mm-2169 === Subject: Re: Implicit surfaces and curves question. >Let we have two implicit convex surfaces F(r) = 0 and G(r) = 0. >How can we determine a intersection curve (implicit) I(r) if they do intersect? >The only way I can think of is I(r) = (F(r))^2 + (G(r))^2. That will work if (1) you are working over the real numbers, and (2) your interest in the intersection set (which is not necessarily a curve) is limited to knowing it as a topological subspace of 3-space. Since you use the word convex I assume that (1) is true, and *probably* (2) is good enough for you too. >Then, how I can take an integral on such implicitly represented closed curve >(excuse my non-professional slang)? How to determine a length of the curve? In general, both those problems might be very difficult (or impossible), depending on what you mean by take an integral and determine [the] length. Is a numerical approximation good enough for you, or do you insist on an expression `in closed form'? What do you know about F and G (other than convexity of their sets of zeros)? What, in fact, is the context where your problem arises? Lee Rudolph === Subject: JSH: Brainstorming Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY Years ago I started out trying to figure out some simple proof of Fermat's Last Theorem, almost as a lark, and found myself obsessed with the problem for several years. During that time I employed various problem solving techniques, including brainstorming, where you come up with a LOT of ideas, critique lightly at first, and hope to find something in the mess that's useful. Repeatedly I'd find ideas that I thought were worth talking out, which I strongly hoped would solve the problem and repeatedly I was wrong. Ok. I'm looking at a very old problem, trying to find something that many in the field no longer thinks exists, and I'm falling down a lot. No surprise. What was a surprise was the reaction of Usenet, where people said just about anything they could to try and control my posts, and behavior. Sure, it can be annoying that some guy is loudly proclaiming his own genius, and screwing up all the time, or making claims that turn out to be false, as he makes mistake after mistake after mistake working on what I think those of you who are sane would call a difficult problem. But it's Usenet. You can just ignore such people. But instead I had people calling me subhuman, continually questioning my sanity, TELLING me over and over again to stop posting. Talking sbout trying to change the group to allow greater censorship. Going on and on in vicious attacks meant to be personal, meant to cause emotional hurt. There was something wrong with math society as shown on Usenet. If you people loved mathematics, or even liked it, then you'd know that mathematical proofs are absolutes. They are perfect. You can't fake a math proof, as there is always an error in there somewhere if it's not an actual proof. Of all the disciplines, mathematics should be one of the most calm. Crazy claims can safely be ignored, as they'll either fall away as false, and not able to be proven, or they'll turn out to be absolutely true. There is no middle ground in mathematics. But instead, posters reacted as if mathematics were iffy, as if it needed protection. Some would even claim their vicious behavior, endless personal attacks, and postings aimed at censorship were for the GOOD of mathematics, as if maybe false beliefs could take hold, as if the field were vulnerable. It has taken some time for me to understand that the field did become vulnerable, and false beliefs *did* take hold--over a hundred years ago. In watching the reactions and lack of reactions of mathematicians world wide when presented with my work, I've come to the conclusion that on some level, they knew. The attacks I faced came from people who had a sense of their world as fragile. Deep down, I think they felt that everything they'd been taught didn't all make mathematical sense; that error lurked. My brainstorming efforts eventually yielded avenues of pursuit and discovery, and now I find myself not even interested in Fermat's Last Theorem, with major results in number theory, specifically in algebraic number theory, and in the area of primes, and now in the area of integer factorization. But oddly, people arguing with me, keep relying on social games, innuendo, and out and out lies, and it appears to work. I remain on the fringes, despite dramatic occurrences like the publications and retraction of my paper Advanced Polynomial Factorization, by the now defunct Southwest Journal of Pure and Applied Mathematics. You people behave like people living a lie. It makes sense to me now how vicious some of you have been, as that's all you have. You can't win with the mathematics. You can't rely on logic, objectivity and rational discourse, as then it's clear that I'm right. So you cheat. The circle is closing on your society, as a major paper of mine is now at the Annals, and as I find, to my own shock, that my fears about the factoring problem were true, and it is solvable, in a way that mathematicians said was impossible. And I feel the risk and vulnerability of living in a world endangered by the lies of people who've proven they cannot be trusted. I am trying to understand how this happened. How could such a minority potentially cause so much trouble? I think it is that mathematics is a hard discipline, but human beings can be too soft in many ways, and not understand just how hard it is. I think soon many more of you will understand. There is no forgiveness in Mathematics. It is not soft in any way, at all. There is no mercy in Mathematics. It doesn't understand the concept. In Mathematics, there is only truth. James Harris === Subject: Re: Brainstorming > Crazy claims can safely be ignored, as they'll either fall away as > false, and not able to be proven, Kind of like your claims. > But instead, posters reacted as if mathematics were iffy, as if it > needed protection. Some would even claim their vicious behavior, > endless personal attacks, and postings aimed at censorship were for the > GOOD of mathematics, as if maybe false beliefs could take hold, as if > the field were vulnerable. > It has taken some time for me to understand that the field did become > vulnerable, and false beliefs *did* take hold-- at the Southwest Journal of Pure and Applied Mathematics. By the way James, when is the next issue due out? === Subject: Re: Brainstorming > In Mathematics, there is only truth. Can you prove that? Or is it your faith? === Subject: Re: JSH: Brainstorming >[...] >But oddly, people arguing with me, keep relying on social games, >innuendo, and out and out lies, [...] >You people behave like people living a lie. >It makes sense to me now how vicious some of you have been, as that's >all you have. You can't win with the mathematics. You can't rely on >logic, objectivity and rational discourse, as then it's clear that I'm >right. >So you cheat. >The circle is closing on your society, as a major paper of mine is now >at the Annals, Let us know when it's accepted. >and as I find, to my own shock, that my fears about the >factoring problem were true, and it is solvable, in a way that >mathematicians said was impossible. You talk a lot about people lying, but curiously you lie a lot yourself. Take this statement about factoring. You know that nobody's ever said it was impossible. And you know that you're lying when you claim that you've shown it's possible - in your paper all you say is that a certain silly idea _might_ work. What, you're not lying? Then why haven't you solved the RSA challenge yet? You get actual money for that, you know. >And I feel the risk and vulnerability of living in a world endangered >by the lies of people who've proven they cannot be trusted. >I am trying to understand how this happened. How could such a minority >potentially cause so much trouble? >I think it is that mathematics is a hard discipline, but human beings >can be too soft in many ways, and not understand just how hard it is. >I think soon many more of you will understand. >There is no forgiveness in Mathematics. It is not soft in any way, at >all. >There is no mercy in Mathematics. It doesn't understand the concept. >In Mathematics, there is only truth. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Brainstorming David C. Ullrich skrev i melding >>[...] >>But oddly, people arguing with me, keep relying on social games, >>innuendo, and out and out lies, [...] >>You people behave like people living a lie. >>It makes sense to me now how vicious some of you have been, as that's >>all you have. You can't win with the mathematics. You can't rely on >>logic, objectivity and rational discourse, as then it's clear that I'm >>right. >>So you cheat. >>The circle is closing on your society, as a major paper of mine is now >>at the Annals, > Let us know when it's accepted. >>and as I find, to my own shock, that my fears about the >>factoring problem were true, and it is solvable, in a way that >>mathematicians said was impossible. > You talk a lot about people lying, but curiously you lie a lot > yourself. Take this statement about factoring. You know that nobody's > ever said it was impossible. And you know that you're lying when you > claim that you've shown it's possible - in your paper all you > say is that a certain silly idea _might_ work. > What, you're not lying? Then why haven't you solved the RSA > challenge yet? You get actual money for that, you know. >>And I feel the risk and vulnerability of living in a world endangered >>by the lies of people who've proven they cannot be trusted. >>I am trying to understand how this happened. How could such a minority >>potentially cause so much trouble? >>I think it is that mathematics is a hard discipline, but human beings >>can be too soft in many ways, and not understand just how hard it is. >>I think soon many more of you will understand. >>There is no forgiveness in Mathematics. It is not soft in any way, at >>all. >>There is no mercy in Mathematics. It doesn't understand the concept. >>In Mathematics, there is only truth. >>James Harris > ************************ > David C. Ullrich Why can I not see the original post, only answers scattered all ower this group? KON === Subject: Re: JSH: Brainstorming [Karl-Olav Nyberg] > Why can I not see the original post, only answers scattered all ower this > group? > KON Sorry, I don't know. You can still see the original via Google groups, with message id: I first saw it via Giganews. === Subject: Re: JSH: Brainstorming Discussion, linux) > Why can I not see the original post, only answers scattered all ower > this group? I'm not sure, but I think we can blame Google. Its propagation has been flaky since they gave us the ty new interface. Like you, I haven't seen the originals (except on Google). -- Who knows, maybe that may be the only way to settle this crap. It's not like it'd be that hard for me to go back and get a math degree. I can penetrate the math social group and then finish the takedown from inside. -- James S. Harris contemplates a new strategy. === Subject: Re: JSH: Brainstorming Discussion, linux) Damn Google groups. I haven't seen any of James's posts today. Only the followups. James: do your audience a favor. Use a less flaky interface to Usenet. Since the switch to the ugly and obfuscatory Google Groups interface, their posts have been a bit spotty in propagating. -- Jesse F. Hughes [Iota]'s the smallest infinitesimal, Russell, there are smaller infinitesimals. -- Ross Finlayson === Subject: Re: JSH: Brainstorming >Damn Google groups. I haven't seen any of James's posts today. Only >the followups. >James: do your audience a favor. Use a less flaky interface to >Usenet. Since the switch to the ugly and obfuscatory Google Groups >interface, their posts have been a bit spotty in propagating. it contains a claim that he knows is not true, to wit that he's _solved_ the factoring problem. something he knew was false right next to complaints about our lies. But he's _said_ that he doens't know that his method works.) ************************ David C. Ullrich === Subject: Re: JSH: Brainstorming <87652ii13s.fsf@phiwumbda.org> Discussion, linux) >>Damn Google groups. I haven't seen any of James's posts today. Only >>the followups. >>James: do your audience a favor. Use a less flaky interface to >>Usenet. Since the switch to the ugly and obfuscatory Google Groups >>interface, their posts have been a bit spotty in propagating. > it contains a claim that he knows is not true, to wit that > he's _solved_ the factoring problem. I don't think that's it. I saw his post on Google, so it hasn't been removed from the archive. I don't even know if removing posts from You've seen it on your nntp server, right? In that case, I really have no idea why it never appeared on my server. Once it's out in the wild, I thought that it should propagate just like every other post, regardless of its ultimate origin. So, it's a bit of a mystery to me. this isn't the case. Besides, everyone loves JSH. Who would do this? -- And yes, I will be darkening the doors of some of you, sooner than you think, even if it is going to be a couple of years, and when you look in my eyes on that last day of work at your school, then maybe you'll understand mathematics. -- James S. Harris on Judgment Day === Subject: Re: JSH: Brainstorming ... > You've seen it on your nntp server, right? In that case, I really > have no idea why it never appeared on my server. Once it's out in the > wild, I thought that it should propagate just like every other post, > regardless of its ultimate origin. So, it's a bit of a mystery to me. That is known as a Usenet glitch. If some major backbone server fails to receive it for some reason, most systems downstream will not receive it either. And there are many possible reasons why a server fails to receive reached Deja.) -- 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: JSH: Brainstorming > ... > > You've seen it on your nntp server, right? In that case, I really > > have no idea why it never appeared on my server. Once it's out in the > > wild, I thought that it should propagate just like every other post, > > regardless of its ultimate origin. So, it's a bit of a mystery to me. > That is known as a Usenet glitch. If some major backbone server fails to > receive it for some reason, most systems downstream will not receive it > either. And there are many possible reasons why a server fails to receive > reached Deja.) I now know why the original header did not make it to many sites. The at many places. -- 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: JSH: Brainstorming Discussion, linux) > > ... > > > You've seen it on your nntp server, right? In that case, I really > > > have no idea why it never appeared on my server. Once it's out in the > > > wild, I thought that it should propagate just like every other post, > > > regardless of its ultimate origin. So, it's a bit of a mystery to me. > > That is known as a Usenet glitch. If some major backbone server fails to > > receive it for some reason, most systems downstream will not receive it > > either. And there are many possible reasons why a server fails to receive > > reached Deja.) > I now know why the original header did not make it to many sites. The > at many places. Aha! But how was it invalid? It looks normal on Google Groups. Did they fix it before archiving it, but send it to other sites unedited? Oh. Or is the problem the space between the comma and alt.math.undergrad? -- Sale or rental of this disc is ILLEGAL. If you have rented or purchased this disc, please call the MPAA at 1-800-NO-COPYS. -- The MPAA begins a new anti-piracy program, found on a DVD purchased in China === Subject: Re: JSH: Brainstorming >> > ... >> > > You've seen it on your nntp server, right? In that case, I really >> > > have no idea why it never appeared on my server. Once it's out in the >> > > wild, I thought that it should propagate just like every other post, >> > > regardless of its ultimate origin. So, it's a bit of a mystery to me. >> > >> > That is known as a Usenet glitch. If some major backbone server fails to >> > receive it for some reason, most systems downstream will not receive it >> > either. And there are many possible reasons why a server fails to receive >> > reached Deja.) >> I now know why the original header did not make it to many sites. The >> at many places. >Aha! But how was it invalid? It looks normal on Google Groups. Did >they fix it before archiving it, but send it to other sites unedited? >Oh. Or is the problem the space between the comma and >alt.math.undergrad? That was my guess - since we both seem to recall that there's a no-space clause there probably is. ************************ David C. Ullrich === Subject: Re: JSH: Brainstorming >>Oh. Or is the problem the space between the comma and >>alt.math.undergrad? > That was my guess - since we both seem to recall that there's a > no-space clause there probably is. Yes, that's the problem. I didn't notice it myself until I tried to post about the invalid Newsgroups header. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Brainstorming > Years ago I started out trying to figure out some simple proof of > Fermat's Last Theorem, almost as a lark, and found myself obsessed > with the problem for several years. You still are. Bet? > There is no mercy in Mathematics. It doesn't understand the concept. > In Mathematics, there is only truth. See? I told you. Dirk Vdm === Subject: World Hologram and Dark Energy woke up fresh with a start with the correct way to think about the problem - after taking a 600 mg ibuprofen for shoulder ache from over exercise at the health club. Here is how Lenny Susskind's world hologram really does work and how it explains the dark energy! 1. Trash H. We don't need it. 2. The FRW space expansion factor R(t) is dimensionless. Everything is in units of Lp. 3. The Susskind-Hawking-Bekenstein-t'Hooft hologram entropy of the Universe is simply S/kB = (1/4)R(t)^2 I have written this before of course. R(now) = 10^61 This is a nice formula because it also explains the Arrow of Time since the entropy of the universe is 0 at the initial singularity here, i.e. R(initial singularity) = 0 or, if you prefer Finkelstein's chronons it is 1-bit at t = 0. 4. OK roughly model everything as photons. The photon thermal distribution is hf/(e^hf/kBT - 1) + hf/2 But the dark energy is, to first approximation virtual photons hf/2 whose mean value is hc/LpR(t). 5. Therefore we need virtual energy hf/2 to erase each bit every time the cosmic quantum computer clears its register to step forward another chronon to compute the history of the universe Therefore, the dark energy density the vacuum fabric of curved spacetime needs to compute itself is (R(t)^2/4)(hc/LpR(t))(1/Lp^3R(t))^3 = (hc/Lp^4)1/4R(t)^2 = (hc/4Lp^2)/ Einstein's cosmological constant / = 1/Lp^2R(t)^2 = 10^-56 cm^-2 NOW hc/Lp^2 = c^4/G 6. So this says that Einstein's cosmological constant / is getting smaller as the Universe expands. On the other hand, w = pressure/(energy density) = -1 for random micro-quantum zero point energy. The energy density of cosmic stuff scales as R(t)^-3(1 + w) For example w = 0 for ordinary matter so that Omega(matter) ~ R(t)^-3 w = +1/3 for cosmic black body real photons so that Omega(CMB) ~ R(t)^-4 w = -1 for RANDOM micro-quantum vacuum zero point fluctuations (from covariance) so that Omega(ZPF) ~ R(t)^-0 = constant. 7. However in my macro-quantum theory of the world hologram / = (1/Lp^2)[Lp^3|Vacuum Coherence|^2 -1] Equating this with the world hologram formula in the large-scale FRW metric limit 1/Lp^2R(t)^2 = (1/Lp^2)[Lp^3|Vacuum Coherence|^2 -1] 1/R(t)^2 = [Lp^3|Vacuum Coherence|^2 -1] R(t)^2 = [Lp^3|Vacuum Coherence|^2 -1]^-1 Where from inflation Omega(random ZPF) + Omega(matter) + Omega(radiation) + Omega(vacuum coherence) = 1 === Subject: Re: Lexicons JdyOWwwAAAC8AhuHITILJrgANkHxqjXJ >> somewhere in the digits of Pi, not only does the complete works of >> Shakespeare occur, using >letter of the English alphabet. > All sorts of encodings are possible. See e.g. . > The first occurrence of the 25-bit binary equivalent of to_be (in the > encoding used there) in the binary digits of pi is at index 1919281060. > However, to_be_or does not occur in the first four billion binary > digits. Neither of these facts is at all surprising. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada Of course, pi might be a lexicon even if the digits are nor random. But the existence of spigot algorithms for the digits of pi that give the digits one at a time would seem to contradict randomness. Maybe, however, this is not so if one needs to have the number pi completely to use the algorithm since then, in essence, one would need to have all the digits to determine the next one. It still amazes me how it can be so hard to determine whether a specific number has a certain property when almost all of them have it. Numbers normal to every base have measure one but no one knows whether the sraure root of 2 is normal, for example. === Subject: Re: Lexicons >But the existence of spigot algorithms for the digits of pi that give >the digits one at a time would seem to contradict randomness. What do you (think you) mean by that sentence? Lee Rudolph === Subject: Re: Lexicons JdyOWwwAAAC8AhuHITILJrgANkHxqjXJ >But the existence of spigot algorithms for the digits of pi that give >the digits one at a time would seem to contradict randomness. > What do you (think you) mean by that sentence? > Lee Rudolph I might not have the concept very clearly in mind. I remember an topic. As I recall, a spigot algorithm is a method of computing the digits of a number that produces them in order, one at a time, i.e. an algorithm that takes, say, pi, and produces, one by one, the terms of a sequence such that a_n is the nth digit of pi. Not all algorithms for computing numbers are spigot algorithms. Newton-Raphson, for example, produces decimal approximations that become more accurate by more than one digit at each iteration. As for the quoted sentence from my earlier post, on re-examining it I think my later sentence as to needing to have pi completely accurately to compute the next digit is the correct way to look at it. I was thinking that the next digit would always be determined, hence not random. === Subject: Re: Lexicons http://mygate.mailgate.org/mynews/sci/sci.math/c0c7a09ea9251771a699bca429776 e 98.48257%40mygate.mailgate.org > I was thinking that the next digit would always be > determined, hence not random. But you've been working from an entirely false premise all along: there's absolutely nothing random about the digits of pi. The are absolutely predictable, and come out the same way each and every time you compute them. Not only that, but there's at least dozens of different ways that you _can_ compute them, making them the very antithesis of random. There are probably some splendidly interesting issues about the difficulty of predicting where a given string of digits will first occur without doing a search from the beginning of the number pi's representation until you encounter that string of digits, but that doesn't make the string of digits of pi the least bit random, just not wonderfully well indexed, a very different thing. HTH xanthian. -- === Subject: Re: Lexicons >> I was thinking that the next digit would always be >> determined, hence not random. >But you've been working from an entirely false >premise all along: there's absolutely nothing >random about the digits of pi. >The are absolutely predictable, and come out the >same way each and every time you compute them. >Not only that, but there's at least dozens of >different ways that you _can_ compute them, making >them the very antithesis of random. That there is a algorithm that produces the digits of pi makes them the very anithesis of random? I don't understand this. Is there an algorithm for computing the shortest representation of a given sequence of digits? I.e. given a potential canidate for random sequence of digits is there some algorithm that outputs YES if there is a short algorithm that produces the digits and NO otherwise? If not, why use this test as *the* distinguishing characteristic of random? More important, perhaps, are there any results about numbers without short representations always being absolutely normal? Or is it possible that there is some real number x, whose digits have no short explanation, and yet x is not normal? I am confused. Rich === Subject: Re: Lexicons >>But you've been working from an entirely false >>premise all along: there's absolutely nothing >>random about the digits of pi. >>The are absolutely predictable, and come out the >>same way each and every time you compute them. >>Not only that, but there's at least dozens of >>different ways that you _can_ compute them, making >>them the very antithesis of random. >That there is a algorithm that produces the digits of pi makes them the very >anithesis of random? I don't understand this. Is there an algorithm for >computing the shortest representation of a given sequence of digits? First of all, in order for a sequence of digits to be given there must be a finite representation of them (i.e. some statement that distinguishes this sequence of digits from all other sequences). Since there are only countably many possible statements, only countably many sequences can be given. One might say that no sequence that can be given is random. But as to your question, the answer is no. Consider any statement P(n) about natural numbers, and the corresponding sequence of digits (s_n) where s_n = 1 if P(n) is true and 0 if it is false. If it happens that P(n) is always true, there is a very short representation for this sequence: s_n = 1 for all n. So in order to compute the shortest representation for such a sequence, you would have to be able to decide whether the statement for all n P(n) is true or false. And there is no algorithm for doing that, even if P(n) is recursive. > I.e. >given a potential canidate for random sequence of digits is there some >algorithm that outputs YES if there is a short algorithm that produces the >digits and NO otherwise? The I.e. is wrong: this is a different question. I assume that short means less than a given length N. There are only finitely many possible algorithms of length < N (in a given language with a finite alphabet), so there is some M such that any sequence that starts with more than M 1's but is not all 1's can't be given by an algorithm of length < N. So again, your desired algorithm would have to be able to decide whether for all n P(n) is true or false. > If not, why use this test as *the* distinguishing >characteristic of random? I don't think I'd call it *the* distinguishing characteristic. >More important, perhaps, are there any results about numbers without short >representations always being absolutely normal? Or is it possible that there >is some real number x, whose digits have no short explanation, and yet x is not >normal? As Timothy Little remarked, there certainly are such numbers x. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Lexicons http://mygate.mailgate.org/mynews/sci/sci.math/1b6bdee9f699e0ff4369d2a70090e 8 6e.48257%40mygate.mailgate.org > I am confused. Yes, you are. Something is random when you have no mechanism better than chance by which to predict it. Since the digits of pi can be predicted with perfect success simply by computing them, in any of dozens of ways, they are the very antithesis of random. HTH xanthian. -- === Subject: Re: Lexicons > Is there an algorithm for computing the shortest representation of a > given sequence of digits? No, there is not. > Or is it possible that there is some real number x, whose digits > have no short explanation, and yet x is not normal? It is not only possible, it is certain. The set of non-normal reals is uncountable. The set of explanations is countable. Normality is not necessarily connected with randomness, except in the sense that if X is a random real variable from a uniform distribution on a given interval, then P(X is normal) = 1. Pi itself is not random; no number considered by itself is random. Even an algorithmic complexity definition does not say that any number in itself is random, it only defines randomness in the context of a given labelling system. - Tim === Subject: Fish pond population problem... interesting by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBVEMtf07759; Assume that a fish population grows according to the equation: p = m/(1+ ae^-kt) p is the initial population, m is the carrying capacity of the pond (maximum number of fish it can sustain), k = 0.9, and t is the number of years. a) given that you have 1000 fish initially, what is the value of a? (I obtained 24.40878378 by substituting 0 for t) b) How many years will it take for your pond to reach the carrying capacity? (Does this mean that p is now equal to m??) === Subject: Re: Fish pond population problem... interesting > Assume that a fish population grows according to the equation: p = m/(1+ > ae^-kt) > p is the initial population, m is the carrying capacity of the pond > (maximum number of fish it can sustain), k = 0.9, and t is the number of > years. > a) given that you have 1000 fish initially, what is the value of a? (I > obtained 24.40878378 by substituting 0 for t) > b) How many years will it take for your pond to reach the carrying capacity? > (Does this mean that p is now equal to m??) It means for what value of t does p(t) = m? === Subject: Re: Fish pond population problem... interesting http://mygate.mailgate.org/mynews/sci/sci.math/c9a1bee74b2b7c1b3fffd9df49ba8 a 22.48257%40mygate.mailgate.org > Assume that a fish population grows according to > the equation: p = m/(1+ ae^-kt) > p is the initial population, m is the carrying > capacity of the pond (maximum number of fish it > can sustain), k = 0.9, and t is the number of > years. > a) given that you have 1000 fish initially, what > is the value of a? (I obtained 24.40878378 by > substituting 0 for t) > b) How many years will it take for your pond to > reach the carrying capacity? (Does this mean that > p is now equal to m??) 1) Simply by its expression, this is obviously homework. 2) Asking other people to do you homework means that you don't learn the skills needed to succeed either at the exams, or at finding a job in your field after graduation. This is, after all, a really trivial problem, you should be able to solve it without assistance if you've bothered to keep up in class at all, and read the appropriate parts of the textbook. If you've handed off so much of your homework that this problem is one on which you require help, you have doomed yourself already. 3) Therefore, by answering anyway, I am not trying to help you, just gratifying my own ego at your expense. a) Your approach was correct. b) The answer depends on how you do the rounding of p to an integer, since real fish are fairly obviously only swimming in integer quantities. If you round down (use a floor function), then at no finite time can p exceed (m - 1), so in some sense (m - 1), not m, _is_ the carrying capacity, and you should solve for t such that p = (m - 1). If you choose instead to round to the nearest integer, then you can find t such that p = (m - 0.5), and choose to believe that beyond that value of t, the pond is at its carrying capacity. Either approach is potentially correct, so in your submitted homework, you should specifically explain which one you chose and why. HTH xanthian. -- === Subject: bernoulli numbers (combinatorics) I'm trying to find an expression for the Sum (from k=0 to n) nCk * B k * B n-k where B k is the kth bernoulli number. i know f(x) = x/ (exp(x) -1)) have tried using the differential equation f '(x)= f(x)- f(x)*f(x)*exp(x)/x but cant seem to find a closed expression. Any pointers would be much appreciated Trent === Subject: Re: bernoulli numbers (combinatorics) Jngi7wwAAAD2WLn2V2E6Gh2GXydPdCaE > I'm trying to find an expression for the Sum (from k=0 to n) nCk * B k * B > n-k > where B k is the kth bernoulli number. > i know f(x) = x/ (exp(x) -1)) > have tried using the differential equation > f '(x)= f(x)- f(x)*f(x)*exp(x)/x > but cant seem to find a closed expression. > Any pointers would be much appreciated > Trent I get sum{k=0 to n} binomial(n,k) B(k) B(n-k) is equal to - B(m) (m-1) - B(m-1) m. I used a generating function to get the result. (Keep in mind that for m >= 3 and m odd, B(m) = 0. So the above result, for m >= 3, is either -B(m)*(m-1) or is -B(m-1)*m.) Leroy Quet === Subject: Re: bernoulli numbers (combinatorics) Jngi7wwAAAD2WLn2V2E6Gh2GXydPdCaE > I'm trying to find an expression for the Sum (from k=0 to n) nCk * B k * B > n-k > where B k is the kth bernoulli number. > i know f(x) = x/ (exp(x) -1)) > have tried using the differential equation > f '(x)= f(x)- f(x)*f(x)*exp(x)/x > but cant seem to find a closed expression. > Any pointers would be much appreciated > Trent I get sum{k=0 to n} binomial(n,k) B(k) B(n-k) is equal to - B(m) (m-1) - B(m-1) m. I used a generating function to get the result. (Keep in mind that for m >= 3 and m odd, B(m) = 0. So the above result, for m >= 3, is either B(m)*(m-1) or is B(m-1)*m.) Leroy Quet === Subject: Re: bernoulli numbers (combinatorics) >I'm trying to find an expression for the Sum (from k=0 to n) nCk * B k * B >n-k >where B k is the kth bernoulli number. >i know f(x) = x/ (exp(x) -1)) >have tried using the differential equation >f '(x)= f(x)- f(x)*f(x)*exp(x)/x >but cant seem to find a closed expression. The general solution to that differential equation is f(x) = exp(x)/(C + Ei(2*x)) where C is an arbitrary constant, and Ei is the exponential integral function Ei(x) = int_{-infty}^x exp(t)/t dt I'm not sure how that helps you with your first question. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: bernoulli numbers (combinatorics) > The general solution to that differential equation is > f(x) = exp(x)/(C + Ei(2*x)) > where C is an arbitrary constant, and Ei is the exponential integral > function > Ei(x) = int_{-infty}^x exp(t)/t dt > I'm not sure how that helps you with your first question. not exactly sure what Ei(x) means? === Subject: Re: bernoulli numbers (combinatorics) >> The general solution to that differential equation is >> f(x) = exp(x)/(C + Ei(2*x)) >> where C is an arbitrary constant, and Ei is the exponential integral >> function >> Ei(x) = int_{-infty}^x exp(t)/t dt >> I'm not sure how that helps you with your first question. >not exactly sure what Ei(x) means? from -infinity to x. For x > 0 this is interpreted as a Cauchy principal value integral. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: How many lights in a generation ship? In sci.math, Javid http://uk.geocities.com/aa_spaceagent/firstarktoalphacentauri.html > The interior biosphere floor is some 600 square kilometres. If he > mounts each sun-simulating light on top of a tall post which lights up > an effective ground area of radius 10 metres in a cone of light rays > coming down, how many lights will he need for the entire 600 sq. km? > Millions...? However, within the original problem, assuming each lamp generates the right inlampation (about 1,350 W/m^2), each lamp is responsible for an area of 314.1 m^2 and therefore a shade less than 2 million lamps would be required, for a total power consumption of 850 gigaWatts. There's a small corrective factor for the post, if anyone cares. :-) > cheers! > JH -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: World Hologram Vacuum Dark Energy Formula Typo in sign fixed below i.e. the correct formula is R(t)^2 = [1 - Lp^3|Vacuum Coherence|^2]^-1 woke up fresh with a start with the correct way to think about the problem - after taking a 600 mg ibuprofen for shoulder ache from over exercise at the health club. Here is how Lenny Susskind's world hologram really does work and how it explains the dark energy! 1. Trash H. We don't need it. 2. The FRW space expansion factor R(t) is dimensionless. Everything is in units of Lp. 3. The Susskind-Hawking-Bekenstein-t'Hooft hologram entropy of the Universe is simply S/kB = (1/4)R(t)^2 I have written this before of course. R(now) = 10^61 This is a nice formula because it also explains the Arrow of Time since the entropy of the universe is 0 at the initial singularity here, i.e. R(initial singularity) = 0 or, if you prefer Finkelstein's chronons it is 1-bit at t = 0. 4. OK roughly model everything as photons. The photon thermal distribution is hf/(e^hf/kBT - 1) + hf/2 But the dark energy is, to first approximation virtual photons hf/2 whose mean value is hc/LpR(t). 5. Therefore we need virtual energy hf/2 to erase each bit every time the cosmic quantum computer clears its register to step forward another chronon to compute the history of the universe Therefore, the dark energy density the vacuum fabric of curved spacetime needs to compute itself is (R(t)^2/4)(hc/LpR(t))(1/Lp^3R(t))^3 = (hc/Lp^4)1/4R(t)^2 = (hc/4Lp^2)/ Einstein's cosmological constant / = 1/Lp^2R(t)^2 = 10^-56 cm^-2 NOW hc/Lp^2 = c^4/G 6. So this says that Einstein's cosmological constant / is getting smaller as the Universe expands. On the other hand, w = pressure/(energy density) = -1 for random micro-quantum zero point energy. The energy density of cosmic stuff scales as R(t)^-3(1 + w) For example w = 0 for ordinary matter so that Omega(matter) ~ R(t)^-3 w = +1/3 for cosmic black body real photons so that Omega(CMB) ~ R(t)^-4 w = -1 for RANDOM micro-quantum vacuum zero point fluctuations (from covariance) so that Omega(ZPF) ~ R(t)^-0 = constant. 7. However in my macro-quantum theory of the world hologram / = (1/Lp^2)[1 - Lp^3|Vacuum Coherence|^2] Equating this with the world hologram formula in the large-scale FRW metric limit 1/Lp^2R(t)^2 = (1/Lp^2)[1 - Lp^3|Vacuum Coherence|^2] 1/R(t)^2 = [1 - Lp^3|Vacuum Coherence|^2] R(t)^2 = [1 - Lp^3|Vacuum Coherence|^2]^-1 Where from inflation Omega(random ZPF) + Omega(matter) + Omega(radiation) + Omega(vacuum coherence) = 1 === Subject: Can anyone recommend good resources for learning maths please? I'm planning to give a set of the Feymann lectures vol 1-3 as a present and would like to complement that with resources for learning the maths required to do the physics. What I'd like to do ideally is include a book (or books) that assumes _almost_ no mathematical knowledge (ok, high shool maths) and yet imparts enough knowledge to cover the maths in Feymann (and physics at undergrad level in general). If you recommend something, please indicate if you've read the book (or used the resource personally). Other resources such as websites and CDs/DVDs are also welcome. Kunle === Subject: Re: Can anyone recommend good resources for learning maths please? FTuAOw0AAABvanTywnBw2b5Ylxy2JOBx > I'm planning to give a set of the Feymann lectures vol 1-3 as a present and > would like to complement that with resources for learning the maths required > to do the physics. > What I'd like to do ideally is include a book (or books) that assumes > _almost_ no mathematical knowledge (ok, high shool maths) and yet imparts > enough knowledge to cover the maths in Feymann (and physics at undergrad > level in general). Mathematical Methods in the Physical Sciences, 2nd Edition by Mary L. Boas > If you recommend something, please indicate if you've read the book (or used > the resource personally). Other resources such as websites and CDs/DVDs are > also welcome. Yes to both. > Kunle === Subject: Re: Can anyone recommend good resources for learning maths please? > I'm planning to give a set of the Feymann lectures vol 1-3 as a present and > would like to complement that with resources for learning the maths required > to do the physics. > What I'd like to do ideally is include a book (or books) that assumes > _almost_ no mathematical knowledge (ok, high shool maths) and yet imparts > enough knowledge to cover the maths in Feymann (and physics at undergrad > level in general). > If you recommend something, please indicate if you've read the book (or used > the resource personally). Other resources such as websites and CDs/DVDs are > also welcome. > Kunle Actually, high school math should be sufficient to read the lectures. Whenever a subject is touched that *might* not have been treated adequately in high school, Feynman nicely provides the background in a separate lecture. I don't think you need to add any math resources in the present. Excellent idea by the way! Happy 2005, Dirk Vdm === Subject: Re: Can anyone recommend good resources for learning maths please? > I'm planning to give a set of the Feynman lectures vol 1-3 as a present and > would like to complement that with resources for learning the maths required > to do the physics. > What I'd like to do ideally is include a book (or books) that assumes > _almost_ no mathematical knowledge (ok, high school maths) and yet imparts > enough knowledge to cover the maths in Feynman (and physics at undergrad > level in general). > If you recommend something, please indicate if you've read the book (or used > the resource personally). Other resources such as websites and CDs/DVDs are > also welcome. After you master high school algebra, geometry, trigonometry, then at the college level, calculus, matrix theory, differential equations, Laplace transforms, Real Analysis, etc., then look at http://math.ucr.edu/home/baez/physics/Administrivia/booklist.html#mathematic al-methods === Subject: Differentiating e^(-(x^2)) boIyvA0AAABj2nj2r8relC4JpM3EN2sb Does anyone know if the nth derivative of e^(-(x^2)) has a simple representation? John. === Subject: Re: Differentiating e^(-(x^2)) > Does anyone know if the nth derivative of e^(-(x^2)) has a simple > representation? > John. Hermmite polynomials... see equation (17) at http://mathworld.wolfram.com/HermitePolynomial.html -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Differentiating e^(-(x^2)) > Does anyone know if the nth derivative of e^(-(x^2)) has a simple > representation? It's more-or-less the n-th Hermite polynomial times exp(-x^2). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Differentiating e^(-(x^2)) > Does anyone know if the nth derivative of e^(-(x^2)) has a simple > representation? There was a thread concerning this subject in sci.math.research eight years ago, called HELP: e^(-x^2). See: Jose Carlos Santos === Subject: Re: open maps and base change by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBVG1l616609; >I'm trying to prove that topological open maps are stable under base >change: that is, if f : X -> Y is open, h : Z -> Y is continuous and g >: P -> Z is the pullback map of f over h then g is open. >So far, no luck >Any hint? Yes: combine so-called Frobenius Reciprocity together with a Beck-Chevalley equation. If you don't know those buzzwords, then FR: g(g^{-1}W / V) = W / g(V) BC: g(j^{-1}(U)) = h^{-1}(f(U)) where j lies over h in the pullback diagram, and things like f(U), g(V) denote direct images. The equation FR is a general set-theoretic fact (which is important in more general studies of categorical logic), and so is BC which applies to pullback diagrams. Now let W be open in Z, and V be of the form j^{-1}(U) where U is open in X, and note that sets of the form g^{-1}(W) / j^{-1}(U) form a basis of open sets in P. The rest should then be clear. Todd Trimble === Subject: Re: for the categorists and sheafifiers (was, Re: So I lied about epsilon-0) [Strangely (since panix.com is usually very well connected), Michael Barr's post to which this is a response never showed up here; so I've pasted the relevant parts of it in from Google, adding extra attribution chevrons as required.] > ... >The ordinals < omega_1 are uncountable, however, and while >each initial segment shorter than that has in principle a naming >system, there isn't one that covers the whole thing. I suppose >this seems a little paradoxical, but there's no real reason to >think that from the existence of (increasingly hard to define) >naming systems on the initial segments falling short of it one >can patch together a naming system for the whole thing. > That sounds like just the sort of thing one of sci.math's resident > categorists and/or sheafifiers should be able to rephrase (impenetrably > to lower-rank mortals), perhaps in terms of some homology theory or other. ... >This really has nothing to do with either sheaves or categories, but >from my very general knowledge of set theory, I think I can make a >stab. [proof of Keith Ramsay's first quoted sentence omitted] You miss my point (such as it was). I had and have not much interest in the truth of Keith's sentence, much less a proof of it. But when I see the phrase patch together (especially following the phrase systems on the initial segments), my mouth waters in true Pavlovian style, and I expect a sheaf to appear immediately. For instance, there's a copy of the long line lurking in the background of the ordinals < omega_1, and we know that there are interesting and unexpected facts about the topology of that manifold; I would think that james dolan, for instance, would be more than happy to throw together an interpretation of Keith's there isn't one that covers the whole thing as the non-existence of a global section of *something* over the long line, and then to interpret that interpretation as the non-vanishing of some (amazingly abstract) (co)homology group. But what do I know? Lee Rudolph === Subject: Re: for the categorists and sheafifiers (was, Re: So I lied about epsilon-0) ixf50QwAAABFW6BQ4fQR2F0f49XaC1bE |You miss my point (such as it was). I had and have not much interest |in the truth of Keith's sentence, much less a proof of it. But |when I see the phrase patch together (especially following |the phrase systems on the initial segments), my mouth waters |in true Pavlovian style, and I expect a sheaf to appear immediately. I decided not to reply before, since it wasn't clear there was anything interesting to say. Some holiday hiccup also led to this posting of yours only showing up for me today. I guess it's still not clear that this is anything interesting to say, but that's usenet for you.... The unfortunate thing here is that my statement was a denial of the possibility of patching some structures together, which suggests the appearance of a pre-sheaf that is *not* a sheaf, rather than of a sheaf. I was using naming system in a very weak sense, as just a bijection with a subset of some (let's say fixed) countable set of expressions. For each set and a naming system on it, and each subset, there's a natural notion of restriction of the naming system to the subset. On a given set X with any topology on it, this defines a pre-sheaf, which is seldom a sheaf. I was hoping that some variation on the idea would produce a pre-sheaf that would sheafify in an interesting way, but I haven't managed to come up with one. In the discrete topology the set can be covered with singleton sets. The resulting sheaf is just the set of functions to names of elements. I wouldn't be surprised if there were some way to contrive a homology or cohomology theory that detected uncountability. Keith Ramsay === Subject: question: Lebesgue measure > I'm not sure as to the source of your confusion. Are you under the > impression that an uncountable set must be dense? Dense in what? Dense in itself, presumably. The problem arises wen I try to picture the set. It looks quite empty. Happy new year everyone! === Subject: Re: question: Lebesgue measure > I'm not sure as to the source of your confusion. Are you under the > impression that an uncountable set must be dense? Dense in what? > Dense in itself, presumably. The problem arises wen I try to picture > the set. It looks quite empty. > Happy new year everyone! You may be interested in measuring how thinly Cantor set D is scattered. The tool is Hausdorff dimension, and the classical copy of it (the middle thirds removed) has dinemsion log(2)/log(3), a number between 5/8 and 2/3. There are homeomorphic copies of D of any dimension from 0 to 1, depending on how much you remove at the intermediate steps, instead of one-third. === Subject: Re: question: Lebesgue measure >>I'm not sure as to the source of your confusion. Are you under the >>impression that an uncountable set must be dense? Dense in what? >> >Dense in itself, presumably. The problem arises wen I try to picture >the set. It looks quite empty. I don't think you mean that - any set is dense in itself. In any case, it is not surprising that an uncountable subset of [0,1] is not dense there; for example, consider any subinterval with more than one point. What is slightly surprising about the Cantor set is that it provides an example of a set that is uncountable and nowhere dense; i.e., any open interval has a subinterval that does not intersect the Cantor set. -- Stephen J. Herschkorn sjherschko@netscape.net === Subject: Re: Numbers >Is mathematics a natural language? >Are there any book or website that explains the magic even if the reader is not a math expert? Mathematics itself is not a natural language, or even a language, except formally. However, the language of mathematics, which is the general use of variables, is the magic involved. On the other hand, mathematics is essentially pure grammar. There are rules for deciding what is a formal expression in mathematics, and which of these are sentences, theorems, and proofs. Something is a theorem if and only if there is a proof of it. These rules allow the making of assumptions, so that non-mathematical objects can be formalized in mathematics. If the formalization is correct, the use of the theorems to make conclusions from them will be correct. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: The state-of-the-art in mathematics > The other standard nonconstructive axiom is the axiom of choice. > There are slightly more people with qualms about it than there > are constructivists, but the mainstream point of view about it is > also that it's perfectly okay. You may read about the Banach- > Tarski paradox and so on, but that's not treated as a serious > problem generally. The Banach-Tarski paradox is not a true antimony. It is a proof that non-measurable sets exist. Bob Kolker === Subject: Re: The state-of-the-art in mathematics >> The other standard nonconstructive axiom is the axiom of choice. >> There are slightly more people with qualms about it than there >> are constructivists, but the mainstream point of view about it is >> also that it's perfectly okay. You may read about the Banach- >> Tarski paradox and so on, but that's not treated as a serious >> problem generally. >The Banach-Tarski paradox is not a true antimony. It is a proof that >non-measurable sets exist. Actually, it is a proof that there is no finitely additive measure in 3d which is invariant under rigid motion. Such finitely additive measures exist in 2d, but the existence of sets which are non-measurable with respect to countably additive measures is the same in both 2d and 3d. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: The state-of-the-art in mathematics > Actually, it is a proof that there is no finitely additive > measure in 3d which is invariant under rigid motion. Such > finitely additive measures exist in 2d, but the existence > of sets which are non-measurable with respect to countably > additive measures is the same in both 2d and 3d. I sit corrected. However the B-T theorem is still not a paradox. It is just counter intuitive. Bob Kolker === Subject: prime=4k+1=x^2+y^2 e =is an element of E =exist N ={0, 1, 2 ...} /=and E!=exist only one How can I prove this? Theorem If p=4k+1 prime => (E!x,y x,yeN/p=x^2+y^2) DIM ... === Subject: Re: prime=4k+1=x^2+y^2 > e =is an element of > E =exist > N ={0, 1, 2 ...} > /=and > E!=exist only one > How can I prove this? > Theorem > If p=4k+1 prime => (E!x,y x,yeN/p=x^2+y^2) > DIM For the existence, there are two proofs here: http://planetmath.org/?op=getobj&from=objects&id=3827 For the uniqueness, you can use the fact that a polynomial mod p, in this case 1 + uu = 0 mod p can have no more zeros than its degree. === Subject: Re: prime=4k+1=x^2+y^2 >e =is an element of >E =exist >N ={0, 1, 2 ...} >/=and >E!=exist only one >How can I prove this? >Theorem >If p=4k+1 prime => (E!x,y x,yeN/p=x^2+y^2) Unique up to interchange of x and y, of course... It's usually done using the ring of Gaussian integers. Uniqueness is a consequence of unique factorization in that ring. For existence, show that if x = ((p-1)/2)!, then p divides x^2+1 = (x+i)(x-i), and conclude that p is not a prime in the Gaussian integers. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: prime=4k+1=x^2+y^2 >>e =is an element of >>E =exist A =for all >>N ={0, 1, 2 ...} >>/=and >>E!=exist only one >>How can I prove this? >>Theorem >>If p=4k+1 prime => (E!x,y x,yeN/p=x^2+y^2) >Unique up to interchange of x and y, of course... >It's usually done using the ring of Gaussian integers. >Uniqueness is a consequence of unique factorization in that >ring. For existence, show that if x = ((p-1)/2)!, then p >divides x^2+1 = (x+i)(x-i), and conclude that p is not a >prime in the Gaussian integers. I think I don't have understood the existence part We have to say peN prime => (Ex,y x,yeN => x^2+y^2=p) DIM Is it for absurd? if Ep peN prime / (Ax,y x,yeN=>x^2+y^2!=p) if x'=((p-1)/2)!, y'=1 Hp=> x'^2+y'^2!=p so how prove p is not prime? === Subject: Re: prime=4k+1=x^2+y^2 >>It's usually done using the ring of Gaussian integers. >>Uniqueness is a consequence of unique factorization in that >>ring. For existence, show that if x = ((p-1)/2)!, then p >>divides x^2+1 = (x+i)(x-i), and conclude that p is not a >>prime in the Gaussian integers. >I think I don't have understood the existence part >We have to say >peN prime => (Ex,y x,yeN => x^2+y^2=p) > DIM >Is it for absurd? >if Ep peN prime / (Ax,y x,yeN=>x^2+y^2!=p) >if x'=((p-1)/2)!, y'=1 Hp=> x'^2+y'^2!=p Nothing like that. If p is a prime congruent to 1 mod 4, p divides x^2+1 = (x+i)(x-i). If p was a prime Gaussian integer, it would have to divide at least one of x+i and x-i. But the imaginary part of p (a+bi) is divisible by p for any Gaussian integer a+bi. So p can't be a prime Gaussian integer, and thus p = (a+bi)(c+di) for some non-unit Gaussian integers a+bi and c+di. Then p^2 = (a^2+b^2)(c^2+d^2), and neither a^2+b^2 nor c^2+d^2 can be 1 so they must be p. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: P prime and P=6k+1=x^2+3y^2 e =is an element of E =exist N ={0, 1, 2 ...} /=and E!=exist only one How can I prove this? Theorem If P=6k+1 is prime => Ex,y x,yeN / P=x^2+3y^2 === Subject: a series of unfortunate textbook economics questions http://mygate.mailgate.org/mynews/sci/sci.math/97717d935a50de29696dd1da4cf5b 2 82.48257%40mygate.mailgate.org If I understand the complaints I'm reading, to most students, the price of textbooks now (in some cases vastly) exceeds the price of tuition, at least at a public college. Yet I'm wondering if the textbook authors see much of this. I know, for stuff they publish in journals, though it boggles the mind, they are being asked to subsidize publishing what the journal turns around and charges a small fortune to readers to access. I'm guessing textbook publishing must be similarly marginal for most authors. Is the return on authoring textbooks sufficiently above minimum wage that the authors wouldn't be better rewarded in reputation and future job upgrades by making the book widely used as a free softcopy publication, than by the current paper publishing economic bottleneck to their books being widely used? Is the vetting and improving by the publication industry of the author's original product really of sufficient benefit to the end user to justify the markup over original author proceeds? I've recently been grinding my teeth to shreds at being blockaded learning stuff I need to write code, by IEEE, ACM, and similar pay-to-play journal prices are being charged for stuff that can often be downloaded free from the author's university homepage, if one knows to look there, is what motivates the question, really, but it has a larger backdrop: the whole technical publishing industry seems to have gone outside the usual laws of economics via a professor decreed monopoly class textbook selection mechanism leaving students with no choice in which books to buy. Are efforts to do peer reviewed publishing at free but for navigating past the advertising prices bearing any useful fruit? I know arXive or however that's spelled is at least an attempt along those lines. xanthian. An alternate solution, of course, would be for professors to start assigning one from this list of textbooks for classes, with reviews and prices prominently on display, to put a little competition back in the marketplace, but I suspect the Christian Hell would come to have the icy qualities of the Norse Hel before that change occurred. -- === Subject: how is SMS capacity measured? A7jCjQwAAACCoY2tvqvTTopdhR8F75Ts voice-throughput capacity of a telephone network, is measured in Erlangs. how is SMS-throughput capacity (which is different, because it's out-of-band) of a cellular-phone network measured? === Subject: Re: matrices by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBVHbMg25148; >hi, just a quick question. >with matrix A say : >|1 3| | 2| >|4 5| X = | 3| > >if i want to determine the rank of augment A, A' .. do i have to find >the rank of >|4 5 3| |4 3| I am not quite sure what you are asking here. What is augment A? In the euqation Ax = b, x and b have nothing to do with the rank of A. A good question regarding b is whether it is in the column space of A. If it is then the equation has a solution. And b is surely in the column space of A if A is of full rank. - MO p.s. Some advice about posting. It is best to post a new question, like yours, under a new posting rather than as a response to an old, unrelated posting. - MO === Subject: Re: Inverse laplace of sqrt(s) - Half differentiation? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBVHre026494; >I always thought integration and differentiation were a bit discreet. >Surely there must be some form of half integration. >So anyway, if integrating is equivalent to multiplication by s, what >is the inverse laplace of multiplication by sqrt(s)? >-Tim H Google for Weyl fractional calculus === Subject: Re: Inverse laplace of sqrt(s) - Half differentiation? L Should then be half-differentiation of one. Please explain why it is not a constant. Geert-Jan === Subject: outer measure of the Vitali non-measurable set ZGy-nQ0AAACjyyWiLftgioL-H_FOBVqf Hi all, Just out of curiosity, does anyone have any idea regarding the (Lebesgue) outer measure of the Vitali non-measurable set? Pouya === Subject: Re: outer measure of the Vitali non-measurable set >Just out of curiosity, does anyone have any idea regarding the >(Lebesgue) outer measure of the Vitali non-measurable set? See the thread Vitali nonmeasurable from January 1997. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: outer measure of the Vitali non-measurable set ZGy-nQ0AAACjyyWiLftgioL-H_FOBVqf have limited mathematical background. The example I had in mind was actually a set constructed by taking one representative from each class defined by the equivalence x~y <=> x-y=nalpha (modulo-1 summation/subtraction) with alpha irrational, although I now understand that the construction you discussed above is also valid. If I have understood your comments correctly, the outer measure of such a set can be any number in (0,1] __depending on how the representatives are chosen__. Is this right? -- Pouya D. Tafti, McMaster University, Hamilton, Canada http://grads.ece.mcmaster.ca/~pouya p dot d dot tafti at ieee dot org === Subject: Re: outer measure of the Vitali non-measurable set > Hi all, > Just out of curiosity, does anyone have any idea regarding the > (Lebesgue) outer measure of the Vitali non-measurable set? > Pouya Do you mean a set V in [0,1] containing one element from each equivalence class for the relation: x-y is rational ? Because of the arbitrary choices involved, it is best to call it a Vitali non-measurable set, not the Vitali non-measurable set. The outer measure could be any number in (0,1] . === Subject: walks in graphs e6YlJQ0AAAAvv4g651mtwLq3UFaamCS6 Is there a special term for a closed walk in a graph that visits each node? Is there an algorithm for finding the smallest such walk? === Subject: Re: walks in graphs 4lZjEA0AAACcI9lfzifBAQlMFCToSJI4 > Is there a special term for a closed walk in a graph that visits each > node? > Is there an algorithm for finding the smallest such walk? The term you're looking for is a Hamiltonian Circuit. It is defined as a closed path through a graph that visits each node at least once. Attempting to find an algorithm that discovers that smallest such walk is called the Travelling Salesman Problem; it is NP-Complete. For more, loop up the above terms at http://mathworld.wolfram.com. Hope this helps. Breckin === Subject: Re: walks in graphs 4lZjEA0AAACcI9lfzifBAQlMFCToSJI4 Ack! I forgot my most important correction: A Hamiltonian Circuit is a closed path through a graph that visits each node EXACTLY once, not at least once. Arrgh, I think it's time to get off the Internets and enjoy New Years Eve. === Subject: Re: walks in graphs e6YlJQ0AAAAvv4g651mtwLq3UFaamCS6 Anyone else want to help me out? I'm only really interested in unweighted graphs here, so length = # of edges on the walk. === Subject: Re: walks in graphs 4lZjEA0AAACcI9lfzifBAQlMFCToSJI4 A correction and a few additions to my last post: A Hamiltonian Circuit is sometimes called a Hamiltonian Cycle. Not all graphs have such a cycle, those that do are called Hamiltonian Graphs. Finding such a cycle is NP-Complete. The Travelling Salesman Problem is NP-Hard. Sorry for the screw-ups. === Subject: Re: walks in graphs > Is there a special term for a closed walk in a graph that visits each > node? > Is there an algorithm for finding the smallest such walk? Is walks on graphs anything like dances with wolves? Bob Kolker === Subject: Re: ga8- Integers found that form the sqrt(2) or transcendental numbers. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBVIVHJ29524; >> 14142135623730950488016887242097... / 10000000000000000000000000000000... = ? >> If you do not accept the two numbers that I have written down as >>integers, you must proof why they are not Integers to you. >Actually, the burden of proof is on you, since you're the one claiming >that such integers exist. >However, it is simple to prove that they are not integers. One proof: >every integer is finite. For any finite number, your representation >denotes something larger. Therefore it denotes something infinite, >and hence does not represent an integer. >Or another proof: No integer is both even and odd. By the proof you >quoted, your numbers must be both even and odd. Therefore they are >not integers. >Either one suffices, and if you don't accept those there are plenty >more. >There is an out though: you could define a set consisting of both >finite and infinite decimal representations, together with arithmetic >operations based on manipulating the digits much as usual integers are >added, subtracted, divided and so forth. This seems to be what you're >trying to do. However, you can't expect the world to accept calling >your objects integers, since that name has already been used for >hundreds of years to name objects with quite different properties. >Your objects would include the integers, but would not be the same >thing. >You'll have to watch out for traps inherent in expecting such objects >to obey the usual laws of arithmetic for integers, since they don't. >You'll have to decide for objects like a = 1000... , whether a+1 or >10a are different numbers from a or not. Any choice leads to >complications, which is fine so long as you're aware of them and >ensure that your definitions don't lead to self-contradiction. >Once you've worked out the arithmetic of the objects you've defined, >you can then define a set of objects representing ratios of them, in >much the same way as the rationals are developed from the integers. >Then you can try to find an isomorphism between a subset of your >ratios and a subset of the reals. At that point, you can say these >ratios are the same things as these real numbers. If you made good >choices earlier, you might even be able to show that sqrt(2) is a >member. >That's a lot of work. Maybe some of it has been done already, you >could search the Web or written literature to see if anyone else has >done similar work you could build on. In the end, it's up to you to >decide whether it's interesting enough to pursue. >- Tim ANSWER: Responding to your observations: However, it is simple to prove that they are not integers. One proof: every integer is finite. For any finite number, your representation denotes something larger. Therefore it denotes something infinite, and hence does not represent an integer. The definition every integer is finite means that integers have no decimal expansions and that they are exact as I understand it. However, the set of integers is infinite, this means there is no last integer, therefore the number of digits can be infinite but on the real line one integer is smaller or bigger than another, making them finite. The integers I find using the formula are finite in the number of digits they have. Since they construct irrational numbers such the sqrt(2) in the formula, gives me a reason to say this is why I am saying that irrational numbers are also finite. which is what I am trying to prove. your other observation: Or another proof: No integer is both even and odd. By the proof you quoted, your numbers must be both even and odd. Therefore they are not integers. Not at all, they are either even or odd. The confusion comes when you multiply the initial integer by a common integer for the numerator and denominator. At this point the numerator changes depending if the common integer for numerator and denominator is even or odd. but each numerator in each fraction is even or odd, never both. The reason why I multiply the fraction was to show that there are many fractions that make an irrational number, but only the irreducible fraction complies to prove the rationality of the real numbers. thank you again for your observations, you have hit it in the direction I am trying to get. Guillermo === Subject: Re: ga8- Integers found that form the sqrt(2) or transcendental numbers. > The definition every integer is finite means that integers have no > decimal expansions and that they are exact as I understand > it. Finite has nothing to do with decimal expansions. It is usually expressed in terms of a bijection property in set theory, based on the axiomatic definition of the natural numbers in terms of specific sets. It is extended to the integers as a consequence of their defintion in terms of the natural numbers. > However, the set of integers is infinite, this means there is no > last integer, therefore the number of digits can be infinite The *set* of integers is infinite, but every specific integer from that set is finite and has a finite number of digits. This should not be surprising: the size of a set need not have anything to do with the size of its elements. For example, every element in {2,4,6} is even, but the number of elements is odd. Likewise every element in {1,2,3,...} is finite, but the number of elements is infinite. As I said in my post, you can extend the definition of the integers to include new objects with infinite numbers of digits, but then you have to provide new rules for manipulating them and make sure those rules are consistent. > The integers I find using the formula are finite in the number of > digits they have. Your formula for sqrt(2) gives an infinite number of digits in both the numerator and denominator. They do not represent integers in the sense that everyone else means the word. You can treat such objects mathematically, and call them what you like. It's just that if you use a name that is well-established as meaning something else, you're just going to cause confusion whenever you talk to anyone else about them. I do have one question: what is your representation for 10 sqrt(2), and how does it differ from that for sqrt(2)? - Tim === Subject: Re: ga7- Integers found that form the sqrt(2) or transcendental numbers. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBVIVHo29520; ----- Original Message ----- === Subject: Re: My answer about infinite numbers are quotients of two integers > Well, it really just sounds like you want something to be true, so then you > call it true, and then make some odd arguments to convince yourself it is > true. Sorry, but that has no place in mathematics. What convinces you does > not necessarily convince others. Math requires proofs, not just I said > so's. > I like your interest in math. And it is natural for new persons to question > things they may not fully understand. However, unlike other disciplines (IE > physics, biology, etc.) the fundamentals of math are extremely well-defined > and explored. > While I do not know you very well, your sqrt(2) statement is often the type > of creation by those who think they are better in math than they really > are. The I've discovered something missed by everyone principle. > Those types rarely amount to much in math until they mature some. Of course, > they are also the type to think that they are the exception to that rule. > Hum, that's sad! > Hopefully you will not take that overly traveled path. > One of the steps in becoming a better mathematician (or person in general) > is to recognize your errors, oversights, and limitations of your knowledge. > Then, learn from those items and move forward. > You're stuck in the realm of rational numbers. Sorry, but there are more > numbers than those. A lot more. In fact, rational numbers are one of the > smallest groups of numbers. > If you think you have a fraction of two integers that is equal to sqrt(2) > then provide it. Not a sequence of numbers that depend on sqrt(2) but just > one integer numerator and one integer denominator. You can't do it. Not even > if your numbers have a 100 trillion trillion trillion trillion trillion > trillion trillion trillion trillion trillion trillion trillion trillion > trillion trillion trillion trillion trillion trillion trillion trillion > trillion trillion trillion trillion trillion trillion trillion trillion > trillion trillion trillion trillion trillion trillion trillion trillion > trillion trillion trillion trillion trillion trillion trillion trillion > digits. > Sqrt(2) exactly solves the equation X^2=2. You cannot. > Period. > You quoted one of the more common simple proofs for why sqrt(2) is > irrational (IE not a quotient of two integers). I'm left to assume that you > do not believe it only because you cannot follow it. > If that is the case, then maybe a little help to explain the proof is in > order. > Mathematics is built upon a few axioms (very simple statements that form its > foundation), definitions (which are never right or wrong, just useful or not > useful), and then theorems based on the axioms, definitions and previously > proved theorems. This is what makes math so incredibly powerful, repeatable, > and logical. > Dwayne ANSWER: Dwayne: Unfortunately your comments are based more on the personal side than in the math side I invite you to send more math based comments so we can get the truth at the end. Whether I am wrong or not is to be mathematically proven. To achieve this we must do it mathematicaly, sending arguments back and forth in a constructive way. Guillermo === Subject: Re: P prime and P=6k+1=x^2+3y^2 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBVItlA31299; >e =is an element of >E =exist >N ={0, 1, 2 ...} >/=and >E!=exist only one >How can I prove this? >Theorem >If P=6k+1 is prime => Ex,y x,yeN / P=x^2+3y^2 You want to show you can write P = (x + y(-3)^.5)(x - y(-3)^.5) if P is of this form, which amounts to saying that Z[(-3)^.5]/(P) ~ Z[x]/(x^2 + 3, P) ~ Z_P[x]/(x^2 + 3) splits as a direct product of rings (note in passing that (3) is the only prime which ramifies in Z[(-3)^.5]) if P is of this form, or that x^2 + 3 = 0 mod P is solvable for such P. Then the statement follows from quadratic reciprocity. Todd Trimble === Subject: Re: f( x +2f(y) ) = f(x) + y + f(y) ,is my solving correct? . >I've been considering the case where f is not assumed continuous. I >haven't yet been able to construct an example that isn't f(x) = x or >f(x) = -x/2, but haven't been able to prove they're the only >solutions. I know f(0) = 0, and with h(x) = x + 2 f(x), f(h(x)) = h(x) >for all x. >There are many other restrictions on f(x) and the related function >g(x) = f(x) - x such as periodicity and so forth. Also, if f(x) is >continuous anywhere, then it is continuous everywhere. In fact, the most important property is additivity: f(x+y) = f(x) + f(y). This allows all solutions to be found - see below. The general solution below is not constructive - in fact the only constructive solutions I know (i.e. not using AC) are f(x) = x and f(x) = -x/2. I'd be interested in any other constructive examples. Here's the details: Setting x = y gives f(x + 2f(x)) = x + 2f(x). Replacing y by y + 2f(y) gives f(x) + 2y + 4f(y) = f(x + 2y + 4f(y)) = f(x + 2y + 2f(y)) + y + f(y) = f(x + 2y) + 2y + 2f(y) Hence f(x + 2y) = f(x) + 2f(y) Setting x = y = 0 gives f(0) = 0, and x = 0 gives f(2y) = 2f(y) Hence f(x + y) = f(x) + f(y) Define g(y) = (x + 2f(x))/3. Then g() is also additive, and the given equation is equivalent to the two conditions g(x + y) = g(x) + g(y) and g(g(x)) = g(x). These in turn are equivalent to: The set A = { x in R | g(x) = x } is an additive subgroup of R, and g is an additive homomorphism R -> A. So the general solution is: take an arbitrary additive subgroup A of R. Take an arbitrary homomorphism g: R -> QA such that g(x) = x for x in A. Then f(x) = (3g(x) - x)/2 is a solution. If A = {0}, we get the solution f(x) = -x/2; if A = R we get the solution f(x) = x. But I don't know how to define any homomorphism R -> A when {0} < A < R without using AC. Mike Guy === Subject: Re: f( x +2f(y) ) = f(x) + y + f(y) ,is my solving correct? . > Replacing y by y + 2f(y) gives > f(x) + 2y + 4f(y) = f(x + 2y + 4f(y)) > = f(x + 2y + 2f(y)) + y + f(y) > = f(x + 2y) + 2y + 2f(y) > Hence f(x + 2y) = f(x) + 2f(y) Nice work! > Define g(y) = (x + 2f(x))/3. Then g() is also additive, and the > given equation is equivalent to the two conditions > g(x + y) = g(x) + g(y) and g(g(x)) = g(x). I managed to derive that through a much more tortuous route but didn't know where to go from there. > If A = {0}, we get the solution f(x) = -x/2; if A = R we get the solution > f(x) = x. But I don't know how to define any homomorphism R -> A > when {0} < A < R without using AC. If we do assume AC, does that guarantee more solutions? I'm not familiar with how to use AC. - Tim === Subject: Re: f( x +2f(y) ) = f(x) + y + f(y) ,is my solving correct? . >> Define g(y) = (x + 2f(x))/3. Then g() is also additive, and the >> given equation is equivalent to the two conditions >> g(x + y) = g(x) + g(y) and g(g(x)) = g(x). >I managed to derive that through a much more tortuous route but didn't >know where to go from there. My initial route there was also very tortuous. But once you know where you're going, it's fairly easy to find a short route. >> If A = {0}, we get the solution f(x) = -x/2; if A = R we get the solution >> f(x) = x. But I don't know how to define any homomorphism R -> A >> when {0} < A < R without using AC. >If we do assume AC, does that guarantee more solutions? I'm not >familiar with how to use AC. The general homomorphism R -> A can be constructed by choosing representatives for each of the cosets of A in R, then assigning arbitrary values for those representatives. AC tells such sets of representatives exist, and therefore that other solutions exist. But it's perfectly possible that one can choose representatives in some constructive way, not using AC. I just don't know how to. Mike Guy === Subject: Re: Analyitic functions that preserve the rationals jnanCwwAAADhdDGKIZxmMT8cdQqnhCJe >Let I be a non-empty open interval of the reals and let f be an >analytic function from I into the reals such that, for every rational >number q in I, f(q) is rational. Is it true that f must then be a >rational function? > No. This has come up a number of times before. See e.g. my posting > of August 15 1992 in the thread f: Q -> Q, f analytic, Israel's proof is interesting in that it uses a clever idea which seems to be completely inessential. We don't need an enumeration of the polynomials with rational coefficients. Let {q n} be an enumeration of the rationals in I such that each rational appears exactly once. Let p n(x) = product (x-q k) for k=1..n Let f(x) = sum a k p k(x) for k=0..oo where {a n} is a sequence of rapidly decaying non-zero rational numbers. Then I'll leave it to the reader to show 1) If {|a n|} decays sufficiently rapidly, then f(x) is entire 2) f(q n) is rational for all n 3) f is not a polynomial or a ratio of polynomials (this requires nothing more than basic facts about interpolating polynomials) === Subject: Re: Analyitic functions that preserve the rationals >Let I be a non-empty open interval of the reals and let f be an >analytic function from I into the reals such that, for every > rational >number q in I, f(q) is rational. Is it true that f must then be a >rational function? > No. This has come up a number of times before. See e.g. my posting > of August 15 1992 in the thread f: Q -> Q, f analytic, > Israel's proof is interesting in that it uses a clever idea which > seems to be completely inessential. We don't need an enumeration > of the polynomials with rational coefficients. > Let {q n} be an enumeration of the rationals in I such that each > rational appears exactly once. > Let p n(x) = product (x-q k) for k=1..n > Let f(x) = sum a k p k(x) for k=0..oo where {a n} is a sequence > of rapidly decaying non-zero rational numbers. > Then I'll leave it to the reader to show > 1) If {|a n|} decays sufficiently rapidly, then f(x) is entire > 2) f(q n) is rational for all n > 3) f is not a polynomial or a ratio of polynomials (this requires > nothing more than basic facts about interpolating polynomials) f is entire so a nontrivial ratio of polynomials is not a possibility. === Subject: Re: Analyitic functions that preserve the rationals >Let I be a non-empty open interval of the reals and let f be an >analytic function from I into the reals such that, for every > rational >number q in I, f(q) is rational. Is it true that f must then be a >rational function? > No. This has come up a number of times before. See e.g. my posting > of August 15 1992 in the thread f: Q -> Q, f analytic, > Israel's proof is interesting in that it uses a clever idea which > seems to be completely inessential. We don't need an enumeration > of the polynomials with rational coefficients. > Let {q n} be an enumeration of the rationals in I such that each > rational appears exactly once. > Let p n(x) = product (x-q k) for k=1..n > Let f(x) = sum a k p k(x) for k=0..oo where {a n} is a sequence > of rapidly decaying non-zero rational numbers. > Then I'll leave it to the reader to show > 1) If {|a n|} decays sufficiently rapidly, then f(x) is entire > 2) f(q n) is rational for all n > 3) f is not a polynomial or a ratio of polynomials (this requires > nothing more than basic facts about interpolating polynomials) A nontrivial ratio of polynomials is automatically ruled out; f is entire. === Subject: Re: Analyitic functions that preserve the rationals >>Let I be a non-empty open interval of the reals and let f be an >>analytic function from I into the reals such that, for every >rational >>number q in I, f(q) is rational. Is it true that f must then be a >>rational function? >> No. This has come up a number of times before. See e.g. my posting >> of August 15 1992 in the thread f: Q -> Q, f analytic, >Israel's proof is interesting in that it uses a clever idea which >seems to be completely inessential. We don't need an enumeration >of the polynomials with rational coefficients. >Let {q_n} be an enumeration of the rationals in I such that each >rational appears exactly once. >Let p_n(x) = product (x-q_k) for k=1..n >Let f(x) = sum a_k p_k(x) for k=0..oo where {a_n} is a sequence >of rapidly decaying non-zero rational numbers. >Then I'll leave it to the reader to show >1) If {|a_n|} decays sufficiently rapidly, then f(x) is entire >2) f(q_n) is rational for all n >3) f is not a polynomial or a ratio of polynomials (this requires >nothing more than basic facts about interpolating polynomials) Would you mind giving some details on how you do (3)? I may be missing something obvious, but as I recall (3) was not all that simple, which is why I introduced the enumeration of the polynomials with rational coefficients. Of course there are other ways of doing it, e.g. you can choose a_n to ensure that the denominator of f(q_{n+1}) is divisible by a prime > n that is not a factor of the denominator of q_{n+1}. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Analyitic functions that preserve the rationals >>Let I be a non-empty open interval of the reals and let f be an >>analytic function from I into the reals such that, for every >rational >>number q in I, f(q) is rational. Is it true that f must then be a >>rational function? >> No. This has come up a number of times before. See e.g. my posting >> of August 15 1992 in the thread f: Q -> Q, f analytic, >Israel's proof is interesting in that it uses a clever idea which >seems to be completely inessential. We don't need an enumeration >of the polynomials with rational coefficients. >Let {q_n} be an enumeration of the rationals in I such that each >rational appears exactly once. >Let p_n(x) = product (x-q_k) for k=1..n >Let f(x) = sum a_k p_k(x) for k=0..oo where {a_n} is a sequence >of rapidly decaying non-zero rational numbers. >Then I'll leave it to the reader to show >1) If {|a_n|} decays sufficiently rapidly, then f(x) is entire >2) f(q_n) is rational for all n >3) f is not a polynomial or a ratio of polynomials (this requires >nothing more than basic facts about interpolating polynomials) > Would you mind giving some details on how you do (3)? > I may be missing something obvious, but as I recall (3) was not > all that simple, which is why I introduced the enumeration of the > polynomials with rational coefficients. Of course there are other > ways of doing it, e.g. you can choose a_n to ensure that the > denominator of f(q_{n+1}) is divisible by a prime > n that is not > a factor of the denominator of q_{n+1}. The nth coefficient in the power series expansion of f about 0 is the sum of all nth coefficients from the summands. This will be a_n + contribution from sum (k=n+1,oo) a_k*p_k. So we can proceed inductively: let a_0 = 1, choose a_1 small relative to a_0, choose a_2 small relative to a_0 and a_1, etc. In this way, f can be designed to have all of its power series coefficients positive; hence f is not a polynomial. === Subject: Re: Analyitic functions that preserve the rationals , > >>Let I be a non-empty open interval of the reals and let f be an >>analytic function from I into the reals such that, for every >rational >>number q in I, f(q) is rational. Is it true that f must then be a >>rational function? > >> No. This has come up a number of times before. See e.g. my posting >> of August 15 1992 in the thread f: Q -> Q, f analytic, > >Israel's proof is interesting in that it uses a clever idea which >seems to be completely inessential. We don't need an enumeration >of the polynomials with rational coefficients. > >Let {q_n} be an enumeration of the rationals in I such that each >rational appears exactly once. > >Let p_n(x) = product (x-q_k) for k=1..n > >Let f(x) = sum a_k p_k(x) for k=0..oo where {a_n} is a sequence >of rapidly decaying non-zero rational numbers. > >Then I'll leave it to the reader to show > >1) If {|a_n|} decays sufficiently rapidly, then f(x) is entire >2) f(q_n) is rational for all n >3) f is not a polynomial or a ratio of polynomials (this requires >nothing more than basic facts about interpolating polynomials) Would you mind giving some details on how you do (3)? > I may be missing something obvious, but as I recall (3) was not > all that simple, which is why I introduced the enumeration of the > polynomials with rational coefficients. Of course there are other > ways of doing it, e.g. you can choose a_n to ensure that the > denominator of f(q_{n+1}) is divisible by a prime > n that is not > a factor of the denominator of q_{n+1}. > The nth coefficient in the power series expansion of f about 0 is > the sum of all nth coefficients from the summands. This will be > a_n + contribution from sum (k=n+1,oo) a_k*p_k. So we can proceed > inductively: let a_0 = 1, choose a_1 small relative to a_0, > choose a_2 small relative to a_0 and a_1, etc. In this way, f can > be designed to have all of its power series coefficients > positive; hence f is not a polynomial. Here's a more explicit example. Define P_n(z) = product(m=1,n) (z-q_m)^2/(n+|q_m|)^2. Note that |z| < n implies |P_n(z)| < 1. Therefore sum(n=1,oo) P_n/2^n converges uniformly on compact subsets of C to an entire function f that takes Q to Q. On R, positivity shows f >= P_n for each n, which implies f is not a polynomial. === Subject: Re: Analyitic functions that preserve the rationals jnanCwwAAADhdDGKIZxmMT8cdQqnhCJe >>Let I be a non-empty open interval of the reals and let f be an >>analytic function from I into the reals such that, for every >rational >>number q in I, f(q) is rational. Is it true that f must then be a >>rational function? >Israel's proof is interesting in that it uses a clever idea which >seems to be completely inessential. We don't need an enumeration >of the polynomials with rational coefficients. >Let {q n} be an enumeration of the rationals in I such that each >rational appears exactly once. >Let p n(x) = product (x-q k) for k=1..n >Let f(x) = sum a k p k(x) for k=0..oo where {a n} is a sequence >of rapidly decaying non-zero rational numbers. >Then I'll leave it to the reader to show >1) If {|a n|} decays sufficiently rapidly, then f(x) is entire >2) f(q n) is rational for all n >3) f is not a polynomial or a ratio of polynomials (this requires >nothing more than basic facts about interpolating polynomials) > Would you mind giving some details on how you do (3)? If two n'th degree polynomials have the same value for n+1 values of the variable, they are the same polynomial. So if f(x) is a polynomial of degree N, it must equal f N(x) = sum a n p n(x) , n = 1..N because f(q n) = f N(q n) for n = 1..N+1, but clearly f(q {N+2}) =/= f N(q {N+2}) === Subject: Re: Analyitic functions that preserve the rationals >>Let I be a non-empty open interval of the reals and let f be an >>analytic function from I into the reals such that, for every >rational >>number q in I, f(q) is rational. Is it true that f must then be a >>rational function? >Israel's proof is interesting in that it uses a clever idea which >seems to be completely inessential. We don't need an enumeration >of the polynomials with rational coefficients. >Let {q n} be an enumeration of the rationals in I such that each >rational appears exactly once. >Let p n(x) = product (x-q k) for k=1..n >Let f(x) = sum a k p k(x) for k=0..oo where {a n} is a sequence >of rapidly decaying non-zero rational numbers. >Then I'll leave it to the reader to show >1) If {|a n|} decays sufficiently rapidly, then f(x) is entire >2) f(q n) is rational for all n >3) f is not a polynomial or a ratio of polynomials (this requires >nothing more than basic facts about interpolating polynomials) > Would you mind giving some details on how you do (3)? > If two n'th degree polynomials have the same value for n+1 values > of the variable, they are the same polynomial. > So if f(x) is a polynomial of degree N, it must equal > f N(x) = sum a n p n(x) , n = 1..N because f(q n) = f N(q n) > for n = 1..N+1, but clearly f(q {N+2}) =/= f N(q {N+2}) I don't understand your notation. === Subject: Re: Analyitic functions that preserve the rationals jnanCwwAAADhdDGKIZxmMT8cdQqnhCJe > So if f(x) is a polynomial of degree N, it must equal > f N(x) = sum a n p n(x) , n = 1..N because f(q n) = f N(q n) > for n = 1..N+1, but clearly f(q {N+2}) =/= f N(q {N+2}) > I don't understand your notation. Your newsreader software seems to have removed all underscores. Without the underscores, I don't understand my notation. === Subject: Re: beanie > how old are you beanie? > you cause a lot of problems. thats my observation > interesting indeed. whats in it for you? do you learn from this group? do > you provide input that others appreciate? > dr. x Unheard of for Gareth to post anything helpful. === Subject: Re: beanie : Unheard of for Gareth to post anything helpful. : : 5.6 on the cluckometer === Subject: lebesgue integrable? why is the derivative of f(x) := x^2 sin(1/x^2) for x != 0 and 0 for x = 0 not Lebesgue integrable over the interval [-1, 1]?? === Subject: Last post The last post listed on the forum has an earlier time (4:42 pm) than the actual last post (4:52 pm). Any particular reason? *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Last post >The last post listed on the forum has an earlier time (4:42 pm) than >the actual last post (4:52 pm). Any particular reason? Usenet is a distributed medium. Participating servers usually list the messages in the order they received them, not in any sorted order. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Last post > The last post listed on the forum has an earlier time (4:42 pm) than > the actual last post (4:52 pm). Any particular reason? > *-----------------------* > www.GroupSrv.com > *-----------------------* This thread has the subject last post, but the previous post is Problem with dates. Any particular reason? Herc === Subject: re:lebesgue integrable? > You're one of these people who can never admit an error, aren't > you? The function does not not have a 1/x singularity in the > neighborhood of 0. It looks roughly like |cos(1/x^2)|/x near 0. I don't see what you are complaining about? I am not concerned with the general question of f(x)/x as an integrand, only with the specific case of |cos(1/x^2)|/x. As I've already indicated, it behaves approximately like 2/(pi*x) near 0, i.e a constant times 1/x. Therefore the integral behaves like lnx near 0, hence divergent. I am not asserting anything further! *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: lebesgue integrable? > why is the derivative of f(x) := x^2 sin(1/x^2) for x != 0 and 0 for x = 0 > not Lebesgue integrable over the interval [-1, 1]?? First, split [-1,1] into [-1,0] and [0, 1] (using symmetry). Estimate (from below) the integral of the absolute value of the integrand between two subsequent positive roots of cos(1/x^2) -- by what? By the absolute value of the integral. That can be calculated explicitly. These absolute values of integrals will form terms of a divergent series (about as quickly divergent as the harmonic series). This is an example when improper Riemann integral converges for a Lebesgue non-integrable function. This shortcoming of Lebesgue integral can be overcome by Kurzweil-Henstock integral, also known as the gauge integral, which closes the gap: if f has a derivative everywhere on [a,b] (one-sided derivatives at the endponts) then (gauge)integral(f' over [a,b]) = f(b)-f(a). Exercise: Let h>0, then x^(2+h)*sin(1/x^2) is Lebesgue integrable. (This will illustrate, for 0, |cos(1/x^2)| oscillates very rapidly between 0 and 1, with an average value approx 2/pi. Therefore we can approximate the integrand (for integration purposes as x->0) by 2/(pi*x). *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: lebesgue integrable? > As x ->, |cos(1/x^2)| oscillates very rapidly between 0 and 1, with > an average value approx 2/pi. Therefore we can approximate the > integrand (for integration purposes as x->0) by 2/(pi*x). You're one of these people who can never admit an error, aren't you? The function does not not have a 1/x singularity in the neighborhood of 0. It looks roughly like |cos(1/x^2)|/x near 0. Yes, int_(0,1) |cos(1/x^2)|/x dx diverges, but there are positive functions f that oscillate very rapidly between 0 and 1, with a positive average value on (0,1), such that int_(0,1) f(x)/x dx converges. So your reasoning is incorrect. === Subject: re:lebesgue integrable? Integral of 1/x is lnx which blows up as x->0. 1/x is not integrable when the domain includes 0! *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: lebesgue integrable? > Integral of 1/x is lnx which blows up as x->0. 1/x is not > integrable when the domain includes 0! We're not integrating 1/x; we're integrating |cos(1/x^2)|/x. === Subject: re:lebesgue integrable? The derivative has a 1/x singularity in the neighborhood of 0. If you try to integrate the absolute value of the derivative, it blows up in this neighborhood. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: lebesgue integrable? > The derivative has a 1/x singularity in the neighborhood of 0. Not exactly. > If you > try to integrate the absolute value of the derivative, it blows up in > this neighborhood. Lots of functions blow up and are still integrable. === Subject: Cosmological Constant from World Hologram Inflation demands total stuff is at critical density for flat 3D space, i.e. k = 0 in large-scale FRW metric Omega(random ZPF) + Omega(matter) + Omega(radiation) + Omega(vacuum coherence) = 1 Omega(on-mass-shell matter) ~ 1/R(t)^3, i.e. w = 0 in R(t)^-3(1 + w) for v/c << 1 Omega(on-mass-shell radiation) ~ 1/R(t)^4, i.e. w = +1/3 with Omega(random zpf) + Omega(vacuum coherence) ~ 1/R(t)^0 independent of R(t) since w = -1 for both random normal fluid zpf and condensate (vacuum coherence) / is from random zpf causing both dark energy (negative pressure) and dark matter (positive pressure) exotic vacuum phases. Clumps of w = -1 positive pressure, possibly like the Galactic Halo mimic w = 0 dark matter in their gravity lensing. Positive and negative pressure exotic vacua can each either universally attract or repel However localized negative pressure exotic vacuum clumps anti-gravitate support. The effective gravity strength G* of clumps of exotic vacuum regions can be larger than Newton's G. Guv + /zpfguv = 0 is the zero torsion field GR Poisson equation for exotic vacua. In the Newtonian weak-field slow-speed limit, neglecting gravimagnetism, this is Grad^2(Potential Energy per unit test mass of exotic vacuum field) ~ c^2/zpf Susskind's World Hologram Conjecture requires Einstein's cosmological constant / ~ [LpR(t)]^-2 With critical energy density hc/Lp^4R(t)^2 = (mpc^2/Lp^3)R(t)^-2 Therefore Omega(/) is independent of R(t) consistent with w = -1 Lorentz covariance and EEP of GR. Similarly for Omega(Vacuum Coherence), which is my new term. woke up fresh with a start with the correct way to think about the problem - after taking a 600 mg ibuprofen for shoulder ache from over exercise at the health club. Here is how Lenny Susskind's world hologram really does work and how it explains the dark energy! 1. Trash H. We don't need it. 2. The FRW space expansion factor R(t) is dimensionless. Everything is in units of Lp. 3. The Susskind-Hawking-Bekenstein-t'Hooft hologram entropy of the Universe is simply S/kB = (1/4)R(t)^2 I have written this before of course. R(now) = 10^61 This is a nice formula because it also explains the Arrow of Time since the entropy of the universe is 0 at the initial singularity here, i.e. R(initial singularity) = 0 or, if you prefer Finkelstein's chronons it is 1-bit at t = 0. 4. OK roughly model everything as photons. The photon thermal distribution is hf/(e^hf/kBT - 1) + hf/2 But the dark energy is, to first approximation virtual photons hf/2 whose mean value is hc/LpR(t). 5. Therefore we need virtual energy hf/2 to erase each bit every time the cosmic quantum computer clears its register to step forward another chronon to compute the history of the universe Therefore, the dark energy density the vacuum fabric of curved spacetime needs to compute itself is (R(t)^2/4)(hc/LpR(t))(1/Lp^3R(t))^3 = (hc/Lp^4)1/4R(t)^2 = (hc/4Lp^2)/ Einstein's cosmological constant / = 1/Lp^2R(t)^2 = 10^-56 cm^-2 NOW hc/Lp^2 = c^4/G 6. So this says that Einstein's cosmological constant / is getting smaller as the Universe expands. On the other hand, w = pressure/(energy density) = -1 for random micro-quantum zero point energy. The energy density of cosmic stuff scales as R(t)^-3(1 + w) For example w = 0 for ordinary matter so that Omega(matter) ~ R(t)^-3 w = +1/3 for cosmic black body real photons so that Omega(CMB) ~ R(t)^-4 w = -1 for RANDOM micro-quantum vacuum zero point fluctuations (from covariance) so that Omega(ZPF) ~ R(t)^-0 = constant. 7. However in my macro-quantum theory of the world hologram / = (1/Lp^2)[Lp^3|Vacuum Coherence|^2 -1] Equating this with the world hologram formula in the large-scale FRW metric limit 1/Lp^2R(t)^2 = (1/Lp^2)[Lp^3|Vacuum Coherence|^2 -1] 1/R(t)^2 = [Lp^3|Vacuum Coherence|^2 -1] R(t)^2 = [Lp^3|Vacuum Coherence|^2 -1]^-1 Where from inflation Omega(random ZPF) + Omega(matter) + Omega(radiation) + Omega(vacuum coherence) = 1 === Subject: Re: reply: question: automorphism group of the dihedral group You're welcome. > It was illuminating. > My early reply apparently didn't get through the server. > fact, Aut(D_{2p}) is the holomorph of C_p, a semidirect product > Im not quite sure what you mean by holomorph. The holomorph of a (finite?) group G is the semidirect product G x| Aut(G). For G = C_p, Aut(G) =~ C_{p-1}. -- Jim Heckman === Subject: Re: Math and the baby ixf50QwAAABFW6BQ4fQR2F0f49XaC1bE | My wife and I are expecting our 1st baby by the end of January and we were | wondering if anyone can recommend anything that will stimulate our baby | mathematically? I knew a woman who asked a math professor for advice on how not to pass on her math anxiety to her son. He had two pieces of advice: one, don't tell him you found it frightening, and two, count things. As her son grew up, she said that this had worked out well. I think spending time reading with your kid is a good idea once they're to that age. It may not seem especially mathematical, but it makes a lot of bookish things, including mathematics, seem like normal. Having different kinds of math books simply on the shelf, along with all the novels and so on, made it seem more like a normal part of life. I always used to like looking things up in the dictionary and encyclopedia we got for the family when I was about 9. This may sound odd, but I think a tolerance for reading things that you don't understand was helpful to me. Other kids seemed to regard encounters with stuff that they didn't yet understand with more alarm. I liked sort of geometric puzzles as a kid. I don't know whether this had much to do with my interest in mathematics later. My brother and I both liked playing with wooden blocks, Legos [tm] and so on. As far as actually going into mathematics, one of the attitudes that seems to have helped me was a sense that some things can be cool without being popular. I think we must lose a lot of the potential talent in all the areas deemed nerdy by teenagers just because teenagers who would otherwise find them interesting decide to conform. I remember in 9th grade algebra how the best students besides myself were mostly girls, but then over the next few years it was as if some kind of switch was pulled and by 12th grade calculus only a much smaller share of the class was girls. I don't very often feel the urge to talk about the things that are *not* my cup of tea. I don't go around explaining how botany seems okay but I never got very interested in it, and so on. But for whatever reason people now and then feel like explaining that math is not their thing, and it seems to have gotten started during those teen years. A lot of the people I've known who've gone into mathematics have shown a sense of independence about their interests. If the kid is a girl, a little bit of pragmatic feminism strikes me as a good idea-- those kids say these things are not for girls, but they're confused. The way I loosely categorize topics in my own mind, there's a broad category of things that are interesting because they are esoteric. They appeal to me a bit like the idea of adventures in exotic locations appeals to a lot of people. The world has all kinds of striking things going on in it.... Mathematics that I don't know often has this kind of appeal. People I know who've had kids have said they are impressed by how much of their kids' development occurs spontaneously. People in this country talk baby talk to their kids, and they make encouraging gestures as their babies are learning to walk, but there are cultures where people don't talk differently to their babies and where no attempt is made to encourage walking, and it seems not to make much difference. If I ever have a kid, I'll probably do that kind of stuff but mainly as just a way of spending time with the kid, not in the hope of speeding the process. I suspect the biggest influence of all is just by example. My parents both had a positive attitude to math, and it seems like this liking of mathematics passes pretty well from generation to generation without any deliberate attempt to pass it on, just like interests in music, religion, sports, and so on tend to. I think just the fact that the parents treat these things as a natural part of life, without especially pushing their kids into it, makes it easy for their kids to pick up a taste for it. Even when parents are fans of some activity and their kid is not, the kid will tend to think of it as one of those things that some people like, rather than as one of those weird things that nobody likes. Keith Ramsay === Subject: Re: Math and the baby > My wife and I are expecting our 1st baby by the end of January and we were > wondering if anyone can recommend anything that will stimulate our baby > mathematically? If you child is normal (or will be normal, may God so grant) he comes prewired to do mathematics, music and art. It takes parenting to beat the urge to invent and fantisize out of a kid. If you do not kill your child's adventurous spirit (I am sure you do not intend to and I hope you will not inadvertantly) he/she will grow up with a head for numbers and a good mathematical and artistic intuition. All human beings are geniuses by the age of three and few are by the age of fifteen. Bob Kolker === Subject: Re: Math and the baby >> My wife and I are expecting our 1st baby by the end of January and we were >> wondering if anyone can recommend anything that will stimulate our baby >> mathematically? >If you child is normal (or will be normal, may God so grant) he comes >prewired to do mathematics, music and art. It takes parenting to beat >the urge to invent and fantisize out of a kid. Oh, come now. Schooling is also important in that task! >If you do not kill your >child's adventurous spirit (I am sure you do not intend to It is hopeful, or polite, to have that certainty. Many parents, and some entire subcultures, equate a child's adventurous spirit with something (sometimes old-fashioned deviltry, sometimes new-fangled hyperactivity, and so on) that *should* be killed. We have little evidence that Worried isn't of that sort. Lee Rudolph === Subject: Re: Math and the baby > Oh, come now. Schooling is also important in that task! Schooling is surrogate parenting. It takes a parent to deliver his children into the hands of public educators > It is hopeful, or polite, to have that certainty. Many parents, > and some entire subcultures, equate a child's adventurous spirit > with something (sometimes old-fashioned deviltry, sometimes > new-fangled hyperactivity, and so on) that *should* be killed. > We have little evidence that Worried isn't of that sort. I count myself fortunate to be brought up as I was. I was a smart-ass when I was young and I am a smart-ass now. It is the smart-asses and the cheeky insolent youth of the world that moves it forward. Bob Kolker === Subject: Re: Math and the baby SvMIqg0AAAAH6exCIp08M-zGENb3kCtd You could paper the wall of the nursery with calculus notes. See http://great.russian-women.net/Sofia_Kovalevskaya.shtml === Subject: Re: Math and the baby nrmAfQ0AAAAVEvs440MQyHc_cv3FW6ZR Does anyone know of neighbors or relatives who have actually tried something like this, i.e. mathematical symbols as nursery wallpaper, to ameliorate potential notational anxiety later? I have only heard of this one well documented story. What eventually happened to the kids? Did they turn out happy? We have all also heard horror stories of overbearing parents in any field, and the great psychological damage they have done (I think of the autobiographies of Norbert Weiner, as well as numerous show business personalities), but what about experiences with a much lighter, less neurotic touch. above that her real awakening into math began with conversations she had with her uncle, while still a child. Sofia credits her uncle Peter for first sparking her curiosity in mathematics. He took an interest in Sofia and made time to discuss numerous abstractions and mathematical concepts with her (Rappaport 564). It is silly to simply ridicule this common desire of parents to strike a good balance between overeducating their kids, and leaving their development to chance. We all do it. I thank the poster for bringing it up. === Subject: Re: Math and the baby > We have all also heard horror stories of overbearing parents in any > field, and the great psychological damage they have done There's a book called something like The Drama of the Gifted Child by a psychoanalyst named something like Miller. She outlines the chief trouble here in that precocious kids almost automatically tie their performance to the reception of parental love. After all, they get the most and best attention when they have achieved things. It's easy for us mathematicians to look around at our colleages and see ample examples of such children as adults. How many folks do y'all know who recount some small mental victory for you and then stare at you waiting for their praise? (At least half of all math departments, eh?) I think the bottom line here is that it's crucial, but incredibly difficult, to convey to a child that your love is not conditioned upon their academic performance. It incredibly hard because a parent can give such subtle signals of disappointment, but the child is hyper-tuned in to those signals. Bart === Subject: Re: Math and the baby ixf50QwAAABFW6BQ4fQR2F0f49XaC1bE | | > We have all also heard horror stories of overbearing parents in any | > field, and the great psychological damage they have done | | There's a book called something like The Drama of the Gifted | Child by a psychoanalyst named something like Miller. | She outlines the chief trouble here in that precocious kids | almost automatically tie their performance to the reception | of parental love. After all, they get the most and best | attention when they have achieved things. When I used the word gifted in the title, I had in mind neither children who receive high grades in school nor children talented in a special way. I simply meant all of us who have even to unspeakable cruelty by becoming numb... Without this gift offered us by nature, we would not have survived. | It's easy for us mathematicians to look around at our | colleages and see ample examples of such children as | adults. How many folks do y'all know who recount some | small mental victory for you and then stare at you | waiting for their praise? (At least half of all math | departments, eh?) Yes, I have seen that; it can be fairly silly. One guy I knew had a habit of answering questions in such a way as to maneuver you into asking him more questions. (ObMath: Once someone studying for an exam asked him for help with the following problem. How many subgroups of index 2 are there in the free group on n generators? His response was to think a bit, then smile and say, Ah, trees! Then he stood there smiling, waiting to be prompted to go on. (Points for solving the problem; bonus points for figuring out how he was thinking the solution would go.)) I don't know anybody who doesn't sometimes want to share their small victories, however, and it would be peculiar not to want to, sometimes. I think that it is difficult for people to distinguish between such healthy and unhealthy celebrations of success, especially when it comes to forms of success that aren't in one's own range of interests. So what seems to some people like a perfectly natural sharing of a success to some will sometimes seem like a neurotic search for approval to others. It's easier to see celebrating as silly if it's a small victory on a field that neither you nor your friends play. For a lot of people mathematics, of the kind mathematicians do, pretty much does not exist. There's a stereotype of the academic with his narrow, artificial interests, failing to appreciate more normal aspects of life. I think however people with less unusual interests are on average at least as prone to presume that the hobbies and career goals that they themselves seek out as cool, admirable, worth being proud of and so on are what are really cool, admirable, worth being proud of, and so on, and others' are relatively small, superficial ones. There's this way that some people have of assuming that successes in some areas are compensated somehow. They seize on it as a kind of consolation prize. Athletes sometimes are assumed to be less intelligent. So are the good-looking, sometimes. The wealthy are sometimes thought to be less happy in various ways. If one has power, one must be less moral. The intelligent are thought of as being less well adjusted socially. And so on. Not everybody thinks this way, and people think this way about different areas of success, but there are a lot of stereotypes of this form. | I think the bottom line here is that it's crucial, but | incredibly difficult, to convey to a child that your | love is not conditioned upon their academic performance. True, not just of academic but other sorts of performance too. | It incredibly hard because a parent can give such subtle | signals of disappointment, but the child is hyper-tuned in to | those signals. One of the paradoxes described by mystics of some traditions is that abiding in the unconditional requires sensitivity to all conditions. Noumena and phenomena are not separate.... Some parents make a kind of dual mistake, sending the message that their children *shouldn't* be too excited about the things that interest them. Overly careful indifference is not the way to go either. I also don't know of anyone whose parents didn't sometimes show disappointment. But the sanest people I know are generally people whose parents paid attention to their ups and downs, and not just in narrowly selected parts of life, while helping to keep the ups and downs in a broad perspective. Keith Ramsay === Subject: Re: Math and the baby > I think the bottom line here is that it's crucial, but > incredibly difficult, to convey to a child that your > love is not conditioned upon their academic performance. And if your homework brings you down Then we'll throw it on the fire And take the car downtown Kooks, by David Bowie === Subject: Re: Math and the baby >My wife and I are expecting our 1st baby by the end of January and we were >wondering if anyone can recommend anything that will stimulate our baby >mathematically? Find a copy of Wiles' proof of Fermat's Last Theorem on the internet, print it out and read it to the kid every night. It's important to start now - if you wait until the kid's actually born it may be too late (everyone knows that mathematicians do their best work when young.) >Mkajuma ************************ David C. Ullrich === Subject: Re: Math and the baby > It's important to start now - if you wait until the kid's actually > born it may be too late (everyone knows that mathematicians do their > best work when young.) You are an evil man, did you know that? Bob Kolker