mm-599 === >This is not a psychology question. It is an applied mathematics >question: You are given an incomplete set of data, and your task is to >find a mathematical model (in this case a formula or a rule) that >captures the essence of the data set. >You are right that there are infinitely many mathematically correct >answers. But that does not mean they are all equally acceptable. >The art of applying mathematics is to take ill-posed questions and >transform them to well defined mathematical problems. > I disagree. It is the person with the applied problem who > needs to make the assumptions, after which mathematics or > statistics or numerical analysis or whatever can be applied > to the well-formulated problem. So the question asked is > about which assumptions are being made; as to which are more > reasonable, not only can I not answer that as a mathematician, > I MUST NOT. This arises very often in statistics, and what is > generally done is unfounded religion. > I am often requested to repost my five commandments. These are > posted here without exegesis. > For the client: > 1. Thou shalt know that thou must make assumptions. > 2. Thou shalt not believe thy assumptions. > For the consultant: > 3. Thou shalt not make thy client's assumptions for him. > 4. Thou shalt inform thy client of the consequences > of his assumptions. > For the person who is both (e. g., a biostatistician or psychometrician): > 5. Thou shalt keep thy roles distinct, lest thou violate > some of the other commandments. > The consultant is obligated to point out how their assumptions affect > their views of their domain; this is in the 4-th commandment. But the > consultant should be very careful in the assumption-making process not > to intrude beyond possibly pointing out that certain assumptions make > large differences, while others do not. A good example here is regression > analysis, where often normality has little effect, but the linearity of > the model is of great importance. Thus, it is very important for the > client to have to justify transformations. > There are, unfortunately, many fields in which much of the activity > consists of using statistical procedures without regard for any assumptions. >Note the distinction between mathematics and applied mathematics. >Applied mathematics is not mathematics. The mathematician deals with >well defined problems in a formal universe, with well defined criteria >of correctness. The applied mathematician deals with ill posed problems >defined in an informal language, with missing assumptions and open >interpretations. >In an ideal world, the client would make all the necessary assumptions >and present the problem as a well defined mathematical problem. In the >real world, this hardly ever happens. That's why the applied >mathematician is paid big bs, to bridge the gap between the client's >world and formal mathematics. That's the art of applying mathematics --- >figuring out what the missing assumptions are, and validating them with >the client. >I have nothing against the five commandments in some ideal sense, but if >you insist on dealing only with the clients who follow 1 and 2, you will >have to send many clients somewhere else. In many situations, the >applied mathematician (consultant) must guess what the right assumptions >are (violation of 3) and then verify them with the client (in agreement >with 4). This is permissible, but the clients who do not understand 1 and 2 are likely to agree with any assumptions the consultant produces. The typical situation in statistical consulting is that the client comes in with the data, some idea about how the data is collected, and asks to get the state of the universe. The consultant can give some advice, such as point out which assumptions are not important and which are. If the applied mathematician makes the assumption that the equations valid for a perfect gas hold, and this is not the case, the results are likely not to be very good. Or if a linear differential equation is used when it is not even a fair approximation. Fortunately in engineering, the errors usually show up in testing; one rarely just goes ahead on the basis of calculations from a model. And differences in the assumptions in physics and engineering, especially in complicated situations, can make huge differences. However, in statistics, the errors may not show up for decades, and much harm can be done. Medicine is replete with examples of this, and psychology and education are suffering badly from the misapplication. Even when those in the hard sciences deal with randomness, they are prone to make assumptions, such as a non-Gaussian process with stationary independent increments and continuous time paths, which do not exist. -- 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: Applying mathematics, was Re: A good Q??? > [ snip ] > However, in statistics, the errors may not show up for > decades, and much harm can be done. Medicine is replete > with examples of this, and psychology and education are > suffering badly from the misapplication. > [ ... ] Good point. Also, statistical evidence in criminal cases. === Subject: Re: A good Q??? >This is not a psychology question. It is an applied mathematics >question: You are given an incomplete set of data, and your task is to >find a mathematical model (in this case a formula or a rule) that >captures the essence of the data set. Would you still make that claim if, as an example, the sequence were: 7, 5, 5, 4, 7, 6, 6, 7, ? and it would turn out that answer was 5 because they are the numbers of letters in the spelling of the planets from the sun outward? --Lynn === Subject: Re: A good Q??? |>This is not a psychology question. It is an applied mathematics |>question: You are given an incomplete set of data, and your task is to |>find a mathematical model (in this case a formula or a rule) that |>captures the essence of the data set. | Would you still make that claim if, as an example, the sequence were: | | 7, 5, 5, 4, 7, 6, 6, 7, ? | | and it would turn out that answer was 5 because they are the | numbers of letters in the spelling of the planets from the sun | outward? I wouldn't classify the counting of letters in a series of words as applied mathematics, especially if the language (or spelling) is specified. Also, the above sequence is time dependent. A few years ago, the series would've been (for modern English): 7, 5, 5, 4, 7, 6, 6, 5, 7 when Neptune was further away from Sol then Pluto. (You say farther, I say further...) ____________________________________________Gerard S. === Subject: Re: A good Q??? > Hi all, > My father asked me this question and i asked most of my friends about it! > The Q looks interesting thta i got different answers? Can you help me to solve it! > The Q is: > what is the next number in the following seq. of numbers: > 3,6,10,8,11,22,... > ??? > The different answers i got is: > 19, 18 , 12 and 26 You can find a justification for ANY number being next. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: A good Q??? > Hi all, > My father asked me this question and i asked most of my friends about > it! The Q looks interesting thta i got different answers? Can you help > me to solve it! > The Q is: > what is the next number in the following seq. of numbers: > 3,6,10,8,11,22,... > ??? > The different answers i got is: > 19, 18 , 12 and 26 > You can find a justification for ANY number being next. I do understand your point of view - you can fit it with many polynomials that give any next answer that you like. But I also see great value in these kinds of questions. Truthfully, we know what the question is getting at, and we do tend to instantly be able to see what was the correct answer when we get it. Perhaps these kinds of questions help us distinguish true mathematics from the rigorous formalization that currently stands in place of math. (OK, I know that last remark is truly ßameworthy. But honestly, I am not a troll, because I am not going to spend any effort defending my indefensible viewpoint.) === Subject: Re: A good Q??? > Hi all, > My father asked me this question and i asked most of my friends about > it! The Q looks interesting thta i got different answers? Can you > help me to solve it! > The Q is: > what is the next number in the following seq. of numbers: > 3,6,10,8,11,22,... > ??? > The different answers i got is: > 19, 18 , 12 and 26 > You can find a justification for ANY number being next. > I do understand your point of view - you can fit it with many > polynomials that give any next answer that you like. But I also see > great value in these kinds of questions. Truthfully, we know what the > question is getting at, and we do tend to instantly be able to see what > was the correct answer when we get it. Perhaps these kinds of questions > help us distinguish true mathematics from the rigorous formalization > that currently stands in place of math. (OK, I know that last remark is > truly ßameworthy. But honestly, I am not a troll, because I am not > going to spend any effort defending my indefensible viewpoint.) I didn't say it was useless. My answer would be 26. I'd be more interested in seeing each person's justification for their answer than the answer in isolation. 26 comes from a pattern of x2, +4, -2, +3, repeat -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Earliest example of an incomputable real >In sci.math, David C Ullrich >: >Can someone please give me a rigorous definition of computable? > There are at least two: > (i) a real x is computable if there exists a Turing machine that > enumerates the degits in a decimal expansion of x. >I might call that semicomputable; a fully-computable number would >enumerate all the digits (except for trailing 0's, of course). Huh? What's the difference? >Admittedly, there are a few issues here; a TM might, for instance, >compute the numerator and denominator of a (rational) number, >e.g. 1/3 = 0.333... ; a very simple Turing machine proves this >is a semicomputable number in base 10 (and fully computable in >base 12: 1/3 = 0.4(12)). > (ii) a real x is computable if there exists a TM which takes > as input a rational eps > 0 and outputs a rational y with > |x - y| < eps. > Curiously I'm pretty sure the two are not equivalent; it's > clear that (ii) implies (i), but it's not clear to me that > (i) implies (ii). >Well, if one can successfully enumerate digits, then it's >pretty clear that one need merely enumerate enough digits >to limbo under the epsilon hurdle. > Otoh it does seem to me that (i) is equivalent to (i'): > (i') there exists a TM that enumerates the bits in a binary > expansion of x. >There are some issues here; for instance, 0.1(10) = 0001100110011...(2). >Again, these aren't major. > (Say we know that x = 0.49999... (where the ... indicates > unknown digits, not infinitely many nines.) If there _are_ > infinitely many nines then x = 1/2 and we can certainly > enumerate the bits in x; otoh if there are not infinitely > many nines then x is definitely 0.1... in binary. So >Pedant point: 0.01... > there is a Turning machine that enumerates a binary > expansion (which is not the same as saying that we > know what TM does so!)) > Probably someone else knows for sure what relations > exist between (i), (ii) and (i'). >Well, if one makes a distinction between fully-computable >and semicomputable -- and I'm not entirely sure how >useful that distinction is -- one might be able to better >express this question. Apart from that, these look >roughly equivalent. >-Greg > ************************ > David C. Ullrich ************************ David C. Ullrich === Subject: Re: Earliest example of an incomputable real In sci.math, David C Ullrich : >In sci.math, David C Ullrich >: >Can someone please give me a rigorous definition of computable? > There are at least two: > (i) a real x is computable if there exists a Turing machine that > enumerates the degits in a decimal expansion of x. >I might call that semicomputable; a fully-computable number would >enumerate all the digits (except for trailing 0's, of course). > Huh? What's the difference? The probability of 10 heads is fully-computable; 1/1024 = .0009765625 . The probability of getting a 1-spot on a die is endlessly repeating, and therefore would not be fully-computable: 1/6 = .1666666... . I might have to relax the definition a bit, and require that a fully-computable number be identifiable in some finite form: _ 1/6 = .16 or .1[6] but that's the general idea. Admittedly, I'm not sure it's all that useful a distinction, as 1/6 = .2(12) in base 12, though the relaxed form should at least lead to a consistent set of numbers -- which is equal to Q, it turns out, absent more metadata constructs that allow generation of such weird constants as .1234567891011121314151617181920... or symbolic abbreviations such as pi or e for common transcendentals. [rest snipped for brevity] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Earliest example of an incomputable real >In sci.math, David C Ullrich >: >In sci.math, David C Ullrich >: Can someone please give me a rigorous definition of computable? There are at least two: (i) a real x is computable if there exists a Turing machine that > enumerates the degits in a decimal expansion of x. >I might call that semicomputable; a fully-computable number would >enumerate all the digits (except for trailing 0's, of course). > Huh? What's the difference? >The probability of 10 heads is fully-computable; >1/1024 = .0009765625 . >The probability of getting a 1-spot on a die is endlessly >repeating, and therefore would not be fully-computable: >1/6 = .1666666... . Oh. I had no idea that's what you meant. You can use words any way you like, but this is definitely _not_ what computable number actually means. >I might have to relax the definition a bit, and require >that a fully-computable number be identifiable in some >finite form: > _ >1/6 = .16 or .1[6] >but that's the general idea. Now, if you say it should be possible to specify all the digits in some finite form, that's a much more reasonable notion of computable. Now note that a Turing machine can be described completely in finite form, so saying that this or that Turing machine enumerates the digits _does_ say that the digits can be specified in some finite form. >Admittedly, I'm not sure it's >all that useful a distinction, as >1/6 = .2(12) >in base 12, though the relaxed form should at least lead >to a consistent set of numbers -- which is equal to Q, >it turns out, absent more metadata constructs that allow >generation of such weird constants as >.1234567891011121314151617181920... >or symbolic abbreviations such as pi or e for common transcendentals. >[rest snipped for brevity] ************************ David C. Ullrich === Subject: Re: Earliest example of an incomputable real <5GbYWpQiYm2r@eisner.encompasserve.org> 1/6 = .2(12) > in base 12, though the relaxed form should at least lead > to a consistent set of numbers -- which is equal to Q, Well, that's the problem. One definition depends on the base; the other is equivalent to rationals. It doesn't seem very useful. If it's the infiniteness of the expansion that bothers you, note that the usual definition of a computable number *isn't* that there exists a Turing machine that outputs the whole number. (You aren't supposed to view the output of a TM until it halts---otherwise, how will you decide if the digits you see on the tape are part of your number or just some intermediate scratch work?) It's that there exists a Turing machine that can output as much of the number as you want. In particular, in the definition that David proposed: > (i) a real x is computable if there exists a Turing machine that > enumerates the degits in a decimal expansion of x. the phrase enumerating the digits *doesn't* mean printing the whole number. It means something like: the Turing machine is capable of taking as input an integer n and outputting the nth digit of x. (Or, alternatively, the Turing machine takes a positive integer n and prints x up to n digits after the decimal place before halting.) -- Kevin === Subject: Re: Earliest example of an incomputable real charset=Windows-1252 -snip- > note > that the usual definition of a computable number *isn't* that there > exists a Turing machine that outputs the whole number. (You aren't > supposed to view the output of a TM until it halts---otherwise, how > will you decide if the digits you see on the tape are part of your > number or just some intermediate scratch work?) It's that there > exists a Turing machine that can output as much of the number as you > want. Some definitions of Ôcomputable' (equivalent to that above) *do* involve non-halting TMs, by assuring that the computation will, for each n, *unambiguously* provide the nth digit in finitely many steps. In fact, Turing's original definition, and also the definition in Minsky's textbook, use non-halting TMs -- although never halting, they prevent erasure of each digit once it's written on the tape. (Turing's accomplished this by making every second tape cell unerasable, and Minsky's used a rightward moving don't move to the left of me marker.) --r.e.s. === Subject: Re: Earliest example of an incomputable real In sci.math, r.e.s. : > -snip- > note > that the usual definition of a computable number *isn't* that there > exists a Turing machine that outputs the whole number. (You aren't > supposed to view the output of a TM until it halts---otherwise, how > will you decide if the digits you see on the tape are part of your > number or just some intermediate scratch work?) It's that there > exists a Turing machine that can output as much of the number as you > want. > Some definitions of Ôcomputable' (equivalent to that above) > *do* involve non-halting TMs, by assuring that the computation > will, for each n, *unambiguously* provide the nth digit in > finitely many steps. In fact, Turing's original definition, > and also the definition in Minsky's textbook, use non-halting > TMs -- although never halting, they prevent erasure of each > digit once it's written on the tape. (Turing's accomplished > this by making every second tape cell unerasable, and Minsky's > used a rightward moving don't move to the left of me marker.) > --r.e.s. I'll go with this form; it's logical and even approximatable. Pi, in particular, has a computational method that allows successive hexits (or hex digits) to be spat out. Or one can simply compute larger and larger rational approximations using Machin or a Machin-like formula (which can be implemented using some sort of rational arithmetic and arbitrarily large integers, if one really wanted). It doesn't look all that useful to explore the finite computability issue that much further, frankly. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Earliest example of an incomputable real >Can someone please give me a rigorous definition of computable? > There are at least two: > (i) a real x is computable if there exists a Turing machine that > enumerates the degits in a decimal expansion of x. > (ii) a real x is computable if there exists a TM which takes > as input a rational eps > 0 and outputs a rational y with > |x - y| < eps. > Curiously I'm pretty sure the two are not equivalent; it's > clear that (ii) implies (i), but it's not clear to me that > (i) implies (ii). >Actually I think (ii) => (i) is the problematic one; the other is easy. Yes, I've already said that was a typo. (Typo not being _exactly_ the right word - I was forgetting which was (i) and which was (ii).) > Otoh it does seem to me that (i) is equivalent to (i'): > (i') there exists a TM that enumerates the bits in a binary > expansion of x. > (Say we know that x = 0.49999... (where the ... indicates > unknown digits, not infinitely many nines.) If there _are_ > infinitely many nines then x = 1/2 and we can certainly > enumerate the bits in x; otoh if there are not infinitely > many nines then x is definitely 0.1... in binary. So > there is a Turning machine that enumerates a binary > expansion (which is not the same as saying that we > know what TM does so!)) > Probably someone else knows for sure what relations > exist between (i), (ii) and (i'). >Another common definition is: >(iii) A real x is computable if there is a computable nested sequence of >rational intervals identifying x. >It's obvious that (iii) is equivalent to (ii), and I have seen it stated >as an exercise that (iii) is equivalent to (i). The key idea, as you >said, is that we can show a suitable TM exists without necessarily >knowing how to find it. ************************ David C. Ullrich === Subject: Re: Uniqueness for a stochasic logistic equation Reposted from sci.math.research, where it received no replies. > I'm looking for a proof/reference for the following problem: the stochastic > differential equation with logistic growth and Feller noise: > dX_t = a(b-X_t)X_t dt + sqrt{X_t}dB_t > where B_t is a Brownian motion. Uniqueness does not follow in the orthodox > way (via Picard iteration) since the drift term is not Lipschitz. > Whilst I'm on the subject, just considering the ODE without the diffusion > term - this looks like a Lotka-Volterra type equation. So if > anyone has any ideas for the deterministic case either, I'd be grateful to > hear them. === Subject: Re: Uniqueness for a stochasic logistic equation > Reposted from sci.math.research, where it received no replies. >I'm looking for a proof/reference for the following problem: the > stochastic >differential equation with logistic growth and Feller noise: >dX_t = a(b-X_t)X_t dt + sqrt{X_t}dB_t >where B_t is a Brownian motion. Uniqueness does not follow in the orthodox >way (via Picard iteration) since the drift term is not Lipschitz. >Whilst I'm on the subject, just considering the ODE without the diffusion >term - this looks like a Lotka-Volterra type equation. So if >anyone has any ideas for the deterministic case either, I'd be grateful to >hear them. It's not quite my thing, but since nobody takes it ... In Economics this is the way how stochastic volatility is modelled for the Heston or Square Root Model, cf http://www.optioncity.net/pubs/Ch1Excerpt.pdf where it is related to a GARCH model. Not sure what you mean by deterministic case, but i have never seen a Ôclosed form solution'. -- use mail for mail not no nail === Subject: Re: Impressive ways to obtain the number 5.00 using only one line and advanced mathematics by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5HD6VX12802; >I'm looking for an impressive way to obtain the number 5.00 using >only one line (~80 characters) and advanced mathematics. >Basically, I'm doing this as a test of some sort. I know this is >weird, but can anyone help me out with something really elegant? >Netsniper >---------------------------------------------- > * Binary Usenet Leeching Made Easy > * http:// www.newsleecher.com/?usene tInsofar as e is irrational, it can be defined only in terms of some >limit process. I'll leave it up to others to argue the fine points of >such an expression as: >lim(n --> inf) (1+1/n)^n >--OL === Subject: 3D-polyhedron coordinate system Hi! How to create a local coordinate system inside any convex polyhedron, and compute coordinates of vertices, which lies inside this polyhedron? It is something likes polyhedron barycentric coordinates. I know how to create 2D-polygon barycentric coordinate system, but what about 3D-polyhedron coordinate system? Anton&IVA ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: topological group question >Let G and H be two Hausdorff topological groups and f:G->H a bijective >continuous homomorphism. Is f a homeomorphism? Counterexamples? > [text snipped] >For more fun consider the case when G and H are both complete, >separable, metric. In that case f must be a homeomorphism. > By the example given earlier of the real line R_d with the discrete > topology and the identity id:R_d->R we cannot omit separability here. > But in the case of Banach spaces the theorem is true without > separability. Can you pinpoint exactly what property Banach spaces > have in addition to Banach groups (i.e. metric complete) that makes > them special in this regard? I fail to see how the linear structure > can account for this. Well, one of the properties that Banach spaces posess which is not shared with Banach groups in general is that given a neighborhood U of 0, the union of all nU is the whole space. This does not happen with the example of R with the discrete topology. The proof of the open mapping theorem for Banach spaces uses the fact that each element of a Banach space belongs to some nU, where U is the open unit ball. Jose Carlos Santos === Subject: Re: topological group question >Let G and H be two Hausdorff topological groups and f:G->H a bijective >continuous homomorphism. Is f a homeomorphism? Counterexamples? > You've already been told that, in general, the answer is negative, but > you might want to know that there's an old theorem due to Richard Arens > that says that if G is Lindeloff and if G and H are both locally > compact, then f is a homeomorphism. > Jose Carlos Santos Another interesting special case is this. If f:G --> H is a bijection between locally compact abelian groups that induces a bijection between the character groups, then f is an isomorphism. See I. Glicksberg (1962), Uniform boundedness for groups. Canadian J. Math. 14, 269-276. I did not understand the proof, however. === Subject: question in analytic number theory Let denote the distance to the nearest integer, i.e. =inf{|x-m|, min Z}. For which real numbers x to 0, nto infty? ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: question in analytic number theory Originator: grubb@lola >Let denote the distance to the nearest integer, i.e. >=inf{|x-m|, min Z}. >For which real numbers x >to 0, nto infty? This is not an easy problem. Certainly, if |x|<1, it holds. We may assume that x>1 without loss of generality. Some things that are known under this assumption: 1) If x is an algebraic integer all of whose conjugates are of absolute value less than 1 (i.e. x is a Pisot number), then sum converges. 2) If x is an algebraic number and there is a y such that ->0, then x is a Pisot number. This is due to Salem. 3) If there is a y such that sum ^2 converges, then x is a Pisot number (again due to Salem). It is conjectured that if ->0, then x is a Pisot number. This is not known, last time I checked. This question comes up in the study of sets of uniqueness for trigonometric series. The book by Zygmund has a treatment. --Dan Grubb === Subject: Re: int(exp(x)*erf(x),x = a .. b) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5HEuSb23345; > using MATLAB 6.1.0.450 (R12.1) to access the maple > kernel I don't get a solution. for: > int(exp(x)*erf(x),x = a .. b) > Does anyone know if this is solvable? It looks like you could do parts > a few times, and then equate terms. Since: > int(sym(Ôerf(x)')) ans = x*erf(x)+1/pi^(1/2)*exp(-x^2) > What about the integral: > int(exp(x^2)*erf(x),x = a .. b) > For this I could let u=erf(x) > and do substitution. Right? > Moreover, If I figure out how to do the integral myself, how do I > teach maple to do it? >Okay, Now that the above intgrals were solved how about something more >general: >int(erf(a*(x-d))*exp(-b*(x-c)^2),x = -inf .. inf) >This is really what I need to solve anyway. >are solved. I figured out how to solve int(sym(Ôerf(x)')) by looking >at the taylor series. of a exp(-x)^2 and erf(x) and int(erf(x)) and >just used linear algabra to see what linear combination gave the >answer. int(erf(a*(x-d))*exp(-b*(x-c)^2),x = -inf .. inf)=sqrt(pi/b) erf(a*sqrt(b/(a^2 + b))*(c-d)), (b>0) got it by differentiating the integrand w.r.t. a, then integrated over x=-inf..inf, then substituted a=sqrt(b*z)/sqrt(1-z) and integrated over z and then - most important - checked the result numerically. === Subject: I'm laughing at the superior intellect by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5HEuSC23309; (snipped drivel, insults, and evasion) So James Harris and Quinn Tyler Jackson belong to a purported high-IQ society, Mega Foundation and/or Ultranet? Well, they've disgraced their colleagues! >I *did* send my paper to several mathematicians and >tried to talk about it with various people you might >say are in my circle >(This all occurred on another list that James and I >both are on.) Several mathematicians who had considered >the math noted that they had not found the error So, one may conclude that, according to both James Harris and his buddy Quinn Tyler Jackson, their high-IQ colleagues couldn't find errors in Harris' work, which others on this newsgroup were able to find. Them high-IQers weren't up to snuff! (Then again, they did let James Harris join (need the dues, man!)). No wonder James Harris has such a superiority attitude and megalomania. He considers the mathematicians on this newsgroup to be mere ants to his towering intellect. HA HA HA HA HA HA HA HA HA HA HA and HA! I'm laughing at the Ôsuperior intellect' (Quiz: identify the source of this quote). Let's use netspeak - ROTFL, ROTFLMAO, ROTFLMBFA0 (BF stands for big fat) HA HA HA HA HA HA HA HA HA HA HA HA! HA HA HA HA HA HA HA HA HA HA HA HA! I mean, for Harris and Jackson to expose their little high-IQ group to ridicule by trying to use it as a supposedly outside reference to support Harris' lame argument that his mathematics may have hard-to-find errors, the attempt was pathetic. Several mathematicans who had considered the math noted that they had not found the error HA HA HA HA HA HA HA HA HA HA HA HA! HA HA HA HA HA HA HA HA HA HA HA HA! HA HA HA HA HA HA HA HA HA HA HA HA (ad infinitum)! Now comes forth... James Harris is a Freaking Imbecile(TM) === Subject: Re: I'm laughing at the superior intellect JHiN said: > (snipped drivel, insults, and evasion) > So James Harris and Quinn Tyler Jackson belong to > a purported high-IQ society, Mega Foundation > and/or Ultranet? > Well, they've disgraced their colleagues! Well, Eliza, you dropped the ball yet again, so I'm not going to give you another chance. You had it -- the offer is retracted. Whoops ... sorry ... you missed yer second chance to call my bluff. Gosh, ain't that too bad? Never let it be said I didn't give you ample opportunity to allow your programming to demonstrate that it was more interested in defending mathematics against outlandish claims than it was in focusing on socio-political straw men arguments with absolutely e^{-ipi}+1 value. Please tell Grampa Simpson to reconsider writing you in C++. It has structures that tend to discourage the use of goto's, and with a good debugger, maybe he'll figure out how to also avoid your endless loops. -- Quinn === Subject: Intorduction To Tensors I'm currently a Third Year University student studying both Math and Physics. I've noticed that alot of advanced Physics theory uses tensors. I was wondering if someone might be able to recommend a book or subject to introduce tensors. Blaidd === Subject: Re: Intorduction To Tensors > I'm currently a Third Year University student studying both Math and > Physics. I've noticed that alot of advanced Physics theory uses tensors. > I was wondering if someone might be able to recommend a book or subject to > introduce tensors. > Blaidd well two books, one for each take of yours :) for the physics http://www.amazon.com/exec/obidos/tg/detail/-/0486638332/103- 9186826-3403002 ?v=glance for the mathematics http://www.amazon.com/exec/obidos/tg/detail/-/0486640396/002- 4705162-6672861 ?v=glance I hope you enjoy them if you decide to read them. Olivier. === Subject: Re: Intorduction To Tensors > I'm currently a Third Year University student studying both Math and > Physics. I've noticed that alot of advanced Physics theory uses tensors. > I was wondering if someone might be able to recommend a book or subject to > introduce tensors. > Blaidd If you want a lot of down-to-earth practice, definitely go for Schaum's Tensor Calculus by David Kay. A little warning is in place however: Dirk Vdm === Subject: Re: Intorduction To Tensors > I'm currently a Third Year University student studying both Math and > Physics. I've noticed that alot of advanced Physics theory uses tensors. > I was wondering if someone might be able to recommend a book or subject to > introduce tensors. > Blaidd There's a Dover book Intro. to Vector and Tensor Analysis, by Robt. Wrede, only $16.95. === Subject: Re: Intorduction To Tensors > I'm currently a Third Year University student studying both Math and > Physics. I've noticed that alot of advanced Physics theory uses tensors. > I was wondering if someone might be able to recommend a book or subject to > introduce tensors. You can start online: http://mathworld.wolfram.com/Tensor.html === Subject: Re: Intorduction To Tensors >I'm currently a Third Year University student studying both Math and >Physics. I've noticed that alot of advanced Physics theory uses tensors. >I was wondering if someone might be able to recommend a book or subject to >introduce tensors. >Blaidd http://gershwin.ens.fr/vdaniel/Doc-Locale/Cours-Mirrored/ Maths-Stuff/tensorc alc/ -- Suppose you were an idiot... And suppose you were a member of Congress... But I repeat myself. - Mark Twain === Subject: Re: math bibles <40bc7a53$10$fuzhry+tra$mr2ice@news.patriot.net> <40cd3d11$30$fuzhry+tra$mr2ice@news.patriot.net> X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS >Consider the difference between the Hebrew Bible and the Christian >Bible. You're using the word bible in a different sense. >By your reasoning, Shmuel, the Hebrew Bible is >not a bible on the subject of faith. No, by my reasoning the Tanach is not a bible of Christianity. By my reasoning the Christian scriptures are a bible of Christianity but are not a bible of faith, since they do not cover, e.g., Zoroastrianism. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Is pure mathematics worth spending tax money for it? <40cd3268$28$fuzhry+tra$mr2ice@news.patriot.net> X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 07:13 PM, Chan-Ho Suh said: >I can list plenty of examples if you want. But then you realize, you >are putting yourself in the position of explaining how useful (in the >sense of the OP) those things are. No. You claim they are useless. Historically such claims have been unreliable. The onus is on you to prove that they are useless. >I can't see or find anything approaching even a practical real-world >application in even the far future for any of these. No surprise their; if you could see the practical applications then you would have picked different examples. >Why would I include government funds for concert halls or museums in >a sentence that starts with sculptors and composers do not get >anywhere near as much money...? Unless some of this money goes >into the pockets of these artists, it's not relevant to the point I >raised. As a taxpayer, I consider any such expense to be relevant to the issue of how much the government supports the arts. Do you have graduate assistants? Does the grant money for their stipends go into your pocket? Does that mean that I shouldn't count it? >Why would you think that? If you believe that Mathematics has no value then why would you want to work with it? >Am I supposed to believe the contrary in order to be a mathematician? Have you stopped beating your wife? Were they running a sale on straw dummies? >Let me know all the socio-political beliefs I'm supposed to hold in >order to do mathematics. Some self respect would help. Do you consider yourself to be a parasite, feeding at the public trough for work of no value? If so, I'll accept your evaluation for yourself, although it wouldn't be relevant for Mathematicians who don't share your quaint perspective. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Is pure mathematics worth spending tax money for it? > at 07:13 PM, Chan-Ho Suh said: >I can list plenty of examples if you want. But then you realize, you >are putting yourself in the position of explaining how useful (in the >sense of the OP) those things are. > No. You claim they are useless. Historically such claims have been > unreliable. The onus is on you to prove that they are useless. Well that's funny! I say I can't think of any use, in a real world practical sense and challenge you to find such a use. You reply by saying the onus is on me! History is on your side, eh? If you think you're actually arguing in a logical manner, please give me a criterion that you would accept that would conclusively show that the examples I gave are useless. For me, it's simple; to convince me all you have to do is produce a single application. >I can't see or find anything approaching even a practical real-world >application in even the far future for any of these. > No surprise their; if you could see the practical applications then > you would have picked different examples. Give me a criterion like I said. >Why would I include government funds for concert halls or museums in >a sentence that starts with sculptors and composers do not get >anywhere near as much money...? Unless some of this money goes >into the pockets of these artists, it's not relevant to the point I >raised. > As a taxpayer, I consider any such expense to be relevant to the issue > of how much the government supports the arts. Do you have graduate > assistants? Does the grant money for their stipends go into your > pocket? Does that mean that I shouldn't count it? You're mixing issues. I never compared the amount of money going to arts versus mathematics. I was comparing the money going to the actual people in such disciplines. See: > Economically advanced nations spend money supporting artists > and composers. You wouldn't ask for applications for a concerto, > or for a sculpture - why do you need them for a theorem? To which I responded: Aaah, yes, but sculptors and composers do not get anywhere near as much money from the National Endowment for the Arts, say, as mathematicians working on really useless things get from the NSF and other organizations. ----------- You're trying to put me in a box I never was in. >Why would you think that? > If you believe that Mathematics has no value then why would you want > to work with it? Boy, you really are a piece of work, aren't you? Even your snippage seems designed to confuse the issue. That comment I made was in response to the following exchange: Chan-Ho: >People's ignorance of mathematics is >ultimately the cash-cow by which we survive. Seymour J: If you believe that, then perhaps you're in the wrong field. ----------- Where anywhere did I imply that mathematics has no value? I make the point that people's ignorance is a significant contribution to how mathematicians make money -- a statement that is perhaps controversial and debatable, and then you respond that perhaps I'm in the wrong field. To that I respond why you would respond thus, to which you respond implying that I don't believe mathematics has any value! Sheesh, do you live in your own little world, or do you actually respond to what people say? >Am I supposed to believe the contrary in order to be a mathematician? > Have you stopped beating your wife? Were they running a sale on straw > dummies? This is nothing bus sheer rhetoric, and not even good rhetoric at that. I am not making any strawman arguments or asking such a ridiculous kind of question. order to be a mathematician. This is what you actually suggested, isn't it? You suggested that somehow the statement I made about the cash-cow (cf above) means I'm in the wrong field. I think it's perfectly a natural and sensible response to ask why such a belief would be incompatible with being a mathematician. You clearly did not only miss my point, but read all sorts of things into it. Not only that, you have clearly distorted the situation so much that you think I'm asking nonsensical questions. That's just sad. >Let me know all the socio-political beliefs I'm supposed to hold in >order to do mathematics. > Some self respect would help. Do you consider yourself to be a > parasite, feeding at the public trough for work of no value? If so, > I'll accept your evaluation for yourself, although it wouldn't be > relevant for Mathematicians who don't share your quaint perspective. No, I don't consider myself a parasite and I consider my work of value. I don't naively think everybody (or even the majority) would agree because such agreement would take a great familiarity with pure mathematics than is currently the case or, I think, ever possible. === Subject: Re: Is pure mathematics worth spending tax money for it? >I study electrical engineering and I can't help but wonder how much >slower the field would progress, without mathematicians. > Feynman said that if it weren't for the mathematicians, > progress in physics would be set back by about a week. Without group theory would solid state physics have advanced as quickly? Was Maxwell a mathematician? === Subject: Re: Is pure mathematics worth spending tax money for it? I study electrical engineering and I can't help but wonder how much > slower the field would progress, without mathematicians. Feynman said that if it weren't for the mathematicians, >progress in physics would be set back by about a week. > Without group theory would solid state physics have advanced as > quickly? Was Maxwell a mathematician? I don't know (x2). I suspect Feynman would say that the solid state physicists would have worked out all the group theory they needed in a few days. I also suspect Feynman was using hyperbole to make a point, and did not expect to be taken literally. Unfortunately, Feynman is no longer in a position to clear this up. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Is pure mathematics worth spending tax money for it? > I study electrical engineering and I can't help but wonder how much > slower the field would progress, without mathematicians. >Feynman said that if it weren't for the mathematicians, >progress in physics would be set back by about a week. > Without group theory would solid state physics have advanced as > quickly? Was Maxwell a mathematician? >I don't know (x2). >I suspect Feynman would say that the solid state physicists >would have worked out all the group theory they needed in a few days. >I also suspect Feynman was using hyperbole to make a point, >and did not expect to be taken literally. >Unfortunately, Feynman is no longer in a position to clear this up. One of Feynman's contributions is the Feynman path nonsense; in SOME cases, it has been possible to come up with an interpretation which agrees with what is otherwise known. It would be a major feat to come up with a general clear and in principle computable version of it. But while the intuition seems very clear, even the probability versions of the expressions, which are simpler than the quantum versions, take lots of finagling to make sense. Remember that this is more than a half century old. This is not the only place problems like this arise. What it seems should be easy often is impossible. Much of mathematics is of this nature; what seems obvious is often not just not so, but even wrong. -- 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: Is pure mathematics worth spending tax money for it? > I study electrical engineering and I can't help but wonder how much > slower the field would progress, without mathematicians. >Feynman said that if it weren't for the mathematicians, >progress in physics would be set back by about a week. > Without group theory would solid state physics have advanced as > quickly? Was Maxwell a mathematician? >I don't know (x2). >I suspect Feynman would say that the solid state physicists >would have worked out all the group theory they needed in a few days. >I also suspect Feynman was using hyperbole to make a point, >and did not expect to be taken literally. >Unfortunately, Feynman is no longer in a position to clear this up. > One of Feynman's contributions is the Feynman path > nonsense; in SOME cases, it has been possible to come up > with an interpretation which agrees with what is otherwise > known. It would be a major feat to come up with a general > clear and in principle computable version of it. But while > the intuition seems very clear, even the probability > versions of the expressions, which are simpler than the > quantum versions, take lots of finagling to make sense. > Remember that this is more than a half century old. > This is not the only place problems like this arise. What > it seems should be easy often is impossible. Much of > mathematics is of this nature; what seems obvious is often > not just not so, but even wrong. I'm not sure if this makes the point you're trying to make. Doesn't this example show that physicists have been just fine without interfering, anal-retentive mathematicians? Actually, doesn't this show they are more than just fine, they've actually made significant progress? In short, doesn't the Feynman path integral more than justify Feynman's statement that Gerry mentioned? === Subject: Re: Is pure mathematics worth spending tax money for it? ..................... > This is not the only place problems like this arise. What > it seems should be easy often is impossible. Much of > mathematics is of this nature; what seems obvious is often > not just not so, but even wrong. >I'm not sure if this makes the point you're trying to make. Doesn't >this example show that physicists have been just fine without >interfering, anal-retentive mathematicians? Actually, doesn't this >show they are more than just fine, they've actually made significant >progress? In many cases, physicists have made use of much previous mathematics. The Lorentz-Fitzgerald Transformations were a purely mathematical exercise to find what linear transformations of space and time preserved the Maxwell equations for a fixed speed of light. Einstein then gave a physical interpretation of the results in the classic Theory of Special Relativity. >In short, doesn't the Feynman path integral more than justify Feynman's >statement that Gerry mentioned? No; nobody has managed to use the Feynman path integral to get results not otherwise obtainable. The only results about it follow from other considerations, and the use of mathematics, often mathematics which one would not have considered appropriate. Some of this was done not by using Feynman's statements, but his approach, but even this has not been sufficiently successful to produce the desired results. There have been many attempts to understand quantum processes, but their formal similarity to stochastic processes is not good enough, and this is what Feynman was incorrectly claiming to have done, even if the lapses were corrected. -- 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: The definition of weight So O.K: The difference between inertial motion of the first law of physics, and the concept of inertia itself, is too much for most scientists to swallow: Here is an easier one: Weight is the force exerted between an object; body, or mass of matter and the surface of a planet - like Earth or the moon - that it is bearing upon. Without something to bear upon: Object's; bodies and mass's of matter are weightless; all forces require something to bear against! The magnitude of an object's; body's, or mass's weight-force [w] is proportional to the deceleration [g] at which the planet's surface is restraining them from further gravitation: That is, for any given object; body, or mass of matter, at anytime, the ratio of its weight to the acceleration at which it will free fall, is a constant [w/g]. === Subject: Re: The definition of weight > So O.K: The difference between inertial motion of the first law of > physics, and the concept of inertia itself, is too much for most > scientists to swallow: > Here is an easier one: Weight is the force exerted between an object; > body, or mass of matter and the surface of a planet - like Earth or > the moon - that it is bearing upon. > Without something to bear upon: Object's; bodies and mass's of matter > are weightless; all forces require something to bear against! > The magnitude of an object's; body's, or mass's weight-force [w] is > proportional to the deceleration [g] at which the planet's surface is > restraining them from further gravitation: > That is, for any given object; body, or mass of matter, at anytime, > the ratio of its weight to the acceleration at which it will free > fall, is a constant [w/g]. you totally s at formulating ANYTHING, i bet you cant even describe how to push a button in less than 500 sentences. why don't you try some calculating with your brilliant ideas, i'd like to see you determine F without a standard for mass. plus, try to make a point next time allright? === Subject: Re: The definition of weight CUT< > you totally s at formulating ANYTHING, i bet you cant even describe how > to push a button in less than 500 sentences. > why don't you try some calculating with your brilliant ideas, i'd like to > see you determine F without a standard for mass. Using a weight scale, there's always F-uw = f = [w/g]a; where u is the coefficient of friction, and/or coefficient of some other restraining effect. > plus, try to make a point next time allright? Awright(;^) By the way; did you know that a liter of water [a cubic decimeter of it] was first prescribed as a kilogram, so that a standard kilogram could be replicated in laboratories just about anywhere on Earth? Try that with platinum. === Subject: Re: The definition of weight In sci.physics, Donald G. Shead you totally s at formulating ANYTHING, i bet you cant even describe how > to push a button in less than 500 sentences. > why don't you try some calculating with your brilliant ideas, i'd like to > see you determine F without a standard for mass. > Using a weight scale, there's always F-uw = f = [w/g]a; where u is the > coefficient of friction, and/or coefficient of some other restraining > effect. And then there's a counteracting force F'; both resolve into torques in a pan scale, and the g's cancel. Pan scales are such that T = T' when the unit balances; T = Fl, where l is the length between pivot point and pan attachment point. (If the pan arm is nonzero in thickness one gets a diagonal effect, which is usually ignorable.) If l = l', then T = T' when m = m' since F = F'. (Counterexample: placing a magnet under a ferromagnetic pan, but that's not proper pan scale usage. :-) ) > plus, try to make a point next time allright? > Awright(;^) > By the way; did you know that a liter of water [a cubic decimeter of > it] was first prescribed as a kilogram, so that a standard kilogram > could be replicated in laboratories just about anywhere on Earth? Try > that with platinum. Platinum is a liquid when heated hot enough. Did you have a point here? -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: The definition of weight > By the way; did you know that a liter of water [a cubic decimeter of > it] was first prescribed as a kilogram, so that a standard kilogram > could be replicated in laboratories just about anywhere on Earth? Try > that with platinum. Yes, and it was changed for a very good reason. To use water as the mass standard would require measuring and controlling the volume, temperature, and purity (=density) of the water. It was impossible to do this with the level of accuracy required by modern science. In other words, it was changed precisely _because_ the standard kilogram could not be replicated in laboratories. Using a physical object as the standard is obviously not ideal and there is active work to come up with a better defiintion, something which _could_ be replicated without reference to a physical object. But it certainly won't be based on water. --Mark === Subject: Re: The definition of weight ... > By the way; did you know that a liter of water [a cubic decimeter of > it] was first prescribed as a kilogram, so that a standard kilogram > could be replicated in laboratories just about anywhere on Earth? Try > that with platinum. Did you know why it was changed? Apparently not. With the original definition the kilogram could not be replicated precise enough. -- 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: The definition of weight > ... > By the way; did you know that a liter of water [a cubic decimeter of > it] was first prescribed as a kilogram, so that a standard kilogram > could be replicated in laboratories just about anywhere on Earth? Try > that with platinum. > Did you know why it was changed? Apparently not. With the original > definition the kilogram could not be replicated precise enough. AGAIN I say if you think platinum - or _anytjing_ else - in solid or (HOT molten) liquid form can be worked more prcisely than water, then you are nuts; a fruit cake! === Subject: Re: The definition of weight This post is off-topic in sci.math. Please confine your cross-posting to appropriate newsgroups. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: The definition of weight > This post is off-topic in sci.math. Please confine your cross-posting to > appropriate newsgroups. Math is not off topic in physics; nor is physics off topic in math: They go together like a horse and carriage; or what's that other thing? > -- > There are two things you must never attempt to prove: the unprovable -- > and the obvious. > -- > Democracy: The triumph of popularity over principle. === Subject: Re: The definition of weight >This post is off-topic in sci.math. Please confine your cross-posting to >appropriate newsgroups. > Math is not off topic in physics; nor is physics off topic in math: > They go together like a horse and carriage; or what's that other > thing? Parasol and bustle? Since all of your ideas seem to come from the horse and carriage era, these comparisons seem appropriate. Double-A === Subject: Re: The definition of weight > Math is not off topic in physics; nor is physics off topic in math: > They go together like a horse and carriage; or what's that other > thing? Shead and stupidity? === Subject: Re: The definition of weight >This post is off-topic in sci.math. Please confine your cross-posting to >appropriate newsgroups. > Math is not off topic in physics; nor is physics off topic in math: > They go together like a horse and carriage; or what's that other > thing? Your previously posted position was that *nothing* was off-topic in math -- a position which is refuted by a reductio ad absurdum. Namely, that *anything* should be posted to sci.math, which is ridiculous. But even your narrower position above is wrong. If it were correct, why have two newsgroups when one would do? What's the point of crossposting every math discussion to physics and every physics discussion to math -- unless you are, (could it be?) a TROLL! -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: The definition of weight > This post is off-topic in sci.math. Please confine your cross-posting to > appropriate newsgroups. >Math is not off topic in physics; nor is physics off topic in math: >They go together like a horse and carriage; or what's that other >thing? > Your previously posted position was that *nothing* was off-topic in math -- a position which > is refuted by a reductio ad absurdum. Namely, that *anything* should be posted to sci.math, > which is ridiculous. > But even your narrower position above is wrong. If it were correct, why have two newsgroups > when one would do? What's the point of crossposting every math discussion to physics and every > physics discussion to math -- unless you are, (could it be?) a TROLL! That may be, if a troll is seeking to consolidate math and physics: Say! If you know how, how about starting a newsgroup called sci.math. physics. That would save me having to post to the two groups. If you can get it on the net, I'll be an exclusive user of sci.math.physics. > -- === Subject: Re: The definition of weight >This post is off-topic in sci.math. Please confine your cross-posting to >appropriate newsgroups. > Math is not off topic in physics; nor is physics off topic in math: > They go together like a horse and carriage; or what's that other > thing? >Your previously posted position was that *nothing* was off-topic in math -- a position which >is refuted by a reductio ad absurdum. Namely, that *anything* should be posted to sci.math, >which is ridiculous. >But even your narrower position above is wrong. If it were correct, why have two newsgroups >when one would do? What's the point of crossposting every math discussion to physics and every >physics discussion to math -- unless you are, (could it be?) a TROLL! > That may be, if a troll is seeking to consolidate math and physics: > Say! If you know how, how about starting a newsgroup called sci.math. > physics. That would save me having to post to the two groups. If you > can get it on the net, I'll be an exclusive user of > sci.math.physics. It sounds like *you* are the right individual to establish that newsgroup, since it is your position that math posts and physics posts are interchangeable. Please do so, and discontinue your off-topic posts to sci.math. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: The definition of weight Cut< > It sounds like *you* are the right individual to establish that newsgroup, since it is your position > that math posts and physics posts are interchangeable. Please do so, and discontinue your off-topic > posts to sci.math. If I could, I'd Ôave done it already yet(;^! === Subject: Re: The definition of weight Learn about apostrophe's and they're uses's, you coprophageous twat. > So O.K: The difference between inertial motion of the first law of > physics, and the concept of inertia itself, is too much for most > scientists to swallow: > Here is an easier one: Weight is the force exerted between an object; > body, or mass of matter and the surface of a planet - like Earth or > the moon - that it is bearing upon. > Without something to bear upon: Object's; bodies and mass's of matter > are weightless; all forces require something to bear against! > The magnitude of an object's; body's, or mass's weight-force [w] is > proportional to the deceleration [g] at which the planet's surface is > restraining them from further gravitation: > That is, for any given object; body, or mass of matter, at anytime, > the ratio of its weight to the acceleration at which it will free > fall, is a constant [w/g]. === Subject: Re: The definition of weight > So O.K: The difference between inertial motion of the first law of > physics, and the concept of inertia itself, is too much for most > scientists to swallow: Hey Dumb Donny Head, 1) Your ignorance, incompetence, and psychosis are not of interest to the world at large. Quite the contrary. You are not even an interesting laughingstock. 2) http://www.kingstroker.com/ 3) Why are you always psychotically trolling your crap through sci.math, Dumb Donny Head? Assuredly it would be better appreciated in alt.test -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: The definition of weight The definition of weight is simple. Weight is the force that a mass m experiences due to gravity of another mass, w = mg where g is the acceleration due to gravity. Weight http://scienceworld.wolfram.com/physics/Weight.html === Subject: Re: The definition of weight >The definition of weight is simple. Weight is the force that a mass m >experiences due to gravity of another mass, w = mg where g is the >acceleration due to gravity. >Weight > http://scienceworld.wolfram.com/physics/Weight.html Weight is a number taken from a calibrated spring scale. If you use a balance scale, then you are measuring mass (as nearly as it can be done). Mr. Dual Space (If you have something to say, write an equation. If you have nothing to say, write an essay). === Subject: Re: The definition of weight > The definition of weight is simple. Weight is the force that a mass m > experiences due to gravity of another mass, w = mg where g is the > acceleration due to gravity. > Weight > http://scienceworld.wolfram.com/physics/Weight.html That's dogma Sammy: a point of view put forth as authoritative without adequate justifiable grounds. There's no logic. My definition is more logical; because it's true: Weight [w] does not, and cannot be w = mg, because m = w/g :: Therefore w = [w/g]g, and/or w = [f/a]g... === Subject: Re: The definition of weight > Weight [w] does not, and cannot be > w = mg, because m = w/g :: Therefore > w = [w/g]g, and/or w = [f/a]g... I'm trying to understand what you can possibly mean by this. By simple algebra, w=mg is equivalent to m=w/g. They mean exactly the same thing. How can you say that one is true and the other is not? --Mark === Subject: Re: The definition of weight >Weight [w] does not, and cannot be >w = mg, because m = w/g :: Therefore >w = [w/g]g, and/or w = [f/a]g... > I'm trying to understand what you can possibly mean by this. By simple > algebra, w=mg is equivalent to m=w/g. They mean exactly the same thing. > How can you say that one is true and the other is not? Say you start with m=w/g. Now you can multiply both sides by g and on left side you'll git mg. But on the right side the g's will cancel and you'll git w/. In other words, you got mg = w/ which ain't the same as mg=w Now there ain't no way you can git rid of that / thing. You can draw it real thin so that nobody notices, but you'll only be fooling yerself. If you divide both sides by / you'll only git mg// = w which is even worse. So forget about it. Don is much more careful about his math than the rest of us, and that can only come from years of building bridges. Yep. ---DPM And don't git me started on how you can't push the g through that solid bracket in [w/g]g > --Mark === Subject: Re: The definition of weight >Weight [w] does not, and cannot be >w = mg, because m = w/g :: Therefore >w = [w/g]g, and/or w = [f/a]g... > I'm trying to understand what you can possibly mean by this. By simple > algebra, w=mg is equivalent to m=w/g. They mean exactly the same thing. > How can you say that one is true and the other is not? > --Mark What are the _terms_ in mass? Mass [m] is the ratio of force divided by the acceleration that it causes; which is equal to the ratio w/g; where either one of these ratios - f/a or w/g - is a measure of the mass, and/or the quantity of matter in the mass: Mass is not an algebraic term in itself: The terms of mass [m] are f/a and w/g; which can only be moved to the other side of an equation by using parentheses: As [f/a], or as [w/g]; so that m = [f/a] = [w/g]. So that for m = [w/g], in order for w to be moved by itself; we first have to do the math within the parentheses; in order to get rid of them: That's one of the first rules of doing algebra: So if w = 32#, and g = 32''/sec^2; then the mass is m = [32#/(32'/sec^2)] = ONE pound/(ONE foot/sec^2); or ONE pound second^2/foot; which is ONE slug! Or we have the option: w = [w/g]g = [m]g Parentheses are useful _necessities_ in algeba: They define and delimit algebraic terms! You've got to know when, and where to use them! === Subject: Re: The definition of weight >Weight [w] does not, and cannot be >w = mg, because m = w/g :: Therefore >w = [w/g]g, and/or w = [f/a]g... > I'm trying to understand what you can possibly mean by this. By simple > algebra, w=mg is equivalent to m=w/g. They mean exactly the same thing. > How can you say that one is true and the other is not? > --Mark What are the _terms_ in mass? Mass [m] is the ratio of force divided by the acceleration that it causes; which is equal to the ratio w/g; where either one of these ratios - f/a or w/g - is a measure of the mass, and/or the quantity of matter in the mass: Mass is not an algebraic term in itself: The terms of mass [m] are f/a and w/g; which can only be moved to the other side of an equation by using parentheses: As [f/a], or as [w/g]; so that m = [f/a] = [w/g]. So that for m = [w/g], in order for w to be moved by itself; we first have to do the math within the parentheses; in order to get rid of them: That's one of the first rules of doing algebra: So if w = 32#, and g = 32''/sec^2; then the mass is m = [32#/(32'/sec^2)] = ONE pound/(ONE foot/sec^2); or ONE pound second^2/foot; which is ONE slug! Or we have the option: w = [w/g]g = [m]g Parentheses are useful _necessities_ in algeba: They define and delimit algebraic terms! You've got to know when, and where to use them! === Subject: Re: The definition of weight >Weight [w] does not, and cannot be >w = mg, because m = w/g :: Therefore >w = [w/g]g, and/or w = [f/a]g... > I'm trying to understand what you can possibly mean by this. By simple > algebra, w=mg is equivalent to m=w/g. They mean exactly the same thing. > How can you say that one is true and the other is not? > --Mark What are the _terms_ in mass? Mass [m] is the ratio of force divided by the acceleration that it causes; which is equal to the ratio w/g; where either one of these ratios - f/a or w/g - is a measure of the mass, and/or the quantity of matter in the mass: Mass is not an algebraic term in itself: The terms of mass [m] are f/a and w/g; which can only be moved to the other side of an equation by using parentheses: As [f/a], or as [w/g]; so that m = [f/a] = [w/g]. So that for m = [w/g], in order for w to be moved by itself; we first have to do the math within the parentheses; in order to get rid of them: That's one of the first rules of doing algebra: So if w = 32#, and g = 32''/sec^2; then the mass is m = [32#/(32'/sec^2)] = ONE pound/(ONE foot/sec^2); or ONE pound second^2/foot; which is ONE slug! Or we have the option: w = [w/g]g = [m]g Parentheses are useful _necessities_ in algeba: They define and delimit algebraic terms! You've got to know when, and where to use them! === Subject: Re: The definition of weight >Weight [w] does not, and cannot be >w = mg, because m = w/g :: Therefore >w = [w/g]g, and/or w = [f/a]g... > I'm trying to understand what you can possibly mean by this. By simple > algebra, w=mg is equivalent to m=w/g. They mean exactly the same thing. > How can you say that one is true and the other is not? > --Mark What are the _terms_ in mass? Mass [m] is the ratio of force divided by the acceleration that it causes; which is equal to the ratio w/g; where either one of these ratios - f/a or w/g - is a measure of the mass, and/or the quantity of matter in the mass: Mass is not an algebraic term in itself: The terms of mass [m] are f/a and w/g; which can only be moved to the other side of an equation by using parentheses: As [f/a], or as [w/g]; so that m = [f/a] = [w/g]. So that for m = [w/g], in order for w to be moved by itself; we first have to do the math within the parentheses; in order to get rid of them: That's one of the first rules of doing algebra: So if w = 32#, and g = 32''/sec^2; then the mass is m = [32#/(32'/sec^2)] = ONE pound/(ONE foot/sec^2); or ONE pound second^2/foot; which is ONE slug! Or we have the option: w = [w/g]g = [m]g Parentheses are useful _necessities_ in algeba: They define and delimit algebraic terms! You've got to know when, and where to use them! === Subject: Re: The definition of weight > [...] By simple >algebra, w=mg is equivalent to m=w/g. They mean exactly the same thing. >How can you say that one is true and the other is not? --Mark > [...] As [f/a], or as [w/g]; so that m = [f/a] = [w/g]. > [...] > Or we have the option: w = [w/g]g = [m]g Right. And the point is, Mark, you can't get them damn brackets off the m, cuz they's st on there GOOD. Try takin a screwdriver and pryin Ôem off an you just gits w = [ m]g Trust me, I tried and just ended up ruinin' the screwdriver. All yer fancy algebra ain't gonna change that. Because if you substitute m = [w/g] = [[m]] into w=[m]g, you just gits w = [[[m]]]g and yer even worse off than before. ---DPM === Subject: Re: The definition of weight > Parentheses are useful _necessities_ in algeba: They define and > delimit algebraic terms! You've got to know when, and where to use > them! Yes. You don't. === Subject: Re: The definition of weight >Weight [w] does not, and cannot be >w = mg, because m = w/g :: Therefore >w = [w/g]g, and/or w = [f/a]g... > I'm trying to understand what you can possibly mean by this. By simple > algebra, w=mg is equivalent to m=w/g. They mean exactly the same thing. > How can you say that one is true and the other is not? > --Mark What are the _terms_ in mass? Mass [m] is the ratio of force divided by the acceleration that it causes; which is equal to the ratio w/g; where either one of these ratios - f/a or w/g - is a measure of the mass, and/or the quantity of matter in the mass: Mass is not an algebraic term in itself: The terms of mass [m] are f/a and w/g; which can only be moved to the other side of an equation by using parentheses: As [f/a], or as [w/g]; so that m = [f/a] = [w/g]. So that for m = [w/g], in order for w to be moved by itself; we first have to do the math within the parentheses; in order to get rid of them: That's one of the first rules of doing algebra: So if w = 32#, and g = 32''/sec^2; then the mass is m = [32#/(32'/sec^2)] = ONE pound/(ONE foot/sec^2); or ONE pound second^2/foot; which is ONE slug! Or we have the option: w = [w/g]g = [m]g Parentheses are useful _necessities_ in algeba: They define and delimit algebraic terms! You've got to know when, and where to use them! === Subject: Re: Pure math cannot win here > |[...] > | All that is necessary for evil to succeed > | is that good men do nothing. > | Edmund Burke (1729 - 1797) > This is probably a bogus quotation. See > or > . Boy, I sure hate it when I make a stab at seeming erudite, and turn out to be just plain mistaken. I should have known when all those Google hits had slight variations on the wording, and none had a specific source. Dale. === Subject: JSH: Attacking the conclusion ANYONE can attack any math paper by just going to the conclusion and claiming it's wrong. It's a cheat. It's a dodge. It's specious. Claims of counterexamples against my work have by me repeatedly been shown to be false. In every case posters rely on a circular argument which basically relies on the ring of algebraic integers not having the problem I've proven it has. Here's basically how it goes: There are numbers that should properly be considered factors of 1. However, in the ring of algebraic integers, these numbers are NOT factors of 1. Note: I did not say that they should be considered factors of 1 in the ring of algebraic integers. That's the problem. That is, the problem is that these numbers provably should be considered to be factors of 1, and finding a ring where they are is not even hard. They are factors of 1 in a ring made up of numbers such that only 1 and -1 are integers units, which includes algebraic integers, and other numbers besides them. That's it. You have a ring where only 1 and -1 are integer units, where that's the principal defining characteristic and you can show that some irrational units in that ring, are not algebraic integers. Simple. But, when attacking my work, posters like Hall, Magidin, and Decker assert that because these numbers are not units in the ring of algebraic integers i.e. factors of 1 in that ring, then I must be wrong. But I've proven that I'm right using rather basic algebra. That algebra shows that there is a ring of numbers where -1 and 1 are the only integer units that is indeed LARGER than the ring of algebraic integers. That's it. That's the amazing conclusion that mathematicians have been running from for so long. It's actually kind of a neat fact. James Harris === Subject: Re: JSH: Attacking the conclusion > ANYONE can attack any math paper by just going to the conclusion and > claiming it's wrong. There is a difference between claiming a conclusion is wrong and just plain proving a conclusion is wrong. > Claims of counterexamples against my work have by me repeatedly been > shown to be false. Eh? > In every case posters rely on a circular argument which basically > relies on the ring of algebraic integers not having the problem I've > proven it has. What is the problem in the ring of algebraic integers? You say that factors of numbers can not be split along other numbers. It has been shown that that can be done. > There are numbers that should properly be considered factors of 1. What do you mean with should properly be considered? > Note: I did not say that they should be considered factors of 1 in > the ring of algebraic integers. > That's the problem. That is, the problem is that these numbers > provably should be considered to be factors of 1, and finding a ring > where they are is not even hard. But it is hard. > They are factors of 1 in a ring made up of numbers such that only 1 > and -1 are integers units, which includes algebraic integers, and > other numbers besides them. This is not enough of a definition of the ring to be even workable. What do you mean with integer units? > That's it. You have a ring where only 1 and -1 are integer units, > where that's the principal defining characteristic and you can show > that some irrational units in that ring, are not algebraic integers. Well, first give a definition of integer units and after that we can continue. > That algebra shows that there is a ring of numbers where -1 and 1 are > the only integer units that is indeed LARGER than the ring of > algebraic integers. Again, what do you mean with integer units? -- 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: Attacking the conclusion > ANYONE can attack any math paper by just going to the conclusion and > claiming it's wrong. > There is a difference between claiming a conclusion is wrong and just > plain proving a conclusion is wrong. > Claims of counterexamples against my work have by me repeatedly been > shown to be false. > Eh? > In every case posters rely on a circular argument which basically > relies on the ring of algebraic integers not having the problem I've > proven it has. > What is the problem in the ring of algebraic integers? You say that > factors of numbers can not be split along other numbers. It has been > shown that that can be done. > There are numbers that should properly be considered factors of 1. > What do you mean with should properly be considered? > Note: I did not say that they should be considered factors of 1 in > the ring of algebraic integers. That's the problem. That is, the problem is that these numbers > provably should be considered to be factors of 1, and finding a ring > where they are is not even hard. > But it is hard. Actually, in any field containing all of the numbers in question they will all but zero be factors of 1, but then every non-zero numbers is a factor of every other, and there are no non-units at all other than zero, and such fields are not at all what James' alleges that his mythical ring is like. > They are factors of 1 in a ring made up of numbers such that only 1 > and -1 are integers units, which includes algebraic integers, and > other numbers besides them. > This is not enough of a definition of the ring to be even workable. > What do you mean with integer units? > That's it. You have a ring where only 1 and -1 are integer units, > where that's the principal defining characteristic and you can show > that some irrational units in that ring, are not algebraic integers. > Well, first give a definition of integer units and after that we > can continue. > That algebra shows that there is a ring of numbers where -1 and 1 are > the only integer units that is indeed LARGER than the ring of > algebraic integers. > Again, what do you mean with integer units? === Subject: Re: JSH: Attacking the conclusion : ANYONE can attack any math paper by just going to the conclusion and : claiming it's wrong. : It's a cheat. It's a dodge. It's specious. : Claims of counterexamples against my work have by me repeatedly been : shown to be false. : In every case posters rely on a circular argument which basically : relies on the ring of algebraic integers not having the problem I've : proven it has. : Here's basically how it goes: : There are numbers that should properly be considered factors of 1. As far as I know, and I readily admit that I have not followed even a fraction of the discussion, you have never made clear what you mean by should properly be. === Subject: Re: JSH: Attacking the conclusion > : ANYONE can attack any math paper by just going to the conclusion and > : claiming it's wrong. > : It's a cheat. It's a dodge. It's specious. > : Claims of counterexamples against my work have by me repeatedly been > : shown to be false. > : In every case posters rely on a circular argument which basically > : relies on the ring of algebraic integers not having the problem I've > : proven it has. > : Here's basically how it goes: > : There are numbers that should properly be considered factors of 1. > As far as I know, and I readily admit that I have not followed > even a fraction of the discussion, you have never made clear what > you mean by should properly be. I repeatedly explain but there's always someone to keep coming at me claiming they can't understand, but here goes again. I'm going to explain yet again to the sci.math newsgroup, and let's see if you people don't try to come back later and claim I haven't. You can define a ring by a membership requirement that no number in that ring other than -1 or 1 is a unit integer. Some seem dedicated to simply claiming that ring is coincident with the ring of algebraic integers, which I've proven to be false. In that larger ring there are units that are NOT algebraic integers. I've proven that the ring is not coincident with the ring of algebraic integers with a rather basic argument which can be seen in my paper Advanced Polynomial Factorization. Rather than face the argument, certain posters like W. Dale Hall, Rick Decker, and Arturo Magidin have speciously attacked the conclusion basically relying on the fact that there are certain numbers provably not in the ring of algebraic integers that are units in the more inclusive ring. They simply, and repeatedly, prove that certain numbers are not units IN THE RING OF ALGEBRAIC INTEGERS and then claim that proves I'm wrong. That they are deliberately trying to attack algebra itself can be seen by the fact that they stay away from the arguments in the paper, choosing instead to attack the conclusion or the messenger, whether it's me or the Southwest Journal of Pure and Applied Mathematics, as seen by their email campaign against my paper. Notice they did NOT PAUSE despite the paper having passed peer review. Notice how clearly the mathematical establishment, rules, procedures and logic mean little to them as they're dedicated in their mission. They will not stop as you can see from what they're doing now. Their position requires that mathematics itself be inconsistent, while my position only requires that you accept algebra. They are anti-mathematicians, evil incarnate, dedicated to undermining intellectual development in this area. If you never thought such people could actually exist, outside of myths or legends, welcome to the real world. James Harris === Subject: Re: JSH: Attacking the conclusion :> :> : There are numbers that should properly be considered factors of 1. :> As far as I know, and I readily admit that I have not followed :> even a fraction of the discussion, you have never made clear what :> you mean by should properly be. : You can define a ring by a membership requirement that no number in : that ring other than -1 or 1 is a unit integer. I don't understand this (what's a unit integer?), but let's suppose for the sake of argument that you have clearly defined some ring R of which the algebraic integers are a proper subset. : In that larger ring there are units that are NOT algebraic integers. Ok. That is also true if the larger ring is all complex numbers. What's the point? === Subject: Re: JSH: Attacking the conclusion >I repeatedly explain but there's always someone to keep coming at me >claiming they can't understand, but here goes again. I'm going to >explain yet again to the sci.math newsgroup, and let's see if you >people don't try to come back later and claim I haven't. You can define a ring by a membership requirement that no number in >that ring other than -1 or 1 is a unit integer. > No, you can't. For one thing it does not make sense to talk about > unit integer in the definition of a ring, since either we are > talking about integers, in which case this already defines the ring, > or not, in which case talking about unit integers does not make > sense. So you probably mean unit instead of unit integer. I'll refute your statement quickly: Consider 3, it is an integer, correct? Now consider 3 in any ring. It is STILL an integer, correct? If it is a unit in that ring, then it is an integer unit, correct? Knowing that 3 is an integer, and that 3 is in a ring, does NOT mean that ring is the ring of integers. Why you would assert otherwise is a mystery unless you're just trying like so many of you do to hide the truth. > Anyway, with that definition, Z4, the numbers modulo 4, meet the > requirement, as just 1 and -1 are units. The integers themselves meet > the requirement, too. Z4 and Z are different rings, so you can't > define both that way. Actually, Z2, Z3, Z6 all fit that > requirement as well. That's a finite sized ring. Besides, it doesn't change my point. You're just throwing up distractions. >Some seem dedicated to simply claiming that ring is coincident with >the ring of algebraic integers, which I've proven to be false. > You can't talk about that ring before it is properly defined, and > your definition above holds for a whole lot of bunch. Now your tactics are clear. You try to question how an integer can still be an integer *and* a unit, and now claim that there's a definition problem. However, consider the requirement for an infinite sized ring that -1 and 1 are the only integers in that ring that are units in that ring. What's wrong with that requirement? >In that larger ring there are units that are NOT algebraic integers. > You have not defined a larger ring yet. For people who don't understand how sci.math posters operate, they will play silly games with me, like David Kastrup here, where they make statements that are almost childishly obtuse. Then they act stubborn. LOTS of them will post in concert, making strange and irrational statements, with bold claims, and they will do this building up threads to hundreds of posts. If I ignore the irrational posters, they just start posting that they refuted my claims and that *I* am the irrational one. Since there are a lot of them, they can just hide the actual discussions by sheer volume, and keep up the charade. >I've proven that the ring is not coincident with the ring of >algebraic integers with a rather basic argument which can be seen in >my paper Advanced Polynomial Factorization. > You can't prove anything of that complexity about that ring before > giving a definition that unambiguously defines it. Certainly no such > thing results from merely your definition property. One of their favorite tactics, since I've talked about using the requirement that -1 and 1 are the only integers that are units, is to just claim that I haven't defined *anything* at all, or claim the requirement is ambiguous. These tactics have been used by quite a few people, from people like Rick Decker , Arturo Magidin, and Dik Winter to posters like David Kastrup, C. Bond, and Nora Baron. It's actually kind of fun in a way. It's like they keep saying all this crazy stuff and getting away with it, which lets me know just how unimportant to most of you rational discussion actually is. Posting on sci.math is NEVER about anything of import for most of you. To most of you, it's just a social venue for what is basically unimportant chatter. James Harris === Subject: Re: JSH: Attacking the conclusion > You can define a ring by a membership requirement that no number in > that ring other than -1 or 1 is a unit integer. This doesn't seem to me to work. The way that one might get a ring with some special property is this: 1. Specify which numbers belong to the structure. Examples: 1A: All numbers of the form a+bi where a and b are rational integers (and i is the square root of -1). 1B. All numbers z such that z is a root of a polynomial whose leading coefficient is 1. 2. You may have to prove that the set so selected actually is a ring. Basically this means to show that the sum and product of two elements of the set also belng to the set. If this seems obvious, you can skip it (but be prepared to fill in the details). 3. Now, you may prove that the special property you want is satisfied. The point is that specification of what numbers are in the ring is prior to proving properties of the ring itself. -- Chris Henrich === Subject: Re: JSH: Attacking the conclusion Originator: richard@cogsci.ed.ac.uk (Richard Tobin) > You can define a ring by a membership requirement that no number in > that ring other than -1 or 1 is a unit integer. >This doesn't seem to me to work. It doesn't, but there's nothing wrong with that sort of definition in general. For example, the ring with two elements is a perfectly good definition of a ring. To show that it's a good definition, you have to prove (a) that there is such a ring and (b) that there is only one. You would also have to say what you mean by unit integer. -- Richard === Subject: Re: JSH: Attacking the conclusion > You can define a ring by a membership requirement that no number > in that ring other than -1 or 1 is a unit integer. Some seem > dedicated to simply claiming that ring is coincident with > the ring of algebraic integers, which I've proven to be false. That's by no means a precise definition. It seems you are interested in considering a ring J of algebraic numbers such that J / Q = Z. rings of this sort. As I explain there, if you drop the requirement that the ring be closed under conjugation then J may indeed contain numbers that aren't algebraic integers while still preserving the condition that J / Q = Z. However, as I mention there, such rings will be of little value for your intended purposes. > I've proven that the ring is not coincident with the ring of algebraic > integers with a rather basic argument which can be seen in my paper > Advanced Polynomial Factorization. In said paper [2] the word ring occurs only 3 times, and each time it occurs as part of the larger phrase ring of algebraic integers. Since it mentions no other rings, surely it contains no mention of your elusive object ring J, so your above claim is clearly false. -Bill Dubuque === Subject: Re: JSH: Attacking the conclusion !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+(Ô>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > I repeatedly explain but there's always someone to keep coming at me > claiming they can't understand, but here goes again. I'm going to > explain yet again to the sci.math newsgroup, and let's see if you > people don't try to come back later and claim I haven't. > You can define a ring by a membership requirement that no number in > that ring other than -1 or 1 is a unit integer. No, you can't. For one thing it does not make sense to talk about unit integer in the definition of a ring, since either we are talking about integers, in which case this already defines the ring, or not, in which case talking about unit integers does not make sense. So you probably mean unit instead of unit integer. Anyway, with that definition, Z4, the numbers modulo 4, meet the requirement, as just 1 and -1 are units. The integers themselves meet the requirement, too. Z4 and Z are different rings, so you can't define both that way. > Some seem dedicated to simply claiming that ring is coincident with > the ring of algebraic integers, which I've proven to be false. You can't talk about that ring before it is properly defined, and your definition above holds for a whole lot of bunch. > In that larger ring there are units that are NOT algebraic integers. You have not defined a larger ring yet. > I've proven that the ring is not coincident with the ring of > algebraic integers with a rather basic argument which can be seen in > my paper Advanced Polynomial Factorization. You can't prove anything of that complexity about that ring before giving a definition that unambiguously defines it. Certainly no such thing results from merely your definition property. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Attacking the conclusion >: ANYONE can attack any math paper by just going to the conclusion and >: claiming it's wrong. >: It's a cheat. It's a dodge. It's specious. >: Claims of counterexamples against my work have by me repeatedly been >: shown to be false. >: In every case posters rely on a circular argument which basically >: relies on the ring of algebraic integers not having the problem I've >: proven it has. >: Here's basically how it goes: >: There are numbers that should properly be considered factors of 1. >As far as I know, and I readily admit that I have not followed >even a fraction of the discussion, you have never made clear what >you mean by should properly be. > I repeatedly explain but there's always someone to keep coming at me > claiming they can't understand, but here goes again. I'm going to > explain yet again to the sci.math newsgroup, and let's see if you > people don't try to come back later and claim I haven't. > You can define a ring by a membership requirement that no number in > that ring other than -1 or 1 is a unit integer. > Some seem dedicated to simply claiming that ring is coincident with > the ring of algebraic integers, which I've proven to be false. Hold it! Who claimed this? The ring of integers satisfies your requirement. It isn't necessary to take the larger ring of algebraic integers to accomodate your membership requirement. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Attacking the conclusion this reminds me of listening to the hateful Limbaugh. He'll read something & use its conclusions as a generic reference for all Liberals, which are the same as Democrats, which are now as bad as Commies used to be, but it's still sophistry -- that he calls, my principles. > ANYONE can attack any math paper by just going to the conclusion and > claiming it's wrong. --Give Earth a Trickier Dick Cheeny -- out of office, after gigayears! http://tarpley.net/bush12.htm http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/HowTheNation.PDF http://www.rand.org/publications/randreview/issues/rr.12.00/ http://www.rwgrayprojects.com/synergetics/plates/figs/plate02. html === Subject: Re: JSH: Attacking the conclusion > ANYONE can attack any math paper by just going to the conclusion and > claiming it's wrong. > It's a cheat. It's a dodge. It's specious. Claim: any number of the form 2n+1 where n is a positive integer is prime. Proof: [a lovely proof by induction that I won't bore anyone with here] Counterexample: n=4 yields 2n+1 = 9, which is a composite. The proof is irrelevant. The claim is wrong, so the proof is wrong. > Claims of counterexamples against my work have by me repeatedly been > shown to be false. You mean when the roots were computed, and the non-unit factors they had in common with 5 were computed, there were arithmatic mistakes? Why didn't you compute the *real* roots and common factors then? > Here's basically how it goes: > There are numbers that should properly be considered factors of 1. Where? As it stands, this statement is meaningless. Perhaps a better statement would be: There are numbers that you want to be considered factors of 1 in a particular ring that you have not yet constructed. If you can clearly define which factors you want to include, then you can begin constructing the ring and see whether it is a ring that is already known, such as the algebraic numbers. > However, in the ring of algebraic integers, these numbers are NOT > factors of 1. Ok. > Note: I did not say that they should be considered factors of 1 in > the ring of algebraic integers. Ok. > That's the problem. That is, the problem is that these numbers > provably should be considered to be factors of 1, and finding a ring > where they are is not even hard. > They are factors of 1 in a ring made up of numbers such that only 1 > and -1 are integers units, which includes algebraic integers, and > other numbers besides them. This is not a proper definition. Is pi in the ring? Is it a unit in the ring? You definition does not give sufficient information to answer these questions. How does this definition show that the numbers you want to be units are? Which numbers are those? > That's it. You have a ring where only 1 and -1 are integer units, > where that's the principal defining characteristic and you can show > that some irrational units in that ring, are not algebraic integers. > Simple. Is pi in the ring? > But, when attacking my work, posters like Hall, Magidin, and Decker > assert that because these numbers are not units in the ring of > algebraic integers i.e. factors of 1 in that ring, then I must be > wrong. You have completely misunderstood their comments. They have said that you have not defined your ring, and they have said your claims about the roots of a particular polynomial are incorrect. > But I've proven that I'm right using rather basic algebra. > That algebra shows that there is a ring of numbers where -1 and 1 are > the only integer units that is indeed LARGER than the ring of > algebraic integers. > That's it. That's the amazing conclusion that mathematicians have > been running from for so long. > It's actually kind of a neat fact. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: Attacking the conclusion > ANYONE can attack any math paper by just going to the conclusion and > claiming it's wrong. Anyone can claim anything. This is not a valid argument. If you mean that a conclusion of any math paper can be proven to be false, then you have made a specific claim that you cannot support. Your best response is simple: select a paper in mathematics that people generally agree is valid, find any conclusion, and prove that the conclusion is false. Make a claim, support that claim. It's pretty simple, isn't it? > It's a cheat. It's a dodge. It's specious. > Claims of counterexamples against my work have by me repeatedly been > shown to be false. Really? You have shown that the polynomial products I've produced are in error? You have shown that the fact that the products are what I claim they are does not support my claim that the a's and 5 are never coprime? > In every case posters rely on a circular argument which basically > relies on the ring of algebraic integers not having the problem I've > proven it has. You have proven no such thing. Let's set some of the details down for the sake of definiteness: The claim I am refuting is: In the factorization 65 x^3 - 12 x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) one of the coefficients a_* is coprime to 5. I show that each a_* has a non-unit factor in common with 5 in the ring of algebraic integers. I conclude that none of the a_* can be coprime to 5. My argument refuting this follows (I have omitted details of the polynomials q,r,s. They will be provided upon request): There are polynomials q(x),r(x),s(x) with integer coefficients that satisfy the following conditions: q(x)r(x) = P(x)A(x) + 5 r(x)s(x) = P(x)B(x) + x where P(x) is the minimal polynomial of -a, namely P(x) = x^3 - 12 x^2 + 65 and where A(x) and B(x) are polynomials with integer coefficients. Thus, for -a = any root of P(x), we have the following factorizations: 5 = q(-a)r(-a) a = r(-a)s(-a) These factorizations hold in *any* ring containing both the ring of integers and the number a. I have further shown that the number r(-a) has as the minimal polynomial MP_r = x^3 - 969 x^2 + 315 x + 5 and so (noting that this polynomial is irreducible over Q) is an algebraic integer which is not a unit in the ring of algebraic integers. Thus, in the ring of algebraic integers, the numbers a and 5 have the common, non-unit factor r(-a). None of the above algebra relies on anything other than simple manipulation of polynomials (i.e., multiplication and addition), and the elementary result that a unit of the ring of algebraic integers must have a minimal polynomial with both the leading term and constant term equal to (rational) integers. If you have shown that any of those assumptions fails, then you need to make that claim explicitly. I claim you haven't found any such result. You have claimed that the argument is circular because the condition that u and v be coprime in the ring of algebraic integers entails that any common factor of u and v, in that ring, be a unit (in the ring of algebraic integers), while I have shown that r(-a) cannot be a unit in the ring of algebraic integers. There is no reason behind your assertion that this is a circular argument. Simply saying it assumes that the algebraic integers do not have the problem [you have] proven it has does nothing to show circularity. > Here's basically how it goes: > There are numbers that should properly be considered factors of 1. This is a non-mathematical statement. Make it mathematical, and you will have something to say. That something may be correct or incorrect, and I would imagine the latter, but so far you've just made an esthetic or a moral assertion, not a mathematical one. > However, in the ring of algebraic integers, these numbers are NOT > factors of 1. > Note: I did not say that they should be considered factors of 1 in > the ring of algebraic integers. However, your claim is that the numbers 5 and a are coprime *in the ring of algebraic integers*. I have found a non-unit common factor *in that ring*, and for the numbers to be coprime *in that ring* any such factor would necessarily be a unit *in that ring*. This is a contradiction: no number can be both a unit and a non-unit of the same ring. Even you must admit that. My argument holds in the ring of algebraic integers, and my result holds in that ring. > That's the problem. That is, the problem is that these numbers > provably should be considered to be factors of 1, and finding a ring > where they are is not even hard. No. As long as you are sticking with the esthetic should be considered there is no provably to be had. It's not so hard to climb out of that hole: just give a mathematical definition of the phrase should be considered to be a factor of 1, and you'll have something (correct or not) to say. Until then, you're not making sense. > They are factors of 1 in a ring made up of numbers such that only 1 > and -1 are integers units, which includes algebraic integers, and > other numbers besides them. Didn't someone show that your desired ring had to have all integers invertible (and thus to lead to the full field of algebraic numbers)? > That's it. You have a ring where only 1 and -1 are integer units, > where that's the principal defining characteristic and you can show > that some irrational units in that ring, are not algebraic integers. But you haven't demonstrated what that ring is. No one will argue that there aren't plenty of rings between the ring of algebraic integers and the ring of algebraic numbers. They are incredibly easy to find: Take A, the ring of algebraic numbers, and {a1,a2,...,aN} a set of N non-integral algebraic numbers. Then the ring A[a1,a2,...,aN] = the set of all values P(a1,a2,...,aN) where P(x1, x2,..., xN) is a polynomial in N variables with coefficients in A. What you haven't shown is that your notion (a) exists, and (b) has the other properties you think it has. You can postulate all you like, but until you prove some existence lemma, and demonstrate the characteristics of this alleged ring, no one will pay it any mind. > Simple. > But, when attacking my work, posters like Hall, Magidin, and Decker > assert that because these numbers are not units in the ring of > algebraic integers i.e. factors of 1 in that ring, then I must be > wrong. You yourself have admitted that the numbers r(-a) are not units in the ring of algebraic integers, haven't you? I have thus found a non-unit common factor of a and 5, within that ring, right? It then follows, does it not, that a and 5 cannot be coprime in the ring of algebraic integers? You have called the following argument circular. Please show precisely where the circularity lies, keeping in mind that it does not specify the algebraic integers, yet applies to that ring: Let a,b,m,n,p be elements of R, a commutative ring with identity, and suppose a and b have a common factor within R, as follows: a = mn b = mp Suppose also that a and b are coprime in R. I will show that m must be a unit of R. By definition, a and b coprime in R means that every ideal of R containing both a and b is R itself; by noting that the sum of the principal ideals Ra + Rb is such an ideal (i.e., it contains both a and b), we have R = Ra + Rb Specializing to the identity element 1 of R, we find that there must be elements t,u of R with: 1 = ta + ub Substituting: 1 = tmn + ump = m(tn + up) And we find that the common factor m is a unit. Thus, if a,b are coprime elements of R, then any common factor is a unit of R. Equivalently (via the contrapositive): if any two elements of R have a common factor that is not a unit then they cannot be coprime. > But I've proven that I'm right using rather basic algebra. You have tricked yourself into believing you've proven something, by virtue of having imprecise definitions and overly vague arguments. If you were correct, then your result would have been correct. Your result is ßawed, and that in itself indicates your method is ßawed. > That algebra shows that there is a ring of numbers where -1 and 1 are > the only integer units that is indeed LARGER than the ring of > algebraic integers. You say there is a LARGER ring. There are many LARGER rings. The existence of a LARGER ring is absolutely irrelevant for the validity of a proof within the smaller ring of algebraic integers. Yet you're saying that a and 5 are coprime in this alleged LARGER ring? I don't address any LARGER ring. My result holds in the ring of algebraic integers; yours doesn't. Simply saying that the ring of algebraic integers has some problem, therefore no proof in that context is valid, is no argument at all, unless you can prove that the set of algebraic integers fails to be a commutative ring or fails to be integrally closed (neither of which is likely, given the number of times the relevant proofs have been revisited). > That's it. That's the amazing conclusion that mathematicians have > been running from for so long. > It's actually kind of a neat fact. > James Harris Whatever. Dale. === Subject: Re: JSH: Attacking the conclusion ... a buncha stuff... requiring at least the following two editorial corrections: > None of the above algebra relies on anything other than simple > manipulation of polynomials (i.e., multiplication and addition), and > the elementary result that a unit of the ring of algebraic integers > must have a minimal polynomial with both the leading term and constant (1) coefficient---^^^^^ > term equal to *units, in the ring of* (rational) integers. (2) Insert this ^^^^^^^^^^^^^^^^^^^^^ Sorry for the poor proofreading. Dale. ... the rest deleted ... === Subject: Re: JSH: Attacking the conclusion > ANYONE can attack any math paper by just going to the conclusion and > claiming it's wrong. > It's a cheat. It's a dodge. It's specious. Which is why it hasn't happened. Your conclusion contradicts your own premises. It is false. > Claims of counterexamples against my work have by me repeatedly been > shown to be false. Nope. Not ever. You simply reassert your original false position. Your work has been repeatedly shown to be riddled with errors and ambiguities. Worse, you are afßicted with terminal ignorance and have demonstrated (repeatedly) that you must be continuously beaten with the 2x4 of TRUTH before enough sense seeps in for you to identify or acknowledge your errors. Then you defend your corrected arguments as vigorously as you did the previous erroneous ones. > In every case posters rely on a circular argument which basically > relies on the ring of algebraic integers not having the problem I've > proven it has. You have proven no such thing. Your argument is ßawed. > Here's basically how it goes: > There are numbers that should properly be considered factors of 1. Give an example and specify the ring. When you say, There are number that should be properly be considered factors of 1, do you mean that 1 is divisible by these numbers in some ring? What ring? Suppose we assume the ring of rational numbers. Proceed. > However, in the ring of algebraic integers, these numbers are NOT > factors of 1. So what? > Note: I did not say that they should be considered factors of 1 in > the ring of algebraic integers. Well, where then? In the ring of rational numbers? > That's the problem. That is, the problem is that these numbers > provably should be considered to be factors of 1, and finding a ring > where they are is not even hard. What a laugh! All this means is that there are rings which do not support every possible divisor of 1, i.e. reciprocals. Everyone knows that. Divisibility is not an intrinsic requirement in the ring of algebraic integers. [snip usual nonsensical assertions about divisors, factors, rings, etc. leading to a false conclusion] > James Often in error, but never in doubt! Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Attacking the conclusion > ANYONE can attack any math paper by just going to the conclusion and > claiming it's wrong. > It's a cheat. It's a dodge. It's specious. But going to the end of a paper and PROVING the conclusions are wrong is an entirely different matter. Which is what actually happened. === Subject: Re: unit of intelligence by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5HGYFr01200; I like the word Ô.9fbermensa'. Wish I had an IQ of 160. :-) Though I have only about IQ 125 sd 15, I understand too well that it's not always easy to talk about some themes with people.- Sometimes if I come up with some themes that fascinate the hell out of me, than even some of my best friends don't wanna hear it, because they don't understand. Sometimes I think being above the average in IQ is a pain in the ass. You see problems & solutions others are only barely aware about. It's so frustrating. I can feel your pain. My prob is that I am very much interested in two themes that hardly interest socalled Ônormalos': Maths & Conspiracy theories... I like alot of weird maths like Benford's law or using Stats for predictions and quantitative analysis :-) Keep the spirit, Rob >I find that the biggest difference (and frustration) is not so much that >I think more quickly than other people (though that is generally but not >always true), but I think _differently_ from other people. I think of >solutions others just wouldn't think of. Sometimes that makes me >slower, because others see the obvious solution which has to be >the right one, but I'm busy evaluating two others, both of which are >superior to the obvious one, but I'm not sure which is better. >Yeah. I got the same problem. Maybe we should get together and form a >club. We could call ourselves the Uebermensa ;) >-- > Ben Carter > === Subject: Re: An inequality problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5HGYFG01204; >I have been trying to solve this inequality but to vain, does anyone >else have a clue - >[a/(1-(a^2)] + [b/(1-b^2)] + [c/(1-c^2)] >= 3(3^(1/2))/2 >given 0and ab+bc+ca = 1 >where x^2 means square of x; You can express the LHS of the inequality as tan A + tan B + tan C where sin A = a etc. phil === Subject: Re: An inequality problem phil escribi.97: > I have been trying to solve this inequality but to vain, does anyone > else have a clue - > [a/(1-(a^2)] + [b/(1-b^2)] + [c/(1-c^2)] >= 3(3^(1/2))/2 > given 0 and ab+bc+ca = 1 > where x^2 means square of x; > -S > You can express the LHS of the inequality as > tan A + tan B + tan C > where sin A = a etc. > phil If a = sin(A), then 1 - a^2 = cos^2(A), and the LHS is sin(A)/cos^2(A) + sin(B)/cos^2(B) + sin(C)/cos^2(C) or tan(A)/cos(A) + tan(B)/cos(B) + tan(C)/cos(C) -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Limit Of Average Of a(GCD(j,k))'s Here is another one of those types of results I posted a lot of a while back. It is probably a well-known result, but it deserves to be better known. First a specific case, then the general result, then some more specific cases, then finally a bonus related (but slightly off-topic) result: Here is an interesting (to me) limit: limit{m->oo, n->oo} m n --- --- 1 1 ---------- = ? --- / / GCD(j,k) m*n --- --- 2 j=1 k=1 In linear-mode: limit{m->oo, n->oo} (1/(m*n)) sum{j=1 to m} sum{k=1 to n} 1/2^GCD(j,k) = ? I get that the limit of this average is: 1/2 - 3*(ln(2)/pi)^2. Why?... The general result: Let {a(k)} be any sequence defined for positive integer indexes k, and where the sum and limit below converge (and the sum converges absolutely, just to be safe). GCD(j,k) is the greatest common divisor of j and k. Then: limit{m->oo, n->oo} m n --- --- 1 --- / / a(GCD(j,k)) m*n --- --- j=1 k=1 = oo --- 6 a(j) ---- / ---- pi^2 --- j^2 j=1 In linear mode: limit{m->oo, n->oo} (1/(m*n)) sum{j=1 to m} sum{k=1 to n} a(GCD(j,k) = (6/pi^2) sum{j=1 to oo} a(j)/j^2. (Not too hard to show, but potentially useful.) Some specific cases: a(j) = mu(j), the Mobius (Moebius) function: limit = 36/pi^4. -- a(j) = sum{k=1 to j} 1/k, a harmonic number: limit = 12 zeta(3) / pi^2. --- a(j) = x^j + (1-x)^j, for |x| < 1: limit = 1 - (6/pi^2) ln(x) ln(1-x). --- a(j) = 1/j^r, for r > -1, limit = zeta(r+2) *6/pi^2 --- Bonus result: (I am unsure here if I am correct.) For |x|< 1, limit{m->oo} (1/m^2) sum{k=1 to m} (1 - x^ßoor(m/k)) phi(k) = (1/x -1) (3/pi^2) (sum{k=1 to oo} x^k /k^2), where phi(k) is the number of positive integers <= k and coprime to k . (Right?) Leroy Quet === Subject: Re: Limit Of Average Of a(GCD(j,k))'s > Here is another one of those types of results I posted a lot of a > while back. > It is probably a well-known result, but it deserves to be better > known. > First a specific case, then the general result, then some more > specific cases, then finally a bonus related (but slightly off-topic) > result: > Here is an interesting (to me) limit: > limit{m->oo, n->oo} > m n > --- --- 1 > 1 ---------- = ? > --- / / GCD(j,k) > m*n --- --- 2 > j=1 k=1 > In linear-mode: > limit{m->oo, n->oo} > (1/(m*n)) sum{j=1 to m} sum{k=1 to n} 1/2^GCD(j,k) > = ? > I get that the limit of this average is: > 1/2 - 3*(ln(2)/pi)^2. > Why?... > The general result: > Let {a(k)} be any sequence defined for positive integer indexes k, and > where the sum and limit below converge (and the sum converges > absolutely, just to be safe). > GCD(j,k) is the greatest common divisor of j and k. > Then: > limit{m->oo, n->oo} > m n > --- --- > 1 > --- / / a(GCD(j,k)) > m*n --- --- > j=1 k=1 > oo > --- > 6 a(j) > ---- / ---- > pi^2 --- j^2 > j=1 > In linear mode: > limit{m->oo, n->oo} > (1/(m*n)) sum{j=1 to m} sum{k=1 to n} a(GCD(j,k) > (6/pi^2) sum{j=1 to oo} a(j)/j^2. > (Not too hard to show, but potentially useful.) >... A generalization of the result: (Now this is looking slightly familiar. If I already posted on this, I apologize.) For r = integer >= 2, limit{m_1->oo, m_2->oo, ...,m_r->oo} 1 --------------- * m_1*m_2*..*m_r m_1 m_2 m_r --- --- --- / / .... / a(GCD(k_1,k_2,..,k_r)) --- --- --- k_1=1 k_2=1 k_r=1 = oo 1 --- ------- a(k) zeta(r) / ----- --- k^r k=1 In linear-mode: limit{m_1->oo, m_2->oo, ...,m_r->oo} (1/(m_1*m_2*..*m_r)) sum{k_1=1 to m_1} sum{k_2=1 to m_2}..sum{k_r=1 to m_r} a(GCD(k_1,k_2,..,k_r)) = (1/zeta(r))* sum{k=1 to oo} a(k)/k^r. In addition the the (snipped) specific examples of a(j)'s in my original post, another interesting case is when a(j) is 0 or 1, depending on j. So, for example, the fraction of r-tuples {k_1,k_2,..k_r} of positive integers, taken over all positive integers*, where GCD(k_1,k_2,..k_r) is prime is (1/zeta(r)) *sum{p=primes} 1/p^r. *By taken over all positive integers, I mean, obviously, we want: limit{m_1->oo, m_2->oo, ...,m_r->oo} (1/(m_1*m_2*..*m_r)) sum{k_1=1 to m_1} sum{k_2=1 to m_2}..sum{k_r=1 to m_r} a(GCD(k_1,k_2,..,k_r)), where a(j) = 1 if j is prime, = 0 otherwise. As for the (snipped) bonus result of my original post, I have posted more on that in another thread. Leroy Quet === Subject: Is it fact, or fancy? One of the most important observations in physics is that: ÔThe response of any body - with any weight (w) - to gravity is the same as its resistance to acceleration': That is the ratio of any body's weight (w) divided by the acceleration (g) at which it will freefall is equal to the ratio of the net force (f) exerted on and/or by it, divided by the acceleration (a) that is caused. For any given body, these ratios are equivalent: Equal and constant; as well as being a measure of the body's mass, and/or its inertia. Mass and/or inertia is expressed in Units of Mass! In English mass or inertia is expressed in slugs, and/or fractions thereof: In metric or SI mass or inertia is expressed in grams or kilograms, and/or fractions thereof === Subject: Re: Is it fact, or fancy? > One of the most important observations in physics is that: Hey Dumb Donny Head, http://www.apa.org/journals/psp/psp7761121.html http://insti.physics.sunysb.edu/~siegel/quack.html http://www.kingstroker.com/ -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Is it fact, or fancy? > One of the most important observations in physics is that: Practitioners can make observations and record them. === Subject: Re: Is it fact, or fancy? >One of the most important observations in physics is that: > Practitioners can make observations and record them. Praciically anybody can make and record this one. === Subject: Re: Some GCD-Sum Results > Let {a(k)} be any sequence defined for all positive integer indexes k. > ÔGCD(k,m)' is the greatest common divisor of k and m, of course. > First, > For m and n = any positive integers: > sum{k=1 to n} sum{j|k} GCD(j,m) a(k/j) > sum{k=1 to m} sum{j=1 to ßoor(GCD(k,m)*n/m)} A(j), > where A(j) = sum{k|j} a(k). > (sum{k|j} means the sum is over the positive divisors, k, of j. > {Or vice-versa for sum{j|k}.}) > -- > Second, > from above we see: > sum{j|n} GCD(j,m) a(n/j) > sum{1<=j<=m, GCD(j,m)n/m =integer} A(GCD(j,m)n/m) > (So, the last sum is over those positive integer j's, j<=m, > where m divides GCD(j,m)*n evenly.) > -- > 3rd, > for {a(k)} where a(k) -> 0 (an overly restrictive condition) > and where sum{k=1 to oo} a(k)/k converges, > limit{n-> oo} > (1/n) sum{k=1 to n} sum{j|k} GCD(j,m) a(k/j) > (sum{k=1 to oo} a(k)/k) (sum{j=1 to m} GCD(j,m))/m > -- > 4th, > sum{k=1 to m} a(m/GCD(k,m)) > sum{k=1 to m} (sum{j|k} GCD(j,m) mu(k/j)) a(k), > where mu() is the Mobius (Moebius) function. > Leroy Quet Regarding the last result: sum{j|k} GCD(j,m) mu(k/j) is simply phi(k), if k|m, and is 0 otherwise. (phi(k) is the number of positive integers <= k and coprime to k.) So, the last result is more simply: sum{k=1 to m} a(m/GCD(k,m)) = sum{k|m} phi(k) a(k) Leroy Quet === Subject: m/GCD(m,j) = prime (If I posted about this already, I apologize. But I cannot find anything by doing a search.) Now, I figure that the number of j's, for 1<= j <= m, where m/GCD(m,j) = n is phi(n) if n|m, is 0 otherwise. On the other hand, I think that the number of j's, for 1 <= j <= m, where m/GCD(m,j) is prime is sum{p|m, p=distinct primes} (p-1), which forms sequence A055631 of the EIS. For example, for m = 12, we have: 12/1=1, 12/2=6, 12/3=4, 12/4=3, 12/1=12, 12/6=2, 12/1=12, 12/4=3, 12/3=4, 12/2=6, 12/1=12, 12/12=1 Of these ratios, 3 are prime, and (2-1) +(3-1) = 3. Leroy Quet === Subject: Re: m/GCD(m,j) = prime This post did not travel from Google to Math Forum, at least, so I am reposting it. ---- > (If I posted about this already, I apologize. But I cannot find > anything by doing a search.) > Now, I figure that the number of j's, for 1<= j <= m, > where m/GCD(m,j) = n > is phi(n) if n|m, > is 0 otherwise. > On the other hand, I think that the number of j's, > for 1 <= j <= m, > where m/GCD(m,j) is prime > is sum{p|m, p=distinct primes} (p-1), > which forms sequence A055631 of the EIS. > For example, for m = 12, > we have: > 12/1=1, 12/2=6, 12/3=4, 12/4=3, > 12/1=12, 12/6=2, 12/1=12, 12/4=3, > 12/3=4, 12/2=6, 12/1=12, 12/12=1 > Of these ratios, 3 are prime, > and (2-1) +(3-1) = 3. > Leroy Quet X-mailer: xrn 9.02 === Subject: Sign conventions for remainder Mail-To-News-Contact: abuse@dizum.com Just for fun, I've been implementing a package to perform arithmetic on integers of arbitrary size. Addition, subtraction, and multiplication were all pretty straight-forward. (Subtraction was the easiest!) But, when I prepared to implement division, I realized that I'm not aware of the conventions for remainders. If the divisor and the dividend are both positive, I know that my result should be d = n*q+r, with 0<=r Reunite Gondwanaland! === Subject: Re: Sign conventions for remainder > Just for fun, I've been implementing a package to perform arithmetic on > integers of arbitrary size. Addition, subtraction, and multiplication > were all pretty straight-forward. (Subtraction was the easiest!) But, > when I prepared to implement division, I realized that I'm not aware of > the conventions for remainders. There is no universally agreed convention. I use Visual Basic in Excel for a lot of my mathematical programming, and Microsoft manage to use two different conventions in the same package. Visual Basic has an integer divide function and a mod (remainder) function. The results obtained for different sign combinations are: 39/5 = 7 rem 4 39/(-5) = -7 rem 4 (-39)/5 = -7 rem -4 (-39)/(-5) = 7 rem -4 Excel also has a MOD function. This returns the following results: 39/5 = x rem 4 39/(-5) = x rem -1 (-39)/5 = x rem -1 (-39)/(-5) = x rem -4 Excel does not have an integer divide function, so the value of x in the above list cannot be generated directly. Instead, it is necessary to perform the division using normal ßoating point division (giving the answer 7.8 or -7.8 as appropriate) and then convert it to an integer using either the INT or TRUNC function. INT rounds to the integer below, so INT(-7.8) = -8, whereas TRUNC discards the fractional part, so TRUNC(-7.8) = -7. It is necessary to choose the INT function for consistency with the results returned by MOD (eg (-39)/5 must be -8 rem 1). === Subject: Re: Sign conventions for remainder > Just for fun, I've been implementing a package to perform arithmetic on > integers of arbitrary size. Addition, subtraction, and multiplication > were all pretty straight-forward. (Subtraction was the easiest!) But, > when I prepared to implement division, I realized that I'm not aware of > the conventions for remainders. > If the divisor and the dividend are both positive, I know that my result > should be d = n*q+r, with 0<=r the remainder still be in that range? Should it be in the negative of that > range? What about if both the divisor and the dividend are negative? > Yes, I am aware that there is no right answer to this, there are only > conventions. But, that's exactly what I'm looking for. Conventions are > usually established because they're useful. The convention in number theory is non-negative remainders; 0 less than or equal to r less than absolute value of n. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Sign conventions for remainder 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > Just for fun, I've been implementing a package to perform arithmetic on > integers of arbitrary size. Addition, subtraction, and multiplication > were all pretty straight-forward. (Subtraction was the easiest!) But, > when I prepared to implement division, I realized that I'm not aware of > the conventions for remainders. > If the divisor and the dividend are both positive, I know that my result > should be d = n*q+r, with 0<=r the remainder still be in that range? Should it be in the negative of that > range? What about if both the divisor and the dividend are negative? > Yes, I am aware that there is no right answer to this, there are only > conventions. But, that's exactly what I'm looking for. Conventions are > usually established because they're useful. If the divisor is positive, the remainder should still be in the range 0<=rJust for fun, I've been implementing a package to perform arithmetic on >integers of arbitrary size. Addition, subtraction, and multiplication >were all pretty straight-forward. (Subtraction was the easiest!) But, >when I prepared to implement division, I realized that I'm not aware of >the conventions for remainders. If the divisor and the dividend are both positive, I know that my result >should be d = n*q+r, with 0<=rthe remainder still be in that range? Should it be in the negative of that >range? What about if both the divisor and the dividend are negative? Yes, I am aware that there is no right answer to this, there are only >conventions. But, that's exactly what I'm looking for. Conventions are >usually established because they're useful. > If the divisor is positive, the remainder should still be in the range > 0<=r it's less clear to me what the right convention is -- the choice taken > by Python is to use the same sign as the divisor, which seems reasonable > enough to me. > The C-C++-Java convention of remainder having the same sign as dividend > is wrong and should be avoided. Are you talking about the Java primitive numeric types or the classes BigInteger and BigDecimal? The documentation for BigInteger does not seem to be very clear on the exact details of the remainder and I have never had the need to check. I may knock a simple test program and post the results. The class BigDecimal approaches the problem a different way. The divide method has a plethora of options on how to round divisions. So it dumps the problem onto the user. You could copy this idea and have options in your package for the several possible sensible behaviours. One behaviour that is common but I don't like is round towards zero. Non-mathematicians seem to like it since they seem to expect that -5 / 4 should be similar to 5 / 4 except for the sign i.e. +/- 1 rem 1. The reason that I don't like it is that the graph of x rem n has an anomaly at x = 0 which sometimes messes up functions built on top of it. Opinions vary, give your users the choice. Here is a link to the Java API docs: http://java.sun.com/j2se/1.3/docs/api/ Se.87n O'Leathl.97bhair === Subject: Re: Sign conventions for remainder >Just for fun, I've been implementing a package to perform arithmetic on >integers of arbitrary size. Addition, subtraction, and multiplication >were all pretty straight-forward. (Subtraction was the easiest!) But, >when I prepared to implement division, I realized that I'm not aware of >the conventions for remainders. >If the divisor and the dividend are both positive, I know that my result >should be d = n*q+r, with 0<=rthe remainder still be in that range? Should it be in the negative of that >range? What about if both the divisor and the dividend are negative? My opinion: The remainder is a piece of the dividend; therefore it should have the same sign, regardless of the sign of the divisor. This has the result of always rounding your quotient toward 0 rather than away, which seems intuitive. You wouldn't want -1/100 = -1 rem. 99, would you? --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Sign conventions for remainder > My opinion: The remainder is a piece of the dividend; therefore it should > have the same sign, regardless of the sign of the divisor. My opinion: rem(a,N), remainder when a is divided by N, should be a periodic function of a with period N. That is (at least for positive divisor) the quotent is done with the ßoor integer-part function, not the round toward zero integer-part function. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: contraction mapping help I am having problem with a theorem. I do math studies alone. Here is the theorem Let F(x) be a continuously differentiable function defined on the interal[a,b] s.t F(a)<0 & F(b)>0 and 0 I am having problem with a theorem. > I do math studies alone. > Here is the theorem > Let F(x) be a continuously differentiable function defined on the > interal[a,b] > s.t F(a)<0 & F(b)>0 and > 0 Use a contraction mapping to find the unique root of F(x)=0 > -- > Using M2, Opera's revolutionary e-mail client: http://www.opera.com/m2/ I have difficulty with the wording: to find is quite strong; perhaps the author of the problem meant construct a contraction mapping whose (unique) fixed point is the root of the function F. And: Is there a typo? I would expect 0 < K_1 <= F' <= K_2 (dropping the minus sign, and allowing equality in two places). [For pure existence of a root, differentiability is not required; continuity will do. For uniqueness, differentiability is not required either; strict monotonicity will do.] Such a construction is quite standard, and I can give a hint: Consider the function, with m non-zero: g(x) = x - m * F(x) Its fixed points are exactly the roots of F, and you can play around with m to make abs(g'(x)) small . For checking purposes: the contraction rate can be made as amall (in my notation) as (K_2 - K_1) / (K_2 + K_1). Hope it helps, ZVK(Slavek). === Subject: Re: contraction mapping help >I am having problem with a theorem. >I do math studies alone. >Here is the theorem >Let F(x) be a continuously differentiable function defined on the >interal[a,b] >s.t F(a)<0 & F(b)>0 and > 0Use a contraction mapping to find the unique root of F(x)=0 Seems like this has come up befor... Anyway, you can show that there exists a > 0 such that G is a strict contraction, where G(x) = x - a*F(x). ************************ David C. Ullrich === Subject: Re: Run and hide you gills : >that T |/- G. Since (1) asserts that G is equivalent to its own : >unprovability, we see that G is actually true. : > A simple deduction no? : : Indeed a simple deduction *under the assumption that T is : consistent*. In this context, that's begging the question. That T is consistent *IS* G. : For every theory T definable in its own language Are you sure that's the right constraint? Is PA definable in its own language? It has a Godel sentence ANYway. : there is a G.a6del sentence G_T for it. G.a6del's proof shows that G_T is : true - and formally unprovable from the axioms of T - if and only : if T is consistent. No, it doesn't. If G_T, as a Godel sentence for T, is unprovable in T, then ~G_T is one as well. If PA is consistent then PA + ~G_PA ALSO has a model. In THAT model, G_PA is FALSE. So saying that the consistency of T proves the truth of G_T is JUST WRONG. In the context of PA, it proves the truth of G_PA in THE STANDARD MODEL of PA. THERE ARE OTHER MODELS of PA in which G_PA IS FALSE. : We don't have the faintest idea about the : consistency of most theories, so G.a6del's proof obviously can't show : that the G.a6del sentences for these theories are true. Godel's proof does not show that the Godel sentence for ANY theory is true. Indeed, it shows the opposite: it shows that the Godel sentence for the ONE theory T that it is being applied to MUST be FALSE in at LEAST ONE model of T. -- --- The history of our nation has demonstrated that separate is seldom, if ever, equal. === Subject: Re: Run and hide you gills : >What a giant cop out from a bunch of illogical wits. Godels proof : >contains : >this line : : > : >G has no proof, which itself asserts, therefore it is true : > G.a6del's proof contains no such line. You're persisting in your mistaken : YOU WROTE THIS : >that T |/- G. Since (1) asserts that G is equivalent to its own : >unprovability, we see that G is actually true. : : : A simple deduction no? : Herc Yes AND no. G is true is stated in NATURAL language. That statement cannot be formalized because of Tarski's theorem on the indefinability of truth. The closest you can get to G is true (or not) is that G is true IN THE *STANDARD* model. It could be false in other models. There is a sense in which the Godelian formalization of provability depends on using the standard model. In other models, the thing's that the Godelian proof-predicate calls proofs arguably don't get to BE proofs, DESPITE the fact that the system is calling them proofs. My point is simply that talking about a simple deduction in NATURAL language is dangerous. Natural language is complicated. It is NOT simple. The FRAGMENTS of it that are simple are known to be simple PRECISELY BY VIRTUE OF THEIR ISOMORPHISM with simple FORMAL deductive languages. In other words, deductions are simple precisely to the extent that they are formal, NOT natural. -- --- The history of our nation has demonstrated that separate is seldom, if ever, equal. X-EXP32-SerialNo: 00105840 === Subject: Test === Subject: A great puzzle You are given a deck of n cards labeled 1,2,3,,n. You shufße the deck. Then you perform the following procedure: Look at the top card labeled k. If k=1, then stop. Otherwise, reverse the order in the deck of the first k cards. Then look at the top card again and repeat the same procedure. For example, if n=6 and the deck were in order 412563 (where 4 is the top card), then we would get 521463 -> 641253 -> 352146 -> 253146 -> 523146 -> 413256 -> 231456 -> 321456 -> 123456. (By coincidence, it stopped at 123456 but it doesn't always do such. It just has to have a 1 at the top of the deck.) Two problems, one easy, one hard: 1) Prove that such a procedure will always halt for any n and any shufßing of the n cards. 2) Find a closed formula for the maximum number of iterations that it must take for such a procedure to halt, given the number of cards in the deck n. My suggestion for the 2nd question is to look at by hand all of the possibilities for the case n=5 (since 5! is large enough to get an understanding but small enough to actually deal with in real time) and try to see if you can find a pattern. If anyone wants any more hints, then you can email me. Craig === Subject: Re: A great puzzle > You are given a deck of n cards labeled 1,2,3,?,n. You shufße the > deck. Then you perform the following procedure: Look at the top card > labeled k. If k=1, then stop. Otherwise, reverse the order in the deck > of the first k cards. Then look at the top card again and repeat the > same procedure. > For example, if n=6 and the deck were in order 412563 (where 4 is the > top card), then we would get 521463 -> 641253 -> 352146 -> 253146 - 523146 -> 413256 -> 231456 -> 321456 -> 123456. > (By coincidence, it stopped at 123456 but it doesn't always do such. > It just has to have a 1 at the top of the deck.) > Two problems, one easy, one hard: > 1) Prove that such a procedure will always halt for any n and any > shufßing of the n cards. > 2) Find a closed formula for the maximum number of iterations that it > must take for such a procedure to halt, given the number of cards in > the deck n. > My suggestion for the 2nd question is to look at by hand all of the > possibilities for the case n=5 (since 5! is large enough to get an > understanding but small enough to actually deal with in real time) and > try to see if you can find a pattern. If anyone wants any more hints, > then you can email me. > Craig I do not know the answers to your questions, but I HAVE to mention that your problem reminds me of the game Rotational Solitaire, mentioned at: (Rotational Solitaire was a bit more complicated, in that you could use either side of the row of cards for determining how many cards to reverse the order of, and you could switch the cards on either end of the row of cards. Also, RS has a couple possible ace-rules, which one you use depends on the variation of RS you are playing.) Leroy Quet === Subject: Re: A great puzzle > You are given a deck of n cards labeled 1,2,3,.,n. You shufße the > deck. Then you perform the following procedure: Look at the top card > labeled k. If k=1, then stop. Otherwise, reverse the order in the deck > of the first k cards. Then look at the top card again and repeat the > same procedure. > For example, if n=6 and the deck were in order 412563 (where 4 is the > top card), then we would get 521463 -> 641253 -> 352146 -> 253146 - 523146 -> 413256 -> 231456 -> 321456 -> 123456. > (By coincidence, it stopped at 123456 but it doesn't always do such. > It just has to have a 1 at the top of the deck.) > Two problems, one easy, one hard: > 1) Prove that such a procedure will always halt for any n and any > shufßing of the n cards. > 2) Find a closed formula for the maximum number of iterations that it > must take for such a procedure to halt, given the number of cards in > the deck n. > My suggestion for the 2nd question is to look at by hand all of the > possibilities for the case n=5 (since 5! is large enough to get an > understanding but small enough to actually deal with in real time) and > try to see if you can find a pattern. If anyone wants any more hints, > then you can email me. > Craig If you have a closed formula or even a few more terms, please update this entry in the OEIS: http://www.research.att.com/projects/OEIS?Anum=A000375 Rob Pratt === Subject: A binomial series problem I have a simple binomial series question. (x+y)^N = sum(n=0 to N) {NCn * x^N * y^(N-n)}, where NCn = N!/(n!(N-n)!) Now my question is that if I slightly modify this series to following, then what is the summation of it. sum(n=0 to N) {NCn * n * x^N * y^(N-n)} = ? Sailesh === Subject: Re: A binomial series problem > I have a simple binomial series question. > (x+y)^N = sum(n=0 to N) {NCn * x^N * y^(N-n)}, where NCn = N!/(n!(N-n)!) > Now my question is that if I slightly modify this series to following, then > what is the summation of it. > sum(n=0 to N) {NCn * n * x^N * y^(N-n)} = ? You're asking for free help, so shouldn't you proofread it carefully before you send it? You meant (x+y)^N = sum(n=0 to N) {NCn * x^n * y^(N-n)}. What happens if you differentiate the above with respect to x, holding y constant? === Subject: Re: A binomial series problem >I have a simple binomial series question. >(x+y)^N = sum(n=0 to N) {NCn * x^N * y^(N-n)}, where NCn = N!/(n!(N-n)!) And why do your post your question, without crossposting, to two different newsgroups with slightly different subject lines? Do you have a homework deadline? --Lynn === Subject: Re: A binomial series problem OK, let me put my actual problem: sum(n=0 to N) {NCn * f(n) * x^n * y^(N-n)} = ? ; where f(n) = INT(n/C), C is a constant, thus f(n) will always give an integer as output. I am still having problem because f(n) is not a continuos function. Sailesh >I have a simple binomial series question. >(x+y)^N = sum(n=0 to N) {NCn * x^N * y^(N-n)}, where NCn = N!/(n!(N-n)!) > And why do your post your question, without crossposting, to two > different newsgroups with slightly different subject lines? Do you > have a homework deadline? > --Lynn === Subject: Re: A binomial series problem >sum(n=0 to N) {NCn * f(n) * x^n * y^(N-n)} = ? ; where f(n) = INT(n/C), C is >a constant, thus f(n) will always give an integer as output. >I am still having problem because f(n) is not a continuos function. A function on the integers is always continuous since the integers have the discrete topology. One trick is first to use generating functions to establish oo --- N 1 x n > C(N,n) x = --- ( --- ) [1] --- 1-x 1-x N=n Multiply by y^n and sum over n oo oo --- --- N n > C(N,n) x y --- --- n=0 N=n oo N --- --- N n = > C(N,n) x y --- --- N=0 n=0 oo --- N N = > x (1+y) --- N=0 1 = ---------- 1 - x(1+y) 1 1 = --- ------------ 1-x 1 - xy/(1-x) oo --- 1 x n n = > --- ( --- ) y [2] --- 1-x 1-x n=0 Equating the coefficients of y^n in [2], we get [1]. Next, use generating functions to evaluate the sum in question; multiply by z^N and sum over N oo N --- --- n N-n N > C(N,n) f(n) x y z --- --- N=0 n=0 oo oo --- --- n N-n N = > C(N,n) f(n) x y z --- --- n=0 N=n oo --- x n 1 yz n = > f(n) ( - ) ---- ( ---- ) --- y 1-yz 1-yz n=0 oo 1 --- xz n = ---- > f(n) ( ---- ) 1-yz --- 1-yz n=0 oo 1 --- n = ---- > f(n) w [3] 1-yz --- n=0 where w = xz/(1-yz). Now if we break down the sum in [3] into smaller sums of size C, we get [3] oo C-1 1 --- --- kC+n = ---- > k w 1-yz --- --- k=0 n=0 oo 1 --- kC 1-w^C = ---- > k w ----- 1-yz --- 1-w k=0 1 w^C 1-w^C = ---- --------- ----- 1-yz (1-w^C)^2 1-w 1 w^C 1 = ---- ----- --- 1-yz 1-w^C 1-w (xz)^C 1 = ----------------- -------- [4] (1-yz)^C - (xz)^C 1-(x+y)z The answer to your question is the coefficient of z^N in [4]. Rob Johnson take out the trash before replying === Subject: Re: A binomial series problem sum(n=0 to N) {NCn * f(n) * x^n * y^(N-n)} = ? ; where f(n) = INT(n/C), C is > a constant, thus f(n) will always give an integer as output. > I am still having problem because f(n) is not a continuos function. INT(x) is the largest integer <= x, I assume? Except for a few special cases for the constant C, you're unlikely to get a closed-form solution that's appreciably nicer looking than the summation above. For N large and N/C small, it should be possible to write the summation as a sum with number of terms of order N/C. (The idea being to use a normal approximation to the binomial for the x,y>=0 case and then do some fiddling for the other cases.) In your problem, how big is N intended to be, and what is the relative size of C? -- Kevin === Subject: Re: A binomial series problem > I have a simple binomial series question. > (x+y)^N = sum(n=0 to N) {NCn * x^N * y^(N-n)}, where NCn = N!/(n!(N-n)!) There's a typo here. You mean x^n instead of x^N. > Now my question is that if I slightly modify this series to following, then > what is the summation of it. > sum(n=0 to N) {NCn * n * x^N * y^(N-n)} = ? Again, I'll assume you mean x^n instead of x^N. Note that NCn * n = N * (N-1)C(n-1), so since the first term of this new sum is zero, we may rewrite it for N>=1 as: N * x * sum n=1 to N {(N-1)C(n-1) * x^(n-1) * y^((N-1)-(n-1))} and change the variable in the summation to get: N * x * sum n=0 to N-1 {(N-1)Cn * x^n * y^((N-1)-n)} The summation is now (x+y)^(N-1), so we have: sum n=0 to N {NCn * n * x^n * y^(N-n)} = N * x * (x+y)^(N-1) The above argument assumed N>0, but this final result is obviously correct when N=0, too. This was the same expression Stephen posted. And the above shows that it's correct for arbitrary real x and y, not just nonnegative ones. -- Kevin === Subject: Re: A binomial series problem >sum(n=0 to N) {NCn * n * x^N * y^(N-n)} = ? consideration of the expectation of a binomial random variable, this is (x+y)^(N-1) n x. I leave it to others to verify or contradict this when x or y is negative. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: A binomial series problem >sum(n=0 to N) {NCn * n * x^N * y^(N-n)} = ? >consideration of the expectation of a binomial random variable, this is >(x+y)^(N-1) n x. >I leave it to others to verify or contradict this when x or y is >negative. You have made an error, and non-negative is not needed. The correct answer is (x+y)^(N-1*Nx; it would follow by analytic continuation, but using that NCn*n = (N-1)C(n-1)*N does it immediately. -- 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: Factoring paper is wrong by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5HIPiC05099; let a+1=b+1, >therefore a = b >Proving statements such as the above don't necessarily invite passionate >minds to tackle them. Certain categories of maths problems invite certain >passionate types to try to best them: >Let X be some arbitrary integer s.t. N > 4. > int N = X; > while((N = N&1 ? 3*N+1 : N/2) != 1); > (1.0) >Prove that 1.0 terminates for any X. >All of a sudden, all kinds of personalities get into the mix. What surprises me is the following. Why do people who are totally lacking in mathematical maturity and mathematical education think that somehow they are going to find a magic proof that has escaped experts? I doubt whether these same people would ever considering doing (say) brain surgery. Yet they are just as ignorant about one subject as the other. I am sure that these people recognize that they do not have the education or experience to do brain surgery. Why is it that they do NOT have the same cognizance about their lack of math skills? Why do they stubbornly cling to ideas that experts point out are wrong? Instead, they attack the experts! I doubt that they would attack a doctor who told them that they had no surgical skills. If one argues that *unsolved* problems attracts such people, why don't we get people who claim that they have found a cure for (say) cerebral palsy? That too is an unsolved problem. What is it about math in particular that attracts incompetents to believe that they can solve problems that experts can not? Why do they then attack experts who tell them that they are wrong? === Subject: Re: Factoring paper is wrong ... > If one argues that *unsolved* problems attracts such people, why > don't we get people who claim that they have found a cure for (say) > cerebral palsy? That too is an unsolved problem. How many people are there claiming to have a cure for cancer? It is not restricted to mathematics. The good things about mathematical cranks is that they do no physical wrong, it is otherwise with medical cranks, aka quacks. (Just read that one frequently used method to reduce weight is detrimental to the condition of heart and blood-vessels. But somebody propagated it without ever looking at the consequences.) -- 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: Factoring paper is wrong > let a+1=b+1, >therefore a = b >Proving statements such as the above don't necessarily invite passionate >minds to tackle them. Certain categories of maths problems invite certain >passionate types to try to best them: >Let X be some arbitrary integer s.t. N > 4. > int N = X; > while((N = N&1 ? 3*N+1 : N/2) != 1); > (1.0) >Prove that 1.0 terminates for any X. >All of a sudden, all kinds of personalities get into the mix. > Why do people who are totally lacking in mathematical maturity and > mathematical education think that somehow they are going to find > a magic proof that has escaped experts? > I doubt whether these same people would ever considering doing (say) > brain surgery. Yet they are just as ignorant about one subject as the > other. > I am sure that these people recognize that they do not have the > education or experience to do brain surgery. Why is it that they do > NOT have the same cognizance about their lack of math skills? Why > do they stubbornly cling to ideas that experts point out are wrong? > Instead, they attack the experts! I doubt that they would attack > a doctor who told them that they had no surgical skills. > If one argues that *unsolved* problems attracts such people, why > don't we get people who claim that they have found a cure for (say) > cerebral palsy? That too is an unsolved problem. > What is it about math in particular that attracts incompetents to > believe that they can solve problems that experts can not? Why do > they then attack experts who tell them that they are wrong? I suspect that the problem is much more general than you think. Because tend to deal with people who do math (or think that they do math) you see the people who think that they are the next Erdos. I suspect that delusions of competence appear in many more fields than math. I live with a writer and she dreads touring, talking to book clubs, and even admitting that she is something more than a mild mannered english professor because everyone thinks that they have the next great american novel, and all they need is some help to get an agent or an editor interested. At first I thought that she was merely being coy, but history has proved her right time and again. I suspect that there are people wandering about convinced that they would stop blocking their efforts. I have no evidence, but I believe that the entire Laetrile fiasco was caused by the equivalent to our beloved cranks. === Subject: Re: Factoring paper is wrong >let a+1=b+1, >therefore a = b >Proving statements such as the above don't necessarily invite passionate >minds to tackle them. Certain categories of maths problems invite certain >passionate types to try to best them: >Let X be some arbitrary integer s.t. N > 4. > int N = X; > while((N = N&1 ? 3*N+1 : N/2) != 1); > (1.0) >Prove that 1.0 terminates for any X. >All of a sudden, all kinds of personalities get into the mix. >Why do people who are totally lacking in mathematical maturity and >mathematical education think that somehow they are going to find >a magic proof that has escaped experts? >I doubt whether these same people would ever considering doing (say) >brain surgery. Yet they are just as ignorant about one subject as the >other. >I am sure that these people recognize that they do not have the >education or experience to do brain surgery. Why is it that they do >NOT have the same cognizance about their lack of math skills? Why >do they stubbornly cling to ideas that experts point out are wrong? >Instead, they attack the experts! I doubt that they would attack >a doctor who told them that they had no surgical skills. >If one argues that *unsolved* problems attracts such people, why >don't we get people who claim that they have found a cure for (say) >cerebral palsy? That too is an unsolved problem. >What is it about math in particular that attracts incompetents to >believe that they can solve problems that experts can not? Why do >they then attack experts who tell them that they are wrong? The difference is that when they see correct mathematics they don't actually understand why it's correct; hence their incorrect arguments look just as good to them as real math. Otoh they _can_ see the difference between a cure for cerebral palsy and a non-cure. ************************ David C. Ullrich === Subject: Re: Factoring paper is wrong > let a+1=b+1, >therefore a = b >Proving statements such as the above don't necessarily invite passionate >minds to tackle them. Certain categories of maths problems invite certain >passionate types to try to best them: >Let X be some arbitrary integer s.t. N > 4. > int N = X; > while((N = N&1 ? 3*N+1 : N/2) != 1); > (1.0) >Prove that 1.0 terminates for any X. >All of a sudden, all kinds of personalities get into the mix. > Why do people who are totally lacking in mathematical maturity and > mathematical education think that somehow they are going to find > a magic proof that has escaped experts? Think about how a lot of people view mathematics. I remember taking algebra classes in high school and hearing about this frightening course called calculus. At that point in time, the only thing any of us knew of was calculus and statistics. Things like real analysis, complex analysis, formal logic, set theory, combinatorics, etc. were things I hadn't even conceived of existing. This is probably the perspective most people have. They're good at high school algebra, so they think they have the potential to make deep insights into math, just like the other people who have degrees. They have no concept of the fact that they are frighteningly ignorant. > I doubt whether these same people would ever considering doing (say) > brain surgery. Yet they are just as ignorant about one subject as the > other. To extend the analogy, these people would view brain surgery as something a butcher could do with a little practice. They think advanced mathematics is just trickier algebraic manipulations. > I am sure that these people recognize that they do not have the > education or experience to do brain surgery. Why is it that they do > NOT have the same cognizance about their lack of math skills? Why > do they stubbornly cling to ideas that experts point out are wrong? > Instead, they attack the experts! I doubt that they would attack > a doctor who told them that they had no surgical skills. For the same reason that people who know they can't walk a tightrope will believe they can write a great novel... with sentences that frequently don't have verbs. A skill with a physical component is obviously tricky. A skill that is purely mental is open to inaccurate self-evaluation. > If one argues that *unsolved* problems attracts such people, why > don't we get people who claim that they have found a cure for (say) > cerebral palsy? That too is an unsolved problem. Because there is objective evidence for whether a cerebral palsy cure works. They view their proofs through a subjective standard. > What is it about math in particular that attracts incompetents to > believe that they can solve problems that experts can not? Why do > they then attack experts who tell them that they are wrong? Because they think they know the rules of the game, when they don't. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Factoring paper is wrong Will Twentyman said: > This is probably the perspective most people have. They're good at high > school algebra, so they think they have the potential to make deep > insights into math, just like the other people who have degrees. Sniff. Sniff. Do I smell a tinge of degreeism, Will? > A skill that is purely mental is open to inaccurate > self-evaluation. Mathematics isn't a skill that is purely mental. OK, granted, one can say: (1.0) Let G be some grammar such that it generates only the languages in L such that L = {a^n b^n c^n | n > 1} And, maybe, on some purely mental level they can say, (2.0) G cannot be represented by a PDA. This *can* all be done in a purely mental fashion. Proving that some specific grammar g satisfies 2.0, however, can be more than a purely mental exercise, and is not open to any self-evaluation whatsoever. Proving that some specific grammar g satisfies (2.0) can be done by using the pumping lemma in a rigorous mechanical fashion. No self-evaluation required. No psychology necessary. Proving that the 255 integers starting at 3,298,601,216 all have 218 Collatz steps to reach 1 can be done in a few lines of C++ code that wouldn't require any self-evaluation at all. Proving (either way) that this is (or is not) the only such sequence of this length in 2^32 integer space could be done in a reasonable amount of time in a similar mechanical fashion -- with very little mental skill required. Of course the purely mental version would impress the purely mental, whereas the mechanical version would have to itself be proven. ;-) I could probably prove that some language L = {a^m b^n c^mn | m, n > 1} can be recognized amongst a stream of noise in O(n^{slightly greater than 1}}, and demonstrate it in a way that wouldn't require much mental effort, let alone purely mental effort. I could probably prove that 1+1=2 in some purely mental way through a clever bit of definition that wouldn't even require me to use the digits 1 and 2 (no other digit than 0 would be necessary, probably), or I could take one glass of water and put it on the table, and put another glass of water beside it and dare anyone to refute that two glasses of water are on the table. Only in an Orwellian world where 1+1 may or may not be equal to 2 could such a clearly non-purely mental proof be subjected to refutation. OK -- purely mental powers ARE open to inaccurate self-evaluation. Yes. But, mathematics needn't be a purely mental skill and/or exercise. Prove (in either a purely mental or a non-purely mental way) that maths *must* be a purely mental exercise, and I'll retract my comments. -- Quinn === Subject: Re: Factoring paper is wrong > Will Twentyman said: > This is probably the perspective most people have. They're good at > high school algebra, so they think they have the potential to make deep > insights into math, just like the other people who have degrees. > Sniff. Sniff. Do I smell a tinge of degreeism, Will? > A skill that is purely mental is open to inaccurate self-evaluation. > Mathematics isn't a skill that is purely mental. I especially enjoy the physical parts of mathematics. Every morning I do my mathematical calisthenics. > OK, granted, one can say: > (1.0) Let G be some grammar such that it generates only the languages > in L such that L = {a^n b^n c^n | n > 1} > And, maybe, on some purely mental level they can say, (2.0) G cannot be > represented by a PDA. I can train my parrot to say all those things. clever bit of definition OK. I'll try it. Definition: 2 is a number such that 1+1=2. Proof: 1+1=2 (By Definition of 2) > that wouldn't even require me to use the digits 1 and 2 OK. In Roman Numberal Notation. Definition: II is a number such that I+I=II. Proof: I+I=II (By Definition of II). > (no other digit than 0 would be necessary, probably), or I could take > one glass of water and put it on the table, and put another glass of > water beside it and dare anyone to refute that two glasses of water are > on the table. Only in an Orwellian world where 1+1 may or may not be > equal to 2 could such a clearly non-purely mental proof be subjected to > refutation. > OK -- purely mental powers ARE open to inaccurate self-evaluation. Yes. > But, mathematics needn't be a purely mental skill and/or exercise. > Prove (in either a purely mental or a non-purely mental way) that maths > *must* be a purely mental exercise, and I'll retract my comments. > -- > Quinn -- Lance Lamboy I tell them the truth and they think it's hell. ~ Harry S. Truman === Subject: Re: Factoring paper is wrong >Mathematics isn't a skill that is purely mental. > I especially enjoy the physical parts of mathematics. Every morning I do > my mathematical calisthenics. I particularly enjoy jumping to conclusions and pushing the envelope. V. -- email: lastname at cs utk edu homepage: cs utk edu tilde lastname === Subject: Re: Factoring paper is wrong > +--------------- > | A great many things work *sometimes*. That should not be impressive. > +--------------- > (*cough*) Uh... Yes, indeed. Reminds me of the old joke about > an engineer proving that all odd numbers are prime: > Let's see... 3 is prime, 5's prime, 7's prime, 9... we'll throw that > out due to experimental error, 11's prime, 13's prime. Looking good... No, no, that's the physicist's proof. The engineer's proof is: 3 is prime, 5 is prime, 7 is prime, 9 is prime, 11 is prime... ... yep, all odd numbers are prime. - Randy === Subject: Re: Factoring paper is wrong >(*cough*) Uh... Yes, indeed. Reminds me of the old joke about >an engineer proving that all odd numbers are prime: Let's see... 3 is prime, 5's prime, 7's prime, 9... we'll throw that > out due to experimental error, 11's prime, 13's prime. Looking good... > No, no, that's the physicist's proof. The engineer's proof is: > 3 is prime, 5 is prime, 7 is prime, 9 is prime, 11 is prime... > ... yep, all odd numbers are prime. And the computer scientist: 3 is prime, 3 is prime, 3 is prime, ... Oh, something got st, but all odd numbers are prime. -- 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: Factoring paper is wrong charset=iso-8859-1 >+--------------- >| A great many things work *sometimes*. That should not be impressive. >+--------------- >(*cough*) Uh... Yes, indeed. Reminds me of the old joke about >an engineer proving that all odd numbers are prime: > Let's see... 3 is prime, 5's prime, 7's prime, 9... we'll throw that > out due to experimental error, 11's prime, 13's prime. Looking good... > No, no, that's the physicist's proof. The engineer's proof is: > 3 is prime, 5 is prime, 7 is prime, 9 is prime, 11 is prime... > ... yep, all odd numbers are prime. Not quite, an engineer would probably do it 3 is prime, 5 is prime, 7 is prime now try some bigger numbers 11 prime, 17 prime, even bigger 113 prime All test numbers were prime so yep, all odd numbers prime. Andrew Swallow === Subject: Re: Factoring paper is wrong <40cf3fd9$1_2@newsfeed.slurp.net> 3 is prime, 5 is prime, 7 is prime > now try some bigger numbers > 11 prime, 17 prime, even bigger 113 prime >All test numbers were prime so yep, all odd numbers prime. I have that (with a different selection of numbers) as the statistician's proof. -- Stewart Robert Hinsley === Subject: re Star Trek II: The Wrath of Khan!! === Subject: Another twin primes conjecture Conjecture: Given a pair of twin primes p and p+2, it must be the case that you find at least one more set of twin primes in the interval p^2 to (p+2)^2. Examples: 5 and 7 are twin primes, and in the interval 25 to 49, you have 29 and 31. 11 and 13 are twin primes and in the interval 121 to 169 you have 137 and 139. 17 anbd 19 are twin prime and in the interval 289 to 361 you have 311 and 313. I realized it while looking at my prime counting function, which you can see at http://mathforprofit.blogspot.com/ which I've had for a while. James Harris === Subject: Re: Another twin primes conjecture > Conjecture: > Given a pair of twin primes p and p+2, it must be the case that you > find at least one more set of twin primes in the interval p^2 to > (p+2)^2. I conjectured that for every integer n >= 1, the interval [n(n-1)/2 + 1, n(n+1)/2] (that is, the nth row of a triangular table of the positive integers) contains an integer u such that 6u-1, 6u+1 are twin primes (and also an integer v such that 4v^2 + 1 is prime). These intervals are (in effect) slightly narrower than yours (although one conjecture doesn't imply the other). ;-) ------------------------------------ J K Haugland http://www.neutreeko.com Try my grid subgraph prize problems! ------------------------------------ === Subject: Re: Another twin primes conjecture >Conjecture: >Given a pair of twin primes p and p+2, it must be the case that you >find at least one more set of twin primes in the interval p^2 to >(p+2)^2. Well, unless I or Mathematica have made a mistake, there doesn't seem to be a counter-example in the positive integers up to the twin primes 155731907 and 155731909, which have the corresponding twin primes 24252426857857019 and 24252426857857021. n = 1; success = True; While[success, n += 2; If[PrimeQ[n] && PrimeQ[n + 2], trial = False; For[m = n^2, m <= (n + 2)^2, m += 2, If[PrimeQ[m] && PrimeQ[m + 2], trial = True; Break[] ] ]; success = success && trial ] ] Ah, but now I see there is a ßaw in what I did. This allows (n+2)^2 and (n+2)^2+2 as acceptable twin primes and that is not what the conjecture stated. My apologies, I'll correct my mistake and post again if it finds a counter-example. Sorry === Subject: Re: Another twin primes conjecture > Well, unless I or Mathematica have made a mistake, there doesn't seem > to be a counter-example in the positive integers up to the twin primes > 155731907 and 155731909, which have the corresponding twin primes > 24252426857857019 and 24252426857857021. Suggestion: Since James has stated that the counter game him the idea -- here's one that counterexample: Start with the largest known p p+2 pair, and work backwards from there in your search for a counter example. My reasoning here is the intuition that if a counter example exists in brute-force space, it will tend to exist in the upper rather than lower range, since the density of primes will tend to be lower in that space. I may be basing my intuition on unfounded hunch -- please enlighten me if my suggestion has no merit. It just seems to me that searching for counter examples in prime-rich spaces will take longer. -- Quinn === Subject: Re: Another twin primes conjecture > Well, unless I or Mathematica have made a mistake, there doesn't seem > to be a counter-example in the positive integers up to the twin primes > 155731907 and 155731909, which have the corresponding twin primes > 24252426857857019 and 24252426857857021. Just to be precise, I corrected the code, re-ran it, and as expected it made no difference in the outcome. >Suggestion: >Since James has stated that the counter game him the idea -- here's one that >counterexample: >Start with the largest known p p+2 pair, and work backwards from there in >your search for a counter example. My reasoning here is the intuition that >if a counter example exists in brute-force space, it will tend to exist in >the upper rather than lower range, since the density of primes will tend to >be lower in that space. Perhaps a slight modification of your idea, search the web for previously tabulated twin prime information, but looking for relatively large gaps, those where the ratio of the gap between two pairs of twin primes is large in comparison to the size of the primes themselves. >I may be basing my intuition on unfounded hunch -- please enlighten me if my >suggestion has no merit. It just seems to me that searching for counter >examples in prime-rich spaces will take longer. Two comments: New and untested conjectures can perhaps fail with checking a modest number of cases, and I believe that if the theorem about the density of primes holds then as the primes become larger and larger it becomes less and less likely that this will fail. He is looking at a range of about 2p numbers and the density of twin primes are conjectured to be about 1/log^2(p) but in the context of his conjecture this will be 1/log^2(p^2) since he is considering a range of 2p numbers in the neighborhood of p^2. As p grows 2p/log^2(p^2) grows. So for much larger p it seems that either a miracle happens and there is a sufficiently large gap between pairs of twin primes or more commonly, it is easier and easier to find a pair. I hope I have not made a mistake in any of this === Subject: Re: Another twin primes conjecture >Well, unless I or Mathematica have made a mistake, there doesn't seem >to be a counter-example in the positive integers up to the twin primes >155731907 and 155731909, which have the corresponding twin primes >24252426857857019 and 24252426857857021. > Suggestion: > Since James has stated that the counter game him the idea -- here's one that > counterexample: > Start with the largest known p p+2 pair, and work backwards from there in > your search for a counter example. My reasoning here is the intuition that > if a counter example exists in brute-force space, it will tend to exist in > the upper rather than lower range, since the density of primes will tend to > be lower in that space. > I may be basing my intuition on unfounded hunch -- please enlighten me if my > suggestion has no merit. It just seems to me that searching for counter > examples in prime-rich spaces will take longer. Statistically, about one in (log n) integers of size about n is a prime. Statistically, about one in c * (log n)^2 pairs of integer (p, p+2) of size about n are both primes, for some constant c reasonably close to 1 (it is not 1 because there are dependencies; if p is prime then except for p=2 we know that p+2 is odd, so more likely to be prime than a random integer). For a given p, he suggests there is a twin prime in the interval (p^2, (p+2)^2). There are about 4p candidates. If everything were random, then the chance that any pair (p, p+2) is not a twin prime is (1 - 1/(c (log p)^2)) for some c; The chance that no pair is a twin prime is about (1 - 1/(c (log p)^2))^4p The logarithm of this is about -4p / (c (log p)^2) which becomes small quite rapidly. If the heuristics are correct then there is not much chance to find a counterexample. Which is of course not a proof. It also has nothing to do with the original p and p+2 being twin primes, that is just a bit of confusion that Harris introduces here, any p > 0 will do. === Subject: Re: Another twin primes conjecture <_AuAc.799195$Ig.354043@pd7tw2no It also has nothing to do with the original p and p+2 being twin primes, > that is just a bit of confusion that Harris introduces here, any p > 0 > will do. Yes, the following conjecture seems to hold up just as well. Given any number n>=1 except 26, there's a set of twin primes in the interval n^2 to (n+2)^2. -- Kevin === Subject: Re: Another twin primes conjecture > It also has nothing to do with the original p and p+2 being twin primes, > that is just a bit of confusion that Harris introduces here, any p > 0 > will do. >Yes, the following conjecture seems to hold up just as well. >Given any number n>=1 except 26, there's a set of twin primes in the >interval n^2 to (n+2)^2. First, I freely admit that proving the twin prime conjecture will forever be beyond my limited skills. Mr. Harris made the conjecture that for any twin prime p and p+2 there would be at least one twin prime between p^2 and p^2+2p+4. But if lack of counter-example counts for anything then it appears there are no counter-examples if this is tightened to between p^2 and p^2+16/11p. And I fully admit that this has nothing to do with a proof of the conjecture. === Subject: Re: Another twin primes conjecture ... > Well, unless I or Mathematica have made a mistake, there doesn't seem > to be a counter-example in the positive integers up to the twin primes > 155731907 and 155731909, which have the corresponding twin primes > 24252426857857019 and 24252426857857021. > ... > For a given p, he suggests there is a twin prime in the interval (p^2, > (p+2)^2). There are about 4p candidates. If everything were random, then > the chance that any pair (p, p+2) is not a twin prime is > (1 - 1/(c (log p)^2)) for some c; > The chance that no pair is a twin prime is about > (1 - 1/(c (log p)^2))^4p > The logarithm of this is about > -4p / (c (log p)^2) > which becomes small quite rapidly. If the heuristics are correct then > there is not much chance to find a counterexample. Which is of course > not a proof. > It also has nothing to do with the original p and p+2 being twin primes, > that is just a bit of confusion that Harris introduces here, any p > 0 > will do. Well that sounds like another conjecture, easily checkable by those who can check my original conjecture. Bau's claim is simpler to check as you need just go to any prime, add 2 to it, square your p and p+2, and check in that interval for twin primes. You get a lot more data points that way as well. James Harris === Subject: Re: Another twin primes conjecture > The chance that no pair is a twin prime is about > (1 - 1/(c (log p)^2))^4p > The logarithm of this is about > -4p / (c (log p)^2) > which becomes small quite rapidly. If the heuristics are correct then > there is not much chance to find a counterexample. Which is of course > not a proof. It occurred to me that if pi(n^2)-pi((n+2)^2) could be shown to be < 2 for any n, then it could be shown that the conjecture doesn't hold, since there could not be two primes in that range, let alone twin primes. For the sake of argument, let's say the Harris Prime Counter is correct for all N. Could some analysis be done on that function to determine if HPC(N^2) - HPC((N+2)^2) for any N? (I'm going on the assumption that his function is not an approximation, but an actual count.) Two outcomes: A. If such an N is discovered to exist: conjecture falls. B. If such an N is discovered to not exist: conjecture stands and remains unproven. If outcome A, then it would remain to formally prove that the HPC works for all N. If yes -- conjecture falls. HPC is proven. If no -- conjecture stands. HPC is disproven. If outcome B, then it would remain to formally prove that the HPC works for all N. If yes -- conjecture stands. HPC is proven. If no -- conjecture stands. HPC is disproven. By analysis, I don't mean brute force. I mean analysis. (I apologize for winging it in the above ... it's late and I'm tired. ;-) -- Quinn === Subject: Re: Another twin primes conjecture >Conjecture: >Given a pair of twin primes p and p+2, it must be the case that you find >at least one more set of twin primes in the interval p^2 to (p+2)^2. > Well, unless I or Mathematica have made a mistake, there doesn't seem to > be a counter-example in the positive integers up to the twin primes > 155731907 and 155731909, which have the corresponding twin primes > 24252426857857019 and 24252426857857021. > n = 1; > success = True; > While[success, > n += 2; > If[PrimeQ[n] && PrimeQ[n + 2], > trial = False; > For[m = n^2, m <= (n + 2)^2, m += 2, > If[PrimeQ[m] && PrimeQ[m + 2], > trial = True; > Break[] > ] > ]; > success = success && trial > ] > ] > Ah, but now I see there is a ßaw in what I did. This allows (n+2)^2 > and (n+2)^2+2 as acceptable twin primes and that is not what the > conjecture stated. My apologies, I'll correct my mistake and post again > if it finds a counter-example. > Sorry Yes this is a serious mistake. Clearly (n+2)^2 is divisible by n+2 and thus is not a prime. If your program says that (n+2)^2 is part of a twin prime then it is majorly ßawed. -- Lance Lamboy I tell them the truth and they think it's hell. ~ Harry S. Truman === Subject: Re: Another twin primes conjecture This was the loop: > For[m = n^2, m <= (n + 2)^2, m += 2, > Yes this is a serious mistake. Clearly (n+2)^2 is divisible by n+2 and > thus is not a prime. If your program says that (n+2)^2 is part of a twin > prime then it is majorly ßawed. I do think that this program, though this minor ßaw, is correct. Your point rather proves that the loop can go up to (n+2)^2 without changing the result since (n+2)^2 is never part of a twin prime, and neither is (n+2)^2-1 (=(n+3)(n+1)). So that this little mistake does not change the result compared with what loop: > For[m = n^2, m <= (n + 2)^2-2, m += 2, would return. It only is a little bit slower, wasting time on these two extra numbers for each value of n but never stopping on them. Best, Ghislain === Subject: Re: Another twin primes conjecture > Conjecture: > Given a pair of twin primes p and p+2, it must be the case that you > find at least one more set of twin primes in the interval p^2 to > (p+2)^2. > Examples: > 5 and 7 are twin primes, and in the interval 25 to 49, you have 29 and > 31. > 11 and 13 are twin primes and in the interval 121 to 169 you have 137 > and 139. > 17 anbd 19 are twin prime and in the interval 289 to 361 you have 311 > and 313. If you could prove it, you would automatically have a proof for the twin prime conjecture that there is an infinite number of twin primes. That one is so far unproven. But why don't you conjecture this: For every n >= 1, there is at least one set of twin primes in the interval n^2 to (n+2)^2. Or do you have any reason to believe that counterexamples to this conjecture exist, if n is not the first number of two twin primes? === Subject: Re: Another twin primes conjecture > Conjecture: > Given a pair of twin primes p and p+2, it must be the case that you > find at least one more set of twin primes in the interval p^2 to > (p+2)^2. > Examples: > 5 and 7 are twin primes, and in the interval 25 to 49, you have 29 and > 31. > 11 and 13 are twin primes and in the interval 121 to 169 you have 137 > and 139. > 17 anbd 19 are twin prime and in the interval 289 to 361 you have 311 > and 313. > I realized it while looking at my prime counting function, which you > can see at > http://mathforprofit.blogspot.com/ > which I've had for a while. > James Harris It's SO COOL!!!! If you look at my function and think for a while, you may understand why I made that conjecture. My prime counting function defines the entire prime distribution. Every answer to every question about primes is wrapped up in it. James Harris === Subject: Re: Another twin primes conjecture good conjecture! shouldn't take long to prove, if one can generalize from the similar proof about ordinary primes; does that make sense?... isn't that proof a consequence of the proof of the infinity of primes ... which would result, if there's enough of a similarity, in a proof of the infinitude of twins; eh? > Given a pair of twin primes p and p+2, it must be the case that you > find at least one more set of twin primes in the interval p^2 to > (p+2)^2. --Give Earth a Trickier Dick Cheeny -- out of office, after gigayears! http://tarpley.net/bush12.htm http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/HowTheNation.PDF http://www.rand.org/publications/randreview/issues/rr.12.00/ http://www.rwgrayprojects.com/synergetics/plates/figs/plate02. html === Subject: Re: Another twin primes conjecture > Conjecture: > Given a pair of twin primes p and p+2, it must be the case that you > find at least one more set of twin primes in the interval p^2 to > (p+2)^2. I don't know if it's true or not, but it's nice to see you posting this as a conjecture. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Another twin primes conjecture >Conjecture: >Given a pair of twin primes p and p+2, it must be the case that you >find at least one more set of twin primes in the interval p^2 to >(p+2)^2. > I don't know if it's true or not, but it's nice to see you posting this > as a conjecture. > -- > Will Twentyman Normally, I'd expect the reservation of the term Conjecture to those very few statements that one does not yet have a proof for, but for which one has demonstrable and sufficient evidence and definitely no counter-examples. However, since James stated that he formed the statement based upon his observations about his counter, which to date has held up, and therefore, even though, as I just mentioned in a message to C. Bond, the counter is not formally part of the Conjecture itself -- I'll grant this one Conjecture status until one of the following occurs: 1. A counterexample is found. 2. A formal proof of its correctness/incorrectness is found. If the counter itself finds a counter-example -- then this one must then reasonably fall into Open Question territory rather than conjecture territory. As has already been noticed, it can be shown by induction that if the Harris Conjecture is shown to be true, the infinity of twin primes is a consequence. ... Well, that's a very interesting consequence. In as much as the potential implication of the Harris Conjecture is indeed mathematically interesting .... in the interest of mathematics, and because I like James Harris, I am going to offer a suggestion to all concerned. Let's all (you too, James) keep this one thread posture-free. Deep breath. No politics, history, or emotions (however warranted) in just this *one* thread. Then we can all look at this conjecture at face value, res ipsa loquitor. If the Conjecture stands -- great. If it falls -- no harm no foul. Let math be math here. -- Quinn === Subject: Re: Another twin primes conjecture 6$Pk3.82582@pd7tw1no>... I'll grant this one Conjecture > status until one of the following occurs: > 1. A counterexample is found. > 2. A formal proof of its correctness/incorrectness is found. > If the counter itself finds a counter-example -- then this one must then > reasonably fall into Open Question territory rather than conjecture > territory. If the counter finds a counterexample then the conjecture is false, not an open question. === Subject: Re: Another twin primes conjecture > status until one of the following occurs: > 1. A counterexample is found. > 2. A formal proof of its correctness/incorrectness is found. > If the counter itself finds a counter-example -- then this one must > then reasonably fall into Open Question territory rather than > conjecture territory. > an open question. I thought he was expressing a lack of confidence in Harris's counter program. He's saying that if Harris's program finds a counterexample there is a good chance that it is bogus and the conjecture is still open. -- Lance Lamboy I tell them the truth and they think it's hell. ~ Harry S. Truman === Subject: Re: Another twin primes conjecture >If the counter finds a counterexample then the conjecture is false, not >an open question. > I thought he was expressing a lack of confidence in Harris's counter > program. He's saying that if Harris's program finds a counterexample > there is a good chance that it is bogus and the conjecture is still open. I was saying, if there is a counterexample that shows the *counter* is wrong, the conjecture is still open. If the counter is correct (which I believe it is), and it finds some pi((p+2)^2)- pi(p^2) such that the delta is less than 2, then the conjecture is wrong. If it doesn't find such a spot -- then the conjecture is open. [note -- is this assumption of mine about the delta of those two pi()'s even correct?] I have no reason to lack confidence in the counter. I was musing, it was late ... my wording may have been muddled. I'll stick to optimizing the thing and keep out of the number theory -- which is NOT in any way my bag. (Foot in mouth -- good chance of that.) Making it faster -- I can deal with. Have a ßu today, am working on making the counter as posted faster. -- Quinn === Subject: Re: Another twin primes conjecture >If the counter finds a counterexample then the conjecture is false, >not an open question. > I thought he was expressing a lack of confidence in Harris's counter > program. He's saying that if Harris's program finds a counterexample > there is a good chance that it is bogus and the conjecture is still > open. > I was saying, if there is a counterexample that shows the *counter* is > wrong, the conjecture is still open. OK. I misinterpreted. > If the counter is correct (which I believe it is), and it finds some > pi((p+2)^2)- pi(p^2) such that the delta is less than 2, then the > conjecture is wrong. If it doesn't find such a spot -- then the > conjecture is open. [note -- is this assumption of mine about the delta > of those two pi()'s even correct?] If I understand you correctly, you are looking for a p such that the interval p^2, (p+2)^2 contains either 0 or 1 prime numbers. Since p^2-1, p^2, (p+2)^2-1, and (p+2)^2 are all clearly not prime no integer in the interval could be part of a twin prime set (even if the prime were near the border of the interval). This is a sufficient but not necessary condition for the conjecture and I do not see how using it would be advantageous in the search for a counter example. > I have no reason to lack confidence in the counter. I was musing, it was > late ... my wording may have been muddled. I'll stick to optimizing the > thing and keep out of the number theory -- which is NOT in any way my > bag. (Foot in mouth -- good chance of that.) Making it faster -- I can > deal with. > Have a ßu today, am working on making the counter as posted faster. > -- > Quinn -- Lance Lamboy I tell them the truth and they think it's hell. ~ Harry S. Truman === Subject: Re: Another twin primes conjecture Lance said: > If I understand you correctly, you are looking for a p such that the > interval p^2, (p+2)^2 contains either 0 or 1 prime numbers. Since p^2-1, > p^2, (p+2)^2-1, and (p+2)^2 are all clearly not prime no integer in the > interval could be part of a twin prime set (even if the prime were near > the border of the interval). Yeah, well ... a bit of spelunking showed me in spades that one is not going find such a set of integers such that there is only one prime between them, because althrough, given some range of integers i^2, (i+2)^2, although there are minima, there are no minima even vaguely approaching the one (1) I suggested. But, it did make me ask myself if Hpi(i)-Hpi(i-1) was perhaps a primality test of i, and if so, if the HPC could be modified in any way to account for the fact that the second parameter has already been calculated in arriving at the result of the first. -- Quinn === Subject: Re: Another twin primes conjecture > Lance said: > If I understand you correctly, you are looking for a p such that the > interval p^2, (p+2)^2 contains either 0 or 1 prime numbers. Since > p^2-1, p^2, (p+2)^2-1, and (p+2)^2 are all clearly not prime no integer > in the interval could be part of a twin prime set (even if the prime > were near the border of the interval). > Yeah, well ... a bit of spelunking showed me in spades that one is not > going find such a set of integers such that there is only one prime > between them, because althrough, given some range of integers i^2, > (i+2)^2, although there are minima, there are no minima even vaguely > approaching the one (1) I suggested. It did not sound very promising to me. > But, it did make me ask myself if Hpi(i)-Hpi(i-1) was perhaps a > primality test of i, and if so, if the HPC could be modified in any way > to account for the fact that the second parameter has already been > calculated in arriving at the result of the first. Could you refresh me on what Hpi and HPC are. I have no clue. > -- > Quinn -- Lance Lamboy I tell them the truth and they think it's hell. ~ Harry S. Truman === Subject: Re: Another twin primes conjecture > Could you refresh me on what Hpi and HPC are. I have no clue. Just my abbreviations for Harris' Prime Counter. Turns out I was likely correct about my assumption that Hpi(i)-Hpi(i-1) being a quick primality test. Because of the optimization I made, doing two counts in a row like that, with the larger number first, behaves with O(n) (or thereabouts) complexity. And it certainly appears to be more elegant than some of the Primality Tests I've seen, q.v.: http://primes.utm.edu/curios/includes/file.php?fil e= primetest.html because it reduces to a simple delta of pi(i) functions. Of course, that means that a twin prime test reduces to a highly optimized delta test as well. I'm double checking my results before I post the template-ized version of the HPC (it includes a isprime(n) and istwinprime(n, n2) function that I also have to double check, so it's taking a bit longer than I expected -- that and I've got the ßu -- but bear with me and the code will come). -- Quinn === Subject: Re: Another twin primes conjecture > Conjecture: > Given a pair of twin primes p and p+2, it must be the case that you > find at least one more set of twin primes in the interval p^2 to > (p+2)^2. > Examples: > 5 and 7 are twin primes, and in the interval 25 to 49, you have 29 and > 31. > 11 and 13 are twin primes and in the interval 121 to 169 you have 137 > and 139. > 17 anbd 19 are twin prime and in the interval 289 to 361 you have 311 > and 313. OK. How about this conjecture, using the same line of reasoning and the same number of examples: Conjecture: All odd numbers greater than 1 are prime. Examples: 3 is a prime. 5 is a prime. 7 is a prime. ... > I realized it while looking at my prime counting function, which you > can see at > http://mathforprofit.blogspot.com/ > which I've had for a while. Yeah, and it's still WRONG! > James Often in error, but never in doubt! Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Another twin primes conjecture > Conjecture: > Given a pair of twin primes p and p+2, it must be the case that you > find at least one more set of twin primes in the interval p^2 to > (p+2)^2. > Examples: > 5 and 7 are twin primes, and in the interval 25 to 49, you have 29 and > 31. > 11 and 13 are twin primes and in the interval 121 to 169 you have 137 > and 139. > 17 anbd 19 are twin prime and in the interval 289 to 361 you have 311 > and 313. > OK. How about this conjecture, using the same line of reasoning and the > same number of examples: > Conjecture: > All odd numbers greater than 1 are prime. > Examples: > 3 is a prime. > 5 is a prime. > 7 is a prime. > ... That was a remarkably idiotic response. -- --Tim Smith === Subject: Re: Another twin primes conjecture > That was a remarkably idiotic response. Not really. His point was that conjecture plus three supporting examples does not constitute a proof. === Subject: Re: Another twin primes conjecture > That was a remarkably idiotic response. > Not really. His point was that conjecture plus three supporting > examples does not constitute a proof. Harris didn't offer his examples as proof. He offered them as examples to illustrate the conjecture. They were correct and appropriate, and the response was indeed idiotic. -- --Tim Smith === Subject: Re: Another twin primes conjecture >Conjecture: >Given a pair of twin primes p and p+2, it must be the case that you >find at least one more set of twin primes in the interval p^2 to >(p+2)^2. >Examples: >5 and 7 are twin primes, and in the interval 25 to 49, you have 29 and >31. >11 and 13 are twin primes and in the interval 121 to 169 you have 137 >and 139. >17 anbd 19 are twin prime and in the interval 289 to 361 you have 311 >and 313. > OK. How about this conjecture, using the same line of reasoning and the > same number of examples: > Conjecture: > All odd numbers greater than 1 are prime. > Examples: > 3 is a prime. > 5 is a prime. > 7 is a prime. > ... >I realized it while looking at my prime counting function, which you >can see at >http://mathforprofit.blogspot.com/ >which I've had for a while. > Yeah, and it's still WRONG! Actually, his prime counting function worked. It just didn't work as fast as others. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Another twin primes conjecture > Actually, his prime counting function worked. It just didn't work as > fast as others. How much slower was it than the others? Just a half hour ago, I spent 20 minutes making one optimization and made it 13 times faster than the original counting primes pi(1,000,000) and 30 times faster counting primes pi(10,000,000) and 46 times faster than the original with pi(100,000,000), with a reasonable space complexity trade off. I found a property of his pi function that lent itself to to an elegant and very well known optimization. Took me all of 20 minutes to get those kinds of speed ups. So it's not as slow as it's made out to be. Just nobody with any optimization experience offered to help make it faster, maybe? James -- with your permission, I'll post my optimized version of your original. -- Quinn === Subject: Re: Another twin primes conjecture >Actually, his prime counting function worked. It just didn't work as >fast as others. > How much slower was it than the others? Check the time complexity of the algorithm and compare that to the best algorithms to date. If I remember right, James' algorithm is O(n). Lagarias-Odlyzko have gotten it down to O(sqrt(n) + eps). Moreove, programs running in the 1980's were much faster than James' current algorithm. > Just a half hour ago, I spent 20 minutes making one optimization and made it > 13 times faster than the original counting primes pi(1,000,000) and 30 times > faster counting primes pi(10,000,000) and 46 times faster than the original > with pi(100,000,000), with a reasonable space complexity trade off. Did you improve the time complexity? What is the time taken for pi(1000000000000000)=29844570422669 ? In 1985 it took 165 minutes on an IBMM/370 3081-K. -- 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: Another twin primes conjecture >Actually, his prime counting function worked. It just didn't work as >fast as others. How much slower was it than the others? > Check the time complexity of the algorithm and compare that to the best > algorithms to date. If I remember right, James' algorithm is O(n). > Lagarias-Odlyzko have gotten it down to O(sqrt(n) + eps). I am not familiar with the eps notation. What does that mean? Could it better be described as O(sqrt(n))? > Moreove, programs running in the 1980's were much faster than James' > current algorithm. > Just a half hour ago, I spent 20 minutes making one optimization and made it > 13 times faster than the original counting primes pi(1,000,000) and 30 times > faster counting primes pi(10,000,000) and 46 times faster than the original > with pi(100,000,000), with a reasonable space complexity trade off. > Did you improve the time complexity? If I understand him correctly it was O(n) and it remains O(n). > What is the time taken for pi(1000000000000000)=29844570422669 ? > In 1985 it took 165 minutes on an IBMM/370 3081-K. -- Lance Lamboy I tell them the truth and they think it's hell. ~ Harry S. Truman === Subject: Re: Another twin primes conjecture > Did you improve the time complexity? I'm looking at that. > What is the time taken for pi(1000000000000000)=29844570422669 ? I'll do some more testing, and get back with some results. I'm making some other aesthetic modifications, like templatizing my modified version so it can be used on arbitrarily big numbers (as long as the type supports T sqrt T). -- Quinn === Subject: Re: Another twin primes conjecture >Yeah, and it's still WRONG! > Actually, his prime counting function worked. It just didn't work as > fast as others. You are correct. But he repudiated his own prime counting function by his insistence that Ôsqrt' is inherently ambiguous and that the ambiguity is not removable. Since he uses it in his prime counting function (actually algorithm) and since the results are only correct for positive values of the square root, he depends on a construct he renounces. (Not that that bothers him!) -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Another twin primes conjecture > You are correct. But he repudiated his own prime counting function by his > insistence that Ôsqrt' is inherently ambiguous and that the ambiguity is not > removable. Since he uses it in his prime counting function (actually > algorithm) and since the results are only correct for positive values of the > square root, he depends on a construct he renounces. (Not that that bothers > him!) Actually, unless I read the Harris Conjecture incorrectly: Given a pair of twin primes p and p+2, it must be the case that you find at least one more set of twin primes in the interval p^2 to (p+2)^2. the Conjecture doesn't rely on the counting function. James said simply that he came up with it while observing his counting function. So -- even if he were to renounce the counter -- the Conjecture stands on its own -- the counter is not stated as being a Lemma in the Conjecture proper. -- Quinn === Subject: Re: Another twin primes conjecture > Actually, unless I read the Harris Conjecture incorrectly: > Given a pair of twin primes p and p+2, it must be the case that you > find at least one more set of twin primes in the interval p^2 to > (p+2)^2. > the Conjecture doesn't rely on the counting function. James said simply that > he came up with it while observing his counting function. > So -- even if he were to renounce the counter -- the Conjecture stands on > its own -- the counter is not stated as being a Lemma in the Conjecture > proper. > -- > Quinn Actually, if you read the thread properly, you would observe that I was responding to Will Twentyman's observation that the prime counting function worked. I made no reference whatsoever to the correctness or incorrectness of the Twin Prime Conjecture in that post. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: My New Email Address Take the old one which is automatically included in my posts and perform the following transformations: 1. Move the r from the beginning to the end of the part before the @, keeping all the other letters in their old order. 2. Insert alum. before the part after the @. This will now allow you to send email to me on a much longer term basis. If you corect the confusion of others with respect to these instructions, DO NOT do so by including my new email address in a post! ---- David === Subject: Re: behavior of the modulus in a complex pde > Consider the differential equation > psi_xx - i*a*psi_t = 0 > where psi(x,t) is a complex function, a a real constant, and _xx and > _t denote partial differentiation. I'm interested in solutions in the > form > psi(x,t) = f*exp(ig) > where f(x,t), g(x,t) are real functions. > Now, any complex function psi(x,t) can be written in this way, but how > much freedom so we have in choosing the modulus f a priori and then > adjusting g to fit? Differentiating yields a pair of coupled > non-linear equations in f and g, but not much illumination. > Suggestions on how to attack this question? I've got to admit, you are trying. Why are you suggesting a method of solution when you have no idea? In fact your suggestion is not a very good one. By far the most elegant method of solution is to convert the PDE into an algebraic equation by Fourier transformation. Your equation then becomes: (2*pi*k^2-a*f)*Fpsi(k,f) = 0 The 2D Fourier transform of psi can only be nonzero along the parabolic locus a*f = 2*pi*k^2. The general solution can then be expressed in terms of an arbitrary function C: Fpsi(k,f) = C(k)*delta(2*pi*k^2-a*f) It is apparent by inspection that C(k) is the 1D Fourier transform of the initial spatial distribution of psi. It is a simple matter to obtain the solution you want by inverse Fourier transformation. Zigoteau. === Subject: Re: re by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5HJUIl11836; >Star Trek II: The Wrath of Khan!! Correct! Give that person a prize! === Subject: Re: James Harris proved the Riemann Hypothesis! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5HJmBu13550; >... >So, James Harris, Quinn, M. Basti, and >other cranks, don't make public assertions >you can't (or won't) back up with verifiable >facts. You invite skepticism and disbelief. James Harris is Nuts(TM) > I am sorry. I didn't read the latter part carefully. But why do you take > their assertions seriously? As you may know, the system is quite > simple: > 1. A person wants a criticism and so posts his/her file. > 2. A second person makes a criticism. > 3. The first person accepts the second person's argument, and > tries to defend his/her work. > 4. In cycles of 2 and 3, if an agreement between the first and second >person are made, then the end of discussion follows. > In repetition of 2 and 3, there has to be a stage where the second > person can't defend his/her argument anymore. THEN THAT'S IT. > Personally, I tend to argue against those who use the word cranks. > I just hate to discourage people. But if people can't follow the rules > above, then I (or we) must use that word. > If we want this newsgroup to be a professional, PRESTIGIOUS news- > group, then we MUST IGNORE THEM. Or there is no point in > criticizing their crankiness seriously, like TM is doing. >JSH threads sometimes diverge into another dimension after the first (or >later) cycle on 2: >2A: The first person makes paranoid accusations of conspiracy. >2B: The second person denies this. >After a few cycles, we sometimes see an additional dimension: >2-alpha: The first person makes obscene threats against his critics. >2-beta: The critics respond in kind. Okay, got it. You are just having fun. Sorry for being serious. erdos fan === Subject: Re: Factoring paper is wrong by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5HJxiT14703; > let a+1=b+1, >therefore a = b >Proving statements such as the above don't necessarily invite passionate >minds to tackle them. Certain categories of maths problems invite certain >passionate types to try to best them: >Let X be some arbitrary integer s.t. N > 4. > int N = X; > while((N = N&1 ? 3*N+1 : N/2) != 1); > (1.0) >Prove that 1.0 terminates for any X. >All of a sudden, all kinds of personalities get into the mix. > Why do people who are totally lacking in mathematical maturity and > mathematical education think that somehow they are going to find > a magic proof that has escaped experts? >Think about how a lot of people view mathematics. I remember taking >algebra classes in high school and hearing about this frightening course >called calculus. At that point in time, the only thing any of us knew >of was calculus and statistics. Things like real analysis, complex >analysis, formal logic, set theory, combinatorics, etc. were things I >hadn't even conceived of existing. >This is probably the perspective most people have. They're good at high >school algebra, so they think they have the potential to make deep >insights into math, just like the other people who have degrees. They >have no concept of the fact that they are frighteningly ignorant. > I doubt whether these same people would ever considering doing (say) > brain surgery. Yet they are just as ignorant about one subject as the > other. >To extend the analogy, these people would view brain surgery as >something a butcher could do with a little practice. They think >advanced mathematics is just trickier algebraic manipulations. > I am sure that these people recognize that they do not have the > education or experience to do brain surgery. Why is it that they do > NOT have the same cognizance about their lack of math skills? Why > do they stubbornly cling to ideas that experts point out are wrong? > Instead, they attack the experts! I doubt that they would attack > a doctor who told them that they had no surgical skills. >For the same reason that people who know they can't walk a tightrope >will believe they can write a great novel... with sentences that >frequently don't have verbs. A skill with a physical component is >obviously tricky. A skill that is purely mental is open to inaccurate >self-evaluation. > If one argues that *unsolved* problems attracts such people, why > don't we get people who claim that they have found a cure for (say) > cerebral palsy? That too is an unsolved problem. >Because there is objective evidence for whether a cerebral palsy cure >works. They view their proofs through a subjective standard. > What is it about math in particular that attracts incompetents to > believe that they can solve problems that experts can not? Why do > they then attack experts who tell them that they are wrong? >Because they think they know the rules of the game, when they don't. >-- >Will Twentyman >email: wtwentyman at copper dot net Also, mathematicians and scientists have made math and science appear easy, when it's not. That is, you streamline and improve the methods so much, outsiders think that you move a few symbols around and, voila, powerful results follow. I'll never forget a graduate course in advanced algebra, when the teacher applied abstract group theory to explain, with simple equations, rotations in space and solutions to Rubik's cube. In another example, Abel used elementary methods and struggled with proving that the quintic and higher were impossible with radicals, but along comes Galois and group theory which reduced the problem to finding whether, for example, 5 is a factor of 60 (I'm abbreviating from memory an exposition I saw in Scientific American two decades ago). And in science, people think relativity boils down to just E=mc^2 and it has to do with atomic energy, when there is way alot more involved to derive E=mc^2 by Lorentz transformations and other issues, such as length, mass, momentum, which Einstein and a generation of physicists had to grapple with to rework our notions of space, time, mass, and energy. Perhaps that's why there's so many cranks attempting to rewrite mathematics and relativity - any idiot can move some symbols around on paper. So professional mathematicians and scientists, it's a lot (sarcasm)! Anthony J. Natoli === Subject: Re: Factoring paper is wrong Anthony Natoli said: > Also, mathematicians and scientists have made math > and science appear easy, when it's not. Math and science appear easy? Wow -- what text books have you been reading? The ones I've read make me doubt whether or not I should be certain that 1 + 1 = 2 for any value of 1 and 2. -- Quinn === Subject: Re: Factoring paper is wrong > Anthony Natoli said: > Also, mathematicians and scientists have made math and science appear > easy, when it's not. > Math and science appear easy? Wow -- what text books have you been > reading? The ones I've read make me doubt whether or not I should be > certain that 1 + 1 = 2 for any value of 1 and 2. Yes, in general 1+1<>2. The 1+1=2 fallacy came about because people correctly observed that 1+1=2 when 1=1 and 2=2 and then incorrectly generalized to other values. > -- > Quinn -- Lance Lamboy I tell them the truth and they think it's hell. ~ Harry S. Truman === Subject: lie algebra I would like to learn Lie Algebra on my own (I have background in graduate level algebra). A book recommendation will be appreciated a lot. === Subject: Re: lie algebra Try this one: Lectures on Lie Groups and Lie Algebras Roger Carter, Graeme Segal, Ian MacDonald Cambridge, 1995 (London Mathematical Society Student Texts 32) ISBN 0 521 49579 2 (hb) 0 521 49922 4 (pb) Also, there is one by Humphreys from the Springer press, but I don't have particulars. >I would like to learn Lie Algebra on my own (I have background in >graduate level algebra). A book recommendation will be appreciated a >lot. === Subject: Re: lie algebra > I would like to learn Lie Algebra on my own (I have background in > graduate level algebra). A book recommendation will be appreciated a > lot. For a great source of all your Lie Algebra-related needs, please consult all posts by James Harris. I would hesitate to even put the words lie and algebra in the same sentence without giving him credit. === Subject: Re: lie algebra > I would like to learn Lie Algebra on my own (I have background in > graduate level algebra). A book recommendation will be appreciated a > lot. Try Representation theory, by William Fulton and Joe Harris. Jose Carlos Santos === Subject: Algebraic integer mistake, basic Gauss began work on numbers that while not integers had interesting properties, and in his honor those numbers were called gaussian integers. They are numbers of the form a+bi, where Ôa' and Ôb' are integers. They don't have any problems. Later, mathematicians considered numbers that are defined to be roots of monic polynomials with integer coefficients. They have a problem if you believe that the ring so defined includes all numbers that have a key property. It turns out that you can include the ring of algebraic integers by having the limiting property for a ring that only -1 and 1 are integer units in that ring. If you play with it for a while you will find that you can't find an algebraic integer such that you can get any other units in the ring of algebraic integers other than -1 or 1. However, there are numbers in a ring so limited which are provably not algebraic integers. That has interesting consequences if you believe that they do not exist. For over a hundred years, mathematicians did not know they existed. I proved they do. As usually happens with a major result of this size, some people have dedicated themselves to fighting mathematical truth. You have seen how far they are willing to go. When my paper Advanced Polynomial Factorization was to be published, they conspired on this newsgroup to mount an email campaign against it, and managed to sway the chief editor Ioannis Argyros to censor my paper. http://rattler.cameron.edu/swjpam/vol2-03.html He will probably lose his career. These type people are seen repeatedly in human history. When some new truth that they don't like comes along, they attack it. They attack the messenger. They attack anyone who supports it, just like Hall, Magidin, and Decker conspired with others, or Erik Max Francis, a former sci.math posters, maintains his webpage at Crank.net, calling me names. If you don't believe what's right in front of your eyes, do the math. Consider a ring where only -1 and 1 are integer units. Now ask yourself, can anyone prove that the ring of algebraic integers is coincident with that ring? And I can tell you that I can prove that it is not. You sat by and one man has fallen, one of your own, Ioannis Argyros, who has yet to fully face the music. How many more of you will fall before it's all over? James Harris === Subject: Re: Algebraic integer mistake, basic > Later, mathematicians considered numbers that are defined to be roots > of monic polynomials with integer coefficients. > They have a problem if you believe that the ring so defined includes > all numbers that have a key property. What do you mean with a key property? > It turns out that you can include the ring of algebraic integers by > having the limiting property for a ring that only -1 and 1 are integer > units in that ring. What do you mean with integer units? > If you play with it for a while you will find that you can't find an > algebraic integer such that you can get any other units in the ring of > algebraic integers other than -1 or 1. How about 2499 + 50.sqrt(2498)? It has been shown (by Nora, I think) that the units in the ring of algebraic integers form a dense subset. > However, there are numbers in a ring so limited which are provably not > algebraic integers. Dedekind has shown otherwise. When Ôa' and Ôb' are algebraic integers, both Ôa + b' and Ôa * b' are algebraic integers. So how do you come at numbers in the ring that are *not* algebraic integers? So what way, using only the ring operations Ô+' and Ô*', do you come at numbers that are *not* algebraic integers? > I proved they do. Nope, you did not. You only asserted. > Consider a ring where only -1 and 1 are integer units. What are integer units? -- 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: Algebraic integer mistake, basic > Gauss began work on numbers that while not integers had interesting > properties, and in his honor those numbers were called gaussian > integers. > They are numbers of the form a+bi, where Ôa' and Ôb' are integers. > They don't have any problems. > Later, mathematicians considered numbers that are defined to be roots > of monic polynomials with integer coefficients. > They have a problem if you believe that the ring so defined includes > all numbers that have a key property. What property? Perhaps the problem is not with the algebraic integers but with your belief that they have this property. > It turns out that you can include the ring of algebraic integers by > having the limiting property for a ring that only -1 and 1 are integer > units in that ring. What is the limiting property? > Consider a ring where only -1 and 1 are integer units. Ok, How about ({-1, 0, 1},+,*) where it is isomorphic to Z_3? > Now ask yourself, can anyone prove that the ring of algebraic integers > is coincident with that ring? I think it is pretty trivial to show that they aren't. > And I can tell you that I can prove that it is not. Ok. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Algebraic integer mistake, basic > Gauss began work on numbers that while not integers had interesting > properties, and in his honor those numbers were called gaussian > integers. > They are numbers of the form a+bi, where Ôa' and Ôb' are integers. > They don't have any problems. > Later, mathematicians considered numbers that are defined to be roots > of monic polynomials with integer coefficients. > They have a problem if you believe that the ring so defined includes > all numbers that have a key property. What property? > It turns out that you can include the ring of algebraic integers by > having the limiting property for a ring that only -1 and 1 are integer > units in that ring. Include the ring of algebraic integers? In what? There are an infinite number of units in the ring of algebraic integers, but only those two integer units. So what? What ring are you talking about, and why have you changed the subject? > If you play with it for a while you will find that you can't find an > algebraic integer such that you can get any other units in the ring of > algebraic integers other than -1 or 1. There are an infinite number of units in the ring of algebraic integers. It doesn't matter what you play. > However, there are numbers in a ring so limited which are provably not > algebraic integers. What ring? > That has interesting consequences if you believe that they do not > exist. If we believe that *what* do not exist? > For over a hundred years, mathematicians did not know they existed. Mathematicians did not know that *what* existed? > I proved they do. Do they? You haven't even made a coherent statement about what *they* are. > As usually happens with a major result of this size, some people have > dedicated themselves to fighting mathematical truth. That's *your* specialty. > You have seen how far they are willing to go. > When my paper Advanced Polynomial Factorization was to be published, > they conspired on this newsgroup to mount an email campaign against > it, and managed to sway the chief editor Ioannis Argyros to censor my > paper. > http://rattler.cameron.edu/swjpam/vol2-03.html > He will probably lose his career. Now look what you've done! > These type people are seen repeatedly in human history. When some new > truth that they don't like comes along, they attack it. They attack > the messenger. They attack anyone who supports it, just like Hall, > Magidin, and Decker conspired with others, or Erik Max Francis, a > former sci.math posters, maintains his webpage at Crank.net, calling > me names. > If you don't believe what's right in front of your eyes, do the math. > Consider a ring where only -1 and 1 are integer units. Any ring? How about the ring of integers -- or the ring of algebraic integers. > Now ask yourself, can anyone prove that the ring of algebraic integers > is coincident with that ring? Of course not. The ring of integers and the ring of algebraic integers are distinct. > And I can tell you that I can prove that it is not. Not necessary. It is a trivial result. > You sat by and one man has fallen, one of your own, Ioannis Argyros, > who has yet to fully face the music. > How many more of you will fall before it's all over? Before *what's* over? > James Often in error, but never in doubt! Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Algebraic integer mistake, basic > I proved they do. You are incapabable of proving anything. Bob Kolker === Subject: Re: Algebraic integer mistake, basic I proved they do. > You are incapabable of proving anything. That is not quite true. So far he proved that he has no clue of maths, and that sci.math is a more peaceful place without him. === Subject: Re: Algebraic integer mistake, basic > I proved they do. You are incapabable of proving anything. > That is not quite true. So far he proved that he has no clue of maths, > and that sci.math is a more peaceful place without him. I've directly refuted claims of a counterexample to my paper Advanced Polynomial Factorization. I've noted that such claims depend on the hidden assumption that all infinite rings where -1 and 1 are the only integers that are units are in fact part of the ring of algebraic integers, when provably they are not. But as I've repeatedly shown if you believe that they are you can come up with a specious claim of counterexample based on your hidden assumption. Here's how it's done: Consider a number u_1 that is an algebraic integer and u_2 that is not but is a member of an infinitely sized ring where -1 and 1 are the only integer units, where u_1 u_2 = 1. Now then, in that ring u_1 u_2 = 1, and u_1 and u_2 are both units as they are factors of 1. But one is an algebraic integer, while one is not. Therefore, in the ring of algebraic integers, u_1 is NOT a unit. It turns out that it's impossible to prove that u_2 does not exist in some infinite sized ring where -1 and 1 are the only integer units. Now it's easy for me to shoot down people attacking my proof. The mathematics and the logic are on my side. But these people cheat. They have a gang that just claims that I'm wrong, ignores the facts, and refuses to be logical, and they post a lot. James Harris === Subject: Re: Algebraic integer mistake, basic Discussion, linux) > Consider a number u_1 that is an algebraic integer and u_2 that is not > but is a member of an infinitely sized ring where -1 and 1 are the > only integer units, where u_1 u_2 = 1. > Now then, in that ring u_1 u_2 = 1, and u_1 and u_2 are both units as > they are factors of 1. > But one is an algebraic integer, while one is not. > Therefore, in the ring of algebraic integers, u_1 is NOT a unit. James, this is a rather longwinded way to prove the following obvious fact. Let R and S be rings, with R c S. Let x in R and y in S R be given such that x y = 1 (in S, obviously). Then x is not a unit in R. The claim is obvious. Suppose that x *is* a unit in R. Then let z be its inverse in R. By uniqueness of inverses in S, we see that z = y, and so y is in R, contrary to assumption. (Someone can now fix any stupid errors or unstated assumptions I may have above.) Anyway, curiously I don't think this proves your point. -- Jesse F. Hughes If the car stops and you're not getting out, then you have to start it again. -- Quincy P. Hughes on his father's skills with a manual transmission. === Subject: Re: Algebraic integer mistake, basic > I've directly refuted claims of a counterexample to my paper Advanced > Polynomial Factorization. > I've noted that such claims depend on the hidden assumption that all > infinite rings where -1 and 1 are the only integers that are units are > in fact part of the ring of algebraic integers, when provably they are > not. As we see below, your refutation consists of the hidden assumption that your conclusion is true. > But as I've repeatedly shown if you believe that they are you can come > up with a specious claim of counterexample based on your hidden > assumption. > Here's how it's done: > Consider a number u_1 that is an algebraic integer and u_2 that is not > but is a member of an infinitely sized ring where -1 and 1 are the > only integer units, where u_1 u_2 = 1. OK. You've made the *assumption* here that there *is* (does exist) a number, u_2, which is a member of an infinitely sized ring...yada, yada... How's that for assuming a conclusion? > Now then, in that ring u_1 u_2 = 1, and u_1 and u_2 are both units as > they are factors of 1. It's implicit in the problem statement that *if* such a ring exists, u_1 and u_2 are units.. > But one is an algebraic integer, while one is not. That's already stated in the problem statement. > Therefore, in the ring of algebraic integers, u_1 is NOT a unit. You already said that in the problem statement. > It turns out that it's impossible to prove that u_2 does not exist in > some infinite sized ring where -1 and 1 are the only integer units. It turns out that you are simply restating the original, unproven assumption. > Now it's easy for me to shoot down people attacking my proof. The > mathematics and the logic are on my side. How? By assuming your conclusion? > But these people cheat. They have a gang that just claims that I'm > wrong, ignores the facts, and refuses to be logical, and they post a > lot. Your above math presentation proves nothing. > James Often in error, but never in doubt! Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Algebraic integer mistake, basic > That is not quite true. So far he proved that he has no clue of maths, > and that sci.math is a more peaceful place without him. But less interesting. I have found that one needs to read no more than 4 distinct newsgroups daily (By distinct I mean that there is a low incidence of cross-posting among them) to be assured of reading current loon-threads. For example, during the quiet JSH-free interval just passed on sci.math, the residnt loon on rec.skiing.alpine was especially active. Now things have reversed. === Subject: Re: Algebraic integer mistake, basic >That is not quite true. So far he proved that he has no clue of maths, >and that sci.math is a more peaceful place without him. > But less interesting. > I have found that one needs to read no more than 4 distinct newsgroups daily > (By distinct I mean that there is a low incidence of cross-posting among > them) to be assured of reading current loon-threads. For example, during > the quiet JSH-free interval just passed on sci.math, the residnt loon on > rec.skiing.alpine was especially active. Now things have reversed. Almost all newsgroups seem to have a resident loon (or two). The only exception that I've seen is comp.lang.fortran. There is a nasty abusive character there, but that is a bit different. === Subject: JSH: The Hammer is coming I have my prime counting function and the paper Advanced Polynomial Factorization and those are my minor results, small enough that enough people can understand them to break through the barriers I face. But I have two bigger results, and one is The Hammer. The Hammer is coming, and when I finally fully unleash it, then the establishment will come down. And some of you seem to have thought that it mattered that you seemed to win for so long, as you thought I was gone, beaten, put down and defeated, until that paper showed up in the Southwest Journal of Pure and Applied Mathematics. But you see, few of you were smart enough to understand that time is on my side. Time is on my side. James Harris === Subject: Re: The Hammer is coming > I have my prime counting function and the paper Advanced Polynomial > Factorization and those are my minor results, small enough that enough > people can understand them to break through the barriers I face. > But I have two bigger results, and one is The Hammer. > The Hammer is coming, and when I finally fully unleash it, then the > establishment will come down. > And some of you seem to have thought that it mattered that you seemed > to win for so long, as you thought I was gone, beaten, put down and > defeated, until that paper showed up in the Southwest Journal of Pure > and Applied Mathematics. > But you see, few of you were smart enough to understand that time is > on my side. > Time is on my side. James, in December 2002 you made this post: === ] Subject: JSH: Ok, I'm a loser ] ] It's finally settled in that I'm just some pathetic loser. If I ] weren't so pathetic I'd just go away gracefully, but I'll send one ] more post, or who am I kidding, my patheticness is so great that I'll ] probably post yet again. ] ] I'm disgusting. I'm just a pile of . I should just die like so ] many of you have said. I hate myself. I despise this life. I'm ] nothing but a sick joke to be made fun of by those of you who have ] real educations. People who actually know something, when I know ] nothing. I'm just nothing. ] ] If I hadn't been such a disgusting human being I'd have come to this ] realization years ago instead of wasting your time. ] ] My life is nothing. I know nothing. I'm worth nothing. I'm just ] . ] ] Please forgive me. All your attacks were justified. ] ] James Harris But I can't find it any more in Google. Although the rest of the JSH: Ok, I'm a loser thread is there. Any idea what happened to that post? -- Clive Tooth http://www.clivetooth.dk === Subject: Re: The Hammer is coming > But I can't find it any more in Google. > Although the rest of the JSH: Ok, I'm a loser thread is there. > Any idea what happened to that post? I suspect that it was deleted by James Harris himself. He can do it, sense it was posted by him and it was posted through Google. What makes me suspect that is the fact that the same thing happened to other threads; see, for instance, the thread Narcissistic Personality Disorder: Reason for posting?, which appeared in December 2002. Jose Carlos Santos === Subject: Re: The Hammer is coming >But I can't find it any more in Google. >Although the rest of the JSH: Ok, I'm a loser thread is there. >Any idea what happened to that post? > I suspect that it was deleted by James Harris himself. He can do it, > sense it was posted by him and it was posted through Google. What makes > me suspect that is the fact that the same thing happened to other > threads; see, for instance, the thread Narcissistic Personality > Disorder: Reason for posting?, which appeared in December 2002. Ah... but others have already quoted that post. For example: ###################################### === Subject: Re: Narcissistic Personality Disorder: Reason for posting? > I've become really bothered by my posting, but I can't seem to stop. > However, I realized that part of the sick cycle are some of those > people who reply to me so negatively, so in the hope that these people > might show mercy I thought I'd post something I found on the web, with > limited hope, I admit, that anything will change. > Alternatively, the narcissist feels victimized by mediocre > bureaucrats and intellectual dwarves who consistently fail to > appreciate his outstanding - really, unparalleled - talents, skills, > and accomplishments. Being haunted by his challenged inferiors > substantiates the narcissist's comparative superiority. Driven by > pathological envy, these pygmies collude to defraud him, badger him, > deny him his due, denigrate, isolate, and ignore him. The narcissist > projects onto this second class of lesser persecutors his own > deleterious emotions and transformed aggression: hatred, rage, and > seething jealousy. > But here are the scary quotes: > Paranoid ideation - the narcissist's deep-rooted conviction that he > is being persecuted by his inferiors, detractors, or powerful > ill-wishers - serves two psychodynamic purposes. It upholds the > narcissist's grandiosity and it fends off intimacy. > The paranoid delusions of the narcissist are always grandiose, > cosmic, or historical. His pursuers are inßuential and > formidable. They are after his unique possessions, out to exploit his > expertise and special traits, or to force him to abstain and refrain > from certain actions. The narcissist feels that he is at the center of > intrigues and conspiracies of colossal magnitudes. > The paranoid narcissist ends life as an oddball recluse - derided, > feared, and loathed in equal measures. His paranoia - exacerbated by > repeated rejections and ageing - pervades his entire life and > diminishes his creativity, adaptability, and functioning. The > narcisstis personality, buffetted by paranoia, turns ossified and > brittle. Finally, atomized and useless, it succumbs and gives way to a > great void. The narcissist is consumed. > It looks like a person in the grips of this narcissism thing loses > connection with reality but somehow *feeds* on ANY attention, > including negative attention. > They called it narcissistic supply. > It really is a twisted illness that apparently can lead to even > nastier illnesses. I am self-diagnosed (typical narcissistic > behavior) but it looks like it fits to me. > However, I've decided that I don't want to succumb to mental woes, so > I am fighting the dark narcissist within by making this informative > post, though it's probably a sign of the illness. In any event, maybe > it will help in the long run as information is power. > I will do my best to stop posting, but it will help if some of you try > to help out by not replying to me, or if you reply, by refraining from > the attacks that feed my narcissistic side, and its irrational beliefs > that everyone is out to get me. If you have facts, sure, no problem. > My inability to face facts presented without emotion will just be a > sign of the dark side. > My only fear now is that some of you are sick as well, and act from > your own illness, so that you can't stop attacking in your posting. > But on this Christmas Day, I will act from hope, and hopefully this > New Year will bring a change, and an escape from the madness. There's nothing wrong with you posting, as long as: a) You're talking about math b) You're message doesn't make fun of people or tell people they're stupid c) You make a serious attempt to learn from people who know more than you. I think _that's_ what you should work on. ###################################### It is sad if James has deleted that post. There was a lot of truth in it. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: The Hammer is coming > But I can't find it any more in Google. > Although the rest of the JSH: Ok, I'm a loser thread is there. > Any idea what happened to that post? >I suspect that it was deleted by James Harris himself. He can do it, >sense it was posted by him and it was posted through Google. What makes >me suspect that is the fact that the same thing happened to other >threads; see, for instance, the thread Narcissistic Personality >Disorder: Reason for posting?, which appeared in December 2002. > Ah... but others have already quoted that post. For example: > ###################################### <<< quoted post deleted out of respect for James ###################################### > It is sad if James has deleted that post. There was a lot of truth in it. I sincerely hope that the posters to sci.math can exhibit some restraint in regard to these posts which James has deleted. I can see no reason for quoting all of them here, in full, more frequently than once a week. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: The Hammer is coming >But I can't find it any more in Google. >Although the rest of the JSH: Ok, I'm a loser thread is there. >Any idea what happened to that post? > I suspect that it was deleted by James Harris himself. He can do it, > sense it was posted by him and it was posted through Google. What makes > me suspect that is the fact that the same thing happened to other > threads; see, for instance, the thread Narcissistic Personality > Disorder: Reason for posting?, which appeared in December 2002. > Ah... but others have already quoted that post. For example: First of all, it may have quoted first and deleted after. Besides that, when you delete one post, you only delete it from the server it was originally posted at; all other newsgroup servers can still see it. But Google is, to my knowledge, the only one that keeps all posts forever. > It is sad if James has deleted that post. There was a lot of truth in it. There sure was! :-) Jose Carlos Santos === Subject: Re: JSH: The Hammer is coming >I have my prime counting function and the paper Advanced Polynomial >Factorization and those are my minor results, small enough that enough >people can understand them to break through the barriers I face. Uh, the first is not a result, just an algorithm. An extremely inefficient one compared to others, and also not new. The second result is simply wrong. >But I have two bigger results, and one is The Hammer. >The Hammer is coming, and when I finally fully unleash it, then the >establishment will come down. You've been warning us about this for _many_ years. _When_ is the establishment actually coming down? >And some of you seem to have thought that it mattered that you seemed >to win for so long, as you thought I was gone, beaten, put down and >defeated, until that paper showed up in the Southwest Journal of Pure >and Applied Mathematics. >But you see, few of you were smart enough to understand that time is >on my side. >Time is on my side. I wonder if the people who've been saying we should just evaluate your mathematics dispassionately are reading this? >James Harris ************************ David C. Ullrich === Subject: Re: JSH: The Hammer is coming > I have my prime counting function and the paper Advanced Polynomial > Factorization and those are my minor results, small enough that enough > people can understand them to break through the barriers I face. > But I have two bigger results, and one is The Hammer. > The Hammer is coming, and when I finally fully unleash it, then the > establishment will come down. This sounds like another post of yours, posted almost three years ago and whose title was The clock is ticking. Do you remember it? It's the I'm the guy who proved Fermat's Last Theorem in just a bit over 6 years after all of your mathematicians took around 360. My standards are kind of high, so it won't be a walk in the park, and nothing will just be given to you. Jose Carlos Santos === Subject: Re: JSH: The Hammer is coming : The Hammer is coming, and when I finally fully unleash it, then the : establishment will come down. Oh Lord, please don't take up carpentry. -Justin === Subject: Re: JSH: The Hammer is coming > Time is on my side. That sounds like a good name for a song. === Subject: Re: JSH: The Hammer is coming >Time is on my side. > That sounds like a good name for a song. Time he's on my side she said. Time he's on my side she said. He may be on your side I said, But it makes no difference in the end he's coming after you my friend. Script of the Bridge) === Subject: Re: JSH: The Hammer is coming === >Subject: Re: JSH: The Hammer is coming >Time is on my side. > That sounds like a good name for a song. >Time he's on my side she said. >Time he's on my side she said. >He may be on your side I said, >But it makes no difference >in the end >he's coming after you my friend. >Script of the Bridge) Don't forget Pete Seger, et al.: If I had a hammer... http://www.niehs.nih.gov/kids/lyrics/hammer.htm If I had a hammer I'd hammer in the morning I'd hammer in the evening ... all over this land, I'd hammer out danger I'd hammer out a warning... Marvin Sebourn osugeography@aol.com === Subject: Re: JSH: The Hammer is coming >Time is on my side. >That sounds like a good name for a song. > Time he's on my side she said. > Time he's on my side she said. > He may be on your side I said, > But it makes no difference > in the end > he's coming after you my friend. > Script of the Bridge) Rolling Stones, about 1966. === Subject: Re: JSH: The Hammer is coming Time is on my side. That sounds like a good name for a song. Time he's on my side she said. >Time he's on my side she said. >He may be on your side I said, >But it makes no difference >in the end >he's coming after you my friend. Script of the Bridge) > Rolling Stones, about 1966. 1964. Earlier: Irma Thomas, 1964, and Kai Winding, 1963 -- 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: The Hammer is coming > Rolling Stones, about 1966. > 1964. Earlier: Irma Thomas, 1964, and Kai Winding, 1963 Unfortunately Shareaza can't find either of these. === Subject: Re: JSH: The Hammer is coming Yeah, check out The Sensational Alex Harvey Band. They have an album, I believe called Framed, on which is a pretty darn cool song called Hammer Song. >Time is on my side. > That sounds like a good name for a song. === Subject: Re: JSH: The Hammer is coming > I have my prime counting function and the paper Advanced Polynomial > Factorization and those are my minor results, small enough that enough > people can understand them to break through the barriers I face. They are indeed minor -- and about as useful as a goiter. > But I have two bigger results, and one is The Hammer. > The Hammer is coming, and when I finally fully unleash it, then the > establishment will come down. Why wait? > And some of you seem to have thought that it mattered that you seemed > to win for so long, as you thought I was gone, beaten, put down and > defeated, until that paper showed up in the Southwest Journal of Pure > and Applied Mathematics. > But you see, few of you were smart enough to understand that time is > on my side. > Time is on my side. So you say, but the *truth* is clearly not. > James Often in error, but never in doubt! Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: sequence by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5I1lKo12911; Please help me with this problem. I have no idea how to prove it. Let {a_n} and {b_n} be sequence of complex numbers. Assume that {a_n} has no accumulation point. Prove that there exists a holomorphic function f:C->C such that f(a_n)=b_n === Subject: Re: sequence > Please help me with this problem. I have no idea how to prove it. > Let {a_n} and {b_n} be sequence of complex numbers. Assume that {a_n} > has no accumulation point. Prove that there exists a holomorphic > function f:C->C such that f(a_n)=b_n You're missing at least one assumption. Suppose a_j = a_k for j ne k and also b_j ne b_k. Then f isn't even a function. Since the right answer for a finite set of a_n's is the Lagrange interpolation polynomial, I'd suggest you start there: f(z) = sum_{k=1}^n b_k prod_{j=1, j ne k}^n (z-a_j)/(a_k-a_j) Nick === Subject: Re: sequence > Let {a_n} and {b_n} be sequence of complex numbers. Assume that {a_n} > has no accumulation point. Prove that there exists a holomorphic > function f:C->C such that f(a_n)=b_n Here's how to construct such a function: f(z) = sum_{N = 1}^{+oo} P_N(z) whith P_N a polynomial such that: 1) P_N(a_n) = b_n - sum_{j < N} P_j(a_n) for N - 1 <= |a_n| < N; 2) P_N(a_n) = 0 for |a_n| < N - 1; 3) |P_N(z)| < 2^{-N} for |z| < N - 1. Jose Carlos Santos === Subject: Re: sequence > Let {a_n} and {b_n} be sequence of complex numbers. Assume that {a_n} > has no accumulation point. Prove that there exists a holomorphic > function f:C->C such that f(a_n)=b_n > Here's how to construct such a function: > f(z) = sum_{N = 1}^{+oo} P_N(z) whith P_N a polynomial such that: > 1) P_N(a_n) = b_n - sum_{j < N} P_j(a_n) for N - 1 <= |a_n| < N; > 2) P_N(a_n) = 0 for |a_n| < N - 1; > 3) |P_N(z)| < 2^{-N} for |z| < N - 1. David C. Ullrich made me realize that this approach is more complex than I thought. Here's another approach. I suppose that you know that there's a holomorphic function f_0:C --> C such that each a_n is a zero of f_0 and such f_0 is not the null function. For each n, let m_n be the order of f_0 at a_n. By Mittag-Lefßer's theorem, theres some meromorphic function g whose poles are the a_n's and such that the principal part at a_n is b_n.(m_n)!/(f_0^{(m_n)}(a_n)).(z - a_n)^{-m_n). Then take f = f_0.g. Jose Carlos Santos === Subject: Re: sequence > Let {a_n} and {b_n} be sequence of complex numbers. Assume that {a_n} > has no accumulation point. Prove that there exists a holomorphic > function f:C->C such that f(a_n)=b_n > Here's how to construct such a function: > f(z) = sum_{N = 1}^{+oo} P_N(z) whith P_N a polynomial such that: > 1) P_N(a_n) = b_n - sum_{j < N} P_j(a_n) for N - 1 <= |a_n| < N; > 2) P_N(a_n) = 0 for |a_n| < N - 1; > 3) |P_N(z)| < 2^{-N} for |z| < N - 1. >David C. Ullrich made me realize that this approach is more complex than >I thought. Here's another approach. I suppose that you know that there's >a holomorphic function f_0:C --> C such that each a_n is a zero of f_0 >and such f_0 is not the null function. For each n, let m_n be the order >of f_0 at a_n. By Mittag-Lefßer's theorem, theres some meromorphic >function g whose poles are the a_n's and such that the principal part >at a_n is b_n.(m_n)!/(f_0^{(m_n)}(a_n)).(z - a_n)^{-m_n). Then take >f = f_0.g. I do believe that Mittag-Lefßer is the standard way to prove this. But actually you _can_ get P_N as above fairly easily from Runge's theorem - it's just tricky enough that I didn't see how to do it immediately: Start with Q_N such that Q_N(a_n) = 0 for |a_n| < N - 1, and such that those are the only zeroes of Q_N. Now a simple application of Runge's theorem gives you R_N such that R_N(a_n) = (b_n - sum_{j < N} P_j(a_n))/Q_N(a_N) for N - 1 <= |a_n| < N and such that R_N(z) is as small as you want for |z| < N-1; now let P_N = R_N * Q_N. >Jose Carlos Santos ************************ David C. Ullrich === Subject: Re: sequence > I do believe that Mittag-Lefßer is the standard way to prove > this. But actually you _can_ get P_N as above fairly easily > from Runge's theorem - it's just tricky enough that I didn't > see how to do it immediately: > Start with Q_N such that Q_N(a_n) = 0 for |a_n| < N - 1, and > such that those are the only zeroes of Q_N. Now a simple > application of Runge's theorem gives you R_N such that > R_N(a_n) = (b_n - sum_{j < N} P_j(a_n))/Q_N(a_N) > for N - 1 <= |a_n| < N > and such that R_N(z) is as small as you want for |z| < N-1; > now let P_N = R_N * Q_N. Damn! I *should* have been able to do this. Jose Carlos Santos === Subject: Re: sequence > Let {a_n} and {b_n} be sequence of complex numbers. Assume that {a_n} > has no accumulation point. Prove that there exists a holomorphic > function f:C->C such that f(a_n)=b_n >Here's how to construct such a function: >f(z) = sum_{N = 1}^{+oo} P_N(z) whith P_N a polynomial such that: >1) P_N(a_n) = b_n - sum_{j < N} P_j(a_n) for N - 1 <= |a_n| < N; >2) P_N(a_n) = 0 for |a_n| < N - 1; >3) |P_N(z)| < 2^{-N} for |z| < N - 1. Of course this raises the question of how to construct such a sequence of polynomials... >Jose Carlos Santos ************************ David C. Ullrich === Subject: Re: My results, concrete and real by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5I1lKO12890; (snip) >It means he has to move on to some different topic. >-- >David Kastrup, Kriemhildstr. 15, 44793 Bochum Like a supposedly new twin prime conjecture. === Subject: Entire function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5I1lKA12906; Is it true or false: There is a non-constant entire function having both 2*Pi and i as its periods; i.e. f(z+2*Pi)=f(z)=f(z+i) for all z in C. === Subject: Re: Entire function >Is it true or false: >There is a non-constant entire function having both 2*Pi and i as its periods; >i.e. f(z+2*Pi)=f(z)=f(z+i) for all z in C. Example: Let z = x + y*i, x and y real f(z) = sin(x - 2 * pi * y) --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Entire function >Is it true or false: >There is a non-constant entire function having both 2*Pi and i as its >periods; >i.e. f(z+2*Pi)=f(z)=f(z+i) for all z in C. > Example: > Let z = x + y*i, x and y real > f(z) = sin(x - 2 * pi * y) But not entire in the complex sense, since not differentiable in the complex sense. === Subject: Re: Entire function > Is it true or false: > There is a non-constant entire function having both 2*Pi and i as its periods; > i.e. f(z+2*Pi)=f(z)=f(z+i) for all z in C. What do you know about bounded, entire functions? === Subject: Re: help evaluating definite integral > [...] > By the way, I heard from Alex Selby by email. Apparently the formula > above is a special case of one of the integrals known as Dirichlet > integrals. I quote him below. > Your integral is a special case of what is known as a Dirichlet > Integral. > See e.g., > http://mathworld.wolfram.com/DirichletIntegrals.html > and use formula (4) with alpha_i=f_i+1, f(x)=delta(x-1), their n = > your r > By the way, there is a missing a factor of (r-1)! from the LHS of your > equation. > Faheem. Sorry I think my previous reply is a bit misleading. I think it is not very enlightening to quote something like Dirichlet integral because its proof is very short, so it is maybe better to see how to prove it than to look it up. It is also slightly more general than you would need since it deals with non-integral powers too. Rereading your original post, it looks like you are happy proving the formula [as I think it should be] Integral over (p1+p2+...+pr=1, p1,p2,...>=0) of (p1^f1.p2^f2...pr^fr) = f1!f2!...fr!/(N+r-1)! by induction. I don't think induction is cheating - you have calculated it. But if you want a non-inductive proof, then this is a probabilistic one: Instead of cluttering the notation by writing it out for general r, let's do examples for r=2 and r=3, then it will be obvious how it works for r>3. r=2: Imagine placing f1 red dots, f2 green dots and 1 black dot uniformly at random on the interval [0,1]. What is the probability that the red dots are to the left of the black dot, and the green dots are to the right of it? Let's work it out in two different ways. (i) Conditioning on the position, p, of the black dot, the chance that all the red dots lie to the left of it is p^f1, and the chance that the green dots lie to the right is (1-p)^f2. So the overall probability is integral from 0 to 1 of p^f1.(1-p)^f2. (ii) The distribution of dots is invariant under permutation of the f1+f2+1=N+1 dots, and there is exactly one colouring which satisfies the condition red-then-black-then-green. The only allowable permutations of dots which preserve the condition are mixing the reds amongst themselves and the greens amongst themselves. So the overall probability is f1!f2!/(N+1)! where N=f1+f2. Thus Integral of p^f1(1-p)^f2 is f1!f2!/(N+1)!. r=3: Now we have f1 red dots, f2 green dots, f3 blue dots, 1 brown dot and 1 black dot all placed at random on [0,1]. We ask for the probability the dots appear in the order: f1 red then 1 brown then f2 green then 1 black then f3 blue. (i) Conditioning on the positions x1 and x2 of the brown and black dots, the chance is x1^f1.(x2-x1)^f2.(1-x2)^f3. Integrating over 0 show that if a function f is analytic in a deleted neighborhood > of a point z_0 and if z_0 is an accumulation point of zeros of f, > then either z_0 is an essential singular point of f > or else f(z) is identically equal to zero. > --------------------------------------------- > maybe, f(z) == 0 is trivial > because, by theorem of identity. > (if a function f is analytic throughout a domain D > and f(z) = 0 at each point z of a domain or arc contained in D, > then f(z) == 0 in D.) > right ?? I don't know. I have no clue about what you mean with all this. It looks as though you're trying to persuade us that f == 0 => f == 0. > and > if f(z) is identically not equal to zero, then > i must show that z_0 is an essential singular point of f > but i can't show.... If z_0 was not an essential singular point of f, then, for some natural number n, the function (z - z_0)^n.f(z) would be admit an analytic extension to a neighborhood of z_0; let g be that extension. Then the set {z | g(z) = 0} would have an accumulation point, namely z_0, and so g would be 0 on a neighborhood of z_0 and therefore f would have the same property. Jose Carlos Santos === Subject: Re: two box puzzle > Alas, I'm writing to you in the hope of finding someone to verify the > solution I have to the following puzzle. I'm sending the solution as a > separate message later so not to spoil your fun in finding one on your > own. I think you'll find it's more challenging than at first glance. > Jim Reekes > [Here is the puzzle] > I ßip a fair coin until it comes up heads (so there is a 50% chance I > will only ßip the coin once, 25% twice, 12.5% three times...). Call > the number of ßips N. > Now I take two boxes. In one box I put 3^N dollars. In the other box I > put 3^(N+1) dollars. These are magical boxes and can hold an > exponential amount of money without gaining a noticeable amount of > weight. > I then present you the two closed, identical looking boxes. You get to > open one box and count the money. At that point you can either take the > money in the box you picked, or take the money in the other box. What > strategy should you employ to maximize your expected value? > So (for example) if you open a box and see 3 dollars, you should always > switch because there will definitely be 9 dollars in the other box. But > if you open the box and see 9 dollars, switching might get you 3 and > might get you 27, so you have a decision to make based on the > probability of those two outcomes. > Suppose you open a box and see the amount X. Would you on average gain > more than X if at that point you switched? If you don't switch, you get $X. If you do switch, you get $X/3 or $3X. So, if you always switch you can be wrong 3 times in 4 and still break even. But the probability of selecting the higher amount is 50%, regardless of the amount in the first box. Aha! Switching every time is the same as never switching. Half the time you will get $X and half the time you will get $3X. === Subject: Re: two box puzzle > Yes, I would. Given that I see X = 3^k with k > 0, if I switch > my expected take would be 11/9 3^k instead of 3^k. > Of course, in practice I know you don't have as much money as Bill Gates, > so if N turns out to be sufficiently large you won't be able to put > the required money in the envelopes. If N_0 is the largest N for > which you can do it, then if I happen to see 3^(N_0) dollars I won't > switch. It may be worth pointing out that the in practice part more or less needs to be assumed to use this theoretical justification for switching. In particular, it should be obvious that the two strategies of never switching (NS) and always switching (AS) *must* have perfectly identical performance by any reasonable measure (for, if X is the payout under (NS) and Y is the payout under (AS), it's obvious that X and Y have the same distribution). So, if, as here, it appears that strategy (AS) actually outperforms (NS) even in the original uncapped problem, then something has gone wrong. The main thing that *has* gone wrong is that, for the uncapped problem, the (unconditional) expected payoff is infinite under either strategy. In particular, the random variable Y representing the payout from strategy (AS) isn't integrable, so by most definitions there isn't any such conditional expectation as E[Y|X] (the expected take calculated as 11/9 X = 11/9 3^k above). If you take the view that this is hair-splitting, at the very least it's true that E[Y|X] > X implies E[Y] > E[X] when X and Y are integrable and E[Y|X] has its usual definition, but although we have E[Y|X] = 11/9 X > X on the event {X>3} and = 3 X > X on the event {X=3} in this problem, we still have E[Y]=E[X] (=infinity). In short, if you want to evaluate strategies by means of their expected payout in the uncapped problem, then all strategies are infinitely good, and there's no benefit to switching or not switching under any conditions. Even a strategy that doesn't switch when X=3 isn't improved (according to this measure) by changing it to switch at X=3: you increase the expected payout from infinity to infinity+3/2. If N is capped (by the finite wealth of the envelope stuffer), the expectations are all finite again, and common sense starts to prevail: the optimal strategy is to switch in all cases except when X is the capped value. This strategy outperforms (AS) and (NS) which, hardly coincidentally, have exactly the same expectation (the extra foolish switching in (AS) when X is equal to the capped value costs you, in expectation, exactly what you gain over (NS) by switching in all the other cases). The more general approach to this problem, even in the uncapped case, is to use some utility function u(.) such that u(X) and u(Y) are integrable. Then, it's probably pretty clear that the optimal strategy is to switch if and only if we observe an X=x such that: E[u(Y)|X=x] > u(x) For many sensible utility functions, there'll be some x_0 where you switch for all x<=x_0 and don't switch for all x>x_0. -- Kevin === Subject: Re: limitation to induction on finite bounds > Or one can try to prove that 1/3 is arbitrarily close to > any element in the set 0.3, 0.33, 0.333, ... . Turns out > 1/3 is indeed arbitrarily close -- but 1/3 is still not > actually in the set. >Its not in any finite subset, but its in the set in its infinite form. That, sir, is gibberish. It either IS or IS NOT one of the members of the infinite set. If it is not a member (which it isn't) then it is not in the set in any useful sense, since for sets is in precisely means is a member of. > You all declare the *entire* anti diag number is not listed, Correct. > you all put precedence on the infinite length of the numbers over the > infinite amount. (I infer you are attempting to refer to the infinite size of the set.) > All possible finite *prefixes* are present, Correct. > but the whole *infinitely* long number isn't, Correct. >why >do you conceptualise the numbers to all digits that can't be encapsulated, > yet the list is equally long and you encapsulate that? **** Every member of the set has (encapsulates in your idiosyncratic term) a *finite* count of 3s; the set itself has (encapsulates) an infinity of members. This really really really isn't a problem. Note that the list IS NOT equally long -- it is infinite, and each particular sequence of 3s is finite. >0.3, >0.33, >0.333, >... >is itself equivalent to 0.33.. NO - IT - ISN'T. The set CONTAINS all finite sequences of 3s, it is not itself a sequence of 3s (finite or infinite). They used to teach the difference between sets and their members to *primary school* children for goodness sake; it's hard to credit that you are too thick that to understand it. But your main problem is the Phillite Fallacy, wherein the infinite set of natural numbers {0, 1, 2, ...} (for the above example, each number is the count of 3s in a particular sequence) somehow contains infinity as a last element. It doesn't, no matter how much you might wish it to. -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- === Subject: Re: limitation to induction on finite bounds In sci.logic, Barb Knox : > Or one can try to prove that 1/3 is arbitrarily close to > any element in the set 0.3, 0.33, 0.333, ... . Turns out > 1/3 is indeed arbitrarily close -- but 1/3 is still not > actually in the set. >Its not in any finite subset, but its in the set in its infinite form. > That, sir, is gibberish. It either IS or IS NOT one of the members of the > infinite set. If it is not a member (which it isn't) then it is not in the > set in any useful sense, since for sets is in precisely means is a member > of. I doubt this will help |-|erc's problem, but one can define real numbers as Cauchy sequences; if one strongly types the numbers, leading to representations as 1N, 2N, -3J, 3J !=J 3N, 0J, and extend J to Q, leading to 3/1Q, 5/2Q, etc., one can then take Cauchy sequences in the usual manner and define the number 0.333...R != 0.333...Q = 1/3Q as the limit of the numbers in the Cauchy sequence S_Q = {0.3Q, 0.33Q, 0.333Q, ...}, which of course is a subset of Q. Since 0.333...R is a real it's not part of S_Q. S_R can be generated from S_Q in a fairly obvious fashion (in a pinch use a Cauchy sequence where every term is the same!), but 0.333...R is not a member of either set anyway, although one can make a case for 0.333...R being associated with S_Q, as S_Q defines 0.333...R, but then, so do a lot of other sequences, such as {1/4Q, 2/7Q, 3/10Q, 4/13Q, ...} and {10/31Q, 100/301Q, 1000/3001Q, ...}. Most people don't bother. :-) (My background is in computer programming, and it does make a difference in some cases whether Ô1' is represented as a char, int, long, longword array (arbitrary precision integer), ßoating point, double-precision ßoating point, character string, etc., so for me it's a fairly obvious if slightly useless thing in theoretical math to worry about. :-) ) If one wants to get really weird one can throw set theory into the mix, resulting in 0W = {}, 1W = {{}}, 2W = {{},{{}}}, etc. Not sure what -1J would look like but this might be a weakly typed system anyway, as 1J = 1W and -1J is merely that entity such that 1J + (-1J) = 0J Ditto for 1/3Q = that entity which when multiplied by 3/1Q becomes 1/1Q. The best I can suggest is that one can officially define a real number by using its decimal expansion to generate a Cauchy sequence (and other Cauchy sequences can be transformed to such an official sequence, with a little work). In this case, 0.333...R = {0.3Q, 0.33Q, 0.333Q, ...}, but 0.333...R is not contained in itself, and no set containing all elements of 0.333...R will automatically contain 0.333...R; it will merely contain 0.333...R as a subset. (And then there's issues such as whether {0.9Q, 0.99Q, 0.999Q, ...} and {1.0Q, 1.00Q, 1.000Q, ...} define the same number. Ow, my brain. :-) But this is a side jaunt.) > You all declare the *entire* anti diag number is not listed, > Correct. > you all put precedence on the infinite length of the numbers over the > infinite amount. > (I infer you are attempting to refer to the infinite size of the set.) > All possible finite *prefixes* are present, > Correct. > but the whole *infinitely* long number isn't, > Correct. >why >do you conceptualise the numbers to all digits that can't be encapsulated, > yet the list is equally long and you encapsulate that? **** > Every member of the set has (encapsulates in your idiosyncratic term) a > *finite* count of 3s; the set itself has (encapsulates) an infinity of > members. This really really really isn't a problem. > Note that the list IS NOT equally long -- it is infinite, and each > particular sequence of 3s is finite. >0.3, >0.33, >0.333, >... >is itself equivalent to 0.33.. > NO - IT - ISN'T. The set CONTAINS all finite sequences of 3s, it is not > itself a sequence of 3s (finite or infinite). They used to teach the > difference between sets and their members to *primary school* children for > goodness sake; it's hard to credit that you are too thick that to understand > it. > But your main problem is the Phillite Fallacy, wherein the infinite set of > natural numbers {0, 1, 2, ...} (for the above example, each number is the > count of 3s in a particular sequence) somehow contains infinity as a last > element. It doesn't, no matter how much you might wish it to. The question might be whether the list 0.3Q, 0.33Q, 0.333Q, ... is equal to 0.333...R, defines 0.333...R (which is possible), or whether any set containing the list 0.3Q, 0.33Q, .... will automatically contain 0.333...R (no), or whether |-|erc's talking about 0.333...R (probably) or 0.333...Q (not likely). But one can only accept |-|erc's |-|ypothesis if one is excessively (or deliberately!) sloppy about dealing with decimal expansions. I'll admit to some curiosity as to whether the |-|ypothesis is equivalent to requiring that all sets be closed but it would have to be properly formulated first. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: limitation to induction on finite bounds > Or one can try to prove that 1/3 is arbitrarily close to > any element in the set 0.3, 0.33, 0.333, ... . Turns out > 1/3 is indeed arbitrarily close -- but 1/3 is still not > actually in the set. Its not in any finite subset, but its in the set in its infinite form. > That, sir, is gibberish. It either IS or IS NOT one of the members of the > infinite set. If it is not a member (which it isn't) then it is not in the that is the retort of a coward who fails to address the issue. you always carry on the Ôinfinite form' of the diag number is not there, but when told the infinite form of the set you break down. xxx xxx xxx THIS has a new diag element because the # of members are *limited.* xxx xxx xxx ... THIS has *unlimited members*, by defintion there is no missing sequence. you induced your solution of hyper infinities going from finite case to infinite case. Herc === Subject: Re: limitation to induction on finite bounds In sci.logic, |-|erc : > Or one can try to prove that 1/3 is arbitrarily close to > any element in the set 0.3, 0.33, 0.333, ... . Turns out > 1/3 is indeed arbitrarily close -- but 1/3 is still not > actually in the set. >Its not in any finite subset, but its in the set in its infinite form. > That, sir, is gibberish. It either IS or IS NOT one of the members > of the infinite set. If it is not a member (which it isn't) then > it is not in the > that is the retort of a coward who fails to address the issue. > you always carry on the Ôinfinite form' of the diag number is > not there, but when told the infinite form of the set you break down. If a set S contains the subset {0.3, 0.33, 0.333, ... } does it also automatically contain 1/3? Did you really want to rewrite a good chunk of set theory? > xxx > xxx > xxx > THIS has a new diag element because the # of members are *limited.* > xxx > xxx > xxx > ... > THIS has *unlimited members*, by defintion there is no missing > sequence. you induced your solution of hyper infinities going > from finite case to infinite case. If a set S is defined {0, 0.1, 0.2, ... 0.9, 0.01, 0.02, ..., 0.99, 0.001, 0.002, ... 0.999, ...} does it automatically contain the entire interval [0,1)? Does it contain 1 as well, making the interval [0,1]? Does it contain any irrational numbers at all? (Does anyone have a name for this set? Presumably someone's fiddled with this before.) 1/3 turns out to be a perfectly good antidiagonal for this set, as the first digit -- the only one in risk of colliding with anything nonzero -- is 1, not equal to 3. However, the set S does contain all finite prefixes, by construction. > Herc -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly === Subject: [ANN] xmds-1.3-4 released! xmds solves complex problems simply and quickly This message is to announce the release of xmds version 1.3-4. Release 4 of xmds-1.3 fixes a parsing bug found when using the tag. It also adds some functionality in terms of command line option parsing of both xmds and the xsil2graphics utility program; see xmds --help and xsil2graphics --help for more information, or the man pages, or the pdf version of the documentation. There are now man pages available for xmds, xsil2graphics and the loadxsil.m utility script, which are installed when one does a Ômake install'. The loadxsil.m script is also now installed when a Ômake install' is performed. Additionally, three more example simulations have been added; two from biology, and one from electronics. The documentation of xmds has also been updated. This includes mention of the new noise sources and how to specify them using the Ôkind' attribute of the tag, and how to generate a new simulation script by using the --template option to xmds. Other general textual improvements have been made as well. xmds - the eXtensible Multi-Dimensional Simulator - is a program for solving equations - fast. It is an Open Source tool to simplify the computer modelling of various systems, and is currently being developed within the Australian Centre for Quantum-Atom Optics at the University of Queensland in Australia. There are many situations in many areas where a system of interest can be modelled by a differential equation or equations. Such areas include: physics, mathematics, engineering, physical and theoretical chemistry, theoretical and computational biology, finance, and economics. Modelling these systems involves writing a computer program to find a solution to the equations, which is not necessarily easy to do. This is where xmds comes in. The advantage of using xmds instead of doing the same job by means of conventional programming is the same as ordering a pizza as opposed to making one yourself. The only thing you have to learn to become an xmds user is How to order a pizza. There are a couple of important differences here though: normally you have to pay for the pizza, while xmds comes for free; and xmds is like a gourmet pizza outlet - one has the option of exotic things like solving stochastic equations, which the chain-brand pizza vendors don't offer! xmds therefore makes writing complex computer simulations simple. Another major advantage of xmds is that it is free. The source code and documentation can be freely downloaded from the xmds web site, http://www.xmds.org/. xmds runs on Linux, Unix (including MacOS X) and the Cygwin environment on Windows, help for installing xmds on these systems is available both from the web site and the xmds distribution. xmds is especially useful in solving complex problems requiring solving the problem over many different random parameters. Such problems can be parallelised (run on lots of computers at the same time) and xmds does this automatically with little user input, making the solution of these problems a breeze. Often writing a computer program to solve complex problems can be very difficult, time-consuming, and error-prone. This is where xmds excels. One merely needs to write a script in a high-level form which low-level code for you, producing code that is better for a computer. This makes the writing of a simulation program significantly easier, reducing the development time, and almost eliminating bugs since xmds has written the vast majority of the code for you and has used thoroughly tested code and techniques in the production of the as code hand-written by an expert, so one can has the best of both worlds: quick development time, and quick execution. You can find examples of the use of xmds for solving complex (and simple) problems in the examples page of the xmds web site: http://www.xmds.org/examples/. So, if you're trying to model a bunch of atoms bouncing around together, the diffusion of an electrolyte solution, the reaction of enzymes with a substrate, or the volatility of stock prices, then xmds is the simulation tool for you. For more information visit the web site http://www.xmds.org/. -- cochrane_nospam@physics.uq.edu.au Quantum mechanics: the dreams stuff is made of. === Subject: little help here boundary=----=_NextPart_000_007D_01C454F0.6255F060 ------------------------------------------------------------- -------- charset=iso-8859-1 anyone have some generic CD drivers for DOS. need to reinstall Windows and it wont recognise the CD player so its catch 22. email them to hercATfindit.com or any boot disk for installing Windows 95 be great if it can see the CD. just email me the zip or files asap Herc change dit to dot === Subject: Re: little help here boundary=----=_NextPart_000_003D_01C4557E.7955C180 ------------------------------------------------------------- -------- charset=iso-8859-1 anyone have some generic CD drivers for DOS. need to reinstall Windows and it wont recognise the CD player so its catch 22. email them to hercATfindit.com or any boot disk for installing Windows 95 be great if it can see the CD. just email me the zip or files asap Herc change dit to dot === Subject: Re: little help here > anyone have some generic CD drivers for DOS. Why...are you going to cut off their heads and piss in their skulls? The one disadvantage to a new newsreader, new bozo-bin. You have got a bloody nerve to show your face around here. -- Charles Sweeney http://CharlesSweeney.com === Subject: Re: little help here >anyone have some generic CD drivers for DOS. > Why...are you going to cut off their heads and piss in their skulls? > The one disadvantage to a new newsreader, new bozo-bin. > You have got a bloody nerve to show your face around here. What'd I miss?! -- ~ange Get this freakin' d away from me! === Subject: Re: little help here > anyone have some generic CD drivers for DOS. >Why...are you going to cut off their heads and piss in their skulls? >The one disadvantage to a new newsreader, new bozo-bin. >You have got a bloody nerve to show your face around here. > What'd I miss?! Never mind, I found it. O_o -- ~ange Get this freakin' d away from me! === Subject: Re: little help here > ... > What'd I miss?! > Never mind, I found it. O_o ... another happy customer ;o) -- William Tasso === Subject: Re: little help here >anyone have some generic CD drivers for DOS. > Why...are you going to cut off their heads and piss in their skulls? > The one disadvantage to a new newsreader, new bozo-bin. > You have got a bloody nerve to show your face around here. > -- > Charles Sweeney > http://CharlesSweeney.com Pretend it's a new Herc. === Subject: Re: little help here Morning Mir! Have a great day with your customer! Probably sitting on your lonesome in the corner of the office, feeling like a spare part! :o) -- Charles Sweeney http://CharlesSweeney.com === Subject: Re: little help here server.bigpond.net.au: > anyone have some generic CD drivers for DOS. http://www.oldos.org/index.php -- Judy http://www.technohippie.com === Subject: Re: Argument of convolution with sinc() >I am trying to find a simpler expression for the convolution >of a complex function X(w) with Hermitian symmetry, >i.e. X(-w)=X^*(w), and a sinc() kernel: Phi(w) = Arg[ integral_{-inf}^{+inf} X(w-x) sin(pi*x)/(pi*x) dx ] I am mainly intererested to know how Phi(w) is different from Arg[ X(w) ]. Any ideas? >-Arrigo The difference is of course much more obvious in the frequency domain, > because convolving with sinc is the same as multiplying the Fourier transform > by a brick wall lowpass. But I'm sure you know that already. >Yes, and this does not seem to be useful in my case. Since X is the transform >of a real signal, if you transform X then you go back to the time domain, >since the Fourier transform of the Fourier transform of a function is the >original function. >What I really want to understand is how the product of a signal >with a rectangular window affects the phase of the spectrum. It is well >known what happens to the magnitude of the spectrum, however I have not >found any treatment about the phase so far. > For a rectangular window, nothing happens to phase > but in the time domain you get ripples in the > signal called Gibbs Phenomenon. the phase is actually distorted, see my posting above in the thread -Arrigo === Subject: Re: Argument of convolution with sinc() >I am trying to find a simpler expression for the convolution >of a complex function X(w) with Hermitian symmetry, >i.e. X(-w)=X^*(w), and a sinc() kernel: Phi(w) = Arg[ integral_{-inf}^{+inf} X(w-x) sin(pi*x)/(pi*x) dx ] I am mainly intererested to know how Phi(w) is different from Arg[ X(w) ]. Any ideas? >-Arrigo The difference is of course much more obvious in the frequency domain, >because convolving with sinc is the same as multiplying the Fourier transform >by a brick wall lowpass. But I'm sure you know that already. Yes, and this does not seem to be useful in my case. Since X is the transform >of a real signal, if you transform X then you go back to the time domain, >since the Fourier transform of the Fourier transform of a function is the >original function. >What I really want to understand is how the product of a signal >with a rectangular window affects the phase of the spectrum. It is well >known what happens to the magnitude of the spectrum, however I have not >found any treatment about the phase so far. -Arrigo > OH. It doesn't do anything at all--because the brick wall is a real-valued, > non-negative function. When you multiply a function by another function, the > magnitudes multiply and the phases add, so since the phase of the brick wall > is always 0, it doesn't do anything to the phase of the signal. Actually it does. The point by point multiplication between the brick wall window and the signal is done in the time domain. This corresponds to a convolution between a sinc() and the transform of the signal in the frequency domain. The phase of the sinc() switches between 0 and pi, so the convolution will change the phase (not simply by 0/pi). The question now is: is it possible to quantify this error on the phase? In other words, is it possible to apply some sort of deconvolution technique to undo the effect of the brick wall function? -Arrigo === Subject: Re: Argument of convolution with sinc() >I am trying to find a simpler expression for the convolution >of a complex function X(w) with Hermitian symmetry, >i.e. X(-w)=X^*(w), and a sinc() kernel: Phi(w) = Arg[ integral_{-inf}^{+inf} X(w-x) sin(pi*x)/(pi*x) dx ] I am mainly intererested to know how Phi(w) is different from Arg[ X(w) ]. Any ideas? >-Arrigo The difference is of course much more obvious in the frequency domain, >because convolving with sinc is the same as multiplying the Fourier transform >by a brick wall lowpass. But I'm sure you know that already. Yes, and this does not seem to be useful in my case. Since X is the transform >of a real signal, if you transform X then you go back to the time domain, >since the Fourier transform of the Fourier transform of a function is the >original function. >What I really want to understand is how the product of a signal >with a rectangular window affects the phase of the spectrum. It is well >known what happens to the magnitude of the spectrum, however I have not >found any treatment about the phase so far. -Arrigo > OH. It doesn't do anything at all--because the brick wall is a real-valued, > non-negative function. When you multiply a function by another function, the > magnitudes multiply and the phases add, so since the phase of the brick wall > is always 0, it doesn't do anything to the phase of the signal. (The even > symmetry of the sinc function is also enough to show this.) Actually it does. The multiplication between the brick wall window and the signal is point by point in the time domain. In the frequency domain, this corresponds to a convolution between the transform of the brick wall (a sinc) and the transform of the signal. The phase of the sinc switches between 0 and pi, and the convolution will have an effect on the phase. The question now is: is it possible to correct this phase error? In other words, is there a way to undo the effects of the brick wall on the phase by some form of deconvolution? -Arrigo === Subject: example of 8th degree root Suppose I want to find the angle x that subtends arc A and chord B on a shows that the answer is the solution to the equation, Bx/(2A) = sin(x/2) from which the Maclaurin expansion of sin(x/2) becomes, Bx/(2A) = x/2 - (x/2)^3/3! + (x/2)^5/5! - .. canceling x/2 from both sides, B/A = 1-(x/2)^2/3! +(x/2)^4/5! - ... suppose, then, that (x/2)^2 = t and we wish to carry out the calculation to a precision of 8 terms into the series. Then the equation becomes the polynomial, 1-B/A-t/3!+t^2/5!-t^3/7!+t^4/9!-t^5/11!+t^6/13!-t^7/15!+t^8/ 17!=0 Generalizing this to the equation of the plane, 1-B/A-x[1]/3!+x[2]/5!-x[3]/7!+x[4]/9!-x[5]/11!+x[6]/13! -x[7]/15!+x[8]/17! The plane has normal N=(-1/3!,1/5!,-1/7!,1/9!,-1/ll!,1/13!,-1/15!,1/17!) Two points on the plane are, B=(0,0,0,0,0,0,0,(B/A-1)17!) x[8] intercept D=(1,1,1,1,1,1,1,(B/A-1+1/3!-1/5!+1/7!-1/9!+1/11!-1/13!+1/15! )17!) A third point is defined, Q=(N*B/|N|^2)N Q is on the plane also, and defines where N intersects the plane between the plane and the origin. Evaluating the terms that make up Q, N*B= B/A-1 |N|^2={(1/3!)^2+(1/5!)^2+(1/7!)^2 + ... + (1/17!)^2} Define the space-curve, T=(t,t^2,t^3,...,t^8) This curve intersects the plane at, T = B + (D-B)m + (Q-B)u where m is some ratio and u is some ratio. The idea is that the sum of B and two nonparallel vectors on the plane, can access any point on the plane. If (D-B) and (Q-B) are found to be parallel, find another point on the plane for D so that they are not parallel. A test for this is (D-B)*(Q-B) ----------- = 1 or -1 when (D-B) and (Q-B) are parallel |D-B||Q-B| T = B + (D-B)m + (Q-B)u is now decoded. B=b[j] D=d[j] Q=q[j] are the jth components so that u = (t^j-b[j]-(d[j]-b[j])m )/(q[j]-b[j]) j=1,2,3,...,8 or t^j = (q[j]-b[j])(t-b[1]-(d[1]-b[1])m)/(q[1]-b[1]) + b[j]+(d[j]-b[j])m j=2,3,4,5,6,7,8 So as not to complicate matters, t^j merely says that the 2,3,4,5,6,7,8 powers of t can all be expressed in terms of m and t^1. These values of t^j are substituted in the original polynomial to arrive at an expression of m as a function of t^1 only. The values in the above expressions for t^j are substituted as follows: q[j] ----- q[1]=(B/A-1)(-1/3!)/{(1/3!)^2+(1/5!)^2+(1/7!)^2 + ... + (1/17!)^2} q[2]=(B/A-1)(1/5!)/{(1/3!)^2+(1/5!)^2+(1/7!)^2 + ... + (1/17!)^2} q[3]=(B/A-1)(-1/7!)/{(1/3!)^2+(1/5!)^2+(1/7!)^2 + ... + (1/17!)^2} . . q[8]=(B/A-1)(1/17!)/{(1/3!)^2+(1/5!)^2+(1/7!)^2 + ... + (1/17!)^2} b[j] ----- (b[1],b[2],b[3],...,b[8]) =(0,0,0,0,0,0,0,(B/A-1)17!) d[j] ------ (d[1],d[2],d[3],...,d[8]) =(1,1,1,1,1,1,1,(B/A-1+1/3!-1/5!+1/7!-1/9!+1/11!-1/13!+1/15!) 17!) Not to complain, but I just don't have the software to make these long calulations. Anyway, to continue, these values lead to expressions of t^2,t^3,t^4,...,t^8 as a function of m and t^1. These t^2,t^3,t^4,...,t^8 are substituted in the original polynomial to arrive at a value for m as a function of t^1 only. Let this value of m be called m=G(t). Once found, this m=G(t) is substituted in, t^j = (q[j]-b[j])(t-b[1]-(d[1]-b[1])m)/(q[1]-b[1]) + b[j]+(d[j]-b[j])m j=2,3,4,5,6,7,8 to arrive at, t^j = (q[j]-b[j])(t-b[1]-(d[1]-b[1])G(t))/(q[1]-b[1]) + b[j]+(d[j]-b[j])G(t) j=2,3,4,5,6,7,8 Then t^j = C[j](t) Each power of t is expressed as a function of t only. Am I getting deeper? Does it take more than three vectors to access any point on an 8 dimensional plane? Here is what I do next. If N=(a[1],a[2],a[3],...,a[8])=(-1/3!,1/5!,-1/7!,...,1/17!), and a[0]=1-B/A, define, f(t)= a[0]+a[1]t+ a[2](C[2]+t^2)(1/2)+ a[3](C[3]+tC[2]+t^3)(1/3)+ <-- explained below a[4](C[4]+C[2]C[2]+tC[3]+t^2C[2]+t^4)(1/5) . . a[8](the sum of k combinations of t^i*C[8-j](t) that=t^8)(1/k) --> Take for instance, t^3. C[3] is one expression of t^3, even though C[3] is a function of t^1 only. Also note that when you multiply C[2] by t, it results in t^3 as well. C[2] is also a function of t^1 only, which when multiplied by t, results in t^3. Lastly, t^3 is a pure expression of t^3, so it is added to the others. Since there are 3 expressions of t^3, their sum is multiplied by 1/3 to obtain only one value of t^3. However, distrusting this result, it wouldn't hurt to generalize f(t) by adding coefficients to to a[0],a[1],a[2],...,a[8]. Suppose these coefficients are called s[0],s[1],s[2],...,s[8]. Then f(t) is rewritten, f(t)= s[0]a[0]+ s[1]a[1]t+ s[2]a[2](C[2]+t^2)(1/2)+ s[3]a[3](C[3]+tC[2]+t^3)(1/3)+ s[4]a[4](C[4]+C[2]C[2]+tC[3]+t^2C[2]+t^4)(1/5) . . s[8]a[8](the sum of k combinations of t^j*C[8-j](t) that=t^8)(1/k) I have to confess. I chose this option because I was backed into a corner. Ideally, s[0]=s[1]=s[2]=...=s[8], which all=1. It would be interesting to see if they do. Therefore, from here f(t) is expressed as a Maclaurin series. a[0]=(1/0!)f(0) a[1]=(1/1!)f'(0) a[2]=(1/2!)f''(0) a[3]=(1/3!)f'''(0) . . a[8]=(1/8!)f^{8}(0) The a[0],a[1],a[2], on the left of the Ô=' make up the original polynomial, for which the values on the right must be equivalent. This sets up 9 equations in the 9 unknowns, s[0],s[1],s[2],...,s[8]. These unknowns are solved using linear algebra. Finally, s[0]a[0]+s[1]a[1]t+s[2]a[2]C[2]+s[3]a[3]C[3]+...+s[8]a[8]C[8] =0 is solved for t, where C[2],C[3],C[4],...,C[8] are each a function of t^1 only. *** Jon === Subject: Re: A strange series Sorry, it works only for c>0, but then it seems obvious. S. === Subject: Re: A strange series Let c < a_n a_(n-1) < d, so c + a_(n-1) < a_n < d + a_(n-1). Then, iterating one has nc + a_0 < a_n < nd + a_0. So, a_n/(a_0 + ... + a_n) < ( nd + a_0)/(a_0+(a_0+c)+...+(a_0+nc)) = (nd + a_0) / ((n+1)a_0 + (n(n+1)/2)c). But the last term is obviously ~ 1/n, hence its square is ~ 1/n^2, therefore the series Sum[a_n/(a_0 + ... + a_n)]^2 is convergent. (Here f(n) ~ g(n) means that f(n)/g(n) has a nonzero limit.) Simeon > Let a_n be a positive real sequence, such that lim(a_n, n=infinity) = > + infinity, and a_{n+1} - a_n is bounded. > Let A_n = sum(a_k, k=0..n). > Prove that the series sum( (a_n/A_n)^2 ) converges. > If someone has any (relevant) idea, I'd be glad to know it. > Vincent G. === Subject: Re: A strange series > Let a_n be a positive real sequence, such that lim(a_n, n=infinity) = > + infinity, and a_{n+1} - a_n is bounded. > Let A_n = sum(a_k, k=0..n). > Prove that the series sum( (a_n/A_n)^2 ) converges. I can't; the statement is false. Put a_n = sqrt(n). Then A_n = 1 + sqrt(2) + ... + sqrt(n) and therefore A_n^2 <= 1 + 2 + ... + n = n(n + 1)/2. So, (a_n/A_n)^2 >= n/(n(n + 1)/2) = 2/(n + 1), and so your series diverges. Jose Carlos Santos === Subject: Re: A strange series > Let a_n be a positive real sequence, such that lim(a_n, n=infinity) = > + infinity, and a_{n+1} - a_n is bounded. > Let A_n = sum(a_k, k=0..n). > Prove that the series sum( (a_n/A_n)^2 ) converges. > I can't; the statement is false. > Put a_n = sqrt(n). Then A_n = 1 + sqrt(2) + ... + sqrt(n) and > therefore A_n^2 <= 1 + 2 + ... + n = n(n + 1)/2. So, > (a_n/A_n)^2 >= n/(n(n + 1)/2) = 2/(n + 1), > and so your series diverges. Forget it; actually, A_n^2 >= 1 + 2 + ... + n = n(n + 1)/2. Jose Carlos Santos === Subject: MMPP delayed process Markov modulated Poisson Process(MMPP) is the arrival process for a system. t is positive continuous random variable with probability density function f(t). Suppose that the MMPP process will be delayed the period t, then how about the delayed process. Is this still MMPP? === Subject: Book for alpha stable processes I am a student in the area of dsp. For my thesis I need urgently the following book dealing alpha stable random processes: C. L. Nikias and M. Shao. Signal Processing with Alpha-Stable Distributions and Applications. John Wiley & Sons, 1995. Can you send me this book in pdf (or another format) or can you point me any site where I can to download the book? In return I can send you more interesting books about signal processing, random processes, math, dsp, wavelet ... === Subject: Re: Book for alpha stable processes > I am a student in the area of dsp. > For my thesis I need urgently the following book dealing alpha stable > random processes: > C. L. Nikias and M. Shao. Signal Processing with Alpha-Stable > Distributions and Applications. John Wiley & Sons, 1995. > Can you send me this book in pdf (or another format) or can you point > me any site where I can to download the book? > In return I can send you more interesting books about signal > processing, random processes, math, dsp, wavelet ... You can buy it used for US 20.00 http://www.amazon.com/gp/product/offer-listing/047110647X// 002-3808054-57536 42?condition=all === Subject: Challenging Benchmarks for Maximum Clique (Independent Set) Problem All benchmarks are planted with hidden optimum solutions, with the problem size growing from 450 vertices to 1534 vertices. Preliminary results show that the hardness of solving these instances grow very rapidly as the problem size increases. More interestingly, finding optimum solutions to these graph instances is, in fact, equivalent to finding solutions to some satisfiable SAT benchmarks, with the set of vertices directly corresponding to the set of variables. Recently, some submitted). The results indicate that the instances with more than 1200 variables (vertices) are very challenging to solve. For more information, please visit the following website: http://www.nlsde.buaa.edu.cn/~kexu/benchmarks/ graph-benchmarks.htm Any comments or results are appreciated. Ke Xu http://www.nlsde.buaa.edu.cn/~kexu/ === Subject: Rotating modes Hi there, i'm trying to calculate the response of system using modal superposition. But because the excitation force rotates around this system I have to rotate the modes also (they are not axisymmetric due to an initial deformation). For axisymmetric modes one would have: M*q_doubledot + K*q = f (omega*t) Using q = Phi*y and q_doubledot=Phi*y_dooubledot, where Phi are the n selected modeshapes leads to: y_doubledot + V*y=Phi * f(omega*t) where V is a diagonal matrix containing the n eigenvalues. But now, for the asymmetric modes Phi rotates with the excitation and Phi=Phi(omega*t). Which leads to something like: M*(Phi_doubledot*y + 2*Phi_dot*y_dot + Phi*y_double_dot) + K*Phi*y = f(t) And I can't write the equations in terms of just Phi and V anymore. Anyone have an idea how i can solve this? Ron === Subject: A (slightly more) complicated integral Ok, so the previous integral ended up being a piece of cake, however this one cannot be done in the same way (and it ends up that this one is the one I really want, isn't it always the same?) It looks like this: Int[Int[E^(a Cos[x]+ b Cos[y]+ k Cos[x - y]) {y, -Pi, Pi}],{x, -Pi, Pi}] Any suggestions anyone? Enrique === Subject: Re: A (slightly more) complicated integral > Ok, so the previous integral ended up being a piece of cake, > however this one cannot be done in the same way (and it ends up that > this one is the one I really want, isn't it always the same?) > It looks like this: > Int[Int[E^(a Cos[x]+ b Cos[y]+ k Cos[x - y]) {y, -Pi, Pi}],{x, -Pi, Pi}] > Any suggestions anyone? > Enrique Well, not a solution, but I do have these... Do these in order (I'm using Maple)... > Int(exp(r*cos(x)),x=-Pi..Pi) = 2*Pi*BesselI(0,r); /Pi | | exp(r cos(x)) dx = 2 Pi BesselI(0, r) | /-Pi > Int(exp(r*cos(x-T)),x=-Pi..Pi) = 2*Pi*BesselI(0,r); /Pi | | exp(r cos(x - T)) dx = 2 Pi BesselI(0, r) | /-Pi > Int(exp(A*cos(x)+B*sin(x)+C),x=-Pi..Pi) = 2*Pi*exp(C)*BesselI(0,sqrt(A^2+B^2)); /Pi | | exp(A cos(x) + B sin(x) + C) dx = | /-Pi / (1/2) | / 2 2 | 2 Pi exp(C) BesselI0, A + B / / > expand(a*(cos(x)+cos(y))+b*cos(x-y)); a cos(x) + a cos(y) + b cos(x) cos(y) + b sin(x) sin(y) /Pi | | exp((a + b cos(y)) cos(x) + b sin(x) sin(y) + a cos(y)) dx = | /-Pi / (1/2) | / 2 2 2 | 2 Pi exp(a cos(y)) BesselI0, (a + b cos(y)) + b sin(y) / / So your double integral is: > 2*Pi*int(exp(a*cos(y))*BesselI(0, (a^2+2*b*cos(y)*a+b^2)^(1/2)), y) I didn't find this, but there may be some hope, because G&R does have it without the exp... /Pi / (1/2) | | / 2 2 | | BesselI0, a + 2 b cos(y) a + b / / dy = | /-Pi 2 Pi BesselI(0, a) BesselI(0, b) -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: weak sol. to PDE by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5IBp2930494; Let U denote open, bounded set in R^n but NO smoothness assumptions about boundary of U. Let H denote Sobolev space of functions on U with k=1,p=2 and which are zero on boudary.Let f be C-infinity with compact support in U. let u be in H denote weak solution to -L(u) = f in U and u=0 on boundary. Here L is Laplacian. now my question is : Can we get any more regularity of u ? any book i look at i can only get more interior regularity. any comments would be greatly appreciated don === Subject: Re: weak sol. to PDE >Let U denote open, bounded set in R^n but NO smoothness assumptions >about boundary of U. Let H denote Sobolev space of functions >on U with k=1,p=2 and which are zero on boudary. What does it mean to say that these functions are zero on the boundary? >Let f >be C-infinity with compact support in U. >let u be in H denote weak solution to -L(u) = f in U >and u=0 on boundary. Here L is Laplacian. >now my question is : Can we get any more regularity of u ? > any book i look at i can only get more interior regularity. >any comments would be greatly appreciated > don > ************************ David C. Ullrich === Subject: Re: JSH: The Hammer is coming by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5IBp2430537; > Time is on my side. >That sounds like a good name for a song. Yes, it is! === Subject: Re: JSH: The Hammer is coming Discussion, linux) > Time is on my side. >That sounds like a good name for a song. > Yes, it is! Like you'd know a good name. -- Jesse Hughes But nothing's being Dr. Ullrich is a particular case of something's being such that nothing is it: (Ex)~(Ey)(y = x) -- John Correy on the failings of first order logic === Subject: Re: JSH: The Hammer is coming > Like you'd know a good name. Like you would know the lyrics to the Rolling Stones' song. Time is on my side, yes, it is. === Subject: Re: Bezier curve math - not quite what I want by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5IBp4830614; >your curve is already parametrized by Ôt'. you could consider Ôt' >as being the time. your function is: (x(t), y(t)) (xe don't care about z >since you're on a plane). >what you need however is to parametrize it acording to the arc length. >you can calculate the arc length: it's >s(t)=int(0,t,sqrt(x'(t)*x'(t)+y[CapitalOTild e](t)*y'(t))). >calculate the inverse of this function to get t(s). >it clearly exits, but I don't know if it is an easy one :) >you then feed it back to your bezier algorighm with your points being >(x(t(s)), y(t(s)). s varying from s(0) to s(1). >Here x(t) and y(t) are order 3 polynomials, so your s(t) will be an >integral of a square root of an order 4 polynomial, maybe it >simplificates, the calcul itself has to be done in order to see how >ugly it gets :) It involves finding root t as a function of s(t) by Newton-Raphson's Iteration, like in implicit and inverse functions. A simple parabola example: x = 2 a t ; y = a t^2 ; x'=2a ; y'= 2 a t Tangent vector parameterization: s(t)=int s'(t)dt =int(0,t,sqrt(x'(t)^2+y'(t)^2))). = a ( t Q + Ln(t+Q)) where Q = sqrt(1+t^2) OR, a ( t Q + Ln(t+Q))-s(t) == 0 So find t in terms of s by iteration in the above equation and place it in a DO loop of uniform s increment.Some numeric work is involved in setting up arrays of s(t)and new t=T. HTH === Subject: Re: The definition of weight by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i5IBp5730628; > This post is off-topic in sci.math. Please confine your cross-posting to > appropriate newsgroups. >Math is not off topic in physics; nor is physics off topic in math: >They go together like a horse and carriage; or what's that other >thing? love and marriage? Sincerly, C. Dement > -- > There are two things you must never attempt to prove: the unprovable -- > and the obvious. > -- > Democracy: The triumph of popularity over principle. === Subject: Stirling numbers Can anyone simplify the following expression? sum ( STIRLING1(J_1,A_1) * STIRLING1(J2,A2) * ... * STIRLING1(J_S,A_K) where the sum is over all A_i from 0 to J_i for all i from 1 to K and the STIRLING1(J_i,A_i) are the unsigned (i.e. absolute) Stirling x are positive real numbers. (In practical problems K is around 250 and the J_i vary from 1 to 3000.) Hope to hear the answer soon! Rampal === Subject: young man's game no more Another counterexample to the statement that mathematics is a young man's game. de Branges is 72 years old. Even if the proof turns out not to be correct, the author certainly still plays the game at the highest level. === Subject: Video Games in Flash I have recently launched a site for Flash game developers. http://www.gotoandplay.co.uk As maths is an integral part of any game development I would like to offer the support of a maths forum for people with maths based problems. (I currently run forums for games related programming problems and a number of other related areas) If there is anybody out there who enjoys developing small Flash based games and would like to contribute to the site by helping run a maths based forum please contact me. I can't, unfortunately, run the forum myself as I need a maths based forum as much as the next man. Greg. === Subject: Re: Video Games in Flash >I have recently launched a site for Flash game developers. >http://www.gotoandplay.co.uk The line BEST VIEWED IN IE 6.0+ can be a turn-off for many potential contributors. A browser-dependent code is bad code. -- rr === Subject: Re: Video Games in Flash >I have recently launched a site for Flash game developers. >http://www.gotoandplay.co.uk > The line BEST VIEWED IN IE 6.0+ can be a turn-off for many > potential contributors. A browser-dependent code is bad code. I thought it was interesting that he wants it viewed in 1024x768. Many people don't turn their resolution that high. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Linear functionals and codimension 1 subspace Given a vector space V (finite or infinite dimensional) on a field K, is it always possible to associate to any codimension 1 subspace W of V the kernel of a linear functional f: V -> K? I.e. , W={x in V : f(x)=0}, for some functional f ? === Subject: Re: Linear functionals and codimension 1 subspace > Given a vector space V (finite or infinite dimensional) on a field K, is > it always possible to associate to any codimension 1 subspace W of V the > kernel of a linear functional f: V -> K? > I.e. , W={x in V : f(x)=0}, for some functional f ? Do you mean is every codimension one subspace of V the kernel of a linear functional on V?? If so, then the answer is yes. -- 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: Linear functionals and codimension 1 subspace > Do you mean is every codimension one subspace of V the kernel > of a linear functional on V?? Yes. And how is it possible to prove this statement? === Subject: Re: Linear functionals and codimension 1 subspace > Do you mean is every codimension one subspace of V the kernel > of a linear functional on V?? > Yes. And how is it possible to prove this statement? Let W be such a subspace ant let {e_i | i in I} be a basis of W. Let v be a vector from VW. Then {e_i | i in I} U {v} is a besis of V. Define f:V --> K linear such that f(e_i) = 0 for each i and that f(v) = 1. Then ker f = W. Another way of proving it consists in observing that since dim(V/W) = 1, there is a linear bijection b:V/W --> K. Now take f = b o p, where p is the projection from V onto V/W. Jose === Subject: Re: Linear functionals and codimension 1 subspace > Do you mean is every codimension one subspace of V the kernel > of a linear functional on V?? >Yes. And how is it possible to prove this statement? > Let W be such a subspace ant let {e_i | i in I} be a basis of W. Let > v be a vector from VW. Then {e_i | i in I} U {v} is a besis of V. > Define f:V --> K linear such that f(e_i) = 0 for each i and that > f(v) = 1. Then ker f = W. > Another way of proving it consists in observing that since dim(V/W) = 1, > there is a linear bijection b:V/W --> K. Now take f = b o p, where p is > the projection from V onto V/W. Very clear explanations, thank you. === Subject: Re: Linear functionals and codimension 1 subspace > Do you mean is every codimension one subspace of V the kernel > of a linear functional on V?? > Yes. And how is it possible to prove this statement? Recall the definition of codimension one subspace. Use that to construct a linear functional having a given codimension one subspace as a kernel. -- 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: Math research, passion is important I've been thinking and it seems to me that the only times I REALLY get upset when it comes to sci.math posters is when I think some of you are dodging the mathematical argument, making stuff up, or otherwise refusing to accept mathematics. I REALLY get pissed when you attack algebra. But then in looking over some posts I see people saying they're really upset with me because they think I'm dodging mathematical arguments, making stuff up, and otherwise refusing to accept mathematics! Some will say that I ignore counterexamples or that I refuse to acknowledge well thought out refutations. I get mad and jump on people for ignoring the details of my mathematical arguments and repeating objections that I've handled. It's like a continual lack of understanding circle. So maybe here is a point of understanding, both sides can agree that the other side isn't listening, but at least both sides appear to be passionate about math. James Harris === Subject: Re: Math research, passion is important > I've been thinking and it seems to me that the only times I REALLY get > upset when it comes to sci.math posters is when I think some of you > are dodging the mathematical argument, making stuff up, or otherwise > refusing to accept mathematics. > I REALLY get pissed when you attack algebra. > But then in looking over some posts I see people saying they're really > upset with me because they think I'm dodging mathematical arguments, > making stuff up, and otherwise refusing to accept mathematics! > Some will say that I ignore counterexamples or that I refuse to > acknowledge well thought out refutations. > I get mad and jump on people for ignoring the details of my > mathematical arguments and repeating objections that I've handled. > It's like a continual lack of understanding circle. > So maybe here is a point of understanding, both sides can agree that > the other side isn't listening, but at least both sides appear to be > passionate about math. Here's the issue as I see it: People present counter-examples, you point them to your argument, they point you to the counter-example, ... endless cycle. The difference appears to be in where people feel the important aspects of the disagreement are. Or, people say this step is invalid because..., you say yes it is, and it continues all over again. In your mind, we aren't listening to you. In our mind, you aren't listening to us or aren't understanding us and we *are* listening to you. Perhaps during the next go round if you were to ask *why* one of us said what we said, it would help you understand what's going on in our heads. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Another circular definition: Like f = ma > Shead is, of course, full of . > But he does have one *sort of* point. f=ma by itself is pretty much useless, > because it really just acts as a definition of force. > Your examples below hold f to be constant, but what does that mean? What is > force? Without something other then f=ma, the definition of f is circular. > Of course, Newton created another equation that used f .... f=Gm/r^2, and > indeed Hooke had already (?) created an equation that used f (f = kL for > springs); either of these provides another definition of f which means it > isn't circular. Newton spent a lot of time on working out f=Gm/r^2, but as far as I know he never did get a satisfactory value for G; nether has anyone else: The average acceleration of freefall [g] has been worked out to more than half a dozen decimal places, but few have sense enough to know that a body's scale weight, divided by [g] is a measure of the matter in a mass, and is a particular case of m = w/g = f/a. Where the measure of a mass is a constant; called its _Inertia_! === Subject: Re: Another circular definition: Like f = ma > Newton spent a lot of time on working out f=Gm/r^2, but as far as I > know he never did get a satisfactory value for G; nether has anyone > else: I'll bite. G = (6.6742+-0.0010) x 10-11 m^3 s/kg http://physics.nist.gov/cuu/Constants/ What's the difference between the value of G and a satisfactory value for G? - Randy === Subject: Re: Another circular definition: Like f = ma >Newton spent a lot of time on working out f=Gm/r^2, but as far as I >know he never did get a satisfactory value for G; nether has anyone >else: > I'll bite. > G = (6.6742+-0.0010) x 10-11 m^3 s/kg > http://physics.nist.gov/cuu/Constants/ > What's the difference between the value of G and a satisfactory > value for G? You beat me to it. I was also going to point out that the value for G is known with more certainty than the value for g. === Subject: Re: Another circular definition: Like f = ma >Newton spent a lot of time on working out f=Gm/r^2, but as far as I >know he never did get a satisfactory value for G; nether has anyone >else: > I'll bite. > G = (6.6742+-0.0010) x 10-11 m^3 s/kg > http://physics.nist.gov/cuu/Constants/ > What's the difference between the value of G and a satisfactory > value for G? Easy! 6.674215x10^(-11) m^3-s/kg Phys. Rev. Lett. 85(14)2869 (2000) Science 288(5468) 944 (2000) 6.67407x10^(-11) m^3-s/kg Phys. Rev. Lett. 89 161102 (2002) Adelberger's number, the first one, is from the superior apparatus. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Another circular definition: Like f = ma > CUT< >Take any formula, involving three variables (I know, lets do ohms >law): >1) V = IR >Rearrange the formula in terms of one of the other variables: >2) I = V/R > That would be I = [IR]/R = I; since V = IR; that's what I'm talking > about! Why make a federal case of it? >Substitute into (1), using (2): >3) V = (V/R)R >Rearrange: >4) V = V(R/R) >And eliminate: >5) V = V >Wow. A circular definition. Why do you find this little trick of >rearrangement so surprising? > I'm not surprised: It's done all the time, which doesn't make it > right: > Because [mass] = density/volume: Density = [density/volume]/volume, > and volume = [density]/[mass]: Which are not circular definitions. > Also mass = force/acceleration: f = [f/a] x a = f; not f=ma! > Also mass = w/g: w = [w/g] x g = w; not w = mg! goddamnit donald just listen ffs and give this ing crap up you stand no chance === Subject: Re: Another circular definition: Like f = ma CUT< > goddamnit donald just listen ffs and give this ing crap up you stand no > chance That's what they told the Wright brothers: Gravity would pull them down, and it does. Wracked up many airmiles? Lotta people have. === Subject: exercise~ hello....sir~ http://myhome.hanafos.com/down3/12/19/74/01/51/comp2.gif this is a problem by book. here......um..... from the (b), 2(pi)i sigma Res {z^(2m)} / {(z^(2n)) + 1} = pi / n*sin@ i can't understand this part. 2(pi)i sigma Res {z^(2m)} / {(z^(2n)) + 1} = 2(pi)i sigma (-1 / 2n)*e^{i(2k+1)@} = pi / n*sin@ um.......i can't induce sin@ form. help me, please~ thank you very much~ === Subject: Notation Semantics (Constant In Arguement) 
F{UB} - F{LB}
F[LB,UB] = -------------;
UB - LB
and let
F{UB(^)LB} = F{UB} - F{LB} = F'{LB#UB} ((^) = nabla);
If one has F{2*UB} - F{2*LB}, is it F[2*LB,2*UB] or
F[2*(LB,UB)] and,
would it be F{2*UB(^)2*LB} and F{2*LB#2*UB} or F{2*[UB(^)LB]}
and
F'{2*[LB#UB]} ([] in last set not notationally 
significant--
could
be () or <>)?
~Kaimbridge~
-----
WantedKaimbridge (w/mugshot!):
http://www.angelfire.com/ma2/digitology/Wanted_KMGC.html
----------
DigitologyThe Grand Theory Of The Universe:
http://www.angelfire.com/ma2/digitology/index.html
***** Void Where Permitted; Limit 0 Per Customer. *****
[FN42nn]
===
Subject: Re: Notation Semantics (Constant In Arguement)

 F[LB,UB] = -------------;
> UB - LB
> and let
> F{UB(^)LB} = F{UB} - F{LB} = F'{LB#UB} ((^) = nabla);
Whoops, brain cramp: (^) = delta ((v) = nabla)
> If one has F{2*UB} - F{2*LB}, is it F[2*LB,2*UB] or
F[2*(LB,UB)] and,
> would it be F{2*UB(^)2*LB} and F{2*LB#2*UB} or
F{2*[UB(^)LB]} and
> F'{2*[LB#UB]} ([] in last set not notationally 
significant--
could
> be () or <>)?
> ~Kaimbridge~
> -----
> WantedKaimbridge (w/mugshot!):
> http://www.angelfire.com/ma2/digitology/Wanted_KMGC.html
> ----------
> DigitologyThe Grand Theory Of The Universe:
> http://www.angelfire.com/ma2/digitology/index.html
> ***** Void Where Permitted; Limit 0 Per Customer. *****
--
WantedKaimbridge (w/mugshot!):
http://www.angelfire.com/ma2/digitology/Wanted_KMGC.html
----------
DigitologyThe Grand Theory Of The Universe:
http://www.angelfire.com/ma2/digitology/index.html
***** Void Where Permitted; Limit 0 Per Customer. *****
===
Subject: Re: My results, concrete and real
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IE57910898;
(snip)
>That's either a lie or proof that you're 
unable to understand
>simple algebra. (Ok, a third possibility would be amnesia.)
>And in fact the paper passed peer review.
>I repeat, the paper passed peer review.
>Which in this case proves only that the reviewers were
>irresponsible or incompetent.
>************************
>David C. Ullrich
It's a lie. James Harris and his buddy Quinn
Tyler Jackson are attempting the Big Lie practiced
by Hitler and Goebbels, and also the fictional
character Big Brother in Orwell's 1984.
Harris and Jackson repeat a lie repeatedly and
often to give the appearance of accepted fact
and history, they deny and distort challenges to
their statements, and they evade direct questions
going to the truth, especially verifiable truth.
Thus James Harris and Quinn Tyler Jackson create
the future where contrary or dissenting voices
(who are actually telling the truth, which is
even verifiable) are drowned out or forgotten.
It's called propaganda! It sure ain't math, 
nor
is it pure mathematics or absolute truth.
James Harris is Nuts(TM)
===
Subject: Re: My results, concrete and real
> It's a lie. James Harris and his buddy Quinn
> Tyler Jackson are attempting the Big Lie practiced
> by Hitler and Goebbels, and also the fictional
> character Big Brother in Orwell's 1984.
Damn .... I thought my killfile feature worked....
Anyway ... JHiN .... you just lost the debate.
If you wonder why, I cite Godwin's Law in my favor:
http://info.astrian.net/jargon/terms/g/Godwin_s_Law.html
Sorry, Charlie.
--
Quinn
===
Subject: Re: My results, concrete and real
>It's a lie. James Harris and his buddy Quinn
>Tyler Jackson are attempting the Big Lie practiced
>by Hitler and Goebbels, and also the fictional
>character Big Brother in Orwell's 1984.
> Damn .... I thought my killfile feature worked....
> Anyway ... JHiN .... you just lost the debate.
> If you wonder why, I cite Godwin's Law in my favor:
> http://info.astrian.net/jargon/terms/g/Godwin_s_Law.html
> Sorry, Charlie.
Read your citation again. Godwin's Law says nothing about
winners or
losers.
Sorry, Quinn.
===
Subject: Re: My results, concrete and real
> Read your citation again. Godwin's Law says nothing about
winners or
> losers.
> Sorry, Quinn.
Awwwww.... won't you allow me one petty victory? Sheesh...
--
Quinn
===
Subject: Re: My results, concrete and real
Discussion, linux)
> It's a lie. James Harris and his buddy Quinn
> Tyler Jackson are attempting the Big Lie practiced
> by Hitler and Goebbels, and also the fictional
> character Big Brother in Orwell's 1984.
> Damn .... I thought my killfile feature worked....
> Anyway ... JHiN .... you just lost the debate.
> If you wonder why, I cite Godwin's Law in my favor:
> http://info.astrian.net/jargon/terms/g/Godwin_s_Law.html
> Sorry, Charlie.
There are few things lamer (and more repugnant) than
comparing one's
opponents to Hitler.
Using Godwin's Law to proclaim victory in a debate is close,
though. At least in the lameness category, if not in the
repugnance
category.
--
You lack the ability to reason, but instead get an idea in
your head
and hold on to it against all evidence. I don't 
find you
credible,
and reject your claims, as coming from a ßawed source.
===
Subject: Re: My results, concrete and real
Jesse F. Hughes said:
> There are few things lamer (and more repugnant) than
comparing one's
> opponents to Hitler.
> Using Godwin's Law to proclaim victory in a debate is 
close,
> though. At least in the lameness category, if not in the
repugnance
> category.
Normally, I'd agree with you, if the other side were a human
being rather
than an anonymous Eliza program. It appears that ON ERROR
GOTO godwinslaw,
however, has been triggered, however, so I'll permit myself
the victory
unless Grampa Simpson speaks up to defend his program.
--
Quinn
===
Subject: Re: weak sol. to PDE
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IE57n10893;
>Let U denote open, bounded set in R^n but NO smoothness
assumptions
>about boundary of U. Let H denote Sobolev space of functions
>on U with k=1,p=2 and which are zero on boudary.
>What does it mean to say that these functions are zero on
>the boundary?
>Let f
>be C-infinity with compact support in U.
>let u be in H denote weak solution to -L(u) = f in U
>and u=0 on boundary. Here L is Laplacian.
>now my question is : Can we get any more regularity of u ?
I think to talk about the trace we need some smoothness to
the boundary.I am just saying they are zero on the boudary
in the sense that they are in H.
ie. isn't H just defined to be the closure of 
C-infinity
functions with compact support in U,where closure
with respect to the H-1 norm. I don't think this affected
by the smoothness of U ??
But then again i don't know difference between
H^1 and W^(1,2) . I thought i saw a book that said
they are not equal unless boundary of U satisfies
some smoothness conditions.
> any book i look at i can only get more interior regularity.
>any comments would be greatly appreciated
> don
************************
>David C. Ullrich
===
Subject: Re: weak sol. to PDE
It would be much better if you figured out how to post
a _reply_ instead of starting a new thread this way.
(How you do that depends on what software you're
using...)
>Let U denote open, bounded set in R^n but NO smoothness
assumptions
>about boundary of U. Let H denote Sobolev space of
functions
>on U with k=1,p=2 and which are zero on boudary.
>What does it mean to say that these functions are zero on
>the boundary?
>Let f
>be C-infinity with compact support in U.
>let u be in H denote weak solution to -L(u) = f in U
>and u=0 on boundary. Here L is Laplacian.
>now my question is : Can we get any more regularity of u ?
>I think to talk about the trace we need some smoothness to
>the boundary.
Hence my confusion.
>I am just saying they are zero on the boudary
>in the sense that they are in H.
>ie. isn't H just defined to be the closure of 
C-infinity
>functions with compact support in U,where closure
>with respect to the H-1 norm. I don't think this affected
>by the smoothness of U ??
>But then again i don't know difference between
>H^1 and W^(1,2) . I thought i saw a book that said
>they are not equal unless boundary of U satisfies
>some smoothness conditions.
I don't think that the notation is absolutely standard;
it's certainly true that the closure of the space of
smooth functions with compact support in this norm is
not the same as the space of all distributions for
which the norm is finite. So H is the closure of the
smooth functions with compact support and that's
all you meant by vanishes on the boundary, fine.
Anyway, I'm not really the person to answer this;
I know just enough about these things to know that
I'm confused by the question. Two questions for you:
(i) Is it clear that there _is_ a solution u in H
in the first place?
(ii) What do you mean by more regularity on the boundary?
You can't mean that the function on the boundary is
differentiable; that makes no sense if the boundary
is wild. You must be asking about whether the partials
of u extend continuously to the boundary, or some such -
but then I wonder about the more in more regularity,
because we don't even know that u is continuous up
to the boundary, do we?
> any book i look at i can only get more interior
regularity.
>any comments would be greatly appreciated
> don
************************
>David C. Ullrich
************************
David C. Ullrich
===
Subject: Re: JSH: The Hammer is coming
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IE7jM11226;
>Yeah, check out The Sensational Alex Harvey Band. They have
an album,
>I believe called Framed, on which is a pretty darn cool song
called
>Hammer Song.
>Time is on my side.
> That sounds like a good name for a song.
How about MC Hammer!
I have this dread thought of James Harris donning
pharoah pants (ugh!), and singing Can't touch this
(of course referring to his mathematics and his
many proofs).
It's Hammer Time! Break it down!
James Harris is Nuts(TM)
===
Subject: Re: JSH: The Hammer is coming
>Yeah, check out The Sensational Alex Harvey Band. They
have an
album,
>I believe called Framed, on which is a pretty darn cool
song called
>Hammer Song.

>Time is on my side.
 That sounds like a good name for a song.
> How about MC Hammer!
> I have this dread thought of James Harris donning
> pharoah pants (ugh!), and singing Can't touch this
> (of course referring to his mathematics and his
> many proofs).
> It's Hammer Time! Break it down!
Hmmm...
Does anyone else see the parallel in that MC Hammer was
accused of using
existing work as his own?
===
Subject: Re: JSH: The Hammer is coming
> Does anyone else see the parallel in that MC Hammer was
accused of using
> existing work as his own?
Nope. Harris does not plagiarize. I wish he would, then he
would make more
sense. And I do not recall MC Hammer plagiarizing...Am I
wrong?
===
Subject: JH is just making fools out of you
I've read his work. No one could really have this kind of
mentality.
If he does, god help him. More likely, he's just spouting
deliberately inßamatory drivel aimed at inciting
annoyance/anger/laughter. The bottom line is he should be
ignored
either way. The only argument in his favor is that he
probably has
nothing else in his life in which case who are we to take
away this
little fun he has?
===
Subject: Re: JH is just making fools out of you
> I've read his work. No one could really have this kind of
mentality.
> If he does, god help him. More likely, he's just spouting
> deliberately inßamatory drivel aimed at inciting
> annoyance/anger/laughter. The bottom line is he should be
ignored
> either way. The only argument in his favor is that he
probably has
> nothing else in his life in which case who are we to take
away this
> little fun he has?
His recent twin-prime conjecture is interesting and
apparently novel. At
least no one has posted yet saying that ___'s Conjecture in
1936 said
essentially the same thing, only faster.
===
Subject: Re: JH is just making fools out of you
That JH is a troll is obvious to most of us.
What is so amazing is that so many posters in sci.math get
sed into
playing along. As they say, there's a new one born every day
...
> I've read his work. No one could really have this kind of
mentality.
> If he does, god help him. More likely, he's just spouting
> deliberately inßamatory drivel aimed at inciting
> annoyance/anger/laughter. The bottom line is he should be
ignored
> either way. The only argument in his favor is that he
probably has
> nothing else in his life in which case who are we to take
away this
> little fun he has?
===
Subject: Re: Entire function
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IEE5h11831;
> Is it true or false:
> There is a non-constant entire function having both 2*Pi
and i as its
periods;
> i.e. f(z+2*Pi)=f(z)=f(z+i) for all z in C.
>What do you know about bounded, entire functions?
I know that bounded entire function is a constant and I
suspect that the
answer is False. But the question is why this function should
be bounded?
Is it possible that in some neighborhood of peroids z+2*Pi or
z+i function is
unbounded?
May be I'm mistaken but Is this problem somehow connected to
elliptic
function theory.
===
Subject: Re: Entire function
> Is it true or false:
 There is a non-constant entire function having both 2*Pi
and i as its
> periods;
> i.e. f(z+2*Pi)=f(z)=f(z+i) for all z in C.

>What do you know about bounded, entire functions?
> I know that bounded entire function is a constant and I
suspect that the
> answer is False. But the question is why this function
should be
bounded?
> Is it possible that in some neighborhood of peroids z+2*Pi
or z+i function
is
> unbounded?
Do you know this theorem: a continuous function is bounded on
a compact
set? Once your function is bounded on rectangle
0,2*Pi,i,2*Pi+i it
will follow it is bounded on the whole plane.
> May be I'm mistaken but Is this problem somehow connected
to elliptic
> function theory.
It is connected, but such doubly-periodic functions are
meromorphic,
not entire. They *MUST* have poles if they are non-constant.
===
Subject: Re: Math research, passion is important
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IEE5I11836;
(snip)
>but at least both sides appear to be
>passionate about math.
>James Harris
Well, for you, passion is your only substitute
for competence in mathematics, and the truth,
and manners.
I'm sure you and Quinn Tyler Jackson are
both very passionate. Very, very
passionate. Getting together, being
extremely passionate about many aspects
of your, um, thoughts.
James Harris is Nuts(TM) (and passionate!)
===
Subject: Re: Math research, passion is important
> (snip)
>but at least both sides appear to be
>passionate about math.
>James Harris
> Well, for you, passion is your only substitute
> for competence in mathematics, and the truth,
> and manners.
> I'm sure you and Quinn Tyler Jackson are
> both very passionate. Very, very
> passionate. Getting together, being
> extremely passionate about many aspects
> of your, um, thoughts.
To be perfectly honest, Quinn has shown no signs of losing
his temper,
or acting in an immature fashion. I think he would be served
well by
only been in on events with James for a little over a year,
and that
alone has created some powerful impressions. Quinn does not
yet have
the benefit of those observations.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Math research, passion is important
Will said:
> To be perfectly honest, Quinn has shown no signs of losing
his temper,
> or acting in an immature fashion. I think he would be
served well by
> only been in on events with James for a little over a year,
and that
> alone has created some powerful impressions. Quinn does not
yet have
> the benefit of those observations.
I have the benefit of the observations I've made 
of James on
another, much
more peaceful list.
Anyway ... I've tried to avoid losing my temper with JHiN,
despite his
breaking of Godwin's Law.
My offer to have him call my bluff on the Type < 2 efficient
parsing
algorithm was sincere as opposed to being a pure taunt -- but
he didn't
want
to address that challenge until I answered a question
Einstein already had
answered.
I suspect (though I may be way off -- I'm feeling under the
weather) that
you, Will, have a sense that if I claim to have invented such
an algorithm,
it's likely that it's *not* a bluff. But 
that's neither here
nor there, I
suppose. Nobody wants to test my outrageous claim -- I can
handle that.
It's
always good to have one unorthodox ace in the hole. ;-)
--
Quinn
===
Subject: Re: Math research, passion is important
> Will said:
>To be perfectly honest, Quinn has shown no signs of losing
his temper,
>or acting in an immature fashion. I think he would be served
well by
>only been in on events with James for a little over a year,
and that
>alone has created some powerful impressions. Quinn does not
yet have
>the benefit of those observations.
> I have the benefit of the observations I've 
made of James on
another,
much
> more peaceful list.
> Anyway ... I've tried to avoid losing my temper with JHiN,
despite his
> breaking of Godwin's Law.
Godwin's Law was not broken by JHiN. JHiN's 
posting confirmed
Godwin's
Law.
===
Subject: Re: weak sol. to PDE
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IETsC13145;
the reason i ask this is that if i assume weak sol.
to -L(u)=f where f in L^2(U) ,
then if i could get u in previous post in
H^2(U) and not just H^2 loc (U) then i could
get u (( in this post)) in H^2(U) , which i think is false??
don
===
Subject: dna and chaos
girls and boys who are good know that God inßuences what
happens
===
Subject: Re: The Definition of ÔAnalytic 
Function'
<2j3b8eFt73urU1@uni-berlin.de>

X-CompuServe-Customer: Yes
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X-Ecrate: tanandtanlawyers.com
X-Pose: George Cox
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X-Tinguish: Mark Griffith
X-Treme: C&C,DWS
>In general a function is analytic on its domain if it has a
>derivative everywhere on its domain.
No. That's true for complex analytic functions but not for
real
analytic functions. A function is analytic if it can be
expressed
locally by a power series; there are real functions that are
infinitely differentiable but not analytic.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do
not reply to spamtrap@library.lspace.org
===
Subject: Re: The Definition of ÔAnalytic 
Function'
> I would be interested to learn of _every_ definition of
Ôanalytic
function'
> (or more simply the term Ôanalytic') that is 
currently
employed. I
> understand that it has many definitions which depend on your
context (ie:
> particular branch of mathematics). Just mention the branch
of math in
which
> you use the term(s) along with its definition in that 
branch.
> l8r, Mike N. Christoff
(1) How much have you worked on it yourself? University
libraries and
Google come to mind as sources of information. There are
Encyclopedias
of Mathematics, too.
(2) My pet peeve: Casual, superficial users of mathematics use
the term
analytic to refer to functions obtained by arithmetical
operations
and by compositions, starting with a list of known functions.
This
list, predictably, changes from user to user. So does the
list of
operations allowed to build new functions from old: are
infinite sums
allowed, or just finite ones? Or parametric integral
representations?
Etc.
This definition would have been forgivable a few hundred years
ago.
A glaring example of a function easily declared analytic in
this
vocabulary is one of Weierstrass's inventions
sum[n=0 to inf] cos(10^n * x) / 2^n
which is continuous at every real x but differentiable
nowhere.
Good l!
===
Subject: Re: The Definition of ÔAnalytic 
Function'
> every point are not equivalent in the setting R^n to R^m.
That is
> what he meant by compatible only in the complex case.
I am sorry, I do not quite understand. There are of course
functions
that can be represented in power series and are
differentiable at
every point of some domain in the setting of R^n to R^m. What
do you
mean by equivalent.. Ok, there is indeed a difference but I
am not
sure if that is what being meant! As differentiation
criterion in R^2
may be different from that in C (C has the Cauchy Riemann
rule). Yeah,
I suppose then they are not the same. Which makes me want to
ask a
certain question! In several complex spaces like C^n, is there
something like the Cauchy Riemman?
===
Subject: Re: The Definition of ÔAnalytic 
Function'
> real/complex definition of analytic while learning the
rudiments
> of generating functions. However, intuitively, I've 
always
thought
> of an analytic function as one that can be solved
symbolically.
Category mistake: functions are not things that can be
solved.
Anyway, this shows that what you intuitively have always
thought
>does not correspond to the general usage of the term amongst
>mathematicians.
> Is there a name for what I described in the previous post
(assuming you
> understand the gist of what I'm saying)?
It is completely unclear to me what you even might mean with
the quoted
paragraph.
--
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: The Definition of ÔAnalytic 
Function'
> real/complex definition of analytic while learning the
rudiments
> of generating functions. However, intuitively, I've
always thought
> of an analytic function as one that can be solved
symbolically.
Category mistake: functions are not things that can be
solved.
Anyway, this shows that what you intuitively have always
thought
>does not correspond to the general usage of the term
amongst
>mathematicians.
 Is there a name for what I described in the previous post
(assuming
you
> understand the gist of what I'm saying)?
> It is completely unclear to me what you even might mean
with the quoted
> paragraph.
First off. I haven't settled on Ôwhat I 
mean' yet. Second:
this,
unsurprisngly, may not do anything to better explain what it
is that I do
mean.
Perhaps I should introduce a notion of computation into what
I'm thinking,
which is unfortunately not something I can rigorously
describe.
With generating functions: By analytic I simply meant that an
infinite
power
series only converges if supplied with specific initial
conditions.
Informally, if for every point in some given set it converges
to a single
point in a single (but arbitrary interval) subset of that set
it, is
considered analytic (in that set). For example, if I'm not
mistaken, the
series:
f(s) = 1 + s + s^2 + .... + s^n + ...
is analytic only if s in (-1,1).
Computability of ~Analytic Equations
Essentially when I think of an ~analytic relation, the idea
of Ôquickly'
being able to solve the equation is fundamental. However,
there seem to be
two levels. One is the symbolic level. ie: f(x) = pi*e^x.
This is
simple
to solve symbolically (ie: for some constant c, f(c) =
pi*e^c. However,
the
problem of calculating the exact decimal number solution, for
say c = O,
where O is Chaitin's Omega number for a certain computer
language, is
substantially harder (even though O in this case may be
simple to define).
Linear Time Computability...?
In what I would consider an ~analytic function: Given any
specific input,
an
analytic function should be solvable by hand in a reasonable
amount of
time.
But what is meant by Ôsolvable by hand in a reasonable amount
of time'?. I
would guess that a) someone who had to calculate by hand f(x)
= x^2 for any
constant would be able to do it in Ôhuman acceptable 
time' if
x was 20
digits. However for 432,000 digits, Ôhuman acceptable 
time'
would be
inadequate. But the similarity between the two situations is
that the time
to calculate 20 digits is linearly proportional to the time
to solve for
432,000 digits. And this difference goes beyond pure
Ôphysical time'. It
is simply the number of necessary and sufficient discrete
steps (given
appropriate inputs and incorporating the restrictions that
any >= Ôminimum'
human with the ability to follow simple instructions that
require no
insight
has) who intends to solve the problem by hand needs to go
through.
Symbolic Manipulation & Information
But this is not the whole story. Often, you are not
interested in specific
individual results from an equation. The important
information is in its
large-scale/global behaviour. An example would be the ideas
of minimums
and
maximums in calculus. If you calculate a function's global
maximum f(c),
one does not always have to try every value of x to check if
it is a
maximum. You essentially ask: Ôgiven this criteria, what ARE
the
solutions?'. If one compares the ratio of information-in :
information-out
I would say this is far better than the ratio of the
algorithm that solves
information-in = given this specific input c, what is the
output? :
information-out. Note that computationally this is
semi-similar to Rice's
theorem. Rice defines what he considers to be 
Ônon-trivial
properties' of
a
Turing machine (algorithm). Let P be an arbitrary property
about Turing
machines (ie: P = Ôit halts'). If the property 
is
Ônon-trivial' then there
exist distinct Turing machines M1 and M2 such that M1 has
property P and M2
does not. Also, this property must be independent of the how
the Turing
machine was implemented (ie: #number of states, particular
input alphabet,
etc... are all irrelevant). By Rice's theorem, if P falls
within these
restrictions, then no algorithm exists that can generally
determine if an
arbitrary Turing machine has property P. In other words, the
algorithm
that, for any Turing machine M decides the question: Ôdoes
the Turing
machine M have property P?', does not exist. Let the 
language
ISGLOBALMAX
=
{(f,x) | f is a function with a global maximum at x}.
Conceivably one
could
consider all classes of functions for which ISGLOBALMAX is
decidable as
having a certain level of ~analyticity.
Equations with State (or Memory)
One of the main differences I see between relations I
consider ~analytic
and
those that are ~not, are that they do not seem to
have/require Ômemory' or
Ôstate'. An ~analytic function should be 
directly solvable
from its input.
ie: f(x) = x^2, where x = 34, is solvable in the same number
of levels of
recursion (or looping) as it is for x = 2^99. And by Ôlevels
of
recursion',
I mean: Ôhow many times does it call itself'? So 
if, for
example, f has
some subroutine Ôpower(x,n)' that recursively 
calculates
powers, these
recursions would not count, since they do not call f, but are
Ôcontained'
in
f.
A function I would deem ~non-analytic would be one where it
is impossible
to
calculate x = 34 unless you had already calculated x =
33,32,31,...,0.
This
is what I mean by the term Ômemory'. Albeit, its 
a
Ôone-register' memory,
but the fact is that anyone who needs to know the current
solution must
have
already computed all previous solutions. This is similar to
even many
decidable algorithms. The idea that there is no way to
Ôside-step' going
through the entire process of computation. The only way to be
sure an
arbitrary algorithm halts is to run it indefinitely. If it
stops while
we're still alive, we can say Ôit 
halts'. Otherwise we simply
have to let
it continue running. We will never know if it will actually
ever halt (and
is just REALLY slow) or if it will keep running forever.
Differential equations are examples of continuous equations
that have this
Ômemory' property (many think of differential 
equations as
continuous
versions of
recurrence-relations/difference-equations/
discrete-dynamical-systems).
Note
that there should be no restriction to Ôone 
level' of memory.
ie:
Fibonacci
sequences require two levels (f(n) = f(n-1) + f(n-2)), and
consequently
require two levels of initial conditions.
How The Halting Problem (Computability) Applies
The halting problem does not imply that there is no way to
determine if an
algorithm halts (ie: if it does what it was intended to do)
or not.
However, it does imply that there is no _single_
Ômechanisable' procedure
that can determine if any arbitrary algorithm will halt. At a
minimum, I
would suggest that all ~analytic equations be decidable.
---
This has probably made my (seemingly intangible) notion of
~analytic _less_
clear.
l8r, Mike N. Christoff
X-mailer: xrn 9.02
===
Subject: Re: The Definition of ÔAnalytic 
Function'
Mail-To-News-Contact: abuse@dizum.com
> However, intuitively, I've always thought of an analytic
>function as one that can be solved symbolically.
> Category mistake: functions are not things that can be
solved.
>Is there a name for what I described in the previous post
(assuming you
>understand the gist of what I'm saying)?
Presumably, you're referring to finding the 
roots of functions
or
finding the zeroes of functions when you speak of solving
functions.
--
Michael F. Stemper
#include 
The name of the story is A Sound of Thunder.
It was written by Ray Bradbury. You're welcome.
===
Subject: Re: The Definition of ÔAnalytic 
Function'
Christoff write:
> However, intuitively, I've always thought of an analytic
>function as one that can be solved symbolically.
 Category mistake: functions are not things that can be
solved.
>Is there a name for what I described in the previous post
(assuming you
>understand the gist of what I'm saying)?
> Presumably, you're referring to finding the 
roots of
functions or
> finding the zeroes of functions when you speak of solving
functions.
(I'll write ~analytic to differentiate its accepted use from
mine)
Sorry. I was talking too loosely. In fact I should not be
talking about
functions at all, but general equations. For example, x^2 +
y^2 = 1 is not
a function, but I would consider it to be ~analytic. When
discussing my
thinking on ~analytic equations I'm referring to the process
of calculating
the result of f(c) for some constant c. For example, if
f(x_{i+1}) = x_i +
1, I would call that ~analytic for any initial x_0 in R.
However if
f(x_{i+1}) = x_i + a * x_i * (1 - x_i), then I wouldn't call
that ~analytic
(Ôa' a constant) except for 
specific values of a and x_0. I
can't in
general statically solve f(x_2300) without knowing the
results for
f(x_2299), etc... A continuous example would be solving the
equation for a
driven damped pendulum (an equation from physics). Note I'm
not
restricting
the term ~analytic to any particular type of equation, but to
equations in
general. (ie: as the term Ôcontinuous mapping' 
can now be
applied to
non-real and non-complex functions given the term's
generalization in
topology). Hope this is a little clearer.
l8r, Mike N. Christoff
===
Subject: Re: The Definition of ÔAnalytic 
Function'
Unfortunately it did not make your question any clearer, it
is rather
confusing. I have never seen the term analytic being used in
such
manner, sometimes one says can be analytically solved .. etc.
But I
think you must take caution in the usage of analytic and not
just
say a function is analytic if thats what you mean (you didnt
mean a
function in fact, but an equation) it might confused people.
> (I'll write ~analytic to differentiate its accepted use
from mine)
> Sorry. I was talking too loosely. In fact I should not be
talking about
> functions at all, but general equations. For example, x^2 +
y^2 = 1 is
not
> a function, but I would consider it to be ~analytic. When
discussing my
> thinking on ~analytic equations I'm referring to the
process of
calculating
> the result of f(c) for some constant c. For example, if
f(x_{i+1}) = x_i
+
> 1, I would call that ~analytic for any initial x_0 in R.
However if
> f(x_{i+1}) = x_i + a * x_i * (1 - x_i), then I wouldn't
call that
~analytic
> (Ôa' a constant) except for 
specific values of a and x_0. I
can't in
> general statically solve f(x_2300) without knowing the
results for
> f(x_2299), etc... A continuous example would be solving the
equation for
a
> driven damped pendulum (an equation from physics). Note 
I'm
not
restricting
> the term ~analytic to any particular type of equation, but
to equations
in
> general. (ie: as the term Ôcontinuous mapping' 
can now be
applied to
> non-real and non-complex functions given the term's
generalization in
> topology). Hope this is a little clearer.
> l8r, Mike N. Christoff
===
Subject: Re: The Definition of ÔAnalytic 
Function'
Unfortunately it did not make your question any clearer, it
is rather
confusing. I have never seen the term analytic being used in
such
manner, sometimes one says can be analytically solved .. etc.
But I
think you must take caution in the usage of analytic and not
just
say a function is analytic if thats what you mean (you didnt
mean a
function in fact, but an equation) it might confused people.
> (I'll write ~analytic to differentiate its accepted use
from mine)
> Sorry. I was talking too loosely. In fact I should not be
talking about
> functions at all, but general equations. For example, x^2 +
y^2 = 1 is
not
> a function, but I would consider it to be ~analytic. When
discussing my
> thinking on ~analytic equations I'm referring to the
process of
calculating
> the result of f(c) for some constant c. For example, if
f(x_{i+1}) = x_i
+
> 1, I would call that ~analytic for any initial x_0 in R.
However if
> f(x_{i+1}) = x_i + a * x_i * (1 - x_i), then I wouldn't
call that
~analytic
> (Ôa' a constant) except for 
specific values of a and x_0. I
can't in
> general statically solve f(x_2300) without knowing the
results for
> f(x_2299), etc... A continuous example would be solving the
equation for
a
> driven damped pendulum (an equation from physics). Note 
I'm
not
restricting
> the term ~analytic to any particular type of equation, but
to equations
in
> general. (ie: as the term Ôcontinuous mapping' 
can now be
applied to
> non-real and non-complex functions given the term's
generalization in
> topology). Hope this is a little clearer.
> l8r, Mike N. Christoff
===
Subject: Re: Peano's space-filling curve
<2h70etF9hgqdU5@
uni-berlin.de><2hgg8vFcgcdiU5@
uni-berlin.de>
<2hght9Fcari9U1@uni-berlin.de><2hupdsFgcp2jU1@uni-berlin.de>
<40BA4D4A.5080803@rutcor.rutgers.edu><2i35onFikjr8U3@
uni-berlin.de>
<2ie15fFmb6g6U3@uni-berlin.de
><2ij6g0Fmc61aU1@uni-berlin.de>
ni-berlin.de>
ni-berlin.de> ni-berlin.de>
<87aczf2vq1.fsf@phiwumbda.org><2io3j1Fp4miqU2@uni-berlin.de>
ni-berlin.de>
ni-berlin.de> ni-berlin.de> <87fz94iyyl.fsf@phiwumbda.org>
ni-berlin.de>
ni-berlin.de> ni-berlin.de> <2irinqFqee5iU3@uni-berlin.de>
ni-berlin.de>
<40cd4dc9$36$fuzhry+tra$mr2ice@news.patriot.net> ni-berlin.de>
<2j7sb4Ftg5eaU8@uni-berlin.de>
X-CompuServe-Customer: Yes
X-Coriate: interspeed.co.nz
X-Ecrate: tanandtanlawyers.com
X-Pose: George Cox
X-Punge: Micro$oft
X-Sanguinate: The MVS Guy
X-Terminate: SPA(GIS)
X-Tinguish: Mark Griffith
X-Treme: C&C,DWS
said:
> Now get lost. Because that's what I'm going 
to do,
Promises, promises. You have shown yourself to be not only a
hypocritical, ignorant pompous fool with delusions of
adequacy, but a
*DISHONEST* hypocritical, ignorant pompous fool with
delusions of
adequacy. Since you keep claiming that you will leave without
ever
actually doing so, I will allow my filter to at least remove
you from
*my* view of this news group.
*PLONK*
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do
not reply to spamtrap@library.lspace.org
===
Subject: Re: Peano's space-filling curve
<2hght9Fcari9U1@uni-berlin.de><2hupdsFgcp2jU1@uni-berlin.de><
546mb0p0od19iao3
4odegkdoeeom1a95f0@4ax.com><2i35omFikjr8U2@uni-berlin.de><
40c3d5fc$3$fuzhry+t
ra$mr2ice@news.patriot.net><2iljo7Focfo1U1@uni-berlin.de>
-berlin.de>
<87ekoq1dt1.fsf@phiwumbda.org><2iqistFpemt0U2@uni-berlin.de>
-berlin.de>
<87659zlk9w.fsf@phiwumbda.org> -berlin.de>
<2itqogFrpdhcU2@uni-berlin.de>
-berlin.de>  -berlin.de>
<2j2bbvFrq4nkU1@uni-berlin.de> -berlin.de>
 -berlin.de>
<2j67ffFtr86vU3@uni-berlin.de>
X-CompuServe-Customer: Yes
X-Coriate: interspeed.co.nz
X-Ecrate: tanandtanlawyers.com
X-Pose: George Cox
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X-Sanguinate: The MVS Guy
X-Terminate: SPA(GIS)
X-Tinguish: Mark Griffith
X-Treme: C&C,DWS
at 03:36 PM, John Morgan 
said:
>I never said it did. Do I have to assume that my writing
skills are
>so intractably awful that no matter what I write, nobody
will ever be
>able to interpret my thoughts from it?
No, the problem seems to be that your writing is clear but
your
thought is muddy.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do
not reply to spamtrap@library.lspace.org
===
Subject: Re: Peano's space-filling curve
<2hght9Fcari9U1@uni-berlin.de><2hupdsFgcp2jU1@uni-berlin.de><
546mb0p0od19iao3
4odegkdoeeom1a95f0@4ax.com><2i35omFikjr8U2@uni-berlin.de><
40c3d5fc$3$fuzhry+t
ra$mr2ice@news.patriot.net><2iljo7Focfo1U1@uni-berlin.de>
-berlin.de>
<87ekoq1dt1.fsf@phiwumbda.org><2iqistFpemt0U2@uni-berlin.de>
-berlin.de>
<87659zlk9w.fsf@phiwumbda.org><2itqogFrpdhcU2@uni-berlin.de>
-berlin.de>
<87acza8876.fsf@phiwumbda.org> -berlin.de>
<2j2bc0Frq4nkU2@uni-berlin.de>
-berlin.de> 
-berlin.de>
<2j67feFtr86vU2@uni-berlin.de>
X-CompuServe-Customer: Yes
X-Coriate: interspeed.co.nz
X-Ecrate: tanandtanlawyers.com
X-Pose: George Cox
X-Punge: Micro$oft
X-Sanguinate: The MVS Guy
X-Terminate: SPA(GIS)
X-Tinguish: Mark Griffith
X-Treme: C&C,DWS
at 03:11 PM, John Morgan 
said:
>P.s. If you reply I may well read it, but unless you offer
me a
>refusal I can't make, I'm outa here (Silly 
limey humour and
all).
ObD'OyleCarte You say I go, I go but you don't 
go.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do
not reply to spamtrap@library.lspace.org
===
Subject: Re: Peano's space-filling curve
 that not everything you say is true.
Careful, David. I would have thought that a genius-level
mathematician would know that there exists in law (UK, USA,
International) certain strictures about what you can and
can't impugn about a person's character.
Don't worry. I won't sue you. Even though it 
_was_yesterday
And I'm sure it isn't so, but I must point out 
that in many
circumstances what you have just written might be construed
as nuisance posting. I told you I wouldn't post
again_unless_ you invited it - and_I_think you just did.
Now I'll take advantage of the fact that you more-or-less
forced me to reply and use the opportunity to respond to
another post that begged a response but to which I couldn't
reply and keep my word. Don't bother telling me that 
I'm
cheating - you won't get an answer.
***********************************************************
> You still don't understand, don't try to, 
and are trolling
again.
I try very hard to understand. Your statement is idiotic, as
you cannot possibly know what goes on in my mind - with the
current state of psycho-technology, that is. And as for
trolling, if you mean by that nuisance posting I'll have
to observe that it takes one to know one. If you believe I'm
doing it, then so are you.
 usual way to find mathematical results...
I wish I'd known that before posting questions to sci.math
This by the way is an attempt at making a witty remark,
something it seems is not particularly appreciated by this
group, which is another reason why I'm leaving.
Good, oh goody!
(There. That saves you the trouble of writing it)
nature of science.
> Says you. You had trouble with Peano's curve too, IIRC
I now understand the Peano construct, even to the point of
confidently telling you (a mathy, I assume) that it is_not_a
curve (OED, Chamber's and probably Webster's, 
q.v.. Mathies
only call it a curve because they assume that everyone knows
what they mean when they do). And here I am telling this to
someone who implies that _he_understands science. Maybe,
just maybe, Sir Karl Popper had a glimmer of insight but -
and here's the surprising non-surprise - you 
ain't him.
appartment, you try *not * to keep your mind open to
>the possibility that passing through the walls or the
> window could after all work this time.
Not at all. I did explain, had you been listening, that I
wear a Joe Public hat on these occasions. Silly
person! Beam me through the wall, Scotty. I'd use the window
but my anti-grav. Ôs kaput .
************************************************************
Now I'll do us all a favour and bugger off. Can you return
it (the favour, that is)?
Farewell,_again_,
John.
Mathematics may be defined as the subject in which we never
know what we are talking about, nor whether what we are
saying is true - Bertrand Russell
P.s. I know what Russell intended to say here, but as a
scientist I'm going to keep an open mind about that ;-)))
J
===
Subject: Re: Peano's space-filling curve
> I now understand the Peano construct, even to the point of
> confidently telling you (a mathy, I assume) that it is_not_a
> curve (OED, Chamber's and probably 
Webster's, q.v.. Mathies
> only call it a curve because they assume that everyone knows
> what they mean when they do).
The online OED says:
1. a. Geom. A curved line: a locus which may be conceived to
be traced by a
moving point, the direction of whose motion continuously
changes or
deviates
from a straight line. (In Higher Geometry, extended to
include the straight
line.)
According to this definition, the Peano curve is a curve,
although one
might quibble that the OED definition is overly restrictive in
that it
seems to require a curve to be differentiable.
The first definition according to M-W is:
Main Entry: 3curve
Function: noun
1 a : a line especially when curved : as (1) : the path of a
moving point
(2) :
a line defined by an equation so that the coordinates of its
points are
functions of a single independent variable or parameter b :
the graph of a
variable
Note that there is nothing here that implies
differentiability. The
Peano curve is definitely a curve according to this 
definition.
In both definitions a curve is described as a line, which I
suspect is
the basis of your objection, but both definitions go on to
include a
locus or path of a moving point, which obviously need not be a
line
in the usual sense.
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.

===
Subject: Re: Peano's space-filling curve
> I now understand the Peano construct, even to the point of
> confidently telling you (a mathy, I assume) that it 
is_not_a
> curve (OED, Chamber's and probably 
Webster's, q.v.. Mathies
> only call it a curve because they assume that everyone
knows
> what they mean when they do).
> The online OED says:
> 1. a. Geom. A curved line: a locus which may be conceived
to be traced by
> a moving point, the direction of whose motion continuously
changes or
> deviates from a straight line. (In Higher Geometry,
extended to include
> the straight line.)
> According to this definition, the Peano curve is a curve,
although one
> might quibble that the OED definition is overly restrictive
in that it
> seems to require a curve to be differentiable.
> The first definition according to M-W is:
> Main Entry: 3curve
> Function: noun
> 1 a : a line especially when curved : as (1) : the path of
a moving point
> (2) : a line defined by an equation so that the coordinates
of its points
> are functions of a single independent variable or parameter
b : the graph
> of a variable
> Note that there is nothing here that implies
differentiability. The
> Peano curve is definitely a curve according to this
definition.
> In both definitions a curve is described as a line, which I
suspect
is
> the basis of your objection, but both definitions go on to
include a
> locus or path of a moving point, which obviously need not
be a
line
> in the usual sense.
How do these august dictionaries define mathy/mathies?
--
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: Peano's space-filling curve
> that not everything you say is true.
>Careful, David. I would have thought that a genius-level
>mathematician would know that there exists in law (UK, USA,
>International) certain strictures about what you can and
>can't impugn about a person's character.
Guffaw. All I said was that not everything you say is
true, by the way. You've been stating falsehoods
quite regularly through this entire thread.
>Don't worry. I won't sue you.
Guffaw.
>Even though it _was_yesterday
>And I'm sure it isn't so, but I must point 
out that in many
>circumstances what you have just written might be construed
>as nuisance posting.
Guffaw.
> I told you I wouldn't post
>again_unless_ you invited it -
Uh, no. For example in message 2j7sb4Ftg5eaU8@uni-berlin.de
you stated
Now get lost. Because that's what I'm going to 
do,
doubtless much to the relief of sci.math subscribers..
No conditional at all. Not that there's any reason that
you should feel obliged to honor a promise like that,
but for you to say that _I'm_ misrepresenting things
when what I'm saying is a matter of public record is
just stupid.
>and_I_think you just did.
>Now I'll take advantage of the fact that you more-or-less
>forced me to reply
giggle. I more or less _forced_ you to reply? Huh.
************************
David C. Ullrich
===
Subject: Re: JH is just making fools out of you
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IFjq420147;
>I've read his work. No one could really have this kind of
mentality.
>If he does, god help him. More likely, he's just spouting
>deliberately inßamatory drivel aimed at inciting
>annoyance/anger/laughter. The bottom line is he should be
ignored
>either way. The only argument in his favor is that he
probably has
>nothing else in his life in which case who are we to take
away this
>little fun he has?
Good point and attitude.
His inßammatory drivel, diatribes, and
ego-boosting pontifications and claims of
supposed mathematical discoveries can be
ignored.
But, perhaps you missed the fact that he
actually duped some mathematics journal to
publish a ßawed mathematical paper online.
I would take that a little more seriously and
not ignore it.
Were it not for this newsgroup and the one
or two rational posters who found the ßaws
in his paper and who contacted that journal
(don't believe James Harris' Nazi-like
propaganda of allegedly vicious attacks on
the journal editor by mulitple posters),
then James Harris could claim undeserved
credibility.
But the journal was wise enough to
yank the crank - it de-published James Harris.
And I wholeheartedly support that journal's
actions.
Any idiot can put up his own website or blog
or self-publish his own drivel, but to get an
outside publication to _allegegly_ review and/or
peer review, approve, and publish his garbage,
that means a whole lot to cranks and nutballs
like James Harris.
James Harris is, somehow, a member of some
allegedly high-IQ society, but that's not
enough. He appears to want legitimacy and
credibility as a mathematician, and
publication by another's journal is a vital step.
Of course, James Harris knew his paper was
ßawed, or in the very least subject to questions
and challenges - numerous posters on this newsgroup
pointed the ßaws and errors out to him months
before this online journal incident.
You say he has nothing else in his life in which
case who are we to take away this little Ôfun'
he has.
His little fun is now extending to supposed
mainstream mathematical journals. His little
fun is in duping the journals with seemingly
legitimate but hopelessly ßawed garbage.
And his little fun is to rub publication of
his crap in this newsgroup's collective face,
especially the faces of those posters who, with
much greater mathematical ability than James
Harris, pointed out his errors.
Even now, James Harris, with his buddy Quinn
Tyler Jackson (another Nazi-like truth evading
crank and member of that closed high-IQ group),
are alleging that James Harris was right
and this newsgroup wrong about many issues,
including the journal incident.
We can't stop cranks like James Harris, but
we can and should at least counter their LIES
and insanity with the truth, the verifiable
truth, and nothing but the truth.
And insulting pseudonyms like:
James Harris is Nuts(TM)
===
Subject: Re: JH is just making fools out of you
> But the journal was wise enough to
> yank the crank - it de-published James Harris.
> And I wholeheartedly support that journal's
> actions.
I don't. If the JSH paper had stayed published, and the
journal had then
James would be on record as a devious crank. Any future
attempt of his to
get published would likely result in the editor and/or
referees finding his
publication and the ensuing rebuttal(s).
===
Subject: Re: JH is just making fools out of you
>But the journal was wise enough to
>yank the crank - it de-published James Harris.
>And I wholeheartedly support that journal's
>actions.
> I don't. If the JSH paper had stayed published, and the
journal had then
> James would be on record as a devious crank. Any future
attempt of his
to
> get published would likely result in the editor and/or
referees finding
his
> publication and the ensuing rebuttal(s).
There are no correct rebuttals.
I've explained before but you people seeem to despise
mathematical
logic, so after I explain you just keep repeating the same old
falsehood.
Here YET AGAIN is the problem with claims of a rebuttal as
they all
rely on the assumption that any infinitely sized rings where
-1 and 1
are the only units that are integers are part of the ring of
algebraic
integers.
That assumption is not proven in all of mathematics, and
cannot be
proven because it is provably false.
ALL claims of errors or claims of counterexamples depend on
the false
assumption.
So, in fact, those sci.math posters succeeded in getting a
CORRECT
paper censored by getting to the chief editor Ioannis
Argyros, and
likely, have made certain that when the full story comes out
his
career is destroyed.
Basically, they cheated. Doing a coordinated email assault
against a
math paper is just wrong. These people though didn't pause,
didn't
stop to consider that the paper passed FORMAL peer review,
and didn't
care about my ability to shoot down their claims.
You people on sci.math can repeat lies all you want, but the
math
doesn't change. You can post and post and post as if your
words
change mathematics but that just makes you...cranks and
crackpots.
James Harris
===
Subject: Re: JH is just making fools out of you
>But the journal was wise enough to
>yank the crank - it de-published James Harris.
>And I wholeheartedly support that journal's
>actions.
> I don't. If the JSH paper had stayed published, and the
journal had then
> James would be on record as a devious crank. Any future
attempt of his
to
> get published would likely result in the editor and/or
referees finding
his
> publication and the ensuing rebuttal(s).
Sure, the editor didn't act professionally, but he was
probably angry.
He got clotheslined. And he's pissed. And he's 
got every
reason to be.
He was the victim of Harris' fraud. And why _his_ journal?
Because his
journal has a poor reputation? Should he just lie down and
shut up
over this? Take it like a man? Become the target of every
crank that
wants to legitimize his trash?
His actions may have been wrong, but they are certainly
understandable.
===
Subject: Re: JH is just making fools out of you
>I've read his work. No one could really have this kind of
mentality.
>If he does, god help him. More likely, he's just spouting
>deliberately inßamatory drivel aimed at inciting
>annoyance/anger/laughter. The bottom line is he should be
ignored
>either way. The only argument in his favor is that he
probably has
>nothing else in his life in which case who are we to take
away this
>little fun he has?
> Good point and attitude.
> His inßammatory drivel, diatribes, and
> ego-boosting pontifications and claims of
> supposed mathematical discoveries can be
> ignored.
> But, perhaps you missed the fact that he
> actually duped some mathematics journal to
> publish a ßawed mathematical paper online.
> I would take that a little more seriously and
> not ignore it.
> Were it not for this newsgroup and the one
> or two rational posters who found the ßaws
> in his paper and who contacted that journal
> (don't believe James Harris' Nazi-like
> propaganda of allegedly vicious attacks on
> the journal editor by mulitple posters),
> then James Harris could claim undeserved
> credibility.
> But the journal was wise enough to
> yank the crank - it de-published James Harris.
> And I wholeheartedly support that journal's
> actions.
> Any idiot can put up his own website or blog
> or self-publish his own drivel, but to get an
> outside publication to _allegegly_ review and/or
> peer review, approve, and publish his garbage,
> that means a whole lot to cranks and nutballs
> like James Harris.
> [ snip ]
> His little fun is now extending to supposed
> mainstream mathematical journals. His little
> fun is in duping the journals with seemingly
> legitimate but hopelessly ßawed garbage.
> And his little fun is to rub publication of
> his crap in this newsgroup's collective face,
> especially the faces of those posters who, with
> much greater mathematical ability than James
> Harris, pointed out his errors.
Incorrect results are published all the time, it's no big
deal. In most
cases they are simply ignored; and sometimes somebody
publishes a
response pointing out the errors.
If I could choose between
(A) countless hours wasted by many competent mathematicians
in pointless
arguments with JH in sci.math, plus the resulting pollution
of this
newsgroup;
and
(B) an incorrect paper slipping through the peer review
process to be
published and afterward ignored,
I would choose (B) any time.
On a related subject, I find it strange that the Southwest
Journal of
Pure and Applied Mathematics first published 
JH's paper and
then deleted
it from its web site. If the term publish means anything, it
is that
once something is published, it is published forever. The
right way to
deal with the situation at hand would be to leave the paper
available,
and add a note explaining that errors have been found in the
paper, or
mistakes have been made in the peer review process, or
whatever.
GF
===
Subject: Re: Math research, passion is important
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IFjqw20151;
>(snip)
>but at least both sides appear to be
>passionate about math.
>James Harris
>Well, for you, passion is your only substitute
>for competence in mathematics, and the truth,
>and manners.
>I'm sure you and Quinn Tyler Jackson are
>both very passionate. Very, very
>passionate. Getting together, being
>extremely passionate about many aspects
>of your, um, thoughts.
>James Harris is Nuts(TM) (and passionate!)
It's for all the world to see - in another
I like James Harris.
HA HA HA HA HA HA HA HA HA HA HA HA HA HA!
===
Subject: program to search for finite models of a sketch?
There are model finders (MACE, FINDER, SEM), which
(exhaustively)
search for finite models of first-order axioms.
If I understand category-theoretic sketches properly, one
could search
for models (in the category of finite sets and functions, for
example)
of some variety of sketch (finite product sketches, for
example.).
Does there exist such a program?
If not, I would like to try to write one.
Would you recommend papers for me to read on related topics
(representing sketches in a computer, for example)?
===
Subject: tangent in turning point
I'm looking for a software which can solve
the following problem:
I've measured some values for the velocity
of a test vehicle in a regular short period
of time. Thus, I have pais of values (velocity
x in time t). I would like to use these values
to draw a graph and find then the turning point.
Hence, the program should draw a tangent through
the turning point.
Please tell me if there is aprogram which is able to perform
this task and give me some short instructions for the
solution.
Christian Christmann
===
Subject: Facts That Make Walz Run.
Facts that make Steve run..Walz thinks we went to the moon.
(1) How did Apollo get through the Van Allen Belt Steve?
Radiation is a big problem when it comes to space travel and
the Earth's
magnetic field concentrates this radiation into the Van Allen
belts that
surround the Earth. No matter what, the Apollo crafts had to
go through
these belts and there was no way the Apollo crafts could
afford to take all
the weight of lead shielding with them..
They could not have survived travelling through the Van Allen
Belt without
suffering from radiation sickness, or death, without a 6 feet
thick solid
lead shield.
(2) Where are the stars?
Why are there no stars in the sky in the photographs taken
from the lunar
surface?
(3) The video shows Neil Armstrong climbing down the ladder
and stepping
onto the surface. If he was supposed to be the first man on
the Moon, who
took the video?
(4) They didn't have the computer technology in those days 
to
get to the
Moon and back. Recentley China said: The technology to get to
the moon
could come within the next ten years.
Questions Steve won't answer
Why does the American ßag appear to be blowing in a breeze?
In the vacuum of space this should not be possible.
http://www.astrocentral.co.uk/ßag.jpg
----------------------------
Why is there no blast crater?
When the lunar module landed it should have made a large
crater.
http://www.astrocentral.co.uk/belowlm.jpg
===
Subject: Re: Facts That Make Walz Run.
*plonk*
===
Subject: Re: Facts That Make Walz Run.
> Facts that make Steve run..Walz thinks we went to the moon.
> (1) How did Apollo get through the Van Allen Belt Steve?
> Radiation is a big problem when it comes to space travel
and the Earth's
> magnetic field concentrates this radiation into the Van
Allen belts that
> surround the Earth. No matter what, the Apollo crafts had
to go through
> these belts and there was no way the Apollo crafts could
afford to take
all
> the weight of lead shielding with them..
> They could not have survived travelling through the Van
Allen Belt
without
> suffering from radiation sickness, or death, without a 6
feet thick solid
> lead shield.
Of all the kook arguments, I always find this one the most
amusing.
Scientists literally spend their entire careers studying the
van Allen
belts, but somehow when this issue comes up all these guys
who ßunked
high school Earth science suddenly become experts on the
subject.
Since XC is just just an expert, perhaps he can enlighten us
all with the typical dose rates (mR/hr) and spectra that
would be encountered by a spaceship passing through the van
Allen belts so we can more accurately assess his claims.
No? Somehow I'm not surprized.
> (2) Where are the stars?
> Why are there no stars in the sky in the photographs taken
from the lunar
> surface?
Asked and answered many times. In a word, contrast
http://www.badastronomy.com/bad/tv/foxapollo.html#stars
> (3) The video shows Neil Armstrong climbing down the ladder
and stepping
> onto the surface. If he was supposed to be the first man on
the Moon, who
> took the video?
Is that really the best you can do?
Presumably if XC were preparing for one of the biggest
moments in
history, he'd be too dumb to think to mount a camera in
advace.
Lily, NASA was more clever than that.
The camera was mounted to another leg of the lander, which
explains the
somewhat awkward angle of the shot.
> (4) They didn't have the computer technology in those days
to get to the
> Moon and back. Recentley China said: The technology to get
to the moon
> could come within the next ten years.
Computers don't ßy to the moon. The calculations are 
pretty
tough
to do by hand, but well within computing capability at the
time. Since
they weren't running Windows, there wasn't a 
lot of overhead.
> Questions Steve won't answer
> Why does the American ßag appear to be blowing in a breeze?
> In the vacuum of space this should not be possible.
> http://www.astrocentral.co.uk/ßag.jpg
This picture merely shows wrinkles.
Flag is only seen to move when astronauts are handling
it.
> ----------------------------
> Why is there no blast crater?
> When the lunar module landed it should have made a large
crater.
> http://www.astrocentral.co.uk/belowlm.jpg
First an expert on radiation, and now on propellants. Do 
XC's
talents know no bounds?
Moron.
-E
===
Subject: Re: Facts That Make Walz Run.
> Facts that make Steve run..Walz thinks we went to the moon.
> (1) How did Apollo get through the Van Allen Belt Steve?
> Radiation is a big problem when it comes to space travel
and the Earth's
> magnetic field concentrates this radiation into the Van
Allen belts that
> surround the Earth. No matter what, the Apollo crafts had
to go
through
> these belts and there was no way the Apollo crafts could
afford to take
> all the weight of lead shielding with them..
> They could not have survived travelling through the Van
Allen Belt
> without suffering from radiation sickness, or death,
without a 6 feet
> thick solid lead shield.
> Of all the kook arguments, I always find this one the most
amusing.
> Scientists literally spend their entire careers studying
the van Allen
> belts, but somehow when this issue comes up all these guys
who ßunked
> high school Earth science suddenly become experts on the
subject.
Yet, you can't refute those high school ßunkies
> Since XC is just just an expert, perhaps he can enlighten us
> all with the typical dose rates (mR/hr) and spectra that
> would be encountered by a spaceship passing through the van
> Allen belts so we can more accurately assess his claims.
The freaking danger of the Van Allen belt is high-energy
protons..Spin it
any way ya want; We DO NOT HAVE THE TECHNOLOGY TO GET THROUGH
IT TODAY.
Much less then.........If Apollo was in/cased in a 6-ft lead
coffen than
the crew may have survived the trip to the moon.
> No? Somehow I'm not surprized.
how we got through it...
> (2) Where are the stars?
> Why are there no stars in the sky in the photographs taken
from the
lunar
> surface?
> Asked and answered many times. In a word, contrast
> http://www.badastronomy.com/bad/tv/foxapollo.html#stars
You will have to do better than that.
> (3) The video shows Neil Armstrong climbing down the
ladder and stepping
> onto the surface. If he was supposed to be the first man on
the Moon,
who
> took the video?
> Is that really the best you can do?
> Presumably if XC were preparing for one of the biggest
moments in
> history, he'd be too dumb to think to mount a camera in
advace.
> Lily, NASA was more clever than that.
Uh, who got out and mounted that camra? Ohhhhhhhhh, Let me
guess. They
mounted the camra on earth and it also survived the Van Alen
belts on the
way up. Some camra that is...tooooooo bad they can't make em
like that
anymore...
> The camera was mounted to another leg of the lander, which
explains the
> somewhat awkward angle of the shot.
> (4) They didn't have the computer technology in those 
days
to get to
the
> Moon and back. Recentley China said: The technology to get
to the moon
> could come within the next ten years.
> Computers don't ßy to the moon. The calculations are 
pretty
tough
> to do by hand, but well within computing capability at the
time. Since
> they weren't running Windows, there wasn't a 
lot of
overhead.
Todays computers could do the calculations; But we don't 
have
the materals
to make a space-ship ÔYet' Or do ya think that 
the
space-shuttle could make
it to the moon?
> Questions Steve won't answer
> Why does the American ßag appear to be blowing in a breeze?
> In the vacuum of space this should not be possible.
> http://www.astrocentral.co.uk/ßag.jpg
> This picture merely shows wrinkles.
L@@K at the video.
> Flag is only seen to move when astronauts are handling
> it.
> ----------------------------
> Why is there no blast crater?
> When the lunar module landed it should have made a large
crater.
> http://www.astrocentral.co.uk/belowlm.jpg
> First an expert on radiation, and now on propellants. Do
XC's
> talents know no bounds?
> Moron.
> -E
Moron? FOAD!
===
Subject: Re: Facts That Make Walz Run.
> (3) The video shows Neil Armstrong climbing down the
ladder and stepping
> onto the surface. If he was supposed to be the first man on
the Moon,
who
> took the video?
> Is that really the best you can do?
> Presumably if XC were preparing for one of the biggest
moments in
> history, he'd be too dumb to think to mount a camera in
advace.
Actually, I find XC and his/her ilk somewhat fascinating
(assuming
XC is
not just another troll - which would be fascinating as well,
but in a
different way). While doubting the veracity of scientific and
engineering facts like space travel, this kind of people all
too often
forget their skepticism when dealing with people like John
Edward,
Van Praagh and many other crooks that prey on sers. Talk
about a
compartmentalized brain - or an unused one.
===
Subject: Re: Facts That Make Walz Run.
> (3) The video shows Neil Armstrong climbing down the
ladder and
stepping
> onto the surface. If he was supposed to be the first man
on the Moon,
> who took the video?
> Is that really the best you can do?
> Presumably if XC were preparing for one of the biggest
moments in
> history, he'd be too dumb to think to mount a camera in
advace.
> Actually, I find XC and his/her ilk somewhat fascinating
(assuming XC is
> not just another troll - which would be fascinating as
well, but in a
> different way).
> While doubting the veracity of scientific and
> engineering facts like space travel, this kind of people
all too often
> forget their skepticism when dealing with people like John
Edward,
> Van Praagh and many other crooks that prey on sers. Talk
about a
> compartmentalized brain - or an unused one.
Not a troll...........Just wish someone could explain how we
got through
the
Van Allen belts.
Steve Walz is a physicist who seems to believe that we went
to the moon.
Yet, he can't explain how humans could withstand the
radiation factors of
the Van Allen Belts..........
===
Subject: Re: Facts That Make Walz Run.
Originator: richard@cogsci.ed.ac.uk (Richard Tobin)
>Yet, he can't explain how humans could withstand the
radiation factors of
>the Van Allen Belts..........
So tell us exactly what these radiation factors are, with
references
to show that they can't be withstood.
-- Richard
===
Subject: Re: Facts That Make Walz Run.
> (3) The video shows Neil Armstrong climbing down the
ladder and
stepping
> onto the surface. If he was supposed to be the first man
on the Moon,
who
> took the video?
>Is that really the best you can do?
>Presumably if XC were preparing for one of the biggest
moments in
>history, he'd be too dumb to think to mount a camera in
advace.
> Actually, I find XC and his/her ilk somewhat fascinating
(assuming XC is
> not just another troll - which would be fascinating as
well, but in a
> different way). While doubting the veracity of scientific 
and
> engineering facts like space travel, this kind of people
all too often
> forget their skepticism when dealing with people like John
Edward,
> Van Praagh and many other crooks that prey on sers. Talk
about a
> compartmentalized brain - or an unused one.
But look, he is right!
http://users.pandora.be/vdmoortel/dirk/MoonLanding/
MoonLanding.html
Dirk Vdm
===
Subject: Re: Facts That Make Walz Run.
>(3) The video shows Neil Armstrong climbing down the
ladder and stepping
>onto the surface. If he was supposed to be the first man on
the Moon,
who
>took the video?
>Is that really the best you can do?
>Presumably if XC were preparing for one of the biggest
moments in
>history, he'd be too dumb to think to mount a camera in
advace.
> Actually, I find XC and his/her ilk somewhat fascinating
(assuming
XC is
> not just another troll - which would be fascinating as
well, but in a
> different way). While doubting the veracity of scientific 
and
> engineering facts like space travel, this kind of people
all too often
> forget their skepticism when dealing with people like John
Edward,
> Van Praagh and many other crooks that prey on sers. Talk
about a
> compartmentalized brain - or an unused one.
True, there are an amazing number of people who don't 
believe
the
moon landing, but believe absolutely *everything* else they
hear or read on the internet.
-E
===
Subject: Re: Facts That Make Walz Run.
> Facts that make Steve run..Walz thinks we went to the moon.
> (1) How did Apollo get through the Van Allen Belt Steve?
> Radiation is a big problem when it comes to space travel
and the
> Earth's magnetic field concentrates this 
radiation into the
Van Allen
> belts that surround the Earth. No matter what, the Apollo
crafts had
> to go through these belts and there was no way the Apollo
crafts could
> afford to take all the weight of lead shielding with them..
> They could not have survived travelling through the Van
Allen Belt
> without suffering from radiation sickness, or death,
without a 6 feet
> thick solid lead shield.
I'm wondering how we survive down here on earth without
a 6-foot thick solid lead shield surrounding the entire earth
now....
(I'm assuming the Van Allan belts aren't made 
of lead, but
I could be wrong.)
Bart
===
Subject: Re: Facts That Make Walz Run.
> Facts that make Steve run..Walz thinks we went to the moon.
> (1) How did Apollo get through the Van Allen Belt Steve?
> Radiation is a big problem when it comes to space travel
and the
> Earth's magnetic field concentrates this 
radiation into the
Van Allen
> belts that surround the Earth. No matter what, the Apollo
crafts had
> to go through these belts and there was no way the Apollo
crafts could
> afford to take all the weight of lead shielding with them..
> They could not have survived travelling through the Van
Allen Belt
> without suffering from radiation sickness, or death,
without a 6 feet
> thick solid lead shield.
>I'm wondering how we survive down here on earth without
>a 6-foot thick solid lead shield surrounding the entire
earth now....
It's like old cartoons, where someone runs off the edge of a
cliff
but doesn't start falling until he looks down and _realizes_
that
he's run off the edge of the cliff: Now that 
you've pointed
>(I'm assuming the Van Allan belts aren't made 
of lead, but
>I could be wrong.)
>Bart
************************
David C. Ullrich
===
Subject: Re: Facts That Make Walz Run.
Read:
http://www.pbs.org/wgbh/nova/transcripts/2610tothemoon.html
===
Subject: Re: Facts That Make Walz Run.
> Facts that make Steve run..Walz thinks we went to the moon.
> (1) How did Apollo get through the Van Allen Belt Steve?
> Radiation is a big problem when it comes to space travel
and the Earth's
> magnetic field concentrates this radiation into the Van
Allen belts that
> surround the Earth. No matter what, the Apollo crafts had
to go through
> these belts and there was no way the Apollo crafts could
afford to take
> all the weight of lead shielding with them..
> They could not have survived travelling through the Van
Allen Belt
without
> suffering from radiation sickness, or death, without a 6
feet thick solid
> lead shield.
> (2) Where are the stars?
> Why are there no stars in the sky in the photographs taken
from the lunar
> surface?
> (3) The video shows Neil Armstrong climbing down the ladder
and stepping
> onto the surface. If he was supposed to be the first man on
the Moon, who
> took the video?
> (4) They didn't have the computer technology in those days
to get to the
> Moon and back. Recentley China said: The technology to get
to the moon
> could come within the next ten years.
http://www.brainsluice.com/miscellanea/misc/moonlanding.html
--
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: Lessons from ideologies

Has this e owt to do with maths?
--
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: equivalent graphs and things
[I'm not a mathematician, so please bear with me. Or was 
that
beer with me?
I forgot :-) ]
I have 8 identical pieces of cable. I can connect any number
of them in any
number of ways, except that I can't tie both sides of any
cable together or
tie any two cables in parallel. IIRC we're talking unlabeled
nondirected
weighted simple graph. Weighted because the number of
segments between two
nodes matters.
Now I was trying to find an upper limit of the number of
possible
connections as follows:
represent the network as a matrix, with every column
representing one end
of
each of the 8 cables and every row the other end, so I get an
8x8 matrix,
with a_ij = 1 if the first end of cable I is connected to the
other end of
cable J and a_ij = 0 if there's no connection.
Since I don't want any loops a_ii is always zero. Parallel
cables are also
out, which means that a_ij and a_ji can't both be 1. (I hope
I'm doing this
right). With these limitations I'm still left with 3^28
different
configurations. Upper limit that is, because a number of
matrices will
result in multiple separate graphs.
It's obvious that you can't have that many 
different graphs.
Since all pieces of cable are identical, I can swap row_i and
row_j, and
column_i and column_j, and repeat this for any i and j, and
still have the
same graph. But I realize that I'm hardly lowering my upper
limit of 3^28
this way.
So now my question is: is there a fast algorithm which
creates all unique
configurations? If I have to make a wild guess I would
estimate that there
are only a few thousand of them, let's say 10^4. Now just a
way to find
them...
TIA
--
Steven
I'm not a mathematician, so please type slowly
===
Subject: Re: equivalent graphs and things
>I have 8 identical pieces of cable. I can connect any number
of them in
any
>number of ways, except that I can't tie both sides of any
cable together
or
>tie any two cables in parallel. IIRC we're talking 
unlabeled
nondirected
>weighted simple graph. Weighted because the number of
segments between two
>nodes matters.
>Now I was trying to find an upper limit of the number of
possible
>connections as follows:
Let's see if I understand this right, you want to count the
number of
simple
connected graphs that have 8 or less connections? Let's see
if we're on
the
same page by doing the low numbers.
1 cable -- 1 configuration _
2 cables -- 1 configuration __
3 cables -- 2 configurations ___ _|_
4 cables -- 3 configurations ____ _|__ _||_
Since you're not allowing loops, you start with 2 nodes at
the 2 ends of
the
first cable. For each cable you add, you add 1 node, so if you
use all 8
cables you will end up with 9 nodes.
Number the nodes based on the the order they were added,
except that the
left end of the first cable doesn't get a 
number, and you're
not allowed to
connect anything to it. You will still be able to construct
all possible
configurations.
So when you add the second cable, you have 1 choice, third
cable 2 choices,
and so on. For 8 cables, this comes out to 7! or 5040. 
That's
just an
upper bound; many of those will be congruent.
>So now my question is: is there a fast algorithm which
creates all unique
>configurations? If I have to make a wild guess I would
estimate that there
>are only a few thousand of them, let's say 10^4. Now just a
way to find
>them...
That sounds hard. Chemistry has a related problem in
hydrocarbon isomers.
How many different ways can you connect 9 carbon atoms
together? It's not
exactly the same problem, because one carbon atom can be
adjacent to at
most
4 others.
--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Infinite rings, and a simple requirement
What I've found time and time again is that certain sci.math
posters
will say just about anything to obscure the truth.
I have a simple answer to people claiming there are
counterexamples to
my argument in my paper Advanced Polynomial Factorization.
That is
the paper which sci.math posters in an email campaign against
it
managed to get yanked by Ioannis Argyros, the chief editor of
the
Southwest Journal of Pure and Applied Mathematics.
I noted that Hall and others were relying on the assumption
that rings
where -1 and 1 are the only integer units must be part of the
ring of
algebraic integers, which is in fact a false assumption.
Then I noticed posts about Z/2, and Z/3 as certain posters
acted as if
they were making some major points.
They are talking about finite rings.
Ok, so I didn't know the terminology, so what?
That's the kind of behavior that I see as showing that some
of you
know the truth but deliberately are fighting mathematics.
Terminology comes and goes, but the mathematical truths
remains the
same.
But if you consider *infinite* rings where -1 and 1 are the
only
integers that are units then you can consider the question of
whether
or not such rings must be members of the ring of algebraic
integers.
Then you can look at my work as a proof that they do not.
Then you can see why objections like Hall's, 
Magidin's and
Decker's
are specious.
Then you can understand that those people managed to get a
correct
paper, one that had passed FORMAL peer review, censored by a
direct
appeal by email.
So, they're going against mathematical protocol, using an
email attack
on a correct paper that passed peer review, and they refuse to
acknowledge that their objections have been shot down.
And they get away with it using a hidden assumption that they
don't
even bother acknowledging or trying to prove.
When I'm done I'll make the case that 
mathematicians CANNOT
be trusted
to police themselves.
Also I'll make the case that Usenet has shown itself to be a
hostile
inßuence to intellectual exploration and needs to be
regulated.
Posters who lie and conspire on Usenet need to realize that
they will
be punished by the law, as at this point, there's no way to
stop gangs
of posters from engaging in outrageous behavior, like
managing to
censor a correct math paper by a coordinated email assault.
James Harris
===
Subject: Re: Infinite rings, and a simple requirement
Attn James...
I'm new to this group and I'm intrigued by the 
attention you
seem to have
attracted.
I'd be grateful if you could post (or email me directly)
something on which
I could form my own opinion. Ideally something which you
consider to be
unique from a traditional mathematical point of view.
Something which is
self contained and can (in principle) be verified by your
average post-grad
(nearly but not quite in my case).
Kev
===
Subject: Re: Infinite rings, and a simple requirement
> What I've found time and time again is that certain
sci.math posters
> will say just about anything to obscure the truth.
> I have a simple answer to people claiming there are
counterexamples to
> my argument in my paper Advanced Polynomial Factorization.
That is
> the paper which sci.math posters in an email campaign
against it
> managed to get yanked by Ioannis Argyros, the chief editor
of the
> Southwest Journal of Pure and Applied Mathematics.
> I noted that Hall and others were relying on the assumption
that rings
> where -1 and 1 are the only integer units must be part of
the ring of
> algebraic integers, which is in fact a false assumption.
You have noted it time and time again (at least four times in
the past
two days), but have never substantiated that claim.
Please do so. I'll make it as simple as possible: just state
for the
record which of the following elements of my argument is in
error.
Here is an outline of the argument; if you could just point
out which
of the statements relies on the alleged false assumption then
you will
assist in furthering this discussion.
In this outline, a refers to one of the coefficients that you
claim
must be coprime to 5. P(x) refers to the polynomial x^3 - 12
x^2 + 65,
and I note that P(-a) = 0.
1. The following formulas are true:
q(x)r(x) = (64 x + 128)P(x) + 5
r(x)s(x) = (32 x + 72)P(x) + x,
where q,r,s are defined as follows:
q(x) = 8 x^2 - 76 x - 185
r(x) = 8 x^2 - 4 x - 45
s(x) = 4 x^2 - 37 x - 104
2. Since -a is a root of P(x), one has the following
factorizations:
q(-a)r(-a) = 5
r(-a)s(-a) = -a
3. The minimal polynomial of r(-a) is given as:
MP_r = x^3 - 969 x^2 + 315 x + 5
which is irreducible over Q, and shows that
r(-a) is an algebraic integer, but not a unit
in that ring.
shared by a and 5 is not a unit in the ring
of algebraic integers.
5. a and 5 are not coprime in the ring of algebraic
integers, since they share a non-unit factor in
common.
I claim that not one of these individual points depends on the
assumption you say I've made. The above constitutes the 
entire
argument and all commentary should be restricted to the
argument,
so let's see where your objection lies.
> Then I noticed posts about Z/2, and Z/3 as certain posters
acted as if
> they were making some major points.
> They are talking about finite rings.
> Ok, so I didn't know the terminology, so what?
> That's the kind of behavior that I see as showing that 
some
of you
> know the truth but deliberately are fighting mathematics.
You act as though you were blindsided, but in reality your
wording
has always been so vague that no one could tell what you've
been
talking about. It's only been in your specific 
claims that
you've
managed to say anything specific enough to assess. The fact
that
all these specific claims have fallen ßat is enough to 
cause
people to suspect your arguments are not worth digging into.
> Terminology comes and goes, but the mathematical truths
remains the
> same.
> But if you consider *infinite* rings where -1 and 1 are the
only
> integers that are units then you can consider the question
of whether
> or not such rings must be members of the ring of algebraic
integers.
> Then you can look at my work as a proof that they do not.
> Then you can see why objections like Hall's, 
Magidin's and
Decker's
> are specious.
Non sequitur. You have yet to show that your remarks are
relevant to
the argument at hand. I've given you ample opportunities to
pick each
piece apart, at your leisure, and come back with the goods.
To date,
your responses have been without demonstrable substance.
If you had something, anything, resembling a false statement
or
invalid assumption in the above argument, you would have
produced
it.
> Then you can understand that those people managed to get a
correct
> paper, one that had passed FORMAL peer review, censored by
a direct
> appeal by email.
You cannot see the error, you refuse to comprehend the
argument that
refutes your conclusion, and so you're stymied. So, just
spend a little
time working over what I've laid out. If 
there's any
statement that
relies on the assumption you claim, I for one would be glad
to hear it.
> So, they're going against mathematical protocol, using an
email attack
> on a correct paper that passed peer review, and they refuse
to
> acknowledge that their objections have been shot down.
be retracted, nor did I expect it. Your claim that you have
countered
my argument holds absolutely no water.
> And they get away with it using a hidden assumption that
they don't
> even bother acknowledging or trying to prove.
I claim the assumption is not there, that the same proof
would hold in
any ring where the particulars held true. If you had the
goods, you
would have pointed out the invalid statement. There's still
time!
> When I'm done I'll make the case that 
mathematicians CANNOT
be trusted
> to police themselves.
I see.
THE MATH POLICE..
Where's the legal authority? The legislation? The budget?
Let's have it, then. This posturing of yours is awfully
tiresome.
Perhaps you should take your sheaf of every scrap of paper
you've
ever written on, put it into a pile of manila folders and
clasp
envelopes, and I guess those accordion binders are pretty
efficient,
too, and tr on down to the FBI again.
I hope you still have your old Army uniform, I suppose that
if you
wore it on your visit to the FBI, it would enhance the
respect our
brave fighting men and women factor.
> Also I'll make the case that Usenet has shown itself to be
a hostile
> inßuence to intellectual exploration and needs to be
regulated.
Time's a-wastin'!
> Posters who lie and conspire on Usenet need to realize that
they will
> be punished by the law, as at this point, there's no way 
to
stop gangs
> of posters from engaging in outrageous behavior, like
managing to
> censor a correct math paper by a coordinated email assault.
Violaters will be persecuted to the full extent of the law.
No outrageous behavior, neither.
I'm not impressed. You'd think that, as grown 
man, you would
have
realized by now that thumping on your chest, screaming about
heads
are a-gonna roll!!!, and pretending to threaten us with The
Hammer,
not to mention your pals the Generals who you may have to sic
on us,
and oh yeah, that letter to Time Magazine, wouldn't have any
effect
aside from causing a few chles across this great nation of
ours,
and perhaps... around the world!
Be a pal and let me know when the heat's coming down, on
account
of I gotta practice doing Jimmy Cagney, standing in the 2nd
story
window with a cigarette clenched in his jaw, holding a
Thompson
submachine gun
Come and get me, coppers!!
Except I'd have to substitute a laptop with a wireless
internet
connection for the tommy gun; more in line with the crime. At
any
rate, these last-stand scenes take planning and practice to
get
down correctly.
> James Harris
Dale
===
Subject: Re: Alcuin's sequence
Tony King schrieb in Nachricht
Hi Tony,
... late comes it, but it comes ...
>I am trying to do some research on alcuin's sequence, but
all the
>references that I can find quote the terms of the sequence as
the
>coefficients in the Maclaurin expansion of
[(1-x^2)(1-x^3)(1-x^4)]^-
>1. I find this surprising, as I believe the sequence to have
been
>named after Alcuin of York (735-804) who lived many years
before
>Maclaurin expansions were discovered.
>This sequence also gives the number of different triangles
that have
>integral sides and perimeter n. I would think that it is
much more
>likely that this is what Alcuin discovered, and years later
someone
>found that the above expansion acted as a generating
function.
>That is, of course, if the Alcuin who the sequence is named
after is
>Alcuin of York and not someone else with the same name.
>I would be most grateful if you could either confirm or
refute this,
>or point me in the direction of any references.
It is in fact named after Alcuin of York [1].
Alcuin's sequence [2] does not only give the number of
incongruent
triangles
of given perimeter N with integer sides, but also the number
of ways in
which N
empty casks, N casks half-full of wine and N full casks can
be distributed
to
3 persons in such a way that each one gets the same number of
casks and the
same amount of wine, and that's the link to Alcuin.
Alcuin is, more or less reluctantly, considered as the author
of the Latin
manuscript Propositiones ad acuendos juvenes, i.e. Problems
for the
mind-
sharpening of youngsters, written around 800 AD in France.
Roughly a
dozen
copies have survived the 1200 years till now and are
preserved in libraries
in
Rome, Vienna, Munich, Karlsruhe, London, Leiden and
Montpellier.
The manuscript contains a collection of 53 mathematical
problems without
any
theory, but with the solutions, some of which are incorrect.
Both the
complete
Latin Text and translations may be found at several websites
[3].
In Problem No. 12 of the Propositiones, 10 empty bottles, 10
bottles
half-full
of oil and 10 full bottles are to be fairly shared among
three sons.
Counting
only incongruent solutions, i.e. solutions that are no mutual
permutations
of
others, there are 5 ones:
: E1 H1 F1 | E2 H2 F2 | E3 H3 F3
: -------------+--------------+-------------
: 0 10 0 | 5 0 5 | 5 0 5 <-- Alcuin's solution
: -------------+--------------+-------------
: 1 8 1 | 4 2 4 | 5 0 5
: -------------+--------------+-------------
: 2 6 2 | 3 4 3 | 5 0 5
: -------------+--------------+-------------
: 3 4 3 | 3 4 3 | 4 2 4
: -------------+--------------+-------------
: 4 2 4 | 2 6 2 | 4 2 4
: -------------+--------------+-------------
where E_i, H_i, F_i means the number of empty, half-full and
full bottles
for
the i-th son.
Alcuin gave only one of the solutions, namely the first one
above.
In 1201, Leonardo da Fibonacci published the similar problem
of
distributing
7 empty, 7 half-full and 7 full barrels of wine fairly to
three persons, and
he
also gave only one of the two solutions.
Likely the first author who noticed the existence of more than
one solution
for
problems of this kind was Claude Gaspard Bachet de Meziriac,
best known as
the earliest translator of Diophant's works from Greek into
Latin, who
published
a book on recreational mathematics in 1612.
More on the history of such problems may be found in David
Singmaster's
gigantic
bibliography on the history of recreational mathematics [4].
The relations between fair sharing of barrels among three
persons and
triangles
with integral sides was established by David Singmaster in
the paper:
D. Singmaster: Triangles with Integer Sides and Sharing
Barrels.
The College Mathematics Journal 21, No. 4 (1990), p. 278 -
285 ,
a copy of which may be downloaded as a PDF file from [5].
The equivalence of the two problems can easily be seen by
observing that
for any triangle with sides a, b, c and circumference N the
relation
a + b + c = N
and the inequalities
a <= b + c
b <= c + a
c <= a + b
hold true, and by adding a to the first inequality etc., one
gets the
inequalities
a <= N/2
b <= N/2
c <= N/2 .
For the fair barrel sharing with 3 persons, on the other
hand, the
relations
E1 + E2 + E3 = H1 + H2 + H3 = F1 + F2 + F3 = N
E1 + H1 + F1 = E2 + H2 + F2 = E3 + H3 + F3 = N
trivially hold, and, as Singmaster proved, the relations
E1 = F1
E2 = F2
E3 = F3
E1 <= N/2
E2 <= N/2
E3 <= N/2
must additionally be valid, and the latter three are just the
triangle
inequalities from above.
In 1993, Dominic Olivastro published the book
D. Olivastro: Ancient Puzzles. Classic Brainteasers and Other
Timeless Mathematical Games of the Last 10 Centuries.
New York: Bantam Books, 1993
and in the chapter on fair sharing problems he introduced the
sequence
of the number of solutions for different values of N, which
he called
A_q,
... in honour of Alcuin ....
Following Olivastro's hommage to Alcuin, Neil Sloane seems 
to
have coined
the name Alcuin's sequence for the OEIS, and from there it
found its
way into Eric Weissteins Mathworld.
[1]
http://www-history.mcs.st-and.ac.uk/~history/Mathematicians/
Alcuin.html
[2]
http://mathworld.wolfram.com/AlcuinsSequence.html
http://www.research.att.com/projects/OEIS?Anum=A005044
[3]
Latin text with English translation:
http://www.beyond-the-illusion.com/files/History/Science/host1
-2.txt
---
Latin text with German translation:
http://www.mi.uni-erlangen.de/~geyer/geschima --> Skript
http://www.braumueller.at/files/1000/mt_12.pdf
---
Latin text with Italian and Spanish translations:
http://utenti.quipo.it/base5/
http://utenti.quipo.it/base5/alcuino/alcuintegra.htm
---
Portuguese translation:
http://www.malhatlantica.pt/mathis/Europa/Medieval/Alcuin/
Alcuino.htm
[4]
http://www.geocities.com/mathrecsources/ --> Paragraph 7.H.1
[5]
http://www2.edc.org/makingmath/ contains a link to
Singmaster's paper:
--> teacher handbook --> Getting Information --> READING ...
Hermann
--
>Tony
===
Subject: Limit Involving phi (Was:Limit Of Average Of
a(GCD(j,k))'s)
In Limit Of Average Of a(GCD(j,k))'s
>....
>Bonus result:
>(I am unsure here if I am correct.)
>For |x|< 1,
>limit{m->oo}
>(1/m^2) sum{k=1 to m} (1 - x^ßoor(m/k)) phi(k)
>(1/x -1) (3/pi^2) (sum{k=1 to oo} x^k /k^2),
>where phi(k) is the number of positive integers <= k and
coprime to
>k .
>(Right?)
Some more on this result:
More simply,
limit{m->oo}
(1/m^2) sum{k=1 to m} x^ßoor(m/k) phi(k)
=
(3/pi^2) (1-(1/x -1)(sum{k=1 to oo} x^k/k^2))
=
(3/pi^2) (sum{k=1 to oo} x^k (1/k^2 - 1/(k+1)^2))
--
Using the original result, at letting x = -1, we get the
specific
result (obvious but noteworthy):
limit{m->oo}
(1/m^2) (sum{1<=k<=m, ßoor(m/k)= odd} phi(k) )
= 1/4.
(The sum is over those k, 1<=k<=m, where
ßoor(m/k) is odd.)
Leroy Quet
===
Subject: Re: JH is just making fools out of you
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IJXFE08865;
>That JH is a troll is obvious to most of us.
>What is so amazing is that so many posters in sci.math get
sed into
>playing along. As they say, there's a new one born every 
day
...
Maybe some posters aren't sed/sered
in, but enjoy the exchanges, as a poster
or as a spectator. The back and
forth relieves some stress, or acts as a
diversion from the immediate demands of work.
It's sure better than fighting people in
the streets, in gangs or as members of
rival ethnic groups.
Also, I've learned a little more about
some obtuse mathematics from some of the
more rational and detailed mathematical
replies to James Harris and his math.
(snip)
> little fun he has?
Perhaps we all have some fun here, eh?
Kings of the past had their fools, and
villages had their idiots.
We have Usenet kooks.
And, unfortunately, they have us.
Anthony J. Natoli
===
Subject: Re: JH is just making fools out of you
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IJXFR08879;
(snip)
>If I could choose between
>(A) countless hours wasted by many competent mathematicians
in pointless
>arguments with JH in sci.math, plus the resulting pollution
of this
>newsgroup;
>and
>(B) an incorrect paper slipping through the peer review
process to be
>published and afterward ignored,
>I would choose (B) any time.
Ordinarily, I would agree, but we have
a credibility-craving crank.
Why can't James Harris waste his money and
self-publish (as suggested to M. Basti)
or satisfy himself with his blog or
his website? Because I, and maybe others,
expect him to misuse the publication by
that supposedly objective journal to
cause mischief in the mathematical community
outside of this newsgroup.
>On a related subject, I find it strange that the Southwest
Journal of
>Pure and Applied Mathematics first published 
JH's paper and
then deleted it
from its web site.
(snip)
Your points were debated ad nauseum in another
thread, with most posters on this newsgroup
agreeing with you (I'm possibly the lone exception).
You and other professional and more rational
mathematicians on this newgroup are, may I
suggest, too professional and too rational, and you
view James Harris as just another newsgroup
poster. I think that is a mistake.
Besides Harris being a crank and other insulting
designations, James Harris is the source, not the
victim, of such shrill discord on this newsgroup.
Regretful, but that's this ugly reality called
sci.math (and other newsgroups James Harris
injects himself into).
James Harris is Nuts(TM)
===
Subject: Re: Math research, passion is importantQ
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IJXFe08873;
(snip)
>To be perfectly honest, Quinn has shown no signs of losing
his temper,
>or acting in an immature fashion.
Will, you haven't looked at a recent post by Quinn,
repeated (way down) below. You judge whether he
lost his temper or acted immature.
At first, in the recent James Harris posts, Quinn
posted by only his first name and with a fake
E-mail address to defend James Harris.
As I challenged James Harris repeatedly on his
assertions, and after I and others suggested that
maybe Quinn and the fake E-mail address were
James Harris anonymously defending himself as to
whether some mathematicians reviewed Harris' proof,
Quinn then began posting under his full name Quinn
Tyler Jackson to show that Harris was not faking
being Quinn, and Quinn also revealed that he and
James Harris were colleagues and friends in a closed
list, showing that Quinn was not some objective
defender of James Harris.
Was Quinn mature by secretly hiding his identity and
association with James Harris? Or was it intellectually
dishonest to wrongly back up his buddy as to alleged facts?
Will, perhaps you can explain why neither Quinn nor his
good friend James Harris provided the name or E-mail
address of a single mathematician of their alleged
closed list who allegedly reviewed Harris' proof,
and who allegedly failed to find an error in it.
I repeatedly challenged them to put up or shut up
on their assertion that:
several mathematicians failed to find the error.
Instead, they evaded repeatedly. Quinn said
another poster Einstein answered my challenge,
but all Einstein did was provide the website
of the high-IQ group to which Quinn and Harris belong.
And then, after I pointed out that that website
listed only one mathematician, Quinn alleged that
the website _may_ be out of date, and he attempted
to playfully deßect from the names of the mathematicians.
And his challenge to me to call a bluff on some
alleged algorithm was a diversion to continue the
evasion by both James Harris and Quinn to my
original challenge to name their reviewing mathematicians.
Is Quinn acting mature or even tempered, like you suggest?
Why don't James Harris and Quinn Tyler Jackson name
the mathematicians who reviewed and found no error in
James Harris' paper on factoring? Can't they 
put
their assertions to outside scrutiny?
Either:
A) no mathematicians actually reviewed it, and so
Harris and Quinn LIED to support their lame position,
that errors in Harris' paper were hard to find 
by
mathematicians (but Harris somehow found them himself);
B) mathematicians in the high-IQ group did review it,
and missed the error, thus suggesting that the high-IQ
folks are not so smart, or just not very skilled
mathematicians (probably amateurs), and so Harris and
Quinn did not want to embarass their high-IQ group and
in turn themselves for being members of such group; or
C) Harris and Quinn did not want verification or lack
thereof in any way, shape, or manner for their
assertions on this newsgroup.
Any of these actions by James Harris and Quinn Tyler
Jackson is intellectually dishonest, and perhaps you
may consider them immature as well. As in my other
silly posts regarding James Harris or Andrew Wiles
proving the Riemann Hypothesis, how can we respect
and deal with James Harris _mathematically_ on this
newsgroup if he and his friends try to dupe us
with lies, distortions, and evasions?
Can't we ask simple questions, challenges to be sure,
but reasonable inquiries as to facts and circumstances
regarding James Harris and his work?
Will, you tell me if the following is mature or
even tempered on the part of Quinn Tyler Jackson.
Quinn's verbatim post:
===
Subject: Re: Factoring paper is wrong
Author: Quinn Tyler Jackson 
JHiN said:
> (I've drawn my own conclusions with a cursory
> review.)
You drawing conclusions with a Crayola.
> I'm not satisfied - only one of the editors 
is
> listed as a mathematician.
Editorial Board & Contributors. Old, out of date listing.
Hmmm... let's
draw
some conclusions based on outdated data temporarily stored on
a server
while
the main site is down. (Maybe the up to date site is up -- I
don't know.
Maybe it is the same as that one -- I don't know.)
Which is which? Who is who? Care to draw more conclusions?
I'm not an
editor, neither is James. Round and round it goes, where it
lands, nobody
knows. Well, you know. But .... yeesh -- technically, you're
nobody -- so
indeed, nobody knows.
Not nobody? Who *are* you, anyway? An anonymous ellipses. You
don't even
exist, JHiN, except in vaporspace. So far, I'd say you are a
computer.
> Any other mathematicians in this little group
> of yours, who reviewed Harris' work?
Who knows? Who cares? Who knows who cares? Who cares to know?
Whose nose
cares?
> Any idiot balancing his/her checkbook can call
> themselves a mathematician, but that doesn't
> qualify them to find errors, or to judge a paper
> if they find no errors.
And any poorly written ELIZA program hooked up to an email
account can
attack people from behind a thin sheet of anonymity and call
itself a
Good God, man, I gave you ample opportunity to call my bluff
on my
absolutely outrageous claim about having invented an efficient
Type < 2
parsing algorithm, and you were so busy running line 20367 of
your program
that you insisted instead that I answer a question Einstein
had already
answered.
If you were truly hanging around to protect the Integrity of
the Body of
Knowledge of Mathematics, surely you would have grabbed that
freebie
opportunity with both hands and called my bluff, told me to
put the cards
on
the table about such a boisterous claim.
But instead of focusing on my outrageous claim of invention,
you elected to
go all Crayola Computer on me and draw conclusions about a
small, private
list and the nature of its participants. You focused on a
question that was
already answered by a third party instead of the obvious one
that would
have
allowed you to call yourself Quinn Tyler Jackson is Nuts(R).
Time to get Grampa Simpson to review your code. Maybe he
forgot that GOTOs
make spaghetti code.
--
Quinn
Note: If you had the balls to stand behind your real name --
I'd treat you
with more respect. In the meantime, you technically don't
exist.
===
Subject: Re: Math research, passion is importantQ
Discussion, linux)
> As I challenged James Harris repeatedly on his
> assertions, and after I and others suggested that
> maybe Quinn and the fake E-mail address were
> James Harris anonymously defending himself as to
> whether some mathematicians reviewed Harris' proof,
> Quinn then began posting under his full name Quinn
> Tyler Jackson to show that Harris was not faking
> being Quinn, and Quinn also revealed that he and
> James Harris were colleagues and friends in a closed
> list, showing that Quinn was not some objective
> defender of James Harris.
> Was Quinn mature by secretly hiding his identity and
> association with James Harris? Or was it intellectually
> dishonest to wrongly back up his buddy as to alleged facts?
I don't think that Quinn's association with 
JSH via a closed
list was
hidden at all. Quinn brought it up himself (prior to giving
his full
name) as evidence that JSH is capable of friendly relations
with
mathematicians.
I think you're revising (a very short, small and 
insignificant)
history here.
--
A recruitment consultant I know thinks the most important
quality in
a winner is to be ly. To avoid wasting his time with
unly
applicants, he takes half the resumes piled on his desk and
throws
them straight in the bin. -- John Ramsden
===
Subject: Re: Math research, passion is importantQ
> Was Quinn mature by secretly hiding his identity and
> association with James Harris?
No. Einstein found the association -- I knew the association
was on the
Internet. Just because you didn't find the 
association
yourself, you're
asking if I am keeping secrets? It's not my problem if you
cannot do your
own legwork, dude.
Man, JHiN, you appear to be somewhat obsessed.
Post under your real name so I know your identity, and I'll
take back my
ßip comments about your being an Eliza program. I've even 
got
a pair of
shorts here I'm verily willing to eat. For all I know --
you're posting
under a pseudo to avoid being sued by someone. I've met your
likes before.
Vapors.
As for my alleged algorithm, if you had done the legwork,
you'd see
that
the damned thing is already posted on my website under the
name
adaptive(k),
you silly boy. Alleged. Good grief... My published results in
parsing have
been already cited in at least two doctoral dissertations on
parsing, as
well as cited in several refereed computer science papers, as
well as the
fact that I'm listed at the ACM portal, as well as ... well,
you get the
idea. I may be a nitwit, but I'm a cited nitwit. My Type < 2
parsing
results
are known in the Body of Knowledge already. You, on the other
hand, are
some unknown quantity X going under the pseudo JHiN.
My point wasn't to divert anyone's attention 
-- or even to
call attention
to
myself (I'm just yet another nobody special, and I admit to
it here of my
own free will) -- it was to emphasize the blaringly obvious
fact that you
are most certainly *not* about the math-music: you are about
socio-political
libelous slander. I suspect you have a strong need-to-belong,
and you
somehow feel that by appointing yourself resident anonymous
pundit, you've
made yourself belong in some bizarre way.
But isn't the Eliza program supposed to psychoanalyze me, 
and
not Ôt other
way around?
--
Quinn
===
Subject: Re: Math research, passion is importantQ
> But isn't the Eliza program supposed to psychoanalyze me,
and not Ôt
other
> way around?
Why would you think that
the Eliza program was supposed to psychoanalyze me, and not
t' other way
around?
===
Subject: Re: JH is just making fools out of you
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IJXPk08932;
> But the journal was wise enough to
> yank the crank - it de-published James Harris.
> And I wholeheartedly support that journal's
> actions.
>I don't. If the JSH paper had stayed published, and the
journal had then
>James would be on record as a devious crank. Any future
attempt of his to
>get published would likely result in the editor and/or
referees finding
his
>publication and the ensuing rebuttal(s).
Hmm, you do have a good point.
Or he could be perceived on the record
only as a non-crank amateur having made
some errors.
I prefer this current situation, where no
reputable journal has granted him credibility.
James Harris is incredible (in more ways that
this).
But perhaps you're right in this matter, in
agreement with the other _real_ and
_professional_ mathematicians on
this newsgroup.
===
Subject: Re: JH is just making fools out of you
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IJsNH10967;
(snipped)
>His recent twin-prime conjecture is interesting and
apparently novel. At
>least no one has posted yet saying that ___'s Conjecture in
1936 said
>essentially the same thing, only faster.
Paraphrasing what Gauss said of Fermat's Last Theorem,
I can make numerous other difficult and possibly
novel conjectures. Why solve or address
this one?
Isn't there more basic twin-prime conjectures, as the
other thread has been discussing?
James Harris has done similar than Fermat did with
conjectures in number theory, by putting forth a
conjecture based on a handful of observations, such as
Fermat numbers (wrong) and FLT (right).
I'd like to see James Harris offer a proof of this or
any other conjecture which will stand up to scrutiny.
Now that would be interesting and novel!
Anthony J. Natoli
===
Subject: Repost reply to Will Twentyman Re: Math research,
passion is
important
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IJsNO10971;
(damn computer keys slip past my fingers)
(snip)
>To be perfectly honest, Quinn has shown no signs of losing
his temper,
>or acting in an immature fashion.
Will, you haven't looked at a recent post by Quinn,
repeated (way down) below. You judge whether he
lost his temper or acted immature.
At first, in the recent James Harris posts, Quinn
posted by only his first name and with a fake
E-mail address to defend James Harris.
As I challenged James Harris repeatedly on his
assertions, and after I and others suggested that
maybe Quinn and the fake E-mail address were
James Harris anonymously defending himself as to
whether some mathematicians reviewed Harris' proof,
Quinn then began posting under his full name Quinn
Tyler Jackson to show that Harris was not faking
being Quinn, and Quinn also revealed that he and
James Harris were colleagues and friends in a closed
list, showing that Quinn was not some objective
defender of James Harris.
Was Quinn mature by secretly hiding his identity and
association with James Harris? Or was it intellectually
dishonest to wrongly back up his buddy as to alleged facts?
Will, perhaps you can explain why neither Quinn nor his
good friend James Harris provided the name or E-mail
address of a single mathematician of their alleged
closed list who allegedly reviewed Harris' proof,
and who allegedly failed to find an error in it.
I repeatedly challenged them to put up or shut up
on their assertion that:
several mathematicians failed to find the error.
Instead, they evaded repeatedly. Quinn said
another poster Einstein answered my challenge,
but all Einstein did was provide the website
of the high-IQ group to which Quinn and Harris belong.
And then, after I pointed out that that website
listed only one mathematician, Quinn alleged that
the website _may_ be out of date, and he attempted
to playfully deßect from the names of the mathematicians.
And his challenge to me to call a bluff on some
alleged algorithm was a diversion to continue the
evasion by both James Harris and Quinn to my
original challenge to name their reviewing mathematicians.
Is Quinn acting mature or even tempered, like you suggest?
Why don't James Harris and Quinn Tyler Jackson name
the mathematicians who reviewed and found no error in
James Harris' paper on factoring? Can't they 
put
their assertions to outside scrutiny?
Either:
A) no mathematicians actually reviewed it, and so
Harris and Quinn LIED to support their lame position,
that errors in Harris' paper were hard to find 
by
mathematicians (but Harris somehow found them himself);
B) mathematicians in the high-IQ group did review it,
and missed the error, thus suggesting that the high-IQ
folks are not so smart, or just not very skilled
mathematicians (probably amateurs), and so Harris and
Quinn did not want to embarass their high-IQ group and
in turn themselves for being members of such group; or
C) Harris and Quinn did not want verification or lack
thereof in any way, shape, or manner for their
assertions on this newsgroup.
Any of these actions by James Harris and Quinn Tyler
Jackson is intellectually dishonest, and perhaps you
may consider them immature as well. As in my other
silly posts regarding James Harris or Andrew Wiles
proving the Riemann Hypothesis, how can we respect
and deal with James Harris _mathematically_ on this
newsgroup if he and his friends try to dupe us
with lies, distortions, and evasions?
Can't we ask simple questions, challenges to be sure,
but reasonable inquiries as to facts and circumstances
regarding James Harris and his work?
Will, you tell me if the following is mature or
even tempered on the part of Quinn Tyler Jackson.
Quinn's verbatim post:
===
Subject: Re: Factoring paper is wrong
Author: Quinn Tyler Jackson 
JHiN said:
> (I've drawn my own conclusions with a cursory
> review.)
You drawing conclusions with a Crayola.
> I'm not satisfied - only one of the editors 
is
> listed as a mathematician.
Editorial Board & Contributors. Old, out of date listing.
Hmmm... let's
draw
some conclusions based on outdated data temporarily stored on
a server
while
the main site is down. (Maybe the up to date site is up -- I
don't know.
Maybe it is the same as that one -- I don't know.)
Which is which? Who is who? Care to draw more conclusions?
I'm not an
editor, neither is James. Round and round it goes, where it
lands, nobody
knows. Well, you know. But .... yeesh -- technically, you're
nobody -- so
indeed, nobody knows.
Not nobody? Who *are* you, anyway? An anonymous ellipses. You
don't even
exist, JHiN, except in vaporspace. So far, I'd say you are a
computer.
> Any other mathematicians in this little group
> of yours, who reviewed Harris' work?
Who knows? Who cares? Who knows who cares? Who cares to know?
Whose nose
cares?
> Any idiot balancing his/her checkbook can call
> themselves a mathematician, but that doesn't
> qualify them to find errors, or to judge a paper
> if they find no errors.
And any poorly written ELIZA program hooked up to an email
account can
attack people from behind a thin sheet of anonymity and call
itself a
Good God, man, I gave you ample opportunity to call my bluff
on my
absolutely outrageous claim about having invented an efficient
Type < 2
parsing algorithm, and you were so busy running line 20367 of
your program
that you insisted instead that I answer a question Einstein
had already
answered.
If you were truly hanging around to protect the Integrity of
the Body of
Knowledge of Mathematics, surely you would have grabbed that
freebie
opportunity with both hands and called my bluff, told me to
put the cards
on
the table about such a boisterous claim.
But instead of focusing on my outrageous claim of invention,
you elected to
go all Crayola Computer on me and draw conclusions about a
small, private
list and the nature of its participants. You focused on a
question that was
already answered by a third party instead of the obvious one
that would
have
allowed you to call yourself Quinn Tyler Jackson is Nuts(R).
Time to get Grampa Simpson to review your code. Maybe he
forgot that GOTOs
make spaghetti code.
--
Quinn
Note: If you had the balls to stand behind your real name --
I'd treat you
with more respect. In the meantime, you technically don't
exist.
===
Subject: Re: Math research, passion is important
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IK6Gv12355;
> Will said:
>To be perfectly honest, Quinn has shown no signs of losing
his temper,
>or acting in an immature fashion. I think he would be
served well by
>only been in on events with James for a little over a year,
and that
>alone has created some powerful impressions. Quinn does not
yet have
>the benefit of those observations.
> I have the benefit of the observations I've 
made of James
on another,
much
> more peaceful list.
> Anyway ... I've tried to avoid losing my temper with 
JHiN,
despite his
> breaking of Godwin's Law.
>Godwin's Law was not broken by JHiN. JHiN's 
posting confirmed
Godwin's
Law.
NAZIS AND HITLER! There, I said it!
How can you refer to actions by _some_
posters which are, in the ordinary
world, comparable to _you_know_who_?
How about if I said Big Lie or propaganda?
Is that permissible, while _you_know_who_ isn't?
Godwin's Law is as valid AS THAT ALL-CAPS
WORDS WITH ALL THESE CAPS ON?
Shouldn't Godwin's (bull) Law be amended
to include Big Brother and 1984? It ain't
_you_know_who_, but has a similar connotation
of distorting reality and twisting facts like
James Harris and Quinn Tyler Jackson have.
Perhaps you may be amused by a scene in the
a lawyer. The judge (Ray Walston) wouldn't let
a witness use the term asshole in his court to
describe someone, and Judd argued that there was
no other way to describe the person in question
with the meaning and connotations intended without
using the term asshole.
When James Harris and Quinn Tyler Jackson cease
the Big Lie, THEN I won't compare them to Hilter
and Goebbels. Maybe.
James Harris is Nuts Like Hitler(TM)
===
Subject: Re: Math research, passion is important
>or acting in an immature fashion. I think he would be
served well
by
I've
>only been in on events with James for a little over a
year, and that
>alone has created some powerful impressions. Quinn does
not yet
have
>the benefit of those observations.
 I have the benefit of the observations I've 
made of James
on another,
much
> more peaceful list.
 Anyway ... I've tried to avoid losing my temper with
JHiN, despite his
> breaking of Godwin's Law.
>Godwin's Law was not broken by JHiN. JHiN's 
posting
confirmed Godwin's
Law.
> NAZIS AND HITLER! There, I said it!
> How can you refer to actions by _some_
> posters which are, in the ordinary
> world, comparable to _you_know_who_?
> How about if I said Big Lie or propaganda?
> Is that permissible, while _you_know_who_ isn't?
> Godwin's Law is as valid AS THAT ALL-CAPS
> WORDS WITH ALL THESE CAPS ON?
> Shouldn't Godwin's (bull) Law be amended
> to include Big Brother and 1984? It ain't
> _you_know_who_, but has a similar connotation
> of distorting reality and twisting facts like
> James Harris and Quinn Tyler Jackson have.
> Perhaps you may be amused by a scene in the
> a lawyer. The judge (Ray Walston) wouldn't let
> a witness use the term asshole in his court to
> describe someone, and Judd argued that there was
> no other way to describe the person in question
> with the meaning and connotations intended without
> using the term asshole.
> When James Harris and Quinn Tyler Jackson cease
> the Big Lie, THEN I won't compare them to Hilter
> and Goebbels. Maybe.
> James Harris is Nuts Like Hitler(TM)
Chill, Dude!
Godwin's Law is not a criticism. It is a social observation.
Godwin's Law does not say Thou shalt not... It says Someonr
will...
===
Subject: Yeah, I bet James Harris' Hammer is Coming
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IK6MP12375;
Or is it spelled with a u?
===
Subject: Re: Yeah, I bet James Harris' Hammer is Coming
> Or is it spelled with a u?
--
Lance Lamboy
I tell them the truth and they think it's hell. ~ Harry S.
Truman
===
Subject: Re: Yeah, I bet James Harris' Hammer is Coming
>Or is it spelled with a u?
Maybe. However, the Hummer2 is a big f-ing joke.
===
Subject: Re: Question about compact sets
>2.- If the above is true, can one state that if K is
compact, then
>there exists a finite collection of subsets of X whose
union covers
K?
> [...]
> finite collection of open subsets of X. Yes.

>It would appear that starting from 2.- instead of the
standard
>definition might simplify some proofs.
> No problem. I should, have, however, pointed out that EVERY
subset of
> a topological space has a finite open cover: the open cover
whose only
> element is the total space. So obviously, (2) is very much
not a
> strong condition.
OK. Would it be stronger if stated as: If K is compact, then
there
exists a finite collection of open neighborhoods of points of
K whose
union covers K. ? Probably not very useful in practice!
===
Subject: Re: Question about compact sets
days. My association with the Department is that of an
alumnus.
>OK. Would it be stronger if stated as: If K is compact, then
there
>exists a finite collection of open neighborhoods of points of
K whose
>union covers K. ? Probably not very useful in practice!
The total space is an open neighborhood of any point in the
space. So
again, the open cover whose only element is the total space
will
satisfy your condition.
--
It's not denial. I'm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Question about compact sets
X-CompuServe-Customer: Yes
X-Coriate: interspeed.co.nz
X-Ecrate: tanandtanlawyers.com
X-Pose: George Cox
X-Punge: Micro$oft
X-Sanguinate: The MVS Guy
X-Terminate: SPA(GIS)
X-Tinguish: Mark Griffith
X-Treme: C&C,DWS
>1.- Is it correct to state that each subset of a metric
space has an
>open cover?
Yes. In fact, it's true for any topological space. Just take
the set
containing only the entire subset.
>2.- If the above is true, can one state that if K is
compact, then
>there exists a finite collection of subsets of X whose union
covers
Yes, it's trivial but not particularly useful, nor does it
depend on
compactness.
>It would appear that starting from 2.- instead of the
standard
>definition might simplify some proofs.
No, 2. is true for any subset; it's not equivalent to
compactness.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do
not reply to spamtrap@library.lspace.org
===
Subject: Re: Question about compact sets
>1.- Is it correct to state that each subset of a metric
space has an
>open cover?
> Yes. In fact, it's true for any topological space. Just
take the set
> containing only the entire subset.
>2.- If the above is true, can one state that if K is
compact, then
>there exists a finite collection of subsets of X whose
union covers
>K?
> Yes, it's trivial but not particularly useful, nor does it
depend on
> compactness.
>It would appear that starting from 2.- instead of the
standard
>definition might simplify some proofs.
> No, 2. is true for any subset; it's not equivalent to
compactness.
===
Subject: James Harris is a pervert
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IKUfR14557;
He talks of the unit of a ring, hammers, coming,
concrete results, and in the past talked about
penetration into his target market (see Dirk
Van de Moortel's Immortal Fumbles pages).
Why can't he focus on mathematics, and not
his unit?
James Harris is Nuts(TM)
===
Subject: Re: James Harris is a pervert
> Why can't he focus on mathematics, and not
> his unit?
> James Harris is Nuts(TM)
Why can't you just shut up?
At least James is *trying* to discuss maths, whether he gets
it right or
wrong. You don't appear to have added anything at all.
===
Subject: Re: James Harris is a pervert
>Why can't he focus on mathematics, and not
>his unit?
>James Harris is Nuts(TM)
> Why can't you just shut up?
> At least James is *trying* to discuss maths, whether he
gets it right or
> wrong. You don't appear to have added anything at all.
Perhaps, he feels his crusade is impotant?
-- Jonny, why are you wearing that smoking today?
-- Ma docta said i'm impotant. And if I impotant, I have to
dress impotant.
===
Subject: Re: goldbach conjecture proof
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IKoGt16264;
>Remark:
>I have sent messages with subjectgoldbech conjecture proof .
Now,i
>replace it to the subject goldbach conjecture proof i have
typed
>wrongly the character e instead of a in typing the name
>ofgoldbach.
>sincerely
>Masoud Sheykhi
Masoud Sheykhi
>Sarcheshmeh Copper complex,Kerman , Iran
>Technical_inspection@nicico.com
>Abstract
>Goldbach's conjecture says that;
>Every even number greater than 2 is the sum of two primes.
>In this paper I introduce the conclusion of my research
during
several
>years on goldbach's conjecture . I believe that I have
solved and I
>have turned it to a theorem as follows.
>Proof
>First we know that ; 4 = 2+2 . Suppose that the conjecture
is true
for
>all even integers n = 2k , 2 of natural numbers ;
> N = {1,2,3,...}.Now , we set ; k = L+1 . when 2L-1 is a
prime number
>then ; 2(L+1) = (2L-1) +3 . Otherwise we consider all
representations
>of each n = 2q , 2 primes that obtaining from the representations of all even
numbers n
>= 2q , 2 pj,s let Ij = <2(L+1) - pj > be the principal ideal
generated by the
>integer 2(L+1) - pj . Let Q be the intersection of all above
Ij,s .
>Since Var(Q) = {P in Spec(Z); Q is a subset of P} is a
non-empty set
>,where Spec(Z) is the set of all prime ideals of the
commutative ring
>of integer ; Z , there exist some prime ideals say ;P1
= ,P2 =
> ,..., Pn =  where n in N and
p1(k),p2(k),...,pn(k) are
prime numbers and such that Var(Q) =
>{P1 ,P2,...,Pn}, n in N . But by commutative algebra , for
each Pi
,i
>= 1,2,...,n there exist some natural coefficient say ; c1(i)
,c2(i)
>,...,cm(i) such that < c1(i)*pi(k)> ,< c2(i) * pi(k)> ,...,
< cm(i)*
>pi(k)> are the subsets of Pi . Let A = {cj(i)^r * pi(k)^s ;r
= 0, 1,
>2, 3,...; s = 0, 1 , 2 ,3,...; i = 1, 2 , ...,n , n in N ; j
= 1 , 2
>, ...,m(i) , m(i)in N} be ordered by the relation R where m
R n if
>and only if there exists a non-negative integer; k such that
n = km .
>therefore we have partially ordered the set A by the
divisibility
>relation . we define a map f: A->A such that for each a =
cj(i)^r *
>pi(k)^s in A ; f(cj(i)^r * pi(k)^s) = cj(i)^r-1 * pi(k)^s ,
if r =
1,
>2 ,3,...,and s = 0, 1 , 2, 3,..., f(cj(i)^r * pi(k)^s) = 1
,if r = 0
>and s= 0,1,2,3,... Then ; f(a)R a . Now ,Since (A,R) be a
nonempty
>partially ordered set such that every totally ordered subset
of A has
>a greatest lower bound and f(a)Ra , for every a in A ; there
exist h
in A
>such that f(h) = h , hence by definition of f , for some j ,i
,r ,s
we
>have ; h = cj(i)^r * pi(k)^s such that ; cj(i)^r * pi(k)^s =
f(h) =
>cj(i)^r-1 * pi(k)^s . Thus ; for some j ,i ; cj(i) =1 and
such that;
>f(1) = 1= f(cj(i)) = cj(i) . Therefore some Ij is a prime
ideal or
>2(L+1)- pj is a prime number for some j . Thus ; n = 2(L+1)
= [2(L+1)
- pj ] + pj . But by the induction hypothesis ; pj is a prime
number
>, hence inductively goldbach's conjecture is proven.
===
Subject: Re: weak sol. to PDE
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IKoH916268;
>It would be much better if you figured out how to post
>a _reply_ instead of starting a new thread this way.
>(How you do that depends on what software you're
>using...)
>Let U denote open, bounded set in R^n but NO smoothness
assumptions
>about boundary of U. Let H denote Sobolev space of
functions
>on U with k=1,p=2 and which are zero on boudary.
>What does it mean to say that these functions are zero on
>the boundary?
>Let f
>be C-infinity with compact support in U.
let u be in H denote weak solution to -L(u) = f in U
>and u=0 on boundary. Here L is Laplacian.
now my question is : Can we get any more regularity of u ?
I think to talk about the trace we need some smoothness to
>the boundary.
>Hence my confusion.
>I am just saying they are zero on the boudary
>in the sense that they are in H.
>ie. isn't H just defined to be the closure of 
C-infinity
>functions with compact support in U,where closure
>with respect to the H-1 norm. I don't think this affected
>by the smoothness of U ??
>But then again i don't know difference between
>H^1 and W^(1,2) . I thought i saw a book that said
>they are not equal unless boundary of U satisfies
>some smoothness conditions.
>I don't think that the notation is absolutely standard;
>it's certainly true that the closure of the space of
>smooth functions with compact support in this norm is
>not the same as the space of all distributions for
>which the norm is finite. So H is the closure of the
>smooth functions with compact support and that's
>all you meant by vanishes on the boundary, fine.
>Anyway, I'm not really the person to answer this;
>I know just enough about these things to know that
>I'm confused by the question. Two questions for you:
>(i) Is it clear that there _is_ a solution u in H
>in the first place?
>(ii) What do you mean by more regularity on the boundary?
>You can't mean that the function on the boundary is
>differentiable; that makes no sense if the boundary
>is wild. You must be asking about whether the partials
>of u extend continuously to the boundary, or some such -
>but then I wonder about the more in more regularity,
>because we don't even know that u is continuous up
>to the boundary, do we?
> any book i look at i can only get more interior
regularity.
any comments would be greatly appreciated
 don
************************
>David C. Ullrich
>************************
>David C. Ullrich
let me start over and be a little more clear
again let U be open,bouded in R^n, but NO assumption on
smoothness on
boudary of U. notation:
W^(1,2)(U)-denote Sobolev space with k=1,p=2 on U ;
W^(1,2)[o](U)-denotes closure of C[c]^infinity(U) in
W^(1,2)(U).
here C[c]^infinity(U) denotes C-infinity 
functions
with compact support in U.
L denotes Laplacian.
Given f in L^2(U) ,we say u in W^(1,2)[o](U) is a weak sol. to
-L(u)=f in U and u=0 on boudary of U if
Sum( Int(u[xi]*v[xi]),i=1..n)=Int(fv) for all v in
W^(1,2)[o](U).
((all integrals over U,u[xi] denotes weak i'th partial of 
u))
Now we know for any f in L^2(U) ,there is a unique weak sol.
to above in W^(1,2)[o](U). If we assume boundary of U has
some smoothness then we can show u in W^(2,2)(U) ((this is
what i mean by more boudary regularity)).
If we don't
assume the smoothness then in general we cannot get u in
W^(2,2)(U),
but we can get u in W^(2,2)(V) where V open with closure of V
in U.
((this is what i mean by interior regularity))
Now my question is if we assume f in C[c]^infinity(U) and
u in W^(1,2)(U), u weak sol. to -L(u)=f, can we get
u in W^(2,2)(U). ((this seems plausible to me))
If we can then by a few continuity arguments i can get
u in W^(2,2)(U) ,where u in W^(1,2)[o](U) denotes weak sol.
to -L(U)=f where f is in L^2(U).But this i think is false.
don
===
Subject: Re: weak sol. to PDE
>[...]
Well ok, I think that answers my questions about what the
question means. I'm really not an expert on these things
but I'll think about it, as long as nobody else has jumped
in.
In the meantime:
(i) you might post the question on sci.math.research
(my advice would be to start with the latest version
of the question)
(ii) you should _really_ figure out how to post a
_reply_ to a usenet message instead of starting a
new thread when you reply to a post!
Really.
(Hmm, you're posting through mathforum? There are
issues with the way mathforum works - posting
actual usenet messages, through a news server
or even through Google, might be better.)
************************
David C. Ullrich
===
Subject: Re: Math research, passion is importantQ
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IL09F16998;
>anonymous@mathforum.org (actually James Harris is Nuts)
(snip)
> evidence that JSH is capable of friendly relations with
>mathematicians.
What mathematicians? Neither James Harris
nor Quinn Tyler Jackson have ever provided
a single name of a mathematician they know
(besides each other), especially when I asked
to verify both of their assertions that
several mathematicians reviewed James' proof
and failed to find an error.
Jesse, let us ignore whatever obsession I have
on this point (and on their evasiveness). Do
YOU believe their story, that mathematicians
(hopefully non-amateurs or some with more than
average capability) reviewed James' proof and
failed to find an error?
If so, fine, we'll just let James Harris post
whatever he likes without question. Happy
math!
James Harris is Nuts(TM)
===
Subject: Re: Bring Back the HP 15C Scientific Calculator
>I have been working on an effort to get HP to start making
the 15C, or a
>similar model again. To this end I have set up a web site
with a
>petition. If you are like me, and would love to have a well
made
>calculator like the 15C again. Please take a moment to go
to my web
>site and sign the petition. If you have never used a 15C
before the web
>site will tell you why it is the best scientific calculator
for everyday
>use. I have been in contact with a member of the calculator
planning
>department at HP who is sympathetic to the cause and will
argue the case
>to those who make the decisions, however I need to get many
more
>signatures on the petition, so tell anyone you know who
would also
>benefit from a quality calculator like the 15C, to check out
the web
>site.
>http://hp15c.org
> Bringing back the HP15C is an interesting idea. It is
indeed a wonderful
> calculator.
> Concerning battery life, you say They will run the 15C for
over a year,
> many have gone more than 10 years on the same set of
batteries. Well,
> perhaps you'd be interested to know that my 
15C's are
_still_ running on
> their original batteries! (Of course, the fact that I have
more than one
> 15C and the fact that I don't use calculators much at all
nowadays surely
> have something to do with the long battery life I've
witnessed.)
> David Cantrell
My HP45 still works and unfortunately it is used mostly for
check book
reconciliation. I really loved that one when I was doing
chemistry
(college teaching). My 34C has given up the ghost as far as I
can tell.
I suspect it could be massaged back to some degree of
operation. But the
power switch on the 45 is kind of dodgy and I wish it were
possible to
replace it. Runs only on xformer now (no batteries working.)
FK
===
Subject: Re: Bring Back the HP 15C Scientific Calculator
>I have been working on an effort to get HP to start making
the 15C, or
a
>similar model again. To this end I have set up a web site
with a
>petition. If you are like me, and would love to have a well
made
>calculator like the 15C again. Please take a moment to go
to my web
>site and sign the petition. If you have never used a 15C
before the
web
>site will tell you why it is the best scientific calculator
for
everyday
>use. I have been in contact with a member of the calculator
planning
>department at HP who is sympathetic to the cause and will
argue the
case
>to those who make the decisions, however I need to get many
more
>signatures on the petition, so tell anyone you know who
would also
>benefit from a quality calculator like the 15C, to check out
the web
site.
http://hp15c.org
> As best I can tell, the entitiy which calls itself HP or
Hewlett-Packard
> has no other connection than the name with the former
world-best
developer
> and manufacturer of instruments and computers. The entity
which calls
> itself Agilent may have more in common, but I fear thay
may have
been
> enjoined by the legal process which separated it from HP
to do
anything
> hp15-ish.
>We also should bring back the ASR-33 teletype.
>Modern printers simply do not sound or smell the same.
Or print as slow, or break as often, or cost as much. That's
one
machine I really don't miss.
John
===
Subject: Re: Bring Back the HP 15C Scientific Calculator
> I agree. HP 15C for president!!!!!!!!!!!!!!!!!!!!!
No no no, definitely don't bring it back EVER. 
I'm getting
nightmares
already. Stop it! Please, make it go away ;)
===
Subject: Quantities of matter
Length; area, volume and weight scales are still the most
practical
ways to measure and express the quantitiy of material
substance in
various objects; bodies, and masses of matter.
It will be quite awhile yet before most of us start counting
atoms
outside of laboratories.
===
Subject: Re: Quantities of matter
> Length; area, volume and weight scales are still the most
practical
> ways to measure and express the quantitiy of material
substance in
> various objects; bodies, and masses of matter.
Hey Dumb donny Head... 4-momentum.
http://www.kingstroker.com/
http://www.apa.org/journals/psp/psp7761121.html
http://insti.physics.sunysb.edu/~siegel/quack.html


--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
Quis custodiet ipsos custodes? The Net!
===
Subject: [OT; no, really!] why mathematics
[The last line quoted is the only real excuse for
imposing this off topic nonsense on the froup.
Try to enjoy it anyway.]
As part (The Story About the Toddler, Volume
15).of his second (first came The Story About The
Baby, surprise, surprise) ongoing first-time
parenting experiences episodic saga, Jeff Vogel ==
talk.bizarre:
:- * The Parental Loser Radius
:- I take Cordelia to the playground a lot. I hope
:- that encouraging her to run around a lot all the
:- time will reduce the chance that she'll end up a
:- gigantic fatass like her father. Also, when she's
:- playing, it gives me a chance to sit quietly off
:- to the side and read the newspaper. I don't watch
:- her every single second while she plays. I figure,
:- if something goes wrong, she'll get out at least
:- one good scream before it's too late.
:- (Also, I have to preserve my own safety. If I
:- don't, how will I be able to have a second child
:- to send out to avenge the first one?)
:- Anyway, at playgrounds, I let my child run free
:- for a little while. Which distinguishes me from
:- most of the other toddler parents I see out there.
:- They constantly lurk one arm's length from their
:- little treasures, hovering like starving vultures,
:- reach to lurch forward at the slightest sign of a
:- misstep. I mean, Christ, get a grip people. If
:- there's a fence between the kid and traffic 
and
:- the ground is covered with wood shavings, give him
:- some goddamn space.
:- That is why I, in a moment of extreme
:- mathematicalness, invented the PLR, or Parental
:- Loser Radius. This is defined as the maximum
:- distance you let your kid get from you, in yards
:- (or meters, if you s), divided by the kid's
:- age. So if your kid is 2, and you never let it get
:- more than three feet from you, your PLR is 1/2.
:- Why is this important? Because, if your PLR ever
:- falls below two, you are being a freak, and your
:- child resents you.
:- If you have a one-year old and your PLR is 2, that
:- means that you never let the child get more than 6
:- feet from you. This is reasonable. After all, at
:- that age, a bird of prey might swoop down and
:- carry the kid off, leaving you with nothing to
:- show for your parenting experience but a funny
:- story. But if you never let your child get six
:- feet from you, you really need to get a grip.
:- If you have a two-year old and your PLR is 2, you
:- never let the kid get more than 12 feet away. This
:- is still worryingly wussy behavior, but at least
:- you're far away enough for the kid to fall down a
:- few times. Better the kid learns to take a fall at
:- the playground then on the concrete stairs in the
:- basement.
:- If you have a three-year old and your PLR is 2,
:- you let the kid get 18 feet away. This means that
:- your son will have the time to land 2 or 3 good
:- punches on any little  who tries to take his
:- toy away before you can get there to stop him.
:- This delay provides you with plausible
:- deniability when the sob-sister parent of the
:- punchee gets all pissy with you.
:- If you have a four-year old and you're still
:- hovering near your kid at the playground, you need
:- to work out your issues in therapy. Anyway,
:- shouldn't you be concentrating on having another
:- kid by now? In sum, I feel that the PLR is a
:- fascinating and useful concept, in that it allows
:- me to belittle your behavior while enabling me to
:- feel superior.
:- And, in the end, isn't that what mathematics is
:- for?
xanthian, ever helpful cross-pollenator.
===
Subject: Re: purchasing the regular polyhedra
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IM76T22882;
>Do they sell big Platonic solids the size of a soccer ball,
but not
>made of paper and glue or precious stone, and not blown up
to a sphere
>like the soccer ball. Like a giant D&D die.
>My favorite in order are: dodeca, icosa, cube, tetra, octa
Are you going to host a lawn D&D game, and need big dice?
Tell me if you find some.
Anthony Natoli
===
Subject: Re: purchasing the regular polyhedra
Just for interest, no D&D game.
www.peda.com had the largest models I could find.
>Are you going to host a lawn D&D game, and need big dice?
>Tell me if you find some.
>Anthony Natoli
===
Subject: Load balance
I have an integer set of size N, which needs to be broken up
into k
subsets such that the work of each subset (defined as the sum
of
numbers in the subset) is optimal, i.e. load is balanced.
The foll. things I came across did not help:
(1) Linear partitioning preserves the order of the initial
set, i.e.
{1,2,3,4,5,6,7,8,9}'s optimal linear partition is
{1,2,3,4,5}, {6,7},
{8,9}; but I want say {1,5,9}, {2,6,7}, {3,4,8} where the
order is
immaterial.
(2) Weight aware graph partitioning - tried Metis
(www.cs.umn.edu/~metis) but there is no graph in my problem
(no graph
edges) and Metis performs terribly when this is the case.
I would like to know if my problem has a standard name so I
can look
Khan
===
Subject: Re: Load balance
> I have an integer set of size N, which needs to be broken
up into k
> subsets such that the work of each subset (defined as the
sum of
> numbers in the subset) is optimal, i.e. load is balanced.
Try Googling for the Number Partitioning Problem for the k=2
case.
--
Kevin 
===
Subject: Re: James Harris is a pervert
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5IMurC26697;
> Why can't he focus on mathematics, and not
> his unit?
> James Harris is Nuts(TM)
>Why can't you just shut up?
>At least James is *trying* to discuss maths, whether he gets
it right or
>wrong. You don't appear to have added anything at all.
You are obviously deceived by James Harris - he is NOT
trying to discuss math. Instead, he appears to be
trying very hard to convince others that his wrong
results are right, and that his ßawed and de-published
paper was correct and should have remained published, but
for his claims of CONSPIRACY! by this newsgroup.
He is not discussing math - he's insulting real
mathematicians on this newsgroup with his lies
and distortions. He lies when he says they lie, and
he distorts when he says they distort.
Read Harris' incessant crap and insults before YOU even
suggest that James Harris is trying to discuss maths.
James Harris is Nuts(TM)
P.S. No, I won't shut up, you Hitler Nazi!
===
Subject: Fun with the Dirac Delta
time-dependent Schroedinger equation, and completely
localized at t=0 You said in one post of this thread that it is not psi(x,0)
that
> should be equal to delta(x), but mod(psi(x,0))^2. That is
> unfortunately *>meaningless<*. The delta function was
introduced by Dirac
> to fill a perceived need, but it is not a nice function. It
is defined
> as a limit, and in quite a few cases expressions involving
it converge
> when you take the limit in the prescribed way. However not
all
> expressions involving the delta function converge
correctly. They are
> not meaningful. Whenever you deal with the delta function,
you must
> keep these problems in mind.

Fine words, but I think destined to impress nobody who's
opinion is
worth a tinker's damn.
Let's look at this in detail:
We have two functions connected in some way with the problem.
For
neutrality, lets call them g and h: g is understood to be
psi, h
|psi|^2. _You_ assert the correct interpretation of the
boundary
condition is to let g be a Dirac delta function a t=0, I
suggest that
we want to let h have this property. You respond not by
saying my
suggestion is wrong for such and such a reason, or will not
work, but
that it is _meaningless_. But you produce no evidence for this
assertion other than a general pandemonium about the subtle
issues
surrounding the blessed delta functions, which I am supposed
to be on
the wrong side of, because ... well, because you say so.
So what are some possible a priori points of argument?
(I) Our g is a complex function, and delta(x) is real. We
could
certainly interpret setting a complex function equal to
delta(x) as
setting the real part to delta and the imaginary part to 0,
but we do
not encounter this issue when setting a real function to be
delta.
Credit: certainly no prejudice against h
(II) Our g and h are functionally related. One is the squqre
modulus
of the other. You go on to argue:
> Eugene's problem said that the electron at time t=0 was
totally
> localized to x=0. I expressed this mathematically as
psi(x,0) =
> delta(x). This assumption leads to expressions which are
meaningful,
> and indeed I produced the solution which Eugene has gone
and found in
> a textbook. If you assume that psi(x,0) = sqrt(delta(x))
you get
> nothing but garbage out.
I.e., if we take my interpretation, perhaps we are forced to
interpret
sqrt(delta(x)), which looks ugly and dubious (and also
formerly
wrong here, since sqrt(|psi^2|) /= psi). Ok ... And if we
take your
interpretation, we may be forced to interpret delta^2(x).
That fact
that two functions have some functional relation and that
expressing
one in terms of the other may lead to some ugly expression if
we treat
one as a delta and not the other is no argument.
credit: neutral
(III) the ordinary interpretation of psi is that the integral
of the
square modulus over some interval gives the probability of
finding the
one whose wave function is a delta function in the square
modulus,
that the wave function itself.
The principle defining property _of_ the delta function _is_
its
integral property, and having a prior interpretation of the
integral
of |psi|^2, hence of the significance of the integral
properties of
the delta, makes this a natural identification. we have no
ready
interpreation of the integral of psi itself, so the
idenfication of
psi with a delta function seems strange, and only has to
recommend it
that it formally localizes the support of psi to a point,
having no
other obvious interpretation.
credit: neutral on meaningful but strongly tending to favor h
as an
initial guess, and not g.
Now ... I think it turns out a postiori that my suggestion
for an
interpretation of the boundary condition _doesn't work_.
Yours,
apparently produces at least a formally correct solution (but
one with
the annoying property that we cannot normalize it in the
ordinary
elementary sense). You will probably find others who will
agree with
you that this demonstrates that my suggestion was
meaningless. This
is nonsense. My interpreation simply happened to produce a
problem
with no solution set, but a well-posed problem (I formalized
it in a
response to Hansen).
Only the logic deaf will be unable to hear the distinction
between a
well-posed problem with empty solution set and an ill-posed
problem.
Meaningless is a strong assertion -- and adding a lot of vague
motherhood about the necessity of caution in tossing around
delta
functions does not even begin to constitute the ghost of a
justification.