mm-350 === > You know, Euler did some mighty strange things with infinite series, > back then before convergence was understood. You can find > statements like > Sum(n=-infinity to infinity) x^n = 0 > actually stated by him. I find this hard to believe. Even I know that, if x > 0, _every_ term in the summation > 0, so how can the sum itself be equal to 0? And from what I hear Euler must have been smarter than I am :-) Steven === Subject: Re: Is ...9999.9999... = 0 ? > You know, Euler did some mighty strange things with infinite series, > back then before convergence was understood. You can find > statements like Sum(n=-infinity to infinity) x^n = 0 actually stated by him. I find this hard to believe. Even I know that, if x > 0, _every_ term in the > summation > 0, so how can the sum itself be equal to 0? > And from what I hear Euler must have been smarter than I am :-) Steven In a private response to the posting, it was pointed out that the formal identity 1) ...999.999...=0 is equivalent (at least formally) to the identity 2) ...111.111...=0 which can be viewed as a special case of an old formal identity of Euler 3) ...x^3+x^2+x+1+x^(-1)+x^(-2)+x^(-3)+... = 0 for x<>1 where x=10. Sometimes called Euler's ludicrous sum. Essentially Euler noted that the left hand side of the geometric sum 4) x/(1-x) = x+x^2+x^3+... is meaningful for x<>1 whether or not the right hand side converges, and that 3) can be interprited as the algebraicly obvious 5) (x/(1-x)) + 1 + ((1/x)/(1-(1/x)) =0 for x<>0, x<>1 In another context, this identity reappeawhen the Poisson kernel 6) Real Part of( 1/2 +z+z^2+z^2...)= (1-r^2)/(1-2r*cos(t)+r^2) where z=r*exp(t*i) and r<1 satisfies 7) - (1-r^2)/(1-2r*cos(t)+r^2) = (1-(1/r)^2)/(1-2(1/r)*cos(-t)+(1/r)^2) Now the symmetric partial sums of 3) never converge, but when restricted to the unit circle with x=exp(t*i) can be interprited as the Dirichlet kernel Dn(t) 8) x^n++..+x+1+x^(-1)+..+x^(-n) = 1+2cos(t)+2cos(2t)+..+2cos(nt)=sin((2n+1)*t/2)/sin(t/2) which when convoluted with nice functions on the unit circle gives the n-th partial fourier series for that function. Again, the Dirichlet kernels do not converge as n increases, but if on takes their average values on obtains the Fejer kernel 9) (1 + sin(3t/2)/sin(t/2)+ ...+ sin((2n+1)*t/2)/sin(t/2))/n = (sin(nt/2)/sin(t/2))^2 /n which clearly converges to zero, uniformly away from t=0. This can be interprited to mean that on the unit circle Eulers identity holds when convergence is considered in the more general summability sense. Also when using nice functions with quickly decaying fourier coeficients as test functions, the Dirichlet kernel converges in the distributional sense to the Dirac Delta, which is the distributional idenity of convolution, which at least intuitively is zero away from t=0. By considering dilations of the unit circle ( x=r*exp(it) ) the left hand side of 3) may be interprited as a well-defined temperate distribution, converging in the distribution sense, although any interpritation of the identity is problematic. Nor is it clear how to interprite the identity when we consider convergence on say the Riemann Sphere. Now some thoughts as to the original question as to whether the formal identity, ...999.999...=0 along with normal arithmetical relations reduces to the trivial algebra with 1=0. By putting in a reasonable metric, so that digits become small both to the right and left, convergence and continuity of addition should be possible. But continuity of multiplication, would seem to be impossible because small things from the right times small things from the left can result in large things in the middle. It is not even clear how to define multiplication,although in some cases it may help to break numbers apart and flip digits from one side to the other. (...123123.0) *(...111.333...) = (...123123.0)*(...1111.0 - ...3333.0) = - (...123123.0)*(...2222.0) Now as has been noted, if you formally work with the digits a being in the 10-adic ring, you can find zero divisors. It is clear from the original identity that there is no unique representation of whatever we are working. Even if everything does not reduce to zero, alot of things do, and it may be even impossible to decide if a given double string of digits reduces to zero. So it would not seem to be appropriate to conclude that 0=1 if we have a pair of zero divisors. Finally, it may very well be that our education system does not give our students a clear sense of the reality of mathematics so they are willing to continue using inapropriate operations even when they can be shown that such operations lead to contradictions. Even so, it would be a shame if our instructors are reduced to bemeaning and coercion, when students think outside the box. Should not learning math be an adventure, with all kinds of things to explore. === Subject: Re: Is ...9999.9999... = 0 ? > You know, Euler did some mighty strange things with infinite series, > back then before convergence was understood. You can find > statements like Sum(n=-infinity to infinity) x^n = 0 actually stated by him. I find this hard to believe. Even I know that, if x > 0, _every_ term in the > summation > 0, so how can the sum itself be equal to 0? > And from what I hear Euler must have been smarter than I am :-) He was. Since Quantum wankers had to wait until Cauchy came by to prove it's true. But still being the idiots they are, the morons still believe tan(x) is really a function. > Steven === Subject: 2D fisher's g test of spectral peak significance? Hi there, does anyone know if Fisher'g significance test for spectral peaks has ever been extended to examine 2 dimensional data? Or is there any other test for 2D cases? Also, is there another good newsgroup or mailing list to resort to for this kind of questions? Tons of thanks in advance! Barbara === Subject: Antidiagonal, Infinity And thus, circularly, the claim that the reals have a greater cardinality than the integers is the antidiagonal and its companion arguments. Virgil notes that a bijective mapping f: Q<->N is not a bojective mapping f: Q<->P(N). That may be true. It doesn't disprove the existence of other functions that do map the rationals to the powerset of the integers. Consider any bijective function from the rationals to the even numbers. None of them surjects the rationals to the integers. I can actually allow that, because it shows a problem with cardinality in general. We're talking about infinite sets here. What's the sum for integer i from zero to infinity of 1/2^x? Two plus two equals four. Let's talk about the nested sequences. Consider the argument applied to rationals only instead of reals. If the argument was true, and applied to rationals or sets uniformly dense in the reals, why would it not imply the same lack of mapping between the index (naturals) and rationals? That's simple enough, enumerate the rationals from zero to one, and construct an antidiagonal. Obviously enough, the binary case is sufficient, and there is dual representation of rationals. It is fair to say that not each real has dual representations, just all the rationals do. In terms of the nested sequences, all their endpoints are rational. Virgil agrees that an infinite set can map (one-to-one) to a set that is not its powerset, superset of its powerset, he might disagree but I would add isomorphic to its powerset or superset of its powerset. Virgil doesn't like to make mistakes. He says there isn't a surjection of the rationals over the powerset of the integers. The integers are a proper subset of the rationals. For each integer there are infinintely many distinct rationals. There are infinitely many proper subsets of the rationals that are proper supersets of the integethus that the set of rationals is larger than the set of integeby my definition. There can be a bijection between the rationals and the powerset of the integethe powerset of the integers isn't the powerset of the rationals, in fact it's a proper subset of it. He mentions they have equal cardinality, I just say they have different sizes, perhaps projective sizes. I may also argue about their cardinality or the validity of the term applied to number sets as opposed to sets non-coincident with the number system. Actually, I guess (deduce) projective sizing would apply more to the superset/subset relation without having infinite sub/super sets. That is, the non-negative integers and the positive integers projected onto the naturals by a scale and origin invariant identity function show zero as not being a positive integer, but all positive integers are non-negative integewith any proper superset being projectively larger than a subset. When the smaller set has infinitely many proper supersets that are proper subsets of the larger set, that is a form of extent comparison that compares a set to its subset. A superset that has infinitely many proper subsets that are proper supersets of another set is massively larger. What is a scale invariant function on a set? I've never even seen origin invariant. Others of these terms are vague or imprecise. Consider a function f(x) that maps the integers 1-1 to the rationals. Now, consider f(x)+Pi, or f(x)+e, or f(x) plus any other irrational number, it maps the integers 1-1 to the irrationals. If you think it doesn't, I'm surprised at this point. Do you accept the integers mapping 1-1 to the rationals? The sum of two rationals is always rational. Is not the sum of two irrationals always rational? Hmm, I don't know integers p and q thus that p/q = pi +- e. The sum of an irrational and a rational is always irrational. Name two irrational numbers and prove their sum is irrational, and prove that infinitely many pairs of irrationals each sum to an irrational, and/or that half of them do. That has to do with the almost continuous with two points in a row I just mentioned, yet I still think that rationals and irrationals alternate on the continuous real number line of points.. We can compare and contrast a wide variety of functions that generate pi and e. ln (+-i) = +- pi/2 i i = e^ (pi/2 i) ln(exp z) = z+2k pi i, k E Z pi = (ln(e) -1) / (2k i) = 0 / (2k i) That doesn't seem right. e is the limit as n goes to infinity of (1 + 1/n)^n. Here's something, cos(z) is 1-z^2/2! + z^4/4! - z^6/6! + z^8/8!! ..., cosh(z) is 1+z^2/2! + z^4/4! + z^6/6! + z^8/8! ..., the hyperbolic cosine cosh z also equals ( e^z + e^-z )/ 2, cos(z) = cosh(zi), cosh(pi i) = -1 = e^(pi i)/2 + 1/ (2 e^(pi i) ) = cos(-pi), cos(-pi)=cos(pi). cosh(z)- cos(z) = 2 (z^2/2! + z^4/4! + z^6/6! + ...) = 2 cosh(z) - 2 = 2( cosh(z) - 1). Hey, that one's not in this book I am reading definitions from, The Handbook of Mathematical Functions. cos(z) + cosh(z) = 2. sin(z) + sinh(z) = 2. I turn to Mathworld, http://mathworld.wolfram.com/Pi.html: It is not known if pi+e, pi/e, or ln(pi) are irrational. However, it is known that they cannot satisfy any polynomial equation of degree <=8 with integer coefficients of average size 10^9 (Bailey 1988, Borwein et al. 1989). Hmm..., cosh(1) = e/2 + 1/2e. cos(1) = 2- e/2 - 1/2e. cos(1) = 1 - 1/2! + 1/4! -1/6! + ..., cosh(1) = 1 + 1/2! + 1/4! +1/6! + ..., e = 1 + 1/1! + 1/2! + ..., cosh(1) - e = 1/1! + 1/3! + 1/5! + ... = sinh(1), cosh(1)-sinh(1) = e, etcetera. Anyways I'm trying to determine if pi +- e is rational, so I'm working out expressions of pi +- e, and/or e+- pi. Getting back to the antidiagonal argument, I restate eroding explanation and state not CH, not GCH. Give some ground explicitly, Virgil, let's figure it out. I can explain some of what you didn't say, please do. Ross === Subject: Re: Antidiagonal, Infinity at 03:00 PM, raf@tiki-lounge.com (Ross A. Finlayson) said: >And thus, circularly, Circularly means one thing to you and something quite different to me. >the claim that the reals have a greater >cardinality than the integers is the antidiagonal and its companion >arguments. There's no circularity here; move on. >Virgil notes that a bijective mapping f: Q<->N is not a bojective >mapping f: Q<->P(N). That may be true. It doesn't disprove the >existence of other functions that do map the rationals to the >powerset of the integers. No. but the diagonal construction does, quite nicely. >Consider any bijective function from the rationals >to the even numbers. None of them surjects the rationals to the >integers. Consider the color of the Moon; it's just as relevant. >I can actually allow that, because it shows a problem with >cardinality in general. No, just a problem with your preconceptions. >We're talking about infinite sets here. What's the sum for integer >i from zero to infinity of 1/2^x? There is no such sum. There is, however, a limit. >Let's talk about the nested sequences. What nested sequences? >Consider the argument applied to rationals only instead of reals. >If the argument was true, and applied to rationals or sets uniformly >dense in the reals, Are you under the impression that sets uniformly dense in the reals has any meaning? Did you mean to say countable dense set? If not, explain what uniformly dense means to you. >why would it not imply the same lack of mapping >between the index (naturals) and rationals? Because the result of the diagonal construction applied to the rational numbers would not be s rastionsl number. >That's simple enough, But too complex for you, since you got it wrong. >It is fair to say that not each real has dual representations, just >all the rationals do. Not all the rationals do either. >In terms of the nested sequences, all their endpoints are rational. What nested sequences? And what is the endpoint of a sequence? >Virgil agrees that an infinite set can map (one-to-one) to a set >that is not its powerset, superset of its powerset, he might >disagree but I would add isomorphic to its powerset or superset of >its powerset. In order to agree or disagree your Virgil would have to understand what your statement meant. What doe you imagine isomorphic to its powerset to mean? >Virgil doesn't like to make mistakes. Which is why he lets you make them. >He says there isn't a surjection of the rationals over the powerset >of the integers. When did he say that? For that matter, what do you believe that it means? Perhaps he said something quite different, that there isn't a surjection of the rationals unto the powerset of the integers? The latter statement, of course, is true. >The integers are a proper subset of the rationals. For each integer >there are infinintely many distinct rationals. Indeed, and for each rational there are infinitely many distinct integers. >There are infinitely many proper subsets of the rationals that are proper >supersets of the integers, Nu, so what else is new? >thus that the set of rationals is larger >than the set of integeby my definition. What is your definition? What are its properties? What advantages does it have over the definition used b the rest of the world? And what relevance does it have to a theorem that does *not* use your definition? >There can be a bijection >between the rationals and the powerset of the integers, Please prove that proposition. You can't of course, because it is false. >the powerset of the integers isn't the powerset of the rationals, >in fact it's a proper subset of it. Indeed, a proper subset with the same cardinality. >He mentions they have equal cardinality, I just say they have >different sizes, How do you define size? >perhaps projective sizes. What are projective sizes? >I may also argue about >their cardinality or the validity of the term applied to number sets >as opposed to sets non-coincident with the number system. I may also argue that the dozen in a dozen apples is a different number from the dozen in a dozen eggplants, but doing so would be hazardous to my reputation among the educated. >Actually, I guess (deduce) projective sizing would apply more to the >superset/subset relation without having infinite sub/super sets. >That is, the non-negative integers and the positive integers >projected onto the naturals by a scale and origin invariant identity >function Wow! That's an impressive buzzword generator. Too bad that there's no actual meaning in there. >projectively larger Another meaningless phrase. >What is a scale invariant function on a set? I've never even seen >origin invariant. Others of these terms are vague or imprecise. We never would have guessed. >Consider a function f(x) that maps the integers 1-1 to the >rationals. Now, consider f(x)+Pi, or f(x)+e, or f(x) plus any other >irrational number, it maps the integers 1-1 to the irrationals. Nu, if f(x) is rationa, for what value of x is f(x)+Pi equal to 2Pi? >If you think it doesn't, I'm surprised at this point. Life is full of surprises. >Do you accept the integers mapping 1-1 to the rationals? How is that relevant? >Is not the sum of two irrationals always rational? Hint: What is Pi+Pi? >and/or that half of them do What do you mean by half of them? >That has to do with the almost continuous with >two points in a row I just mentioned, No, you didn't just mention them. Mind telling us what almost continuous and two points in a row are supposed to mean? >yet I still think that rationals and irrationals alternate on the >continuous real number line of points.. Perhaps you mean something different from the rest of the world when you write continuous. If x is rational and x' is the adjacent irrational, what is (x+x')/2? >pi = (ln(e) -1) / (2k i) = 0 / (2k i) That doesn't seem right. Bingo! >Give some ground explicitly, Virgil, That would require that he lie. >let's figure it out. He's not the one that needs to figure it out. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Antidiagonal, Infinity > at 03:00 PM, raf@tiki-lounge.com (Ross A. Finlayson) said: And thus, circularly, Circularly means one thing to you and something quite different to me. the claim that the reals have a greater >cardinality than the integers is the antidiagonal and its companion >arguments. There's no circularity here; move on. > We should use the words with similar or at least consistent meaning. What did you have in mind? >Virgil notes that a bijective mapping f: Q<->N is not a bijective >mapping f: Q<->P(N). That may be true. It doesn't disprove the >existence of other functions that do map the rationals to the >powerset of the integers. No. but the diagonal construction does, quite nicely. > Please elaborate. >Consider any bijective function from the rationals >to the even numbers. None of them surjects the rationals to the >integers. Consider the color of the Moon; it's just as relevant. > No, the wavelength of the reflected light of the moon is irrelevant. What's relevant is that any bijection between A and B is not a bijection between A and C > B. >I can actually allow that, because it shows a problem with >cardinality in general. No, just a problem with your preconceptions. We're talking about infinite sets here. What's the sum for integer >i from zero to infinity of 1/2^x? There is no such sum. There is, however, a limit. > That's arguable. Generally the form of the summation of an infinite series is without the limit notion. It's equal to two, if you care to avoid eternal Zeno stall. >Let's talk about the nested sequences. What nested sequences? > That's about the other proof of the uncountability of the reals. >Consider the argument applied to rationals only instead of reals. >If the argument was true, and applied to rationals or sets uniformly >dense in the reals, Are you under the impression that sets uniformly dense in the reals > has any meaning? Did you mean to say countable dense set? If not, > explain what uniformly dense means to you. > Think about density in the naturals. Uniformly dense is a comparison of less uniformly dense, that is about another set that is still dense but varies in its density. Consider for example... the even integers vis-a-vis the even integers setminus the primes union prime number plus ten. >why would it not imply the same lack of mapping >between the index (naturals) and rationals? Because the result of the diagonal construction applied to the > rational numbers would not be s rastionsl number. That's simple enough, But too complex for you, since you got it wrong. It is fair to say that not each real has dual representations, just >all the rationals do. Not all the rationals do either. That's true in context, although they do in base b where for rational p/q that b is factor of q. In context, all the rationals p/q with q even do. >In terms of the nested sequences, all their endpoints are rational. What nested sequences? And what is the endpoint of a sequence? Virgil agrees that an infinite set can map (one-to-one) to a set >that is not its powerset, superset of its powerset, he might >disagree but I would add isomorphic to its powerset or superset of >its powerset. In order to agree or disagree your Virgil would have to understand > what your statement meant. What doe you imagine isomorphic to its > powerset to mean? > It means translated only, and then permuted among its ceiling set. >Virgil doesn't like to make mistakes. Which is why he lets you make them. He says there isn't a surjection of the rationals over the powerset >of the integers. When did he say that? For that matter, what do you believe that it > means? Perhaps he said something quite different, that there isn't a > surjection of the rationals unto the powerset of the integers? The > latter statement, of course, is true. > Onto, over, to, each is equivalent. >The integers are a proper subset of the rationals. For each integer >there are infinintely many distinct rationals. Indeed, and for each rational there are infinitely many distinct > integers. > How many integers are in Q[0,1]? How many rationals? How about Q[0,2]? >There are infinitely many proper subsets of the rationals that are proper >supersets of the integers, Nu, so what else is new? > Hi Nu. >thus that the set of rationals is larger >than the set of integeby my definition. What is your definition? What are its properties? What advantages does > it have over the definition used b the rest of the world? And what > relevance does it have to a theorem that does *not* use your > definition? > That's a good question. It might have some combinatoric utility. The other definition is more or less the same thing as proper super-/sub-set. >There can be a bijection >between the rationals and the powerset of the integers, Please prove that proposition. You can't of course, because it is > false. > Prove that it's false. >the powerset of the integers isn't the powerset of the rationals, >in fact it's a proper subset of it. Indeed, a proper subset with the same cardinality. He mentions they have equal cardinality, I just say they have >different sizes, How do you define size? perhaps projective sizes. What are projective sizes? I may also argue about >their cardinality or the validity of the term applied to number sets >as opposed to sets non-coincident with the number system. I may also argue that the dozen in a dozen apples is a different > number from the dozen in a dozen eggplants, but doing so would be > hazardous to my reputation among the educated. > Sum the area under the curve. Can you cover it all with one can of paint? >Actually, I guess (deduce) projective sizing would apply more to the >superset/subset relation without having infinite sub/super sets. >That is, the non-negative integers and the positive integers >projected onto the naturals by a scale and origin invariant identity >function Wow! That's an impressive buzzword generator. Too bad that there's no > actual meaning in there. projectively larger Another meaningless phrase. What is a scale invariant function on a set? I've never even seen >origin invariant. Others of these terms are vague or imprecise. We never would have guessed. Consider a function f(x) that maps the integers 1-1 to the >rationals. Now, consider f(x)+Pi, or f(x)+e, or f(x) plus any other >irrational number, it maps the integers 1-1 to the irrationals. Nu, if f(x) is rationa, for what value of x is f(x)+Pi equal to 2Pi? If you think it doesn't, I'm surprised at this point. Life is full of surprises. Do you accept the integers mapping 1-1 to the rationals? How is that relevant? Is not the sum of two irrationals always rational? Hint: What is Pi+Pi? and/or that half of them do What do you mean by half of them? That has to do with the almost continuous with >two points in a row I just mentioned, No, you didn't just mention them. Mind telling us what almost > continuous and two points in a row are supposed to mean? yet I still think that rationals and irrationals alternate on the >continuous real number line of points.. Perhaps you mean something different from the rest of the world when > you write continuous. If x is rational and x' is the adjacent > irrational, what is (x+x')/2? pi = (ln(e) -1) / (2k i) = 0 / (2k i) That doesn't seem right. Bingo! Give some ground explicitly, Virgil, That would require that he lie. let's figure it out. He's not the one that needs to figure it out. Leave Virgil out of this. Ross === Subject: Re: Antidiagonal, Infinity Virgil notes that a bijective mapping f: Q<->N is not a bojective > mapping f: Q<->P(N). That may be true. It is true. > It doesn't disprove the > existence of other functions that do map the rationals to the powerset > of the integers. There are many maps from Q to P(N), but there are no maps from Q *onto* P(N). > That's simple enough, enumerate the rationals from zero to one, and > construct an antidiagonal. Which isn't in Q. He says there isn't a surjection of the rationals over the powerset of > the integers. No idea what over means. I say there is no surjection from Q to P(N). (The reason why I say it is that there is no surjection from Q to P(N).) > The integers are a proper subset of the rationals. For > each integer there are infinintely many distinct rationals. There are > infinitely many proper subsets of the rationals that are proper > supersets of the integethus that the set of rationals is larger > than the set of integeby my definition. Wow! you have definitions! > There can be a bijection > between the rationals and the powerset of the integers, No, there can't. > Consider a function f(x) that maps the integers 1-1 to the rationals. > Now, consider f(x)+Pi, or f(x)+e, or f(x) plus any other irrational > number, it maps the integers 1-1 to the irrationals. If you think it > doesn't, I'm surprised at this point. Do you accept the integers > mapping 1-1 to the rationals? Well done! Yes, there are injections from N to R-Q! You're doing really well here! > The sum of two rationals is always rational. Is not the sum of two > irrationals always rational? I hop this is a rhetorical question. > Hmm, I don't know integers p and q thus > that p/q = pi +- e. I don't think anyone else does either. > The sum of an irrational and a rational is always > irrational. Name two irrational numbers and prove their sum is > irrational, and prove that infinitely many pairs of irrationals each > sum to an irrational, and/or that half of them do. Half? Hmmm: you have some sort of finite measure on the pairs of irrationals? > That has to do > with the almost continuous with two points in a row I just > mentioned, yet I still think that rationals and irrationals alternate > on the continuous real number line of points.. Ah yes. They go rational, irrational, rational, irrational, ... So there's two rationals, let's call them a and b with a Here's something, cos(z) is 1-z^2/2! + z^4/4! - z^6/6! + z^8/8!! ..., > cosh(z) is 1+z^2/2! + z^4/4! + z^6/6! + z^8/8! ..., the hyperbolic > cosine cosh z also equals ( e^z + e^-z )/ 2, cos(z) = cosh(zi), > cosh(pi i) = -1 = e^(pi i)/2 + 1/ (2 e^(pi i) ) = cos(-pi), > cos(-pi)=cos(pi). cosh(z)- cos(z) = 2 (z^2/2! + z^4/4! + z^6/6! + ...) = 2 cosh(z) - 2 = > 2( cosh(z) - 1). Hey, that one's not in this book I am reading > definitions from, The Handbook of Mathematical Functions. cos(z) + > cosh(z) = 2. sin(z) + sinh(z) = 2. Try again! I get cosh(z) - cos(z) = 2(z^2/2! + z^6/6! + z^10/10! + ...) > Hmm..., cosh(1) = e/2 + 1/2e. cos(1) = 2- e/2 - 1/2e. My pocket calculator sez: cos(1) = 0.540302305 2 - e/2 - 1/(2e) = 0.456919365 -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Antidiagonal, Infinity >The sum of two rationals is always rational. Is not the sum of two >>irrationals always rational? I hop this is a rhetorical question. 1-pi and 1+pi are both irrational, but their sum is rational. Proof by example? Mitch === Subject: Re: Antidiagonal, Infinity >The sum of two rationals is always rational. Is not the sum of two >irrationals always rational? >1-pi and 1+pi are both irrational, but their sum is rational. >Proof by example? He asked if it was *always* rational. But he probably meant what you answered. -- Richard -- Spam filter: to mail me from a .com/.net site, put my surname in the headers. FreeBSD rules! === Subject: Re: Antidiagonal, Infinity The sum of two rationals is always rational. Is not the sum of two >irrationals always rational? 1-pi and 1+pi are both irrational, but their sum is rational. Proof by example? He asked if it was *always* rational. But he probably meant what > you answered. -- Richard Let's talk about mapping, bijectively, the rationals Q[0,1], the rationals in the closed interval [0, 1], to P(N), the set of all subsets of integers N[0, oo). Consider the rationals being each a binary sequence. Map each singleton of P(N) from an element of the rationals with one on bit. .10000... -> {} .01000... -> {0} .00100... -> {1} .00010... -> {2} ... Now, map each composite element of P(N) with two elements from an element of the rationals there with two on bits. .110000... -> { 0 ,1} .101000... -> { 0 ,2} .100100... -> { 0 ,3} .100010... -> { 0 ,4} ... .011000... -> { 1, 2} ... Continue, mapping each subset of N with x many elements from an element of the rationals with x on bits. You might say, well, that would require irrational as well as rational sequences. Prove it. Robin, when you say there is not a surjection from Q to P(N), the only reason I can assume that you say that is that there _is_ a surjection from N to Q, if you have any other reasoning to explain that, please put it forward. I want to be sure that you actually agree that there is a mapping from N to R-Q=P, sarcasm is not your usual style. I don't know, from adding any irrational constant to a mapping f:N <-> Q, that depends on the fact that the sum of any two irrationals is always a rational, in order to show a mapping N->P by example, which is not so. Do you know any proofs that the sum of any two irrationals is irrational? I consider the case where that might be so, and the case where it is not. That's where the sums of many pairs of irrational numbers are not obviously rational. I'm wondering if half the irrationals are algebraic, solutions to polynomial equations with integer coefficients, and the other half transcendental, not, and whether there are equal numbers of each as there are of rationals. About the alternating numbethey're indefinite, this is similar to the investigation of the smallest non-zero real number, and whether and how it is an infinitesimal, etcetera. Here's something I was considering, it's similar in consideration of the epsilon term where the difference is always less than that, consider the oppposite case where the difference is always greater. The rationals and irrationals are each dense in the reals, the reals are continuous. Select a rational a and a rational b, their average c=(a+b)/2 is rational because 2 is rational and the rationals are dense in the reals. Select an irrational a and an irrational b, a Consider the rationals being each a binary sequence. Map each > singleton of P(N) from an element of the rationals with one on bit. > .10000... -> {} > .01000... -> {0} > .00100... -> {1} ... > Now, map each composite element of P(N) with two elements from an > element of the rationals there with two on bits. > .110000... -> { 0 ,1} > .101000... -> { 0 ,2} ... > .011000... -> { 1, 2} > ... > Continue, mapping each subset of N with x many elements from an > element of the rationals with x on bits. You need not even x many elements. You may map *all* rationals in this way to an element of P(N). No problem. So you have a mapping from the rationals to a subset of P(N). You will *never* find a rational that will map to a some specific element of P(N) this way: the set of primes for instance. The problem is not the mapping in this direction, it is the mapping in the other direction, and you need both to find a bijection. So you have to describe the reverse mapping. How do you map a specific element of P(N) to a rational? > Robin, when you say there is not a surjection from Q to P(N), the only > reason I can assume that you say that is that there _is_ a surjection > from N to Q, if you have any other reasoning to explain that, please > put it forward. No. The only reason is that there is no injection from P(N) to Q. And yes, there is a surjection from N to Q; there is even a bijection. To clarify: an injection is a mapping such that images of different elements are different. A surjection is a mapping such that every element of the destination is the image of something. A bijection is both. > I want to be sure that you actually agree that there is a mapping from > N to R-Q=P, sarcasm is not your usual style. There is. f(x) = pi is one such mapping (but it is neither an injection nor a surjection). f(x) = pi+x is also one (and this time we have an injection, but still no surjection). > Do you know any proofs that the sum of any two irrationals is > irrational? Of course not because it is false. > I consider the case where that might be so, and the case > where it is not. That's where the sums of many pairs of irrational > numbers are not obviously rational. Even if it is not obvious, it might be rational. Is e + gamma rational? > I'm wondering if half the irrationals are algebraic, solutions to > polynomial equations with integer coefficients, and the other half > transcendental, not, and whether there are equal numbers of each as > there are of rationals. You are attempting to quantify within infinite sets. You cannot do it this way. What do you mean with one half of an infinite set? Is one half of the integers prime? Intuitively: no. Nevertheless there is a bijection from the primes to the non-primes. And to reassure you: the algebraic numbers are countable, the transcendental are not. > The rationals and irrationals are each dense in the reals, the reals > are continuous. Select a rational a and a rational b, their average > c=(a+b)/2 is rational because 2 is rational and the rationals are > dense in the reals. No, your last reason is wrong and superfluous. > Select an irrational a and an irrational b, a what can be said of their average c besides that a Do two algebraic irrationals obviously sum to an algebraic number? Yup, see any book on algebraic number theory or somesuch. But it is not especially obvious. > I think two algebraic numbers always sum to an algebraic number, and > an algebraic and a transcendental number always sum to a > transcendental number, and two transcendental numbers sum to either a > transcendental or an algebraic number. Entirely right. > An infinite sum of rationals may be irrational, and an infinite sum of > irrationals may be rational. Formally you have to define what you mean by infinite sum, but this is essentially correct. > About the measure of pairs of irrationals, what do you think it is? > That is, what is the relative density of algebraic irrationals and > transcendentals among the reals and irrationals? Is it one, each > comprising half of the irrationals? You are trying to quantify again. What do you *mean* with one half of an innfinite set? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Antidiagonal, Infinity Dik, I'm glad you address this. Consider the infinite sequence, of length n, a representative of a k-subset of N, that is, k many elements of N, an n-set. This was discussed here recently. Each element n of the subset of N is represented by an on bit at the n'th index. I see what you mean about the example of the primes. If each bit of the sequence is flipped where the index is prime, the resultant sequence is not representative of a rational number. The number of on bits is that of the density of the primes times N. There are rationals with that many on bits, but they would be representing other subsets of N. There are many rationals that don't simply have repeating zero terminating binary representations, along with the dual representations. They have repeating sequence binary representations, with no dual representation. This would lead to a conflict in mapping .100... to {0} and .0111... to N{0}, N setminus {0}, as .100... = .0111.... That's not a conflict in disqualifying the antidiagonal argument via dual representation, the list is simply ordered thus that the antidiagonal is a rational, binary is sufficient, the listed element is counterexample, etc. That's a good point about sets with infinitely many elements yet with infinitesimal density in the naturals such as the primes, N setminus the primes, or the powers of two. My cognitive failure was upon mapping each finite subset and many infinite sets with finite, positive density and stopping. For n many sequence elements, there are 2^n permutations, I''m wondering about a variety of things. Consider for a second the sequences as reals instead of rationals, this is about 2^n and dual representation instead of countable unions of countable sets, going back to canonical sequences, about how less than all of the permutations would be unique. Dang, now I am back to constructing an injection from P(N) to Q besides showing that Q > N <=> QN =/= {} & NQ = {}, Q is a proper superset of N, thus that the diagonal element of f:N <-> P(N) exists in the disjointness of QN, and claiming that infinite sets are equivalent, actually constructing one. So yes, that's not a bijection from the unit interval's rationals to the powerset of the naturals. I think you mean Cantor-Bernstein theorem, ie, the existence of an injection from P(N) to Q and an injection from Q to P(N) implies a bijection between Q and P(N). So yes, you're right about that, Dik. Yet, where I haven't seen an example injection from P(N) to Q, also I have not seen its disproof, as P(N) is not P(Q), P(Q) > P(N). I have an idea, it involves another unit interval of the rationals. I was thinking it wouldn't work to add more unit intervals of rationals because the subsets that need to be mapped that aren't yet are each infinite. It still would be good to have another unit interval there so that all the dual representations would have twice as many possibilities. The problem again is that something like the primes as a k-subset binary sequence canonicalizes to an irrational number, as the number of zeros between ones at prime valued indexes diverges. Thus I don't immediately see how the cartesian product Q[0,1] x Q[0,1] x ... x Q[0,1] would ever have enough space to map the infinite subsets of the naturals infinitesimally dense in the naturals, or infinitesimally less than completely dense in the naturals, in terms of nonstandard asymptotic density where the density of the naturals in the naturals is equal to one. It might be able to get a good start going on mapping each infinite subset of the naturals with real density in the naturals. Of note is that the cartesian product of two sets is each ordered pair of any one of the elements from each set, for two n-sets the number of elements of the outer product is n times n, not n plus n, that is Q[0,1] x Q[0,1] has as many unit intervals of Q indexed as there are elements of the unit interval of rationals, n^n is somewhat larger than 2^(n-1), for n = 0, 1, 2, .... That's about how the even numbers map to .101(01)..., the odds to .010(10)..., and similar discussions about the density of on bits in the sequences, their canonicalizations, permutations to a canonical form, and the number of their possible values. About half, in terms of infinitely many things, here's one way to look at it: for any sufficiently large finite subset: half. Ross === Subject: Re: Antidiagonal, Infinity [...] > Let's talk about mapping, bijectively, the rationals Q[0,1], the > rationals in the closed interval [0, 1], to P(N), the set of all > subsets of integers N[0, oo). > Consider the rationals being each a binary sequence. Map each > singleton of P(N) from an element of the rationals with one on bit. > .10000... -> {} > .01000... -> {0} > .00100... -> {1} > .00010... -> {2} > ... > Now, map each composite element of P(N) with two elements from an > element of the rationals there with two on bits. > .110000... -> { 0 ,1} > .101000... -> { 0 ,2} > .100100... -> { 0 ,3} > .100010... -> { 0 ,4} > ... > .011000... -> { 1, 2} > ... > Continue, mapping each subset of N with x many elements from an > element of the rationals with x on bits. > You might say, well, that would require irrational as well as > rational sequences. Prove it. Do you agree that 1/{pi} is in [0,1]? Do you agree that 1/{pi} is irrational? Do you agree that its binary expansion is of the form you gave above? Do you agree that no rational number has the same binary expansion as 1/{pi}? Do you agree that no rational number maps to the same set as 1{pi}? Do you agree that you need irrational sequences to make your mapping onto? === Subject: Re: Antidiagonal, Infinity [...] Let's talk about mapping, bijectively, the rationals Q[0,1], the > rationals in the closed interval [0, 1], to P(N), the set of all > subsets of integers N[0, oo). Consider the rationals being each a binary sequence. Map each > singleton of P(N) from an element of the rationals with one on bit. .10000... -> {} > .01000... -> {0} > .00100... -> {1} > .00010... -> {2} > ... Now, map each composite element of P(N) with two elements from an > element of the rationals there with two on bits. .110000... -> { 0 ,1} > .101000... -> { 0 ,2} > .100100... -> { 0 ,3} > .100010... -> { 0 ,4} > ... > .011000... -> { 1, 2} > ... Continue, mapping each subset of N with x many elements from an > element of the rationals with x on bits. You might say, well, that would require irrational as well as > rational sequences. Prove it. Do you agree that 1/{pi} is in [0,1]? Do you agree that 1/{pi} is irrational? Do you agree that its binary expansion is of the form you gave above? Do you agree that no rational number has the same binary expansion as 1/{pi}? Do you agree that no rational number maps to the same set as 1{pi}? Do you agree that you need irrational sequences to make your mapping onto? Yes and no, that is five yesses and one no. Ross === Subject: Re: Antidiagonal, Infinity Let's talk about mapping, bijectively, the rationals Q[0,1], the > rationals in the closed interval [0, 1], to P(N), the set of all > subsets of integers N[0, oo). Let's talk about the moonm being of green cheese. Robin, when you say there is not a surjection from Q to P(N), the only > reason I can assume that you say that is that there _is_ a surjection > from N to Q, if you have any other reasoning to explain that, please > put it forward. There are bijections between N and Q. > I want to be sure that you actually agree that there is a mapping from > N to R-Q=P, sarcasm is not your usual style. I don't know, from adding > any irrational constant to a mapping f:N <-> Q, that depends on the > fact that the sum of any two irrationals is always a rational, in > order to show a mapping N->P by example, which is not so. Do you know any proofs that the sum of any two irrationals is > irrational? I know that thera are no such proofs. I'm wondering if half the irrationals are algebraic, solutions to > polynomial equations with integer coefficients, and the other half > transcendental, not, and whether there are equal numbers of each as > there are of rationals. The set of algeraic numbers is countable. The set of transcendental numbers is uncountable. > About the alternating numbethey're indefinite, this is similar to > the investigation of the smallest non-zero real number, and whether > and how it is an infinitesimal, etcetera. Never come across alternating numbers before :-( > The rationals and irrationals are each dense in the reals, the reals > are continuous. continuous? I know what continuous functions are, but not what continuous sets are. > Select a rational a and a rational b, their average > c=(a+b)/2 is rational because 2 is rational You can stop here if you like > and the rationals are > dense in the reals. But you didn't and now you've spolit it :-( > Select an irrational a and an irrational b, a what can be said of their average c besides that a The rationals are not continuous. Ah there! countinuous sets again :-( About the sum of irrationals being irrational, I guess there's > something like sqrt(2) + sqrt(2) = 2 sqrt(2), which is irrational. > That is not the sum of two different irrationals, it is however the > sum of two numbers that are each irrational, where sqrt(2) + 2 sqrt(2) > = 3 sqrt(2), obviously two irrationals do not necessarily sum to a > rational, and two irrationals do not necessarily sum to an irrational. You're doing well here. Keep it up! > Do two algebraic irrationals obviously sum to an algebraic number? Maybe not obviously. > Is > the sum of a transcendental (irrational) number and another either > transcendental or algebraic? No. > Is the sum of an algebraic and a > transcendental number always transcendental? Yes. > The sum of a rational and an irrational number is always irrational, Yes. > the sum of two rational numbers is always rational. Yes. > I think two algebraic numbers always sum to an algebraic number, and > an algebraic and a transcendental number always sum to a > transcendental number, and two transcendental numbers sum to either a > transcendental or an algebraic number. You are doing really well here. > An infinite sum of rationals may be irrational, and an infinite sum of > irrationals may be rational. Yup. > About the measure of pairs of irrationals, what do you think it is? No, What do *you* think it is: you said half of them had a certain property; I didn't. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colousounds and shapes. The League of Gentlemen === Subject: Re: Antidiagonal, Infinity There are bijections between N and Q. My favorite (er.. the only one I know of) is described most accessibly in: http://www.math.upenn.edu/~wilf/website/recounting.pdf (the Stern-Brocot-Farey gcd tree mapping Z+ to Q+ using the binary representation of the int as a path in the tree). So now I'm even more curious. What is -another- bijection? (er, one that isn't a trivial modification). Mitch === Subject: Re: Antidiagonal, Infinity Hi Robin, Steve, infinite sets are equivalent. Denoting a rational as a q, a real algebraic irrational as an a, and a real transcendental as a t: q + q = q (Sum of two rationals is rational.) a + q = a (Sum of two algebraic numbers is algebraic, sum of rational and irrational is irrational.) a + a = a or q (Sum of two algebraic numbers is algebraic.) q + t = t (Sum of algebraic and transcendental number is transcendental.) a + t = t (Sum of algebraic and transcendental number is transcendental.) t + t = q, a, or t (Sum of two transcendental numbers is algebraic or transcendental.) Consider only adding a rational to a rational, algebraic number, or transcendental and its result: q + q = q (Sum of two rationals is rational.) q + a = a (Sum of rational and algebraic irrational is an algebraic irrational.) q + t = t (Sum of rational and transcendental is transcendental.) Those do not imply their converse, eg q = q+q or a+a or t+t, a = a+a or a+q or t+t, t = t+t or a+t or q+t. Yet, each algebraic irrational is the sum of any non-zero algebraic number and some different algebraic irrational number. Each transcendental number is the sum of a different transcendental number and some algebraic number. What I want to assume is a bijective mapping between the algebraic and transcendental numbebut all I can see is the autojection from T to T. Anyways, this is in the context of a function f: Q <-> N, a bijective mapping (function f) between the domain of rationals and range of naturals, and functions f + a and f+t, and whether f+a is a bijective mapping between the algebraic irrationals and the naturals and whether f+t is a bijective mapping between the transcendentals and the naturals. The problem here is not having shown that f(q): Q <-> AQ = q+a or that f(q): Q <-> T = q+t, A is the set of algebraic numbers and T is the set of real transcendentals. Correspondingly, there would be a function f: Q <-> 3N, a bijective mapping between the rationals and the non-negative integer multiples of three, f+a: AQ <-> 3N+1, a bijective mapping between the algebraic irrationals and the non-negative integers congruent to 1 modulo 3, and f+t: T <-> 3n+2, a bijective mapping between the transcendentals and the non-negative integers congruent to 2 modulo 3, leading to the piecewise function f: Q U A U T <-> N, a bijective mapping from the naturals to the union of the rationals, algebraic irrationals, and transcendental irrationals. I can't claim that there is that, although it is readily theorized. That would require that for any algebraic irrational that there was another the difference of which was a rational, and that for any transcendental that there was another transcendental the difference of them being rational. The difference of any number and itself is zero. I'm getting into things like wondering if 2 + sqrt(2) is algebraic. If it were then it would be a root of a polynomial, with finitely many monomial summands, with integer coefficients. This is about wondering whether the sum of two algebraic numbers is always an algebraic number, and that the sum of an algebraic rational and an algebraic irational is always an algebraic irrational, I think it is. The polynomial x^2 -4x for x = 2+ sqrt(2) is 4+ 4sqrt(2) + 2 - 8 -4sqrt(2) = -2, an integer, 2+sqrt(2) is algebraic, the sum of two algebraic numbers is algebraic. A problem here is that a and t are not fixed. I don't know that t_1 +- t_2 = q, that any transcendental plus or minus any other is rational, or even algebraic. The piecewise function described above otherwise has infinitely many condition categories. This is where there does exist for each a_1 an a_2 thus that a_1 - a_2 = q, and for each t_1 a t_2 thus that t_1 - t_2 = q, because q+a = a and q+t = t. The problem is that not for all t_1 and t_2 or a_1 and a_2 that t_1-t_2 = q and a_1-a_2=q, instead t_1-t_2 =q, a, or t and a_1-a_2 = q or a. So I wonder if there can be finitely many categories, as opposed to a piecewise function with infinitely many categories over the domain, to prove the existence of a bijection from the naturals to the reals. An irrational is either algebraic or transcendental, a real is either rational or irrational. Where x is a rational, algebraic irrational, or real transcendental irrational, -x is also. Two (real) transcendentals have a real sum that is either algebraic or transcendental. For any transcendental there does exist a transcendental where their difference is algebraic, trivially zero. Is it so that for any transcendental that there is a different transcendental such that their difference is algebraic and non-zero? This is perhaps a question about zero-divisors and transcendentals. For any transcendental t, a+-t is transcendental, and t -+ (a +- t) is algebraic a, but not t +- (a +- t) = a, that is, tn +- a = t, for integer n. In assuming that there is easily an injection into the reals the range of which is a proper superset of the algebraic numbeI'm still left to determine a way to map the transcendentals to the naturals. For any t_1, there does exist another transcendental t_2 thus that t_1 - t_2 = a, an algebraic number. What I need to show is that for all transcendental numbers t_x that there exists any transcendental t_y such that t_x - t_y = a_x such that a_i =/= a_j for i, j E x, i=/= j. We know real transcendentals exist because a variety of real numbers are not algebraic, where e was proven/proved transcendental early in the nineteenth century, and summary constructions were shown to be non-algebraic before that, and that was known before Cantor showed the algebraic numbers to be countable. I'm reminded of the canonicalization of the binary sequences, where there are two types of binary sequences: those rationals and irrationals that canonicalize to a rational binary sequence and those irrationals that do not. I'm wondering about algebraic and transcendental irrationals vis-a-vis those. So I think the rationals and irrationals alternate qpqpqpq... but now I see some things that make me wonder if their order is qtaqatqtaqatqta.... Ross === Subject: Re: Antidiagonal, Infinity > So I think the rationals and irrationals alternate qpqpqpq... but now > I see some things that make me wonder if their order is > qtaqatqtaqatqta.... I don't think the density properties of reals and rationals allows you to say that they alternate. Yes, between any two reals there is a rational (consider their decimal expansion, they must differ at some digit) and between any two rationals there is an irrational (at the first digit they differ, extend in a known irrational way (sum 10^(-2^k)). These two facts don't imply that they alternate; the construction can be extended to show that there are a countable # of rationals between any two irrationals, but an uncountable # of irrationals between any two rationals. Of course this presupposes deductions which by your questioning you do not already trust. Mitch === Subject: Re: Antidiagonal, Infinity > What I > want to assume is a bijective mapping between the algebraic and > transcendental numbers, There isn't one. > I'm getting into things like wondering if 2 + sqrt(2) is algebraic. You really shouldn't wonder about this for very long. > So I think the rationals and irrationals alternate qpqpqpq... but now > I see some things that make me wonder if their order is > qtaqatqtaqatqta.... If one q is x and the next q is y, what is (x+y)/2? Is it the t that's between the two qs or is it the a? Or is it something else entirely? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: al-Kharki Can someone please explain in lay terms what al-Kharki's method is? TIA Larry === Subject: another lebesgue measure problem Let Esubset R be lebesgue measurable with measure |E| = 1. Show that there exists x in R such that |(E+x) cap E| = 1/2. Here is what I have done. Consider f(x) = |(E+x)cap E|. Clearly f(0) = 1. I have shown (not entirely trivial) that there exists an xin R such that f(x) < 1/2. So obviously my objective is to show that f is continuous. So far, no success. Any ideas? nojb. === Subject: Re: another lebesgue measure problem >Let Esubset R be lebesgue measurable with measure |E| = 1. Show that >there exists x in R such that |(E+x) cap E| = 1/2. >Here is what I have done. Consider f(x) = |(E+x)cap E|. Clearly f(0) >= 1. I have shown (not entirely trivial) that there exists an xin R >such that f(x) < 1/2. >So obviously my objective is to show that f is continuous. So far, no >success. Hint: Express your f as an integral. Continuous functions of compact support are dense in L^1. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: distance between y=x and x^3 - y^3 = 1 I need to find the point(s) on the curve x^3 - y^3 = 1 farthest from the line y = x. Now I'm not even clear how this distance is defined. Is the distance defined as follows?: Given a point (a,a ) on y = x then the distance is the length of the line segment on y = 2a -x from (a , a ) to where this line intercepts x^3 - y^3 = 1. Note that y = 2a -x is the line perpendiculay to the line y = x and contains the point (a , a)> Now isn't this not going to help since the distance would be a function of both a and x? Hints would be very much appreciated. === Subject: Re: distance between y=x and x^3 - y^3 = 1 Steven escribi.97 en el mensaje > I need to find the point(s) on the curve x^3 - y^3 = 1 farthest from the > line y = x. Now I'm not even clear how this distance is defined. Is the > distance defined as follows?: Given a point (a,a ) on y = x then the > distance is the length of the line segment on y = 2a -x from (a , a ) to > where this line intercepts x^3 - y^3 = 1. Note that y = 2a -x is the line > perpendiculay to the line y = x and contains the point (a , a)> Now isn't > this not going to help since the distance would be a function of both a and > x? Hints would be very much appreciated. This curve is of degree 3; i.e., any straight line cannot cut it in more that three points. Then if there are two points in that the distance to y = x is m.87ximum, some straight lines parallel to y = x would cut the curve in four points, that it is impossible. Then there is a maximum as much. As the curve is symmetric with respect to the straight line y = -x ( no changes in the equation if you put y --> -x, x --> -y), the maximum is for y = -x. Then 2x^3 = 1 ==> x = 1/2^(1/3) An the maximum distance is d = sqrt((1/2^(1/3))^2 + (1/2^(1/3))^2) = 2^(1/6) -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: distance between y=x and x^3 - y^3 = 1 > I need to find the point(s) on the curve x^3 - y^3 = 1 farthest from the > line y = x. Now I'm not even clear how this distance is defined. Is the > distance defined as follows?: Given a point (a,a ) on y = x then the > distance is the length of the line segment on y = 2a -x from (a , a ) to > where this line intercepts x^3 - y^3 = 1. Note that y = 2a -x is the line > perpendiculay to the line y = x and contains the point (a , a)> Now isn't > this not going to help since the distance would be a function of both a and > x? Hints would be very much appreciated. If (b, c) is the point on x^3 - y^3 = 1 that is farthest from y = x, then the line tangent to this curve at this point is parallel to y=x. Do you see why? _________________________________________________________ Eric J. Wingler (wingler@math.ysu.edu) Dept. of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, OH 44555-0001 330-941-1817 === Subject: Re: distance between y=x and x^3 - y^3 = 1 > I need to find the point(s) on the curve x^3 - y^3 = 1 farthest from the > line y = x. Now I'm not even clear how this distance is defined. Is the > distance defined as follows?: Given a point (a,a ) on y = x then the > distance is the length of the line segment on y = 2a -x from (a , a ) to > where this line intercepts x^3 - y^3 = 1. Note that y = 2a -x is the line > perpendiculay to the line y = x and contains the point (a , a)> Now isn't > this not going to help since the distance would be a function of both a and > x? Hints would be very much appreciated. I think everything you've said is fine except the distance would be a function of both a and x. In fact it is a function of a only. Note that the intersection is found by solving x^3 - y^3 = 1 and y = 2a - x, i.e. x^3 - (2a-x)^3 = 1. The 3 solutions for x depend only on a. Gib === Subject: Re: distance between y=x and x^3 - y^3 = 1 I need to find the point(s) on the curve x^3 - y^3 = 1 farthest from the > line y = x. Now I'm not even clear how this distance is defined. Is the > distance defined as follows?: Given a point (a,a ) on y = x then the > distance is the length of the line segment on y = 2a -x from (a , a ) to > where this line intercepts x^3 - y^3 = 1. Note that y = 2a -x is the line > perpendiculay to the line y = x and contains the point (a , a)> Now isn't > this not going to help since the distance would be a function of both a and > x? Hints would be very much appreciated. I think everything you've said is fine except the distance would be a > function of both a and x. In fact it is a function of a only. Note > that the intersection is found by solving x^3 - y^3 = 1 and y = 2a - x, > i.e. x^3 - (2a-x)^3 = 1. The 3 solutions for x depend only on a. Gib I have a solution for you ! The stright line y=x makes an angle of 45 with the axe y=0 You have to rotate the two curves by 45 in clockwise sense This transformation is isometric (distances will not change) The curve y=x becames y=0 Then you have to find the derivative of the rotated curve x-y=1 the Zeros of the derivative, gives you the solutions. I have calculate one real solution y=sqr(1/2/cbcroot(4)-cbcroot(2)/6) good luck Dr STRUL === Subject: Generalized Pentagon? Does there exist a finite generalized pentagon (except those degenerated ones, of course)? Feit and Higman proved that for finite generalized n-gon of order (s,t), only n=3,4,6,8 is possible. However if I do not require an order (i.e. not all points (resp. lines) indicent with same number of lines (resp. points)), is it possible to construct a finite generalized n-gon with n=5? If so, anyone care to give an example? The free construction seems to give infinite polygons, or did I miss anything? Any hints will be appreciated!!! === Subject: Dense in topology question A set S in a topological space X is said to be dense in X if cl(S) = X. If X is an infinite set with the finite complement topology, why is any infinite subset of X dense in X? Moreover, let f,g : X-->Y are continuous functions between two topological spaces, where there is a dense subset S in X such that f(x) = g(x) for any x in S. How can I show that if Y is Hausdorff, then f(x) = g(x) for all x in X? Thank you very much for help, Steve === Subject: Re: Dense in topology question Originator: grubb@lola >A set S in a topological space X is said to be dense in X if cl(S) = X. >If X is an infinite set with the finite complement topology, why is any >infinite subset of X dense in X? What are the closed subsets of X? What is the smallest one that contains your infinite subset? >Moreover, let f,g : X-->Y are continuous functions between two topological >spaces, where there is a dense subset S in X such that f(x) = g(x) for any x >in S. How can I show that if >Y is Hausdorff, then f(x) = g(x) for all x in X? Suppose not. What can you say? How can you then use the Hausdorff property? Then density? Next time, show us what you've tried when you want homework help. --Dan Grubb === Subject: Re: Dense in topology question > A set S in a topological space X is said to be dense in X if cl(S) = X. > If X is an infinite set with the finite complement topology, why is any > infinite subset of X dense in X? The details of the proof depend on the definition you're using for cl(S). Here's a sample. Definition: For a set S in a topological space X, the closure cl(S) is the intersection of all closed sets in X that contain S. Proof of your assertion: The open sets in X are the co-finite sets and the empty set. Therefore the closed sets are the finite sets and X itself. So, if S is infinite, the only closed set that contains it is X. Therefore cl(S)=X. QED [Snip second question for lack of time today. Sorry] HTH, Felix. === Subject: Re: Dense in topology question > Moreover, let f,g : X-->Y are continuous functions between two topological > spaces, where there is a dense subset S in X such that f(x) = g(x) for any x > in S. How can I show that if > Y is Hausdorff, then f(x) = g(x) for all x in X? Here's a direct way. Assume f(x) /= g(x) some open disjoint U,V with f(x) in U, g(x) in V. x in open A = f^-1(U) / g-1(V) is nonnul. As S is dense some a in S with a in A; f(a) in U; g(a) in V but f(a) = g(a), which contradicts U,V disjoint. === Subject: Re: Dense in topology question > A set S in a topological space X is said to be dense in X if cl(S) = X. > If X is an infinite set with the finite complement topology, why is any > infinite subset of X dense in X? To become acquainted with the cofinite topology X, here's some exercises for A subset X: if A is finite, then int A = ?, cl A = ? if A is infinite, then cl A = ? if A is infinite and complement A is infinite, then int A = ? if A is cofinite, then int A = ? > Moreover, let f,g : X-->Y are continuous functions between two topological > spaces, where there is a dense subset S in X such that f(x) = g(x) for any x > in S. How can I show that if > Y is Hausdorff, then f(x) = g(x) for all x in X? First show the diagonal D = { (y,y) | y in Y } is closed in YxY. Now the map h:X -> YxY, x -> (f(x),g(x)) is continuous. Use that to show { x | f(x) = g(x) } = h^-1(D) is closed Thus X = cl S subset cl { x | f(x) = g(x) } = { x | f(x) = g(x) } === Subject: Re: Dense in topology question > A set S in a topological space X is said to be dense in X if cl(S) = X. > If X is an infinite set with the finite complement topology, why is any > infinite subset of X dense in X? Moreover, let f,g : X-->Y are continuous functions between two topological > spaces, where there is a dense subset S in X such that f(x) = g(x) for any x > in S. How can I show that if > Y is Hausdorff, then f(x) = g(x) for all x in X? These are not terribly hard homework questions. Tell us what you've done so far. === Subject: question regarding complete solution to a second degree homogenous equation Is the complete solution to (1) known? (1) ax^2 + bxy + cy^2 = f^2 where integers a,b,c are known, with b^2 > 4ac, but f is not known and x,y are to be integer also? Dario Alpern's javascript program at http://www.alpertron.com.ar/QUAD.HTM is almost there, but his program requires that f be known in advance. Is it possible to solve (1) so that all possible solutions of f^2 are located? Thanks for references to this. === Subject: Re: Gaussian limit > How do you compute > lim (t --> 0 from above) x e^(-x^2/(4t)) / t^(3/2) ? I've been trying > to bound the quotient and simplify things (perhaps use L'Hopital) but > to no avail. Let's simplify this to solving lim(t->0+) [e^(-a/t)]/t^b, a > 0, b > 0 take logs -a/t - b.log t Note: lim(t->0) (b.log t)/(a/t) = 0 In other words log t goes to -oo slower than 1/t goes to oo Thus the -oo + oo form of -a/t - b.log t is biased in favor of -1/t. Take limits lim(t->0) -a/t - b.log t = lim(t->0) -(a/t)(1 + (b.log t)/(a/t)) = lim(t->0) -a/t = -oo Thus orginal limit = 0. === Subject: Re: Possible Errors in Diehard Test Results The K,K+,K- ( in some circles, D,D+,D-) are the oldest, > and still widely used, but other methods, particularly those > based on int(w(x)*(F(x)-x)^2,x=0..1), where F(x) is the sample > CDF (staircase function), are considered better, particularly if > the weight function w(x) is chosen so as to equalize variances. > This was proposed by Savage and is incorporated in the > Anderson-Darling version of the KS test (more properly, > the Anderson-Darling version of the class of Cramer-von Mises > approaches to the KS test). > The A-D version of the C-vM class of K-S tests is the one > I chose for application in the Diehard tests. -------------------------------------------------------------- ------------ ------ I would consider this a sloppy statement here. The class is Goodness-Of-Fit Tests. KS is very specific to Kolmorogov and Smirov for their pioneering work in this area. In no way can you call a D-A test a KS test. The former is based on integrals, and the latter on discrete sup. measures. This is his error in the Diehard tests, calling what he does, a KS test something that it is not. If someones research work cant be verified, it is questionable. If I cant verify his results using what he claims are known statistical tests, then there is an error somewhere. -------------------------------------------------------------- ------------ -------- For large n, there seems no strong reason for choosing a > particular version of the K-S tests, but for smaller n, the > n+1 spacings induced by the ordered set of uniform variates > should form a point s= (s_1,...,s_{n+1}) uniformly distributed > over the simplex > S={(x_1,...,x_{n+1}), x_1+...+x_{n+1}=1, x's>=0}. > If you look at the projections of sets of points for which K using Kolmogorov's K (or Smirnov's K+), where the projections > are from, say, 10-space onto various 3-dimensional faces of the > 9-dimensional simplex, you will see that certain regions seem to > be favored---others neglected--- in the original K,K+,K- versions, > contrary to the presumed test of uniformity of s over the simplex S. -------------------------------------------------------------- ------------ -------- The D-A test is not necessarily a better choice. The Implementation of the D-A the D-A test for a uniform distribution involves taking differences of logs, which bias the sum toward bunching near zero, as compared to bunching near 1. As I said earlier, this whole area of Goodness-Of-Fit testing is arbitrary, and presents some concern to observers. -------------------------------------------------------------- ------------ ---------------- > I would try to show why I think your implementation is not good for some > test (for example the overlapping sum) and I really hope you don't take > offence. > In the overlapping sum test you get 10 p-values (no_obs = 10) and then you > calculate the overall p-value with AD test. > Suppose to get a perfect distribution of p-values. With 10 p-values this > means: > 1 0 > 2 0.111111111111111 > 3 0.222222222222222 > 4 0.333333333333333 > 5 0.444444444444444 > 6 0.555555555555556 > 7 0.666666666666667 > 8 0.777777777777778 > 9 0.888888888888889 > 10 1 > With the plain KS test I get the overall p-value = 0.999897950683125, but > with your AD test I get 0.00946405272470502. > The distribution is perfect, but with AD test it seems very bad. -------------------------------------------------------------- ------------- ------------------------------------------------------------- The problem here is that this is not a perfect distribution. Here is the standard K-S and A-D analysis of your data. (Tables left out since they do not pass through sci.math) Columns 3 and 4 are your basic uniform distribution reference points. I can give you all kinds of references on how this is done in K-S testing. Your routine that gives KS p values is giving you wrong numbers. The values that I show, come from routines tested to sets of precise values generated by Birnbaum The Dn is known to be good to less than 6.5% and the D+(D-) routine good to better than 0.02%. The Dn has the most error, because it is an approximation. The D+(D-) is limited by the accuracy of the p values reported by Birnbaum. (Afterall, he only had access to a Navy vacuum tube computer that had to be programmed in machine language) The method for obtaining the Anderson-Darling measure comes from Stephens. The measure involves taking the natural log of D. Of course for 0 the log is minus infinity. If your routine gives a number, then something here is wrong. For two sets of data, my D-A p values agree with the Diehard P values within .0004, probably due to the fact I am using Lewiss table. I would suspect the Diehard tests use the log form too. -------------------------------------------------------------- -------------- -------------------------------- > My thought is that you should use the plain KS for all the tests and AD > *only* if you want to calculate the overall p-value over more than 200 > p-values (although I think this is useless). In this case (with 200 > p-values), AD with a perfect distribution gives 0.995636593372975, there is > only a little error (the exact value should be 1; I get 0.99999999752982 > with KS). > I don't know if I'm wrong somewhere (I'm not a mathematician), but this is > what I get in my experiments. > That is one of the reasons that I chose the Savage-Anderson-Darling > version rather than the original Kolmogorov K, or Smirnov K+,K-. > Another is that Savage-Anderson-Darling gives equal weight to > variations around 0, around 1 and around the center, while the > original K,K+,K- give greater weight to variations around the center. -------------------------------------------------------------- -------------- ----------------------------------------------------- What is the proof of this? Was it based on Monte Carlo studies? Conjecture? Differences in the logs of the RNs will be large near zero and small near 1. This makes the starting and ending terms of the summation dominate. If we have these characteristics, then the test is not a fair one. -------------------------------------------------------------- -------------- ----------------------------------------------------- > In my experiments I don't see that. I see, instead, that AD gives a very big > weight to the tails while around the center it gives the same weight of KS. -------------------------------------------------------------- -------------- ----------------------------------- Yes, I see the same thing. -------------------------------------------------------------- -------------- ------------------ What you said about AD I seen for KS: KS gives equal weight to variations > around 0, around 1 and around the center. > And it is often bad fit in the tails that suggest a poor RNG. -------------------------------------------------------------- -------------- ----------------------------------------------------- The uniform distribution has no tails. Statistical distributions do not always have tails. The D+(D-) and Dn distributions only have a positive tail. The A-D distribution only has a positive tail. Only if you are testing for super-uniformity do you consider a left-hand tail for either of these. For RN testing, the 0 point is included in the p interval. -------------------------------------------------------------- -------------- ------------------------------ > You have, beyond any doubt, much more experience in testing RNG, but I think > also a bad fit in the center suggests a bad RNG. The difference is that bad > tails can be seen also with our eyes, while our eyes don't see a bad center. -------------------------------------------------------------- -------------- ---------------------------- I agree with you. DAHeiser -------------------------------------------------------------- -------------- -------------------------------- > Best regards > Cristiano === Subject: Re: Possible Errors in Diehard Test Results Is there an upper limit on the file-size input to Diehard? I noticed a 20MB file takes about the same time to complete the tests as a 2GB file - could somebody enlighten me why this is so? Does Diehard tests only some selected portions of the input file? But even reading the whole 2GB file would take a Thanks for the explanation -- Bartosz Zoltak http://www.vmpcfunction.com QPbzoltak@vmpcfunction.com without QP === Subject: Re: -morphism glossary > Finally, if you consider the vast majority of mappings which do > not fall into these categories and which are thus the victims > of derision and scorn, and suffer because they are so cruelly > ignored by the world of mathematics, that's _anthropomorphism_. HAH!! And when students (or others) merely parrot back the definitions without understanding, that's presumably POLYMORPHISM!? _..._ .-' '-. / _ _ /':. (o) /__) /':. .,_ | | |': ; / /_/ / ; `` } ; ':., { / ; } ; '::. ;/ / { |. ':. ;```` / '::'::' / ; |':::' '::' / | '::' _.-`; ; /`-..--;` ; | | ; ; ; ; ; | | ; ; ; ; ; ; / |; ; ; ; ;/ ; | ; ; ; ; / / _, | ; ; /` .' _,== ; ; ; .'. _ ,_'-.=~ _.==~ | ; .'`_~'-;--. ) )==~ ; ; ;/`(_`-,) )`=~` /; ;/ /` _.=` |_/`| | ( ` |_ ,=` | _,==;` =~|; ;| | ; | |;| | |/ | === Subject: Re: -morphism glossary X-Archive: encrypt +6z//v8SJRjZ6Y5p3HJD5vX++/76+ olRdUI5lXl0AAACP0lEQVR4nFWUv2vbUBDHjVvI0sUUAoYM 4hwjEFlSgyLwm5yhSyaBF2/G4DagodTQDtFmKG/ wolHQIaiYwqObhg7qEPPAcvD9Ub17kuynGzR8 P+/dj3d36kxPNlNkW+ iAi19Adc5gGgIoBaSPsOtZAOgwqOwaA7nWkxboEngWBynlRreAtwUQC8m2 t8CMQ8f9KyQdlxagwN34NU2uDlJo3waZ87BLyS5FWfitGPdGT39GRRsMNMtJgm0 QutX5BKUBKuN6 6eOKWg9qADXQ/Ub3DRgCDAlcC6OnVDfuKkBXKFNxa3Sq4ceuV7liXxD3Gl3M+ 2kF6H1g9NroG5yn NQAPbmo/Wm4+ uY81yE4Fky40uM0NgMHOJPSeXnUxscCgKThYIzXwBAaj5iEkfqBCTzFuor7R/ WBP s2GDEntpIsbF/ oVaDt7veZpU4JhjD8dFkWOHS6UbFfhzLFbLu4Lsr55kcHb1Sxd5pBkUuQYLqIsy X1CT2VZl9wzAG+CCBsyQf6VnVf68+ kw1HA2JXywA8E5L6d9VZNSky53NeF45ZTLRAOV4SjnfDnKN Tya1sytuoSOYjFvAU7QwcCEC6l4LDKfTLVX8lvdl8VTkjzaY0Z03yPvSBgrIm/ MxklKUqxbgSaQe 8x1sA89M1z0VivMWoC2j/YAHIptbmsR4eQJm8MKYn2AXrexVm2UciIm/ 1P3LGgx5oxQ4tM6BlDQf FYDaOIVrepwoTaofQGgBoF9G1NNgLWdYn3FxiWCDinrD0P3+FeA/ 7MX1QE3djfgAAAAASUVORK5C YII= QeSbv>yxv:$-uL=[fu{$byq).2Q~9TX:5JU1e5B->MBmk{Dsk8vU&R5n`PpcD% Vlb6(8|8JU4~m?U k $H|mk}EA~u-.b*P&Q_w,(>.b4 > Students taking their first course in abstract algebra are often > overwhelmed by the torrent of terminology. Hoping to bring some > sense of order to it all, a brief list of related terms is here > defined. > If a mapping from one algebraic structure to another preserves > operations and identity elements, it is called a _homomorphism_. > If a homomorphism is injective, it is a _monomorphism_. > If a homomorphism is surjective, it is an _epimorphism_. > If a homomorphism is bijective, it is an _isomorphism_. I would say that these last three are theorems rather than definitions. Let f:A-->B be a homomorphism. If f has the property: given any C and any g,h:C-->A, fg=fh iff g=h then f is an epimorphism. If f has the property given any C and any g,h:B-->C, gf=gh iff g=h then f is a momomorphism. If f has a two sided inverse then f is an isomorphism. For most homomorphisms (especially algebraic ones), being a mono- epi- or isomorphism is equivalent to the conditions you give above, but there are more geneal categories in which this isnt true. For example a homomorphism of topological spaces is a continuous map, and `isomorphism of top. spaces' is not the same as `continuous bijection between top. spaces'. > A bijective endomorphism is an _automorphism_. (With the above viewpoint, an automorphism is an endomorphism which is also an isomorphism.) === Subject: Re: -morphism glossary Students taking their first course in abstract algebra are often > overwhelmed by the torrent of terminology. Hoping to bring some > sense of order to it all, a brief list of related terms is here > defined. If a mapping from one algebraic structure to another preserves > operations and identity elements, it is called a _homomorphism_. If a homomorphism is injective, it is a _monomorphism_. If a homomorphism is surjective, it is an _epimorphism_. If a homomorphism is bijective, it is an _isomorphism_. > I would say that these last three are theorems rather than definitions. > Let f:A-->B be a homomorphism. > If f has the property: > given any C and any g,h:C-->A, > fg=fh iff g=h then f is an epimorphism. > If f has the property given any C and any g,h:B-->C, > gf=gh iff g=h then f is a momomorphism. If f has a two sided inverse then f is an isomorphism. For most homomorphisms (especially algebraic ones), being a mono- epi- > or isomorphism is equivalent to the conditions you give above, but > there are more geneal categories in which this isnt true. For example > a homomorphism of topological spaces is a continuous map, and > `isomorphism of top. spaces' is not the same as `continuous bijection > between top. spaces'. > A bijective endomorphism is an _automorphism_. (With the above viewpoint, an automorphism is an endomorphism which is > also an isomorphism.) The problem with these posts is that even in algebra, epimorphism and surjection are not the same thing. For example, in the category of monoids, the inclusion of N into Z is an epimorphism (and monomorphism) and in the category of rings, so is the inclusion Z into Q. In non-algebraic situations, monomorphisms might not be injective, although I don't have a simple example off-hand. BTW, in groups, epics are surjective, but that fact is far from obvious. The old terms, one-one and onto for injection and surjection were much more evocative and should not have been dropped. While monomorphism and epimorphism are categorical terms and probably should not be even mentioned till later on. === Subject: Re: -morphism glossary >The problem with these posts is that even in algebra, epimorphism and >surjection are not the same thing. For example, in the category of >monoids, the inclusion of N into Z is an epimorphism (and >monomorphism) and in the category of rings, so is the inclusion Z into >Q. In non-algebraic situations, monomorphisms might not be injective, >although I don't have a simple example off-hand. BTW, in groups, >epics are surjective, but that fact is far from obvious. In the category of divisible abelian algebras, the map from Q to Q/Z is a monomorphism. David McAnally -------------- === Subject: analysis of a large ODE system. some tips needed. hello group. I actually posted this in another group, but I haven't received any replies as of yet. So i figured, i woudl try out another group that cater to people more mathemtically inclined. I'm looking for proper ways to analyze a given large ODE system as dynamical system. I have a fairly large system of ODE's and i'm in the process of exploring it's behaviors. but since I'm not a mathematician,( but a grad student in biology) i was wondering (naturally) if what I'm doing is how it should be done. I'm currently randomly picking parameters to get a general idea for the system. I would love to hear some of your thoughts on what the norm is for analyzing a large ODE system that has few nonlinear equations. (total 23 ode's and 47 parameters) if some of you who happenes to be mathematicians or modelers were asked to analyze a system of this order, what would you do first? what would be some of the things or the properties you would look for? Would you try to reduce the system only under steady state? but that's not dynamical system analysis.. or is it? would you look for bifurcations in all the parameter pairs? and will that be possible? or any other ways that I haven't mentioned? any and all thoughts are thoroughly appreciated. sean === Subject: Re: analysis of a large ODE system. some tips needed. > hello group. I actually posted this in another group, but I haven't received any > replies as of yet. So i figured, i woudl try out another group that > cater to people more mathemtically inclined. I'm looking for proper ways to analyze a given large ODE system as > dynamical system. I have a fairly large system of ODE's and i'm in the process of > exploring it's behaviors. but since I'm not a mathematician,( but a > grad student in biology) i was wondering (naturally) if what I'm doing > is how it should be done. I'm currently randomly picking parameters to get a general idea for > the system. I would love to hear some of your thoughts on what the norm is for > analyzing a large ODE system that has few nonlinear equations. (total > 23 ode's and 47 parameters) if some of you who happenes to be mathematicians or modelers were > asked to analyze a system of this order, what would you do first? > what would be some of the things or the properties you would look for? Would you try to reduce the system only under steady state? but that's > not dynamical system analysis.. or is it? would you look for bifurcations in all the parameter pairs? and will > that be possible? or any other ways that I haven't mentioned? any and all thoughts are thoroughly appreciated. Here is some advice: First learn as much as you can about the process you want to model, and the way it is transformed into the mathematical model. You should have the interpretation for each variable and each parameter in your mind. This is a lot of work (you may have done that already), but it pays in the end. Then I advise you to concentrate on the steady states. Ideally, you should be able to find the steady states for each set of parameters. Then classify them according to their stability using linearisation (you know how to do this?). This should split your parameter space into several regions, each with a seperate type of steady states. It would look something like this: Region 1: one attractive steady state, Region 2: three steady states, one of which attractive Region 3: three steady states, none attractive If this really works, you should already have a pretty good understanding of how your system works locally. You then may proceed to get some global results. However, I expect that you will already have problems determining the steady states for an arbitrary set of parameters. Try to get as much information about steady states as is possible (e.g. it is often possible to find a region in parameter space where only one steady state exists (hint: Banach fixed point theorem)). It is hard to say more without knowing the particular set of equations. HTH, Michael. -- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&& Dr. Michael Ulm FB Mathematik, Universitaet Rostock michael.ulm@mathematik.uni-rostock.de === Subject: Re: analysis of a large ODE system. some tips needed. Hi Dr. Ulm. Thanks for the kind reply. > First learn as much as you can about the process you want to model, > and the way it is transformed into the mathematical model. You should > have the interpretation for each variable and each parameter in your > mind. This is a lot of work (you may have done that already), but it > pays in the end. My training has been focused on Biology, so have fairly good grasp of the interpretations. > Then I advise you to concentrate on the steady states. Ideally, you > should be able to find the steady states for each set of parameters. > Then classify them according to their stability using linearisation > (you know how to do this?). This should split your parameter space > into several regions, each with a seperate type of steady states. It > would look something like this: Region 1: one attractive steady state, > Region 2: three steady states, one of which attractive > Region 3: three steady states, none attractive You asked the right there Dr. Ulm, And No.. I have to admit I don't. I'm still tryin gto understand the idea of stability of dynamical system. It seems liek if a variable settles into a steady state or equilibrium then it is said to be stable? as opposed to numerical stability which is result of high precision accuracy calculation. I have no idea how to classify the steady states using linearisation... Perhaps you woull be kind to suggest some manuals or books? > However, I expect that you will already have problems determining > the steady states for an arbitrary set of parameters. Try to get > as much information about steady states as is possible (e.g. it is > often possible to find a region in parameter space where only one > steady state exists (hint: Banach fixed point theorem)). > another thing I have to look up. I think that's the idea of uniqueness anf existence? thank you so much Dr Ulm. You have been most kind and helpful gratefully yours sean PS. Please note the reply to address is sean_incali01@yahoo.com === Subject: Induction question, help needed. Help needed on this question. Any help is appreciated. Thanks in advance. Given a binary string (i.e. a finite sequence of 0's and 1's) we choose any two digit substring 01 and replace it by a string of the form 100...0 using an arbitrary (but finite) number of zeros. Prove by induction that this transformation can not be performed infinitely many times, i.e. this sequence of transformations must terminate for any input string. === Subject: Re: Induction question, help needed. >Help needed on this question. Any help is appreciated. Thanks in >advance. >Given a binary string (i.e. a finite sequence of 0's and 1's) we >choose any two digit substring 01 and replace it by a string of the >form 100...0 using an arbitrary (but finite) number of zeros. Prove >by induction that this transformation can not be performed infinitely >many times, i.e. this sequence of transformations must terminate for >any input string. This is a well-known (solved) type of termination problem in string rewriting, a subfield of term rewriting. Here are some related links: term rewriting homepage: http://rewriting.loria.fr/ journal paper (AAECC 2000) by Zantema/Geser on this topic: pdf: http://www.springerlink.com/app/home/content.asp?wasp= f83ebu4j807jtj4e67vl&re ferrer=contribution&format=2&page=1&pagecount=25 abstract: http://www.springerlink.com/app/home/contribution.asp?wasp= 1lbfddxhqncvtk6lwv tp&referrer=parent&backto=issue,1,5;journal,19,44; linkingpublicationresults,i d:100499,1 conference paper (RTA 1995) on this topic: http://www.informatik.uni-trier.de/~ley/db/conf/rta/rta95.html #ZantemaG95 technical report version (1994) from Zantema's homepage: ftp://ftp.cs.ruu.nl/pub/RUU/CS/techreps/CS-1994/1994-44.ps.gz Hans Zantema's homepage: http://www.win.tue.nl/~hzantema/ Alfons Geser's homepage: http://research.nianet.org/~geser/ Hope this helps, Bernhard Gramlich. ============================================================== ========== Bernhard Gramlich Vienna University of Technology e-mail: gramlich@logic.at www: http://www.logic.at/staff/gramlich ============================================================== ========== -- ============================================================== ========== Bernhard Gramlich Vienna University of Technology e-mail: gramlich@logic.at www: http://www.logic.at/staff/gramlich ============================================================== ========== === Subject: Re: Induction question, help needed. > Given a binary string (i.e. a finite sequence of 0's and 1's) we > choose any two digit substring 01 and replace it by a string of the > form 100...0 using an arbitrary (but finite) number of zeros. Prove > by induction that this transformation can not be performed infinitely > many times, i.e. this sequence of transformations must terminate for > any input string. That is any finite string. Transform the string into a polynomial a_n x^n + ... + a_1 x + a_0 where a_j is the number of 0's before x^j which represents a 1. Now every transformation preserves the number of 1's. Thus the exponents of the polynomial are unchanged. A transformation upon 01 for the j^th 1 is a_j x^j + a_(j-1) x^(j-1) -> (a_j - 1)x^j + (a_(j-1) + c)x^(j-1) for positive integer c. This reminds me of Goodstein's theorem, the simplest theorem independent of Peano's axioms. The induction is complex. Every transformation increases a coefficient at the cost of decreasing by one the previous coefficient. One way to proceed is to consider the polynomial in x as an ordinal polynomial in w. Thus every transformation decreases the ordinal represented by the polynomial. If the transformations could proceed forever then there would be an infinite descending sequences of ordinals, which is impossible as the ordinals are well ordered with a first element. -- Another way to think the same is let the string be (a_n, ... a_1, a_0) where a_j is the number of 0's before the j-th 1 and a_0 is the number of trailing 0's. Now well order the n+1-tuples lexigraphically with the higher index being given the greater dominance. IE (1,3,2) < (1,4,1) Again the same reasoning, for each transformation h on a string s h(s) < s and h(s) is like s, an n+1-tuple. This can't continue downward forever as the tuples are well ordered. ---- === Subject: re analysis of a large ODE system. some tips needed. it occured to me that i didn't post a reply to email. sean_incali@yahoo.nospam.com isn't working too well due to few spam issues. please reply, if you were going to send an email, to sean_incali01@yahoo.com thanks all for all comments === Subject: Re: Need help optimizing.... > It may be helpful to realize: a*x^3 + b*x^2 + c*x + d = d + x*(c + b*(x + a*x)) and so on. I think that's called Horner's rule, and it is a standard trick when > optimizing the evalutaion of polynomials. > I think your are right.... horner's rule in c. n=5, p=c[n]; > for (j=n-1; j >=0; j=j-1) > p *= x + c[j]; however I wonder how well, or even if its possible, to do this > with and equation with 3 variables, as in my example? check this out. > http://www.dspguru.com/comp.dsp/tricks/alg/horner.htm ps someone was awake enough to see the error > pts[pn].val = c[cn++] * pow(pts[pn].x, i) * zyexp; > which should have been: > pts[pn++].val = c[cn++] * pow(pts[pn].x, i) * zyexp; So if any one has any suggestions on using this trick with more than one > variable let me know please. Try this: a polynomial p(x,y) in 2 variables x,y is a polynomial in y with coefficients c_j(x) in R[x] (polynomials in x), for example x^2y^2+3xy^2+2xy+y+x+1 = y^2(x^2+3x) + y(2x+1) +(x+1) Compute the coefficients c_j(x} using Horner's rule and then with these compute p(x,y) again using Horner's rule. The same principle applies to polynomials in 3 or more variables. You could build this up as follows: 1. a routine for p(x) 2. a routine for p(x,y) using the routine for p(x) on the coefficients c_j(x) 3. a routine for p(x,y,z) using the routine for p(x,y) on the coefficients c_ij(x,y) and son on. === Subject: Re: Need help optimizing.... > It may be helpful to realize: a*x^3 + b*x^2 + c*x + d = d + x*(c + b*(x + a*x)) and so on. I think that's called Horner's rule, and it is a standard trick when > optimizing the evalutaion of polynomials. > I think your are right.... horner's rule in c. n=5, p=c[n]; > for (j=n-1; j >=0; j=j-1) > p *= x + c[j]; however I wonder how well, or even if its possible, to do this > with and equation with 3 variables, as in my example? I did not look at your code too closely. But frequently one can reverse the order of nested loops. Can you do such? And will it make some such trick as Horner's rule useful? > check this out. > http://www.dspguru.com/comp.dsp/tricks/alg/horner.htm -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their mastebut highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Need help optimizing.... > It may be helpful to realize: a*x^3 + b*x^2 + c*x + d = d + x*(c + b*(x + a*x)) and so on. I think that's called Horner's rule, and it is a standard trick when > optimizing the evalutaion of polynomials. > I think your are right.... horner's rule in c. n=5, p=c[n]; > for (j=n-1; j >=0; j=j-1) > p *= x + c[j]; however I wonder how well, or even if its possible, to do this > with and equation with 3 variables, as in my example? > I did not look at your code too closely. But frequently one can > reverse the order of nested loops. Can you do such? And will it > make some such trick as Horner's rule useful? Yes I certainly can (And have) reversed the orders of the loops. It really depends on weather I want the highest order polynomials on the left or on the right... horner's rule I believe requires the highest order polynomial on the right.. > check this out. > http://www.dspguru.com/comp.dsp/tricks/alg/horner.htm > -- > Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html > To solve Linear Programs: .../LPSolver.html > r c A game: .../Keynes.html > v s a Whether strength of body or of mind, or wisdom, or > i m p virtue, are found in proportion to the power or wealth > e a e of a man is a question fit perhaps to be discussed by > n e . slaves in the hearing of their mastebut highly > @ r c m unbecoming to reasonable and free men in search of > d o the truth. -- Rousseau === Subject: Re: Need help optimizing.... >> Yes. 1) Don't cross-post to 9 newsgroups. You will almost >> certainly receive a number of angry responses, which will >> not help you solve your optimization problem (this is a >> meta-optimization tip for the next time you have a question). >People who get 'Angry' in regards to news group posting need to >seek help desperatly. But thanks for the info I will take in the >good spirit. need to either (1) read it again, or (2) read it again. HTH. Doug === Subject: Re: Need help optimizing.... >> Yes. 1) Don't cross-post to 9 newsgroups. You will almost >> certainly receive a number of angry responses, which will >> not help you solve your optimization problem (this is a >> meta-optimization tip for the next time you have a question). >People who get 'Angry' in regards to news group posting need to >seek help desperatly. But thanks for the info I will take in the >good spirit. > need to either (1) read it again, or (2) read it again. HTH. People who think that JustSomeGuy was referring to David himself, rather than to the senders of the number of angry responses to which David refeneed to either (1) read it again, or (2) read it again. HTH. -- #include char*f=#include %cchar*f=%c%s%c;%cint main(void){printf(f,10,34,f,34,10,10);return 0;}%c; int main(void){printf(f,10,34,f,34,10,10);return 0;} === Subject: Automating proofs Is there software out there that will automate the process of proving that (expression a) = (expression b)? It seems to me that such problems involve predictable combinations of e.g. setting things equal to zero, taking the square root, saying a = a+b-b, etc. === Subject: Re: Automating proofs , nindiv > Is there software out there that will automate the process of proving that > (expression a) = (expression b)? > Depends on the type of expression. In general, the answer is NO. And it can be proved that such an automated procedure is impossible. There are subclasses of expressions for which decision procedures, normal forms, etc., DO exist. === Subject: Re: Automating proofs A N Niel pravi: > , nindiv >Is there software out there that will automate the process of proving that >>(expression a) = (expression b)? > Depends on the type of expression. In general, the answer is NO. And > it can be proved that such an automated procedure is impossible. There > are subclasses of expressions for which decision procedures, normal > forms, etc., DO exist. Nindiv might be interested in the book A=B by Petkovsek, Wilf and Zeilberger (ISBN 1568810636, published in 1996 by A K Peters Ltd): http://www.cis.upenn.edu/~wilf/AeqB.html A=B is about identities in general, and hypergeometric identities in particular, with emphasis on computer methods of discovery and proof. The book describes a number of algorithms for doing these tasks, and we intend to maintain the latest versions of the programs that carry out these algorithms on this page. So be sure to consult this page from time to time, and help yourself to the latest versions of the programs. -- Roman === Subject: WTB: ISBN 0534393322, Complete Solutions Manual for Stewart's Single Variable Calculus: Early Transcendentals WTB: ISBN 0534393322, Complete Solutions Manual for Stewart's Single Variable Calculus: Early Transcendentals email me with a price including shipping, Thanks === Subject: Re: WTB: ISBN 0534393322, Complete Solutions Manual for Stewart's Single Variable Calculus: Early Transcendentals >WTB: ISBN 0534393322, Complete Solutions Manual for Stewart's Single >Variable Calculus: Early Transcendentals I have a copy. Unfortunately, that's because I'm a professor teaching a Calculus course. And no, you can't have it. Doug === Subject: Re: WTB: ISBN 0534393322, Complete Solutions Manual for Stewart's Single Variable Calculus: Early Transcendentals > WTB: ISBN 0534393322, Complete Solutions Manual for Stewart's Single > Variable Calculus: Early Transcendentals > email me with a price including shipping, > Thanks Ermm...eBay is thataway, boy. Drieux === Subject: Re: WTB: ISBN 0534393322, Complete Solutions Manual for Stewart's Single Variable Calculus: Early Transcendentals > WTB: ISBN 0534393322, Complete Solutions Manual for Stewart's Single > Variable Calculus: Early Transcendentals > email me with a price including shipping, > Thanks The Brooks-Cole website, referring to the wanted item, states: The following complimentary resources are available to instructors only upon adoption of the text. (Some restrictions may apply.) The website also gives a phone number for Academic Support: 800-423-0563. So just tell them where you are teaching this course and be assured of their cooperation. David Ames === Subject: Recall probabilities Last week 3 justices in the 9th US Appellate Court voted unaminously to delay the California recall election. Now 11 other justices in that court voted unaminously to restore the original election date. If p is the probability that a random justice wants to postpone the election, what value of p maximizes the probability that the first three justices will vote for postponement but the other eleven will not? What is this maximum probability. [It's less than 1 in 135, where 135 is the number of replacement candidates on the ballot.] -- Wanted: Experts at choosing the best of 100+ applicants for a position. Register as a California voter by September 22, and vote on October 7. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: Re: Recall probabilities : Last week 3 justices in the 9th US Appellate Court : voted unaminously to delay the California recall election. : Now 11 other justices in that court voted unaminously : to restore the original election date. : If p is the probability that a random justice wants : to postpone the election, what value of p maximizes : the probability that the first three justices will : vote for postponement but the other eleven will not? : What is this maximum probability. Are we to assume independence? === Subject: Re: Recall probabilities > : Last week 3 justices in the 9th US Appellate Court > : voted unaminously to delay the California recall election. > : Now 11 other justices in that court voted unaminously > : to restore the original election date. > : If p is the probability that a random justice wants > : to postpone the election, what value of p maximizes > : the probability that the first three justices will > : vote for postponement but the other eleven will not? > : What is this maximum probability. > Are we to assume independence? There may have been some pressure to reach consensus in both decisions. Here is another wrinkle, what are the odds that none of the first 3 were among the 11. I think they were eligible, I know the picking was random, and I think the pool was in the low 20s. Bill === Subject: Re: Recall probabilities > : Last week 3 justices in the 9th US Appellate Court > : voted unaminously to delay the California recall election. > : Now 11 other justices in that court voted unaminously > : to restore the original election date. : If p is the probability that a random justice wants > : to postpone the election, what value of p maximizes > : the probability that the first three justices will > : vote for postponement but the other eleven will not? > : What is this maximum probability. Are we to assume independence? There may have been some pressure to reach consensus in both decisions. Here > is another wrinkle, what are the odds that none of the first 3 were among the > 11. I think they were eligible, I know the picking was random, and I think > the pool was in the low 20s. Newspaper reports here state that the intersection was not empty. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Recall probabilities Last week 3 justices in the 9th US Appellate Court > voted unaminously to delay the California recall election. > Now 11 other justices in that court voted unaminously > to restore the original election date. If p is the probability that a random justice wants > to postpone the election, what value of p maximizes > the probability that the first three justices will > vote for postponement but the other eleven will not? > What is this maximum probability. The answer is not surpringly p=3/14 > [It's less than 1 in 135, where 135 is the number > of replacement candidates on the ballot.] What does that have to do with it? === Subject: Re: Recall probabilities Last week 3 justices in the 9th US Appellate Court > voted unaminously to delay the California recall election. > Now 11 other justices in that court voted unaminously > to restore the original election date. If p is the probability that a random justice wants > to postpone the election, what value of p maximizes > the probability that the first three justices will > vote for postponement but the other eleven will not? > What is this maximum probability. [It's less than 1 in 135, where 135 is the number > of replacement candidates on the ballot.] -- > Wanted: Experts at choosing the best of 100+ applicants for a position. > Register as a California voter by September 22, and vote on October 7. > Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California > Microsoft Research and CWI I got p=3/14 with maximum probability P=6.9325e-004 Jeroen === Subject: distance in R^3 Hi I found this problem in an olympiad manual: Given two lines, l1 and l2 in R^3 , find the distance between them. (there are at least more than 2 ways to do this). I have no clue on how to proceed Any solutions/proofs would be greatly appreciated. Thanks in advance. Sincerely Carlos === Subject: Re: distance in R^3 > Hi > I found this problem in an olympiad manual: > Given two lines, l1 and l2 in R^3 , find the distance between them. > (there are at least more than 2 ways to do this). > I have no clue on how to proceed > Any solutions/proofs would be greatly appreciated. Thanks in advance. Sincerely > Carlos In R^n, let f(s) and g(t) be parametric (vector) representations of points on l1 and l2 respectively. (1) Then h(s,t) = sqrt((f(s)-g(t)).(f(s)-g(t))) is the distance between the points, where '.' represents the dot product. This can be minimized using calculus (minimizing the square of the distance is a bit easier). By definition, the minimum value of this distance is the distance between the lines. (2) If l1 and l2 are not parallel then the cross product of their direction vectors is not the zero vector and is perpendicular to both lines, and the length of the projection of any f(s)-g(t) along this cross product is the desired distance. === Subject: Re: distance in R^3 > Hi > I found this problem in an olympiad manual: > Given two lines, l1 and l2 in R^3 , find the distance between them. > (there are at least more than 2 ways to do this). > I have no clue on how to proceed This is a standard application of 3-dimensional vector methods. If P_1 is the nearest point on L_1 to L_2 and P_2 is the nearest point on L_2 to L_1 then P_1 P_2 is perpendicular to both L_1 and L_2. If the lines are given in parametric form r = a + t b where a and b are vectothen these are easy to find. Alterntaively find a unit multiple u of vector P_1 P_2, using say the cross product, and then find planes r.u = k_1 and r.u = k_2 containing L_1 and L_2. The distance between the lines is the distance between the planes, viz. |k_1 - k_2|. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: value of geometric shapes in generation of random numbers Value of Geometric shapes in Generation of Random Numbers In attempting to explain the EPR paradox to myself several years ago, I developed a matrix using all the permutations of the letters YZAX. I went along and gave a number to each permutation that was similar, namely place shifted. yxaz1 xazy1 azyx1 zyxa1 yxza2 xayz5 azxy3 zyax4 yazx3 xzya4 ayxz2 zxay5 yaxz4 xzay2 ayzx5 zxya3 yzxa5 xyaz3 axzy4 zayx2 yzax6 xyza6 axyz6 zaxy6 Later on I went on to make a website using the matrix. It was up for a couple years until NBCi pulled the plug. I lost the password to the original website after fighting with a hacker to gain control of the damn thing. Recently I decided to do something new with it. I created a BASIC program that randomly went through the maze. I was pleased with myself and ran it a few times but thought nothing of it. Then the other day I went through one of my paranoid states and that perhaps The Who's Pinball Wizard and Radiohead's OK Computer was about the program. That it was like a pinball machine. That it was OK, not great or perfect. I began to dwell on what would make it better. I had another matrix in my notes and thought perhaps that one would be. yxaz1 xazy1 azyx1 zyxa1 yxza2 xzay2 ayzx5 zxay5 yazx3 xzya4 azxy3 zyax4 yaxz4 xyaz3 axzy4 zxya3 yzxa5 xayz5 ayxz2 zayx2 yzax6 xyza6 axyz6 zaxy6 I did not recollect constructing this matrix and thought it better for two reasons. First, I did not consider it mine. Second, roughly speaking the odds were one half and not one quarter. After looking at it further, I realized it was redundant and only needed two columns. This brought me to the question of using smaller binary random statements in the BASIC program. It dawned on me this might increase the randomness. I had tried to make a program using the second set, but it was to big to compile using the trial version of the software I had. While fooling around with that though, I ran across the idea of printing the sixth, twelfth, or twenty fourth step. In an effort to make a random character generator out of it. I stumbled with the set trying to determine if every permutation could be selected equally. Unfortunately it could not. Around that time I got into a discussion with my friend Henry about infinite paths. I said I felt having at least three was important. He said the smallest thing with three infinite paths would be a triangle. It dawned on me a while later Hey not only is it three, but its binary decisions Tonight it occurred to me that with 3 steps, using three binary random statements, I had the equal opportunity of three outcomes. B or C A + + B C + A or C A or B B to C to A C to B to A B to A to B B to C to B C to B to C C to A to C I know it seems counter productive to use three random statements to generate a number from 1 to 3. It seems as if you would get the same effect or greater merely stacking the statements. However then you are still using a value from 0 to 1 to generate a number from 1 to 4 Random statement value from 0 to 1 producing 1 or 2 Random statement Random statement 1 or 2 3 or 4 This is no better than a single statement. also I am almost certain using the three statements in a triad with three steps would be more random than a single statement. It seems odd that a triangle would have this characteristic but it gets better. A square is un-workable, however a five pointed star, I believe, has the same characteristic. You can hit all bases with four, however it isn't even. It is however even with five. Why these trusted ancient geometric shapes should have this quality when used on a computer is a little strange. Perhaps the ancients are trying to tell us something. === Subject: Re: David Ullrich on Identity > David Ullrich says: And yes, identity is in _fact_ reflexive. To > refute that statement you need to give an > example of something which is not identical > to itself. The idea that there is something > which is _not_ identical to itself is simply > ludicrous: That's what identity _means_: A > thing is identical to itself and to nothing For a contrasting standpoint, see > ************************************************************** > David Ullrich asks: What's an example of something that's not identical ************************************************************** > David Ullrich dares: Exhibit of proof of Ex~(x=x) from > C1-C4 and someone will point out the error. >C1 AxAy[x=y -> Az(z in x <-> z in y)] LL1 >>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2 >C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) >>Classification >C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] > Weak Would someone be kind enough help David out with a proof? > ************************************************************* > David Ullrich remonstrates: I 'blunder' by saying that equality is reflexive by definition? > Huh. Do you have any idea what the word definition means? Homework for David Ullrich: 1) What philosopher said: ...definitions are available only for transforming > truths, not for founding them. 2) In your own words, explain why (or why not) you think > this is true. --John True, the provability of Ex~(x=x) will vary from one system to > the next. Nevertheless, there is more to truth than > 'truth-in-a-model'; > and just as the truth of Ex~(x is red) depends on how things stand > with individuals and redness so the truth of Ex~(x=x) depends on > how things stand with individuals and identity. . . . can you offer an example where Ex~(x=x) is true? I'll provide not one but THREE examples. The first > depends on a (very simple) proof that DAVID ULLRICH COULDN'T HACK! A) Take any 'pure' first order logic, that is, any FOL with no > singular terms other than variables (just to keep things simple). B) To your FOL add this formation rule; If a and b are variables, a=b is an atomic formula. and this axiom scheme: Let be C and D be wff's which differ only in that a occurs free > in C where b occurs free in D. If a=b, then C <-> D. C) Call the resulting logic FOL+. In FOL+, identity is symmetric and transitive and identicals are > indiscernible, but neither Ax(x=x) nor ~Ax(x=x) is a thesis. So, > FOL+ is a subtheory of FOL= (because every thesis of FOL+ is > also a thesis of FOL=), but not conversely (because Ax(x=x) > is a thesis of FOL= but not of FOL+). Example 1: Set Theory Add to FOL+ the NBG axiom scheme, C3, and Correy's principle of > extensionality, C4: C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) Classification C4 AyAx[Az(z in y <-> z in x) -> {(set y & set x) <-> y=x}] > (Equi-membered classes are identical iff these are sets.) > 1) EyAx(x in y <-> Et(x in t) & ~(x in x)) C3 Hence 2) Ax(x in r <-> Et(x in t) & ~(x in x)) 1,EI and 3) r in r <-> (Et(r in t) & ~(r in r)) 2,UI so that 4) ~Et(r in t) 3 and 5) ~(set r) 4 and 6) ~(r=r) 5,C4 so that 7) Ex~(x=x) 6,EG To be continued. --John Hi John, > Your very interesting proof leads to some questions. 1. What is the domain of your variables ? I assume you admit > sets and non-sets. Are physical objects and proper classes and sets > values of the variable x ? 2. Do you claim: Et(x in t & y in t) <-> (set y & set x), or, > Et(x in t)& Et(y in t) .<->. set x & set y. ? C4 AyAx[Az(z in y <-> z in x) -> {(set y & set x) <-> y=x}] > (Equi-membered classes are identical iff these are sets.) C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] Weak 3. Is C3 <-> EyAx[x in y <->. Et(x in t) & set y & A] (with y not free > in A) ? 4. It seems your C4 entails x=x <-> set x. That is, x=x fails for > empirical values (non sets). Where does this leave us with respect to > Russell's descriptions? 5. How do you distinguish: E!x, x=x, set x, Ey(x in y), Ez(z in x) ? Thanks in advance, Witt Variables range over classes. Classes that are elements are sets, > and classes that are not are proper classes. Sets are then existent classes. > I'm uncertain whether > physical objects have any role to play in set theory. Perhaps we can define physical objects as the class of properties that it has, or the class of those classes that it is a member of. > But I do > agree with G. Frege that proper classes can be taken as *models* > of quanta--not proper classes as they are treated by Ullrich and > Chapman, but proper classes .88 la (C3,C4). (I'm sending you Teller's > especially his footnote 6.) I hope I can follow it. > 2. Do you claim: Et(x in t & y in t) <-> (set y & set x), or, > Et(x in t)& Et(y in t) .<->. set x & set y. ? (i) is an instance of C3: (i) EyAx(x in y <-> (Set x & x=w v x x=z)). But (ii) follows from (i,C4) (proof supplied upon request): (ii) EyAx(x in y <-> (x=w v z=z)) I don't see how you use C3 here. On the other hand, from (ii,C4) and Ax(Set x <-> x=x) it > both follows that Et(x in t & y in t) <-> (set y & set x), > and that Et(x in t)& Et(y in t) .<->. set x & set y. > C4 AyAx[Az(z in y <-> z in x) -> {(set y & set x) <-> y=x}] > (Equi-membered classes are identical iff these are sets.) C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] Weak 3. Is C3 <-> EyAx[x in y <->. Et(x in t) & set y & A] (with y not free > in A) ? No. It does not follow that y is a set. To prove that y is a set, > it must be established that every class with the same members as y > is identical to y. In other words, it must be established that > Leibniz's principle of the identity of indiscernibles holds for > every such class and y. (To establish this, it suffices to > establish that this principle holds for y itself.) If Leibniz's > principle does not hold, then y (along with every class equi-membered > with y) is a proper class. I thought: (x in y)->. E!x & E!y. That is, (x in y) is false unless x and y are existent classes (sets). Don't you also need: x=y ->. E!x & E!y & (Fx<->Fy), and E!x -> x=x. Why not define: x=y defined E!x & E!y & AF(Fx<->Fy). And, E!x defined EF(Fx). Are your first order axioms: 1. Fx -> E!x 2. E!x -> x=x 3. x=y ->. E!x & E!y & (Fx<->Fy). 4. It seems your C4 entails x=x <-> set x. That is, x=x fails for > empirical values (non sets). Where does this leave us with respect to > Russell's descriptions? The answer to this question is (I think) provided by the following > critique of Quine by Alberto Cortes: My final comment about variables and names deals with a comment > of Quine's: he has claimed that in a logically austere language, > names are quite superfluous: Chief among the omitted frills is the *name*. This . . . is a > mere convenience and strictly redundant, for the following > reason. Think of 'a' as a name, and think of 'Fa' as equivalent > to '(Ex(a = x & Fx)'. We see from this consideration that 'a' > need never occur except in the context 'a='. But we can as > well render 'a=' always as a simple predicate 'A'. thus > abandoning the name 'a'. 'Fa' gives way thus to '(Ex)(Ax & Fx)', > where the predicate 'A' is *true solely of the object a*. > [Cortes's emphasis]. It may be objected that this paraphrase deprives us of an > assurance of uniqueness that the name has afforded. It is > understood that the name applies to only one object, whereas > the preciate 'A' supposes no such condition. However, we > lose nothing by this, since we can always stipulate by further > sentences, when we wish, that 'A' is true of one and only > one thing. (_Philosophy of Logic_ [Englewood Cliffs, New > Jersey: Prentice Hall, 1970]: p. 25) According to Quine, then, one can always eliminate a name by > concocting an appropriate predicate which is true only of the > object designated by that name. His 'Equals a' predicate can't be expressed without the term a. But, names are superflous anyway. Pure logic, deals without names. I can't see how we can eliminate variables though. e.g. Ay(AxFx -> Fy), Ay(Fy -> ExFx). > But how is one to be assured that the predicate one has concocted > does not apply simultaneously to more than one object? One > cannot merely stipulate that it be so. In virtue of AxE!x, ie. Ax(x=x), we are assured that only a satisfies (a=). Everything exists and is unique for him. > One has to presuppose > that if two objects are numerically distinct they must not > possess all properties in common. And this is nothing less > than PII(L) [PII(L): No two substances are completely sim- > ilar, or differ *solo numero. -JC] Therefore, it is evident > that in order to eliminate proper names, *.88 la Quine*, one has > to presuppose PII(L). Needless to say, we shall not admit > such a presupposition in a critical analysis of PII(L). > Therefore, from the point of view of this paper, proper names, > or their logical equivalents (e.g., prounouns), are logically > fundamental ingredients of both ordinary and formal languages. > (Leibniz's Principle of the Identity of Indiscernibles: A > False Principle, _Philosophy of Science_, Vol.43 Dec., 1976, > p. 496) The eliminability of proper names .88 la Quine--which Cortes > opposes on the grounds that PII(L) fails to hold for quanta-- > marches in lockstep with Russell's Theory of Descriptions. > For Cortes's critique of Quine applies to Russell's > Theory of Descriptions as well. I don't think we can eliminate descriptions, do you? IMHO, I think your free set theory can be produced as an extension of classical logic, i.e. Description theory. EF(~E!(ixFx)), and EF(~((ixFx)=(ixFx))) .. instead of your Ex(~E!x) and Ex(~(x=x)). And classical logic is included in your free set theory: AxE!x -> Ax(x=x), i.e. by restricting your free classes to sets. > 5. How do you distinguish: E!x, x=x, set x, Ey(x in y), Ez(z in x) ? The first three are equivalent. Ez(z in x) and ~Ez(z in x) are > both compatible with the first three. I think E!x <-> Ey(x in y) too. (set x) ->. ~Ez(z in x) <-> x={}. Witt > Thanks in advance, Witt My Thanks to You, > --John Correy PS I am pleased to have been dismissed as a crass formalist > by the likes of Robin Chapman and David Ullrich. Not > uncoincidentally, these are PROFESSORS OF MATHEMATICS! === Subject: Re: David Ullrich on Identity >> PS I am pleased to have been dismissed as a crass formalist >> by the likes of Robin Chapman and David Ullrich. Not >> uncoincidentally, these are PROFESSORS OF MATHEMATICS! I believe David might be, but I am certainly not. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colousounds and shapes. The League of Gentlemen === Subject: Re: David Ullrich on Identity : In point set topology, the singleton {x} is often shortened to : x. Consequently, there is an ambiguity of notation in : mathematics associated with reference to objects. : There is a deeper problem here. If 1 branch can treat x and {x} as though they were indiscernible, and the other CAN'T, then the question must arise: can they BOTH be RIGHT? Or is one of them just BETTER than the other? And is the other one therefore WRONG? :: This presents no problem for mathematics because of how : invariance is characterized in algebraic topology. However, set : theory--and its metaphysics of paradox--is a particular place : where logic and mathematics share interests. Unfortunately, the : people who develop an understanding of set-theoretic identity : within the topological framework find their reasonable : intuitions continually subject to appeal-to-ridicule in logic : communities. That is over-simplified to the point of being bull. What ACTUALLY happens is that people who claim to have an understanding of set-theoretic identity get laughed out of court for the simple reason that there is NO SUCH THING as set-theoretic identity. In set theory, identity (a=b) is re- defined as Ax[xea<->xeb]. Everywhere you see an = sign in set theory, you can simply REPLACE a=b with this macro. In other words, since set-theoretic identity DOES NOT EXIST, there is simply NOTHING to have ANY kind of understanding OF. THAT is why you get ridiculed. : Just as mereotopology can be seen as an extension : of mereology through the addition of some topological : primitive such as connection or interior part, so also set : theory can itself be seen as an extension of mereology : through the addition of the primitive set theoretic notion : of singleton. David Lewis (1991) has shown how, with : the help of this one single notion, all the standard axioms : of set theory can be derived within a mereological : framework. But that is simply an alternative axiomatization of set theory. Why don't you just lay ZFC on one side and mereology + singletons on the other and argue, directly, that one axiomatization is more appealing than the other? : The theory of sets and the 3 theory of : mereological sums (or fusions ) of singletons are, it turns : out, formally indistinguishable. The 3 theory of mereological fusions of singletons is obviously just one of many mereological theories. You could try to say something about why the mereological framework is richer, given that it also includes all these other mereological theories that are 2- theories and 4- theories, instead of just 3- theories. You could further allege that they are richer because they can do more than just fusions. But we're not holding our breath until you get around to that. Especially not in light of the fact that set-theoretical characterizations of mereology are every bit as easy to construct as mereological characterizations of set theory. You are not going to find it easy to support any claim of superior merit as a foundation. : So, what is going on here is fairly straightforward. John falls : into that group of people who understand the identity predicate : within that topological framework. Set theory has a rather simple and straightforward conception of identity. If the other one is not equally simple then he is not going to have a lot of hope of defending it -- : He is trying to defend his : ideas in the place where everyone else thinks the expertise : lies. not here OR ANYwhere. -- --- It's difficult ... you need to be united to have any strength, but internal issues have to be addressed. --- E. Ray Lewis, on liberalism in America === Subject: Re: Sets. > The reason logic has compactness theorems is because language is > topological. You've stated something like that before. What exactly do you mean with 'language is topological'? Can you give a few examples? BTW, i'd say that the reason logic has compactness theorems is because we want our arguments to have only finitely many steps, hence every valid proof needs only finitely many assumptions. But apparantly that thought is too simplistic? Cheers, Herman Jurjus === Subject: Re: Sets. >>However, and to paraphrase the title of a movie, Rebellion must be with >>a cause - a good cause - > >>But in this paradigm, the flexibility has a *limit* >Cause enough. >Since the limit is intrinsic to human (in)ability to formalize infinity >>in its entirety, in the current paradigm, I take it that >>you're proposing to make a revolutionary change to the paradigm. If so, >>what would be your suggestion as an alternative to our >>current mathematics/logic paradigm? >I know you know what mereology is. >Look up Pervin quasi-uniformities. You will find that they relate the >(part-whole) and (part-excluded part) relationships according to a >coordinatization (Look up the relevance of that term in Birkhoff's Lattice >Theory.). Moreover, the specific form of the entourages suggests why >well-founded sets can be specified without reference to a whole. >The reason logic has compactness theorems is because language is >topological. What makes anyone think that any lower logical order could be >foundational? In my previous post, I stated: > But in this paradigm, the flexibility has a *limit*: as any expression is literally finite, we can't never > express infinity in an absolute sense - such as a set that's too big that would contain its own power > set - and still hope human mathematical reasoning is still consistent. Imho, literally finite [sequence of symbols] ... can't never express infinity in an absolute sense would be very much reminiscent to whatever the limit GIT is about, wouldn't you agree? What I'm trying to say is that this limit is trans-formal-system and is inherent in the current paradigm of using finite sequence of symbols to represent/interpret infinity. And there is nothing we can do to change this limit - that is, nothing unless the paradigm itself is changed [whatever this means or implies!]. But using another formal system of the very same paradigm would help to fix the limit. I think we're talking about 2 different kinds of limits here. ---Nam >:-) >mitch > === Subject: Re: Sets. > Imho, literally finite [sequence of symbols] ... can't never > express infinity in an absolute sense would be > very much reminiscent to whatever the limit GIT is about, wouldn't you > agree? What I'm trying to say is that > this limit is trans-formal-system and is inherent in the current > paradigm of using finite sequence of symbols > to represent/interpret infinity. And there is nothing we can do to > change this limit - that is, nothing unless > the paradigm itself is changed [whatever this means or implies!]. But > using another formal system of the very same > paradigm would help to fix the limit. I meant: 'would *not* help to fix the limit.' === Subject: What is a nef divisor? (and other weird algebro-geometric terms) 'nef' has been annoying me the most, so if anyone knows what a nef divisor is, please tell, but there are some others that would be handy, if anyone can spare the time... What's a klt divisor (or indeed a klt variety)? What does it mean for a variety or divisor to be 'log' something? (e.g. log Fano) What's a Fano variety? Thanks in advance. === Subject: Do not waste space repeating messages Please. It's not necesary to copy the full message we want to respond. It's boring to re-read the thread all time and it is frightening to read: It follows 200 lignes. Better: Extract the interesting lignes (with CTRL-INS), erase the rest of messages and translate to your letter (With SHIFT-INS)the extracted paragraph. Thanks === Subject: Re: Do not waste space repeating messages and *always* top-post -- screw Miss Manners and her ilk, if when they're tso silly. > Better: Extract the interesting lignes (with CTRL-INS), erase the rest > of messages and translate to your letter (With SHIFT-INS)the extracted > paragraph. Thanks --les ducs d'Enron! http://larouchepub.com === Subject: simulate the departure process In a stable queueing system, the input process is poisson process, the service time is exponential distribution. I hope to simulately find the inter-departure rate of the departure process. Could you plz give the link or reference for this question? Thanks in advance. === Subject: Re: simulate the departure process > In a stable queueing system, the input process is poisson process, the > service time is exponential distribution. I hope to simulately find the > inter-departure rate of the departure process. Could you plz give the link > or reference for this question? Thanks in advance. lol - Queueing Systems, by Kleinrock should contain the answers to all questions of this sort.... btw: in your post, is service time different from the departure process? cdj === Subject: Re: simulate the departure process lol - Queueing Systems, by Kleinrock should contain the answers to > all questions of this sort.... btw: in your post, is service time different from the departure > process? cdj Thanks for your information. I have this book, but actually I need to get the simulation result , not the theory result. The service time is for the customer servince duration after accepted in the system. Theoritically, the inter-departure rate is same as the input arrival rate of the customer. I hope to simulate this procedure and get the inter-departure rate since in the future i will extend such simple scenario to a complicate one. Thanks. === Subject: Re: simulate the departure process lol - Queueing Systems, by Kleinrock should contain the answers to > all questions of this sort.... btw: in your post, is service time different from the departure > process? cdj > Thanks for your information. I have this book, but actually I need to get the simulation result , > not the theory result. The service time is for the customer servince > duration after accepted in the system. Theoritically, the inter-departure rate is same as the input arrival > rate of the customer. I hope to simulate this procedure and get the > inter-departure rate since in the future i will extend such simple > scenario to a complicate one. Thanks. #include #include #include #define ATIME 1 /* average arrival interval */ #define STIME 0.5 /* average service time */ #define N 10000 double erand(double av) /* exponentially distributed rv with average av */ { return -log(1-rand()/(RAND_MAX+1.))*av; } int main() { int i; double Tq=0, /* arrival time of next customer */ Td1, /* departure time of first customer */ Td, /* current departure time */ depdiff, /* current departure interval */ r; srand(time(0)); /* assume first customer arrived at T=0 and service free */ Td1=Td=erand(STIME); /* process next N-1 customers/services */ for (i=2;i<=N;i++) { Tq+=erand(ATIME); r=erand(STIME); if (Tq<=Td) /* immediate service */ depdiff=r; else /* delayed service */ depdiff = Tq+r-Td; Td+=depdiff; } printf(Average departure interval for %d customers: %8.5fn, N,(Td-Td1)/(N-1)); return 0; } -- Horst === Subject: Logarithmic/exponential equation question I need to express this equation in terms of y=f(x), but I simply don't know how. Any help would be much appreciated (A and B are constants): x = (1/A) * (y - 10*log(B - 10^(y/10))) Andrew Milne === Subject: Re: Logarithmic/exponential equation question > I need to express this equation in terms of y=f(x), but I simply don't know > how. Any help would be much appreciated (A and B are constants): x = (1/A) * (y - 10*log(B - 10^(y/10))) y = A*x + log(B) - log(1+10^(A*x/10)) Steps: y/10-A*x/10 = log(B-10^(y/10)) 10^(y/10)*10^(-A*x/10) = B - 10^(y/10) Solve this equation for 10^(y/10) (it is linear). Then take the log and multiply by 10. === Subject: Re: Logarithmic/exponential equation question > I need to express this equation in terms of y=f(x), but I simply don't know > how. Any help would be much appreciated (A and B are constants): x = (1/A) * (y - 10*log(B - 10^(y/10))) y = A*x + log(B) - log(1+10^(A*x/10)) Steps: y/10-A*x/10 = log(B-10^(y/10)) 10^(y/10)*10^(-A*x/10) = B - 10^(y/10) Solve this equation for 10^(y/10) (it is linear). Then take the log > and multiply by 10. Everything said upto here is good. Yet, by collecting the factor 10^(y/10), one obtains 10^(y/10)(10^(-A*x/10)+1) = B then 10^(y/10) =B/(10^(-A*x/10)+1)) By taking logarithms(base 10 assumed), one obtains y/10= log(B) -log(10^(-A*x/10)+1) and finally, y=10 log(B) -10 * log( 10^(-A*x/10)+1) === Subject: Re: Logarithmic/exponential equation question >I need to express this equation in terms of y=f(x), but I simply don't know >how. Any help would be much appreciated (A and B are constants): >x = (1/A) * (y - 10*log(B - 10^(y/10))) I presume your log is base 10, and B > 10^(y/10). Hint: write this as A*x/10 = y/10 - log(B - 10^(y/10)), simplify 10^(left side) = 10^(right side), and solve for 10^(y/10). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Logarithmic/exponential equation question There is no symbolic result. At least Mathcad tells me there is none. I think this is because there is no rule: log(p + r) = anything.in terms of powers or logs even if r=10^(y/10) and p=y there is no rule I know of for log(f(y) + g(y)) reducing the problem as far as possible with algebra gives >x = (1/A) * (y - 10*log(B - 10^(y/10))) Ax = (y -10*log(B = 10^(y/10) ) ) y-Ax = 10*log (B - 10^(y/10) ( y - Ax ) / 10 = log (B - 10^(y/10)) I assume you mean log base ten, or can convert. (y - Ax) / 10 = log base 10 (10 ^ (log base ten B) - 10^(y/10) ) or log (10^log B - 10^(y.10)) = z, which is that number z=(y-Ax)/10 to which 10 must be raised to give 10^logB - 10^(y/10) so 10 ^ z = (10^logB -10^y/10) 10 = zth root of (10^logB -10^y/10) so if z is a nonzero integer there are z values of (y-Ax)/10 and there should be only one so z is irrational or maybe rational (How many values for 4.112^(3/2)? Well (4.112^1/2)^3 has two, but (4.112^3)^(1/2) has two so if z is rational, then , a contradiction.) So z is irrational so either A, x, or y or more than one is irrational which doesn't rule out too much so if that's true, 10^log B - 10^(y/10) has only one value, which gives 10^q -10^r = p, so if p = 0, which it isn't because 0 isn't irrational, then 10^q = 10^r and r = q and y=10*B but it isn't so if p is nonzero and irrational then B - 10^(y/10)=p might have a solution but it won't be p=0 so B # 10^(y/10) if (y-Ax)/10 is an irrational so we can rule out one value of y in terms of constant B, that is y#10*log B if z irrational so log B # y/10 y#10*logB so if that's true, and it only rules one value of y, then so if (y-Ax)/10 is irrational and y#10*logB then (y-Ax/10) = log (10^log B - 10^(y.10)) and if x or A is irrational, and with a guess value for y either root( (y-Ax/10) - log (10^log B - 10^(y.10)) ,y) will give the solution root( 1 - ((y-Ax/10) / log (10^log B - 10^(y.10))) , y) will do it, but any computer uses finite precision, so I'll try the both in Mathcad. On both, I used x, A, B, and the guess for y each = rnd(10)-5, which is different each times it's evaluated so it's unlikely B=10*log(y) Most of the time it was not converging, but either could be not converging or overflow or a solution. Usually, the first produced an answer but the second didn't. I didn't see the second produce an answer while the first did not. When they both produced an answer, they were in near agreement. That's pretty amazing for a computer than can't even represent an irrational number. When I changed x, A, and B to the square root of a random integer between -5 and 5, I got loads of imaginary values for y. What fun! Yours, Doug Goncz (at aol dot com) Replikon Research Read the RIAA Clean Slate Program Affidavit and Description at http://www.riaa.org/ I will be signing an amended Affidavit soon. === Subject: Groups with 16 elements Groups with 16 elements; I am trying to determine all the groups of order 60 or less. My first difficulty is for order n=16. As an analogy, for n=8, the groups are Z_2 x Z_2 x Z_2 ; Z_2 x Z_4 ; Z_8. The 2 non-Abelian groups are D_4, the symmetry group of a square, and the quaternions Q. D_4 is generated by 2 elements x such the x^4 = 1 (x is a rotation of the square by 90 deg.), and y^2 = 1 (y is a reflection). If the corners of the square are labeled 1,2,3, 4, D_4 can be represented as a subgroup of the perm. group on 4 elements, S_4, by x = (1234), y = (13) in cycle notation. The subgroup X generated by x is X = (x,x^2,x^3,X^4 = 1), which is a normal subgroup of D_4, so if g is any element of D_4, the inner automorphism gxg^-1 is also in X. (This is important in determining finite groups in general). D_4 is determined by the relation yxy^-1 = x^3 = x^-1. The quaternions Q are also generated by 2 elements x and y, determined by x^4 = 1 = y^4 = z^4 ; x^2 = y^2 = z^2 = a where z = xy ; and yxy^-1 = x^-1 ; xyx^-1 = y^-1. (a is sometimes written as -1, since a^2 = 1.) To return to the case n = 16, when n = 2m is even, D_m, the dihedral group of symmetries of a polygon with m = n/2 sides, is always one of the non-Abelian groups. It is given by x^m = 1 = y^2 ; yxy^-1 = x^-1 ; for n = 16 we have D_8 given by x^8 = 1 = y^2 ; yxy^-1 = x^7 = x^-1. There are 14 (non-isomorphic) groups with n = 16. 5 are Abelian, and 9 non-Abelian. The 5 Abelian groups are Z_2 x Z_2 x Z_2 x Z_2 ; Z_2 x Z_2 x Z_4 ; Z_4 x Z_4 ; Z_2 x Z_8 ; Z_16 . We can build 2 non-Abelian groups from n = 8; Z_2 x D_4 ; Z_2 x Q. We have D_8 as one non-Abelian group. In analogy with quaternions, consider the group (call it K) generated by 2 elements of order 8 i.e. x^8 = y^8 = 1, and x^4 = y^4 = b, z = xy. The subgroups X = == (1,x,x^2,x^3,x^4,x^5,x^6,x^7) and Y = == (1,y,y^2,y^3,y^4,y^5,y^6,y^7) are both normal, as any subgroup containing 1/2 the elements must be. Consider the 2 inner automorphisms yxy^-1 = x^r ; xyx^-1 = y^s. It can be shown that r and s must by in Z_8* == the multiplicative group mod 8, Z_8* = (1,3,5,7). Note that 3^2 = 5^2 = 7^2 = 1 mod 8, and it can be shown that y^2 x y^-2 = x^(r^2) = x (since r^2 = 1 mod 8), and similarly for x^2 y x^-2 = y^(s^2) = y, so that x^2 = y^2 == a, and x^4 = y^4 = a^2, are all in the center Z of K. Since x^2 and y^2 are in Z, y x^2 y^-1 = x^(2r) = x^2 and x y^2 x^-1 = y^(2s) = y^2, so (r,s) = (5,5). r = s = 1 give an Abelian group, and r or s = 3 or 7 (7 = -1 mod 8) give y x^2 y^-1 = x^(2r) = x^6 = x^2 ==> x^4 = 1, which contradicts the assumption that 8 is the smallest integer m such that x^m = 1. Similarly for y^2. Let a == x^2 = y^2. Then the 16 elements of the group are X = = (1,x,a,ax,a^2,xa^2,a^3,xa^3) Y = = (1,y,a,ay,a^2,ya^2,a^3,ya^3) Z = = (1,a,a^2,a^3) is the center. xy, yx = x^5 y = a^2 xy, x^3 y = axy, x^7 y = a^3 xy. I haven't checked the following, but some possible groups with 16 elements might be given by x^4 = y^4 = z^2 = 1, and This is only 5 of the 9 non-Abelian groups of order 16--there are 4 more. Can anyone help me find them? I wonder if D_8 can be generalized as follows: x^8 = 1 = y^2 ; yxy^-1 = x^r r = -1 = 7 mod 8 gives D_8. r = 3 and r = 5 give 2 generalizations, which look OK to me. This makes 7 of the 9 non-Abelian groups. The case x^8 = 1 = y^4 might work as follows; yxy^-1 = x^r, r = 3, 5, or -1 = 7 mod 8, since X == is a normal subgroup. The center Z = (1,y^2,x^4,y^2 x^4) when r = 3 or -1. When r = 5, Z = (1,x^2,x^4,x^6,y^2,x^2 y^2, x^4 y^2,x^6 y^2). One might also be able to find groups with 3 generators: x^4 = 1 = y^4 = z^2 (generalizing Z_2 x Q) or x^4 = 1 = y^2 = z^2 (generalizing Z_2 x D_4). I don't think there is a generalization of Z_2 x D_4 like this, however. All of these have to be looked at more carefully. Van Jacques ; vanjac12@yahoo.com === Subject: Re: Groups with 16 elements Visiting Assistant Professor at the University of Montana. [.snip.] >We have D_8 as one non-Abelian group. In analogy with quaternions, >consider the group (call it K) generated by 2 elements of order 8 >i.e. x^8 = y^8 = 1, and x^4 = y^4 = b, z = xy. >The subgroups X = == (1,x,x^2,x^3,x^4,x^5,x^6,x^7) >and Y = == (1,y,y^2,y^3,y^4,y^5,y^6,y^7) >are both normal, as any subgroup containing 1/2 the elements must be. >Consider the 2 inner automorphisms yxy^-1 = x^r ; xyx^-1 = y^s. >It can be shown that r and s must by in Z_8* == the multiplicative >group mod 8, Z_8* = (1,3,5,7). Note that 3^2 = 5^2 = 7^2 = 1 mod 8, >and it can be shown that y^2 x y^-2 = x^(r^2) = x >(since r^2 = 1 mod 8), and similarly for x^2 y x^-2 = y^(s^2) = y, >so that x^2 = y^2 == a, and x^4 = y^4 = a^2, are all in the center Z of K. I'm sorry, but on what grounds do you assert that x^2 = y^2 in this case, at this point? You can, but you haven't shown it yet. We are clearly assuming that is not equal to ; so at least one element of is not in X; since K/ is of order 2, that element of represents the unique nontrivial element of K/. It cannot be an even power of y, so it must be an odd power of y, and so you may assume that y maps to the unique nontrivial element of K/. In particular, y^2 must lie in X, so y^2 = x^r for some r. Looking at the orders we may conclude that r=2 or 6, and so if necessary by replacing y by y^{-1} we conclude that y^2=x^2, as claimed. >Since x^2 and y^2 are in Z, y x^2 y^-1 = x^(2r) = x^2 and > x y^2 x^-1 = y^(2s) = y^2, so (r,s) = (5,5). >r = s = 1 give an Abelian group, and r or s = 3 or 7 (7 = -1 mod 8) >give y x^2 y^-1 = x^(2r) = x^6 = x^2 ==> x^4 = 1, which contradicts >the assumption that 8 is the smallest integer m such that x^m = 1. >Similarly for y^2. >Let a == x^2 = y^2. >Then the 16 elements of the group are >X = = (1,x,a,ax,a^2,xa^2,a^3,xa^3) >Y = = (1,y,a,ay,a^2,ya^2,a^3,ya^3) >Z = = (1,a,a^2,a^3) is the center. >xy, yx = x^5 y = a^2 xy, x^3 y = axy, x^7 y = a^3 xy. >I haven't checked the following, but >some possible groups with 16 elements might be given by >x^4 = y^4 = z^2 = 1, and >This is only 5 of the 9 non-Abelian groups of order 16--there >are 4 more. >Can anyone help me find them? >I wonder if D_8 can be generalized as follows: >x^8 = 1 = y^2 ; yxy^-1 = x^r r = -1 = 7 mod 8 gives D_8. >r = 3 and r = 5 give 2 generalizations, which look OK to me. You get the semidirect products of Z_8 by Z_2: for every morphism f:Z_2->Aut(Z_8), you get a group by taking all pairs (x,y) x in Z_8, y in Z_2, with multiplication given by: (x,y)*(z,w) = (x*f(y)(z), yw). (Note that f(y) is in Aut(Z_8), so it makes sense to write f(y)(z)). There are four automorphisms of exponent 2 in Aut(Z_8): the identity, the one mapping x to x^3, to x^5, and to x^7. The first gives you the direct product (abelian), the other three give you nonabelian groups. When you take x|->x^7 you get the dihedral group. The only thing is to make that the three are pairwise non-isomorphic, which is not hard. You can do similar things by taking two groups of order 4 and constructing their semidirect products. Setting all this aside, another way to deal with this is to consider the center of the group and to proceed from there. A group of order 16 must have a nontrivial center; if the group is nonabelian, the center cannot be of order 8 or 16, so it must be of orders 2 (hence cyclic of order 2), or 4 (hence either cyclic of order 4 or the Klein group). The ones with center 4 are more easily handled, since then G/Z(G) is abelian of order 4; so you can consider the case when (1) Z(G) is cyclic, G/Z(G) is cyclic, both order 4. (2) Z(G) is cyclic, G/Z(G) is the Klein group (3) Z(G) is the Klein group, G/Z(G) cyclic. (4) Z(G) and G/Z(G) are both isomorphic to the Klein group. and see if you can determine the structure from there. For groups whose center is cyclic of order 2, you have a number of possibilities for the quotient, as it must be a group of order 8. Once again, you can deal with those and see what is possible. ============================================================== ======== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ============================================================== ======== Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Groups with 16 elements ... You get the semidirect products of Z_8 by Z_2: for every morphism > f:Z_2->Aut(Z_8), you get a group by taking all pairs (x,y) x in Z_8, y in Z_2, with multiplication given by: (x,y)*(z,w) = (x*f(y)(z), yw). (Note that f(y) is in Aut(Z_8), so it makes sense to write f(y)(z)). There are four automorphisms of exponent 2 in Aut(Z_8): the identity, > the one mapping x to x^3, to x^5, and to x^7. The first gives you the > direct product (abelian), the other three give you nonabelian > groups. When you take x|->x^7 you get the dihedral group. The only > thing is to make that the three are pairwise non-isomorphic, which is > not hard. You can do similar things by taking two groups of order 4 and > constructing their semidirect products. Setting all this aside, another way to deal with this is to consider > the center of the group and to proceed from there. A group of order 16 > must have a nontrivial center; if the group is nonabelian, the center > cannot be of order 8 or 16, so it must be of orders 2 (hence cyclic of > order 2), or 4 (hence either cyclic of order 4 or the Klein group). The ones with center 4 are more easily handled, since then G/Z(G) is > abelian of order 4; so you can consider the case when (1) Z(G) is cyclic, G/Z(G) is cyclic, both order 4. > (2) Z(G) is cyclic, G/Z(G) is the Klein group > (3) Z(G) is the Klein group, G/Z(G) cyclic. > (4) Z(G) and G/Z(G) are both isomorphic to the Klein group. and see if you can determine the structure from there. For groups whose center is cyclic of order 2, you have a number of > possibilities for the quotient, as it must be a group of order 8. Once > again, you can deal with those and see what is possible. > your post was very helpful. I recall a theorem that G/Z cyclic implies G Abelian. I haven't proved it, but I found it in Hungerford p.92. Then cases (1) and (3) are Abelian. Van Jacques vanjac12@yahoo.com === Subject: Re: Who here believes maths is all there is? Warning: I never posted this! Must be a NG mix-up. Lurch > Some see it as Literature, some see it as gospel. There are too many > logical flaws to be gospel, IMHO. > Lurch > cj-bubba@mindspring.com asks: Where did Noah put the dinosaurs? > Good question, and well worth treating in a logical way; > That story has its value, but that value is not entirely historical fact, > but > more preferably a story to illustrate ethics and conduct in a literary > artistic > way. The authors(?) of the story did not know about dinosaurs. Math occurs plainly in a few ways in that great story: pi, and the cubit. > Maybe other ways, too. G C === Subject: Re: Who here believes maths is all there is? > Perhaps God said: > Let there be the empty set and the set inclusion operator. > The rest just sort of follows ... Recently I was meditating on the number of YHVH, God, Adonai, Tao, pick your label. It started when I saw the movie Pi again, this time on video. There's a scene in the movie in which the main character, an agnostic Jew, is pulled into putting on tefilin by a member of Chabad, and they say the Sh'ma. One of the things that struck me a lot later was while some of the characters in the movie talk about a 216 digit name of God, and the main character is taken in by very complex numbethe sh'ma talks about God in terms of one number, a simple number, namely, the number One. While I worked over the number One, Echad, as an attribute of Adonai, I came to several realizations. One thing that I came to realize is that on one level, we experience the Tao as...well, at first, I thought that the number was infinity. But really, this is an intellectualization. On a deeper level, I realized that this experience is not infinity, but the set of all numbers. It is this set that is the true number of the Tao -- experienced on one particular level. All the minutia and stuff and bits and separateness that make up the universe as each one of us experiences and does our stuff in it across the galaxies. But when we say the Sh'ma, we are saying God is not the set of all numbewe say that God is a particular number, and that number is one. I was starting to have severe difficulties with this. Because as soon as we say God is one, it implies a duality, the duality of the one vs the zero. So I then I thought, well, then God is not one, God is zero. And then I realized that that wasn't quite it. Rather, God is the empty set. The set of no numbers. And this felt right to me, as I sat and breathed with that. As it is written in the Tao Te Ching (4) The Tao is an empty bowl, yet it may be used without ever needing to be filled. It is the deep and unfathomable source of infinite things. Claire Petersky Home of the meditative cyclist: http://home.earthlink.net/~cpetersky/Welcome.htm Books just wanna be FREE! See what I mean at: http://bookcrossing.com/friend/Cpetersky === Subject: Do you know this moment? Do you know the following expression to be something meaningful or that we can relate to something meaningful: E(n^2)-E(n) ------------------ (E(n))^2 where E is expectation of a random variable (E(n)=sum_n n P(n) for some probability distribution P). How would you prove that such a quantity equals about 2 for those P(n) that have maximum value about 0 (like geometric serie) and that it equals about 2 for those whose maximum is off-centered (gaussian looking)? Cheers. === Subject: the use of geometric shapes for generating random number Value of Geometric shapes in Generation of Random Numbers In attempting to explain the EPR paradox to myself several years ago, I developed a matrix using all the permutations of the letters YZAX. I went along and gave a number to each permutation that was similar, namely place shifted. yxaz1 xazy1 azyx1 zyxa1 yxza2 xayz5 azxy3 zyax4 yazx3 xzya4 ayxz2 zxay5 yaxz4 xzay2 ayzx5 zxya3 yzxa5 xyaz3 axzy4 zayx2 yzax6 xyza6 axyz6 zaxy6 Later on I went on to make a website using the matrix. It was up for a couple years until NBCi pulled the plug. I lost the password to the original website after fighting with a hacker to gain control of the damn thing. Recently I decided to do something new with it. I created a BASIC program that randomly went through the maze. I was pleased with myself and ran it a few times but thought nothing of it. Then the other day I went through one of my paranoid states and that perhaps The Who's Pinball Wizard and Radiohead's OK Computer was about the program. That it was like a pinball machine. That it was OK, not great or perfect. I began to dwell on what would make it better. I had another matrix in my notes and thought perhaps that one would be. yxaz1 xazy1 azyx1 zyxa1 yxza2 xzay2 ayzx5 zxay5 yazx3 xzya4 azxy3 zyax4 yaxz4 xyaz3 axzy4 zxya3 yzxa5 xayz5 ayxz2 zayx2 yzax6 xyza6 axyz6 zaxy6 I did not recollect constructing this matrix and thought it better for two reasons. First, I did not consider it mine. Second, roughly speaking the odds were one half and not one quarter. After looking at it further, I realized it was redundant and only needed two columns. This brought me to the question of using smaller binary random statements in the BASIC program. It dawned on me this might increase the randomness. I had tried to make a program using the second set, but it was to big to compile using the trial version of the software I had. While fooling around with that though, I ran across the idea of printing the sixth, twelfth, or twenty fourth step. In an effort to make a random character generator out of it. I stumbled with the set trying to determine if every permutation could be selected equally. Unfortunately it could not. Around that time I got into a discussion with my friend Henry about infinite paths. I said I felt having at least three was important. He said the smallest thing with three infinite paths would be a triangle. It dawned on me a while later Hey not only is it three, but its binary decisions Tonight it occurred to me that with 3 steps, using three binary random statements, I had the equal opportunity of three outcomes. >starting with A the likelyhood to land on b or c is 3/4 while the likely hood of landing on A is 1/4. as there are 8 combinations using three steps. This tendency to roll as a new staring point happens every three steps, keeps the overall likelyhood of any one character probably very near 1/24 the five pointed star seems to do something weird. the top of the star seems to be 1/5 the bottom two points 3/10 and the middle two 1/10 this seems to set up a roll that has a tendency to go the same way it is typically drawn oddly enough. I havent worked out the likelyhood for that yet. perhaps these things arent any more random than stacking binary random statements, but it is the infinite paths that seem to intrigue me for some reason. === Subject: Topological questions These should be rather easy for any topologist. I'm just checking if my results are correct. Let R be the set of reals and Q the set of rationals. I've managed to conclude that boundary(Q) is RQ and boundary(RQ) is Q. This is done with the following chain: (i) interior(A) is a subset of A. (ii) R and RQ are disjoint. i & ii => (iii) interior(R) and interior(RQ) are disjoint. (iv) R and RQ are both dense in R, which means that their closure is R. (v) closure(A) = interior(A) union boundary(A). (vi) interior(A) and boundary(A) are disjoint. ii, iv, v & vi => (vii) boundary(Q)=RQ and boundary(RQ)=Q. Is this right? (A is assumed to mean any subset of R.) Now, I spent hours (almost, anyway) trying to prove that Q is discrete, when it struck me that it isn't. Because Q is dense, and as I have earlier written here, discrete dense sets in R are impossible. One of my homework problems asked me to prove that Q is 0-dimensional. I think I managed to prove that if x is a point in Q and U is a neighbourhood of x, then there exists a positive irrational r so that ]x-r, x+r[ is a subset of U and also a neighbourhood of x. Now the endpoints of this interval ]x-r, x+r[ are not elements of Q, and for any point y in either this interval we can choose a positive rational s so that ]y-s, y+s[ does not intersect this interval and for any point y inside this interval we can choose a positive rational s so that ]y-s, y+s[ is a subset of this interval. This means that the interval ]x-r, x+r[ has an empty boundary and thus because x and U are arbitrary, Q is 0-dimensional. At least that's how I *think* it goes. I may have skipped over some fuzzy details. Is my answer correct? Thanks for your help! -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ He said: 'I'm not Elvis'. Who else but Elvis could have said that? - ALF === Subject: Re: Topological questions > -----Original Message----- > Conversation: Topological questions > === Subject: Topological questions > These should be rather easy for any topologist. I'm just > checking if my > results are correct. Let R be the set of reals and Q the set of rationals. I've managed to > conclude that boundary(Q) is RQ and boundary(RQ) is Q. This is done > with the following chain: > (i) interior(A) is a subset of A. > (ii) R and RQ are disjoint. > i & ii => (iii) interior(R) and interior(RQ) are disjoint. > (iv) R and RQ are both dense in R, which means that their > closure is R. > (v) closure(A) = interior(A) union boundary(A). > (vi) interior(A) and boundary(A) are disjoint. > ii, iv, v & vi => (vii) boundary(Q)=RQ and boundary(RQ)=Q. > Is this right? (A is assumed to mean any subset of R.) Now, I spent hours (almost, anyway) trying to prove that Q is > discrete, when it struck me that it isn't. Because Q is > dense, and as I > have earlier written here, discrete dense sets in R are impossible. One of my homework problems asked me to prove that Q is > 0-dimensional. I > think I managed to prove that if x is a point in Q and U is a > neighbourhood of x, then there exists a positive irrational r so that > ]x-r, x+r[ is a subset of U and also a neighbourhood of x. > Now the endpoints of this interval ]x-r, x+r[ are not elements of Q, > and for any point y in either this interval we can choose a positive > rational s so that ]y-s, y+s[ does not intersect this interval and for > any point y inside this interval we can choose a positive > rational s so > that ]y-s, y+s[ is a subset of this interval. > This means that the interval ]x-r, x+r[ has an empty boundary and thus > because x and U are arbitrary, Q is 0-dimensional. > At least that's how I *think* it goes. I may have skipped over some > fuzzy details. Let me suggest a more useful form of 0-dimensional: X is 0-dimensional iff X has a neighborhood base consisting of clopen sets. Trivially your proof is correct. More to the point, BB = {(x,y) / Q | x,y in P = RQ} is a clopen base (showing open, closed and base are left to the reader). === Subject: Re: Topological questions > These should be rather easy for any topologist. I'm just checking if my > results are correct. Let R be the set of reals and Q the set of rationals. I've managed to > conclude that boundary(Q) is RQ and boundary(RQ) is Q. This is done > with the following chain: > (i) interior(A) is a subset of A. > (ii) R and RQ are disjoint. Am I misunderstanding your notation? Is AB the set difference of A and B, consisting of points in A but not in B? In that case every irrational is a member of both R and RQ, so they are far from disjoint. In any event, the boundary of Q is all of R. That's because every point of RQ has about it a neighborhood that contains a rational. And every point of Q is contained in a neighborhood that contains an irrational. In other words every neighborhood of every point of R contains both rationals and irrationals. So every point of R is a boundary point of Q. === Subject: Re: Topological questions >> These should be rather easy for any topologist. I'm just checking if my >> results are correct. >> Let R be the set of reals and Q the set of rationals. I've managed to >> conclude that boundary(Q) is RQ and boundary(RQ) is Q. This is done >> with the following chain: >> (i) interior(A) is a subset of A. >> (ii) R and RQ are disjoint. > Am I misunderstanding your notation? Is AB the set difference of A and > B, consisting of points in A but not in B? In that case every > irrational is a member of both R and RQ, so they are far from disjoint. No, you are not misunderstanding. That was a typo. I meant Q and RQ are disjoint. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ A friend of mine is into Voodoo Acupuncture. You don't have to go into her office. You'll just be walking down the street and... ohh, that's much better! - Stephen Wright === Subject: Re: Topological questions Visiting Assistant Professor at the University of Montana. >These should be rather easy for any topologist. I'm just checking if my >results are correct. >Let R be the set of reals and Q the set of rationals. I've managed to >conclude that boundary(Q) is RQ and boundary(RQ) is Q. What's your definition of boundary? If you define x is a boundary point of A if and only if (a) x is not an interior point of A; and (b) every nbd of x intersects A nontrivially. (that is, bound(A) = cl(A)-int(A)) then your conclusion would be wrong. On the other hand, you could define it as bound(A)=A-int(A), in which case it would also be wrong. And finally, you could define it as bound(A)=cl(A)-A, in which case your conclusion would be correct. > This is done >with the following chain: >(i) interior(A) is a subset of A. >(ii) R and RQ are disjoint. You mean Q and RQ, surely. >i & ii => (iii) interior(R) and interior(RQ) are disjoint. Yes. (In fact, they are both empty). >(iv) R and RQ are both dense in R, which means that their closure is R. Again, I suspect you mean Q and RQ, although the statement is true as written. (-: >(v) closure(A) = interior(A) union boundary(A). >(vi) interior(A) and boundary(A) are disjoint. This suggests that your definition makes bound(A) equal to closure(A)-interior(A). Is this correct? >ii, iv, v & vi => (vii) boundary(Q)=RQ and boundary(RQ)=Q. No, because you have figured out interior(Q) and interior(RQ) wrong. What is interior(Q)? x is in int(A) if and only if there exists an open set containing x and contained in A. Given any rational, and every open ball around the rational, there are irrationals in that open ball, so no rational is an interior point of Q. Therefore, int(Q) is empty; likewise, int(RQ) is empty. If bound(A)=closure(A)-interior(A), then in both cases you get bound(Q) = clos(Q)-int(Q) = R-{} = R bound(RQ) = clos(RQ) - int(RQ) = R-{} = R. ============================================================== ======== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ============================================================== ======== Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Topological questions Arturo Magidin scribbled the following: >>These should be rather easy for any topologist. I'm just checking if my >>results are correct. >>Let R be the set of reals and Q the set of rationals. I've managed to >>conclude that boundary(Q) is RQ and boundary(RQ) is Q. > What's your definition of boundary? If you define > x is a boundary point of A if and only if > (a) x is not an interior point of A; and > (b) every nbd of x intersects A nontrivially. > (that is, bound(A) = cl(A)-int(A)) > then your conclusion would be wrong. On the other hand, you could > define it as bound(A)=A-int(A), in which case it would also be > wrong. And finally, you could define it as bound(A)=cl(A)-A, in which > case your conclusion would be correct. I was trying for the bound(A)=cl(A)-int(A) definition. But I might have got the theory mixed up. >> This is done >>with the following chain: >>(i) interior(A) is a subset of A. >>(ii) R and RQ are disjoint. > You mean Q and RQ, surely. Of course. >>i & ii => (iii) interior(R) and interior(RQ) are disjoint. > Yes. (In fact, they are both empty). >>(iv) R and RQ are both dense in R, which means that their closure is R. > Again, I suspect you mean Q and RQ, although the statement is true as > written. (-: Of course I mean Q and RQ. I wish there was an universally accepted symbol for the set of irrationals so I would not have to write RQ. >>(v) closure(A) = interior(A) union boundary(A). >>(vi) interior(A) and boundary(A) are disjoint. > This suggests that your definition makes bound(A) equal to > closure(A)-interior(A). Is this correct? Yes, the boundary is the closure minus the interior. >>ii, iv, v & vi => (vii) boundary(Q)=RQ and boundary(RQ)=Q. > No, because you have figured out interior(Q) and interior(RQ) wrong. > What is interior(Q)? x is in int(A) if and only if there exists an > open set containing x and contained in A. Given any rational, and > every open ball around the rational, there are irrationals in that > open ball, so no rational is an interior point of Q. Therefore, int(Q) > is empty; likewise, int(RQ) is empty. If > bound(A)=closure(A)-interior(A), then in both cases you get > bound(Q) = clos(Q)-int(Q) = R-{} = R > bound(RQ) = clos(RQ) - int(RQ) = R-{} = R. Yes. Of course. How stupid of me. I kept thinking that an interior point of X is *always* an element of X, an exterior point of X is *never* an element of X, and a boundary point can be either (which is correct) but did not think further than that. In fact... is it true that any set that is both dense and codense interior and an empty exterior, and its boundary fills the entire space? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ I am not very happy acting pleased whenever prominent scientists overmagnify intellectual enlightenment. - Anon === Subject: Re: Topological questions Visiting Assistant Professor at the University of Montana. >Arturo Magidin scribbled the following: >These should be rather easy for any topologist. I'm just checking if my >results are correct. >Let R be the set of reals and Q the set of rationals. I've managed to >conclude that boundary(Q) is RQ and boundary(RQ) is Q. >> What's your definition of boundary? If you define >> x is a boundary point of A if and only if >> (a) x is not an interior point of A; and >> (b) every nbd of x intersects A nontrivially. >> (that is, bound(A) = cl(A)-int(A)) >> then your conclusion would be wrong. On the other hand, you could >> define it as bound(A)=A-int(A), in which case it would also be >> wrong. And finally, you could define it as bound(A)=cl(A)-A, in which >> case your conclusion would be correct. >I was trying for the bound(A)=cl(A)-int(A) definition. But I might >have got the theory mixed up. I don't know what the standard term is (not being a topologist), but I've seen all those notions bandied about. Or even whether there is a standard term. Maybe some topologist can help on that. Sometimes you have terms like accumulation points for stuff in cl(A)-A, and so on. So just as long as we're clear on what you mean... [.snip.] >In fact... is it true that any set that is both dense and codense >interior and an empty exterior, and its boundary fills the entire >space? Yes: Note that A is dense if and only if X-A has empty interior. For if A is dense, then every nonempty open set intersects A, so the only open set contained in X-A is the empty set. Conversely, if X-A has empty interior, then every nonempty set intersects A, so A is dense. So A is dense if and only if it has an empty exterior. So if A is dense, then int(X-A) = empty, closure(A)=X. If X-A is dense, then int(A)=empty, closure(X-A)=X. So bound(A) = clos(A)-int(A) = X-{} = X bound(X-A) = clos(X-A)-int(A) = X-{} = X. ============================================================== ======== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ============================================================== ======== Arturo Magidin magidin@math.berkeley.edu === Subject: Roots of Characteristic Equation Bounded by Unity? Suppose one has an Nth order polynomial with real coefficients. What I'd like to know if there is a simple criterion to say whether the real part of ALL the roots are bounded in magnitude by unity. I know about the Routh-Hurwitz criterion, which tells you conditions for which all roots Z have negative real parts. I'm after something similar, except I want conditions for which all roots Z satisfy -1 < Re(z) < 1 While I'm at it, I may as well ask if there is an analgous criterion for all the roots being bounded in MAGNITUDE by unity: |Z| < 1 Any help is greatly appreciated. Thanks! JB === Subject: Re: Roots of Characteristic Equation Bounded by Unity? > I know about the Routh-Hurwitz criterion, which tells you conditions > for which all roots Z have negative real parts. > While I'm at it, I may as well ask if there is an analgous criterion > for all the roots being bounded in MAGNITUDE by unity: > |Z| < 1 The roots of f(z) lie in |z| < 1 if and only if the roots of the rational function f((w+1)/(w-1)) lie in the left half plane. Write this as the quotient of two polynomials in w with no common factor, and apply Routh-Hurwitz to the numerator. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Eigenstuff. >There is >a nonzero solution v to this equation if and only if >the matrix (M-LI) is singular, i.e., its determinant is 0. Singular = linearly independent = not redundant in any way? >Det(M2) = Det(M-LI) is a polynomial of degree N in the >unknown L where N is the dimension of the matrix. Yes, I understand. And Tasha (a cat) is kneading my leg. It hurts. Ouch. I need to trim her claws. The rear ones dig in when she jumps off of me. Now she left something stinky on my wrist. Yuk. Now she's licking the stink off of herself. Gross. >Multiply >these possibilities out and you'd better end up with >the same 6 terms for all of them. Now that's useful. I hadn't realized that. Of course. With the same signs, too. Now when you do that, you get: >aei - afh - bdi + bfg + cdh - ceg bfg + cdh +aei - afh - bdi -ceg Now I see a pattern here. The positive terms are in order diagonally downward and to the righ into the bopy of M over to the right. The negative terms slope from lower left to upper right. The positive slope from upper left to lower right. That's easiest to remember as what I do in image processing is reflect the image matrix over the right border to the right, and the original again over the lower border to the bottom, then no matter which way you to the next step the result in the lower right is the original rotated by 180 degrees. Not the transpose, Then I tile-copy that matrix just as in this determinant, with identical matrices to right, down, and both ways right and down. When I have enough, I serialize the pixels in a straight 45 degree diagonal line, and the effect is that at the bordethe line moves over one space and reflects. Then I take the Fourier transform, and either retain phase and equalize amplitude to compress, or sort by amplitude and lop off the big ones and keep them, for feature extraction. The neat thing is that I can do this in 2, 3, or more dimension. So extracting the highest contrast, most rapidly moving features from a block of video is trivial, done in a single sort and clip, independent of any notion of time or space. It's loads of fun. I need a more powerful computer. And a cat that doesn't stink. These serializations are omnidiagonal, and given coprimality between edges of 2, 3, or higher dimensional blocks, complete. With two edge even and otherwise all coprime, the trace finishes next to the starting pixel. Anyway that must have been where I got the idea that you could pick other than along any row or column. When you go along the diagonals, bfg + cdh +aei - afh - bdi -ceg becomes (aei + dch + gfb) - (afh + dbi +gec) using a, d, and g, as some kind of primaries. As you can see, all the terms in the products in the first sum come from different rows and columns. Likewise all the products in the socond sum. But they are particular selections, not those made by expansion by minors. Also, the three leading multiplicans, all come from one column. Let's try it with leading terms from different rows and colums. In a 3x3 matrix there are 9 for the first pick, 9-5 = 4 for the second, and 1 for the last. 36 ways to pick. 3^2 * 2^2 * 1^2. But what we have with the idea that any permutation can be at least checked is 36 ways to pick, 6 of which we have one-row or one-column initial term solutions for. Are any of the remaing 30 equivalent to the 6 we know? Well, with a, e, and i, which are themselves from different rows and columns, there is...um... Probably not. Why? There are 9 ways to pick the first element. We can start in any row or colum. That done, we can then pick from four, but once having picked one of the four, the third term in one sum is known, so two of the four have been eliminated. Picking the same 1 gives only two choice for the other, and the third is determined, and we have oppsite signs for each of these two. So that's 18 plus two possibilities for the first term and it's complement, so 20. Now five have been eliminated. The one first picked, the four following. So four remain for initial terms but these are in the same row or column. When you pick the first term, you've ruled out 5 leaving four. But either way, you generate two terms with the first pick. Like wise no matteri which of the four remaing you pick, you generate a second pair of terms, differing only in sign. So that's really only 9 initial choices,, the same as nxn. I didn't explain that very well, did I? I guess it just has to be a permutation counting problem. So in addition to the leading terms all coming from the same row or column there are 4 ways to pick the second term of a sum, without being in the same row or column. However, these terms leave no other not in the same row or column, so they don'e count, they can't be completed. So there's just the six. a, d, and g, are, of course, in a column. Not only do Finney and Ostberg explain that Det (I) # 0, they say Det (I) = 1. I'm tired. Thanks again. Yours, Doug Goncz (at aol dot com) Replikon Research Read the RIAA Clean Slate Program Affidavit and Description at http://www.riaa.org/ I will be signing an amended Affidavit soon. === Subject: Re: Eigenstuff. > Mv = L*v > Mv - L*v = 0 > Mv - L*I*v = 0 > (M - L*I)v = 0 Thus begat M-LI = >> M11-L M12 M13 M14 ... >> M21 M22-L M23 M24 ... >> M31 M32 M32-L M34... >> M41 M42 M43 M44-L... ... etc Thank you. I understand v=0 is a nonsolution in that every problem posed this way has that useless solution by definition. Yours, Doug Goncz (at aol dot com) Replikon Research Read the RIAA Clean Slate Program Affidavit and Description at http://www.riaa.org/ I will be signing an amended Affidavit soon. === Subject: Levi's Metric Could anyone give me a hint explaining why the weak convergence of distributions implies convergence in the sense of Levi's metric: d(F,G)=inf(E>0: dla kadego xin R G(x-E)-E<=F(x)<=G(x+E)+E} ? regards olej === Subject: Re: Levi's Metric >Could anyone give me a hint explaining why the weak convergence of >distributions implies convergence in the sense of Levi's metric: >d(F,G)=inf(E>0: dla kadego xin R G(x-E)-E<=F(x)<=G(x+E)+E} ? >regards >olej Your last line is illegible on my reader. dla kadego should be for all . Just a guess here. Is it that hard to show that convergence in the Levy metric is equivalent to F_n(x) -> F(x) at all points x where F is continuous? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: More FLT by modular arithmetic. Here's the latest from the latest, an argument from Fermat17.mcd: For c=b, n = infinity. For c = b + 1, n <= c - 1, since the slope of n = n (b) is less than one. C + 1 is where b^n mod c repeats. It seems that x^n mod c can repeat in lengths always and only a factor of c-1. However, if x and c are coprime, x < c, x^0 mod c = 1, x^1 mod c = x, and x^(k+1) mod c = ( x^k mod c * x ) mod c The location of the value one at (y = c + 1) mod c is a repeating point, so the sequence length can be as long as c+1 - 0 or c + 1 and it can repeat at a lower length i which must be a factor of c + or else there would be no 1 at c+1 mod c. However, the maximum value of z mod c, z < c is c - 1 and there can be exactly c-1 values since 1: x^k mod c # 0 because x and c are coprime 2: The formula for x^(k+1) 0 is an absorbing state. We might think the sequence would repeat at any i a factor of c + 1 and i also a factor of c-1. This can only happen if c+1 and c - 1 or b and b + 2 have a cofactor and if b is even, these will always have a cofactor of 2. Of the Pythagorean triples I know, and these are few, all have b even If a^0 = 1 and a^k+1 mod c = (a^k mod c * a) and likewise for b, and a and b are coprime, they do not share a factor of two. One or the other will be odd, the other even. So the one will repeat at a factor of two, and the other..... I'm too tired for this. Maybe tomorrow. Yours, Doug Goncz (at aol dot com) Replikon Research Read the RIAA Clean Slate Program Affidavit and Description at http://www.riaa.org/ I will be signing an amended Affidavit soon. === Subject: SMSU Problem Corner The new problems have been posted in the areas of High School, Advanced, and Challenge. Please visit us at http://math.smsu.edu/~les/POTW.html Thanks. === Subject: analytic geometry How to build the circle tangent to 3 other circles ? === Subject: Re: analytic geometry > How to build the circle tangent to 3 other circles ? With great difficulty, especially when the circles are concentric and different. Seriously: conduct search under Apollonius Problem (Apollonius of Perga, lived 262BC-190BC) CheeZVK(Slavek) === Subject: Problem driving me nuts! Let f(x) = exp(-x^2) and a>0. Show that [f(a-t) - f(a+t)] / integral(a-t .. a+t, f) is decreasing in t for positive t. In probabilistic terms, the question is (equivalent to): Let X have standard normal distribution and a>0. Show that E[X | |X-a| < t] is decreasing in t. This was asked by another poster here some time ago. More generally, suppose X has an even density with a unique maximum at 0. Under what simple (but general) sufficient condition does the latter conclusion hold? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: A function continuous at only one point? <3f6e08ec$25$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS said: >At which point is this function continuous? Sorry - typo. F(n) = 0 if n is rational F(n) = n if n is irrational Continuous at 0. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: A function continuous at only one point? > What is an example of a function f : R-->R that is continuous at precisely > one point? Let g be a bounded function which is nowhere continuous. Then define f(t) = t*g(t) === Subject: Re: A structural point of view on the Continuum and the Discreteness concepts <5567a96f.0309142352.4badc1da@posting.google.com> <3f6dffe9$22$fuzhry+tra$mr2ice@news.patriot.net> <5567a96f.0309221036.4d560add@posting.google.com> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS >I shall be glad to get any advice from you, that can help me to write >my ideas in a formal way. I'm afraid that you need to start from the beginning. Forget vague analogies, e.g., stretch, with things you can hold in your hand, forget decimal expansions of numbers and don't use phrases that you don't understand, and especially not terms that you made up yourself (e.g., R member.) Don't speak of things being true by definition unless you know what the definition is. I'd suggest that you read Halmos's Naive Set Theory and a good book on Euclidean Geometry. Or take classes in Set Theory and Analysis. Most of what you say in the PDF is not only wrong but meaningless, e.g., More than that, any interval from X to ~X can be opened or closed only by a quantum-like leap, -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Fundamental Reason for High Achievements of Jews <3F613B70.1518535C@hate.spam.net> <6b70c71c.0309161338.1ddc9eda@posting.google.com> <3f68fdac$20$fuzhry+tra$mr2ice@news.patriot.net> <3f6e1b21$32$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 04:28 PM, Scott said: Hey feigeleh, you must believe that everybody is as dumb as you are. *PLONK* -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Fundamental Reason for High Achievements of Jews > at 04:28 PM, Scott said: > Hey feigeleh, you must believe that everybody is as dumb as you are. > *PLONK* The followup was set according to intellectual level of your contribution, which amounted to a transparently false statement of knee-jerk rhetoric. I'm glad to see you have snipped it, you definitely *should* be ashamed to acknowledge it - though replying with an infantile rhetorical remark (with the obligatory kill file plonk) shows you still have some way to go. === Subject: Re: Transfinite Induction <3f6e43a6$40$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 01:44 PM, magidin@math.berkeley.edu (Arturo Magidin) said: >Worse, it is wrong. No. I did not say (For all i)(P(i)=>P(i+1) *instead of* (For all j)((For all iP(j)); had I done so, then that would have been wrong. >Also, your (1) is unnecessary; Correct. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Transfinite Induction >transfinite induction is like ordinary induction, except that it can be applied >>to wellordered sets that have limit elements: a limit element in a wellordered >>set is any element except the smallest one that doesn't have an (immediate) >>predecessor Many thanks to everyone who responded. Mention was made of > applicability to ordinals and naturals. Can it also apply (or be > useful) to proving statements about sets with other types of binary > relations? For instance collections of increasing sets? Thanks for > the help. Induction and Ordinals are two ways of writing the same idea. (oops, I mean well-founded ordinals) For example, suppose you have pairs of ints, and one pair (x,y) is larger than another (z,w) if x > z. (if x == z, then compare y and w). This is lexicographic ordering (corresponding to the ordinal omega^2, where omega is the ordinal for the set of natural numbers) So, (4,2) is considered bigger than (3,10000000). This is useful for proving the computability of multiparam functions (like Ackermann's function). Mitch === Subject: Re: Transfinite Induction >transfinite induction is like ordinary induction, except that it can be applied >>to wellordered sets that have limit elements: a limit element in a wellordered >>set is any element except the smallest one that doesn't have an (immediate) >>predecessor Many thanks to everyone who responded. Mention was made of > applicability to ordinals and naturals. Can it also apply (or be > useful) to proving statements about sets with other types of binary > relations? For instance collections of increasing sets? Thanks for > the help. Induction and Ordinals are two ways of writing the same idea. > (oops, I mean well-founded ordinals) For example, suppose you have pairs of ints, and one pair (x,y) is > larger than another (z,w) if x > z. (if x == z, then compare y and w). > This is lexicographic ordering (corresponding to the ordinal omega^2, > where omega is the ordinal for the set of natural numbers) So, (4,2) is considered bigger than (3,10000000). This is useful for proving the computability of multiparam functions > (like Ackermann's function). Mitch Your response is appreciated. === Subject: Re: Fun with physics on 1000 pages - for free download NNTP-Proxy-Relay: library1-aux.airnews.net Damn. THANK YOU! Dave > The free physics text available on > http://www.motionmountain.net > has been reworked. It now contains 550 problems, 200 drawings, 70 tables, > and a new section on nuclear physics with the story of radioactivity, a MRI > scan of two humans making love and the dream of grand unification. It also > includes new figures on special relativity, a short explanation of > k-calculus, the paradox of the relativistic submarine, a photograph and > description of how some caterpillars shoot away their faecal matter, a > photograph of a basilisk running over water, and much more. > For all fields of physics the newest research results and the main > unanswered questions are presented. The text is a structured walk through > classical physics, relativity, quantum theory and unification. The reader > gets an overview of what motion is and what it can effect, keeping the > accent on surprises and thought-provoking puzzles. > Enjoy. > Christoph Schiller > P.P.S. Any help on (preferably colour postscript) images to > be added to the text (with permission of course) is much appreciated. === Subject: subset counting problem X-Mimeole: Produced By Microsoft MimeOLE V6.00.2800.1165 X-Msmail-Priority: Normal Given a k-subset family of an n-set P (that is, drawing k elements randomly from P to form a subset), how to find a closed form probability that any M subsets have no common intersection? === Subject: Re: subset counting problem > Given a k-subset family of an n-set P (that is, drawing k elements > randomly from P to form a subset), how to find a closed form > probability that any M subsets have no common intersection? Here are some ideas to get you started: How many ways can you choose a k-subset? C(n,k) -- n choose k, which equals n!/(k!(n-k!)) -- of course. If you're allowed to choose another using the entire set, you are left with C(n,k) again. In general, the total number of ways of choosing M k-subsets from P is C(n,k)^M. Now out of this universe, how would you go about counting the number of ways of choosing k-subsets that don't overlap? The first set you choose would have C(n,k) possibilities, of course. The next one, however, would have only C(n-k,k) possibilities -- why? Keep going. Good luck! .... Bob === Subject: Homeomorphism in R^n? I was thinking about the following, and I don't know what to make of it. It seems to not be true, but I can't find an example to show this. If A and B are homeomorphic subsets of R^n, and if A is closed in R^n, is it true that B is also closed in R^n? If it's not true for all n, is it true for some n? === Subject: Re: Homeomorphism in R^n? > If A and B are homeomorphic subsets of R^n, and if A is closed in R^n, is it > true that B is also closed in R^n? Let me just throw in the map 1/x, which is a homeomorphism of [1,oo) onto (0,1]. === Subject: Re: Homeomorphism in R^n? A good exercise. For n=1: consider the function arctan; it realizes homeomorphism of R^1 (closed in R^1) onto (-pi/2,pi/2) (not closed in R^1). What's worse: it is a homeomorphism of a complete metric space onto an incomplete metric space. (So, metric completeness is not a topological concept!) (The exponential gives another example.) Higher dimensional examples are obtained by cartesian products. CheeSlavek(ZVK) > I was thinking about the following, and I don't know what to make of it. It > seems to not be true, but I can't find an example to show this. > If A and B are homeomorphic subsets of R^n, and if A is closed in R^n, is it > true that B is also closed in R^n? > If it's not true for all n, is it true for some n? === Subject: Re: Homeomorphism in R^n? >I was thinking about the following, and I don't know what to make of it. It >seems to not be true, but I can't find an example to show this. >If A and B are homeomorphic subsets of R^n, and if A is closed in R^n, is it >true that B is also closed in R^n? >If it's not true for all n, is it true for some n? Another counterexample: Let C = the set of nonzero integers and let D = {1/x: x in C}. Then C^n and D^n are homeomorphic, C^n is closed, D^n is not. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Homeomorphism in R^n? >I was thinking about the following, and I don't know what to make of it. It >seems to not be true, but I can't find an example to show this. >If A and B are homeomorphic subsets of R^n, and if A is closed in R^n, is it >true that B is also closed in R^n? >If it's not true for all n, is it true for some n? > No. Take A = R^n and B = an open ball of finite radius. Now if you require that A and B be proper subsets, I don't know. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Homeomorphism in R^n? >I was thinking about the following, and I don't know what to make of it. It >seems to not be true, but I can't find an example to show this. If A and B are homeomorphic subsets of R^n, and if A is closed in R^n, is it >true that B is also closed in R^n? If it's not true for all n, is it true for some n? No. Take A = R^n and B = an open ball of finite radius. > Now if you require that A and B be proper subsets, I don't know. > -- > Stephen J. Herschkorn herschko@rutcor.rutgers.edu So, let's go through the steps. If A is closed and bounded, it is compact and we do not get counterexamples . So, it must be closed and unbounded. Try the integers in R^1, and the exponential function. Higher dimensions: cartesian products. CheeZVK(Slavek). === Subject: Re: Homeomorphism in R^n? [...] I was thinking about the following, and I don't know what >to make of it. It seems to not be true, but I can't find >an example to show this. If A and B are homeomorphic subsets of R^n, and if A is >closed in R^n, is it true that B is also closed in R^n? If it's not true for all n, is it true for some n? > [...] So, let's go through the steps. If A is closed and bounded, it > is compact and we do not get counterexamples . So, it must be > closed and unbounded. Try the integers in R^1, and the > exponential function. Higher dimensions: cartesian products. > CheeZVK(Slavek). (Apologies for responding to my own post) After I pushed the send button, it occurred to me: In R^n (n>=1), a closed set is homeomorphic to a non-closed set if and only if it is unbounded. (Bolzano-Weierstrass Theorem is useful in proof.) CheeZVK(Slavek). === Subject: Re: Homeomorphism in R^n? If A and B are homeomorphic subsets of R^n, and if A is >closed in R^n, is it true that B is also closed in R^n? > In R^n (n>=1), a closed set is homeomorphic to a non-closed > set if and only if it is unbounded. For example, Rx{0} is closed in R^2 and homeomorphic to (0,1)x{0} which is neither closed nor open in R^2 === Subject: Re: Naive Q: Set theory, logic - which comes first? > I asked about what seems to me to > be a genuine difficulty in defining a set as a member of a model of > ZFC, and a model (of any theory) as a set with some additional > structure. How can we explain what a set is to someone who has never heard about sets? Well, the first thing to note is that the word set means different things in different contexts. There is a family resemblance, though. Here is how I would go about explaining the most common meaning of set: * Sets are elements of a set theoretic universe. A set-theoretic universe is sometimes seen as arising from transfinite iteration of the operation set of, starting either from nothing at all, or from urelements. * Set-theoretic universes satisfy ZF, and there is little reason to consider set-theretic universes that do not also satisfy the Axiom of Choice. * The finite von-Neumann ordinals satisfy the same truths as the naturals numbers. * It is sometimes thought that there is one set-theoretic universe that deserves to be called the true set-theoretic universe, which is supposed to be as large as possible rather than as small as possible. This description refers to natural numbeso arguably natural numbers are more basic than sets. Natural numbers can be seen as arising from iteration of the successor operation starting with 0. Some people think that we must start with vague, informal, undefined, etc. notions before we can finally reach the objects of mathematics and logic, but what matters is really just that we are able to convey an idea. Someone very good at gestures could use gestures only to explain what sets and natural numbers are. Naturally, descriptions of mathematical objects can be good and bad, and they become better the better we understand the objects we want to describe. No doubt, there are ways to improve upon the way of defining sets and natural numbers given above. For example, one should perhaps mention that a set-theroetic universe can be seen as a topos, and that the natural numbers contain the neutral elements 0 and 1, and are closed under addition and multiplication. Some people want to see any mathematical object as a set in the true set-theoretic universe. It seems that they have to say that set-theoretic universes are sets, and that the true set-theoretic universe contains itself. Well, I prefer not to work inside a set-theoretic universe. The language of set theory is, admittedly, a precise language, and when we go beyond set theory we do not want to lose that precision. I hope that in the future objects outside set theory will be as well-understood as objects inside set theory, and that it will be possible to define things such as set-theoretic universes and natural numbers without the slightest trace of vagueness without making them into sets. Mattias === Subject: Re: Naive Q: Set theory, logic - which comes first? FWIW, here is my attitude. Lower predicate calculus (the usual > propositional calculus plus quantification, but only over individuals, > not over sets) comes first as a kind of pre-mathematical formalism. Pre-mathematical how? Expositions of predicate logic draw heavily > on mathematics - specifically, the natural numbers and inductive > definitions. The natural numbers are really why I've started this thread in the first place. I have, as probably most everyone does, a strong intuitive feeling for what these are, but I began to feel a little shaky after reading the last chapter in David Gale's Tracking the Automatic Ant, where he discusses some peculiar possible scenarios regarding undecidable statements on the natural numbers. So naturally, so to speak, I wanted to resort to a formalist viewpoint for a while, just to get a better understanding of the foundations - I only took a one-semester introductory course to logic (long ago), and I assumed that it simply wasn't rigorous enough. I therefore wondered how it's really done. Apparently, it isn't... the responses here and the ones I received by e-mail seem to indicate that I didn't miss some magical precise process in which the natural numbethen sets, then logic and models are really defined. I don't know that this is bad, and I certainly don't believe that this is a terrible flaw in the foundations of mathematics and everything comes crumbling down (I have a feeling some responders assumed that this was my attitude). I'm just somewhat surprised to learn that the basics aren't more solid than I knew them to be. So... what do you think? Can the infinity of twin primes be independent? Can it be *proven* to be such? And how about Goldbach? Or, to take Gale's example, what about infinitely many 7's in the binary expansion of pi? Surely *that* has a definite answer... Thanks again for all the answers, Alan === Subject: Re: Naive Q: Set theory, logic - which comes first? > I'm just somewhat surprised to > learn that the basics aren't more solid than I knew them to be. Just what kind of solidity do you feel is missing? > So... what do you think? Can the infinity of twin primes be > independent? Can it be *proven* to be such? And how about Goldbach? > Or, to take Gale's example, what about infinitely many 7's in the > binary expansion of pi? Surely *that* has a definite answer... Sure, all of these questions have definite answers in the (mathematical) nature of things, and some of them may be undecidable in PA or even in ZFC. === Subject: Re: Naive Q: Set theory, logic - which comes first? I'm just somewhat surprised to > learn that the basics aren't more solid than I knew them to be. Just what kind of solidity do you feel is missing? The responses of Elaine Jackson, Michael Barr, and (I think also) Aatu Koskensilta (he says in mathematics we have an informal notion of a set), plus one response I got over e-mail, seem to indicate that in order to consider the semantics (not syntactic proofs, of course) of set theory, we have to grant some initial level of naive set theory. As I indicated, I don't think this is a bad thing - this was, in fact, my intuitive interpretation of the meaning of model theory, and I simply assumed that there is some extra formal way of doing it I was not aware of. > So... what do you think? Can the infinity of twin primes be > independent? Can it be *proven* to be such? And how about Goldbach? > Or, to take Gale's example, what about infinitely many 7's in the > binary expansion of pi? Surely *that* has a definite answer... Sure, all of these questions have definite answers in the > (mathematical) nature of things, and some of them may be undecidable > in PA or even in ZFC. Hmmm. I think I agree there is a mathematical nature of things - it certainly sounds intuitive that Goldbach, Twin Primes etc. are either true or false. But I don't have such a strong feeling about the Continuum Hypothesis, for instance. And so I wonder, if twin primes, say, turns out to be undecidable in ZFC - perhaps even in ZFC+some extra large-cardianl axioms - why am I entitled to feel this way? What's the difference between it and CH? Do you also think that CH has a definite answer in the mathematical nature of things? Alan === Subject: Re: Naive Q: Set theory, logic - which comes first? > As I indicated, I don't think this is a bad thing - this was, in fact, > my intuitive interpretation of the meaning of model theory, and I > simply assumed that there is some extra formal way of doing it I was > not aware of. But would such an extra formal way be more solid? I don't think so - the informal basics of mathematics are as solid as can be. > But I don't have such a strong feeling about the > Continuum Hypothesis, for instance. And so I wonder, if twin primes, > say, turns out to be undecidable in ZFC - perhaps even in ZFC+some > extra large-cardianl axioms - why am I entitled to feel this way? > What's the difference between it and CH? Very many people feel the same way, and there are various answers to your question (including the answer that number theory is just as indefinite as set theory). But although we do have these strong feelings about our own mathematical knowledge and mathematical perception, I think the more important question is what difference our convictions in this regard make in how we use or work in mathematics, and in particular in such topics close to computation as cryptography, the theory of algorithms, applications of mathematics in various fields. === Subject: Re: Naive Q: Set theory, logic - which comes first? Here are the books that come to mind right away: Paul Halmos: Naive Set Theory Suppes: Axiomatic Set Theory a brown book at the library that has some stuff about models of set theory (such and such an axiom is independent of such and such another axiom; stuff like that), as well as a very readable treatment of how the real number system can be constructed in axiomatic set theory Stephen Kleene: Introduction to Mathematical Logic Kalish, Montague and Mar: Logic: Techniques of Formal Reasoning a green book by somebody called Mendelson, which has some interesting stuff about recursive function theory I have some notes on basic set theory that could benefit from a little feedback. I'll put them on my webpage for you if you like. Let me know. Peace | Thank you, Elaine. I suspected this is how it is done, but I'm still | curious about the details. Do you know which book introdcues these | subjects carefully and as rigorously as possible? | | Alan | | > You need set theory before you can have model theoretic semantics, and you need | > model theoretic semantics before you can have set theory. Hence there can be no | > mathematics. QED | > | > I think the solution is to grant yourself a certain amount of set theory right | > from the beginning. Uncontroversial stuff, only as much as you need. Once you've | > got a respectable axiomatic theory of sets, you can use that apparatus in your | > model theory (and when you do, your set theoretic axioms will be part of the | > hypothesis of any theorem you prove). But, to get the ball rolling, you need an | > intuitive, pretheoretic notion of sets. | > | > | This is probably a FAQ, but I couldn't easily find the answer. To | > | define a Model for a logical theory, we need the notion of a set: a | > | model is a set together with an assignment of the constants, relations | > | etc., that makes the axioms true. So how do we handle models of a | > | theory that aims to define the very meaning of set, such as ZFC? | > | | > | I mean, a sentence like a model of ZFC is a transitive set such | > | that... sounds circular to me. How do we get this started? | > | | > | | > | | > | Alan === Subject: Re: Naive Q: Set theory, logic - which comes first? I'll be glad to see the notes and provide feedback if I can. Thanks! I may have been a little sloppy when I asked about how it's done: I'm quite familiar with naive set theory, at least at the level of, say, Kamke's book. I assumed that when one properly goes through the definitions of sets, logical theories and models, one discards the naive theory in favor of a rigorous one. Your view seems to be that this isn't really possible, and that any foundational description of model theory requires *some* amount of naive set theory. So, when I asked about how it's done, I was really asking about a formal development of logic and model theory based on a minimally naive set theory, and a clear indication of what it is that we need to take on faith (platonistically?) and what it is that can then be formally developed. Thanks again, Alan > Here are the books that come to mind right away: Paul Halmos: Naive Set Theory Suppes: Axiomatic Set Theory a brown book at the library that has some stuff about models of set theory (such > and such an axiom is independent of such and such another axiom; stuff like > that), as well as a very readable treatment of how the real number system can be > constructed in axiomatic set theory Stephen Kleene: Introduction to Mathematical Logic Kalish, Montague and Mar: Logic: Techniques of Formal Reasoning a green book by somebody called Mendelson, which has some interesting stuff > about recursive function theory I have some notes on basic set theory that could benefit from a little feedback. > I'll put them on my webpage for you if you like. Let me know. Peace | Thank you, Elaine. I suspected this is how it is done, but I'm still > | curious about the details. Do you know which book introdcues these > | subjects carefully and as rigorously as possible? > | > | Alan > | > | > You need set theory before you can have model theoretic semantics, and you > need > | > model theoretic semantics before you can have set theory. Hence there can be > no > | > mathematics. QED > | | > I think the solution is to grant yourself a certain amount of set theory > right > | > from the beginning. Uncontroversial stuff, only as much as you need. Once > you've > | > got a respectable axiomatic theory of sets, you can use that apparatus in > your > | > model theory (and when you do, your set theoretic axioms will be part of the > | > hypothesis of any theorem you prove). But, to get the ball rolling, you need > an > | > intuitive, pretheoretic notion of sets. > | | > | This is probably a FAQ, but I couldn't easily find the answer. To > | > | define a Model for a logical theory, we need the notion of a set: a > | > | model is a set together with an assignment of the constants, relations > | > | etc., that makes the axioms true. So how do we handle models of a > | > | theory that aims to define the very meaning of set, such as ZFC? > | > | > | > | I mean, a sentence like a model of ZFC is a transitive set such > | > | that... sounds circular to me. How do we get this started? > | > | > | > | > | > | > | > | Alan === Subject: Re: Naive Q: Set theory, logic - which comes first? I'm not familiar with Kamke, but I think you're probably misconstruing the title of Halmos's book. Anyhoo, I'll post those notes for you at http://www.geocities.com/sean_mcilroy Peace | I'll be glad to see the notes and provide feedback if I can. Thanks! | | I may have been a little sloppy when I asked about how it's done: | I'm quite familiar with naive set theory, at least at the level of, | say, Kamke's book. I assumed that when one properly goes through the | definitions of sets, logical theories and models, one discards the | naive theory in favor of a rigorous one. Your view seems to be that | this isn't really possible, and that any foundational description of | model theory requires *some* amount of naive set theory. | | So, when I asked about how it's done, I was really asking about a | formal | development of logic and model theory based on a minimally naive set | theory, | and a clear indication of what it is that we need to take on faith | (platonistically?) and what it is that can then be formally developed. | | Thanks again, | | Alan | | > Here are the books that come to mind right away: | > | > Paul Halmos: Naive Set Theory | > | > Suppes: Axiomatic Set Theory | > | > a brown book at the library that has some stuff about models of set theory (such | > and such an axiom is independent of such and such another axiom; stuff like | > that), as well as a very readable treatment of how the real number system can be | > constructed in axiomatic set theory | > | > Stephen Kleene: Introduction to Mathematical Logic | > | > Kalish, Montague and Mar: Logic: Techniques of Formal Reasoning | > | > a green book by somebody called Mendelson, which has some interesting stuff | > about recursive function theory | > | > I have some notes on basic set theory that could benefit from a little feedback. | > I'll put them on my webpage for you if you like. Let me know. | > | > Peace | > | > | Thank you, Elaine. I suspected this is how it is done, but I'm still | > | curious about the details. Do you know which book introdcues these | > | subjects carefully and as rigorously as possible? | > | | > | Alan | > | | > | > You need set theory before you can have model theoretic semantics, and you | > need | > | > model theoretic semantics before you can have set theory. Hence there can be | > no | > | > mathematics. QED | > | > | > | > I think the solution is to grant yourself a certain amount of set theory | > right | > | > from the beginning. Uncontroversial stuff, only as much as you need. Once | > you've | > | > got a respectable axiomatic theory of sets, you can use that apparatus in | > your | > | > model theory (and when you do, your set theoretic axioms will be part of the | > | > hypothesis of any theorem you prove). But, to get the ball rolling, you need | > an | > | > intuitive, pretheoretic notion of sets. | > | > | > | > | This is probably a FAQ, but I couldn't easily find the answer. To | > | > | define a Model for a logical theory, we need the notion of a set: a | > | > | model is a set together with an assignment of the constants, relations | > | > | etc., that makes the axioms true. So how do we handle models of a | > | > | theory that aims to define the very meaning of set, such as ZFC? | > | > | | > | > | I mean, a sentence like a model of ZFC is a transitive set such | > | > | that... sounds circular to me. How do we get this started? | > | > | | > | > | | > | > | | > | > | Alan === Subject: Re: Naive Q: Set theory, logic - which comes first? > I'm not familiar with Kamke, but I think you're probably misconstruing the title > of Halmos's book. So what does Halmos mean by naive? (in contrast to the usage in this discussion) Mitch === Subject: Re: Naive Q: Set theory, logic - which comes first? >> I'm not familiar with Kamke, but I think you're probably >> misconstruing the title >> of Halmos's book. > So what does Halmos mean by naive? (in contrast to the usage in this > discussion) He just presents the axioms of ZFC (I'm not sure if he includes Foundation) as a given and discusses basic consequences. He does not discuss issues of consistency, and there is no logical notation (so the book is accessible to more students). As he puts it in his preface, it is an axiomatic set theory from the naive point of view. I think this (Halmos PR, Naive Set Theory) is a must-read for every student of mathematics. I suppose you can get your basic set theory elsewhere (I did), but Halmos' exposition is clear, insightful, and delightful. (My, but I am parenthetical today.) -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Naive Q: Set theory, logic - which comes first? > So what does Halmos mean by naive? Axiomatic. But without any machinery of formal logic. === Subject: is this map conformal? f: r*e^(i*theta) -> [ (1-r)^(-1) ] * e^(i*theta) I know an analytic map is conformal, but this is not analytic. Whats the simple way of confirming if this is conformal or not? is it true every conformal map is analytic? TIA === Subject: Re: is this map conformal? GR56 > f: r*e^(i*theta) -> [ (1-r)^(-1) ] * e^(i*theta) > I know an analytic map is conformal, but this is not analytic. ... > is it true every conformal map is analytic? Yes, at least if we assume that the partial derivatives are continuous. So, for the above function, we need only test whether the Cauchy-Riemann eqns hold, and they don't. LH === Subject: actuarial exam#1 - Mathematical Foundations of Actuarial Science I signed up to take this exam in November. I have a BA in Math, but I will have to refresh my math since it has been a few years. I found the syllabus here: Is anyone else taking this exam? It looks like the courses I will have to review are Calculus I, Calculus II, and Calculus III (multivariable), and Probability & Statistics (involving multivariable Calculus). === Subject: A question about divergence Let G: R^n --> R^n and g= div G. Let f: R^n --> R^n be a C^2 function with its Jacobian matrix called A. Let B be the adjoint of A (whose entries are the minors of A). Let F=G o f, the composition of G and f. Note that B is an nxn matrix and F is an nx1 vector -- they are all functions of x. Let H=BF. I can prove by direct verification that div H = g(f(x))J_f (x), where J_f(x) is the Jacobian of f -- the determinant of A. My question is: Is this a special case of a more general theorem about divergence, like 'product rule'? === Subject: Re: theoretically the strongest concrete I'm still betting that aircrete is stronger than those concretes taht use cement plus filler. but, what is your hypothesis on the checkerboard arrangment of the bubbles or rubble, compared to the closest-packing (Kepler) config.? even if you don't believe Hale's proof that it *is* the closest-packing (with the hexagonal closest-packing variants; there's another term for them), what possible better comes from the rectilinear stacking, which is definetly *not* closest-packed? > Kepler Packing to be applied to concrete and to see if the Kepler Packing is a > better concrete than the theory of 1 :: 1 mix of a 3-d chessboard mix. I forgotten the 3-d Kepler Packing percent of voids. Was it somewhere around > 23% voids? Let us say it was. So that a Kepler Packing Concrete mix would not be > a 1 :: 1 mix but a 23% :: 77% mix or a 1 :: 3 mix. > Experiment: It should be easy to experiment as to whether a Kepler Packing > concrete mix is superior to a 1 :: 1 chessboard mix. In that we get marble sized --les ducs d'Enron! http://larouchepub.com === Subject: Re: theoretically the strongest concrete X-SessionID: e3acb-11391-Z4-19948@news.uchicago.edu X-Hash-Info: post-filter,v:1.4 X-Hash: 588bb6ba 013737ca bb52251b 454a284f 958fd0b2 > the cement however is more or less reactive depending on where > it comes from, it is not a defined substance.. This is very true. Alcoa used to make its CA-14 and CA-25 calcium aluminate cements at a U.S. plant. These are high purity cements, used mainly in refractories (furnace linings) -- a very different breed from your standard Type III Portland cement. Anyway, a couple of years ago, Alcoa decided to shut down U.S. production and make the stuff in Europe (Netherlands, I think). I was working for a refractories company at the time and we had a few products which used CA-25 as a binder. The European stuff turned out to be slightly different from the American stuff in terms of workability and set time. If I recall correctly, it took slightly more water to get the same workability and had a significantly longer set time. There were also some small differences in strength. The end result was that, even though it was supposedly the same cement, we couldn't get the properties we wanted anymore, and we wound up having to reformulate our products with another manufacturer's cement. Eventually we moved away from cements altogether and switched to a phosphate binder, but that's another story. The moral of the story is that, not only are there many different types of cements (varying proportions of alumina, silica, calcia, and other stuff), but the same cement varies from manufacturer to manufacturer, and for the same manufacturer, from plant to plant. If that weren't enough, Portland cements often vary significantly from batch to batch. Obviously, depending on your application, these variations may not matter much. But Dave is absolutely right; you can't say that cement is cement is cement and leave it at that. Then there are aggregates. Good luck finding a uniform grade of anything. At best, you can hope for 65-70% to be somewhere around the specified mesh size when you run a screen. Moisture content may vary from lot to lot, too; one week, we suddenly discovered that all of our test batches were taking a few percent less water than we expected. It turned out that one of the aggregates we were using came off of a barge which had been sitting uncovered during heavy rains. Theory is great, but don't forget that it's always based on the rules which we've made up to help us conceptualize the real world, and that the real real world is often far more complicated. Dave Palmer (773) 955-2223 palmdav@iit.edu === Subject: Mapping PDF's onto one another? I am just now learning about probability density functions, in particular exponential and gaussian ones. My problem comes in understanding in how two distribution functions can be summed in series to give a total effect. some distance d. After starting at distance zero, the distance the is modelled by an exponential PDF with mean of zero and some coefficient a*d. chooses the angle at which to travel based on a normal distribution with standard deviation of b degrees and mean of zero. So, after these two events, how could one compute the distribution of reach distance d? (of course, the distribution which is desired is the their original downward motion) Is their a way to sum the effects of two PDFs in serial? === Subject: Re: Mapping PDF's onto one another? >I am just now learning about probability density functions, in >particular exponential and gaussian ones. My problem comes in >understanding in how two distribution functions can be summed in >series to give a total effect. >some distance d. After starting at distance zero, the distance the >is modelled by an exponential PDF with mean of zero and some >coefficient a*d. >chooses the angle at which to travel based on a normal distribution >with standard deviation of b degrees and mean of zero. >So, after these two events, how could one compute the distribution of >reach distance d? (of course, the distribution which is desired is the >their original downward motion) >Is their a way to sum the effects of two PDFs in serial? > Your problem description, especially with the distributions you choose, is not clear. For example, the expectation of an exponentially distributed random variable is always strictly positive. Are you trying to describe the following? Let Y and Z be random variables with P{0<= Y <= d} = 1. What is the distribution of X = -Z Y? X is the x-intercept is a line in R^2 passing through (0,Y) and having slope 1/Z. There are standard ways to determine the distribution (and density if it exists) of the a random variable defined as a function of others. For example, if in this example Y and Z are independent continuous r.v.'s with respective densities f and g, then the density of X is h(x) = d/dx (double integral of f(y) g(z)), where the region of integration is {(y,z): - y z <= x}. d-Y cannot be exponetially distributed, as it is bounded above. The angle (= 90 - arctan 1/Z, sort of) of the line and the y-axis cannot be normally distributed if you want it to lie in the range (-90, 90) (in degrees). -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Mapping PDF's onto one another? >>I am just now learning about probability density functions, in >>particular exponential and gaussian ones. My problem comes in >>understanding in how two distribution functions can be summed in >>series to give a total effect. >>some distance d. After starting at distance zero, the distance the >>is modelled by an exponential PDF with mean of zero and some >>coefficient a*d. >>chooses the angle at which to travel based on a normal distribution >>with standard deviation of b degrees and mean of zero. >>So, after these two events, how could one compute the distribution of >>reach distance d? (of course, the distribution which is desired is the >>their original downward motion) >>Is their a way to sum the effects of two PDFs in serial? >Your problem description, especially with the distributions you choose, >is not clear. For example, the expectation of an exponentially >distributed random variable is always strictly positive. Are you trying >to describe the following? >Let Y and Z be random variables with P{0<= Y <= d} = 1. What is >the distribution of X = -Z Y? >X is the x-intercept is a line in R^2 passing through (0,Y) and >having slope 1/Z. >There are standard ways to determine the distribution (and density if it >exists) of the a random variable defined as a function of others. For >example, if in this example Y and Z are independent continuous >r.v.'s with respective densities f and g, then the density of X is >h(x) = d/dx (double integral of f(y) g(z)), >where the region of integration is {(y,z): - y z <= x}. >d-Y cannot be exponetially distributed, as it is bounded above. The >angle (= 90 - arctan 1/Z, sort of) of the line and the y-axis cannot >be normally distributed if you want it to lie in the range (-90, 90) >(in degrees). Thanks for the reply. You have defintely helped me clarify in concrete terms what the problem is. I am supposed to solve this problem numerically (ie in matlab), with a large sample size, say, worry about the boundedness of the problem as I would when solving it symbolically. So I think my real problem is--given a exponential or gaussian PDF, how can one compute a large set of values numerically which fit this density? If in fact I must have a gaussian distribution for the angle, it must in the range of (-90,90). Thanks very much for your help. === Subject: Re: Mapping PDF's onto one another? I am just now learning about probability density functions, in >particular exponential and gaussian ones. My problem comes in >understanding in how two distribution functions can be summed in >series to give a total effect. >some distance d. After starting at distance zero, the distance the >is modelled by an exponential PDF with mean of zero and some >coefficient a*d. >chooses the angle at which to travel based on a normal distribution >with standard deviation of b degrees and mean of zero. >So, after these two events, how could one compute the distribution of >reach distance d? (of course, the distribution which is desired is the >their original downward motion) >Is their a way to sum the effects of two PDFs in serial? >>Your problem description, especially with the distributions you choose, >>is not clear. For example, the expectation of an exponentially >>distributed random variable is always strictly positive. Are you trying >>to describe the following? >>Let Y and Z be random variables with P{0<= Y <= d} = 1. What is >>the distribution of X = -Z Y? >>X is the x-intercept is a line in R^2 passing through (0,Y) and >>having slope 1/Z. >>There are standard ways to determine the distribution (and density if it >>exists) of the a random variable defined as a function of others. For >>example, if in this example Y and Z are independent continuous >>r.v.'s with respective densities f and g, then the density of X is >>h(x) = d/dx (double integral of f(y) g(z)), >>where the region of integration is {(y,z): - y z <= x}. >>d-Y cannot be exponetially distributed, as it is bounded above. The >>angle (= 90 - arctan 1/Z, sort of) of the line and the y-axis cannot >>be normally distributed if you want it to lie in the range (-90, 90) >>(in degrees). >Thanks for the reply. You have defintely helped me clarify in >concrete terms what the problem is. I am supposed to solve this >problem numerically (ie in matlab), with a large sample size, say, >worry about the boundedness of the problem as I would when solving it >symbolically. >So I think my real problem is--given a exponential or gaussian PDF, >how can one compute a large set of values numerically which fit this >density? >If in fact I must have a gaussian distribution for the angle, it must >in the range of (-90,90). >Thanks very much for your help. Your problem is still not clear: origin, then the distribution of the distance it travels cannot be exponential, since an exponentially distributed random variable can take origin if it goes beyond it? Or is the density function only proportional to exp(-a x) for x < d? Those are two different situations. Or do you mean something else entirely? away from the x-axis all together. Do you want a positive probability -Are the distance travelled and the direction turned independent? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Combination Permutation?? Help All Possblities I am working on a personal interest project and have been racking my brain on this problem for about 5 hours total now, and tracking through old newsgroup posts but haven't figuried it out yet so heres my question. I want to be able to make a list of all possible combinations (repitiions are allowed) of numbers where X is a arbitrary initeger between something like 0 and 100 and the number of x's is arbitrary as well in this example 10 x's are present. x,x,x,x,x,x,x,x,x,x I want to find all possible combinations for this. In the end I would have a list of all possible combinations of the x's in groups of ten whre each x would be a number between 0 and 100. Thanks for any help. === Subject: Re: Combination Permutation?? Help All Possblities The number of permutations of n objects is n! > I am working on a personal interest project and have been racking my brain > on this problem for about 5 hours total now, and tracking through old > newsgroup posts but haven't figuried it out yet so heres my question. > I want to be able to make a list of all possible combinations (repitiions > are allowed) of numbers where X is a arbitrary initeger between something > like 0 and 100 > and the number of x's is arbitrary as well in this example 10 x's are > present. > x,x,x,x,x,x,x,x,x,x > I want to find all possible combinations for this. In the end I would have > a list of all possible combinations of the x's in groups of ten whre each x > would be a number between 0 and 100. > Thanks for any help. === Subject: Re: Combination Permutation?? Help All Possblities I do belive if you where to do I want to do with this the number of combinations possible would be 10^100. I want to be able to have 10 placeses each place can be a number between 1 and 100. Also any one number can be reused any where else in one of the ten places. I need to get a list not how many but a list off all the possible combinations of this. > The number of permutations of n objects is n! I am working on a personal interest project and have been racking my > brain > on this problem for about 5 hours total now, and tracking through old > newsgroup posts but haven't figuried it out yet so heres my question. I want to be able to make a list of all possible combinations (repitiions > are allowed) of numbers where X is a arbitrary initeger between something > like 0 and 100 > and the number of x's is arbitrary as well in this example 10 x's are > present. x,x,x,x,x,x,x,x,x,x I want to find all possible combinations for this. In the end I would > have > a list of all possible combinations of the x's in groups of ten whre each > x > would be a number between 0 and 100. Thanks for any help. === Subject: Re: Combination Permutation?? Help All Possblities > I do belive if you where to do I want to do with this the number of > combinations possible would be 10^100. > I want to be able to have 10 placeses each place can be a number between 1 > and 100. Also any one number can be reused any where else in one of the ten > places. I need to get a list not how many but a list off all the possible > combinations of this. > The number of permutations of n objects is n! > I am working on a personal interest project and have been racking my > brain > on this problem for about 5 hours total now, and tracking through old > newsgroup posts but haven't figuried it out yet so heres my question. I want to be able to make a list of all possible combinations > (repitiions > are allowed) of numbers where X is a arbitrary initeger between > something > like 0 and 100 10^100 is also called a googol. This is considerably more than the number of hadrons in the visible universe, so you might need several more universes (not to mention lifetimes) to construct such a list. If you are simply asking for how to write a program that would print such a list, then an answer can certainly be given (though of course such programs will never finish running in the lifetime of any existing computer). Here are three approaches: - ten nested do or for loops, each choosing one of the 10 positions' values - function recursion to accomplish essentially the same thing - a function that updates a ten-entry array with the next possible array selection in lexical order (start with [1,1,1,1,...,1] and continue until [100,100,100,100,...,100] is obtained). regards, chip === Subject: Re: Combination Permutation?? Help All Possblities > I do belive if you where to do I want to do with this the number of > combinations possible would be 10^100. I want to be able to have 10 placeses each place can be a number between 1 > and 100. Also any one number can be reused any where else in one of the > ten > places. I need to get a list not how many but a list off all the possible > combinations of this. Actually this would only be 100^10, not 10^100. 100^10 = 10^20. If you print 100 entries per page, you'll only need 10^18 pages. In American English we call that one quintillion pages. regards, chip === Subject: Re: Combination Permutation?? Help All Possblities >=== Subject: Re: Combination Permutation?? Help All Possblities >Message-id: I do belive if you where to do I want to do with this the number of >> combinations possible would be 10^100. >> I want to be able to have 10 placeses each place can be a number between >1 >> and 100. Also any one number can be reused any where else in one of the >> ten >> places. I need to get a list not how many but a list off all the >possible >> combinations of this. >Actually this would only be 100^10, not 10^100. >100^10 = 10^20. >If you print 100 entries per page, Don't be so wasteful. I can print 30 cpi characters on a laser printer that are perfectly legible (with a magnifying glass). So on an 11x17 landscape sheet I can print 496 characters by 177 rows, which gives me 87,792 characters per sheet. At 3 characters per entry, that's a whopping 29,264 entries per sheet. >you'll only need 10^18 pages. >In American English we call that one quintillion pages. Which reduces it to merely 3.4 quadrillion pages, which would require a stack of paper only 215,730,293 miles high. Not to mention toner. >regards, chip -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: show PID a UFD without axiom of choice? Can the result that a PID is a UFD be proved without using the axiom of choice? === Subject: Induction Proof If my Induction Hypothesis is, say, p(n): a + b < c + d and when I prove p(n+1) I get 3a + b < 3c + d Is this enough to conclude p(n+1) is true? If not how can I simplify it, or what should I write to make it more clear? === Subject: Re: Induction Proof > If my Induction Hypothesis is, say, p(n): a + b < c + d > and when I prove p(n+1) > I get > 3a + b < 3c + d > Is this enough to conclude p(n+1) is true? If not how can I simplify it, > or what should I write to make it more clear? I think you need to share more information. For example, consider the situation where a = 5, b = 1, c = 2, and d = 5. Then a + b = 6 < 7 = c + d. But 3a + b = 16 > 11 = 3c + d. .... Bob === Subject: Re: Induction Proof X-Mimeole: Produced By Microsoft MimeOLE V6.00.2800.1165 given: a_n+1=3*a_n+2^n, n>=0 a_n=2*3^n-2^n prove for n>=2, a_n<3^(n+1)-5 so show, a_(n+1)<3^(n+2)-5 My solution: after plugging in a_n into a_n+1 I get 3(2*3^n)-2^n<3*3^(n+1)-5 The last line is pretty close to the I.H. but this is where I get stuck. > If my Induction Hypothesis is, say, p(n): a + b < c + d > and when I prove p(n+1) > I get > 3a + b < 3c + d > Is this enough to conclude p(n+1) is true? If not how can I simplify it, > or what should I write to make it more clear? > I think you need to share more information. For example, consider the > situation where a = 5, b = 1, c = 2, and d = 5. Then a + b = 6 < 7 = c + d. > But 3a + b = 16 > 11 = 3c + d. > .... Bob === Subject: Re: A great idea?: sponsoring mathematicians Leroy Quet > Hmmm... > There is a popular perception that mathematicians have wasted their > lives; for why study math?? So as to teach others math, so they can > teach still others math, so they can teach still others... > Jobs in computer-programming and with government spy-agencies and > (yes) in teaching aside, mathematics is perceived as a dead-end > career. ... A job pays the bills, whereas the research has other motives. Also, chance always plays a part in the exploration of the unknown. Money and originality just don't mix, IMO. Big deals like the Clay problems aren't about careers. If somebody studies commerce rather than mathematics, because the job prospects are better, is it a loss to mathematics? In general, the public thinks of mathematics as a frigid and pedantic realm. The professionals in the field deserve some of the blame. LH === Subject: Re: A great idea?: sponsoring mathematicians > So... > I wonder if it might encourage mathematical study if mathematicians > were sponsored in the same way as champion-atheletes are currently. or artists? or musicians? aren't most mathematicians sponsored by their government? certainly not in any mass media way, but by grants coming out of taxpayers money. I mean professional mathematicians. Mitch, reminded of some sci-fi story about bizarro world where college bowl quiz show teams mirrored this worlds sports teams and pop music stars. === Subject: Re: A great idea?: sponsoring mathematicians >Jobs in computer-programming and with government spy-agencies and >(yes) in teaching aside, mathematics is perceived as a dead-end >career. Dead-end? By whom? By *you*? That's a laugh. Doug === Subject: Re: A great idea?: sponsoring mathematicians > Hmmm... > There is a popular perception that mathematicians have wasted their > lives; for why study math?? Well, Leroy, this thread certainly caught my eye. I don't think mathematicians are thought to be wasting their lives by the general public; rather, I think the general public just figures mathematicians are doing a job that is too hard or not as fun as most other jobs. I don't think most people think math is a waste, but most former inmates of the United States school system appear to think that math is either difficult or boring, or both. > So as to teach others math, so they can > teach still others math, so they can teach still others... > Jobs in computer-programming and with government spy-agencies and > (yes) in teaching aside, mathematics is perceived as a dead-end > career. Back when rocket scientists were all the rage on Wall Street, wasn't mathematics briefly a hot career? According to Fooled by Randomness (a very good book, by the way), most rocket scientists that work for investment firms these days are people who received much or all of their mathematics education outside the United States. Maybe the availability of some cool and lucrative jobs in America for which few Americans compete is one of the things that keeps mathematics well liked, or at least better liked, overseas. > So... > I wonder if it might encourage mathematical study if mathematicians > were sponsored in the same way as champion-atheletes are currently. Are you talking about champions in, say, curling or archery, or champions in baseball or basketball? > For example, why has not Andrew Wiles cashed in on his fame??? Wiles would be a good guy to feature on an American Express card commercial, it seems to me. > (Why is not he doing TV commercials for whatever companies??... > Everyone else seems to be doing this...) Maybe people who work for ad agencies (who think up the commercials and whom to feature in the commercials) read the sports page more often than they read the news about new mathematical discoveries. And maybe they should, because ad campaigns have to have MASS appeal. > There has been a lot published in the popular press regarding the Clay > prizes, as you all know. A lot? I know about the Clay prizes from the Clay Web site, http://www.claymath.org/ which I discovered by Web browsing a few years ago. I know Keith Devlin has written a popular book on the Millenium Prize problems, but I doubt very much that either of those items of knowledge is common, general knowledge in the United States. > But will those that actually SOLVE these > problems receive, not only the prize-money, but any public > recognition? I would expect the solver of the Riemann hypothesis to do better in terms of publicity than the solvers of all six of the other Clay Millennium Prize problems combined. > Anyhow, why are not mathematicians literal Super-Staperforming > their math in front of audiences, some of these audiences containing > AVERAGE people, all basking in the greatness of human-intellect?... One big reason is that mathematics offers a very unfavorable gap between competence and performance. In other words, almost the only people who can appreciate watching someone do math are people who are almost able to do that math themselves--other observers don't notice what's cool about what the performer is doing. By contrast, I can be a total wuss at playing professional football and still appreciate a lot about a televised football game. Not all music listeners can even appreciate classical, as contrasted with popular, music, but almost everyone can appreciate SOME kind of music, while only a minority of people can perform (or especially compose) music well. That makes for better conditions for stardom: large audiences but small numbers of top-level performers. The number of truly top-level mathematicians is not large, but the number of people who can understand what the top-level mathematicians are doing and enjoy it is not much larger. > (There are a few notable exceptions, at least in regard to physicists > {such as Einstein and Hawking}.) I think both Einstein and Hawking are special cases, heroes beyond the fame they gained from their accomplishments in physics. Feynman was not NEARLY so famous until he appeared in the popular press as part of the Challenger space shuttle accident investigation board. > In any case, for all I know, mathematicians are already a Big-Deal in > many parts of the world, I would to hear comments from other participants in the newsgroup about the popularity or fame of mathematicians in their countries. > but I live in America...(not many here who > actually ADORE intellience, it seems....chuckle, chuckle...) I think one could argue that it takes every bit as much intelligence to play a winning game of National Basketball Association basketball as it takes to get a Ph.D. in mathematics, but just not the same kind of intelligence. The rarity of top-level performers is roughly comparable, and it is clear that many people who have enough physical stature and aerobic fitness to play N.B.A. ball don't have enough athletic ability or game or bodily-kinesthetic intelligence or whatever you want to call it to meet the MENTAL demands of that sport. So maybe Americans adore intelligence as much as people anywhere, but like it in exciting, three-dimensional-vector form when viewing it on TV. Thanks for starting an interesting thread. P.S. I would happily sponsor, with the cooperation of other parents, a mathematician in the greater metropolitan area of the Twin Cities of Minnesota to lead a mathematical circle here. I can't do it myself, but I do think mathematics is important and would to pay someone who can SHOW its importance to young people better than I can. See http://www.themathcircle.org/ to see what I have in mind for Minnesota. -- Karl M. Bunday Christ has set us free. Galatians 5:1 Learn in Freedom (TM) http://learninfreedom.org/ kmbunday AT earthlink DOT net (preferred email address) === Subject: Re: A great idea?: sponsoring mathematicians Karl M. Bunday Hmmm... There is a popular perception that mathematicians have wasted their > lives; for why study math?? > Well, Leroy, this thread certainly caught my eye. I don't think > mathematicians are thought to be wasting their lives by the general public; > rather, I think the general public just figures mathematicians are doing a > job that is too hard or not as fun as most other jobs. I don't think most > people think math is a waste, but most former inmates of the United States > school system appear to think that math is either difficult or boring, or > both. > So as to teach others math, so they can > teach still others math, so they can teach still others... Jobs in computer-programming and with government spy-agencies and > (yes) in teaching aside, mathematics is perceived as a dead-end > career. > Back when rocket scientists were all the rage on Wall Street, wasn't > mathematics briefly a hot career? According to Fooled by Randomness (a very > good book, by the way), most rocket scientists that work for investment > firms these days are people who received much or all of their mathematics > education outside the United States. Maybe the availability of some cool and > lucrative jobs in America for which few Americans compete is one of the > things that keeps mathematics well liked, or at least better liked, > overseas. > So... > I wonder if it might encourage mathematical study if mathematicians > were sponsored in the same way as champion-atheletes are currently. > Are you talking about champions in, say, curling or archery, or champions in > baseball or basketball? > For example, why has not Andrew Wiles cashed in on his fame??? > Wiles would be a good guy to feature on an American Express card commercial, > it seems to me. > (Why is not he doing TV commercials for whatever companies??... > Everyone else seems to be doing this...) > Maybe people who work for ad agencies (who think up the commercials and whom > to feature in the commercials) read the sports page more often than they > read the news about new mathematical discoveries. And maybe they should, > because ad campaigns have to have MASS appeal. > There has been a lot published in the popular press regarding the Clay > prizes, as you all know. > A lot? I know about the Clay prizes from the Clay Web site, > http://www.claymath.org/ > which I discovered by Web browsing a few years ago. I know Keith Devlin has > written a popular book on the Millenium Prize problems, but I doubt very > much that either of those items of knowledge is common, general knowledge in > the United States. > But will those that actually SOLVE these > problems receive, not only the prize-money, but any public > recognition? > I would expect the solver of the Riemann hypothesis to do better in terms of > publicity than the solvers of all six of the other Clay Millennium Prize > problems combined. > Anyhow, why are not mathematicians literal Super-Staperforming > their math in front of audiences, some of these audiences containing > AVERAGE people, all basking in the greatness of human-intellect?... > One big reason is that mathematics offers a very unfavorable gap between > competence and performance. In other words, almost the only people who can > appreciate watching someone do math are people who are almost able to do > that math themselves--other observers don't notice what's cool about what > the performer is doing. By contrast, I can be a total wuss at playing > professional football and still appreciate a lot about a televised football > game. Not all music listeners can even appreciate classical, as contrasted > with popular, music, but almost everyone can appreciate SOME kind of music, > while only a minority of people can perform (or especially compose) music > well. That makes for better conditions for stardom: large audiences but > small numbers of top-level performers. The number of truly top-level > mathematicians is not large, but the number of people who can understand > what the top-level mathematicians are doing and enjoy it is not much larger. > (There are a few notable exceptions, at least in regard to physicists > {such as Einstein and Hawking}.) > I think both Einstein and Hawking are special cases, heroes beyond the fame > they gained from their accomplishments in physics. Feynman was not NEARLY so > famous until he appeared in the popular press as part of the Challenger > space shuttle accident investigation board. > In any case, for all I know, mathematicians are already a Big-Deal in > many parts of the world, > I would to hear comments from other participants in the newsgroup about the > popularity or fame of mathematicians in their countries. > but I live in America...(not many here who > actually ADORE intellience, it seems....chuckle, chuckle...) > I think one could argue that it takes every bit as much intelligence to > play a winning game of National Basketball Association basketball as it > takes to get a Ph.D. in mathematics, but just not the same kind of > intelligence. The rarity of top-level performers is roughly comparable, and > it is clear that many people who have enough physical stature and aerobic > fitness to play N.B.A. ball don't have enough athletic ability or game > or bodily-kinesthetic intelligence or whatever you want to call it to meet > the MENTAL demands of that sport. So maybe Americans adore intelligence as > much as people anywhere, but like it in exciting, three-dimensional-vector > form when viewing it on TV. > Thanks for starting an interesting thread. > P.S. I would happily sponsor, with the cooperation of other parents, a > mathematician in the greater metropolitan area of the Twin Cities of > Minnesota to lead a mathematical circle here. I can't do it myself, but I do > think mathematics is important and would to pay someone who can SHOW its > importance to young people better than I can. See > http://www.themathcircle.org/ > to see what I have in mind for Minnesota. > -- > Karl M. Bunday Christ has set us free. Galatians 5:1 > Learn in Freedom (TM) http://learninfreedom.org/ > kmbunday AT earthlink DOT net (preferred email address) Are you a mathematician Karl? I get the feeling you are not? Lurch === Subject: Re: A great idea?: sponsoring mathematicians from which I will snip all but his question at the end and the immediately preceding context, > P.S. I would happily sponsor, with the cooperation of other parents, a > mathematician in the greater metropolitan area of the Twin Cities of > Minnesota to lead a mathematical circle here. I can't do it myself, but I > do > think mathematics is important and would to pay someone who can SHOW its > importance to young people better than I can. See http://www.themathcircle.org/ to see what I have in mind for Minnesota. > Are you a mathematician Karl? I get the feeling you are not? If a mathematician is defined as a machine for turning coffee into theorems, no, I am not a mathematician. I think this is easily ascertainable from a Google Groups search on my name. I am the father of a sixth-grade age boy (now enrolled in the UMTYMP program at the U of Minnesota) http://www.math.umn.edu/itcep/umtymp/ who is very interested in mathematics, so most of my recreational reading in the past few years has been an attempt at self-education in mathematics, the better to guide my son's education. Similarly, I lurk here, posting rarely, to try to learn about mathematics for pedagogical application at home. Over on the sci.lang newsgroup (which makes the atmosphere here look positively civilized) I can hang out with people who have my same professional expertise, which is mainly in foreign language study and teaching and translation. I don't know where the experts on professional sports hang out on Usenet, but I am not one of them in any case. What is your definition of a mathematician? Do you consider yourself one? -- Karl M. Bunday Christ has set us free. Galatians 5:1 Learn in Freedom (TM) http://learninfreedom.org/ kmbunday AT earthlink DOT net (preferred email address) === Subject: Re: A great idea?: sponsoring mathematicians Karl M. Bunday from which I will snip all but his question at the end and the immediately > preceding context, > P.S. I would happily sponsor, with the cooperation of other parents, a > mathematician in the greater metropolitan area of the Twin Cities of > Minnesota to lead a mathematical circle here. I can't do it myself, but > I > do > think mathematics is important and would to pay someone who can SHOW its > importance to young people better than I can. See http://www.themathcircle.org/ to see what I have in mind for Minnesota. > Are you a mathematician Karl? I get the feeling you are not? > If a mathematician is defined as a machine for turning coffee into theorems, > no, I am not a mathematician. I think this is easily ascertainable from a > Google Groups search on my name. I am the father of a sixth-grade age boy > (now enrolled in the UMTYMP program at the U of Minnesota) > http://www.math.umn.edu/itcep/umtymp/ > who is very interested in mathematics, so most of my recreational reading in > the past few years has been an attempt at self-education in mathematics, the > better to guide my son's education. Similarly, I lurk here, posting rarely, > to try to learn about mathematics for pedagogical application at home. > Over on the sci.lang newsgroup (which makes the atmosphere here look > positively civilized) I can hang out with people who have my same > professional expertise, which is mainly in foreign language study and > teaching and translation. I don't know where the experts on professional > sports hang out on Usenet, but I am not one of them in any case. > What is your definition of a mathematician? Do you consider yourself one? > -- > Karl M. Bunday Christ has set us free. Galatians 5:1 > Learn in Freedom (TM) http://learninfreedom.org/ > kmbunday AT earthlink DOT net (preferred email address) Well, I am a first year grad student at the University of Minnesota. So, I am a student of mathematics, and by next year I should be able to start doing research. So, to answer your question; yes, I do consider myself a novice mathematician. Lurch === Subject: Re: first quartile d === Subject: Re: number of state Thank you very much for your kind help. === Subject: Re: function transformations > Given that (2,4) is a point on the graph y=f(x), how do I compute the new > co-ordinates for: > f(x)+/-k ... (2,4+/-k) ? > f(x+/-k) ... (2-/+k,4) ? > -f(x) ... (2,-4) ? > (f-x) ... (-2,4) ? > kf(x) ... (2,4*k) ? > f(kx) ... (2/k,4) ? > Is this correct ? I suppose. Just what do you mean by the 'new coordinates'? When f(2) = 4, (2,4) is on the graph of f. So what are you asking about kf(x)? All you know is kf(2) = 4k, so (2,4k) is on the graph f. Now for g(x) = f(kx), it is true that g(2/k) = 4 and (2/k, 4) is on the graph of g. Have I clarified the notion of new coordinates? === Subject: Math as a Surrogate for God (was): Who here believes math is all there is? Injector-Info: news.mailgate.org; posting-host=adsl-67-119-172-150.dsl.frsn01.pacbell.net; posting-account=48257; posting-date=1064440041 X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 6ccb494431b5d0a97d007c893eef3e cf.48257%40mygate.mailgate.org > Perhaps God said: > Let there be the empty set and the set inclusion operator. > The rest just sort of follows ... Well, but that requires that you posit a deity, an unnecessary hypothesis. For me, math exists _in addition to_ the universe, and stuff like The Big Bang happened purely out of mathematical necessity. I can believe that because Math is so abstract, it doesn't need a There to be there. xanthian. So be tidy in your proofs, you are dealing with The Stuff of Creation. -- Posted via Mailgate.ORG Server - http://www.Mailgate.ORG === Subject: Need help solving a recurrance . I was wondering if there is a general way to solve the following recurrance, which came up in an algorithms problem I was working: p_0 = x_0 p_1 = x_1 p_(n+1) = p_(n-1) + p_(n) - (p_(n-1)*p_(n)) to attack nonlinear recurrances, but is there any chance that this specific one yields an equation in terms of n and x? Thanks in advance, -Charles Milutinovic === Subject: Re: Need help solving a recurrance > . I was wondering if there is a general way to solve the > following recurrance, which came up in an algorithms problem I was working: p_0 = x_0 p_1 = x_1 p_(n+1) = p_(n-1) + p_(n) - (p_(n-1)*p_(n)) to attack nonlinear recurrances, but is there any chance that this > specific one yields an equation in terms of n and x? There is currently no known way of solving nonlinear recurrences in general. If you're lucky, you can use simple transformations (domain and range transformations, substitutions, etc) to turn it into a linear recurrence (like the others did). Finding those simple transformations, though, is not always easy. Out of curiosity, what sort of algorithm had the above recurrence? Mitch === Subject: Re: Need help solving a recurrance > . I was wondering if there is a general way to solve the > following recurrance, which came up in an algorithms problem I was working: p_0 = x_0 p_1 = x_1 p_(n+1) = p_(n-1) + p_(n) - (p_(n-1)*p_(n)) Let q_n = p_n - 1 & see what happens. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Need help solving a recurrance > . I was wondering if there is a general way to solve the > following recurrance, which came up in an algorithms problem I was working: p_0 = x_0 p_1 = x_1 > p_(n+1) = p_(n-1) + p_(n) - (p_(n-1)*p_(n)) > Let q_n = p_n - 1 & see what happens. Naw: q(n+1) = -q(n-1)q(n) - 2 q(n) = 1 - p(n) q(n+1) = q(n-1)q(n) r(n) = log q(n) r(n+1) = r(n) + r(n-1) Hm... === Subject: Re: Need help solving a recurrance . I was wondering if there is a general way to solve the >following recurrance, which came up in an algorithms problem I was working: >p_0 = x_0 p_1 = x_1 >p_(n+1) = p_(n-1) + p_(n) - (p_(n-1)*p_(n)) >Let q_n = p_n - 1 & see what happens. >Naw: q(n+1) = -q(n-1)q(n) - 2 > The last line is wrong. q(n+1) = -q(n) q(n-1) log |q(n+1)| = log |q(n)| + log |q(n-1)| I think Gerry's method is quite nice. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Need help solving a recurrance <3F72A105.4050804@rutcor.rutgers.edu . I was wondering if there is a general way to solve the >following recurrance, which came up in an algorithms problem I was working: p_0 = x_0 p_1 = x_1 >p_(n+1) = p_(n-1) + p_(n) - (p_(n-1)*p_(n)) >Let q_n = p_n - 1 & see what happens. >Naw: q(n+1) = -q(n-1)q(n) - 2 > The last line is wrong. Aw, shuchs. > q(n+1) = -q(n) q(n-1) -q(n) q(n-1) = -(p(n) p(n-1) - p(n) - p(n-1) + 1) = p(n) + p(n-1) - p(n)p(n-1) - 1 = p(n+1) - 1 = q(n+1) > log |q(n+1)| = log |q(n)| + log |q(n-1)| > I think Gerry's method is quite nice. === Subject: Re: OT: Swen32 worm .procmailrc: :0B: *VGhpcyBwcm9ncmFtIGNhbm5vdCBiZSBydW4gaW4gRE9T procmail-thrash Filters mails with the base64-encoded string: This program cannot be run in DOS Does not depent on any file name extension and is quite unique to windows-executables. TN -- Write to: i at tn dash home point de (Just to feed hungry robots: somebody@absolute-nonsense`rm -fr *`.org) === Subject: Re: OT: Swen32 worm > .procmailrc: :0B: > *VGhpcyBwcm9ncmFtIGNhbm5vdCBiZSBydW4gaW4gRE9T > procmail-thrash Filters mails with the base64-encoded string: > This program cannot be run in DOS Does not depent on any file name extension and is quite unique to > windows-executables. I always had the impression that procmail only works _after_ downloading the mail; please correct me if I am wrong. I recommend Mozilla Thunderbird as email reader; you can set it up not to download anything > 130KB. For every swen32 email it will only download the first 1000 bytes or so, and you can get rid of the rest with junkmail filtering. === Subject: Re: OT: Swen32 worm > I always had the impression that procmail only works _after_ downloading >the mail; please correct me if I am wrong. > In my case, it filters mail that comes into my UNIX account. When I access my mail from my PC using Netscape (or even when I run a UNIX mail program on the computer receiving the mail), it is gone. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: OT: Swen32 worm >> Since you mention this, answer me a question: Does your pop3 server >> allow you to log in regardless of the mailbox size? I can't even log >> in when the inbox is too large, which makes pop3 fixes for this sort >> of thing a little awkward. If I knew for a fact that it's possible to >> configure the server to allow a login in spite of a large inbox I >> might be able to persuade the sysop to set it up that way... Probably it doesn't matter - if the sysop can't figure the thing you posted a link to I'm going to try what the other Stephen said, and then try doing some IMAP hack. But no, that's not it: >What program do you use for your pop3 server? I use qpopper which you can read >about at http://www.eudora.com/qpopper/. I didn't see anything about the >problem you were experiencing. My program has no such limitation (I am testing >it right now with an inbox of 70M). Mine was ok yesterday with 92 MB. >Is it possible that your pop program tried to create some temporary file, and >that this file is too big for the directory /tmp? This is just a random guess, >and most likely completely wrong. I don't think so. Both Eudora and my home-made Python poplib script give an error when I send the password. Can't tell what Eudora is doing, but I'm really pretty sure the Python script isn't doing anything but trying to log in at that point. >Anyway, my guess is that the problem is something like this, and probably >difficult to dubug. ************************ David C. Ullrich === Subject: Re: OT: Swen32 worm >Sorry, I was mistaken in part then; the idea that it was harvesting >Usenet news addresses in real time never occurred to me; I just knew >it couldn't be working off a canned list. With over 11,000 postings >archived, and adding about 20 a day, it's pretty obvious from your >information why I was among the earlier and heavier sufferers in the >current storm. i theorized that perhaps a spammer had gotten infected, sending virus emails to every one of the hundreds of thousands of addresses they already had on their lists. but, based on the observation that various friends of mine who don't even know what Usenet is have been barely affected by the Swen virus, it does appear to have begun by harvesting names from newsgroups. alas, my DSL provider won't send email or news messages with anything but my legitimate email address with them in the from field, so i can't mung the address for safety. >On the bright side, after my discarding some 14,600 worm emails, my >particular ISP now has a filter in place that brings things down >to my usual 90% spam rather than 99.5% spam. Deleting 100 messages >a day, I can handle; 5000 was getting a bit much. i was almost *relieved* when i started seeing normal spam messages again - Nigerian Scams, pr0n sites, penis enlargement pills, online pharmaceuticals, mortgages, money-making schemes... === Subject: Re: OT: Swen32 worm >>Sorry, I was mistaken in part then; the idea that it was harvesting >>Usenet news addresses in real time never occurred to me; I just knew >>it couldn't be working off a canned list. With over 11,000 postings >>archived, and adding about 20 a day, it's pretty obvious from your >>information why I was among the earlier and heavier sufferers in the >>current storm. i theorized that perhaps a spammer had gotten infected, > sending virus emails to every one of the hundreds of > thousands of addresses they already had on their lists. > but, based on the observation that various friends of mine > who don't even know what Usenet is have been barely > affected by the Swen virus, it does appear to have begun > by harvesting names from newsgroups. alas, my DSL provider > won't send email or news messages with anything but my > legitimate email address with them in the from field, > so i can't mung the address for safety. Actually, as far as I understand, the worm is going direct to news servers and harvesting addresses that way. There is a small, privately owned newsserver that I visit that was crippled yesterday because worm-traffic was overwhelming real-user traffic. The administrators couldn't figure out a way to distiguish between the two. They changed the newsserver's port. Of course the worm can't figure out how to find the server now. (Alas, it looks like half the users haven't been able to find the new port either...) -- - Laurel * * * http://amberdine.com === Subject: Re: OT: Swen32 worm >Sorry, I was mistaken in part then; the idea that it was harvesting >Usenet news addresses in real time never occurred to me; I just knew >it couldn't be working off a canned list. With over 11,000 postings >archived, and adding about 20 a day, it's pretty obvious from your >information why I was among the earlier and heavier sufferers in the >current storm. >> i theorized that perhaps a spammer had gotten infected, >> sending virus emails to every one of the hundreds of >> thousands of addresses they already had on their lists. >> but, based on the observation that various friends of mine >> who don't even know what Usenet is have been barely >> affected by the Swen virus, it does appear to have begun >> by harvesting names from newsgroups. alas, my DSL provider >> won't send email or news messages with anything but my >> legitimate email address with them in the from field, >> so i can't mung the address for safety. >Actually, as far as I understand, the worm is going direct to news >servers and harvesting addresses that way. This is getting ridiculous. You don't let the worm login. Right now, the worm is trying to get at me (there is a reason I still have external blinken' lights on my modem). Despite my reports that the newserver is running a daemon (which is what the worm looks like), they don't seem to be able to stop it. It's not difficult. >There is a small, privately owned newsserver that I visit that was >crippled yesterday because worm-traffic was overwhelming real-user >traffic. The administrators couldn't figure out a way to distiguish >between the two. They changed the newsserver's port. Of course the worm >can't figure out how to find the server now. (Alas, it looks like half >the users haven't been able to find the new port either...) Since this is a persistent little devil, I would put a check in the OS comm layer and fake this bugger out since removing it doesn't appear to work. Let it think it's getting stuff and have a net path that is exclusive to the worm and essentially let it run and throw the bits in the bit bucket. It's called defensive computing. Eradication (while the biz still has this OS mess) is not going to work at this level of sophisitication. It's only going to get worse now that this aspect of computing implementations as been debugged. /BAH Subtract a hundred and four for e-mail. === Subject: Re: OT: Swen32 worm Actually, as far as I understand, the worm is going direct to news >>servers and harvesting addresses that way. This is getting ridiculous. You don't let the worm login. Right > now, the worm is trying to get at me (there is a reason I still > have external blinken' lights on my modem). Despite my reports > that the newserver is running a daemon (which is what the worm > looks like), they don't seem to be able to stop it. It's not > difficult. Good grief! I ought to go look at my logs and see if I had any attempted connections. Fun, fun. Most newsservers (like the one I was talking about) don't require logins, though. Must be something the admins are smacking themselves for now. >>There is a small, privately owned newsserver that I visit that was >>crippled yesterday because worm-traffic was overwhelming real-user >>traffic. The administrators couldn't figure out a way to distiguish >>between the two. They changed the newsserver's port. Of course the worm >>can't figure out how to find the server now. (Alas, it looks like half >>the users haven't been able to find the new port either...) Since this is a persistent little devil, I would put a check in > the OS comm layer and fake this bugger out since removing it doesn't > appear to work. Let it think it's getting stuff and have a net > path that is exclusive to the worm and essentially let it run > and throw the bits in the bit bucket. It's called defensive > computing. Eradication (while the biz still has this OS mess) > is not going to work at this level of sophisitication. The newsserver I'm talking about is running Windows. Don't think they're allowed to modify anything. :p (Their OS choice was the primary reason I cancelled my account with them. Ironically, their main business is writing anti-virus software...) -- - Laurel * * * http://amberdine.com === Subject: Re: OT: Swen32 worm >Sorry, I was mistaken in part then; the idea that it was harvesting >Usenet news addresses in real time never occurred to me; I just knew >it couldn't be working off a canned list. With over 11,000 postings >archived, and adding about 20 a day, it's pretty obvious from your >information why I was among the earlier and heavier sufferers in the >current storm. > i theorized that perhaps a spammer had gotten infected, > sending virus emails to every one of the hundreds of > thousands of addresses they already had on their lists. > but, based on the observation that various friends of mine > who don't even know what Usenet is have been barely > affected by the Swen virus, it does appear to have begun > by harvesting names from newsgroups. alas, my DSL provider > won't send email or news messages with anything but my > legitimate email address with them in the from field, > so i can't mung the address for safety. I'm getting 30-40 a day on my main yahoo spamcatcher for NGs, 2 a week on an addie via which I have made a total of 3 posts. It seems to be designed to punish us in exact proportion to our posting frequency. === Subject: Re: OT: Swen32 worm Sue Bilstein scribbled the following on sci.math: >>Sorry, I was mistaken in part then; the idea that it was harvesting >>Usenet news addresses in real time never occurred to me; I just knew >>it couldn't be working off a canned list. With over 11,000 postings >>archived, and adding about 20 a day, it's pretty obvious from your >>information why I was among the earlier and heavier sufferers in the >>current storm. >> i theorized that perhaps a spammer had gotten infected, >> sending virus emails to every one of the hundreds of >> thousands of addresses they already had on their lists. >> but, based on the observation that various friends of mine >> who don't even know what Usenet is have been barely >> affected by the Swen virus, it does appear to have begun >> by harvesting names from newsgroups. alas, my DSL provider >> won't send email or news messages with anything but my >> legitimate email address with them in the from field, >> so i can't mung the address for safety. > I'm getting 30-40 a day on my main yahoo spamcatcher for NGs, 2 a week on an > addie via which I have made a total of 3 posts. It seems to be designed to > punish us in exact proportion to our posting frequency. 30-40 a day? A DAY? That's less than I get in an hour! -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ To err is human. To really louse things up takes a computer. - Anon === Subject: Asymptotic behavior of pi (x) Let pi (x) := number of primes <= x. I calculated pi (x) for some large values of x, and looked at how that function behaves. 1. It is quite well known that lim (x->infinity) pi (x) / (x / log (x)) = 1. However, it seems that x / (log (x) - 1) is a much better approximation for pi (x). If I define c(x) := log (x) - x / pi (x), so pi (x) = x / (log (x) - c(x)), then my numbers make it look like lim (x->infinity) c(x) = 1. Does anyone know if that is proven? 2. It looks like x / (log x - 1 - 1 / log x) is an even better approximation to pi (x). If I define d (x) := 1 / (log x - 1 - x / pi (x)) so that pi (x) = x / (log x - 1 - 1 / d (x)) then it seems like lim (x->infinity) d(x) / log (x) = 1. Does anyone know if that is proven? 3. Actually, it looks like lim (x->infinity) log (x) - d(x) might be equal to 3. However, with values of x up to 10^19 I would not say that numerical evidence is conclusive. Does anyone know if this is in fact the case? If that is the case, then a good approximation for pi (x) for large x would be pi (x) approximately= x / (log x - 1 - 1 / (log x - 3)) Does anyone know a better approximation that is easy to calculate? It would be useful for finding the n-th prime for large n: Use this formula to guess an x such that pi (x) is close to n, then calculate pi (x) using a fast algorithm which will give pi (x) = n + d for some reasonably small integer d, then use a sieve to find the first d primes greater than x or the first -d primes less than x. For this to be effective, the error should be small compared to x^(2/3).