lower limit doesn't exist intuitively but is it right?> G.C> So you are talking about sequences, not series, strictly speaking. -- > G. A. Edgarhttp://www.math.ohio-state.edu/~edgar/ === =On Thu, 8 Jan 2004 12:13:44 +0900, Georg Cantor>Sorry for using inappropriate terminology.>I noticed that non-existence of the most slowly diverging sequence could be>easily demonstrated using reduction to absurdity>That is:>Assume there be the sequence pn which most slowly diverges.>Then we make log(pn).>lim (log(pn)/pn) ----->zero (n--->in?ite)>So log(pn) is a slower sequence than pn.>This is contradictory to the previous assumption.>So there is not the most slowly diverging sequence.This can be very simply stated as a direct proof of existence for amore slowly diverging sequence given any diverging sequence. You canalso do it without the log(). Show that there exists a function f(n)such that lim(f(a_n)/a_n) = 0 and then show that if a_n diverges, it'spossible that f(a_n) also diverges.>But I have another quetion about the following value related to the above>sequence while I thought about the above.>Let fn(x) be> fn(x) = log(log(log(....log(x))..) n times log nesting.>What is the value of the following suppose an be a diverging sequence?>lim fn(an) =? (n--->in?ite)>Is it 0?or another value?Any ?ite term (other than 0) under repeated logarithms will at somepoint become less than unity, then negative, then complex. That'swhere the fun begins. The complex logarithm is an in?itelymany-valued function, but if we limit to the primary value then therepeated logarithms will converge to about (0.31813, 1.33724). So anynon-zero terms in the sequence will converge to this value as n->oo,producing a converging sequence.Note that this does not mean the existence of a most-slowly divergentfunction since any diverging sequence under ?itely many logarithmswill still diverge faster than the next iteration. === Assume there be the sequence pn which most slowly diverges.>Then we make log(pn).>lim (log(pn)/pn) ----->zero (n--->in?ite)>So log(pn) is a slower sequence than pn.>This is contradictory to the previous assumption.>So there is not the most slowly diverging sequence.>Any ?ite term (other than 0) under repeated logarithms will at some>point become less than unity, then negative, then complex. ?Bout time someone pointed this out!Since we're apparently talking about sequences which go to in?ity,there's no harm making the terms of the new sequence beq_n = max(1,log(p_n)) so that the real logarithm is always de?ed.Of course that means that as the sequences go by you start to seemore and more initial 1s. That suggests alternative explanationsas to why there is no slowest-growing sequence: you can alwaysde?e a new sequence with q_n = inf { p_m, m >= n } to get asmaller divergent sequence which is now monotonic, and then withthings like q_n = inf { p_m, m >= n^2 } to get a more slowly-growingsequence. That is, you can consider sets of sequences like this: 1 2 3 4 5 6 7 8 9 10 ... 1 2 2 3 3 3 4 4 4 4 ... 1 2 2 2 2 3 3 3 3 3 ...etc. And of course you can diagonalize and so on.For some reason, people usually play this game the other way: theylook for the _fastest_-growing sequences. Look for Ackerman functions.(If { A_n } is such a rapidly-growing sequence, you can de?e aslowly-growing sequence by e.g. p_n = sup { m ; A_m < n }: roughly speaking you take A_1 1's followed by A_2 2's and so on.)dave === Since we're apparently talking about sequences which go to in?ity,>there's no harm making the terms of the new sequence be>q_n = max(1,log(p_n)) so that the real logarithm is always de?ed.But what happens when the sequence looks something like:a_n = n + (1/n) * i?Clearly lim(a_n) = lim(n) + lim(i/n) = +inf. Maybe if you change thatto q_n = max(1, log(Re(p_n))) it should work on fully complexsequences but now it gets trickier to prove that lim(q_n/p_n) = 0. === Assume there be the sequence pn which most slowly diverges.>Then we make log(pn).>lim (log(pn)/pn) ----->zero (n--->in?ite)>So log(pn) is a slower sequence than pn.>This is contradictory to the previous assumption.>So there is not the most slowly diverging sequence.Any ?ite term (other than 0) under repeated logarithms will at some>point become less than unity, then negative, then complex. ?Bout time someone pointed this out!Since we're apparently talking about sequences which go to in?ity,>there's no harm making the terms of the new sequence be>q_n = max(1,log(p_n)) so that the real logarithm is always de?ed.>Of course that means that as the sequences go by you start to see>more and more initial 1s. That suggests alternative explanations>as to why there is no slowest-growing sequence: you can always>de?e a new sequence with q_n = inf { p_m, m >= n } to get a>smaller divergent sequence which is now monotonic, and then with>things like q_n = inf { p_m, m >= n^2 } to get a more slowly-growing>sequence. Neither of those is necessarily more slowly-growing (in the sense that the ratio should tend to 0.) This is an advantage that log(p_n)has (with appropriate adjustments for unde?ed logarithms...)it _is_ more slowly growing than p_n.In fact I think it's clear you can't ' do it by composition on theright: If a_n is any sequence of positive integers tending toin?ity then there exists p_n -> in?ity such that if q_n = p_{a_n}then q_n/p_n -> 1.>That is, you can consider sets of sequences like this:> 1 2 3 4 5 6 7 8 9 10 ...> 1 2 2 3 3 3 4 4 4 4 ...> 1 2 2 2 2 3 3 3 3 3 ...>etc. And of course you can diagonalize and so on.For some reason, people usually play this game the other way: they>look for the _fastest_-growing sequences. I imagine that's an American thing - bigger is better and all that...>Look for Ackerman functions.>(If { A_n } is such a rapidly-growing sequence, you can de?e a>slowly-growing sequence by e.g. p_n = sup { m ; A_m < n }: roughly >speaking you take A_1 1's followed by A_2 2's and so on.)daveDavid C. Ullrich === =scattering. The basis function used to expressed the acoustic wave includeda term with cos and sin, where cos is put over sin as a vector but withoutthe parentesis.So, my question is: What does the notation cos over sin (as seen below)means? Note that this is written with no parenthesis what so ever.cossinA.H.8ckansson === =My guess is that this indicates that the use of cos or sin in equation isarbitary within the context that it is used. Thiscan be the case when modelling waves since it is often the case that only thepeak amplitudes and relative phase shifts areare of interest.Have a look and see if any results obtained through use of the equation aredependent on the choice.Dan.> scattering. The basis function used to expressed the acoustic wave included> a term with cos and sin, where cos is put over sin as a vector but without> the parentesis. So, my question is: What does the notation cos over sin (as seen below)> means? Note that this is written with no parenthesis what so ever. cos> sin> A.H.8ckansson === So, my question is: What does the notation cos over sin (as seen below)> means? Note that this is written with no parenthesis what so ever. cos> sin>Usually x over y means x/y orx-ybut your cos over sin has no line. Nor was immediate context included.To puzzle something like that, would you quote the line it's in?What happen to the arguments for cos & sin?X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i088sio09455; === c = {(t_n): t_n in R, (t_n) converges }, >|| ||_c = lim n->infty ||tn||. This is not a norm: there are too many vectors with ||t||_c = 0.>And if you mod out those vectors, you just get R, which isWoops, I was setting sup = lim in the back of my mind.And if we mod out out by requiring the sequences to bestictly increasing or decreasing, we get something that is not a vector space- since it would no longer be closed under addition/subtraction: ex. t = (1, 5, 9, ...), s = (-1, 6, 7, ...). >Robert Israel israel@math.ubc.ca>Department of Mathematics http://www.math.ubc.ca/~ israelUniversity of British Columbia >Vancouver, BC, Canada V6T 1Z2 === =I have an asignment for my automata class, where I have to write a Cprogram which simulates A Turing machine that veri?s whether twodecimal numbers written on input track are divisible. Numbers formatis: ?number1/number2='Please help me with algorithm that calculates that! === I have an asignment for my automata class, where I have to write a C> program which simulates A Turing machine that veri?s whether two> decimal numbers written on input track are divisible. Numbers format> is: ?number1/number2='> So would I be correct in assuming that a typical input tape wouldlook like 762/41bbbbbbb...Regardless, a hint would be to ?st write a TM that subtractstwo numbers and use that as a subroutine. You could writea TM routine that computes a mod b by division, but I wouldn'tadvise it.Rickp.s. You might consider asking this question in comp.theory.The signal/noise ratio there is much higher than here,though the traf? volume is much less.X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === =In <8mqjvvkdapqudjc5v0t8mjrp8cvr8bqrj8@no.spam>, on 01/06/2004 at 01:00 AM, Toni Lassila said:>One of problems of picking up elementary analysis texts is that even>the seasoned mathematicians in this group rarely seem to come>together on which de?itions should be used.Any good Mathematics text will de?e the nomenclature used. It's thestudent's responsibility to read the de?itions. That includes notjust de?itions of terms but also typographical conventions.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org <20040104102842.911$Mv@newsreader.com> <468305d6.0401051618.7985909e@posting.google.com>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === =In <468305d6.0401051618.7985909e@posting.google.com>, on 01/05/2004 at 04:18 PM, artur@opengate.com.br (Artur) said:>I think it's interesting to point out that, when de?ing continuity,>some authors, like Rudin in his book on Analysis, prefer to consider>the domain X of the function as a kind of universal set, without>paying any attention to another space where X might be embedded.Actually, the two de?itions are equivalent. Google for relativetopology.>I've heard some people say that f(x) = 1/x has an essential>discontinuity (opposed to removable discontinuity) at x=0 because it>doesn`t have a limit there. This is a completely wrong statement,>because is simply not de?ed at x=0.No, it is not wrong as long as they de?e their terms poperly. De?ef has an essential singularity at x to mean that either f is de?edat x and not continuous at x or ther does not exist an extension of fthat is continuous at x.There is a more subtle issue as well. A function could be continuous,yet it's restriction to a subspace could have an essentialdiscontinuity, as de?ed above. -- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === In <468305d6.0401051618.7985909e@posting.google.com>, on 01/05/2004> at 04:18 PM, artur@opengate.com.br (Artur) said:>I've heard some people say that f(x) = 1/x has an essential>discontinuity (opposed to removable discontinuity) at x=0 because it>doesn`t have a limit there.Such a singularity is normally called a pole. It is neither essential norremovable.>This is a completely wrong statement, because is simply not de?ed>at x=0.It's not de?ed at x=0 because, I assume, you're taking the codomain ofthe function to be just R. But, if the codomain is taken to be, say, R*,the one-point extension of the reals, then the function is de?ed at x=0.Indeed, f(x) = 1/x is a continuous function from R* onto R*.> No, it is not wrong as long as they de?e their terms poperly.Sure, if they de?e their terms in the manner of the pope, then surelythey're not wrong! Indeed, their de?itions should be catholic, and thus,universally accepted.(Or maybe you meant to say properly, rather than poperly. ;-)David CantrellX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === =In <657d7b28.0401052259.c371383@posting.google.com>, on 01/05/2004> am stationed in Tucson>within biking distance of the University of Arizona which has a>respectable math dept. but it is uncertain whether I will remain at>this station a long time or sent elsewhere with no notice: therefor>as far as school goes I am torn between whether to try and get a BA>before being moved, or concentrate on getting a degree from a>distance learning instituteI would opt for a university with a good graduate school, especiallyif your CO will allow you to attend colloquia and other functions inthe Mathematics department.>In the mean time I'm getting the liberal arts nonsense done through>a community college so at least that'll be out of the way.Some of it is nonsense, but by no means all. Whether you can ?d anyworthwhile courses at a CC, however, is another question.>At the same time, at least until I get in some good math courses, I>continue to teach myself. I have, I think, a very good grasp on>basic abstract algebra. Good; that's essential.>In the self-taught world, the distinctions between analysis and >calculus are too gray!Think of Analysis as Calculus done right; instead of engineering andmetaphysical hand waving you use real de?itions, and you have toactually prove things before you can use them. Real and ComplexAnalysis are essential for a lot of Mathematics.>I am wondering what I should study next.Where do your interests lie?>Galois theory seems like a very good option. It wouldn't hurt, although it's not ' as central as some other topics.>Topology wouldnt hurt,In fact it's essential if you plan to go on into more advancedMathematics. If I had my way they'd teach it prior to the Calculusclasses.>and then there are many, many more minor branches >that I could choose fromAnd not so minor. You'll need to ration your time, because there aremore important and interesting ?lds than you will have time tolearn.>representation theory,Important>measure theory,Important>category theory,Imporatant.>Ultimately I'm interested in it all>and want to learn it all but where to go from here ..........What are your career plans in the AF? There used to be variousprograms whereby the military would assign you to duty at a universityand pay for your tuition; you had to maintain your grades and committo remaining on active duty for several more years after graduation.If you plan to remain in the AF, I'd advice you to look into whetheryou would be eligible for such a program.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === =In , on 01/06/2004 at 07:50 PM, Robin Chapman said:>Differential forms are ?e but when ?zisists start talking about>in?itesimal rotations my mind glazes over....That's just their way of talking about the elements of the Lie Algebrafor SO(3). BTW, how do you write so(3) in pseudo-TeX?>Hmmm. Well my experience was that after a load of hand-wavy calculus>at sixth-form, when I ?st saw an analysis book (full of limits,>epsilons and deltas) I was captivated --- the ?st inkling that the>subject made sense after all!I had a similar experience. As a child Calculus for the PracticalMan made no sense to me. Then I got a copy of Thomas, which at thattime was reasonably rigorous, and everything was obvious. As Einsteinsimpler.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === =In , on 01/06/2004 at 07:45 PM, lbudney@pobox.com said:>Solve an exact differential equation. Note the naked dy and dx>terms. One can justify this in terms of integration by parts, or by>appealing to in?itesimals.Or by differential forms.>Likewise with more general differential forms.Germs of functions are perfectly rigorous, as is the rest of themachinery constructed using them.>Almost any mathematician (including me) will agree with you,Yes, but 1. The language of differential forms is quite rigorous. 2. The language of Nonstandard Analysis is quite rigorous.Admittedly neither is what the engineer or physicist has in mind whenblithely slinging around dx and dy, but often the manipulations can bejusti?d ex post facto.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === =[snip]> Yes, but 1. The language of differential forms is quite rigorous. 2. The language of Nonstandard Analysis is quite rigorous. Admittedly neither is what the engineer or physicist has in mind when> blithely slinging around dx and dy, but often the manipulations can be> justi?d ex post facto. Or maybe dx post facto.David === =Let's see the following de?itionLet (X,d) be a complete metric space. A function T:X->X is said to bea contraction mapping if there is a constant q with 0<=q<1 such thatd(Tx,Ty) <= q d(x,y)for all x,y in X. Why d(Tx,Ty) <= q d(x,y) is not written asd(T(x),T(y)) <= q d(x,y)A lot of math books use the ?st formula... what is the difference between both formulas?Diego Andr.8es === =On 8 Jan 2004 05:43:28 -0800, diegoandresalvarez@lycos.co.uk (DiegoLet's see the following de?itionLet (X,d) be a complete metric space. A function T:X->X is said to be>a contraction mapping if there is a constant q with 0<=q<1 such that>d(Tx,Ty) <= q d(x,y)>for all x,y in X. Why d(Tx,Ty) <= q d(x,y) is not written asd(T(x),T(y)) <= q d(x,y)A lot of math books use the ?st formula... >what is the difference between both formulas?The only difference is Tx instead of T(x), right?Well, they mean the same thing. The notationTx is traditional when T is _linear_, while T(x)is more common for nonlinear functions.(I imagine the reason is that in ?ite-dimensionalspaces a linear map is the same as multiplicationby a matrix, so one thinks of a linear map itself asmultiplication by something.)Diego Andr.8esDavid C. Ullrich === =David C. Ullrich scribbled the following:> On 8 Jan 2004 05:43:28 -0800, diegoandresalvarez@lycos.co.uk (DiegoLet's see the following de?itionLet (X,d) be a complete metric space. A function T:X->X is said to be>a contraction mapping if there is a constant q with 0<=q<1 such that>d(Tx,Ty) <= q d(x,y)>for all x,y in X. Why d(Tx,Ty) <= q d(x,y) is not written asd(T(x),T(y)) <= q d(x,y)A lot of math books use the ?st formula... >what is the difference between both formulas?> The only difference is Tx instead of T(x), right?> Well, they mean the same thing. The notation> Tx is traditional when T is _linear_, while T(x)> is more common for nonlinear functions.> (I imagine the reason is that in ?ite-dimensional> spaces a linear map is the same as multiplication> by a matrix, so one thinks of a linear map itself as> multiplication by something.)At least in the University of Helsinki, if there is a function f:X->Y,and U is a subset of X, then fU means {f(x) | x in U}. That is adifferent thing from f(U), but as such, f(U) would not even be de?ed.I don't know what is used for {{f(x) | x in V} | V in U} but lily Ihave never needed such a set anywhere.Could there be any functions f:X->Y and subsets U of X so that fU andf(U) would both be de?ed but would not be equal?-- /-- Joona Palaste (palaste@cc.helsinki.? ------------- Finland ---------- http://www.helsinki.?~palaste --------------------- rules! --------/'I' is the most beautiful word in the world. ' - John Nordberg === David C. Ullrich scribbled the following:> On 8 Jan 2004 05:43:28 -0800, diegoandresalvarez@lycos.co.uk (Diego>Let's see the following de?ition>Let (X,d) be a complete metric space. A function T:X->X is said to be>a contraction mapping if there is a constant q with 0<=q<1 such that>d(Tx,Ty) <= q d(x,y)>for all x,y in X. >Why d(Tx,Ty) <= q d(x,y) is not written as>d(T(x),T(y)) <= q d(x,y)>A lot of math books use the ?st formula... >what is the difference between both formulas? The only difference is Tx instead of T(x), right?> Well, they mean the same thing. The notation> Tx is traditional when T is _linear_, while T(x)> is more common for nonlinear functions. (I imagine the reason is that in ?ite-dimensional> spaces a linear map is the same as multiplication> by a matrix, so one thinks of a linear map itself as> multiplication by something.)At least in the University of Helsinki, if there is a function f:X->Y,>and U is a subset of X, then fU means {f(x) | x in U}. That is a>different thing from f(U), but as such, f(U) would not even be de?ed.>I don't know what is used for {{f(x) | x in V} | V in U} but lily I>have never needed such a set anywhere.>Could there be any functions f:X->Y and subsets U of X so that fU and>f(U) would both be de?ed but would not be equal?Certainly that can happen, although it's not likely to come upin the math most people do. In logic/set theory people havespecial notations for this - I don't know whether it's standard,but I've seen things like f[[U]] for ' {f(x) : x in U}, which thenallows you to write f[[[U]]] for {f[[x]] : x in U}, etc.David C. Ullrich === =The picture I am working from is shown on p.496 of the Pavia, Lampman,Kriz book I mentioned in my original posting. Something like this: | | I haven't seen Thiele's ' orginal apparatus. | | I haven't seen the one at | http://iweb.tntech.edu/chem311-dc/thiele.htm | before, but I don't know much. However, it | | clari?s why it is that different people have | | different impressions as to where to heat the tube. | | | | / / the right equation to use here. I realized that myself | |/ / after posting, but am not sure yet what the right | / equations are to use or how to deal with them. I'm now | / reading a book on the mathematical theory of ? | dynamics, published by Dover, and may be better | | informed when I've ?ished reading it. Meanwhile, | | it would be helpful to know of examples where the ______/ solutions of practical problems like this have been treated by exact, approximate and numerical methods.Ignorantly,Allan Adlerara@zurich.ai.mit.edu*** ** ** Intelligence Lab. My actions and comments do not re?* in any way on MIT. Moreover, I am nowhere near the Boston ** metropolitan area. ** ***** === =Just out of curiosity, with all these unproven conjectures about...what happens when the conjecture is proven to be correct? Does thenewly declared theorem retain the name of the person originating theconjecture, or will it get named after the person(s) who proves it?I suppose some conjectures are so famous one would think the theoremwould retain the name of the person originating the conjecture. Saymaybe one day we will come to know of the Poincare Theorem.It is interesting to note, that historically there wasn't muchconsistency in the naming rights of various theorems etc. Fermatsfamous Last Theorem, was hardly a theorem at all, but only aconjecture (until it was proven). The fact it was called a theorem, itjust one of those things that st. Now that Wiles had given a proofof the theorem (conjecture) now surely it is of?ially a theorem.This may not be a good example, but it strikes me that theFermat-Wiles Theorem would be a more appropriate name. Maybe it isalready called that, that I am not sure of.In general is there an agreed concensus as to naming rights inmathematics, and/or is there a body that governs naming rights? === In general is there an agreed concensus as to naming rights in>mathematics, and/or is there a body that governs naming rights?You'll have to ask the Of?ial Spokesman The Maths Community andSupreme Ruler Of Sci.Math. === =In general is there an agreed concensus as to naming rights in>mathematics, and/or is there a body that governs naming rights?You'll have to ask the Of?ial Spokesman The Maths Community and>Supreme Ruler Of Sci.Math.Alas frivolous questions like this are beneath him...David C. Ullrich === Just out of curiosity, with all these unproven conjectures about...> what happens when the conjecture is proven to be correct? Does the> newly declared theorem retain the name of the person originating the> conjecture, or will it get named after the person(s) who proves it?I suppose some conjectures are so famous one would think the theorem> would retain the name of the person originating the conjecture. Say> maybe one day we will come to know of the Poincare Theorem.It is interesting to note, that historically there wasn't much> consistency in the naming rights of various theorems etc. Fermats> famous Last Theorem, was hardly a theorem at all, but only a> conjecture (until it was proven). The fact it was called a theorem, it> just one of those things that st. Now that Wiles had given a proof> of the theorem (conjecture) now surely it is of?ially a theorem.> This may not be a good example, but it strikes me that the> Fermat-Wiles Theorem would be a more appropriate name. Maybe it is> already called that, that I am not sure of.In general is there an agreed concensus as to naming rights in> mathematics, and/or is there a body that governs naming rights?Names are used to help us communicate with each other. Even if thepriority is incorrect, if everyone knows what you mean, then the namehas served its purpose.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === In general is there an agreed concensus as to naming rights in> mathematics, and/or is there a body that governs naming rights?No. Most things are named after totally irrelevant people(e.g. Pell's equation) :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === = No. Most things are named after totally> irrelevant people (e.g. Pell's equation) :-)Even worse: It isn't named ' after Pell himselfbut after someone else with the same name :-PRainer Rosenthalr.rosenthal@web.de === =In general is there an agreed concensus as to naming rights in> mathematics, and/or is there a body that governs naming rights?No. Most things are named after totally irrelevant people> (e.g. Pell's equation) :-)And:during a representation theory conference some speaker talks of usingBrauer's trick. A hand goes up. ?What's Brauer's trick?' Asks ' RichardBrauer.third hand anecdote, so open to corrections that actually it was Witten,or someone else entirely. === = > In general is there an agreed concensus as to naming rights in> mathematics, and/or is there a body that governs naming rights?No. Most things are named after totally irrelevant people> (e.g. Pell's equation) :-)And:during a representation theory conference some speaker talks of using>Brauer's trick. A hand goes up. ?What's Brauer's trick?' Asks ' Richard>Brauer.That's not so surprising. I expect Brauer knew many tricks.There was a question on one of our exams a few years ago which simplysaid State and prove Cauchy's theorem. Plenty of scope there foralternative answers!Derek Holt. === =On Thu, 8 Jan 2004 17:20:31 +0000 (UTC),>There was a question on one of our exams a few years ago which simply>said State and prove Cauchy's theorem. Plenty of scope there for>alternative answers!Take your pick:Also, prove the Euler Formula and Gauss' Theorem.But wasn't ' at least the Integral Theorem actually due to Cauchy? Ofcourse his proof wasn't perfect and later ?ed by Goursat. === Last night, at about 00.55 AM, I went up on terrace and with full moon>right over my head, I tossed one rupee coin.It was Head.Today morning, I tossed again one rupee coin in blue sky.Again, it was Head.So God is telling me to ?e one more patent application. I have>already ?ed 7 patent applications.No, God is telling you that Rosencrantz and Guildenstern are dead.--Dr.Postman USPS, MBMC, BsD; Disgruntled, But UnarmedMember,Board of Directors of afa-b, SKEP-TI-CULT member #15-51506-253.Shake it like a polaroid picture. - Andre 3000 of Outkast === On second thought...> Tomorrow morning (07.01.2004), I will go UP on terrace and toss one> Rupee coin in blue sky..> Head: I will prepare and ?e detailed patent application.> Tail: I will NOT ?e any patent application. I will start execution> sequence of this Action Device at the same instantaneous moment I see> Tail.> There will be only one toss..> It is UP to Gravity NOW!> ............> My guess: Head and you can't afford to ?e the patent, therefore you> can hold on to your phantasy a bit longer.Greetings!> VolkerLast night, at about 00.55 AM, I went up on terrace and with full moon> right over my head, I tossed one rupee coin.It was Head.Today morning, I tossed again one rupee coin in blue sky.Again, it was Head.So God is telling me to ?e one more patent application. I have> already ?ed 7 patent applications.He has taught me that Love can trigger Logic. But Logic can not> trigger Love. So my beloved God, Gravity, if you are telling me to> patent You so that I sell You to these people on earth, then it is> simply not acceptable.I will not patent God. I will not patent Gravity, Supreme Force. I> will not patent Future of Mankind, Planets, Stars, Galaxies, this> Universe and I will not patent Love.I am handing over Supreme Force, this Action Device to United Nations> provided that India get Veto Power.See? Allright, not the no-money excuse but something similar unlikely.By the way, I'm as ready as you are to publish the drawings of my latestperpetuum mobile, provided pigs ?e earth turns ? a blue moontells me to.Greetings!Volker === =In Dover's reprint of Robert R. Stoll's _Set_Theory_and_Logic_, thesecond chapter has a section on well ordered sets and ordinal numbers. The ?st theorem presented in that section is:[Theorem] If A is a well ordered set, f an isomorphism from A to A, then a <=f(a) for each a in A.[/Theorem]The book yields the following proof:[Proof]Assume that for some element a in A we have a > f(a). Let B be thesubset of A of all such elements and b its least member. Since b >f(b) it follows that f(b) > f(f(b)). Thus f(b) in B, which is acontradiction.[/Proof]I can't grok how b > f(b) implies f(b) > f(f(b)) with any ofthe given knowledge. I could just not be parsing things right in myhead - so please tell me if I'm being stupid or not :-) === In Dover's reprint of Robert R. Stoll's _Set_Theory_and_Logic_, the> second chapter has a section on well ordered sets and ordinal numbers.> The ?st theorem presented in that section is:> [Theorem]> If A is a well ordered set, f an isomorphism from A to A, then a < f(a) for each a in A.> [/Theorem]> The book yields the following proof:> [Proof]> Assume that for some element a in A we have a > f(a). Let B be the> subset of A of all such elements and b its least member. Since b f(b) it follows that f(b) > f(f(b)). Thus f(b) in B, which is a> contradiction.> [/Proof]> I can't grok how b > f(b) implies f(b) > f(f(b)) with any of> the given knowledge. I could just not be parsing things right in my> head - so please tell me if I'm being stupid or not :-)You're forgetting that f is an isomorphism. That means it preservesorder.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === You're forgetting that f is an isomorphism. That means it preserves> order. === In Dover's reprint of Robert R. Stoll's _Set_Theory_and_Logic_, the> second chapter has a section on well ordered sets and ordinal numbers.> The ?st theorem presented in that section is:[Theorem]> If A is a well ordered set, f an isomorphism from A to A, That is, f is a bijection and c < d iff f(c) < f(d).> I can't grok how b > f(b) implies f(b) > f(f(b)) with any o?= f(b), so b > c implies f(b) > f(c).-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === In Dover's reprint of Robert R. Stoll's _Set_Theory_and_Logic_, the>second chapter has a section on well ordered sets and ordinal numbers.> The ?st theorem presented in that section is:[Theorem]> If A is a well ordered set, f an isomorphism from A to A, then a [/Theorem]The book yields the following proof:[Proof]>Assume that for some element a in A we have a > f(a). Let B be the>subset of A of all such elements and b its least member. Since b f(b) it follows that f(b) > f(f(b)). Thus f(b) in B, which is a>contradiction.>[/Proof]>I can't grok how b > f(b) implies f(b) > f(f(b)) with any of>the given knowledge. You're given that f is an _isomorphism_ from one ordered setto another. What does the word isomorphism mean here?>I could just not be parsing things right in my>head - so please tell me if I'm being stupid or not :-)David C. UllrichCancel-Lock: sha1:wU2ipihIzdv6q6ryq5bhZkcCmag === == I can't grok how b > f(b) implies f(b) > f(f(b)) with any of> the given knowledge. You're given that f is an _isomorphism_ from one ordered set to> another. What does the word isomorphism mean here?David already hit it exactly, but I can't resist piggy-backing in caseit isn't completely obvious. In any given context, an isomorphism isa bijection that preserves the property you care about. Thus, anisomorphism of ordered sets...Len. === GM: General Mathematics> -----------------------math.GM/0312360 we knew that already.> math.GM/0312309 -----------------------math.GM/0312360 <5vSKb.2670$fI5.783@reader1.news.jippii.net> we knew that already.math.GM/0312309 its value for every possible argument. You always learn something new.One learns lots of new things in the GM category. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === =In Hebrew, the feminine plural of the foreign loan word, nilpotent, whenwritten without vowels or other punctation, is writtennylpw_tn_tywtwhere _t is teth. What I would like to know is how it would be writtenif one did write the vowels and other punctation. For example:(0) which of the consonants is dotted?(1) is the y following the ?st n retained and the n endowed with a chireq?(2) is there a schwa under the l?(3) is the w following the p replaced by a cholem or is it replaced by a qamats qatan?(4) what is under the ?st teth: a schwa, a segol or a chataf segol?(5) is there a schwa under the second n?(6) is the y following the second teth retained and the teth endowed with a chireq?I would also like to know how I might ?ure out examples like this, involvingforeign loan words, on my own, instead of having to ask people in everyinstance.Ignorantly,Allan Adlerara@zurich.ai.mit.edu*** ** ** Intelligence Lab. My actions and comments do not re?* in any way on MIT. Moreover, I am nowhere near the Boston ** metropolitan area. ** ***** === In Hebrew, the feminine plural of the foreign loan word,> nilpotent, when written without vowels or other punctation, is> written nylpw_tn_tywt where _t is teth. What I would like to know> is how it would be written if one did write the vowels05E0;HEBREW LETTER NUN;Lo;0;R;;;;;N;;;;; 05B4;HEBREW POINT HIRIQ;Mn;14;NSM;;;;;N;;;;;05D9;HEBREW LETTER YOD;Lo;0;R;;;;;N;;;;;05DC;HEBREW LETTER LAMED;Lo;0;R;;;;;N;;;;;05B0;HEBREW POINT SHEVA;Mn;10;NSM;;;;;N;;;;;05E4;HEBREW LETTER PE;Lo;0;R;;;;;N;;;;;05BC;HEBREW POINT DAGESH OR MAPIQ;Mn;21;NSM;;;;;N;;or shuruq;;;05B9;HEBREW POINT HOLAM;Mn;19;NSM;;;;;N;;;;;05D5;HEBREW LETTER VAV;Lo;0;R;;;;;N;;;;;05D8;HEBREW LETTER TET;Lo;0;R;;;;;N;;;;;05B6;HEBREW POINT SEGOL;Mn;16;NSM;;;;;N;;;;;05E0;HEBREW LETTER NUN;Lo;0;R;;;;;N;;;;; 05B0;HEBREW POINT SHEVA;Mn;10;NSM;;;;;N;;;;;05D8;HEBREW LETTER TET;Lo;0;R;;;;;N;;;;;05B4;HEBREW POINT HIRIQ;Mn;14;NSM;;;;;N;;;;;05D9;HEBREW LETTER YOD;Lo;0;R;;;;;N;;;;;05BC;HEBREW POINT DAGESH OR MAPIQ;Mn;21;NSM;;;;;N;;or shuruq;;;05B9;HEBREW POINT HOLAM;Mn;19;NSM;;;;;N;;;;;05D5;HEBREW LETTER VAV;Lo;0;R;;;;;N;;;;;05EA;HEBREW LETTER TAV;Lo;0;R;;;;;N;;;;;View the following string as part of an HTML page in an internationalizedweb browser to see this in Hebrew:נִילְפּ&# 1465;וטֶנְטִי ֹּות> I would also like to know how I might ?ure out examples like this,> involving foreign loan words, on my own, instead of having to ask> people in every instance.Just keep a few simple equivalences in mind -- [i] vowel = hireq, [e]vowel in closed syllable = segol, [e] vowel in open syllable = tsere,[o] vowel = holem, non-word-?al consonant not followed by a vowel =has shewa under it. Hebrew grammatical endings of course followtraditional Hebrew orthographic rules.-- Some Qur'an quotes: 5:20 qaala muusaa 5:21 yaa qawmi dkhuluu l-'arDal-muqaddasata llatii kataba llaahu lakum 17:104 waqulnaa ... libanii'israa'iila ' skunuu l-'arDa || In English: Moses said, My people,go into the Holy Land which God has assigned to you! And we said to theChildren of Israel, Inhabit the land! http://symbolictruth.fateback.com/ === =You said you are using a dictionary, right? What does it say?If it too has no vowels, you probably ought to invest in onethat has vowels.In Hebrew, the feminine plural of the foreign loan word, nilpotent, when> written without vowels or other punctation, is writtennylpw_tn_tywtwhere _t is teth. What I would like to know is how it would be written> if one did write the vowels and other punctation. For example:> (0) which of the consonants is dotted?> (1) is the y following the ?st n retained and the n endowed with a chireq?> (2) is there a schwa under the l?> (3) is the w following the p replaced by a cholem or is it replaced by> a qamats qatan?> (4) what is under the ?st teth: a schwa, a segol or a chataf segol?> (5) is there a schwa under the second n?> (6) is the y following the second teth retained and the teth endowed with> a chireq?I would also like to know how I might ?ure out examples like this, involving> foreign loan words, on my own, instead of having to ask people in every> instance.Ignorantly,> Allan Adler> ara@zurich.ai.mit.edu ****> * *> * Intelligence Lab. My actions and comments do not re?> * in any way on MIT. Moreover, I am nowhere near the Boston *> * metropolitan area. *> * *> ***** ******* === =On Thu, 08 Jan 2004 12:01:17 -0500, Patricia Heil You said you are using a dictionary, right? What does it say?>If it too has no vowels, you probably ought to invest in one>that has vowels.You silly little convert. Were you asleep during yiddish classes?In Hebrew, the feminine plural of the foreign loan word, nilpotent, when> written without vowels or other punctation, is writtennylpw_tn_tywtwhere _t is teth. What I would like to know is how it would be written> if one did write the vowels and other punctation. For example:> (0) which of the consonants is dotted?> (1) is the y following the ?st n retained and the n endowed with a chireq?> (2) is there a schwa under the l?> (3) is the w following the p replaced by a cholem or is it replaced by> a qamats qatan?> (4) what is under the ?st teth: a schwa, a segol or a chataf segol?> (5) is there a schwa under the second n?> (6) is the y following the second teth retained and the teth endowed with> a chireq?I would also like to know how I might ?ure out examples like this, involving> foreign loan words, on my own, instead of having to ask people in every> instance.Ignorantly,> Allan Adler> ara@zurich.ai.mit.edu ****> * *> * Intelligence Lab. My actions and comments do not re?> * in any way on MIT. Moreover, I am nowhere near the Boston *> * metropolitan area. *> * *> ***** ******* === =Never heard this word before. It must be very modern Hebrew or very AncientAramaic. There is the shoresh LVE - to loan, lend, borrow. But that is all Ican help you with.YS In Hebrew, the feminine plural of the foreign loan word, nilpotent, when> written without vowels or other punctation, is written nylpw_tn_tywt where _t is teth. What I would like to know is how it would be written> if one did write the vowels and other punctation. For example:> (0) which of the consonants is dotted?> (1) is the y following the ?st n retained and the n endowed with achireq?> (2) is there a schwa under the l?> (3) is the w following the p replaced by a cholem or is it replaced by> a qamats qatan?> (4) what is under the ?st teth: a schwa, a segol or a chataf segol?> (5) is there a schwa under the second n?> (6) is the y following the second teth retained and the teth endowed with> a chireq? I would also like to know how I might ?ure out examples like this,involving> foreign loan words, on my own, instead of having to ask people in every> instance. Ignorantly,> Allan Adler> ara@zurich.ai.mit.edu ****> ***> * Intelligence Lab. My actions and comments do not re? * in any way on MIT. Moreover, I am nowhere near the Boston*> * metropolitan area.*> **>** **>---Checked by AVG anti-virus system (http://www.grisoft.com). === Never heard this word before. It must be very modern Hebrew or very Ancient> Aramaic. There is the shoresh LVE - to loan, lend, borrow. But that is all I> can help you with.> YS In Hebrew, the feminine plural of the foreign loan word, nilpotent, when> written without vowels or other punctation, is written nylpw_tn_tywt where _t is teth. What I would like to know is how it would be written> if one did write the vowels and other punctation. For example:> (0) which of the consonants is dotted?> (1) is the y following the ?st n retained and the n endowed with a> chireq?> (2) is there a schwa under the l?> (3) is the w following the p replaced by a cholem or is it replaced by> a qamats qatan?> (4) what is under the ?st teth: a schwa, a segol or a chataf segol?> (5) is there a schwa under the second n?> (6) is the y following the second teth retained and the teth endowed with> a chireq? I would also like to know how I might ?ure out examples like this,> involving> foreign loan words, on my own, instead of having to ask people in every> instance. Ignorantly,> Allan Adler> ara@zurich.ai.mit.edu> ***** *******> *> *> *> * Intelligence Lab. My actions and comments do not re?*> * in any way on MIT. Moreover, I am nowhere near the Boston> *> * metropolitan area.> *> *> * ***** ******* ---> Checked by AVG anti-virus system (http://www.grisoft.com).I cannot help with the Hebrew but it may help those outside ofsci.math to know that this is a mathematical term which I do not thinkhave any use outside of mathematics. It is not in my Concise OxfordDictionary and I would be surprised to see it outside of a maths book.So I doubt that it is ancient. More likely is a modern coining. Thespelling above suggests an attempt to mimic the English word.J === Never heard this word before. It must be very modern Hebrew or very Ancient>Aramaic. There is the shoresh LVE - to loan, lend, borrow. But that is all I>can help you with.It is very modern Hebrew, and almost certainly derivedfrom the English word. I can make guesses based onthis, and most will be correct, but one would have toask an Israeli mathematician who uses the word to besure. I am including what I believe to be correct,but on at least some I am unsure.>YS> In Hebrew, the feminine plural of the foreign loan word, nilpotent, when> written without vowels or other punctation, is written> nylpw_tn_tywt> where _t is teth. What I would like to know is how it would be written> if one did write the vowels and other punctation. For example:> (0) which of the consonants is dotted?The p almost certainly, and possibly no others.> (1) is the y following the ?st n retained and the n endowed with a>chireq?I would say so.> (2) is there a schwa under the l?Yes.> (3) is the w following the p replaced by a cholem or is it replaced by> a qamats qatan?Cholem; the English is essential transliterated.> (4) what is under the ?st teth: a schwa, a segol or a chataf segol?Some type of segol, probably full.> (5) is there a schwa under the second n?Yes.> (6) is the y following the second teth retained and the teth endowed with> a chireq?I believe so; this might be standard Hebrew grammar.> I would also like to know how I might ?ure out examples like this,>involving> foreign loan words, on my own, instead of having to ask people in every> instance.> Ignorantly,> Allan Adler> ara@zurich.ai.mit.eduAs I stated, one would have to consult an Israelimathematician who uses it, and even then there might beambiguity. He may never have seen it with nikodoth.My Hebrew is not great; when it comes to foreign words,direct transliteration, except for grammatical needs,is the general rule.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === Never heard this word before. It must be very modern Hebrew or veryAncient>Aramaic. There is the shoresh LVE - to loan, lend, borrow. But that isall I>can help you with. It is very modern Hebrew, and almost certainly derived> from the English word. I can make guesses based on> this, and most will be correct, but one would have to> ask an Israeli mathematician who uses the word to be> sure. I am including what I believe to be correct,> but on at least some I am unsure.What you say makes sense. Sounds like legalese really.YS >YS In Hebrew, the feminine plural of the foreign loan word, nilpotent,when> written without vowels or other punctation, is written nylpw_tn_tywt where _t is teth. What I would like to know is how it would be written> if one did write the vowels and other punctation. For example:> (0) which of the consonants is dotted? The p almost certainly, and possibly no others. (1) is the y following the ?st n retained and the n endowed with a>chireq? I would say so. (2) is there a schwa under the l? Yes. (3) is the w following the p replaced by a cholem or is it replaced by> a qamats qatan? Cholem; the English is essential transliterated. (4) what is under the ?st teth: a schwa, a segol or a chataf segol? Some type of segol, probably full. (5) is there a schwa under the second n? Yes. (6) is the y following the second teth retained and the teth endowedwith> a chireq? I believe so; this might be standard Hebrew grammar. I would also like to know how I might ?ure out examples like this,>involving> foreign loan words, on my own, instead of having to ask people in every> instance. Ignorantly,> Allan Adler> ara@zurich.ai.mit.edu As I stated, one would have to consult an Israeli> mathematician who uses it, and even then there might be> ambiguity. He may never have seen it with nikodoth. My Hebrew is not great; when it comes to foreign words,> direct transliteration, except for grammatical needs,> is the general rule.> --> This address is for information only. I do not claim that these views> are those of the Statistics Department or of Purdue University.> Herman Rubin, Department of Statistics, Purdue University---Checked by AVG anti-virus system (http://www.grisoft.com). === [cut]>Want more advanced polynomial factorization? [cut]>James Harris > Since you asked, I would like to see your factorization of x^3-x+8> To be fair, I haven't seen any assertion by James that hisnonpolynomial factorization works for all polynomials(indeed, it doesn't seem to). However, it does workfor the cubic polynomials James is considering, whicharise from his FLT investigations. That's why I hadto craft my simpler example with care and couldn't,for example, just pull a polynomial out of thin air.Rick[followups trimmed] === =[cut]>Want more advanced polynomial factorization? [cut]>James Harris > Since you asked, I would like to see your factorization of x^3-x+8To be fair, I haven't seen any assertion by James that hisnonpolynomial factorization works for all polynomials> (indeed, it doesn't seem to). However, it does work> for the cubic polynomials James is considering, which> arise from his FLT investigations. That's why I had> to craft my simpler example with care and couldn't,> for example, just pull a polynomial out of thin air.> You still haven't to my knowledge explained how you came up with yourquadratic example Decker, so it might help if you did so now.Readers wondering should note that my investigations into ?ding ashort proof of Fermat's Last Theorem *did* lead me to the advancedtools for factoring polynomials that I'm using now.That should be something of a clue to at least some of you: my workis original to such an extent because I was pushing against one of thehardest problems in mathematics.The tools I have are advanced on a level that simply leavesmathematicians ? and gasping for air, or running away, as I usethose tools in areas where they *thought* they had everything ?uredout.I am the guy who thought outside of the box, and mathematicians stillhaven't caught up yet.James Harris === =...: short proof of Fermat's Last Theorem *did* lead me to the advanced: tools for factoring polynomials that I'm using now....Oh, for ' God's sake...1) You have developed no new tools for factoring polynomials.2) What you're doing is fairly elementary manipulation of variables.3) You're continually screwing it up anyway.Deep exhale,Justin === =Let S be a partially ordered set with a Hausdorff topology which has a base of convex sets.Let <= be the order of S and let L = { (x,y) | x <= y }.Can it been shown L is closed in SxS ?The converse is easy and doesn't require convex base sets.Assume L is closed and x /= y. Take (wlog) not x <= y.Then (x,y) in open (SxS)L.Some open U,V with (x,y) in UxV subset (SxS)LU,V disjoint. Otherwise: some z in U,V (z,z) in UxV subset (SxS)L; not z <= z---- === In Munkres' undergrad topology book, there's a section that talks>about the strong form of Urysohn's Lemma, which is the statement>In X a normal space, there is a continuous function f: X -> [0,1] such>that f(x) =0 on A and f(x) = 1 on B AND 0 < f(x) <1 OTHERWISE, if and>only if A and B are disjoint closed G-delta sets in X.>Does anybody know where I can ?d a proof of this theorem? I don't>GI have not seen this theorem before, but I ?d it trivial.All that one needs to use about G-delta sets is that they arecountable intersections of open sets, the de?ition.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === In X a normal space, there is a continuous function f: X -> [0,1] such>that f(x) =0 on A and f(x) = 1 on B AND 0 < f(x) <1 OTHERWISE, if and>only if A and B are disjoint closed G-delta sets in X.Does anybody know where I can ?d a proof of this theorem? I don't I have not seen this theorem before, but I ?d it trivial. All that one needs to use about G-delta sets is that they are> countable intersections of open sets, the de?ition.>Well ok, would you detail a quicker proof in a few lines? === Does anybody know where I can ?d a proof of this theorem?> As nonnul A,B are G_delta, they're zero sets >Proof? >Compact G_delta zero-set closed G_delta, >but in general (in not necessarily normal space) the converses are >false. Right?The actual theorems are closed G_delta A subset normal S = A zero set compact G_delta A subset Tychonov S = A zero setI used the ?st as A,B are closed G_delta in normal space.For the converse, use theorem for all spaces zero set A = A closed G_delta> and as X is normal> there's some continuous f,g,h : X -> [0,1] with> f^-1(0) = A, g^-1(0) = B, h(A) = 0, h(B) = 1 let k = f*(1 - h)/2 + (1 - g/2)*h> Show k is the desired map.>useful:k = f*(1 - h)/2 + (1 - g/2)*h = f/2 - f*h/2 + h - g*h/2 = 1 - (1 - f/2)*(1 - h) - g*h/2---- === G. A. Edgar> Also, leave out those tacky measure> spaces that are de?ed irrespective of any topology. Who ever uses>them?> probabilists>Well, okay. But the alchemists don't really count:)>LHOn the contrary, the more restrictions on the space permitted for probability, the harder it is to seewhat is going on. The topology of the space is rarelyof importance, in any case.This is especially true for metric spaces. That a space is metrizable may be important, but if themetric is not a natural metric, it is just confusing.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === =I'm currently working on a new analysis text. It's sort of in the> spirit of the Bourbaki texts: I'm doing a fairly long section on> logic and set theory before moving onto the real and complex analysis> portions--just trying to spread my philosophical biases. :) Anyways,> it's really a personal project, trying to make a reference, and> perhaps writing it will help me internalize some of the nuance that is> so easily forgotten. Perhaps others will ?d it useful.Can anyone suggest topics or exercises for inclusion? (Credit will of> course be given). Or perhaps some pointers for ?ding suitable> exercises? I plan to cover the propositional calculus and ?st order> logic (deductions, some meta-theory, including compactness and how it> relates to compactness in analysis), a brief mention of ZF, pointset> topology, sequences and series, continuity, differentiation, Riemann> integration, sequences and series of functions, Fourier analysis,> p-adic analysis, Lesbesgue integration. (No, I didn't copy the table> of contents from baby Rudin)My overarching idea is to make a text that allows the student to gain> mathematical maturity while working with it. To this end, I'm hoping> to make the introductory chapters gentle, but making them more and> more dif?ult until even I'm challenged by the last few sections. So> even esoteric topics can be included (so long as I can understand them> well enough to write about them). But they had better be beautiful!Anyways, I'm probably rambling now. I'll try to work on that. :) AlexHrrm.That sounds a lot like the analysis textbook I'm writing dammit. :) The > race to publish ?st is on!Not really actually, our topics and reasons overlap somewhat but I think > the focus is probably going to be fairly different. If only because my > perspective on the subject is so bloody strange. :) (Sequences? > Sequences you say? What are they...?)At any rate, here's some suggestions based roughly on what ' I've found > useful.Given that you're starting from a set theoretic point of view, I presume > you're going to be including a proof of the existence and uniqueness of > the reals? In which case you may want to include some chapters on > general abstract and linear algebra.I'm introducing pointset topology early on, and I ?d it's a very good > way to tie the subject together. I've done it by playing around with how > it work in the case of the real line, using motivation of ideas such as > limits, continuity and connectedness and then show why it is useful to > generalise this. I'm very big on providing motivation for the ideas as I > introduce them. Too often analysis textbooks are very arbitrary - Here > is this thing, and it's very useful, but you're not going to ' see why > until we've done 30 pages of work on it.I would ditch the riemann integral. It really is silly and pointless. > I'm starting from the Gauge integral, which you may ?d interesting to > check out, and later introducing the lebesgue integral, but you may want > to start with lebesgue (or ignore my advice on this one entirely; a lot > of people disagree).Out of curiousity, what relations between compactness (in the logical > sense) and analysis were you planning on mentioning? I've never seen any > other than non-standard analysis.> Well, it's possible to show that FOL is compact iff the Heine-Boreltheorem holds. Here's a sketch of the idea: Let L be an arbitraryconsistent in?ite subset of (the wffs of) FOL. Let G(L) be the setof Goedel numbers of elements of L. Stick a decimal in front of eachelement of G(L). Now you have a set G(L)* contained in [0,1]. G(L)*is obviously bounded, and in fact, (modulo some facts aboutdeductions) is closed (proving this is tricky). The rest gets kind ofmessy (and frankly, I haven't thought through it all yet), but I'msure you can see the mechanism through which the proof works. Thereason I'm including this is because ' I'm hoping that exposing astudent to inter-disciplinary mathematics will help him understandthat what he is really studying is not a set of objects, but a setof structures. Learning to identify structural features makes iteasier to ?d connections. Maybe I'll make this a guided exercise.The Lesbesgue integral is just as intuitive really, but introductorymeasure theory is a bit of a pain. I'll look into the gauge integral. Which variant are you working with? (Henstock, Perron, or the otherone?)Alex === [...]> Well, it's possible to show that FOL is compact iff the Heine-Borel>theorem holds. Here's a sketch of the idea: Let L be an arbitrary>consistent in?ite subset of (the wffs of) FOL. Let G(L) be the set>of Goedel numbers of elements of L. Stick a decimal in front of each>element of G(L). Now you have a set G(L)* contained in [0,1]. G(L)*>is obviously bounded, and in fact, (modulo some facts about>deductions) is closed (proving this is tricky). If you can prove this I will be _very_ surprised. For exampleit seems incredibly unlikely that no in?ite decimals appearin the closure.> The rest gets kind of>messy (and frankly, I haven't thought through it all yet), but I'm>sure you can see the mechanism through which the proof works. No, I have no idea how the rest of it is supposed to go either.But that doesn't matter until ' you've proved that G(L)* is closed.> The>reason I'm including this is because I'm hoping that exposing a>student to inter-disciplinary mathematics will help him understand>that what he is really studying is not a set of objects, but a set>of structures. Learning to identify structural features makes it>easier to ?d connections. Maybe I'll make this a guided exercise.The Lesbesgue integral is just as intuitive really, but introductory>measure theory is a bit of a pain. I'll look into the gauge integral.> Which variant are you working with? (Henstock, Perron, or the other>one?)AlexDavid C. Ullrich === =The Lesbesgue integral is just as intuitive really, but introductory> measure theory is a bit of a pain.As I keep pointing out, since so many otherwise well-educated mathematiciansstill believe the contrary, it is possible to de?e the Lebesgueintegral without having de?ed Lebesgue measure. One can use themethod of Daniell: see Alan Weir's books _Lebesgue Integration and Measure_and _General Integration and Measure_.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === The Lesbesgue integral is just as intuitive really, but introductory>measure theory is a bit of a pain.> As I keep pointing out, since so many otherwise well-educated mathematicians> still believe the contrary, it is possible to de?e the Lebesgue> integral without having de?ed Lebesgue measure. One can use the> method of Daniell: see Alan Weir's books _Lebesgue Integration and Measure_> and _General Integration and Measure_.> Roughly what approach does it take? I've seen one or two approaches, but good ones.David === >The Lesbesgue integral is just as intuitive really, but introductory>measure theory is a bit of a pain.> As I keep pointing out, since so many otherwise well-educated> mathematicians still believe the contrary, it is possible to de?e the> Lebesgue integral without having de?ed Lebesgue measure. One can use> the method of Daniell: see Alan Weir's books _Lebesgue Integration and> Measure_ and _General Integration and Measure_.> Roughly what approach does it take? I've seen one or two approaches, but> good ones.Start with step functions, extend to a more general class by takingmonotone limits, then extend by linearity.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === >The Lesbesgue integral is just as intuitive really, but introductory>measure theory is a bit of a pain.> As I keep pointing out, since so many otherwise well-educated> mathematicians still believe the contrary, it is possible to de?e the> Lebesgue integral without having de?ed Lebesgue measure. One can use> the method of Daniell: see Alan Weir's books _Lebesgue Integration and> Measure_ and _General Integration and Measure_.> Roughly what approach does it take? I've seen one or two approaches, but> good ones.>Start with step functions, extend to a more general class by taking>monotone limits, then extend by linearity.This is essentially starting with measure. However, I seeno problem with either approach. In fact, the differencebetween the Riemann and Lebesgue approaches is that in theRiemann approach, a function is integrable if it differs froman elementary function by at most an elementary function, orthe sum of a ?ite number of non-negative functions the sumof whose integrals is arbitrarily small, while for theLebesgue approach, replace ?ite by countable.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === >The Lesbesgue integral is just as intuitive really, but introductory>measure theory is a bit of a pain.> As I keep pointing out, since so many otherwise well-educated> mathematicians still believe the contrary, it is possible to de?e the> Lebesgue integral without having de?ed Lebesgue measure. One can use> the method of Daniell: see Alan Weir's books _Lebesgue Integration and> Measure_ and _General Integration and Measure_.> Roughly what approach does it take? I've seen one or two approaches, but> good ones.>Start with step functions, extend to a more general class by taking>monotone limits, then extend by linearity.This is essentially starting with measure.No, it isn't. Measure comes after integration .... the measureof a set can be de?ed to be the integral of its characteristicfunction.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === >I'm currently working on a new analysis text. It's ' sort of in the>spirit of the Bourbaki texts: I'm doing a fairly long section on>logic and set theory before moving onto the real and complex analysis>portions--just trying to spread my philosophical biases. :) Anyways,>it's really a personal project, trying to make a reference, and>perhaps writing it will help me internalize some of the nuance that is>so easily forgotten. Perhaps others will ?d it useful.>Can anyone suggest topics or exercises for inclusion? (Credit will of>course be given). Or perhaps some pointers for ?ding suitable>exercises? I plan to cover the propositional calculus and ?st order>logic (deductions, some meta-theory, including compactness and how it>relates to compactness in analysis), a brief mention of ZF, pointset>topology, sequences and series, continuity, differentiation, Riemann>integration, sequences and series of functions, Fourier analysis,>p-adic analysis, Lesbesgue integration. (No, I didn't copy the table>of contents from baby Rudin)>My overarching idea is to make a text that allows the student to gain>mathematical maturity while working with it. To this end, I'm hoping>to make the introductory chapters gentle, but making them more and>more dif?ult until even I'm challenged by the last few sections. So>even esoteric topics can be included (so long as I can understand them>well enough to write about them). But they had better be beautiful!>Anyways, I'm probably rambling now. ' I'll try to work on that. :) >AlexHrrm.That sounds a lot like the analysis textbook I'm writing dammit. :) The >race to publish ?st is on!Not really actually, our topics and reasons overlap somewhat but I think >the focus is probably going to be fairly different. If only because my >perspective on the subject is so bloody strange. :) (Sequences? >Sequences you say? What are they...?)At any rate, here's some suggestions based roughly on what I've found >useful.Given that you're ' starting from a set theoretic point of view, I presume >you're going to be including a proof of the existence and uniqueness of >the reals? In which case you may want to include some chapters on >general abstract and linear algebra.I'm introducing pointset topology early on, and I ?d it's a very good >way to tie the subject together. I've done it by playing around with how >it work in the case of the real line, using motivation of ideas such as >limits, continuity and connectedness and then show why it is useful to >generalise this. I'm very big on providing motivation for the ideas as I >introduce them. Too often analysis textbooks are very arbitrary - Here >is this thing, and it's very useful, but you're not going to ' see why >until we've done 30 pages of work on it.I would ditch the riemann integral. It really is silly and pointless. >I'm starting from the Gauge integral, which you may ?d interesting to >check out, and later introducing the lebesgue integral, but you may want >to start with lebesgue (or ignore my advice on this one entirely; a lot >of people disagree).Out of curiousity, what relations between compactness (in the logical >sense) and analysis were you planning on mentioning? I've never seen any >other than non-standard analysis. Well, it's possible to show that FOL is compact iff the Heine-Borel> theorem holds. Here's a sketch of the idea: Let L be an arbitrary> consistent in?ite subset of (the wffs of) FOL. Let G(L) be the set> of Goedel numbers of elements of L. Stick a decimal in front of eachSo you're restricting yourself to the case where the language is countable?> element of G(L). Now you have a set G(L)* contained in [0,1]. G(L)*> is obviously bounded, and in fact, (modulo some facts about> deductions) is closed (proving this is tricky). The rest gets kind of> messy (and frankly, I haven't thought through it all yet), but I'm> sure you ' can see the mechanism through which the proof works. The> reason I'm including this is because I'm hoping that exposing ' a> student to inter-disciplinary mathematics will help him understand> that what he is really studying is not a set of objects, but a set> of structures. Learning to identify structural features makes it> easier to ?d connections. Maybe I'll make this a guided exercise.It certainly sounds plausible, and is quite a neat idea. I'm currently trying to work through my own topological proof of compactness, and may see if I can steal some general ideas from this to feed into the proof I'm working on. :) Possibly some sort of embedding into [0, 1]^X, as I would be willing to bet money that you can't get a simple equivalence between heine-borel and the general compactness of FOL, as I don't think you can apply the axiom of choice in a non-trivial way to make it work, and compactness isn't true in ZF (it's equivalent to the ultra?ter theorem, and the fact that [0, 1]^X is compact uses UF).> The Lesbesgue integral is just as intuitive really, but introductory> measure theory is a bit of a pain. I'll look into the gauge integral.> Which variant are you working with? (Henstock, Perron, or the other> one?)I'm currently using the henstock variant, but I may take a look at some of the others or possibly ?dle the theory into my own bizarre form of the integral. :) Some of it seems a bit messy.Incidentally, the lebesgue integral isn't quite as easy to motivate as the riemann or gauge integrals, which is why I'm not starting with it. The latter two follow naturally from the desire to ?d primitives of functions, but the lebesgue one isn't so obvious. I may actually use the gauge integral to motivate the lebesgue integral - introducing measures as the gauge integral of the indicator function, as this will give lebesgue measure on the real line. I haven't reached that section yet, and haven't quite decided how to procede when I get there.> AlexDavid === =On Thu, 08 Jan 2004 11:09:31 +0000, David R MacIver>[...]>Out of curiousity, what relations between compactness (in the logical >sense) and analysis were you planning on mentioning? I've never seen any >other than non-standard analysis.> Well, it's possible to show that FOL is compact iff the Heine-Borel> theorem holds. Here's a sketch of the idea: Let L be an arbitrary> consistent in?ite subset of (the wffs of) FOL. Let G(L) be the set> of Goedel numbers of elements of L. Stick a decimal in front of eachSo you're restricting yourself to the case where the language is countable? element of G(L). Now you have a set G(L)* contained in [0,1]. G(L)*> is obviously bounded, and in fact, (modulo some facts about> deductions) is closed (proving this is tricky). The rest gets kind of> messy (and frankly, I haven't thought through it all yet), but I'm> sure you can see the mechanism through which the proof works. The> reason I'm including this is because ' I'm hoping that exposing a> student to inter-disciplinary mathematics will help him understand> that what he is really studying is not a set of objects, but a set> of structures. Learning to identify structural features makes it> easier to ?d connections. Maybe I'll make this a guided exercise.It certainly sounds plausible, and is quite a neat idea. Why does the fact that G(L)* seem plausible? It seems very implausibleto me (and for that matter it's clear that if true it depends on exactly what Godel-numbering scheme we use - one could certainlyinvent a scheme where G(L)* is not closed.)David C. UllrichX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === =In , on 01/06/2004 at 08:26 PM, poohonlsd@yahoo.com (Acid Pooh) said:>My overarching idea is to make a text that allows the student to gain>mathematical maturity while working with it. To this end, I'm hoping>to make the introductory chapters gentle, but making them more and>more dif?ult until even I'm challenged by the last few sections. >So even esoteric topics can be includedNonstandard Analysis? You don't mention what topics you plan to cover in Complex Analysis.Riemann surfaces would be nice, especially if you will be coveringhomotopies in the Topology material.Would topics like germs, forms and Homology be way out of scope?-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === In , on 01/06/2004> at 08:26 PM, poohonlsd@yahoo.com (Acid Pooh) said:>My overarching idea is to make a text that allows the student to gain>mathematical maturity while working with it. To this end, I'm hoping>to make the introductory chapters gentle, but making them more and>more dif?ult until even I'm challenged by the last few sections. >So even esoteric topics can be includedNonstandard Analysis? > Well, I'm not sure I really want to do that. It seems like I'd notonly have to develop formidable logical machinery, but I'd have tore-do many results in the classical analysis section. Ideally, I'dlike to cover the proof-theoretic side of logic a lot more than themodel theoretic. Since the logic section is introductory material,I'd like to make it as accessible as possible.> You don't mention what topics you plan to cover in Complex Analysis.> Riemann surfaces would be nice, especially if you will be covering> homotopies in the Topology material.Yes... I was planning on doing complex differentiation, contourintegration (and basic, related results), the residue theorem, andenough harmonic analysis to tie the complex analysis and Fourieranalysis together. But homotopies do sound like fun. I'll have to dosome reading, but this is a great idea.Would topics like germs, forms and Homology be way out of scope?Not if I end up doing Riemann surfaces. :) Generalizing from thecomplex case would be fun.Alex === i.e a rotation of x,y,z about the origin of x^2+y^2+z^2 is a homogeneous> linear transformation. It's the homogeneous bit I'm interested in, since it implies of the same> form. > Probably it means that the transformation ?es the origin. I think some> people used to use the word linear transformation for what people now> usually call an af?e transformation. An af?e transformation has the form> x->Ax +b where x is a dimension n column vector, A is an n x n matrix and b> is a ?ed dimension n column vector. If b = 0 then the af?e> transformation (in the old terminology) was said to be homogeneous, if> memory serves.Just that this topic of af?e transformation comes up and because yousay it is just a different name for linear transformation, can youalso please explain what a projective transformation is.It would be easier to tell if you gave the full statement and context. The> origin of x^2 + y^2+z^2 is a little strange.> Also, if the above transformation is a subset of a set of transformations,> why does this mean all the other transformations can be found?> jmc jmc> === i.e a rotation of x,y,z about the origin of x^2+y^2+z^2 is ahomogeneous> linear transformation.> It's the homogeneous bit ' I'm interested in, since it implies of thesame> form.Probably it means that the transformation ?es the origin. I think some> people used to use the word linear transformation for what people now> usually call an af?e transformation. An af?e transformation has theform> x->Ax +b where x is a dimension n column vector, A is an n x n matrixand b> is a ?ed dimension n column vector. If b = 0 then the af?e> transformation (in the old terminology) was said to be homogeneous, if> memory serves. Just that this topic of af?e transformation comes up and because you> say it is just a different name for linear transformation, can you> also please explain what a projective transformation is.I didn't say that af?e transformation is just a different name for lineartransformation. At least not nowadays. Linear transformations are mapping ffrom a vector space to another such that f(av+bw) = af(v)+bf(w) where a andb are any scalars and v and w are any vectors. An af?e transformation issimilar but instead satis?s the conditionf(av+(1-a)w) = af(v) + (1-a)f(w) where a is any scalar and v and w are anyvectors.A projective transformation is an element of the projective group P(V) whereV is a vector space. Seehttp://turnbull.mcs.st-and.ac.uk/~john/geometry/Lectures/ L20.html fordetails.See also John O'Connor's online ' geometry course for more about linear andaf?e transformation and their geometric signi?ance at http://turnbull.mcs.st-and.ac.uk/~john/geometry/index.html-- Edwin === A projective transformation is an element of the projective group P(V)whereI should have said ,A projective transformation is an element of the projective groupPGL(n,F).See http://turnbull.mcs.st-and.ac.uk/~john/geometry/Lectures/L20. htmlfor details.--Edwin === i.e a rotation of x,y,z about the origin of x^2+y^2+z^2 is a homogeneous> linear transformation.It's the homogeneous bit I'm interested ' in, since it implies of the same> form.> A function f is homogeneous of degree n iff f(t x) = t^n f(x) for all scalars t. A linear function is thus always a homogenousfunction of degree 1 (but not vice versa). As you indicated in otherposts that the quote comes from a rather old paper, my guess is thatthe the terminology was not yet standardized.Stephen J. Herschkorn === = Mathematicians are literate guys, so who ever ?st used homogeneous meant> of the same form i.e. ax +by+cz+b isn't homogoneous since b isn't of the> same ' form as ax,by,cz?According to Jeff Miller's nice webpage at http://members.aol.com/jeff570/mathword.htmlthe ?st use is in:HOMOGENEOUS EQUATIONS is found in 1815 in the second edition of Hutton'smathematics dictionary: Homogeneous Equations ... in which the sum of thedimensions of x and y... rise to the same degree in all the terms (OED2). === =I would like to know where can I ?d a proof of the followingstatement: if M is a real analytic manifold and v is an analyticvector ?ld on M, then the ?sociated with M is alsoanalytic.Jose Carlos SantosX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08DRsi27356; === =The kite equation thread inspired me to think back to another property Ive had in the back of my mind for a while. I think I can be fairly sure someone has thought o? before.Let X be a complex normed space. Let B_n(r) = (x_1, x_2, ...,x_n) (a ?ite sequence of elements of X) be an n-gon of radius r; this means(a) x_i != x_j for all 1 <= i,j <= n(b) dim (lin {x_1, x_2, ...,x_n}) = 2(c) ||x_1|| = ||x_2|| = ... ||x_n|| = r > 0 (d) ||x_1 - x_2|| = ||x_2 - x_3|| = ... = ||x_n - x_1|| = s_n > 0Call (B_n(r)) a sequence of n-gons if: for all n in naturals:(1) B_n(r) subset B_{n+1}(r) (2) B_n(r) != B_{n+1}(r)(3) B_{n+1}(r) is an n-gon of radius rFor all n in naturals, de?e s_{n+1} to be as in (d), above,for B_{n+1}(r). What if all we know about X is that it is a normed space and:lim n -> infty (n*s_n/2r) -> pi = 3,14...) for every n-gon sequence(or use for every two dimensional subspace, there exists an n-gonsequence... - which I believe should be equivilant to the above) Notice that if X is a normed space and ||x|| = ||y|| > 0,then ||x-y||/||x|| <= (||x|| + ||y||)/||x|| = 2||x||/||x|| = 2Thus, by requiring X be normed space, we are actuallystating that lim n -> infty (n*s_n/2r) >= 2 for X. It is alsothe property that there are nonzero vectors orthogonal to each other (-> ||x+y|| = ||x-y||) (this may be the case in every lim n -> infty (n*s_n/2r >= 2,6 just by using the parallelogram equation (hint: make a square). Note: actually, I forgot exactly what the number was (2,6 or something close) since its been a while and I dont know where my notes are on that anymore. However, it shouldnt take long to recheck.For me, it would be interesting to know any of the two C. DementX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08DRud27437; === =i'm writing a little project about groupoids & i'd like to talk a bit about applications to differential geometry and the like. does anyone know of any texts or papers that deal with this kind of thing at a fairly elementary level? (preferably not in french.) X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08DRt427374; === 3.14159265358979323...3,1,4,5,?,?,....>How many digits of pi have you memorized? I did 100. The world recordI knew about 1500 digits at one time. A. Say I HAVE A NUMBER to remember PI as 3.1416 B. Desire MAY I HAVE A LARGE CONTAINER OF COFFEE to remember PI as 3.1415926 C. Chant GOPEEBHAGYA MADHURVAATHA SHRADGI SHODHADHI SANDHIGA I KHALAJEEVITHA KHAATHAAVA GALAHAALAARA SANDHARA II a sanskrit verse from Atharvana veda,which is in praise of Lord Vishnu and Lord Shankara.If this is deciphered as per language of Vedic Mathematics,it denotes the value of PI/10 to 32 decimal places as below: .31415926535897932384626433832792 You need to know/learn which of the alphabets represent each of the numerals 1,2,3,4 .... Krsna Tirthaji Maharaja(1884-1960)Sankaracharya of Govardhana Matha,Puri,Orissa State,INDIA) D. Professor Yasumasa Kanada and nine other researchers at The Information Technology Centre at Tokyo University,calculated the value of PI to 1.2411 trillion places,in September 2002. which computed this value working continuously for over 400 hours. - bsrX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08DRt427395; === = Dear Phorum Participants!Please help to deal with the following sum:S(n) = sum {k=1,...,n} n!/(k*(n-k)!)Need closed form solution or asymptotic estimations for large n. === Please help to deal with the following sum:>S(n) = sum {k=1,...,n} n!/(k*(n-k)!)>Need closed form solution or asymptotic estimations for large n.These are called logarithmic numbers. See sequenceA002104 at the On-Line Encyclopedia of Integer Sequences,,and references given there.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Dear Phorum Participants!> Please help to deal with the following sum:> S(n) = sum {k=1,...,n} n!/(k*(n-k)!)> Need closed form solution or asymptotic estimations for large n.Mathematica gives a closed form for your sum. Not surprisingly, it's interms of a hypergeometric function: n HypergeometricPFQ[{1, 1, 1 - n}, {2}, -1]See for moreinformation, if necessary.HTH,David === Dear Phorum Participants!> Please help to deal with the following sum:> S(n) = sum {k=1,...,n} n!/(k*(n-k)!)> Need closed form solution or asymptotic estimations for large n.Mathematica gives a closed form for your sum. Not surprisingly, it's in> terms of a hypergeometric function: n HypergeometricPFQ[{1, 1, 1 - n}, {2}, -1]See for more> information, if necessary.HTH,> DavidAnd can Mathematica show its asymptotics for large n?-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === = > Dear Phorum Participants!> Please help to deal with the following sum:> S(n) = sum {k=1,...,n} n!/(k*(n-k)!)> Need closed form solution or asymptotic estimations for large n. Mathematica gives a closed form for your sum. Not surprisingly, it's in> terms of a hypergeometric function: n HypergeometricPFQ[{1, 1, 1 - n}, {2}, -1] See for more> information, if necessary. And can Mathematica show its asymptotics for large n?Unfortunately it can't, AFAIK.David === Dear Phorum Participants!> Please help to deal with the following sum:> S(n) = sum {k=1,...,n} n!/(k*(n-k)!)> Need closed form solution or asymptotic estimations for large n.> Mathematica gives a closed form for your sum. Not surprisingly, it's in> terms of a hypergeometric function:> n HypergeometricPFQ[{1, 1, 1 - n}, {2}, -1]> See for more> information, if necessary.>And can Mathematica show its asymptotics for large n?I would not try the closed form for it, however. I seean asymptotic expansion coming out easily.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue UniversityX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08FmcF05762; === =Dear Dr. Israel,There is currently a thread Naming rights:------------------------------------------------------ ---------Just out of curiosity, with all these unproven conjectures about...what happens when the conjecture is proven to be correct? Does thenewly declared theorem retain the name of the person originating theconjecture, or will it get named after the person(s) who proves it?[snip]---------------------------------------------------- --------------Well, as far as I'm concerned, if the kite equation were toever show up in any liturature (For example: as an example),and you feel somehow displeased with that name- just go aheadand rename it something else of your own choosing!C. DementWhat if all that we know about X is that it is a real>normed space and:for all x, y in X, for all t in reals:>||x|| = ||y|| -> ||x + t(x + y)|| = ||y + t(x + y)|| Informally, the structure sais If x and y are of the same norm, >then x and y are the same distance away from each point on the line>passing through 0 - orthogonal to the segment passing through >x and y. It is fairly easy to show that the kite equation above holds>for R^n. But I have not found it easy to show either of the two >general real normed space X. Am I missing something easy?for any scalars a, b. Fix any 2-dimensional subspace V, and this says>that for any x and y in the subspace with ||x|| = ||y||, there is a linear>isometry of V that interchanges x and y. It has eigenvalues -1 and 1>(with eigenvectors x+y and x-y) and thus determinant -1: it is a >re?n. The product of two different re?ns (say T_1 T_2 >where T_1 interchanges y and z and T_2 interchanges x and z) is a linear >isometry T of determinant 1, and eigenvalues in the unit circle, taking>x to y. Taking a basis of V, this corresponds to a matrix.>The eigenvalues of this matrix vary continuously with y, so>it will be possible to have eigenvalues exp((+/-) i theta)>where theta is not a rational multiple of pi. If w = a + i b>(with a and b in R^2, corresponding to vectors in V) is an >eigenvector of the matrix for eigenvalue exp(-i theta), we have >T a = cos(theta) a + sin(theta) b and T b = - sin(theta) a + cos(theta) b.>Moreover, T^n a = cos(n theta) a + sin(n theta) b, and so >||cos(n theta) a + sin(n theta) b|| = ||a||. But the vectors>[cos(n theta), sin(n theta)] are dense in the unit circle, so we>actually have ||cos(t) a + sin(t) b|| = ||a|| for all t. And this>implies that the mapping [x,y] -> x a + y b is an isometry of R^2>onto V. In particular we have the parallelogram law in V. But >since V was an arbitrary two-dimensional subspace of X, you ?d>that the parallelogram law holds in X, which means X can be made>into an inner product space.Robert Israel israel@math.ubc.ca>Department of Mathematics http://www.math.ubc.ca/~ israelUniversity of British Columbia >Vancouver, BC, Canada V6T 1Z2X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08Fmcd05782; === =Three corrections to my ?st message:1. dim X >= 2 should be prerequisited.2. lim n -> infty (n*s_n/2r) >= 2 does not follow from the triangle inequality as shown. It follows from norm homogenity: For x != 0, take the 2-gon (x, -x) Then the distance from x to -x,and from -x back to x, divided by twice the radius r = ||x|| is (||x - (-x)|| + ||-x - x||)/(2||x||) = 4||x||/(2||x||) = 23. Not so much a correction, as (hopefully helpful) statement I forgot to include: should be expected. If I remember correctly from my (lost) notes, one can go from the square 4-gon (x, y, -x, -y) where ||x+y|| = ||x-y|| to an 8-gon (x, (x+y)/2, y, (y-x)/2, -x, (-x - y)/2, -y, (-y +x)/2) Increasing the lower bound estimate of lim n -> infty (n*s_n/2r) from ~2.6 to something closer to 3. Of course, one ?st has to assume that the limit exists, i.e. (n*s_n/2r) is a Cauchy-sequence. >The kite equation thread inspired me to think back >to another property Ive had in the back of my mind for a >while. I think I can be fairly sure someone has thought of>it before.Let X be a complex normed space. Let B_n(r) = (x_1, x_2, ...,x_n) >(a ?ite sequence of elements of X) be an n-gon of radius >r; this means(a) x_i != x_j for all 1 <= i,j <= n(b) dim (lin {x_1, x_2, ...,x_n}) = 2(c) ||x_1|| = ||x_2|| = ... ||x_n|| = r > 0>(d) ||x_1 - x_2|| = ||x_2 - x_3|| = ... = ||x_n - x_1|| = s_n > 0Call (B_n(r)) a sequence of n-gons if: >for all n in naturals:>(1) B_n(r) subset B_{n+1}(r) >(2) B_n(r) != B_{n+1}(r)>(3) B_{n+1}(r) is an n-gon of radius rFor all n in naturals, de?e s_{n+1} to be as in (d), above,>for B_{n+1}(r).>What if all we know about X is that it is a normed space and:lim n -> infty (n*s_n/2r) -> pi = 3,14...) for every n-gon sequence>(or use for every two dimensional subspace, there exists an n-gon>sequence... - which I believe should be equivilant to the above)>Notice that if X is a normed space and ||x|| = ||y|| > 0,>then ||x-y||/||x|| <= (||x|| + ||y||)/||x|| = 2||x||/||x|| = 2>Thus, by requiring X be normed space, we are actually>stating that lim n -> infty (n*s_n/2r) >= 2 for X. It is also>the property that there are nonzero vectors orthogonal to each >other (-> ||x+y|| = ||x-y||) (this may be the case in every >lim n -> infty (n*s_n/2r >= 2,6 just by using the parallelogram >equation (hint: make a square). Note: actually, I forgot exactly what the number was (2,6 or >something close) since its been a while and I dont know where my >notes are on that anymore. However, it shouldnt take long to >recheck.For me, it would be interesting to know any of the two C. Dement>X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08FmcN05758; === = Who proved that an in?ite number of numbers (i.e. in?ite series)> can have a ?ite sum?No one proved that.> An in?ite sum is a de?ition.If you have a series:> t_1, t_2, t_3, ...,t_n, ....> and you make the associated series of partial sums:> s_1, s_2, s_3, ....., s_n, ....> where the partial sums s_j are de?ed as> s_j = sum( t_i, i=1..j ) for all natural j,> then the series is *de?ed* to> have a ?ite sum> if one can prove the following:> There is a real number S, such that> for each positive number e > 0> there is a natural number N, such that> for all natural numbers n:> if n > N then | sn - S | < e.> In this case, the number S is called:> the sum of the seriesNo one proved that.> Someone de?ed it.All right, instead of proved let's substitute observed and answer>the OP's intended question.It goes back at least to the Greeks, who were aware of geometric>series. (And this was the source of Zeno's paradoxes.) I suspect it>goes back still further.Somewhere in the IRS forms [and this is no joke, it's true] there is a>calculation whose exact nature I don't recall, so I'll invent a>parallel scenario using the same idea: let's suppose you live in a>state with 8% sales tax, whose gropenfuhrer, er, excuse me, governor,>wants to increase revenues without raising the sales tax rate. So he>gets the brilliant idea: we'll TAX THE TAX. That is, pre-gropenfuhrer>a $100 purchase would have cost $100 + $100 x 0.08 = $108.But the new proposal ALSO taxes the $8 tax, thus: $100 + $100 x 0.08 + $8 x 0.08 = $108.64.And not to let a good idea go to waste, that tax on a tax is also taxed: $100 + $8 + $0.64 + $0.64 x 0.08 = 108.6912.Then that extra nickel is taxed, etc. etc. until common sense prevails>and the tax is less than half a cent and even the gropenfuhrer takes>his hand out of your, er, pocket.Now, we mathematicians understand that the tax rate is equivalent to a>simple tax of 0.08 + 0.08^2 + 0.08^3 + ... = 0.08/0.92 = 0.0869565... >(slightly less, on a ?ite purchase, because of the roundoff).Well, don't tell that to the gropenfuhrer or he might start the whole process over again.>The IRS, incredibly enough, has the taxpayer ?l out the form>calculating the tax-on-tax, tax-on-tax-on-tax, etc., instead of just>dividing tax by (1-taxrate). Actually, I don't think the real-life calculation was of a tax, but a>credit, but I don't feel like looking it up right now.--Ron Br === =In sci.math, James Harrison 7 Jan 2004 09:48:10 -0800<3c65f87.0401070948.66ddc0fc@posting.google.com>:> I don't know if you noticed or not, but I've been in an argument that> has gone on for quite some time about a problem with the ring of> algebraic integers. Lily for me, I've found the way to end the> arguing, by relying on your understanding of *independent* variables. > Note that you have x and y below where x is INDEPENDENT of y, and vice> versa. My hope is that such a simple concept that should be familiar> to all, will end the arguing.I've been using a modi?ation of an example put forward by Rick> Decker, a professor at Hamilton College.Consider,7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7(xy - y^2)(5) + 7^2 y^2so(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of a^2 - (x - y)a + 7(x^2 + xy).Notice that x and y are independent variables.And their domain is ... ?Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, I leta_2(x,y) = b_2(x,y) - y, so(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2) Which shows that dividing both sides by 7 must result in(5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2as long as x does not equal y, as then some of the terms with y as a> factor that do NOT have x as a factor are clipped out.The point is that y is an independent variable, so it and its> coef?ients CANNOT vary based on the value of x, so my setting x=0 is> NOT a special case!!!Oh yes it is. You've not worked out the roots for the general casea_i(x,y) yet. You've merely shown that a_1(0,y) = 0 and a_2(0,y) = -y,which is true enough as it turns out.It turns out (x - y) + sqrt(-27*x^2 - 30*y*x + y^2)a_1(x,y) = -------------------------------------- 2 (x - y) - sqrt(-27*x^2 - 30*y*x + y^2)a_2(x,y) = -------------------------------------- 2> It's a way to clear x out of the picture to see> what the coef?ients are for the terms with y as a factor that do NOT> have x as a factor as well.Now then, on to what happens if x=y. Then you havea_1(y,y) = 2 * i * sqrt(14) * ya_2(y,y) = -2 * i * sqrt(14) * ya^2 + 7(x^2 + xy),where I've used x-y=0 to cancel out the middle term but left the> variable names as they were.That case IS a special case as notice that it is indistinguishable> from the case if in generala^2 + 7(x^2 + xy) = 0,with independent x, and y.Previous discussions were basically looking over the special case> where y=1.The implications of the result are quite huge, as now you can put in> some numbers like y=1, x=2, to dodge the special case at x=1, to get> a_1(2,1) = (1 + sqrt(-167))/2a_2(2,1) = (1 - sqrt(-167))/2 a^2 - a + 42 = 0, which solves to a = (1+/-sqrt(-167))/2Well, I'm happy. :-) The general equations check out.and, now you know that *one* of the solutions is coprime to 7.If a is a root of a^2 - a + 42 = 0, then b = a/7 is a root of( (7b)^2 - (7b) + 42)/7^2 = 0orb^2 - b/7 + 6/7 = 0Sorry, neither solution is divisible by 7, as b is not analgebraic integer; your logic is ? (I can't pinpointwhere, though.)One thing worth noting is that operator ambiguity means that> sqrt(-167) represents two numbers, as that operator gives *two*> solutions.Therefore, while the sqrt() operator is in place you cannot further> resolve the expression to see *which* of the solutions is coprime to> 7.Those willing to go looking can try to ?d their own quadratics like> Decker did, and ?d one with integer roots, for integer x, and then> you can see the result directly.Of course the problem with the ring of algebraic integers is the> implication from previous interpretations (here I rely on what I've> heard primarily from sci.math posters like Arturo Magidin, Nora> Baron, and Dik winter) is that *both* roots have non-unit factors in> common with 7 in the ring of algebraic integers.Want more advanced polynomial factorization?Then check out my blog archives:James Harris-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === = > The implications of the result are quite huge, as now you can put in > some numbers like y=1, x=2, to dodge the special case at x=1, to get > a^2 - a + 42 = 0, which solves to > a = (1+/-sqrt(-167))/2 > and, now you know that *one* of the solutions is coprime to 7.Now that Keith Ramsey has posted algebraic integers p, q, r and ssuch that p.q = 7^11, r.s = 6^11, p.r = ((1 + sqrt(-167))/2)^11 andq.s = ((1 - sqrt(-167))/2)^11, can we put that nonsense to a rest? > Of course the problem with the ring of algebraic integers is the > implication from previous interpretations (here I rely on what I've > heard primarily from sci.math posters like Arturo Magidin, Nora > Baron, and Dik winter) is that *both* roots have non-unit factors in > common with 7 in the ring of algebraic integers.Well, as Keith Ramsey has shown those factors in the case above, whatnow?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === = [.snip.]Correction, I think:>No. If one is coprime to 7 the other is, too. They are associates,>meaning you can ?d a unit u for which (1 + sqrt(-167)/2 = u(1 - sqrt(-167))/2In other words, as far as divisibility is concerned, the two values>are indistinguishable. > I *think* this is incorrect. Both numbers are conjugate (there is> Galois automorphism of Q(sqrt(-167) sending one to the other, namely> complex conjugation). But I don't think they are associates.>Yes. I noticed that after making my post. They are of course conjugates. Rick === = But a_1(x, y)/7 isn't in general an algebraic integer. You have> several examples of that. You were better off with your argument> involving w_1 and w_2.> Harris knows this much. He just doesn't accept the implications ... as long as x does not equal y, as then some of the terms with y as a> factor that do NOT have x as a factor are clipped out.The point is that y is an independent variable, so it and its> coef?ients CANNOT vary based on the value of x, so my setting x=0 is> NOT a special case!!! It's a way to clear x out of the picture to see> what the coef?ients are for the terms with y as a factor that do NOT> have x as a factor as well.Now then, on to what happens if x=y. Then you havea^2 + 7(x^2 + xy),where I've used x-y=0 to cancel out the middle term but left the> variable names as they were. > You can do that if you want, but in terms of values> you still have a^2 + 7(2x^2), so no matter what you call the> variables, the roots will be +/- (insert some integer here)*sqrt(-14)and unless 7 divides x and y, that critter won't be divisible by 7. > The implications of the result are quite huge, as now you can put in> some numbers like y=1, x=2, to dodge the special case at x=1, to get > a^2 - a + 42 = 0, which solves to a = (1+/-sqrt(-167))/2and, now you know that *one* of the solutions is coprime to 7.> This is of course not true. Let a1 = (1 + sqrt(-167))/2 anda2 = (1 - sqrt(167))/2. Since a1*a2 = 7*6,suppose as Harris claims that a2 (for example) is coprime to 7. Thismeans that a1 is *divisible* by 7. That is, a1 = 7*b where b is an algebraicinteger. Therefore 14*b - 1 = sqrt(-167), or 196*b^2 - 28*b + 1 = -167, or b*(7*b - 1) = 6, or 7*b^2 - b - 6 = 0. This is an irreducible, non-monic, primitive polynomial in b. Therefore b cannot be an algebraic integer. > Why do you know that? Certainly not from your factorization above. > One thing worth noting is that operator ambiguity means that> sqrt(-167) represents two numbers, as that operator gives *two*> solutions.Therefore, while the sqrt() operator is in place you cannot further> resolve the expression to see *which* of the solutions is coprime to> 7. > No. If one is coprime to 7 the other is, too. They are associates,> meaning you can ?d a unit u for which (1 + sqrt(-167)/2 = u(1 - sqrt(-167))/2In other words, as far as divisibility is concerned, the two values> are indistinguishable.> No, this isn't quite right either. Note that u - 1 = (u + 1)*sqrt(-167).Squaring both sides, u^2 - 2*u + 1 = -167*(u^2 + 2*u + 1), or 42*u^2 + 83*u + 42 = 0Thus u is not an algebraic integer and not a unit.It's close to a unit, in the sense that u^2 + (83/42)*u + 1 is close to u^2 + u + 1,whose roots are units. The most you can say, as above, is that both (1 + sqrt(-167))/2 and (1 - sqrt(-167))/2 are notcoprime to 7. It would be nice to know explicitly whatfactors each has in common with 7. It is not the samefactor, since otherwise it would be sqrt(7). Nora B.> Rick === =meaning you can ?d a unit u for which (1 + sqrt(-167)/2 = u(1 - sqrt(-167))/2In other words, as far as divisibility is concerned, the two values>are indistinguishable. No, this isn't quite right either. Note that u - 1 = (u + 1)*sqrt(-167).Squaring both sides, u^2 - 2*u + 1 = -167*(u^2 + 2*u + 1), or 42*u^2 + 83*u + 42 = 0Thus u is not an algebraic integer and not a unit.Yeah. I goofed. I can't even claim that it was a numericalerror. Guess I was just passing a brain stone. The most you can say, as above, is that both > (1 + sqrt(-167))/2 and (1 - sqrt(-167))/2 are not> coprime to 7. It would be nice to know explicitly what> factors each has in common with 7. It is not the same> factor, since otherwise it would be sqrt(7).> It would be nice, but I suspect that there isn'tanything more explicit than expressing gcd((1 + sqrt(-167))/2, 7)as the root of [some high-degree polynomial]though I confess I haven't looked too closely at it.Rick === =The implications of the result are quite huge, as now you can put in>some numbers like y=1, x=2, to dodge the special case at x=1, to get >a^2 - a + 42 = 0, which solves to a = (1+/-sqrt(-167))/2and, now you know that *one* of the solutions is coprime to 7.One thing worth noting is that operator ambiguity means that>sqrt(-167) represents two numbers, as that operator gives *two*>solutions.Therefore, while the sqrt() operator is in place you cannot further>resolve the expression to see *which* of the solutions is coprime to>7.>Let primes denote complex conjugation and let c denote a root ofa^2 - a + 42 = 0 supposed to be coprime to 7 in the ring A ofalgebraic integers. Note that the other root is c'.If c and 7 are coprime in A then there are elements p and q in Asuch that cp + 7q = 1.So 1 = 1' = (cp + 7q)' = (cp)' + ' (7q)' = c'p' + 7'q' = c'p' + 7q',and since p' and ' q' are in A, c' and 7 are coprime in A.Since c and c' are each coprime to 7 in A, so also is theirproduct, cc' = 42. That is absurd, so the original supposition isfalse and neither root is coprime to 7 in A.John Roberts-Jones === = But a_1(x, y)/7 isn't in general an algebraic integer. You have> several examples of that. You were better off with your argument> involving w_1 and w_2.I didn't say that they are in general algebraic integers.But since you brought up them NOT being algebraic integers in generalI think it's a good time to explain again the error in coremathematics.The problem isn't that you're pushed out of the ring of algebraicintegers, but that mathematicians seem intent on refusing toacknowledge that happens.So mathematicians just *decide* that they're still in the ring ofalgebraic integers, and with results irreducible over rationals, theycan't see anything that contradicts them.What I did was use a special technique that I call advanced polynomialfactorization to break through the barrier.If mathematicians accede to mathematical reality, then the limitationsof the ring of algebraic integers are known, and the error is nolonger in core.Algebraic integers are just quirky. > as long as x does not equal y, as then some of the terms with y as a> factor that do NOT have x as a factor are clipped out.The point is that y is an independent variable, so it and its> coef?ients CANNOT vary based on the value of x, so my setting x=0 is> NOT a special case!!! It's a way to clear x out of the picture to see> what the coef?ients are for the terms with y as a factor that do NOT> have x as a factor as well.Now then, on to what happens if x=y. Then you havea^2 + 7(x^2 + xy),where I've used x-y=0 to cancel out the middle term but left the> variable names as they were. > You can do that if you want, but in terms of values> you still have a^2 + 7(2x^2), so no matter what you call the> variables, the roots will be +/- (insert some integer here)*sqrt(-14)and unless 7 divides x and y, that critter won't be divisible by 7.> The thing is that x=y is a special case as (x-y)a goes to 0. Noticethat in ?ding the best picture of the factorization, when I set x=0,-ya is what means that one of the a's is not 0.That is,a^2 - (x - y)a + 7(x^2 + xy) = 0, with x=0, givesa^2 + ay = 0, so a(a+y) = 0, but if that term is NOT there, then youhavea^2 = 0.> The implications of the result are quite huge, as now you can put in> some numbers like y=1, x=2, to dodge the special case at x=1, to get > a^2 - a + 42 = 0, which solves to a = (1+/-sqrt(-167))/2and, now you know that *one* of the solutions is coprime to 7. > Why do you know that? Certainly not from your factorization above.> Well, from(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)it's clear that only one of the factors (5b_2(x,y) + 2y) doesn't ingeneral share any factors in common with 7.Now from that picture the other factor seems like it should, but thering of algebraic integers doesn't allow it, so in fact, no non-unitfactors in common with 7 can be divided from (5a_1(x,y) + 7y) to givean algebraic integer result. > One thing worth noting is that operator ambiguity means that> sqrt(-167) represents two numbers, as that operator gives *two*> solutions.Therefore, while the sqrt() operator is in place you cannot further> resolve the expression to see *which* of the solutions is coprime to> 7. > No. If one is coprime to 7 the other is, too. They are associates,One is coprime to 7, while the other has a muddled picture in the ringof algebraic integers. In a more inclusive ring, the other has 7 as afactor.> meaning you can ?d a unit u for which (1 + sqrt(-167)/2 = u(1 - sqrt(-167))/2In other words, as far as divisibility is concerned, the two values> are indistinguishable.> Why don't you elaborate on why you make that claim?James Harris === = Well, from(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)it's clear that only one of the factors (5b_2(x,y) + 2y) doesn't in> general share any factors in common with 7.>It can. If x = y = 1 you'll have a_1 = sqrt(-14) a_2 = -sqrt(-14)so, leting b_2 = 1 - sqrt(-14) we have (5a_1 + 7)(5b_2 + 2) = (5sqrt(-14) + 7)(5(1 - sqrt(-14) + 2) = (7)(57) = P(1, 1)and the ?st factor is not divisible by 7. However, both ofthe factors are divisible by sqrt(7).Thus, in spite of the constant term, 2, in the second factor,we see that it does indeed have a factor in common with 7.The same sort of thing happens with x = 2, y = 1, though thenumbers are trickier to calculate. Rick === there are no poxitions after in?itely many nines, and particularly, > there is no last one. Thus your easy is easily wrong.Squares of 0.999... tend away from Last One 9^4 = 6561 .9^4 = .6561 99^4 = 96059601 .99^4 = .96059601 999^4 = 996005996001 .999^4 = .996005996001 9999^4 = 9996000599960001 .9999^4 = .9996000599960001 99999^4 = 99996000059999600001 .99999^4 = .99996000059999600001 999999^4 = 999996000005999996000001 .999999^4 = .999996000005999996000001 9999999^4 = 9999996000000599999960000001 .9999999^4 = .9999996000000599999960000001 99999999^4 = 99999996000000059999999600000001 .99999999^4 = .99999998000000059999999600000001 999999999^4 = 999999996000000005999999996000000001 .999999999^4 = .999999996000000005999999996000000001 9999999999^4 = 9999999996000000000599999999960000000001 .9999999999^4 = .9999999996000000000599999999960000000001 99999999999^4 = 99999999996000000000059999999999600000000001 .99999999999^4 = .99999999996000000000059999999999600000000001 999999999999^4 = 999999999996000000000005999999999996000000000001.999999999999^ 4 = .999999999996000000000005999999999996000000000001[snip rest for brevity]Challenge: prove Last One wrong. Garry Denke, GeologistDenoco Inc. of Texas === =*** post for FREE via your newsreader at post.newsfeed.com ***> there are no poxitions after in?itely many nines, and particularly,> there is no last one. Thus your easy is easily wrong. Squares of 0.999... tend away from Last One 9^4 = 6561> .9^4 = .6561 99^4 = 96059601> .99^4 = .96059601 [snip rest for brevity] Challenge: prove Last One wrong.But, Last One is not wrong. It exists for each and every member of yourlist. karl mhttp://www.newsfeed.com - The #1 Newsgroup Service in the World!-----== 100,000 Groups! - 19 Servers! - Unlimited Download! =----- === =In sci.logic, Dave Seamanon Wed, 7 Jan 2004 17:22:28 +0000 (UTC):[3] The digit sequence .999... is meaningless as one must always> use a trailing digit approach, writing it as .999...9, where ...> can be either ?ite (which is allowed under the Standard> Representation) or in?ite (which the Standard Representation> does not cover and dismisses). This means that operations> such as (.999...9)^2 = .999...99800...001 are possible.Note that [3] would require a bit of work rewriting trans?ite> theory, as omega (the ordinal analogue to the cardinal in?ity)> is such that omega+1 = omega in the Standard Representation.Omega+1 has the same cardinality as omega, but they are different> ordinals. If the (w+1)-strings of digits are ordered lexicographically,> then there is no order-preserving map of such strings to the reals.> Interesting. So expressions such as omega+1 and 2*omega actually domake sense? If so, mea culpa.However, if we do allow trans?ites and in?itesimals into ournumber system a la Garry Denke, I'm not sure how the rest of mathematicsis going to deal with it (the concept of limit in particular may haveto be rede?ed).Hopefully it's a simple retro? but I wonder.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === In sci.logic, Dave Seaman>:> [3] The digit sequence .999... is meaningless as one must always> use a trailing digit approach, writing it as .999...9, where ...> can be either ?ite (which is allowed under the Standard> Representation) or in?ite (which the Standard Representation> does not cover and dismisses). This means that operations> such as (.999...9)^2 = .999...99800...001 are possible.> Note that [3] would require a bit of work rewriting trans?ite> theory, as omega (the ordinal analogue to the cardinal in?ity)> is such that omega+1 = omega in the Standard Representation.> Omega+1 has the same cardinality as omega, but they are different> ordinals. If the (w+1)-strings of digits are ordered lexicographically,> then there is no order-preserving map of such strings to the reals.> Interesting. So expressions such as omega+1 and 2*omega actually do> make sense? If so, mea culpa.Yes. In fact, w is identical to N, the set of natural numbers (including 0),and w+1 = w U {w} = { 0, 1, 2, 3, ..., w }. Addition and multiplication are noncommutative: 1+w = w < w+1, (w+1 has a last element but 1+w does not) 2*w = w < w*2 = w+w. (an w of pairs vs. a pair of w's)> However, if we do allow trans?ites and in?itesimals into our> number system a la Garry Denke, I'm not sure how the rest of mathematics> is going to deal with it (the concept of limit in particular may have> to be rede?ed).> Hopefully it's a simple retro? but I wonder.There's nothing wrong with considering trans?ite ordinal sequences ofdigits in general, but as I said, there is no order-preserving map ofsuch objects to the reals for ordinals > w.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === All he ever did was decline to de?e what he meant by 0.999...> He said it wasn't a limit, but never said what it was. karl m1 = 10.999... = 0.999...Squares of 0.999... tend away from 1 9^4 = 6561 .9^4 = .6561 99^4 = 96059601 .99^4 = .96059601 999^4 = 996005996001 .999^4 = .996005996001 9999^4 = 9996000599960001 .9999^4 = .9996000599960001 99999^4 = 99996000059999600001 .99999^4 = .99996000059999600001 999999^4 = 999996000005999996000001 .999999^4 = .999996000005999996000001 9999999^4 = 9999996000000599999960000001 .9999999^4 = .9999996000000599999960000001 99999999^4 = 99999996000000059999999600000001 .99999999^4 = .99999998000000059999999600000001 999999999^4 = 999999996000000005999999996000000001 .999999999^4 = .999999996000000005999999996000000001 9999999999^4 = 9999999996000000000599999999960000000001 .9999999999^4 = .9999999996000000000599999999960000000001 99999999999^4 = 99999999996000000000059999999999600000000001 .99999999999^4 = .99999999996000000000059999999999600000000001 999999999999^4 = 999999999996000000000005999999999996000000000001.999999999999^ 4 = .999999999996000000000005999999999996000000000001[snip rest for brevity]Challenge: Prove 0.999... squares tend toward 1 Garry Denke, GeologistDenoco Inc. of Texas === =*** post for FREE via your newsreader at post.newsfeed.com ***> All he ever did was decline to de?e what he meant by 0.999...> He said it wasn't a limit, but never said what it was. karl m 1 = 1 0.999... = 0.999... Squares of 0.999... tend away from 1 9^4 = 6561> .9^4 = .6561 99^4 = 96059601> .99^4 = .96059601It sure looks to me like it tends toward 1. As has already been pointedout to you, each 9 you add on the left adds another 9 on the right.karl mhttp://www.newsfeed.com - The #1 Newsgroup Service in the World!-----== 100,000 Groups! - 19 Servers! - Unlimited Download! =----- === I did not say that .999... = 1 isn't true; I merely was> pointing out a problem with the proof. As it turns out,> .999... = 1 by any conventional proof (the simplest one> uses Cauchy sequences, as the snipped part indicated)> and therefore the proof works anyway, except for the> circularity problem.[snip non-topic text]Another problem are its squares, like 0.999...^4, which do not tendtowards 1 as falsely stated in this thread, but do tend away from 1 9^4 = 6561 .9^4 = .6561 99^4 = 96059601 .99^4 = .96059601 999^4 = 996005996001 .999^4 = .996005996001 9999^4 = 9996000599960001 .9999^4 = .9996000599960001 99999^4 = 99996000059999600001 .99999^4 = .99996000059999600001 999999^4 = 999996000005999996000001 .999999^4 = .999996000005999996000001 9999999^4 = 9999996000000599999960000001 .9999999^4 = .9999996000000599999960000001 99999999^4 = 99999996000000059999999600000001 .99999999^4 = .99999998000000059999999600000001 999999999^4 = 999999996000000005999999996000000001 .999999999^4 = .999999996000000005999999996000000001 9999999999^4 = 9999999996000000000599999999960000000001 .9999999999^4 = .9999999996000000000599999999960000000001 99999999999^4 = 99999999996000000000059999999999600000000001 .99999999999^4 = .99999999996000000000059999999999600000000001 999999999999^4 = 999999999996000000000005999999999996000000000001.999999999999^ 4 = .999999999996000000000005999999999996000000000001[snip rest for brevity]Challenge: Prove the squares of 0.999... tend toward 1. Garry Denke, GeologistDenoco Inc. of Texas === Another problem are its squares, like 0.999...^4, which do not>tend towards 1 ^^^^^^^^^^^^^^>as falsely stated in this thread, but do>tend away from 1 ^^^^^^^^^^^^^^^^ .9^4 = .6561> .99^4 = .96059601[snip]> .99999999999^4 = .99999999996000000000059999999999600000000001>.999999999999^4 = .999999999996000000000005999999999996000000000001It now seems clear that the reason Garry is talking past everyone else, and vice versa, is that he has a different meaning of tends to 1 in mind. This is illuminated by a subsequent post:> there are no poxitions after in?itely many nines, and particularly, > there is no last one. Thus your easy is easily wrong.Squares of 0.999... tend away from Last One .9^4 = .6561[snip]>.999999999999^4 = .999999999996000000000005999999999996000000000001[snip rest for brevity]Challenge: prove Last One wrong. In the *abstract numerical value* view, the claim that .999999999999^4 is closer to 1 than .9^4 is (i.e. that |1-.999999999999^4| < |1-.9^4|) is (I hope) indisputably true, so that clearly isn't the view that Garry is meaning here.ISTM he (and Charlie Volkstorf) hold a *concrete representational* view. For Garry, the successively longer decimal fractions *look* progressively less like the simple numeral 1.A big problem with Garry's view is that the concrete decimal representation of .999... is unending, yet he hold fast to the notion that there is a last 9 digit (and hence a last 1 digit in (.999...)^2 and (.999...)^4). This reminds me of Phil.It turns out that if one takes the trouble to *precisely de?e* numerical addition and multiplication on potentially in?ite decimals, the ONLY WAY to make it work consistently with the accepted arithmetic for ?ite decimals (which can each be considered to have an in?ite sequence of 0s to the right) results in the concrete decimal representation of (.999...)^2 being exactly .999... There really is no last one (or Last One). ISTM the reason Garry doesn't accept that fact is that it's not immediately obvious, and he hasn't worked ' through the detailed consequences of arithmetic for in?ite decimals.Note that another of these consequences is that in terms of concrete decimals, 1.000... - .999... is exactly equal to 0.000... This is much easier to show than examples involving multiplication, since this can be done one digit at a time.HTH.-- ---------------------------| B B aa rrr b || BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- === But you never de?ed what 0.999... was.The subject 0.999... was, and is, not equal to 1.0.999...^4 is (and was) not equal to 1,therefore .999... is not equal to 1. 9^4 = 6561 .9^4 = .6561 99^4 = 96059601 .99^4 = .96059601 999^4 = 996005996001 .999^4 = .996005996001 9999^4 = 9996000599960001 .9999^4 = .9996000599960001 99999^4 = 99996000059999600001 .99999^4 = .99996000059999600001 999999^4 = 999996000005999996000001 .999999^4 = .999996000005999996000001 9999999^4 = 9999996000000599999960000001 .9999999^4 = .9999996000000599999960000001 99999999^4 = 99999996000000059999999600000001 .99999999^4 = .99999998000000059999999600000001 999999999^4 = 999999996000000005999999996000000001 .999999999^4 = .999999996000000005999999996000000001 9999999999^4 = 9999999996000000000599999999960000000001 .9999999999^4 = .9999999996000000000599999999960000000001 99999999999^4 = 99999999996000000000059999999999600000000001 .99999999999^4 = .99999999996000000000059999999999600000000001 999999999999^4 = 999999999996000000000005999999999996000000000001.999999999999^ 4 = .999999999996000000000005999999999996000000000001[snip rest for brevity]The subject 0.999... = l is a false statement. Garry Denke, GeologistDenoco Inc. of Texas === =*** post for FREE via your newsreader at post.newsfeed.com ***> But you never de?ed what 0.999... was. The subject 0.999... was, and is, not equal to 1. 0.999...^4 is (and was) not equal to 1,> therefore .999... is not equal to 1.Again, you persist saying what 0.999... is NOT. karl mhttp://www.newsfeed.com - The #1 Newsgroup Service in the World!-----== 100,000 Groups! - 19 Servers! - Unlimited Download! =----- === =Well, I don't know whether this argument is already used, but anyway:A mathematical property of the real numbers says that between every twodifferent numbers another third number can be found. (e.g. the mean).Well, if 0.999... doesn't equal 1. What is this third number? ' Isn'tthe mean 0.999...? I know this isn't a full proof. But still, I don'tthink I will ever be able to ?d such a number.Tim Wouters === =* Tim Wouters> Well, I don't know whether this argument is already used, but anyway:A mathematical property of the real numbers says that between every two> different numbers another third number can be found. (e.g. the mean).> Well, if 0.999... doesn't equal 1. What is this third number? Isn't> the mean 0.999...? I know this isn't a full ' proof. But still, I don't> think I will ever be able to ?d such a number.Ah, that's easy:0.999999.....9999 < 10.999999.....9999 < 0.999999.....99995 < 1(To be sure: :-), but I remember seeing this argument before.)-- Jon Haugsand <1073568342.205862@seven.kulnet.kuleuven.ac.be> === 0.999999.....9999 < 1> 0.999999.....9999 < 0.999999.....99995 < 1Hmm, when I use your argumentation, I would also get the following:0.999999.....99 < 0.999999.....995 < 0.999999.....999But isn't 0.999999.....99 = 0.999999.....999=0.99? And thus 0.999999.....995 = 0.999999.....99 = 0.99...Tim Wouters <1073568342.205862@seven.kulnet.kuleuven.ac.be> === = Correction:0.999999.....99 < 0.999999.....995 < 0.999999.....999 But isn't 0.999999.....99 = 0.999999.....999=0.99...? And thus 0.999999.....995 = 0.999999.....99 = 0.99...Tim Wouters === =* Tim Wouters> Correction:0.999999.....99 < 0.999999.....995 < 0.999999.....999 > But isn't 0.999999.....99 = 0.999999.....999=0.99...? > And thus 0.999999.....995 = 0.999999.....99 = 0.99...No, because:0.999...999 = nonsense <> 0.9999 = 1-- Jon Haugsand === =In sci.math, Jon Haugsandon 08 Jan 2004 15:04:57 +0100:> * Tim Wouters> Correction:0.999999.....99 < 0.999999.....995 < 0.999999.....999 > But isn't 0.999999.....99 = 0.999999.....999=0.99...? > And thus 0.999999.....995 = 0.999999.....99 = 0.99...No, because:0.999...999 = nonsense <> 0.9999 = 1> If 0.999... != 1, then obviously 0.999... = 1 - d, for someextremely in?tesimal d. (The obvious question of whetherd is a real number or a ?ment of my imagination we'll tablefor the moment. :-) )Now consider 10 * (0.999...) - 9 or (0.999... / 10) + 0.9.What are they? Well, if one thinks a bit one realizes theyare both 0.999..., although Garry Denke might disputethe ?st as a 0 might be shifted in from the right. Theproblem is the right can't really be shifted into anyway,and much hilarity ensues.But the two expressions are also equal to 1 - 10*d and 1 - d/10,respectively.So we have one of the problems below.[1] .999... = 1 - d = 1 - 10*d = 1 - d/10, and d is a real number. The only way this works in the reals is to equate d with 0, and therefore .999... = 1.[2] d has very strange algebra, where d = 10*d = d/10 != 0.[3] .999... as a numeric speci?ation is ambiguous.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. <1073568342.205862@seven.kulnet.kuleuven.ac.be> === =0.999...999 = nonsense <> 0.9999 = 1Sorry, I didn't get your point. Do you mean you can't add a 9 at the end of 0.99... because then you get 1?Tim Wouters === =* Tim Wouters0.999...999 = nonsense <> 0.9999 = 1Sorry, I didn't get your point. Do you mean you can't add a 9 at the end > of 0.99... because then you get 1?You can't add a 9 at the end of 0.99.. because it has no end.-- Jon Haugsand <1073568342.205862@seven.kulnet.kuleuven.ac.be> === Sorry, I didn't get your point. Do you mean you can't add a 9 at the end > of 0.99... because then you get 1?You can't add a 9 at the end of 0.99.. because it has no end.But you did add a 5 at the end of 0.99.. Why's that allowed? There ain't no difference.Tim Wouters === Sorry, I didn't get your point. Do you mean you can't add a 9 at the end > of 0.99... because then you get 1?You can't add a 9 at the end of 0.99.. because it has no end.But you did add a 5 at the end of 0.99.. Why's that allowed? There ain't > no difference.If you read carefully, he added a ?5' at the end of an arbitrary, butFINITE, sequence of ?9'. The sequence 0.999... has no end. karl m === What is the square of .9999.... then?Pick a vowel, because its not equal to 1. Neither is its cube.The cube of .999... is not equal to 1,therefore .999... is not equal to 1. 9^3 = 729 .9^3 = .729 99^3 = 970299 .99^3 = .970299 999^3 = 997002999 .999^3 = .997002999 9999^3 = 999700029999 .9999^3 = .999700029999 99999^3 = 999970000299999 .99999^3 = .999970000299999 999999^3 = 999997000002999999 .999999^3 = .999997000002999999 9999999^3 = 999999700000029999999 .9999999^3 = .999999700000029999999 99999999^3 = 999999970000000299999999 .99999999^3 = .999999970000000299999999 999999999^3 = 999999997000000002999999999 .999999999^3 = .999999997000000002999999999 9999999999^3 = 999999999700000000029999999999 .9999999999^3 = .999999999700000000029999999999 99999999999^3 = 999999999970000000000299999999999 .99999999999^3 = .999999999970000000000299999999999 999999999999^3 = 999999999997000000000002999999999999 .999999999999^3 = .999999999997000000000002999999999999 9999999999999^3 = 999999999999700000000000029999999999999 .9999999999999^3 = .999999999999700000000000029999999999999 99999999999999^3 = 999999999999970000000000000299999999999999 .99999999999999^3 = .999999999999970000000000000299999999999999 999999999999999^3 = 999999999999997000000000000002999999999999999.999999999999999 ^3 = .999999999999997000000000000002999999999999999[snip the rest for brevity] The subject 0.999... = 1 is a false statement.Garry Denke, GeologistDenoco Inc. of Texas === =In sci.logic, Garry Denkeon 8 Jan 2004 04:40:39 -0800<210a83f4.0401080440.31cfda7e@posting.google.com>:> What is the square of .9999.... then?Pick a vowel, because its not equal to 1. Neither is its cube.The cube of .999... is not equal to 1,> therefore .999... is not equal to 1.If (.999...)^3 = .999...9997000...02999..., where's the 7?A ?ite fractional digit expansion can be treatedas a series:r = sum(i=1,n) (d_i * 10^(-i))where each d_i is an element of the set {0,1,2,3,4,5,6,7,8,9}.An in?ite fractional digit expansion can be treatedas an in?ite series:r = sum(i=1,+oo) (d_i * 10^(-i))It turns out this is the limit of a Cauchy sequence, namelyr_n = sum(i=1,n) (d_i * 10^(-i))and one can cube them, leading to some rather interesting expressions:r_n^3 = sum(i=1,n)(j=1,n)(k=1,n) (d_i * d_j * d_k * 10^(-i-j-k)) = sum(i=1,3*n)(P_{i,n} * 10^(-i))r^3 = sum(i=1,+oo)(P_i * 10^(-i))and it turns out P_i = P_{i,n} if i < n, andP_0 = 0, P_1 = d_1^3, P_2 = 3*d_1^2*d_2,P_3 = 3*d_1*d_2^2 + d_1*d_2*d_3, etc., if I've done this correctly.(There's probably a name for these.)But never mind all that. Where's the 7? For what n is d_n = 7 inthe expansion of r = (.999...)^3 ? Does it make sense to sayn = omega and be done with it (n is an ordinal, after all)? 9^3 = 729> .9^3 = .729 99^3 = 970299> .99^3 = .970299 999^3 = 997002999> .999^3 = .997002999[rest snipped]-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Then, since x=0 solves the equation sin x = 0, you are saying that 0 is> the only answer? And if Garry Denke = Troll solves the thread> 0.999... = 1, then Garry Denke = Troll is the only answer?Yes, I have shown the square of the subject 0.999... as not equal to 1a bit too often proving the subject 0.999... = 1 as a false statement.So I shall move on and cube the 0.999... to see if its result equals 1 9^3 = 729 .9^3 = .729 99^3 = 970299 .99^3 = .970299 999^3 = 997002999 .999^3 = .997002999 9999^3 = 999700029999 .9999^3 = .999700029999 99999^3 = 999970000299999 .99999^3 = .999970000299999 999999^3 = 999997000002999999 .999999^3 = .999997000002999999 9999999^3 = 999999700000029999999 .9999999^3 = .999999700000029999999 99999999^3 = 999999970000000299999999 .99999999^3 = .999999970000000299999999 999999999^3 = 999999997000000002999999999 .999999999^3 = .999999997000000002999999999 9999999999^3 = 999999999700000000029999999999 .9999999999^3 = .999999999700000000029999999999 99999999999^3 = 999999999970000000000299999999999 .99999999999^3 = .999999999970000000000299999999999 999999999999^3 = 999999999997000000000002999999999999 .999999999999^3 = .999999999997000000000002999999999999 9999999999999^3 = 999999999999700000000000029999999999999 .9999999999999^3 = .999999999999700000000000029999999999999 99999999999999^3 = 999999999999970000000000000299999999999999 .99999999999999^3 = .999999999999970000000000000299999999999999 999999999999999^3 = 999999999999997000000000000002999999999999999.999999999999999 ^3 = .999999999999997000000000000002999999999999999[snip the rest for brevity] The cube of 0.999... is not equal to 1,therefore 0.999... is not equal to 1.Garry Denke, GeologistDenoco Inc. of Texas === When did we switch to religious.sci.crap?We didn't, why do you say that?> Because you keep invoking the In?ite One (your capitals)In fact, oblique and continual references to god, texas, naturalresources, and idiotic pronouncements on things you are ignorant of... areyou dubya? > If I'm ever in need of a survey for oil or gas remind me not to contact> your company if this is a fair re?n of the mathematical skills of its > employeesNo worries, you would be instructed to conduct your own surveys.> (it's company policy).Garry Denke, Geologist> Denoco Inc. of Texas === How much oil do you ?d with your sort of algebra?Only the accounting department knows for sure.The cube of .999... is not equal to 1,therefore .999... is not equal to 1. 9^3 = 729 .9^3 = .729 99^3 = 970299 .99^3 = .970299 999^3 = 997002999 .999^3 = .997002999 9999^3 = 999700029999 .9999^3 = .999700029999 99999^3 = 999970000299999 .99999^3 = .999970000299999 999999^3 = 999997000002999999 .999999^3 = .999997000002999999 9999999^3 = 999999700000029999999 .9999999^3 = .999999700000029999999 99999999^3 = 999999970000000299999999 .99999999^3 = .999999970000000299999999 999999999^3 = 999999997000000002999999999 .999999999^3 = .999999997000000002999999999 9999999999^3 = 999999999700000000029999999999 .9999999999^3 = .999999999700000000029999999999 99999999999^3 = 999999999970000000000299999999999 .99999999999^3 = .999999999970000000000299999999999 999999999999^3 = 999999999997000000000002999999999999 .999999999999^3 = .999999999997000000000002999999999999 9999999999999^3 = 999999999999700000000000029999999999999 .9999999999999^3 = .999999999999700000000000029999999999999 99999999999999^3 = 999999999999970000000000000299999999999999 .99999999999999^3 = .999999999999970000000000000299999999999999 999999999999999^3 = 999999999999997000000000000002999999999999999.999999999999999 ^3 = .999999999999997000000000000002999999999999999[snip the rest for brevity] Simple accounting for Virgil.Garry Denke, GeologistDenoco Inc. of Texas === Well, that's a bit more evidence regarding how much you know about how> math works - mathematicians rede?e existing terms all the time.Notice how you transitioned from math to mathematicians. This is oneof the most fundamental problems that I see in Google Groupsdiscussions: treating the publication of an assertion as proof that itis valid. Logical analysis is replaced by citations (and the morefamous the author, the better.) To me, the writings of mathematiciansare a source of ideas and possible results. Mathematics is the subsetthat turns out to be valid.> If you treat 0.999...> as a literal equal to 1.000... then you have two different literals> being equal, which is not a good idea.> you shouldn't make cracks about paying attention to details> if you're going to turn around and babble like this. Neither> 0.999... nor 1.000... is a literal. 0.999... and 1.000...> are literals, and nobody's claimed that they're equal.And now logical analysis is being replaced by quibbling overterminology [ http://dictionary.reference.com/search?q=quibble ]. Byliteral I mean an actual number, as opposed to a general expression(function). BTW: If literal is not the right word for what .999and 1.000 are, then the correction is not to give examples o?ls, but rather to give the correct word for what .999 and1.000 really are.> But if you are clear that you>mean the limits of two series, then there is no problem.The only person to whom this is _not_ clear is you, which> is why you're the only one speaking nonsense in this discussion.> It's clear to anyone who understands the standard meaning of> the notation that the symbol 0.999... denotes the sum of a> certain series. (A series has a sum, not a limit - _sequences_> have limits. Pay attention to details, indeed.)> Now all we have to do is learn what the notation actually> does mean and we'll be set.(More quibbling over terminology.) When a referee tells an authorthat he is not being clear, it doesn't mean that the referee doesn'tunderstand what the author is attempting to say. It means that thepresentation is ?It is not a question of the correct meaning of .999 or 1.000 As Ihave said, you can de?e these symbols as you wish, as long as youare consistent. The author needs to be clear as to what he means. Let me be completely explicit:Case # 1: .999 and 1.000 are numbersThen we have two different numbers (character strings) that are equal. The principle of there being a single literal representation of anygiven number is violated. We can no longer tell, in general, if twonumbers are equal simply by comparing their digits. As I have said,that is not a good idea (IMHO).Case # 2: .999 and 1.000 are symbols for in?ite series/sequences(with limits of 1)The post to which I responded (early in this thread) was:Just try to subtract 0.999... from 1:1 - 0.999... = 0Reason: There is no real number between 0.999... and 1, and,therefore, theymust be one and the same number!'It is clear to me that .999 is being treated as a number, not asymbol for an in?ite series/sequence. You can't subtract a symbolfrom an integer in one step or talk about whether there is a numberbetween a symbol and an integer. Indeed, if .999 is aseries/sequence (with a limit) equal to 1, then the author is sayingthat there is no number between 1 and 1 (rather than simply sayingthat they are equal since they are both equal to 1.)Personally, I prefer case # 2 and dropping the above proof that.999 = 1.Finally, I do appreciate a working mathematician such as yourselfspending time contributing to an unmoderated forum (that doesn'tappear in print.) However, I think that your knowledge of what isbeing said by other mathematicians is more useful as a source of otheropinions (and formal demonstrations) rather than as a justi?ationfor accepting particular hypotheses per se.Charlie Volkstorf> David C. Ullrich === = Well, that's a bit more evidence regarding how much you know about how> math works - mathematicians rede?e existing terms all the time.Notice how you transitioned from math to mathematicians. This is one>of the most fundamental problems that I see in Google Groups>discussions: treating the publication of an assertion as proof that it>is valid. Logical analysis is replaced by citations (and the more>famous the author, the better.) To me, the writings of mathematicians>are a source of ideas and possible results. Mathematics is the subset>that turns out to be valid.Here we were not talking about the validity of a mathematical fact.> If you treat 0.999...> as a literal equal to 1.000... then you have two different literals> being equal, which is not a good idea. you shouldn't make cracks about paying attention to details> if you're going to turn around and babble like this. Neither> 0.999... nor 1.000... is a literal. 0.999... and 1.000...> are literals, and nobody's claimed that they're ' equal.And now logical analysis is being replaced by quibbling over>terminology [ http://dictionary.reference.com/search?q=quibble ]. By>literal I mean an actual number, as opposed to a general expression>(function). This is hilarious. A few posts up you stated that I was beingsloppy about the details - now when I use terms very preciselyI'm quibbling.>BTW: If literal is not the right word for what .999>and 1.000 are, then the correction is not to give examples of>literals, but rather to give the correct word for what .999 and>1.000 really are.I did - you snipped it (or it appears elsewhere, not sure)..999... and 1.000... are real numbers. Except of course theplural is a little problematic because they're the same number.> But if you are clear that you>mean the limits of two series, then there is no problem.The only person to whom this is _not_ clear is you, which> is why you're the only one speaking nonsense in this discussion.> It's clear to anyone who understands the standard meaning of> the notation that the symbol 0.999... denotes the sum of a> certain series. (A series has a sum, not a limit - _sequences_> have limits. Pay attention to details, indeed.) Now all we have to do is learn what the notation actually> does mean and we'll be set.(More quibbling over terminology.) When a referee tells an author>that he is not being clear, it doesn't mean that the referee doesn't>understand what the author is attempting to say. It means that the>presentation is ?It is not a question of the correct meaning of .999 or 1.000 As I>have said, you can de?e these symbols as you wish, as long as you>are consistent. The author needs to be clear as to what he means. In this case there _is_ a _completely_ _standard_ de?ition of whatthese symbols mean already. An author is not required to restateevery standard de?ition he uses. But the de?ition of what thesesymbols mean _has_ been given in this thread, several times byseveral people.>Let me be completely explicit:Case # 1: .999 and 1.000 are numbersThen we have two different numbers (character strings) that are equal.Simply hilarious. You accuse me of being sloppy, and now yousay a number is a character string. A number is not a characterstring - this is not quibbling, this is the whole point. One more time: .999... is not a character string, it is a number.It happens to be equal to the number 1.000... .On the other hand .999... and 1.000... are characterstrings - they are not equal, and nobody has said that theyare.> The principle of there being a single literal representation of any>given number is violated. It is _true_ that a real number can have more than one representationas a character string. So what? The fact that 2 + 2 = 4 _also_violates the supposed principle that a number can have atmost one representation as a character string. Where does thisprinciple come from? This post is the ?st I've heard about it.You want to say 2+2 is not a literal, ?e. The other day youmade a crack about how if I were a programmer I'd get thisstuff straight. In every programming language I know(which is two extremely standard one, Python and ObjectPascal) numbers have more than one literal representation:'10.0' denotes the same number ' as ?10.0e1' (or whatever,the notation varies). And the ' ?10.e1' _is_ a _literal_, nota compound expression.To save time: That doesn't count because those twoliterals denote the same number. Uh, right. Exactlyas the two literals 1.0 and .999... denote the samenumber.> We can no longer tell, in general, if two>numbers are equal simply by comparing their digits. That's correct. So what? (Even if reals had uniquedecimal expansions we could not tell whether twonumbers were equal by comparing ?itely manydigits, so we can't be talking about practicalconcerns here. And if we're allowed to inspect_all_ the digits then we _can_ tell whether twonumbers are equal - the test is just a little morecomplicated than all the digits are the same.)>As I have said,>that is not a good idea (IMHO).So what? The question of what is a good idea in your opinion isevidently very different from the question of what's a good ideain the opinion of the typical mathematician. And _both_questions are 100% completely irrelevant here! Becausewe're not talking about what the notation _should_ mean,we're talking about what it _does_ mean.>Case # 2: .999 and 1.000 are symbols for in?ite series/sequences>(with limits of 1)The post to which I responded (early in this thread) was:Just try to subtract 0.999... from 1:1 - 0.999... = 0Reason: There is no real number between 0.999... and 1, and,>therefore, they>must be one and the same number!'The fact that you can read a lot of crap on the internet does notprove that 0.999... does not equal 1. The reason above saysthey're the same because they're the same. But you've alsoseen _proofs_ that ' they are the same. Here's the proof, incase you forgot:_By de?ition_, 0.999... is the limit of the sequence 0.9, 0.99,0.999, ... . The n-th term in this sequence is a_n = 1 - 1/10^n.So |1 - a_n| = 1/10^n. So for any epsilon > 0 there exists Nsuch that |1 - a_n| < epsilon for all n > N. So lim a_n = 1.So 0.999... = 1, by de?ition.>It is clear to me that .999 is being treated as a number, not a>symbol for an in?ite series/sequence. .999... is not a symbol at all, it is a number. .999... is a symbolfor this number. It _is_ a symbol for the _sum_ of a certain in?ite series. That means that it _is_ a symbol for number.>You can't subtract a symbol>from an integer in one step or talk about whether there is a number>between a symbol and an integer. Indeed, if .999 is a>series/sequence (with a limit) equal to 1, then the author is saying>that there is no number between 1 and 1 (rather than simply saying>that they are equal since they are both equal to 1.)Personally, I prefer case # 2 and dropping the above proof that>.999 = 1.Finally, I do appreciate a working mathematician such as yourself>spending time contributing to an unmoderated forum (that doesn't>appear in print.) However, I think that your knowledge of what is>being said by other mathematicians is more useful as a source of other>opinions (and formal demonstrations) rather than as a justi?ation>for accepting particular hypotheses per se.You have said several times that _if_ 0.999... denotes the sum of acertain in?ite series then there's no problem, 0.999... = 1. So,as I stated some time ago, the only question is whether that really_is_ what it denotes. The question of what a certain symbol denotes is not a mathematical question per se, it has nothingto do with accepting hypotheses. By de?ition, what a symmboldenotes is de?ed by what mathematcians _say_ it denotes._When_ we're talking about the meaning of a certain notation,as opposed to talking about the truth or falsity of an actualfact, then the truth of the matter _is_ decided by concensus.>Charlie Volkstorf David C. UllrichDavid C. Ullrich === =In sci.logic, Garry Denkeon 7 Jan 2004 05:49:13 -0800<210a83f4.0401070549.34581e73@posting.google.com>:> This is a Cauchy sequence. The left side> is equal to (1 - 10^(-n))^2; the right> side is (1 - 2 * 10^(-n) + 10^(-2n)).At the limit, we get (1 - d)^2 = 1 - 2d + d^2,> where 0 < d^2 < d < any positive real.(Or, for those who are more conventional, 1^2 = 1, which> probably makes more sense anyway; otherwise things such> as .999.... and .333... are not uniquely de?ed.)So what you're saying is to forget arithmetic when doing this math? 9^2 = 81> .9^2 = .81 99^2 = 9801> .99^2 = .9801 999^2 = 998001> .999^2 = .998001 9999^2 = 99980001> .9999^2 = .99980001> [rest snipped for brevity]Left to right in arithmetic, right to left in math?So you are suggesting that (.999...[in?ite]...)^2= .999...[in?ite]...998000...[in?ite]...001?An interesting idea, that -- but where is the ?8', precisely?Also, what happens if one subtracts the two?Does one get -0.000...[in?ite]...00199...[in?ite]...999?Where's the ?1'?As for math: math is best done side-crossways. :-) One can grindthrough a proof, or, if one sees the secret passage, cut straightto the heart of the matter and prove it very simply, in many cases.Garry Denke, Geologist> Denoco Inc. of Texas-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Left to right in arithmetic, right to left in math?So you are suggesting that (.999...[in?ite]...)^2> = .999...[in?ite]...998000...[in?ite]...001?No, I'm cubing the subject 0.999... to show the kidsthat the subject 0.999... = 1 is a false statement.The cube of .999... is not equal to 1,therefore .999... is not equal to 1. 9^3 = 729 .9^3 = .729 99^3 = 970299 .99^3 = .970299 999^3 = 997002999 .999^3 = .997002999 9999^3 = 999700029999 .9999^3 = .999700029999 99999^3 = 999970000299999 .99999^3 = .999970000299999 999999^3 = 999997000002999999 .999999^3 = .999997000002999999 9999999^3 = 999999700000029999999 .9999999^3 = .999999700000029999999 99999999^3 = 999999970000000299999999 .99999999^3 = .999999970000000299999999 999999999^3 = 999999997000000002999999999 .999999999^3 = .999999997000000002999999999 9999999999^3 = 999999999700000000029999999999 .9999999999^3 = .999999999700000000029999999999 99999999999^3 = 999999999970000000000299999999999 .99999999999^3 = .999999999970000000000299999999999 999999999999^3 = 999999999997000000000002999999999999 .999999999999^3 = .999999999997000000000002999999999999 9999999999999^3 = 999999999999700000000000029999999999999 .9999999999999^3 = .999999999999700000000000029999999999999 99999999999999^3 = 999999999999970000000000000299999999999999 .99999999999999^3 = .999999999999970000000000000299999999999999 999999999999999^3 = 999999999999997000000000000002999999999999999.999999999999999 ^3 = .999999999999997000000000000002999999999999999[snip the rest for brevity] The subject 0.999... = 1 is a false statement.Please discontinue lying to kids in textbooks.Garry Denke, GeologistDenoco Inc. of Texas === =In sci.logic, Garry Denkeon 8 Jan 2004 04:00:23 -0800<210a83f4.0401080400.4734cb5f@posting.google.com>:> Left to right in arithmetic, right to left in math?So you are suggesting that (.999...[in?ite]...)^2> = .999...[in?ite]...998000...[in?ite]...001?No, I'm cubing the subject 0.999... to show the kids> that the subject 0.999... = 1 is a false statement.The cube of .999... is not equal to 1,> therefore .999... is not equal to 1.No, .999... is not equal to 1, therefore any operationinvolving .999... may not substitute 1 therefor.At least, one can fashion a semi-consistent math systemtherewith, although it would take a lot of work and probablywould not be worth it.(For starters, if .999... != 1, what is .999... - (.999... * 10 - 9)equal to? Answer: 0.000...9. This sort of thing will get ratherweird very quickly.)Of course, if one cubes .999... one can do something like the following:.9 * .9 * .9 = .729.99 * .99 *.99 = .729 + 3*.0729 + 3*.00729 + .000729 = .729 + .2187 + 0.02187 + .000729 = .970299Note the 7 in the 10^-2 spot..999 * .999 * .999 = .970299 + 3*.0088209 + 3*.00008019 + .000000729 = .970299 + .0264627 + .00024057 + .000000729 = .997002999and yet another 7, this time in the 10^-3 spot..9999 * .9999 * .9999 = .997002999 + 3*.0008982009 + 3*.0000008091 + .000000000729 = .997002999 + .0026946027 + 3*.0000024273 + .000000000729 = .999700029999and there's that 7 again. Is that second term always going to be.000...0002something ?Of course it's probably simpler to abstract the problem andequate .999...9 = 1 - 10^(-n), and then(1 - 10^(-n))^3 = 1 - 3*10^(-n) + 3*10^(-2*n) - 10^(-3*n)which proves that (.999...9)^3 = .999...99700...0299...9, in asimpler fashion than your equations below, or for that mattermy algebra.So far, no problem here. But one cannot conclude that(.999...)^3 = .999...99700...0299...unless one ascribes a de?itive meaning to digit expansionswith in?ite ellipses and de?es their proper addition,subtraction, etc. It is naive to conclude that thearithmetic operation10 * (.999...)^3 - 9 - (.999...)^3 = 9.999...97000...2999...990 - 9 - .999...99700...0299...999= -.000...02699...7300...009.makes any sense at all. 9^3 = 729> .9^3 = .729 99^3 = 970299> .99^3 = .970299 999^3 = 997002999> .999^3 = .997002999 9999^3 = 999700029999> .9999^3 = .999700029999[snip for brevity]The subject 0.999... = 1 is a false statement.> Please discontinue lying to kids in textbooks.> I don't write textbooks. Software yes, but not textbooks.Garry Denke, Geologist> Denoco Inc. of Texas-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === =In sci.logic, Garry Denkeon 7 Jan 2004 06:19:03 -0800<210a83f4.0401070619.68279dea@posting.google.com>:> If 1 solves the equation x^2 = x1 = x> x^2 = x> 1^2 = 11 gets 2 values: 1 and 0.nope, 1 is the only answer, sorry ghost0^2 = 0.Now you're getting ridiculous. If 1 solves the equation x^2 = x,> making x = 1, how can x = 0? If 0 solves the equation x^2 = x...0 = x> x^2 = x> 0^2 = 0then 0 is the only answer.The standard question for equations such as x^2 = x is: Give the set of values that x can be set to, such that the equation holds.This set is {0, 1}; both 0 and 1 satisfy the equation.It's a little harder to show that nothing else will,though one can try algebra and get ly in this case:x^2 = xx^2 - x = 0x(x - 1) = 0In an in?ite ?ld such as the reals, it's now obvious that0 and 1 are the only solutions. (In the ring of integersmod 6, though, things get interesting, as the solution setis {0,1,3,4}.}Garry Denke, Geologist> Denoco Inc. of Texas-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === =[snip]> The standard question for equations such as x^2 = x is: Give the set of values that x can be set to, such that the> equation holds. This set is {0, 1}; both 0 and 1 satisfy the equation.> It's a little harder to show that nothing else will,> though one can try algebra and get ly in this case: x^2 = x> x^2 - x = 0> x(x - 1) = 0 In an in?ite ?ld such as the reals, it's now obvious that> 0 and 1 are the only solutions. (In the ring of integers> mod 6, though, things get interesting, as the solution set> is {0,1,3,4}.}FWIW, the original equation also has another solution if our number systemis the extended reals. Then the solution set is {0, 1, +oo}.David === =In sci.logic, David W Cantrell<20040108005431.441$aW@ newsreader.com>:> [snip]> The standard question for equations such as x^2 = x is: Give the set of values that x can be set to, such that the> equation holds. This set is {0, 1}; both 0 and 1 satisfy the equation.> It's a little harder to show that nothing else will,> though one can try algebra and get ly in this case: x^2 = x> x^2 - x = 0> x(x - 1) = 0 In an in?ite ?ld such as the reals, it's now obvious that> 0 and 1 are the only solutions. (In the ring of integers> mod 6, though, things get interesting, as the solution set> is {0,1,3,4}.}FWIW, the original equation also has another solution if our number system> is the extended reals. Then the solution set is {0, 1, +oo}.DavidGood point. :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === In sci.logic, David W Cantrell> :> [snip]> The standard question for equations such as x^2 = x is: Give the set of values that x can be set to, such that the> equation holds. This set is {0, 1}; both 0 and 1 satisfy the equation.> It's a little harder to show that nothing else will,> though one can try algebra and get ly in this case: x^2 = x> x^2 - x = 0> x(x - 1) = 0 In an in?ite ?ld such as the reals, it's now obvious that> 0 and 1 are the only solutions. (In the ring of integers> mod 6, though, things get interesting, as the solution set> is {0,1,3,4}.} FWIW, the original equation also has another solution if our number> system is the extended reals. Then the solution set is {0, 1, +oo}. Good point. :-)Is that the best you can say? Many mathematicians call the point +ooideal, which makes it sound far better than merely good! ;-)It might also be noted that x = +oo is not a solution ofeither x^2 - x = 0 or x(x - 1) = 0. (But that shouldn't be surprising.After all, the extended reals ' don't form a ?ld.)David === it's a very easy thing to notice that for each string of nines of> length n, the square has n-1.. so as n approaches in?ity, so does> the square's number of nines... thus .999....^2 = .999... because the> 8, n-1 0's, and the 1 are somewhere beyond in?ity.. they're> in?itessimal to the in?itessimal's scaleNotice the equal number of 0's between the 8's and the In?ite One> Ha ' ha ha, Good One!Garry Denke, GeologistDenoco Inc. of Texas === =I've recently passed my Dimension Theory course. The grade? A full 3 outof 3.I honestly can't ?ure how I got so good a grade. I didn't understandnearly half of the stuff taught in that course. In the homeworkexercises, I was the third most active one, but after about halfwaythrough, I usually only did one or two problems.At the exam I answered three questions out of ?e, which was theminimum required. One of them asked to de?e the three topologicaldimension concepts. I provided the right de?itions for the inductivedimensions (small and big) but I had forgot the de?ition of thecover dimension. So I took a wild guess at that one.Another was to compute the topological function of Gamma(f) ={(x, f(x)) | x in R^n} where f:R^n->R was continuous. I got the rightanswer, but *way* too complicatedly and confusingly.THe last was to express the main results of my seminar paper. LilyI had read my whole seminar paper mere minutes before the exam, so Ithe exam.Has anyone had similar experiences in their university or college life?I'd be interested in hearing from both students and teachers.-- /-- Joona Palaste (palaste@cc.helsinki.? ------------- Finland ---------- http://www.helsinki.?~palaste --------------------- rules! --------/'I' is the most beautiful word in the world. ' - John Nordberg === =...> Anyhow, for the [5,13] case one can ?d an exact solution as> follows. Suppose a>b>=0 and c>d>=0. Let @ stand for an add> or subtract, and ~ for approximately equal.> If 13 ~ (2^a @ 2^b)/m and 5 = (2^c @ 2^d)/m then> m = (2^c @ 2^d)/5 so 13/5 ~ (2^a @ 2^b)/(2^c @ 2^d).> With some effort one can show that for all a,b,c,d as stated,> 63/24 = 2.625 and 31/13 = 2.5833... are the closest approximates.(Note, 31/13 was a typo for 31/12. 31/12 = 2.5833...)Your search algorithm is smarter than mine. Because you do not search for> the real-number divider directly, instead, you search integer pairs (2^a @> 2^b)/(2^c @ 2^d) that are as close approximations to 13/5 as possible...But once again, we meet with the problem of deciding search algorithm: you> provide two choices, 63/24 and 31/13... how did you get them? If it is> exhaustive search, what are the possible ranges? ...Consider four classes of ratios, (2^a+2^b)/(2^c+2^d), (2^a-2^b)/(2^c+2^d),(2^a-2^b)/(2^c-2^d), and (2^a+2^b)/(2^c-2^d); then in each class ?d limits (depending on the target ratio, T) on a,b,c,d.Here's an example of narrowing the limits in the -,+ case: Let R = (2^a-2^b)/(2^c+2^d) = (2^(a-b)-1)/(2^(c-b)+2^(d-b)) etc. If a < c+1 then R < 2/(1+2^(d-c)) < 2. If a > c+3 then R > (8-4)/(1+1/2) = 2.66..., so if we want R values between 2 and 2.66... then we must have a=c+2. With i=b-c and j=d-cwe have R = (4-2^i)/(1+2^j). If i=1, R<2/(1+2^j)<2 so i<1.If j<-2, R>3/(1.125)=2.66... so 0>j>-3. This narrows us down to(16-2^k)/(4+1) and (16-2^k)/(4+2) with k<4. Obviously these increaseas k decreases so only a small number of values need to be computed.2.583... at k=-1 and 2.625 at k=-2 are the closest approximants for the -,+ case. Other cases have similar treatment except it may beless obvious if a ratio increases. For example, to see that (2^k+2^(k-1))/(2^(k-2)+1) increases as k increases, write this asf(j)=(6*2^j)/(2^j+1) and note that f(k)-f(j)>0 iff(2^j+1)(2^k+1)(f(k)-f(j))>0, ie, iff 2^k-2^j>0, ie, iff k>j.program that in a fraction of a second computed, reduced, and sorted all the values between 2.5 and 2.7 of (2^a @ 2^b)/(2^c @ 2^d) with a,b,c,d each ranging from 1 to 30. Here is a section of its output:2.50489 at 1280 / 5112.50980 at 640 / 2552.51969 at 320 / 1272.53968 at 160 / 632.57143 at 18 / 72.58065 at 80 / 312.58333 at 31 / 122.62500 at 63 / 242.64583 at 127 / 482.65625 at 255 / 962.66146 at 511 / 192-jiw === Anyhow, for the [5,13] case one can ?d an exact solution as> follows. Suppose a>b>=0 and c>d>=0. Let @ stand for an add> or subtract, and ~ for approximately equal.> If 13 ~ (2^a @ 2^b)/m and 5 = (2^c @ 2^d)/m then> m = (2^c @ 2^d)/5 so 13/5 ~ (2^a @ 2^b)/(2^c @ 2^d).> With some effort one can show that for all a,b,c,d as stated,> 63/24 = 2.625 and 31/13 = 2.5833... are the closest approximates.> For the latter, MSE = (12/m-5)^2+(31/m-13)^2 = 194-926/m+1105/m^2> =1105k^2-926k+194 (substituting k for 1/m) which is minimal when> is deriv. is zero, ie, k=926/(2*1105), or m=2210/926=2.3866.> Check: 12/m=5.028054, 31/m=12.98914, similar to the answers> you got. (Note that the numbers in your answer are 128 times> larger.) You can do a similar calculation for 63/24 and compare> results.Dear James,It seems to me that this 2.3866 is a magic number for pair [5, 13], any2.3866*2^i, where i is an integer 0, 1, 2, 3, etc. can be a good divider...It turns out that for any pair of numbers, there is such a magic number, Iam now trying to see if it is possible to determine such magic number forany numbers systematically... Any thoughts?-Walalal === I am soliciting some advice and opinions about how I should interact with my>granddaughter.>First some history. I am a 72 year old Mathematician. I got my degree in>Recursive Function theory in 1955. I was a bit of a prodigy in that by the time>I graduated high school I know nearly all the Mathematics I would need for a>Ph.D (I had done little research; I did not yet have my thesis).>My daughter is now nearly 40 years old. When she was growing up, I taught her>elementary Mathematics, in part, as a form a high quality father-daughter>interaction. The effect of this was that when she was in the 9th grade she took>AP Calculus and aced the test.>I have often wondered whether it would have made much of a difference in her>life if she learned Calculus, three years later, the usual time for bright kids.>She never learned much more Mathematics and eventually went to Caltech, where>she got a degree in Molecular Biology.>Her daughter is now seven years old. She is also intellectually precocious, I>am wondering how I should interact with my granddaughter .>Her arithmetic skills are now good enough that I could begin to teach her>beginning high school Algebra, but I don't know to what end. No matter what I>do or don't do, she will eventually know high school Algebra.>Perhaps the best legacy I could leave her, in the few years I may have left, is>an an understanding and love for the beauty of Mathematics. But I don't know>how to impart that.>Three other things ' I've thought of teaching her are:> Beginning Number Theory (divisibility, primes, modulo arithmetic ...)> Combinatorics (how many subsets does a ?ite set have, how many> permutations, ...).> Very elementary group theory, with an eye toward being able to produce a> rigorous proof.>I would very much appreciate some opinions. Also any suggestions about text>books, outlines of material, etc. would be most welcome.I would suggest teaching really basic mathematics. Thisdoes not include such things as calculus. Also, it wouldbe better that she learn things well, NOT as the schoolsteach them, in classes which are far too slow for anyonewith ability. Is there ANYTHING which the present schoolsteach even reasonably well?Concentrate on concepts, not manipulation. It looks likemost of what you have listed is manipulation; this iseasily learned after the concepts, but the other way isnot; it is HARD. Learning computational calculus doesnot help with the concepts, nor does learning arithmetic,nor algebra as it is usually taught, and the good Euclidcourse almost does not exist.I would suggest you start with logic, including therestricted predicate calculus. This is what my son didat ages 5-6, using the two books of Suppes, while alsostudying algebra. However, I believe that you will ?dmy late wife's logic book easier, although many of theapplications may not be appropriate. Frankly, I believethat this book, somewhat rewritten, should be a part of theregular elementary school curriculum for all those withconceptual ability. However, start with the linguisticnotion of variables, which is more general. If you havedif?ulty in getting a copy (it is out of print), letme know; I am in the process of ?ding out how to makeit generally available. My daughter did it later, andby not using mnemonics, she never had any problem withusing any notation for variables in any ?ld.I would also suggest at least the ?st part of Landau's_Foundations of Analysis_, which consists of deriving theproperties of the positive integers and positive rationalsfrom the Peano Postulates. I believe that this, rewritten,should be no later than the primary grades. But anyone whodoes not understand induction does not understand integers,and this includes most college mathematics majors, and while Landau may be short on verbiage, the approach is very detailed, especially for the positive integers. Thereis a logical lack, which you might catch, but if you assumethat addition and multiplication exist, it goes away.In both of these, if she has dif?ulty with producing proofs, do not make too big an issue early. Learning morewill make proofs easier. Foundations belong in kindergarten, not in graduate school.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === =I think that you should teach her what are the mathematics usefulfor... Simple examples... teach her that maths can be very useful inour lifes...Don't use the mathematician point of view, maybe the engineer point ofview... application.Also, have you considered teach her to program?. She would need alittle algebra, then you could teach her numerical methods... beginingwith QBASIC could be good. === I think that you should teach her what are the mathematics useful> for... Simple examples... teach her that maths can be very useful in> our lifes...Don't use the mathematician point of view, maybe the engineer point of> view... application.Also, have you considered teach her to program?. She would need a> little algebra, then you could teach her numerical methods... begining> with QBASIC could be good.No! I had a go at teaching my 11 year old son to program while on holiday in a rainy caravan last year. He picked up a bit of Perl quite quickly. There are so many problems with BASIC that I don't know where to begin. Try Perl or Java. Or anything but any form of BASIC...Mark Atherton === =Do not Spam> I am soliciting some advice and opinions about how I should interact withmy> granddaughter. ... [good stuff snipped, just to save bandwidth]I would get Ross Honsberger's great little book Ingenuity in Mathematics,and leave it within her reach. Maybe also Dudeney's The Canterbury Puzzlesor one of Martin Gardner's books. There aren't many good popularizers inmath.You don't say that the girl is a math talent speci?ally, but she might be.As I'm sure you know, math talents will appreciate the subject, and willpursue it without being pushed. Ask the girl why, if we add 1+2+...+n, theanswer is always n(n+1)/2. Observe the reaction. Neat phenomena fromgeometry are all around us, as well. Why is a satellite dish shaped justlike so? for example. Calculus, with its in?ite limiting processes, is aless suitable stimulant than ?ite stuff like probability and numbertheory, IMO.One little anecdote. I attended a high school which had about 2300 students.In all that group there were only two girls headed into mathematics. I wassure that both would become career mathematicians. Both started as mathmajors, but both changed their majors before B.Sc. time.LHX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === =In <3ffc236f.3980545@netnews.worldnet.att.net>, on 01/07/2004> Take her to movies, plays, concerts. Take walks with her. Take her>?hing! Spend as much time as she is willing with her. Play games>with her. You were doing all right up through this point.>Do ant=ything BUT mathematics, The schools can teach her math!legitimate one.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.orgX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === =In , on 01/07/2004 at 01:59 AM, spam@dad.dad.org (Do not Spam) said:>I am soliciting some advice and opinions about how I should interact>with my granddaughter.How good is her reading? Could she handle The World of Mathematics orFantasia Mathematica?>The effect of this was that when she was in the 9th grade she took>AP Calculus and aced the test.Not really that early a start.>Her daughter is now seven years old. She is also intellectually>precocious, I am wondering how I should interact with my>granddaughter .Like a child. Show her that Mathematics is fun. Start her on moreabstract concepts as games and puzzles.>Three other things I've thought of teaching her are:> Beginning Number Theory (divisibility, primes, modulo>arithmetic ...)That's probably a good choice.> Combinatorics (how many subsets does a ?ite set have, how> many permutations, ...).Well, I found that to be dull, but YMMV. Certainly a lot of top rateMathematicians found it fascinating.> Very elementary group theory, with an eye toward being able to> produce a rigorous proof.Also a good choice.>I would very much appreciate some opinions. Seven is a bit early, but you might consider in a few years giving hera rigorous book on Euclidean Geometry and perhaps introducing her toProjective Geometry. The latter you could tie into DescriptiveGeometry and Art if she has artistic interests. Show her some Escher and talk to her about perspective drawings o?possible objects. Maybe a picture of the three pronged Blivitwould be appropriate. >Also any suggestions about text books, outlines of material, etc. >would be most welcome.Check with your local Montessori schools about what suitablemanipulables and books they could recommend.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === On 7 Jan 2004 09:03:10 -0800, daryl@atc-nycorp.com (Daryl McCullough)>gowan says...>A rigorous de?ition of a^b for real a and b, a > 0, b irrational can>be done by developing the Riemann integral, de?ing the natural log>via the integral, de?ing e^x as the inverse function of the natural>log, and then de?ing a^b = e^(b ln a). Does anyone know an easier>way?It depends on what you mean by easier. One way to extend a^b to>irrational values of b (and positive values of a) is like this: 1. De?e a^b, with b a nonnegative integer by recursion:> a^0 = 1> a^{n+1} = a * a^n 2. If b is a nonnegative fraction of the form m/n, then a^b = that> number z such that z^n = a^m. 3. If b is a nonnegative irrational number, then let r_1, r_2, ... be> an in?ite sequence of rationals converging to b. Then> a^b = limit n->in?ity of a^r_n. 4. If b is negative, then a^b = (1/a)^{-b}I guess that there is some proof involved showing that if the>sequence r_1, ..., r_n, ... converges, then so does the sequence>a^r_1, a^r_2, ..., a^r_n, ...> Yes there is. It really seems unlikely that this is the easy way> to do it, _if_ we actually include the proofs.David C. UllrichIt's not terribly dif?ult. Using a^x - a^y = a^y[a^(x-y)-1],monotonicity of a^x, 1/a^x = a^(-x) = (1/a)^x, where x, y rational,we can easily reduce this to showing that for a>1 we have lim a^(1/n)=1.To see this note that a^(1/n)-1 = (a-1)/sum(a^(i/n),i=0..n-1)<(a-1)/n === =On Thu, 08 Jan 2004 13:30:16 -0500, Zbigniew Fiedorowicz> On 7 Jan 2004 09:03:10 -0800, daryl@atc-nycorp.com (Daryl McCullough)>gowan says...>A rigorous de?ition of a^b for real a and b, a > 0, b irrational can>be done by developing the Riemann integral, de?ing the natural log>via the integral, de?ing e^x as the inverse function of the natural>log, and then de?ing a^b = e^(b ln a). Does anyone know an easier>way?>It depends on what you mean by easier. One way to extend a^b to>irrational values of b (and positive values of a) is like this:> 1. De?e a^b, with b a nonnegative integer by recursion:> a^0 = 1> a^{n+1} = a * a^n> 2. If b is a nonnegative fraction of the form m/n, then a^b = that> number z such that z^n = a^m.> 3. If b is a nonnegative irrational number, then let r_1, r_2, ... be> an in?ite sequence of rationals converging to b. Then> a^b = limit n->in?ity of a^r_n.> 4. If b is negative, then a^b = (1/a)^{-b}>I guess that there is some proof involved showing that if the>sequence r_1, ..., r_n, ... converges, then so does the sequence>a^r_1, a^r_2, ..., a^r_n, ...> Yes there is. It really seems unlikely that this is the easy way> to do it, _if_ we actually include the proofs.David C. UllrichIt's not terribly dif?ult. Using a^x - a^y = a^y[a^(x-y)-1],>monotonicity of a^x, So you and Herman both say. To tell you the truth, I'dbeen under the impression that showing a^x is a monotonefunction of x (for x rational, de?ed in terms of roots) wastricky, but of course it isn't, it follows from the fact thata^n is a monotone function of n.>1/a^x = a^(-x) = (1/a)^x, where x, y rational,>we can easily reduce this to showing that for a>1 we have lim a^(1/n)=1.>To see this note that> a^(1/n)-1 = (a-1)/sum(a^(i/n),i=0..n-1)<(a-1)/n David C. Ullrich === On 7 Jan 2004 09:03:10 -0800, daryl@atc-nycorp.com (Daryl McCullough)>gowan says...>A rigorous de?ition of a^b for real a and b, a > 0, b irrational can>be done by developing the Riemann integral, de?ing the natural log>via the integral, de?ing e^x as the inverse function of the natural>log, and then de?ing a^b = e^(b ln a). Does anyone know an easier>way?>It depends on what you mean by easier. One way to extend a^b to>irrational values of b (and positive values of a) is like this:> 1. De?e a^b, with b a nonnegative integer by recursion:> a^0 = 1> a^{n+1} = a * a^n> 2. If b is a nonnegative fraction of the form m/n, then a^b = that> number z such that z^n = a^m.> 3. If b is a nonnegative irrational number, then let r_1, r_2, ... be> an in?ite sequence of rationals converging to b. Then> a^b = limit n->in?ity of a^r_n.> 4. If b is negative, then a^b = (1/a)^{-b}>I guess that there is some proof involved showing that if the>sequence r_1, ..., r_n, ... converges, then so does the sequence>a^r_1, a^r_2, ..., a^r_n, ...>Yes there is. It really seems unlikely that this is the easy way>to do it, _if_ we actually include the proofs.It is not that dif?ult, if one uses the monotonicity properties of exponentiation.For example, if u is positive, (1+u/n)^n is increasing,using only algebra. Then if h is a rational number less than 1, and x > 1, x^h - 1 < h(x - 1). Now onehas the needed tools.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === =In sci.math, gowanon 7 Jan 2004 07:22:37 -0800:> A rigorous de?ition of a^b for real a and b, a > 0, b irrational can> be done by developing the Riemann integral, de?ing the natural log> via the integral, de?ing e^x as the inverse function of the natural> log, and then de?ing a^b = e^(b ln a). Does anyone know an easier> way?Why wouldn't a derived Cauchy sequence work?If b = lim(n->+oo) b_n, where b_n is always rational, thenc = a^b = lim(n->+oo) c_n, where c_n = a^b_n, for any a > 0.While c_n is not always rational, it is a Cauchy sequence(the proof thereof I'll leave to the reader :-) ).(I assume the usual real semantics here; if a is complex thingsget complicated. For example, (1 + 0i)^(1/3) has 3 possible values.)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === A rigorous de?ition of a^b for real a and b, a > 0, b irrational can> be done by developing the Riemann integral, de?ing the natural log> via the integral, de?ing e^x as the inverse function of the natural> log, and then de?ing a^b = e^(b ln a). Does anyone know an easier> way?> My earlier post with the details corrected.> For m, n positive integers, de?e a^(m/n) = sup (x in R+: x^n <= a^m} ^ with (m,n) = 1> Then, if a > 1, de?e a^b = sup {a^x: x in Q+, x <= b}> If a < 1, replace sup with inf.> This is the standard de?ition.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === =In , on 01/07/2004>A rigorous de?ition of a^b for real a and b, a > 0, b irrational>can be done by developing the Riemann integral, de?ing the natural>log via the integral, de?ing e^x as the inverse function of the>natural log, and then de?ing a^b = e^(b ln a). Does anyone know an>easier way?Easy is in the eye of the beholder. I'd probably de?e exp ?st,then de?e ln as the inverse. Exp is easily de?ed either by itsproperties or by a power series.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === =Squares of 0.999... tend away from 1 .9^2 = .81 .9^4 = .6561 .99^2 = .9801 .99^4 = .96059601 .999^2 = .998001 .999^4 = .996005996001 .9999^2 = .99980001 .9999^4 = .9996000599960001 .99999^2 = .9999800001 .99999^4 = .99996000059999600001 .999999^2 = .999998000001 .999999^4 = .999996000005999996000001 .9999999^2 = .99999980000001 .9999999^4 = .9999996000000599999960000001 .99999999^2 = .9999999800000001 .99999999^4 = .99999998000000059999999600000001 .999999999^2 = .999999998000000001 .999999999^4 = .999999996000000005999999996000000001 .9999999999^2 = .99999999980000000001 .9999999999^4 = .9999999996000000000599999999960000000001 .99999999999^2 = .9999999999800000000001 .99999999999^4 = .99999999996000000000059999999999600000000001 .999999999999^2 = .999999999998000000000001.999999999999^4 = .999999999996000000000005999999999996000000000001[snip rest for brevity]Challenge: Prove squares of 0.999... tend toward 1 Garry Denke, GeologistDenoco Inc. of Texas === Squares of 0.999... tend away from 1There is only one square of 0.999...; perhaps you can tell me why youused the plural?But your numerical evidence below supports that the squares of0.9, 0.99, 0.999, ... tend TOWARD 1, not AWAY. To see this, all youneed to do is arithmetic. I don't see anything in your post tending AWAY from 1. .9^2 = .81> .9^4 = .6561> .99^2 = .9801> .99^4 = .96059601 .999^2 = .998001> .999^4 = .996005996001> .9999^2 = .99980001> .9999^4 = .9996000599960001 .99999^2 = .9999800001> .99999^4 = .99996000059999600001 .999999^2 = .999998000001> .999999^4 = .999996000005999996000001 .9999999^2 = .99999980000001> .9999999^4 = .9999996000000599999960000001> .99999999^2 = .9999999800000001> .99999999^4 = .99999998000000059999999600000001 .999999999^2 = .999999998000000001> .999999999^4 = .999999996000000005999999996000000001 .9999999999^2 = .99999999980000000001> .9999999999^4 = .9999999996000000000599999999960000000001 .99999999999^2 = .9999999999800000000001> .99999999999^4 = .99999999996000000000059999999999600000000001 .999999999999^2 = .999999999998000000000001> .999999999999^4 = .999999999996000000000005999999999996000000000001[snip rest for brevity]Challenge: Prove squares of 0.999... tend toward 1 > Garry Denke, Geologist> Denoco Inc. of Texas === Squares of 0.999... tend away from 1.99999.... is a single number, so there is only one square of it. Talking about Squares of 0.999... is a bit of a non-starter.You seem to be interested in the sequence .9^2, .99^2, .999^2, ... Clearly these numbers increase, so how can they tend away from 1? Your own computations show they get closer to 1.Assuming your post is not a troll, start with the fact that the sequence .9, .99, .999, ... tends to 1. Next observe that if 0 < x < 1, then 1 - x^2 = (1+x)(1-x) < 2(1-x). Therefore the numbers 1-.9^2, 1-.99^2, 1-.999^2, ... are, respectively, less than 2(1-.9), 2(1-.99), 2(1-.999), ... and the latter sequence tends to 0. Therefore the former sequence tends to 0, which is the same as saying .9^2, .99^2, .999^2, ... tends to 1. === =*** post for FREE via your newsreader at post.newsfeed.com *** Squares of 0.999... tend away from 1 .99999.... is a single number, so there is only one square of it. Talking> about Squares of 0.999... is a bit of a non-starter.Oh, he started that already in another thread. He means squares from a listdrawn from a breakdown of 0.999... .> You seem to be interested in the sequence .9^2, .99^2, .999^2, ... Clearly> these numbers increase, so how can they tend away from 1? Your own> computations show they get closer to 1.In the past, he has meant the Last One, but not as a limit value of thesequence.Hope this helps, karl mhttp://www.newsfeed.com - The #1 Newsgroup Service in the World!-----== 100,000 Groups! - 19 Servers! - Unlimited Download! =----- === =En el mensaje:210a83f4.0401081034.24b607ce@posting.google.com,> Squares of 0.999... tend away from 1[snip rest for brevity] Challenge: Prove squares of 0.999... tend toward 1>You meant prove that (1 - 10^k)^2 --> 1, as k ---> inf ?-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === =*** post for FREE via your newsreader at post.newsfeed.com ***> Squares of 0.999... tend away from 1 .9^2 = .81> .9^4 = .6561>(...)I think you mean that each Square from your list is adding digits away fromits trailing one, which is illustrated by your example. BTW, you stillhaven't said what 0.999... means. karl mhttp://www.newsfeed.com - The #1 Newsgroup Service in the World!-----== 100,000 Groups! - 19 Servers! - Unlimited Download! =----- === = [snip wishy-washy argument, cut to the quick] So once again, I think that Einstein got it right, and left the> physics mainstream behind.> But the mainstream, understandably, take a different position.> =Erk= (Eric Baird)> : Never ascribe to malice that which is adequately explained> : by incompetence.> : -- Napoleon Bonaparte> See> http://www.androc1es.pwp.blueyonder.co.uk/Fundamental_rv_2.0. htm> Einstein was incompetent and blundered. He did not get it right.> AndroclesFortunately Androcles was competent and got it right ;-) http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ TheLiar.html (new!) http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ CrapHuh.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Loadcrap.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Vision.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ HelpPretend.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ ProveProof.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ AndroMMX.html (new!) http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Equation.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Relativist.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Humour.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Chle.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Gibberish.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ AboutTheories.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ ConArtist.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ ProvePostulate.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Abstraction.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ EnergyConservation.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ SpeedInvariant.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Androrgasm.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ AndersenLogic.ht.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ SqrtAnswers.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ PartialDiff.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ AndArg.htmlDirk Vdm === [snip wishy-washy argument, cut to the quick] > So once again, I think that Einstein got it right, and left the> physics mainstream behind.> But the mainstream, understandably, take a different position.> =Erk= (Eric Baird)> : Never ascribe to malice that which is adequately explained> : by incompetence.> : -- Napoleon Bonaparte> See> http://www.androc1es.pwp.blueyonder.co.uk/Fundamental_rv_2.0. htm> Einstein was incompetent and blundered. He did not get it right.> Androcles Fortunately Androcles was competent and got it right ;-)> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ TheLiar.html(new!)> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ CrapHuh.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Loadcrap.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Vision.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ HelpPretend.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ ProveProof.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ AndroMMX.html(new!)> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Equation.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Relativist.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Humour.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Chle.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Gibberish.html>http://users.pandora.be/vdmoortel/dirk/Physics /Fumbles/AboutTheories.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ ConArtist.html>http://users.pandora.be/vdmoortel/dirk/Physics /Fumbles/ProvePostulate.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Abstraction.html>http://users.pandora.be/vdmoortel/dirk/ Physics/Fumbles/EnergyConservation.html>http:// users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ SpeedInvariant.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Androrgasm.html>http://users.pandora.be/vdmoortel/dirk/ Physics/Fumbles/AndersenLogic.ht.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ SqrtAnswers.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ PartialDiff.html> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ AndArg.html>Quite a CV {:-))Franz === =[snip Dinky's trash]> Quite a CV {:-)) FranzApart from your subjective opinions of any people involved, Franz, do youhave anyobjective opinion on the mathematics athttp://www.androc1es.pwp.blueyonder.co.uk/Fundamental_rv_ 2.0.htmor are you completely enraptured by Dinky's silly game of one-up-manship?Androcles === = [snip Dinky's trash] Quite a CV {:-)) Franz> Apart from your subjective opinions of any people involved, Franz, do you> have any> objective opinion on the mathematics at> http://www.androc1es.pwp.blueyonder.co.uk/Fundamental_rv_2.0. htm> or are you completely enraptured by Dinky's silly game of one-up-manship?> Androcles http://groups.google.com/groups?&as_umsgid=TCVIb.109057$ Jl3.4986923@phobos.telenet-ops.beDirk Vdm === =[snip]> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ AndArg.html Quite a CV {:-))Calamitas Vitae indeed.Dirk Vdm === =Hash: SHA1Could anyone please direct me to/supply me with a proof that the binomialtheorem:(1 + f(x))^r = 1 + r*f(x) + r*(r - 1)/2! * (f(x))^2 + ...Works for real exponent, r? It's easy to prove/reasonably obvious that itworks for positive integral exponent, but I've never seen a proof for realr.- -- Richard Hayden (rahaydenukNOSPAM@yahoo.co.uk)Webmaster: http://www.dx-dev.comDXOS (My Operating System): http://www.dx-dev.com/dxosI contend that we are both atheists. I just believe in one fewer god thanyou do. When you understand why you dismiss all the other possible gods,you will understand why I dismiss yours.- -Stephen RobertsComment: KeyID: 0x779D0625Comment: Fingerprint: 5642 852D 88BD BD0B 1A88 94B0 BFD2 C47C 779D 0625iQA/AwUBP/2pAb/ SxHx3nQYlEQIdXwCgsyCXmeVUogDt9xpzpMdidMJHt38An1z4Wu3dS1khRRPjp qSfL3Wrr/8p=uAQS === =I have a pdf that I can put up on my webpage if none of your other answers issatisfactory. Let me know.| Hash: SHA1||| Could anyone please direct me to/supply me with a proof that the binomial| theorem:|| (1 + f(x))^r = 1 + r*f(x) + r*(r - 1)/2! * (f(x))^2 + ...|| Works for real exponent, r? It's easy to prove/reasonably obvious that it| works for positive integral exponent, but I've never seen a proof for real| r.||| - --|| Richard Hayden (rahaydenukNOSPAM@yahoo.co.uk)| Webmaster: http://www.dx-dev.com| DXOS (My Operating System): http://www.dx-dev.com/dxos|| I contend that we are both atheists. I just believe in one fewer god than| you do. When you understand why you dismiss all the other possible gods,| you will understand why I dismiss yours.| - -Stephen Roberts|| Comment: KeyID: 0x779D0625| Comment: Fingerprint: 5642 852D 88BD BD0B 1A88 94B0 BFD2 C47C 779D 0625|| iQA/AwUBP/2pAb/ SxHx3nQYlEQIdXwCgsyCXmeVUogDt9xpzpMdidMJHt38An1z4| Wu3dS1khRRPjpqSfL3Wrr/8p| =uAQS|| === Hash: SHA1> Could anyone please direct me to/supply me with a proof that the binomial> theorem:(1 + f(x))^r = 1 + r*f(x) + r*(r - 1)/2! * (f(x))^2 + ...Works for real exponent, r? It's easy to prove/reasonably obvious that it> works for positive integral exponent, but I've never seen a proof for real> ' r.Taylor's Theorem.(1 + x)^r = 1 + r*x + r*(r - 1)/2! * x^2 + ...The LHS is de?ed for x>-1 and in?itely differentiableat x=0. The RHS is its Taylor series, with radius of convergence 1(except when the series terminates, then it has in?ite radius ofconvergence). Show that the remainder for Taylor's Theoremgoes to zero, and that is it. Certainly true for |x| < 1, andalso true in some boundary cases |x| = 1 depending on r.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === [snip]> So, if Russell's proofs are correct, then I can prove 1 = 2 and any> other statement you could dream of. Sounds like doom to me.Which of Russell's proofs?And which Russell? I don't think Leonard Blackburn was referring to Bertrand.Sorry for the confusion. I was intending to refer to the Russell who began this thread. Also, thank you to R3769, Michel Hack and Jesse F. Hughes for correctingmy sloppy thinking concerning the notion of inconsistency with regard tomathematics. I now understand that an inconsistency in ZFC does notmean an inconsistency in mathematics. But I am now confused as to whatmathematics is. Is it something Platonic waiting to be formalized?Also, I think that if Russell Easterly's proofs about countabilityand the seemingly unrelated Turing Machines are correct, then math maybe in a lot of trouble since his proofs are not in any formal system andsince proofs of the negations of his propositions can be given withoutresort to a formal system. So, the inconsistency would then seem to reside in the realm of Platonic math. Also, I am quite intrigued about the possibility of working withininconsistent systems by restricting the lengths of proofs. Anyway, I willnow bow out of this thread due to a lack of time. Leonard <87smiudtrj.fsf@phiwumbda.org> <3FF9DD0F.CC42A336@mdli.com> === Also, thank you to R3769, Michel Hack and Jesse F. Hughes for correcting> my sloppy thinking concerning the notion of inconsistency with regard to> mathematics. I now understand that an inconsistency in ZFC does not> mean an inconsistency in mathematics. Well, now we must be a bit careful. Shmuel Metz informs me thatGoedel proved the relative consistency of ZFC to PA --- so that *if*PA is consistent, *then* ZFC is consistent. Hence, an inconsistencyin ZFC would in fact yield an inconsistency in PA --- not due to thesimple fact that PA has a model in ZFC, but to the more surprising(reported) fact that ZFC has a model in PA.Now, I am not familiar with this result (I won't say that I've neverseen it, but if ' I've seen it, it musta been some time ago). So, don'ttake my word for it. I'm only reporting what Shmuel said (though, Iassume that he's not mistaken on such an explicit statement of fact).> But I am now confused as to what mathematics is. Is it something> Platonic waiting to be formalized? I won't get into that can of worms.> Also, I think that if Russell Easterly's proofs about countability> and the seemingly unrelated Turing Machines are correct, then math> may be in a lot of trouble since his proofs are not in any formal> system and since proofs of the negations of his propositions can be> given without resort to a formal system. So, the inconsistency> would then seem to reside in the realm of Platonic math.I wouldn't be too concerned about the implications of RussellEasterly's proofs. They might tell us something, but not aboutmathematics.-- Jesse HughesOf course, my ability to admit my mistakes and correct them is atrait that many of you seem to never have properly appreciated. -- JSH, discussing his 1463rd proof of Fermat's Last Theorem.X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oftX-Terminate: SPA(GIS) === =In <20040106145550.12131.00002567@mb-m05.aol.com>, on 01/06/2004 at 07:55 PM, r3769@aol.com (R3769) said:>Pardon my ignorance, but what does modelled ZFC in PA mean?You can encode statement about ZFC as statements about naturals, insuch a way that a proof in ZFC translates to a proof in PA. As aresult, if there is an inconsistency in ZFC then there is also aninconsistency in PA. Google for relative consistency or for G.9adel.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === at 07:55 PM, r3769@aol.com (R3769) said:Pardon my ignorance, but what does modelled ZFC in PA mean?You can encode statement about ZFC as statements about naturals, in>such a way that a proof in ZFC translates to a proof in PA. As a>result, if there is an inconsistency in ZFC then there is also an>inconsistency in PA. Google for relative consistency or for G.9adel.>Ok. Section 3.6 of Enderton's A Mathematical Introduction to Logic describesthis process (near as I can ?ure). However, the *relative* inconsistency ofPA isn't really what we are concerned about. For mathematics to be doomed, youstill need to show the inconsistency of ALL other set theories, ST, where STcan be modelled in PA. Because (we assume) ZFC is inconsistent, there is NO possibility of ?dingsome other consistent ST, such that ST can be modelled in PA? Now I'm really confused.rich <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oftX-Terminate: SPA(GIS) === =In <87smisrg0b.fsf@phiwumbda.org>, on 01/06/2004 at 09:14 PM, jesse@phiwumbda.org (Jesse F. Hughes) said:>No, it can't. Not only can it, but it's old news. Google ' for G.9adel.>Not unless the inconsistency is provable by the axioms of PA itself.Il va sans dire. How could one speak meaningfully of a model in PA ifthe conclusions about the model required more than the axioms of PA?>We have the axioms of PA. We have an interpretation of those axioms>in ZFC. You're looking at it backwards. G.9adel's famous ' result did not model PAin ZFC; such a model would have been obvious and trivial. What he didwas to model ZFC and others in PA in order to prove relativeconsistency.>To contradict my near as I can ?ger, you must show me how a>proof of P & ~P in ZFC leads to some proof of Q & ~Q in PA. No, I need only point you to the existing proofs, which Kurt G.9adelpublished long since. He, of course, had to do exactly what you ask.>I don't see why it should be so.Because he constructed a model using PA. Every inference in ZFCtranslates into an equivalent inference in PA.>Now, if PA is inconsistent, I see how to show that ZFC is, too. But>the other way around seems less than obvious.Well, it takes more machinery. Read a Scienti? American papaercalled Goedel's Proof for background, then read On the Consistencyof the Axiom of Choice and the Generalized Continuum Hypotheses.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org <20040103180216.29591.00001754@mb-m02.aol.com> <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> === In <87smisrg0b.fsf@phiwumbda.org>, on 01/06/2004> at 09:14 PM, jesse@phiwumbda.org (Jesse F. Hughes) said:No, it can't. Not only can it, but it's old news. Google for G.9adel.I know who Goedel is, thanks. This response is unhelpful.>Not unless the inconsistency is provable by the axioms of PA itself. Il va sans dire. How could one speak meaningfully of a model in PA> if the conclusions about the model required more than the axioms of> PA?We have the axioms of PA. We have an interpretation of those axioms>in ZFC. You're looking at it backwards. G.9adel's famous result did not model PA> in ZFC; such a model would have been obvious and trivial. What he did> was to model ZFC and others in PA in order to prove relative> consistency.This response ?ally points out the answer to my question.>To contradict my near as I can ?ger, you must show me how a>proof of P & ~P in ZFC leads to some proof of Q & ~Q in PA. No, I need only point you to the existing proofs, which Kurt G.9adel> published long since. He, of course, had to do exactly what you ask.I don't see why it should be so. Because he constructed a model using PA. Every inference in ZFC> translates into an equivalent inference in PA.Now, if PA is inconsistent, I see how to show that ZFC is, too. But>the other way around seems less than obvious. Well, it takes more machinery. Read a Scienti? American papaer> called Goedel's Proof for background, then read On the Consistency> of the Axiom of Choice and the Generalized Continuum Hypotheses.What does the consistency of the axiom of choice have to do with this?I am aware of these results of Goedel's, of course, but not themethods of the proofs. Is this where he models ZFC in PA? (Sorry,-- Jesse F. HughesAnd hey, if you're moping and miserable because mathematics tests you,then maybe, if you think you're a mathematician, you might want to trya different ?ld. -- Another James S. Harris self-diagnosis. <3ffadb60$18$fuzhry+tra$mr2ice@news.patriot.net> <87ptdwrfzl.fsf@phiwumbda.org>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oftX-Terminate: SPA(GIS) === =In <87ptdwrfzl.fsf@phiwumbda.org>, on 01/06/2004 at 09:15 PM, jesse@phiwumbda.org (Jesse F. Hughes) said:>No, not as far as I can see.Google for G.9adel and for relative consistency.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === =In , on 01/06/2004 at 06:58 PM, Russell Easterly said:>How would a TM know the difference between a ?ite tape and an>in?ite tape?It wouldn't, nor is that relevant. With an in?ite set of numbers onthe tape, the TM never halts, and hence never computes its output.>Say I write a TM that ?ds the smallest rational on a tape.You can't, unless there are only ?itely many.>Now, I give it a tape with the in?ite set of rationals [0,1].Then it never halts.>This TM will correctly determine that 0 is the smallest rational on>the tape.No. See above.>Why wouldn't this same TM ?d the smallest rational in (0,1)?In addition to the above, there's the fact that it doesn't exist.>Computable numbers is not a subset of real numbers if there are>computable numbers that aren't real numbers.What is a computable number? You don't seem to be using the term tomean the same thing as what Mathematicians mean by it.>Many people would argue that every real number is computable.Many people would arge that ?saucers have landed at Are 51.>But, if there computable numbers that look like real numbers, but >aren't, then computable is not a subset of real.Personally I'm sticking to the unicorn story; it's more believable.>I describe a TM that will ?d a natural number larger than any>member of a set of computable natural numbers.You can only do that for a ?ite set.>This larger natural allows me to compute a number of the form above>and prove this number is not in SNo. See above.>It basically means that S is well ordered.Do you know what well ordered means?>I described a well ordering of [0,1).No you didn't.>Assumption (2) is probably the hardest to justify.Who cares?The fault is not in those three assumptions.>I think computable and real diverge because they>have different de?itions of in?ity.Neither one has a de?ition of in?ity.>Say I gave you the in?ite series 1/2! + 1/3! + 1/4! + ... and>asked you to add them together.I'd tell you that it was impossible by de?ition and ask whether youwould be content knowing that it converged and what the limit was.>Eventually you would realize the process will never end. No. I wouldn't start a process that I never was meaningless.>A computer would just keep adding the numbers>together. Only if it was programmed to do that.>Assuming the computer can perform an in?te number of operations >with in?ite prescision,Why not assume that Oberon will present you with the answer? One makesas much sense as the other.>Such a computer could also compute the largest rational number >less than e-2.Yes, the pink Unicorn can do six impossible things before breakfast.it's too bad that what you want the computer to do is not evenimpossible; it's meaningless.In , on 01/06/2004 at 09:54 PM, Russell Easterly said:>This same TM will try to ?d the largest natural on an in?ite>tape.And fail.>The TM has no way of knowing the tape is in?itely long.A TM has no way of knowing anything. All that it does is to perform analgorithm and to produce a result should the algorithm halt. In thecase of in?itely many numbers, the algorithms does not halt andthere is no result.>This TM will always say there is a natural number larger than any>natural it has found on the tape.No, because it will never halt. Don't confuse intermediate steps withoutputs.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === In , on 01/06/2004> at 06:58 PM, Russell Easterly said:>I described a well ordering of [0,1).No you didn't.He described a bijection between the rationals in [0,1) and thenaturals. A relation suitable for well ordering can be easily derivedfrom that. Of course, that particular ordering makes 1/2 the leastrational in (0,1), so it's only going to be relevant to an existenceproof.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes /039 53 36 N / 086 11 55 W === = How would a TM know the difference between a ?ite tape> and an in?ite tape? It would never ?ish processing an in?ite tape in any ?ite time. Say I write a TM that ?ds the smallest rational on a tape.> Now, I give it a tape with the in?ite set of rationals [0,1].> This TM will correctly determine that 0 is the smallest> rational on the tape. Why wouldn't this same TM ?d> the smallest rational in (0,1)? Because there is no smallest rational in (0,1). Proof: Suppose there were a smallest rational number in (0,1). Call it q1.> Consider q2 = q1/2. Since q1 > 0, q2 = q1/2 > 0/2 = 0. Since q1 q2+q2 and q2 is positive, q1 > q2. Since q1 < 1, q2 < q1 < 1. Since 0> < q2 < 1, q2 is in (0,1). Thus, q2 is a rational number smaller than q1> and in (0,1), contradicting the assumption that q1 was the smallest> rational number in (0,1). Q.E.D. Computable numbers is not a subset of real numbers if there> are computable numbers that aren't real numbers. Okay, I have got to know what your de?itions of computable numbers and> real numbers are.Turing de?es a computable number as any number that can be representedby a computable sequence. A computable sequence is the outputtape of a TM that has an in?ite number of 0's and/or 1's.> Many people would argue that every ' real number is computable. Many...Turing believed that numbers like PI and e were computable.> There are people who argue that some natural numbers are not natural> numbers.I never said that many people agree with my arguments.> You asserted that 1/2! + 1/3! + ... + 1/k! is not in S. I describe a TM that will ?d a natural number larger than any> member of a set of computable natural numbers. A natural number larger than every member of a FINITE set of natural> numbers: Trivial, but irrelevant A natural number larger than every member of an INFINITE set of natural> numbers: Impossible True or false:> S((k-1)! + (k-2)! + ... + 2! + 2) = 1/2! + 1/3! + ... + 1/k!Neither true or false. This member of S is not de?ed.> (Remember, the right side of the equation is what you speci?d for x.) If true, this refutes your claim that x is not in S. If false, tell me what S((k-1)! + (k-2)! + ... + 2! + 2) is.Unde?ed.> A valid question. I give 3 assumptions: 1) Every member, S(i), of S can be examined.> 2) I can detemine if S(i) is less than, equal or greater than S(j).> 3) S contains every rational number in [0,1). What does examine mean?> It basically means that S is well ordered.> Saying I can examine member S(i) means> S(i) can be computed from i.> I described a well ordering of [0,1). Assumption (2) is probably the hardest to justify.> It might require that I show S can be well ordered> by value. It only requires the weaker result that S can be totally ordered by> value, a trivial operation. Which is fortunate, because S can not be> well-ordered by value.I never understood how a set can be totally orderedand still not be well orderable.I guess I am arguing that a computer will neverunderstand this, either.> I think computable and real diverge because they> have different de?itions of in?ity. Say I gave you the in?ite series 1/2! + 1/3! + 1/4! + ...> and asked you to add them together.> Eventually you would realize the process will never end.> You would stop adding the numbers together and> use other means to show that the sum must be less> than or equal to e-2. A computer would just keep adding the numbers> together. Assuming the computer can perform> an in?te number of operations with in?ite> prescision, it will get the right answer, e-2. Your assumption leaves out ... in ?ite time. Otherwise, it might> never get an answer, ?ite or otherwise.Yes, I am assuming that the computer can perform an in?itenumber of operations in ?ite time.> Such a computer could also compute the> largest rational number less than e-2. Let q1 be a rational number less than e-2. It is possible to ?d> another rational number q2 which is greater than q1 but less than e-2. Proof:A nice proof.It looks remarkably similar to mine.I am not saying that I can ?d the largest rational lessthan e-2, I am saying I can ?d the smallest rationalnumber that is larger than any member of S that isless than e-2. IOW, S does not contain everyrational number less than e-2.My point is that a computer will not decide to ignore itsprogramming and invent a proof that the number it issearching for doesn't exist.Any reasonable de?ition of computable that I know of willallow me to ?d the smallest member of a ?ite set.Why would a computer assume that an algorithm thatworks on any ?ite set will not work on some in?ite sets?Russell- 2 many 2 count <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <1g75yr4.1si6z861lg1554N%panoptes@iquest.net> === Many people would argue that every real number is computable. Many... Turing believed that numbers like PI and e were computable.Probably, he believed this because numbers like PI and e arecomputable.Now, how does that support the claim, Many people would argue thatevery real number is computable? -- Jesse HughesHow ly we are to be able to hear how miserable Willie Nelson couldimagine himself to be. -- Ken Ter on Fresh Air === Any reasonable de?ition of computable that I know of will> allow me to ?d the smallest member of a ?ite set.> Why would a computer assume that an algorithm that> works on any ?ite set will not work on some in?ite sets?This requires that the computer be able to compare sizes of members of the set, which is not necessarily the case.One can describe sets of numbers in a way that does not allow of comparisons of those numbers for size. === = Any reasonable de?ition of computable that I know of will> allow me to ?d the smallest member of a ?ite set.> Why would a computer assume that an algorithm that> works on any ?ite set will not work on some in?ite sets? This requires that the computer be able to compare> sizes of members of the set, which is not necessarily the case.OK> One can describe sets of numbers in a way that does not allow of> comparisons of those numbers for size.Can you describe such sets to a Turing Machine?I have noticed that many of the refutations other people have givenrely on induction. I am assuming that a TM can perform an in?itenumber of operations in ?ite time. Let's call this in?ite computation.It appears that induction and in?ite computation come to differentconclusions on some problems. Is there any way to formalize thedifferences?Russell- 2 many 2 count === Yes, I am assuming that the computer can perform an in?ite> number of operations in ?ite time.So where can we get one? === = Yes, I am assuming that the computer can perform an in?ite> number of operations in ?ite time. So where can we get one?There is no theoretical limit on how fast something can be computed.(At least there isn't a limit I have seen.)IBM is saying they have already exceeded Moore's law.Russell- 2 many 2 count === =I am looking for a book which would study (quite thoroughly) differentialequations, partial D.E, and differential systems from an almost-onlymathematical (which means rigorous and methodical) point of view (both exactmethods and numerical methods). I have very little knowledge in this area (abasic undergrad level).Would such a pearl exist ??--J.S === =I am trying to get some closure on this debate of at least 2 years that has been interesting in that it has deepened my heuristic understanding of Einstein's magni?ent achievement and of the cheap attempts to replace it as in Hal Puthoff's PV approach to metric engineering (exotic UFO warp drive). I support a lot of Hal's scienti? work BTW, but not this particular item.Paul Zielinski is concerned with distinguishing arti?ial gravity with real gravity. The latter, would only happen when the local 4th rank curvature tensor did not vanish. Paul seems to think there is a logical contradiction in the foundations of Einstein's thinking that is glossed over in MTW (1973) Gravitation. I do not think Paul is correct here.The relevant form of Einstein's equivalence principle is that:weight or g-force proportional to the 3rd-rank non-tensor connection geodesic (neutralize the charge) and the g-force vanishes.In this sense, gravity is locally equivalent to an accelerating frame dependent inertial force.line that break down in quantum gravity, also lack of a torsion-spin coupling.without an electric charge in an external EM ?ld. Neglect weak and strong charges for simplicity for now.Note that I is independent on the contingency that the 4th rank curvature tensor vanishes or does not vanish on the world line of the Therefore, the decomposition suggested by Zielinskiguv = guv(arti?ial) + guv(real)makes no physical sense at all IMHO.II. Measurement of the local 4th rank curvature tensor, outside the regime of quantum gravity, uses the time-like geodesic motion of a pair Einstein's geodesic deviation equation for stretch-strain relative completely independent of the measurement of g-force on a single test Indeed, the two kinds of measurement are incompatible in the sense of Bohr's principle of complementarity.III. What about the nonlocality of energy and momenta of the pure gravity ?ld? This has to do with the issue of gravity waves. You need to split the near ?ld from the far ?ld. You need to de?e a global ?tegral over a space-like slice for total Pu (also Muv for angular momentum), for example, using an effective non-tensor for ONLY the far ?ld dynamical degrees of freedom!Note that the elimination of dynamical degrees of freedom of the electromagnetic ?ld by Wheeler and Feynman, extended to the gravity ?ld by Hoyle and Narlikar, only applies to the far ?ld on the light cone not to the near ?ld off the light cone!What about Freud's identity as in Yilmaz's ' theory? It is not relevant because it attempts to replace curved space-time covariant divergences by global ?ace-time ordinary divergences. It is useful in treating gravity waves in asymptotic ?ace-times, but it is not more than that. Yilmaz is mistaken to try to elevate it as a basis for a new theory.There is a local stress-energy density covariant tensor for the pure gravity ?ld, it is simplytuv(Gravity) = [(Witten's String Tension)/(QED dimensionless coupling)]Guv(Einstein)Guv(Einstein) = Ruv - (1/2)RguvIn the non-exotic vacuum Guv = 0 exactly.Note that if you try to writetuv(Gravity) = tuv(Gravity)near-?ld +tuv(Gravity)far-?ldthe two terms on the RHS are not covariant tensors separately.This may have been part of the confusion.In exotic vacua with positive zero point energy pressure (attractive dark matter Omega ~ 0.3) and negative zero point energy pressure (repulsive dark energy Omega ~ 0.7)tuv(Gravity) = - (Witten String Tension)/(QED dimensionless coupling)/zpfguv/zpf = (Loop Gravity Area Quantum)^-1[(Loop Gravity Area Quantum)^3/2|Vacuum Coherence|^2 - 1]Witten String Tension = hc/(Loop Gravity Area Quantum)The discrete stringy link edge in a spin network is dual to the Loop Gravity Area Quantum) ~ Bekenstein BIT.guv = Einstein's Special Relativity Metric + (Loop Gravity Area Quantum)(Phase of Vacuum Coherence)(,u,v)where passive general coordinate transformations at a ?ed event P emerge from local gauge transformations on the Goldstone Phase of the Vacuum Coherence ?ld.,u is ordinary partial derivative.PSI = |Vacuum Coherence|e^i(Goldstone Phase of Vacuum Coherence)Curvature comes from stringy vortex core topological defects in the U(1) order parameter PSI as shown by Hagen Kleinert.More speci?ally curvature from disclination defects and torsion from dislocation defects in 4D world crystal lattice, whose discrete symmetry groups inside the unit cell on scale (Loop Gravity Area Quantum) are different vacuum phases.This theory can be extended to hyperspace with supersymmetry matrix dimensions of M-theory in which gauge force charges are reduced to Kaluza-Klein geometrodynamics. === =87,160 formulas and 10,828 graphics about mathematical functionsare now available free at The Wolfram Functions Site, animportant new resource for mathematicians, scientists, engineers,and students at:http://functions.wolfram.comIn the applications of mathematics to science and engineering--aswell as in pure mathematics itself--there are several hundred so--called special functions that have been intensively used for acentury or more. These special functions--with names like Besselfunctions, hypergeometric functions, and totient functions--de?e focal points of mathematical knowledge. The WolframFunctions Site provides in a readily accessible way the largestcollection, by far, of such knowledge ever assembled.Several widely used handbooks of mathematical functions have beenpublished, the largest of which contained about 15,000 formulas,meticulously compiled from thousands of technical papers.Traditional handbooks have also included only handfuls ofgraphics illustrating the properties of functions. WithMathematica, a huge number of new visualizations of functionshave become possible. The Wolfram Functions Site assembles over10,000 of these, with many more being planned.A major tour de force of reference website construction, TheWolfram Functions Site contains over 30 gigabytes of data.Material in The Wolfram Functions Site can be downloaded inseveral standard formats, including Mathematica InputForm andStandardForm, MathML, and PDF. Formulas can be copied from thesite and immediately used as input to a computer system. For easeof citation, each formula has been assigned a unique permanent ID.While having already far surpassed previous knowledge bases formathematical functions, continued growth is projected for TheWolfram Functions Site, with new searching capabilities, externalcontributions, and new classes of graphics and information. Visitthe site at:http://functions.wolfram.com === =hi.> As William said, you need to restate this slightly. For example: Let > (M,d_M) and (N,d_N) be metric spaces and f : M -> N a bijection. Assume for > every subset G of M that G is compact in M if and only if f(G) is compact > in N. Then f is a homemorphism.OK, point taken. That is what I ment to write, but I guees that didn'treally succeed.> 1,2,...} U {x} is compact in M.That tells me that K={f(x_n) : n = 1,2,...} is compact which tells methat (f(x_n)) has a convergent subsequence. I also see that there issomething special about x. It is, losely formulated, the only singlepoint which if removed from K will break compactness. This tells me (Ithink) that f(x) is the only possible point of convergence of asequence, that is not eventually constant, of K (I must of cause showthis), and hence that the convergent subsequence of (f(x_n)) -- that Iknow exists -- must converge to f(x). I think I need some sort ofargument which uses the fact that K is bounded to show that (f(x_n))and not only a subsequence converge to f(x).Am I on track or do I miss something important?-- === hi.> As William said, you need to restate this slightly. For example: Let > (M,d_M) and (N,d_N) be metric spaces and f : M -> N a bijection. Assume for > every subset G of M that G is compact in M if and only if f(G) is compact > in N. Then f is a homemorphism.OK, point taken. That is what I ment to write, but I guees that didn't>really succeed.>and that the x_n are distinct>and observe that {x_n : n = > 1,2,...} U {x} is compact in M.That tells me that K={f(x_n) : n = 1,2,...} is compact Or rather that {f(x_n)} union {f(x)} is compact.>which tells me>that (f(x_n)) has a convergent subsequence. I also see that there is>something special about x. It is, losely formulated, the only single>point which if removed from K will break compactness. This tells me (I>think) that f(x) is the only possible point of convergence of a>sequence, that is not eventually constant, of K (I must of cause show>this), Suppose that f(x_n_j) -> y. Then {f(x_n_j)} union {y} is compact,so {x_n_j} union {f^(-1)(y)} is compact. Since x_n_j -> x and thex_n_j are distinct it follows that f^(-1)(y) = x, because otherwise{x_n_j} union {f^(-1)(y)} would not be closed. So y = f(x).>and hence that the convergent subsequence of (f(x_n)) -- that I>know exists -- must converge to f(x). I think I need some sort of>argument which uses the fact that K is bounded to show that (f(x_n))>and not only a subsequence converge to f(x).Show this: If K is compact and every convergent subsequenceof y_n converges to y then y_n -> y.>Am I on track or do I miss something important?David C. Ullrich === =[I by mistake crossposted the message below to both sci.math andsci.math.research.Please ignore the .research thread. It will be replied to from there. Damn!]Assume:f(z)=c^z, (c, z in C, not both 0)f^(n)(z)=[f(z), n=1, f(f^(n-1))(z), if n>1]The sequence {f^(n)(1), n in N} is chaotic for c=e^{t/e^t},t=e^{alpha*Pi*i}, alpha irrational, while it converges (it's a m-cycle,coalescing into a 1-cycle eventually) if t=e^{2*k*Pi*i/n}, k in{0,1,2,...n-1}, some n in N.The process z->c^z always posesses the ?ed point: c_0=e^{-LW(-Log(c))}(where LW is the prinipal branch of the LambertW function), however, in bothcases above, |f ?(c_0)|=1.Does anyone have any idea how one can go about proving convergence ordivergence in the two cases above without resorting to the theory o?eration by Fatou and Julia?--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------ ------------------------Eventually, _everything_ is understandable === =Assume:f(z)=c^z, (c, z in C, not both 0)f^(n)(z)=[f(z), n=1, f(f^(n-1))(z), if n>1]The sequence {f^(n)(1), n in N} is chaotic for c=e^{t/e^t},t=e^{alpha*Pi*i}, alpha irrational, while it converges (it's a m-cycle,coalescing into a 1-cycle eventually) if t=e^{2*k*Pi*i/n}, k in{0,1,2,...n-1}, some n in N.The process z->c^z always posesses the ?ed point: c_0=e^{-LW(-Log(c))}(where LW is the prinipal branch of the LambertW function), however, in bothcases above, |f ?(c_0)|=1.Does anyone have any idea how one can go about proving convergence ordivergence in the two cases above without resorting to the theory o?eration by Fatou and Julia?--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------ ------------------------Eventually, _everything_ is understandable === =Working out a solution to a problem I empirically came across with anumerical series that is given below, for i = 1, 2, 3, ...,(A+B), and where A is a any given integer, B=1, 2,...,A, and the ratios are alltaken as the ceiling values.if i belongs to the interval [1, A-B] then :index[i]= ((A-B)/2) + 1, if (A-B) and i are even;index[i]= ((A-B)/2), otherwise. and if i belongs to the interval [A-B+1, A+B] then :index[i]= ((i-1)/2), if i belongs to the interval [(A-B+1), A];index[i]= (2*A-(i-1))/2), otherwise. Example:Let A = 5 and B = 1, then we have that:index[1]=2index[2]=3index[3]=2index[4]=3index[5]=2index[ 6]=3My question is how I can formally prove that this is true for anyinteger A under the restrictions I posed above.I have tried many values experimentally, namely A = 1,2,3,...50. Theyall worked. But now I need a formal mathematical proof or, at least, agood idea of how I could claim this result without needing this formalmathematical proof.I ?st thought of induction. But the problem is that I have toconsider two independent variables, namely A and B. How can I thenpossibly make the inductive assumption that it holds for A and provethat it holds for A+1, ignoring the value of B? This seems quiteweird.At second, I tried to identify a known series that could ? in theseries I have derived. Unfortunately, it was in vain either. === =Greetings,I am curious as to whether or not there is a fairly standard libraryor package for doing sparse matrix computations (like multiplication,addition, etc. - fairly straightforward stuff). I'd prefer C, butFortran is ?e too. Cordially,Ted === Greetings,I am curious as to whether or not there is a fairly standard library> or package for doing sparse matrix computations (like multiplication,> addition, etc. - fairly straightforward stuff). I'd prefer C, but> Fortran is ?e too. Try the Matrix Template Library (C++); it is free and good, and handles both dense and sparse matrices in a uniform way. For more packages, seethe software section on my home page.Arnold Neumaier === =There is dspslv.f90 from the NSWC Fortran library at my web site (seesignature below).It is for the solution of a set of sparse linear equations.At netlib (www.netlib.org), there is a directory called sparse whichcontains code in C.-- Alan Millerhttp://users.bigpond.net.au/amillerRetired Statistician> Greetings, I am curious as to whether or not there is a fairly standard library> or package for doing sparse matrix computations (like multiplication,> addition, etc. - fairly straightforward stuff). I'd prefer C, but> Fortran is ?e too. Cordially,> Ted === =Let me add to my previous message, a link to sparskit at:http://www-users.cs.umn.edu/~saad/software/SPARSKIT/ sparskit.htmland an ftp link to the same code:ftp.cs.umn.edu/dept/sparse/> -- > Alan Miller> http://users.bigpond.net.au/amiller> Retired Statistician Greetings, I am curious as to whether or not there is a fairly standard library> or package for doing sparse matrix computations (like multiplication,> addition, etc. - fairly straightforward stuff). I'd prefer C, but> Fortran is ?e too. Cordially,> Ted === =Greetings,I am curious as to whether or not there is a fairly standard library> or package for doing sparse matrix computations (like multiplication,> addition, etc. - fairly straightforward stuff). I'd prefer C, but> Fortran is ?e too.Cordially,> TedNIST Sparse BLAS provides computational kernels for fundamental sparsematrix operations: sparse matrix products and solution of triangularsystems. Look at http://math.nist.gov/spblas/Hugo Pfoertner === =I have a large data vector (10E6 pts or so) and I would like to determinehow correlated the noise is. I have a few hundred recorded measurementsof the vector and effectively want to compute the corresponding covariancematrix but the sheer size of it is somewhat problematic for a simplistic S = 1/n (X' (I - 1/n ' 11') Xcalculation. Are there any statistical tricks I can use or am I stcomputing S in chunks? I expect it will be more-or-less diagonal (orsparse banded) but there is a Fourier Transform and a some dodgy lookingalgorithms applied to correct for non-uniform sampling and calibrationissues. I essentially want to quantify how diagonal S is. Theory statesthat each measurement in each data vector is completely independent, butthe resampling and calibration algorithms look like they will correlate it. Would measures of entropy/information content be of use?Many thanks,JamieX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i07GT6902717; Wed, 7 Jan 2004 11:29:06 -0500 === =As you are successfully using Clapack with Visual C++, maybe you couldhelp me...I am having problems using certain functions in visual C++: it returnsLINK errors of the form:error LNK2001: unresolved external symbol __imp__?ve you had such problems and if so how can one resolve them? === As you are successfully using Clapack with Visual C++, maybe you could>help me...>I am having problems using certain functions in visual C++: it returns>LINK errors of the form:>error LNK2001: unresolved external symbol __imp__?ave you had such problems and if so how can one resolve them?I have VC++ 5 and I have used clapack with it, from the version inclaw32.zip downloaded from netlib. I don't remember what I did tomake it work, and I only use it for 1 program to calc eigenvalues.megabytes).I looked in the clapack.h and f2c.h ?es and there is nothing thatcontains the word ?n them, so maybe the error has to do withsome other library you use. === =Miranda, I'm sorry I cannot answer your question directly, but are you using theWindows version of CLAPACK (i.e., CLAPACK3-Windows.zip, fromwww.netlib.org/clapack/index.html)? Have you attempted to use (link against) the precompiled binaries from theCLAPACK3-Windows.zip ?e? Have you tried running the testing programs from CLAPACK3-Windows.zip? You might try contacting Bob Denny rdenny@dc3.com (the guy who puttogether the CLAPACK3-Windows.zip distribution of CLAPACK) if you cannot getthings working.John> As you are successfully using Clapack with Visual C++, maybe you could> help me...> I am having problems using certain functions in visual C++: it returns> LINK errors of the form:> error LNK2001: unresolved external symbol __imp__?Have you had such problems and if so how can one resolve them?>X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i07Lmfh27722; === =I'M LOOKING FOR A MATHCAD PROGRAM TO GRAPH DATA USING PIPER ANDTERNARY DIAGRAMS. === I have asked Rob Stevenson from Utrecht and he refered me to Lemma 6.2.5 of> Hackbusch's multigrid book where one can ?d a more general interpolation> result.Lemma 6.2.5:> Let $Lambda_1, Lambda_2$ be spd and $A$ be an arbitrary matrix. Then> $$> norm{Lambda_2^beta A Lambda_1^{-beta}} leq> norm{Lambda_2^alpha A Lambda_1^{-alpha}}^{frac{gamma-beta}{gamma-alpha}}> norm{Lambda_2^gamma A Lambda_1^{-gamma}}^{frac{beta-alpha}{gamma-alpha}}> $$> for $alpha leq beta leq gamma$.But this reduces to your problem only when the lambda's commute!?Arnold NeumaierNo. Here is some more detailed explanation.The Lemma is the matrix form of an interpolation result between scales ofR^n endowed with the norm |x|_{H_s}=|Lambda_1^s x|, and K_s is R^n withnorm |x|_{K_s}=|Lambda_2^s x| (|.| denotes the Euclidean norm).A:R^ntoR^n is a linear operator.Since||A||_{K_s <- H_s} := sup_{x neq 0} |Ax|_{K_s}/|x|_{H_s}the Lemma says that||A||_{K_beta <- H_beta} <= ||A||_{K_alpha <- H_alpha}^{frac{gamma-beta}{gamma-alpha}} ||A||_{K_gamma <- H_gamma}^{frac{beta-alpha}{gamma-alpha}}The application to the original problem is as follows: Choose alpha=0,beta=s, gamma=1, Lambda_1=A^{1/2} (with A from the original problem),Lambda_2=B^{1/2}, and A (from lemma) = Identity we obtain that||A^{s/2} B^{-s/2}||} <= ||A^{1/2} B^{-1/2}||^sThus, we havesup_{x neq 0} (A^s x,x) / (B^s x,x)= sup_{x neq 0} |A^{s/2} x|^2 / |B^{s/2} x|^2 = sup_{y neq 0} |A^s B^{-s} y|^2 / |y|^2= sup_{y neq 0} (|A^{1/2} B^{-1/2} y|^2 / |y|^2)^s<= sup_{x neq 0} ((Ax,x) / (B x,x))^s<= 1Therefore A <= B implies A^s <= B^s for 0<=s<=1.Nicolas. === =Does anyone know of any shareware that can take a series of (x,y,z) points, plot them and give the best ? equation? For example, lets say I have a ?ting where I can apply Px, Py, and Pz loads, and I determine several allowed combinations of Px, Py, and Pz that give me a margin of safety of zero. If I want to publish an equation that can be used to determine whether they are inside the allowable surface or outside, what is the best way to determine this equation? (i.e. how do I arrive at the best characteristic equation?). I've done this exercise by hand, using a 12th order multivariate regression, but I didn't get a great R-squared value. Is there shareware software out there that tries several characteristic equations? === =Try PolSurfIt gives at most ten polynomial models. The best is mod0.It does not plot. That you must do somewhere else.Arto Huttunen> Does anyone know of any shareware that can take a series of (x,y,z)points,> plot them and give the best ? equation? For example, lets say I have a ?ting where I can apply Px, Py, and Pz> loads, and I determine several allowed combinations of Px, Py, and Pz that> give me a margin of safety of zero. If I want to publish an equation that> can be used to determine whether they are inside the allowable surface or> outside, what is the best way to determine this equation? (i.e. how do I> arrive at the best characteristic equation?). I've done this exercise by> hand, using a 12th order multivariate regression, but I didn't get a great> R-squared value. Is there shareware software out there that tries several> characteristic equations? === =we are planing to start with a project refering imaging software for patternrecognition. In a ?st step we are looking for some good books in thisarea e.g. math-background, standard algorithms.Any help is welcome,Michael -- www.enertex.de - Innovative Systeml.9asungen der Energie- und Elektrotechnik === =we are planing to start with a project refering imaging software for pattern> recognition. In a ?st step we are looking for some good books in this> area e.g. math-background, standard algorithms.By far the best book on general pattern recognition is that by Ripley.It is very well written, both mathematically, didactically,and from the selection of material (modern and just what is needed).Complementing this, the book by Spiegelhalter et alprovides a comprehensive comparison of software packages for classi?ation (which is the same as pattern recognition).In both books, one does not ?d, however, techniques that arespeci? to image interpretation (such as edge detection orimage-related feature extraction techniques).Arnold Neumaier === =87,160 formulas and 10,828 graphics about mathematical functionsare now available free at The Wolfram Functions Site, animportant new resource for mathematicians, scientists, engineers,and students at:http://functions.wolfram.comIn the applications of mathematics to science and engineering--aswell as in pure mathematics itself--there are several hundred so--called special functions that have been intensively used for acentury or more. These special functions--with names like Besselfunctions, hypergeometric functions, and totient functions--de?e focal points of mathematical knowledge. The WolframFunctions Site provides in a readily accessible way the largestcollection, by far, of such knowledge ever assembled.Several widely used handbooks of mathematical functions have beenpublished, the largest of which contained about 15,000 formulas,meticulously compiled from thousands of technical papers.Traditional handbooks have also included only handfuls ofgraphics illustrating the properties of functions. WithMathematica, a huge number of new visualizations of functionshave become possible. The Wolfram Functions Site assembles over10,000 of these, with many more being planned.A major tour de force of reference website construction, TheWolfram Functions Site contains over 30 gigabytes of data.Material in The Wolfram Functions Site can be downloaded inseveral standard formats, including Mathematica InputForm andStandardForm, MathML, and PDF. Formulas can be copied from thesite and immediately used as input to a computer system. For easeof citation, each formula has been assigned a unique permanent ID.While having already far surpassed previous knowledge bases formathematical functions, continued growth is projected for TheWolfram Functions Site, with new searching capabilities, externalcontributions, and new classes of graphics and information. Visitthe site at:http://functions.wolfram.com === 87,160 formulas and 10,828 graphics about mathematical functions87,160 formulas including such wonders as PrimePi[1] = 0 PrimePi[2] = 1 PrimePi[3] = 2 PrimePi[4] = 2 PrimePi[5] = 3(for the numbers 1..50). 10,828 graphics, including 2401showing various statistics on the digits (continued fractioncoef?ients, blah) of pi.The thing about using a package like Mathematica is thatit's very easy to in?hese numbers without actuallyincreasing the amount of useful information.Don't get me wrong; there *is* a lot of useful informationhere, and it's very cool of Wolfram to make it available forfree. But is there 5 times as much as in, say, Abramowitzand Stegun? No way. 5 times as much data, perhaps, but datais not information.> are now available free at The Wolfram Functions Site, an> important new resource for mathematicians, scientists, engineers,> and students at:> http://functions.wolfram.com[etc]-- Gareth McCaughan.sig under construc === =Students are often amazed--well, I've known PROFESSORS who were> amazed--to discover that Mathematica and Maple do not evaluate> (-1)^(1/3) to -1. Instead, they return a complex cube root. The> reason is to maintain consistency in branch cuts...Must be the same lot that's poisoning the students with Matlab!? As tothe above, it's a matter of returning the *principal* root, rather thanthe intuitive one, at least, if De Moivre's teachings still stands. === =Things have got so bad that students need a CALCULATOR to add> fractions?!It seems like it's going from bad to worse -- US graduates now needMatlab to aquare or invert a 2x2 matrix... === Things have got so bad that students need a CALCULATOR to add> fractions?!Few calculators can add fractions, and return a fraction.This should be on all.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === =Has anyone considered publishing in the Public Library of Sciencejournal(s)?http://www.plos.org/Seems a good idea, although they are presently charging $1500 for anpart...Hang on a minute... presently the library pays for journals and theauthors do the work and get it published for free. But with Plos, thelibrary now gets the journal for free and the authors both do the work*and* pay for it!!!Huh? I'm not so sure about it now. === =I believe the sum of the 3x3 square is actually 45. The 4x4square sums to 136.> Orig-To: sacredlandscapelist at yahoogroups.com50031029 viii om?st installment, 0, 1, 2x2, 3x3 (+ 4x4 notes)# ...I propose ...we try to solve the magic squares once> # and for all.I think I know what you mean. I tried something similar with an> 8x8 ?Mercury' Square in order to con?ure it properly for Yijing> divination using it and Chess. I found out that while there are> several solutions (as with many magic squares), there is one most> simple (transposition) solution. there are famous, named squares,> however, with intriguing properties of identical sum throughout> that make it dif?ult to do more than grade such solutions in'solution once and for ' all' is, without further de?ition of what> solution means, too ambiguous to produce. mathematics constitutes> the ?ld of endeavor which includes number arrangement (logic,> puzzles, computers, and, for some, thaumaturgy or theurgy).orders (sizes) of magic square were provided assignment by> occultists to planets, gods, and other quirky principles, its> apparent fact of sequence mapped to the apparent Planets of the> time. with the discovery of 3 more planets, and as I'm beginning> a refashioning for mages of my ilk in the PISCEAN age what should> have been done long ago, I agree that some kind of new Solution> should be devised with speci?d parameters for the construction> of novel magical devices and systems. the tendency to remain in> the past, accept the well-worn constructions of the past, even> though their knowledge-base is superceded by plain contradiction,> is to set into Golden Age remembrance what should be oriented to> the present and personal as a mapping one-to-one correspondence.# I believe that they do represent a higher form of mathematics> # that currently is not in use by our scientist.magic squares are formal structures, not formulae in a sense as> to be recognizable by many modern scientists, and to call them> a higher form of mathematics is to provide them a greater range> of meaning than should include them. yes, the magic square is a> device usable by magicians to in? the universe (they think),> and typically magic is considered a form of science unrecognized> by the witness (esp. for fans of ACClarke). I'm not sure what is> supportable beyond this.# I also believe that we collectively have the ability to solve> # the apparent complexities concerning these squares.very true, many are gifted in the area of mathematics that have> the time and interest in lending a hand. references are also> available, interestingly enough (many online). the best appear> to be long treatises that haven't made it online yet. I'm very> seriously considering transcribing some of Andrews' book if it> ' hasn't yet been done.# Anybody who wants to partake or have suggestions/input is> # more than welcome to join in and help out.agreed! : > # I have read the three books of occult philosophy by cornellius> # agrippa,which translation? I've been looking at it carefully lately and> noting its resonances, geometrics and attributions. the Llewellyn> Tyson version has some helpful but sometimes ambiguous notes that> are supplemented with generous Appendices. to read the thing> straight through is different than using it as a reference> (indexing, or via ToC for topics). his magic squares are perhaps> in need of re?ement (Tyson corrected many errors in translations> at least, a few from the original manuscript, if I recall arightly).# and have gained some insights which I would like to share.thank you! more!!# ...the 3x3 square is probably useless and/or unly in that its> # total result is equal to 666 and any attempt at solving it would> # be futile.LOL! the 3x3 square, also known as the Lo Shu or Lo River> diagram, also known as the Saturn Square, is widely known for> its l, power and the bene?s it affords its magical users.> its actual total is 136 (the ?Sun' square 6x6 adds to 666).> the questions are to what ought the 3x3 be associated and > whether there is more than one solution to it.3x3 is certainly the lowest *order* of magic square, by the> usual minimum de?ition (rows, columns, and major diagonals> composed of a set of sequential numbers adding to the same total).> as I have suggested elsewhere, fundamentals of magic square> construction allow earlier orders by restricted quali?ation.> more below. === === === Magic Squares :> Arcane Speculation on> Mathematical Objects> in Series> by Seyfert-1 (nagasiva)> ------------------------------------------------------------> Proto-Squareswhat follows is a beginning of analysis of magic squares for> arcane reformation purposes and intrinsic morphological content.(order)> === = (1? 0?) 1the smallest order (0?) of magic square? only in a sense> like anything to the 0 power produces the result of 1. : perhaps this could be attributed somehow (with the next> order(s 0/1/2)) with a Without/Limitless/LVX trinity> of pre-Ketherian Qabalistic Nothingness. : > one might also construct the square out of letters, in> which case we're probably talking about Aor perhaps + (t/200? x/600?)------------------------------------------------------ ------(2) 1 2 1 4> A -- 3 4 B -- 2 3no possible solution obtains for a 2x2 square, but there > is a con?uration with the least differential in > summation (sums in parens): (5)(5)> (3) (7)> (4) 1 3 A C> C -- (6) 4 2 D B (Ulian Schemata)the extreme sums displaced to the diagonals. 2x2 pursued> in letters might double upon itself like so: (4)> A B 1 2 (3)> B A 2 1 (3)> (2)> (3)(3)simultaneously referencing paternal authority (ABBA),> or> (1600)> A Z 1 800 (801)> Z A 800 1 (801)> (2)> (801)(801)symbolizing the Beginning and the End (Romanization> of the Greek, popular Alpha et Omega).> General> -------> any letter / number / ?ure might be utilized within> a square of this order to combine 2 (or 4) elements> (e.g. Jupiter(6) and Saturn(7): 6 7 F G> 7 6 G For the use of their sigils inside an amulet or some> ceremonial diagram which calls for their qualities).-------------------------------------------------- ----------> Proper Magic Squares (3x3 13x13)the ?st of the odd magic squares (Andrews divides them> into families of odd and even because of methods of their> derivation and construction) has *1* solution, and is> the only magic square that has a single solution.the variation of this square is not its numerical arrange-> ment so much as its *orientation*. while this is true of> all magic squares (that they have orientations, and also> mirror-images), this is seldom remarked upon by occultists> who identify their selections, since the solution producing> the requisite sums suf?es to assign the whole.for my purposes, and in order to be thorough, I'll provide> what information I am able about traditional cultural usage> of these, assigning each to the recti?d astronomical> standard after the pattern of Agrippa and others in occult> traditions, and requesting additions from the interested.my impression is that few take great care in dealing with> magic squares as a whole, haphazardly presenting what to> them conforms to a logical parameter (true enough, but> insuf?iently-detailed for scrutinizing occult practice).--------------------------------------------------- ---------(3) PLUTO -- traditional: Lo Shu, Saturnthus the Square of Pluto (Lo Shu, formerly Saturn) in its> various possible orientations is as follows: (15) (15) (15)> (15)> (row i) (15) 4 9 2 D I B(row ii) (15) 3 5 7 = C E G(row iii) (15) 8 1 6 = H A F> (15)this is the con?uration presented as both the Lo Shu> (e.g. in Master Huang's book on Numerology) and for the> Planet Saturn (in Tyson's Agrippa).I have labelled the rows here (i->iii) which could also> be arranged in columns or transposed to iii->i. this> latter arrangement (with 8/1/6 at top) is presented in> Andrews for the 3x3 square (page 2, ?ure 1).there are 2 families of 4 orientations, constituting a> total of 8 possible arrangements of this 3x3 square if> one details its rectilinear positions and mirrorings: ----------------------------------> 2 7 6 6 7 2> 9 5 1 1 5 9> 4 3 8 8 3 4> -------------> 4 3 8 8 3 4> 9 5 1 1 5 9> 2 7 6 6 7 2> ----------------------------------> 2 9 4 4 9 2> 7 5 3 3 5 7> 6 1 8 8 1 6> -------------> 6 1 8 8 1 6> 7 5 3 3 5 7> 2 9 4 4 9 2> ----------------------------------these might be associated with 8 Yijing trigrams> for a complex cross-reference. how to begin such> an assignment will depend on what one takes as a> primary sequencing key or mapping translator (e.g.> orienting the corners to a 2-4-6-8 clockwise> progression might initiate the sequence, or the> 2/7/6 at top might come ?st as its numbers in> the upper-left and top-center are lowest).that the circling 8 numbers (1->9 excepting 5) are> *already* given trigrammatic assignments within> the Lo Shu (see Master Huang's Numerology at least> for this) may make identifying the particular squares> within this scheme easier by keying the numbers to> trigrams and then solving for ?net trigram' by> con?urative quality. substituting numbers allows> the following, initially, the following may be used> within some formula to conclude as to ordinal value: ---------------------------------- 2 7 6 6 7 2> __ __ __ __ _____ _____ __ __ __ __> __ __ _____ _____ _____ _____ __ __> __ __ _____ _____ _____ _____ __ __ 9 5 1 1 5 9> _____ __ __ __ __ _____> __ __ . o _____ _____ o . __ __> _____ __ __ __ __ _____ 4 3 8 8 3 4> _____ __ __ _____ _____ __ __ _____> _____ __ __ __ __ __ __ __ __ _____> __ __ _____ __ __ __ __ _____ __ __ ------------- 4 3 8 8 3 4> _____ __ __ _____ _____ __ __ _____> _____ __ __ __ __ __ __ __ __ _____> __ __ _____ __ __ __ __ _____ __ __ 9 5 1 1 5 9> _____ __ __ __ __ _____> __ __ . o _____ _____ o . __ __> _____ __ __ __ __ _____ 2 7 6 6 7 2> __ __ __ __ _____ _____ __ __ __ __> __ __ _____ _____ _____ _____ __ __> __ __ _____ _____ _____ _____ __ __ ---------------------------------- 2 9 4 4 9 2> __ __ _____ _____ _____ _____ __ __> __ __ __ __ _____ _____ __ __ __ __> __ __ _____ __ __ __ __ _____ __ __ 7 5 3 3 5 7> __ __ __ __ __ __ __ __> _____ . o __ __ __ __ o . _____> _____ _____ _____ _____ 6 1 8 8 1 6> _____ __ __ _____ _____ __ __ _____> __ __ _____ __ __ __ __ _____ __ __> __ __ __ __ __ __ __ __ __ __ __ __ ------------- 6 1 8 8 1 6> _____ __ __ _____ _____ __ __ _____> __ __ _____ __ __ __ __ _____ __ __> __ __ __ __ __ __ __ __ __ __ __ __ 7 5 3 3 5 7> __ __ __ __ __ __ __ __> _____ . o __ __ __ __ o . _____> _____ _____ _____ _____ 2 9 4 4 9 2> __ __ _____ _____ _____ _____ __ __> __ __ __ __ _____ _____ __ __ __ __> __ __ _____ __ __ __ __ _____ __ __ ----------------------------------and once a formula of solution is put into operation,> (e.g. root lines value 4, mid-lines value 2, top 1,> left-most value 4, mid-trigram value 2, right 1, solve> for each magic square arrangement to 8 values), then> these may be arranged and identi?d accordingly and> even used for divinatory and magical enterprises in> association with the Planetary in?s of Pluto.Andrews makes plain that one might substitute for the> numbers in any of these squares a formula whereby the> algebra of variables may be used to solve for all the> magic square's contents given any initial ?ure ?X'> which is ' the least number in the square (our solution> merely derives the *least* integer, X=1), such that for ?X'=2 (45) (45) (45)> (45)> (row i) (45) 10 28 07 (row ii) (45) 12 15 18 (row iii) (45) 23 02 20> (45)which will not map to an order A=1...Z=26 system,> requiring dual-letter combinations such as the> most evident Ulian simple combination system: J KH G JB JE JH KC B Kthrough this method one might utilize Plutonian> squares with variable least seed numbers to construct> any number of alternatives, each having 8 sets of> orientations and mirrorings which might be ciphered> through an Ulian combination system to string sets.> Magic Squares Essay terminates.> === == returning to conversation in Sacredlandscapelist, venturing to(4) (notes)# ...most ...fortunate square, ... the square of jupiter> # or 4x4....by traditional Planetary assignment, 4x4 = Jupiter, which is> called in astrology ?The Great Bene?ent'. for this reason> your assertion is valuable within such a context.# 04 14 15 01> # 09 07 06 12> # 05 11 10 08> # 16 02 03 13 = D N O A> I G F L> E K J H> P B C M (A=1... Z=26)through proliferation merely by orientation and mirrorings,> we're talking about at LEAST 4 orientations and their 4 mirrors> once more, *plus* any variations in composition beyond the obvious > row/column reversal (yielding at least 8 possibles) by virtue > of formulation of number positions (again where here X=1).why 04 should be placed at top left (rather than 01) is a> valuable query. we have been presented the ?Jupiter Square'> as selected by Agrippa. in fact, Andrews derives 6 laws > (that I am not reproducing here for the sake of simplicity, > but a thorough investigation of the 4x4 ought include it) > governing all 4x4 magic squares, and produces the 4x4 square > below, indicating that squares of *higher* order are too > complex to reduce to laws. 01 08 12 13> 14 11 07 02> 15 10 06 03> 04 05 09 16as we can see just by these 2 examples, there are a large> number of possible 4x4 arrangements, before even considering> minor variation like orientation or mirroring, which may, as> part of their composition, have the ?associated' quality> of adding to a speci? sum in symmetrical positions (e.g.> here the sums equate to 17 on positions across a diagonal).I'm not aware of many cultural designations of the 4x4 magic > square comparable to Lo Shu. Andrews does mention what he> calls a Jaina Square of 4x4, which is: JAINA SQUARE> -------------> 07 12 01 14> 02 13 08 11> 16 03 10 05> 09 06 15 04from an inscription of the 12th or 13th centuries by Jains> mentioned by a Professor Smith and passed on by Andrews> (pp. 124-128) and apparently of a different quality than> many of the other possibilities (to be explored later).> the qualities of 4x4 squares is only compounded in depth> as one progresses to greater orders, plus it may be> integrated into larger squares as central portions!# According to the philosophers (agrippa,solomon,pythagoras etc)there's a wealth beyond this, apparently, in the world of> philosophy and magic. :> I am only barely touching on this> myself by elaborating on your wonderful post.# there> # is an intelligence or higher spirit if you like that determine the> # outcome of certain events. There is a spirit to what is good and a> # spirit to that which is bad. So I gather that the future prediction> # of number is dependant on which spirit has the most in? at the> # time. However as Agrippa tells us the number of these intelligences> # its not neccessary to know who rules, just that there are 2 possible> # future outcomes when predicting a number between 1-16. The> # intelligences are both allocated 136 (the total of the square) and> # are arrived at by means of the seals and characters of jupiter.my impression (a guess) is that some kind of name was discernable> *within* the square being considered as it was substituted with> letters and that these were ascribed *after the fact*, then> drawn out as sigils accordingly. I see no reason why they would> be necessary in any sense, but am open to hearing arguments for> this. of course the way Agrippa has it these arrive in Hebrew,> but one could construct this within English or other languages.> that is certainly my intent, which is in part why I'm providing> the Ulian and AlphaZed equivalences in this post, for future> reference and construction by the student (myself included!).# the problem has been that no one knows how these seals where arrived> # at and therefore can not be assigned to the squares directly. This> # seems to be the ?st road block in it's translation. Is anyone here> # familiar with the seals and sigils?this constitutes a secondary level of analysis where magic squares> are concerned because they enter into linguistic construction that> has no mathematical basis and therefore few real delimiters of form.a Jovian construct analysis:> # However Agrippa tells us that they are arrived at in this fashion:this actually constitutes a description of what was constructed,> rather than its construction method per se.# Intelligence to what is good:> # Jophiel = IHPHIAL> # I=10 + H=5 + PH=80 + I=10 + A=1 + L=30 TOTAL=136> #> # A spirit to what is bad:> # H=5 + S=60 + M=40 + A=1 + L=30 TOTAL=136one may therefore analyze the *method of determination* > to a degree and construct the rules: a) the total of the names of the Intelligences> and Spirits must equate to the total of the > numbers in the square; b) the letters for these numbers are connected,> creating the Seals for magical operations.# ...> #> # The ?st step in this unsolvable mess was> # the disproval of the 4 elements, our scientits discovered more and> # more, and the basic concepts of the elements and their ratios was> # lost.the re?ement of ?elements' made this possible. modern scientists> are merely reductionists, focussing on subatomic elements, whereas> previous perspectives of elements weren't reductionist at all but> material in a compository manner (examine the early Greeks and we> can see that they were struggling to come up with a ?universal> element' in the same way that today there is a move toward what> has been called ?Uni?d Field Theory'. whether these early Greeks> were actually atomists, precursors to today's materialists is> something of a controversy as I understand it (something which> I think Heidegger argued against, if I recall correctly).# ...if the> # elements where ratios of numbers, like... 1 is to 2 as 2 is to 4 each> # one being it's double, then why not the seven planets also? why could> # the seven planets also not be ratios of 1 being related to one as the> # other?, I mean is it a coincidence that an atom has seven shells, or> # the the sacred tetractys (the foundation of all magic and all form)> # also has seven points in this fashion?> #> # THE SACRED TETRACTYS OF PYTHAGORAS:> #> # POINT 1 UNITY> # LINE 2 3 PRIME> # SURFACE 4 9 SQUARE (2X2) + (3X3)> # SOLID 8 27 CUBE (2X2X2) + (3X3X3)very lovely. the tetraktys is usually: 1> 2 3> 4 5 6> 7 8 9 10and so while I can see the resemblance, I'm not sure > I understand your point here yet. Seyfert-1> lymojo.com@nagasiva> === > refs> Agrippa, H.C.: Three Books of Occult Philosophy,> Llewellyn Publications, tr. Freake,> ed. Tyson, 1997; ISBN 0-87542-832-0 Andrews, W.S.: Magic Squares and Cubes,> Dover Publications, 1960; ISBN 0-486-20658-0 === = === > END -- apologies for any duplications.Can someone provide me a stand-alone snippet that is ready to be embeddedMarcus === =Can someone provide me a stand-alone snippet that is ready to be embeddedMarcusThis should be pretty quick to put into your project. See kiss_fftr.cin the sample_code directory.http://sourceforge.net/projects/kissfft- Mark === =Can someone provide me a stand-alone snippet that is ready to be embeddedMarcusHave you tried Numerical Recipes in ...?Jerry-- Engineering is the art of making what you want from things you can get.[OSlash][OSlash][OSlash][OSlash] === =Can someone provide me a stand-alone snippet that is ready to be embeddedMarcus/*** FFT.C**** Fast Fourier Transform (FFT)**** Forward transform:**** X(n) = fft (x, w, n)**** Inverse transform:**** x(n) = 1/n * fft (X, wi, n)***/#include #de?e TWOPI 6.28318530718struct cmplx{ double x, y;};struct cmplx cconj ();struct cmplx cmult ();struct cmplx csub ();struct cmplx cadd ();/*** FFT initialization*/void fftinit (struct cmplx *w, struct cmplx *wi, int n){int i;double cos (), sin ();double factr, angle; factr = TWOPI / (?n; for (i = 0; i < n; i++) { angle = i * factr; w[i].x = cos (angle); w[i].y = -sin (angle); wi[i] = cconj (w[i]); } return;}/*** FFT function*/void fft (struct cmplx *x, struct cmplx *w, int n){int n1, logn, i, j, k, l, logl, p;struct cmplx s, t; logn = log2 (n); n1 = n > 1; j = logn - 1; k = 0; /* begin transform */ for (logl = 0; logl < logn; logl++) { do { for (i = 0; i < n1; i++) { p = bitr ((k > j), logn); l = k + n1; t = cmult (w[p], x[l]); x[l] = csub (x[k], t); x[k] = cadd (t, x[k]); k++; } k += n1; } while (k < n); k = 0; j--; n1 = n1 > 1; } for (i = 1; i < n; i++) { /* reorder */ k = bitr (i, logn); if (i > k) { /* exchange i,k elements */ s.x = x[i].x; s.y = x[i].y; x[i].x = x[k].x; x[i].y = x[k].y; x[k].x = s.x; x[k].y = s.y; } } return;}/*** complex conjugate function*/struct cmplx cconj (struct cmplx a){struct cmplx c; c.x = a.x; c.y = -a.y; return (c);}/*** complex multiplication function*/struct cmplx cmult (struct cmplx a, struct cmplx b){struct cmplx c; c.x = a.x * b.x - a.y * b.y; c.y = a.x * b.y + a.y * b.x; return (c);}/*** complex addition function*/struct cmplx cadd (struct cmplx a, struct cmplx b){struct cmplx c; c.x = a.x + b.x; c.y = a.y + b.y; return (c);}/*** complex subtraction function*/struct cmplx csub (struct cmplx a, struct cmplx b){struct cmplx c; c.x = a.x - b.x; c.y = a.y - b.y; return (c);}/*** base 2 logarithm function*/int log2 (int n){int i; i = -1; /* will return -1 if n <= 0 */ while (1) { if (n == 0) break; n = n > 1; i++; } return (i);}/*** bit reversal function*/int bitr (int k, int logn){int ans, j, i; ans = 0; j = k; for (i = 0; i < logn; i++) { ans = (ans << 1) + (j & 1); j = j > 1; } return (ans);} === Please clarify doesn't quite work; your equation appears to> match your problem statement. Not quite a match. The equation should be 100x = 2(2x + 100).> The answer is 100/99. Very skinny lot!I suggest that you attempt to _check_ your answer. (Your equation is> incorrect. Dan's equation was correct.)My apologies to the group. It was a computer glitch --what I meant to write! === =Can anyone direct me to a good source of downloadable lectures on numerical methods?Anwer === =Can anyone direct me to a good source of downloadable lectures on numerical ^^^^ beauty is in the eye of the beholder> methods?> Anwer http://www.phys.virginia.edu/classes/551.jvn.fall01/ 551Notes.htmIf I do say so.-- Julian V. NobleProfessor Emeritus of Physicsjvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ Science knows only one commandment: contribute to science. -- Bertolt Brecht, Galileo. === = Can anyone direct me to a good source of downloadable lectures on numerical> ^^^^> beauty is in the eye> of the beholder methods?> Anwer http://www.phys.virginia.edu/classes/551.jvn.fall01/ 551Notes.htm If I do say so. --> Julian V. Noble> Professor Emeritus of Physics> jvn@lessspamformother.virginia.edu> ^^^^^^^^^^^^^^^^^^> http://galileo.phys.virginia.edu/~jvn/ Science knows only one commandment: contribute to science.> -- Bertolt Brecht, Galileo.Prof. Noble,--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 === |> Hmm. Perhaps I can supply some ancient perspective. I was>|> taught the hand method in, IIRC, the 8th or 9th grade. It >|> could be done reasonably rapidly and I don't recall ever having>|> any trouble with it.>I was taught it a bit earlier, and had trouble, probably because I>couldn't work out why it worked, and am bad at learning by rote.>So I invented binary chop. A common reaction.I see no reason why even the present teachers cannotpresent why it works. >|> Why were we not taught any other methods? Because back then>|> in the dim recesses of time there were no other number systems.>|> No binary, no mixed radix, nothing but the occasional Roman>|> numeral. >Eh? None of those are relevant. Binary chop has nothing to do>with binary notation, and neither interpolation nor Newton-Raphson>are discrete methods in that sense.I agree. The successive digit method is well suited to anybase whatever, although it has advantages in binary. TheMonroe desk calculators multiplied the number for which thesquare root was extracted by .5, and used successive oddmultiples of .5 to subtract in the last position, to avoidcertain problems, but it was essentially the divisionalgorithm.>In any case, with UK-based teaching, we had been using mixed base>arithmetic for years. What is the cost of 3 tons, 7 hundredweight,>25 pounds, 6 ounces of something at 2 pounds, 11 shillings and>3 pence 3 farthings a stone. All right, it was usually a BIT>simpler :-)>|> Nobody knew Newton's method and there was no way it could have>|> been taught. Of course there were no computers, electronic>|> calculators, and the mechanical ones capable of taking >|> square roots were very very new and very very expensive.>|> I did not see one of them until I was a graduate student.>|> Slide rules were good only to three digits at best and the>|> common log tables only good to four at best.>Yer whaa? I thought that you said an ANCIENT perspective! What>on earth do you need any of those for? I was using Newton's>method for ages before they were allowed in mathematics teaching>or examinations.I do not know if the Greeks had this or not; they did have means of approximating the square root, includinggoing from one side to the other. They may have beenable to do it even before algebraic notation, as theydid have a good intuitive idea of limit, which our students, taught to manipulate, seem unable to manage.>|> The fun thing about the method was demonstrating algebraically>|> how it worked. If we could do that, we were considered top>|> students.>For that, you need algebra.You need algebraic notation, which belongs in ?st grade,more generally than for numbers. Algebra is trivial i? is considered linguistic, which it is. When people are talking about its>teaching in elementary school, most of us are referring to LONG>before we were taught algebra. If I recall, I was taught it>before I was taught about even the use of variables, though after>I had been started on non-trivial geometry.As did the Greeks.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === =I was taught it a bit earlier, and had trouble, probably because I>couldn't work out why it worked, and am bad at learning by rote.>So I invented binary chop. A common reaction.I see no reason why even the present teachers cannot>present why it works. I agree. But I was taught it before I was taught algebraic notation(a.k.a. the use of variables), and it is not easy to explain withoutthat. Binary chop and interpolation are.>I do not know if the Greeks had this or not; they did >have means of approximating the square root, including>going from one side to the other. They may have been>able to do it even before algebraic notation, as they>did have a good intuitive idea of limit, which our >students, taught to manipulate, seem unable to manage.Yes :-(>|> The fun thing about the method was demonstrating algebraically>|> how it worked. If we could do that, we were considered top>|> students.For that, you need algebra.You need algebraic notation, which belongs in ?st grade,>more generally than for numbers. Algebra is trivial if>it is considered linguistic, which it is.That is true, but I was referring to traditional UK-style teaching(and I went to some very backwards schools, so think 1930s), wherethe concept of a variable was introduced LONG after geometry. Yes,I do mean that I was introduced to the 9-point circle BEFORE I wastaught to use variables, as such.In fact, I was never taught how to perform arithmetic on variables,using only the axioms of arithmetic, which caused me hell when I wentto university and was dropped into group theory. It took 6 monthsfor that concept to ?click', and now I can't ' understand how I can'thave seen it. The course didn't teach it, ' as it assumed that allmathematics undergraduates had already been taught it.However, that is by the way. The reason that I don't think that thetraditional method of square root should be taught early is that itleads almost nowhere. Whereas the concepts of N-fold chop andinterpolation (including Newton-Raphson) do.Nick Maclaren. === =It connects the area under the curve of a function with its anti-derivative.You can read it at:www.precalculus.net?ms.com/#3TR