mm-2729 === Subject: Re: Continuum > Methinks I finally disproved CH by finding something with cardinality of C, > all think by looking in places noone else looks. You would need to find something of cardinality strictly greater than Aleph-0 and strictly less than C to disprove CH. If all you did was find something of card C, that's not very impressive. For example the reals have card C. === Subject: Re: Continuum by C i ment continuum, as in CH. Goedel would be proud. But since I am not a mathematician, I will take the proof with me to my grave. > Methinks I finally disproved CH by finding something with cardinality of C, > all think by looking in places noone else looks. > You would need to find something of cardinality strictly greater than > Aleph-0 and strictly less than C to disprove CH. If all you did was > find something of card C, that's not very impressive. For example the > reals have card C. === Subject: Re: Continuum > by C i ment continuum, as in CH. Goedel would be proud. But since I am not > a mathematician, I will take the proof with me to my grave. Or you could write an SMS saying you have found A Marvelous Proof, but that there is not enough characters available in the message to give any details... Then for the next 300 years, mathematicians will try to re-create your proof. Of course, it all depends on your reputation as a mathematician...... -Michael. === Subject: Re: Continuum yours truly can promptly word it into a hypothesis : there exist an _infinite_ (technically wrong term, i know) amount of non-interequivalent axioms, each of them completing ZF; cardinality of this _set_ of axioms is strictly greater then card(N) and strictly lower then card(Z). Not of much use without a proof, which took inventing another level of abstraction, but who cares ? yours, amateur > by C i ment continuum, as in CH. Goedel would be proud. But since I am > not > a mathematician, I will take the proof with me to my grave. > Or you could write an SMS saying you have found A Marvelous Proof, but > that there is not enough characters available in the message to give any > details... > Then for the next 300 years, mathematicians will try to re-create your > proof. > Of course, it all depends on your reputation as a mathematician...... > -Michael. === Subject: Re: Continuum > by C i ment continuum, as in CH. Goedel would be proud. But since I am not > a mathematician, I will take the proof with me to my grave. You can't know how much I approve of your well-mannered decision. Rick === Subject: Re: Continuum > by C i ment continuum, as in CH. Goedel would be proud. But since I am not > a mathematician, I will take the proof with me to my grave. > You can't know how much I approve of your well-mannered decision. > Rick Lucky me, living in bliss ... === Subject: Re: Continuum >> Methinks I finally disproved CH by finding something with cardinality of C, >> all think by looking in places noone else looks. >You would need to find something of cardinality strictly greater than >Aleph-0 and strictly less than C to disprove CH. If all you did was >find something of card C, that's not very impressive. For example the >reals have card C. Yeah, but *everyone* looks *there*. Lee Rudolph === Subject: Re: Continuum >> Methinks I finally disproved CH by finding something with cardinality of C, >> all think by looking in places noone else looks. >> >You would need to find something of cardinality strictly greater than >Aleph-0 and strictly less than C to disprove CH. If all you did was >find something of card C, that's not very impressive. For example the >reals have card C. > Yeah, but *everyone* looks *there*. > Lee Rudolph I looked where everyone already looked. === Subject: Re: Was Lucy the split image - Jabriol is a creationist. Jabriol is indeed a creationist as are all Jehovah's Witnesses. He's been trying to disprove evolution for the past several years here on Usenet, using what he's learned from the Watchtower publications, with no success. You can't reason with a religious fanatic. Zulo....... Jesus, save me from your followers. > Sorry, I first thought you were a creationist. > nope I am not === Subject: Re: Was Lucy the split image - Jabriol is a creationist. > Jabriol is indeed a creationist as are all Jehovah's Witnesses. He's been > trying to disprove evolution for the past several years here on Usenet, > using what he's learned from the Watchtower publications, with no success. > You can't reason with a religious fanatic. and of course they need you to tell them this? tell them also how you love non jw-pedophiles. and how you agree with sex among children, when you send your own son to wild parties at motels.. oh by the way since you are in an informative mood.. tell them how you have been caught lying so much.. that nobody belives you anymore.. calling me a creationist does not make me one... and in fact you being a JW groupie should know that JW's are not creationist... However we also know, thta you need to lie, because, you cant get your point across by telling the truth.. now go get a new husband and service him like a house wife should, because the first one bashed you like a romper room punching bag, and the second run you out of Nashville, because you send to much time on the net.. === Subject: Re: Was Lucy the split image - Jabriol is a creationist. > tell them also how you love non jw-pedophiles. > and how you agree with sex among children, when you send your own son > to wild parties at motels.. > oh by the way since you are in an informative mood.. tell them how you > have been caught lying so much.. that nobody belives you anymore.. > calling me a creationist does not make me one... and in fact you being > a JW groupie should know that JW's are not creationist... > However we also know, thta you need to lie, because, you cant get your > point across by telling the truth.. > now go get a new husband and service him like a house wife should, > because the first one bashed you like a romper room punching bag, and > the second run you out of Nashville, because you send to much time on > the net.. === Subject: Re: Riemann's Zeta Function by H. M. Edwards > > OK, so backtracking a few lines: Take Gamma(1-s)(-2*pi*n)^s-1. Summing > over all integers n other than n=0 and using 3) -Zeta(s)-Sigma(Gamma > 1-s/2*pi*i)Int_|x+-2*pi*n|=Epsilon ((-x)^s)/((e^x)-1) * dx/x=0 then > gives > > Zeta(s)=Sigma_n=1^infinity(Gamma(1-s)[(-2*pi*n)^s-1 + (2*pi*n)^s-1]. I > don't see how 3) is implemented to give this result. Sorry ahead of > time for any Ughs. > > Well can I present you with an ugh for each asterisk above :-) > > Let's translate this. Does > Gamma(1-s)(-2*pi*n)^s-1 > mean > Gamma(1-s)(-2pi n)^s-1 > or > Gamma(1-s)(-2pi n)^{s-1}. > The latter seems to make more sense in this context. > You are correct > I can't make head nor tail of what you write for 3). > Is n the summation variable? what has happened to the first integral. > Is the equals sign before the epsilon really meant to be there? > Yes. I lifted this verbatim from p 13. OK, maybe not quite verbatim! > > I presume you are using the contour integral argument for continuing > Gamma(s)zeta(s). One introduces a countour C_e in three parts: > imaginary axis from -infinity to -e, circle radius e about origin, > imaginary axis from -e to -infinity and take the integral > of z^{s-1} e^z on C_e (with branch cut on negative real axis). > Thse integral f(s) is independent of e by Cauchy's theorem. > It is also an entire function of s: convegence is nice since e^t -> 0 > rapidly as t -> -infinity. Really dumb question. is e in the contour integral an epsilon, or the basis of the natural logarithm. (I am getting confused by both uses of e.) > The integral of z^{s-1} e^z on the first part of the contour is > integral_e^infinity t^{s-1} exp(-pi i(s-1)) e^{-t} dt > = - integral_e^infinity t^{s-1} exp(-pi is) e^{-t} dt. > Similarly on the third part of the contour it is > integral_e^infinity t^{s-1} exp(pi is) e^{-t} dt. > These add to > 2i integral_e^infinity t^{s-1} sin(pi s) e^{-t} dt. > If Re(s) > 0 the integral over the circle of radius e is > O(e^Re(s)). Letting e -> 0 we get that What is this O Is this Big-O notation? > f(s) = 2i sin(pi s) Gamma(s) > for Re(s) > 0. > > Now we consider > g(s) = integral_{C_e} z^{s-1} e^z/(1-e^z) dz > where we insist e < 2pi (so that e^z =/= 1 for 0 < |z| < e ). > For Re(s) > 1, as with f(s), we can take e -> 0 so that > g(s) = 2i sin(pi s) integral_0^infinity t^{s-1} e^{-t}/(1-e^{-t}) dt > = 2i sin(pi s) Gamma(s)zeta(s). Got it. > > Consider > G_N(s) = integral_{C_{(2N+1)pi}} z^{s-1} e^z/(1-e^z) dz. > By Cauchy's theorem the difference G_N(s) - g_e(s) is 2pi i times > the sum of the residues of the poles of the integrand > at +-2pi i, +- 4pi i, ..., +- 2Npi i, that is > 2pi i sum_{n=1}^N [-(2pi ni)^{s-1} - (-2pi ni)^{s-1}] > = 2pi i (2pi)^{1-s} sum_{n=1} (-2) cos(pi(s-1)/2)/n^{1-s} > = something nasty times sum_{n=1}^N 1/n^{1-s} Hmm. Edwards gets sin(s pi/2). However I have seen the functional equation written both with cos and sin, so I am sure you must be right. Now at least I see what [-(2pi ni)^{s-1} - (-2pi ni)^{s-1}] is all about > > If Re(s) < 0, as N -> infinity, G_N(s) -> 0. This is because > |e^z/(1-e^z)| = 1/(1-e^{-z}) and on the contours C_{(2N+1)pi} > e^{-z} is bounded away from 1. Thus for Re(s) < 0 > Gamma(s)zeta(s) = something nasty times zeta(1-s). OK === Subject: Re: Riemann's Zeta Function by H. M. Edwards >> I presume you are using the contour integral argument for continuing >> Gamma(s)zeta(s). One introduces a countour C_e in three parts: >> imaginary axis from -infinity to -e, circle radius e about origin, >> imaginary axis from -e to -infinity and take the integral >> of z^{s-1} e^z on C_e (with branch cut on negative real axis). >> Thse integral f(s) is independent of e by Cauchy's theorem. >> It is also an entire function of s: convegence is nice since e^t -> 0 >> rapidly as t -> -infinity. > Really dumb question. is e in the contour integral an epsilon, or the > basis > of the natural logarithm. (I am getting confused by both uses of e.) No Real Mathematician thinks of the logarithm having a basis:-) If you want to use epsilon instead of e you're welcome: using e does save typing. >> The integral of z^{s-1} e^z on the first part of the contour is >> integral_e^infinity t^{s-1} exp(-pi i(s-1)) e^{-t} dt >> = - integral_e^infinity t^{s-1} exp(-pi is) e^{-t} dt. >> Similarly on the third part of the contour it is >> integral_e^infinity t^{s-1} exp(pi is) e^{-t} dt. >> These add to >> 2i integral_e^infinity t^{s-1} sin(pi s) e^{-t} dt. >> If Re(s) > 0 the integral over the circle of radius e is >> O(e^Re(s)). Letting e -> 0 we get that > What is this O Is this Big-O notation? Yes. Very useful! >> >> Consider >> G_N(s) = integral_{C_{(2N+1)pi}} z^{s-1} e^z/(1-e^z) dz. >> By Cauchy's theorem the difference G_N(s) - g_e(s) is 2pi i times >> the sum of the residues of the poles of the integrand >> at +-2pi i, +- 4pi i, ..., +- 2Npi i, that is >> 2pi i sum_{n=1}^N [-(2pi ni)^{s-1} - (-2pi ni)^{s-1}] >> = 2pi i (2pi)^{1-s} sum_{n=1} (-2) cos(pi(s-1)/2)/n^{1-s} >> = something nasty times sum_{n=1}^N 1/n^{1-s} > Hmm. Edwards gets sin(s pi/2). How does sin(s pi/2) relate to cos((s-1)pi/2)? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Riemann's Zeta Function by H. M. Edwards >> I presume you are using the contour integral argument for continuing >> Gamma(s)zeta(s). One introduces a countour C_e in three parts: >> imaginary axis from -infinity to -e, circle radius e about origin, >> imaginary axis from -e to -infinity and take the integral >> of z^{s-1} e^z on C_e (with branch cut on negative real axis). >> Thse integral f(s) is independent of e by Cauchy's theorem. >> It is also an entire function of s: convegence is nice since e^t -> 0 >> rapidly as t -> -infinity. > > Really dumb question. is e in the contour integral an epsilon, or the > basis > of the natural logarithm. (I am getting confused by both uses of e.) > No Real Mathematician thinks of the logarithm having a basis:-) Oops I meant base :-( > If you want to use epsilon instead of e you're welcome: using e > does save typing. True. :-) >> The integral of z^{s-1} e^z on the first part of the contour is >> integral_e^infinity t^{s-1} exp(-pi i(s-1)) e^{-t} dt >> = - integral_e^infinity t^{s-1} exp(-pi is) e^{-t} dt. >> Similarly on the third part of the contour it is >> integral_e^infinity t^{s-1} exp(pi is) e^{-t} dt. >> These add to >> 2i integral_e^infinity t^{s-1} sin(pi s) e^{-t} dt. >> If Re(s) > 0 the integral over the circle of radius e is >> O(e^Re(s)). Letting e -> 0 we get that > > What is this O Is this Big-O notation? > Yes. Very useful! >> >> Consider >> G_N(s) = integral_{C_{(2N+1)pi}} z^{s-1} e^z/(1-e^z) dz. >> By Cauchy's theorem the difference G_N(s) - g_e(s) is 2pi i times >> the sum of the residues of the poles of the integrand >> at +-2pi i, +- 4pi i, ..., +- 2Npi i, that is >> 2pi i sum_{n=1}^N [-(2pi ni)^{s-1} - (-2pi ni)^{s-1}] >> = 2pi i (2pi)^{1-s} sum_{n=1} (-2) cos(pi(s-1)/2)/n^{1-s} >> = something nasty times sum_{n=1}^N 1/n^{1-s} > > Hmm. Edwards gets sin(s pi/2). > How does sin(s pi/2) relate to cos((s-1)pi/2)? Oh yeah. They're equal! I think I am going to look at the Davenport book, Chapter 8. If I can find it for less than $58.00! This Edwards book is over my head. :-o === Subject: Re: Skeptical Inquirer UFO ah, so; am I to infer that PBK works both sides of the conspiracy?... and here, I thought it was Skull and Bones. do you know of the cache taht Roswell has with WW2?... I know of two things, that Art Bell (et al ad vomitorium) never mention (although others have mentioned one of them, on his show ... I stopped listening to taht **** over 3 years ago .-) as the co-author exposed him during the book tour, Corso's thesis is that human beings cannot create ideas, withe corrolary that they were not made in the image of God. (I didn't know, when he came to borders without Corso, that he was deathly ill; probably made it easier to drop his little clues; are they in the book?... the funny things was, I was the only one in the audience who asked any hard questions; everyone else must have been above top secret !-) > mysterious aerospace reports and events (see www.jamesoberg.com) and share > what I've found, for discussion, with no directives, constraints, or other > Jim Oberg > Phi Beta Kappa, Ohio Wesleyan University, 1966 > on by systematically ridiculing those of us who suspect the truth about this > phenomenon, all the while being privy to above-top-secret knowledge that > would prove us right? One wonders.... > C. SCOTT LITTLETON > President, Phi Beta Kappa Alumni in Southern California --ils duces d'Enron! http://tarpley.net/bush8.htm http://www.wlym.com/PDF-SpReps/SPRP13.pdf === Subject: Re: Skeptical Inquirer UFO I'm sensing that you're offering some interesting insights to this discussion, but frankly I can't figure out what it is you're trying to say. Sorry! > ah, so; am I to infer that PBK works both sides > of the conspiracy?... and here, > I thought it was Skull and Bones. > do you know of the cache taht Roswell has with WW2?... > I know of two things, that Art Bell (et al ad vomitorium) never mention > (although others have mentioned one of them, > on his show ... I stopped listening to taht **** > over 3 years ago .-) > as the co-author exposed him during the book tour, > Corso's thesis is that human beings cannot create ideas, > withe corrolary that they were not made in the image of God. > (I didn't know, when he came to borders without Corso, > that he was deathly ill; probably made it easier > to drop his little clues; are they in the book?... the funny things was, > I was the only one in the audience who asked any hard questions; > everyone else must have been above top secret !-) > mysterious aerospace reports and events (see www.jamesoberg.com) and share > what I've found, for discussion, with no directives, constraints, or other > Jim Oberg > Phi Beta Kappa, Ohio Wesleyan University, 1966 > on by systematically ridiculing those of us who suspect the truth about this > phenomenon, all the while being privy to above-top-secret knowledge that > would prove us right? One wonders.... > C. SCOTT LITTLETON > President, Phi Beta Kappa Alumni in Southern California > --ils duces d'Enron! > http://tarpley.net/bush8.htm > http://www.wlym.com/PDF-SpReps/SPRP13.pdf === Subject: Re: Ring problem > Suppose R is a ring with s = s ^ 2 for each s in R. Why s + s = 0? how about this one? (-a)^2 = (-a) (-a)^2 = (-a)(-a) = a^2 = a Therefore, a = (-a), or a+a = 0 === Subject: Re: Ring problem Justin Young grava .88 la saucisse et au marteau: > how about this one? > (-a)^2 = (-a) > (-a)^2 = (-a)(-a) = a^2 = a > Therefore, a = (-a), or a+a = 0 Actually, that's false, because you say that (-1)(-1) = 1. It should be -1. -- Nicolas === Subject: Re: Ring problem >> how about this one? >> (-a)^2 = (-a) >> (-a)^2 = (-a)(-a) = a^2 = a >> Therefore, a = (-a), or a+a = 0 > Actually, that's false, because you say that (-1)(-1) = 1. > It should be -1. No. (-a)(-b) = ab in any Ring R (with 1) due to the Law of Signs: ----------- a b + a(-b) + (-a)(-b) since over/underlined = 0 ---------------- by the distributive law => a b equals (-a)(-b) -Bill Dubuque === Subject: Re: Ring problem > Justin Young grava .88 la saucisse et au marteau: > how about this one? > (-a)^2 = (-a) > (-a)^2 = (-a)(-a) = a^2 = a > Therefore, a = (-a), or a+a = 0 > Actually, that's false, because you say that (-1)(-1) = 1. It should be > -1. He didn't say that (-1)(-1) = 1. In fact, 1 may not even be in the ring. (-a)(-a) = a^2 is true in *all* rings, not just the ring mentioned in the original problem. -- Bill Hale === Subject: Re: Ring problem William Hale grava .88 la saucisse et au marteau: > (-a)(-a) = a^2 is true in *all* rings, not just the ring mentioned > in the original problem. -- Nicolas, who didn't use rings nor fields since 3 years. === Subject: Re: Using De L'Hopital for solving equations The question now is : Does exist a particular function g(x) where : lim (x->t) g(x) = 0 and lim (x->t) g ' (x) = 0 ? === Subject: Re: Using De L'Hopital for solving equations > The question now is : > Does exist a particular function g(x) where : > lim (x->t) g(x) = 0 and lim (x->t) g ' (x) = 0 ? Certainly, no problem. g(x) = (x-t)^2 comes to mind. Or g(x) = cos(x) - 1. Any function that touches or crosses the but also f' must be zero. In other words, your method won't work unless the function f *and* its derivative have the same root t. And I don't see how you could know that without solving both of them. === Subject: Re: Using De L'Hopital for solving equations > but also f' must be zero. In other words, your method won't > work unless the function f *and* its derivative have the > same root t. And I don't see how you could know that without > solving both of them. ''(x) ) won't work. But if I consider only the first derivate of the functions? So I have the limit: lim (x->t) ( f (x) / g (x) - f '(x) / g '(x) ) = 0 f (x) and g (x) are 0 per hypotesis, and the first derivates of functions are not necessarily to be 0, because their quotient represents the result of the limit not in form 0/0. === Subject: Re: JSH: Advanced Polynomial Factorization the FBI is probably going to pay him through his blog, before they dump him in the river. [sorry, if it's not you folks .-] > Who is the one who calls people liars for disagreeing? Who is the one > who resorts to cursing? Who is being childish? > http://mathforprofit.blogspot.com/ > Have you made any profit yet? --ils duces d'Enron! http://tarpley.net/bush8.htm http://www.wlym.com/PDF-SpReps/SPRP13.pdf === Subject: Averaging Errors: Restricted #-of-Divisors Function Although this post is regarding a topic of its own, it is a continuation, in a way, of the thread at: Let d(r,m) = the number of divisors, k, of m, where 1 <= k <= m^(1/r) for each k. (So, d(1,m) = d(m) {sometimes called tau(m)}, the standard number-of-divisors function.) So, let e(r,m) = d(r,m) - d(m)/r. Then, for r = any fixed integer >=2, is: limit{m -> oo} (1/m) sum{k=1 to m} e(r,k) = c *(1 -2/r), where c is Euler's constant (.5772...)?? (Maybe this WOULD have been the limit if a limit existed...) Leroy Quet === Subject: UFO Bogus Physics re: Nick Cook's Hunt for the Zero Point http://www.salon.com/books/review/2002/08/05/zero_gravity/ --- In ItalianPhysicsCenter@yahoogroups.com, Jack Sarfatti How about this from a PhD physicist, in the minority, who has studied the UFO subject: I did some investigating and interviewed retired senior USG executives who laid the Nazi flying saucer (disguised as UFOs) claim (a la Cook's book) to rest - it didn't exist and it didn't fly. Which is pretty much what I have also been saying about Cook's book BTW. I'm interested in any debunkery of Cook's book. What did this Ph.D. fellow discover precisely? You need to ask him. I have bcc'd him on this. He can reveal his ID if he wishes. He has USG Intelligence connections for information. The point is that we at ISSO 1999-2000 at checked out many of the bogus claims on zero point energy by fringe people and found them to be essentially worthless. Creon Levit on leave from NASA Ames did a very thorough job on several of the claims. More information on this is in my two books from 2002 Destiny Matrix and Space-Time and Beyond II and in my semi-popular Society for Literature and Science http://qedcorp.com/APS/StarGate1.mov Note of clarification I do not include the Haisch-Puthoff zero point energy program nor Puthoff's PV gravity program in the same crackpot category as ALL the other work mentioned by Nick Cook in his book. My objections to the HRP program are that they do not ask the right questions, are basically superficial in their formulations, make some errors of interpretation of their formalisms and most importantly have not led to any testable predictions nor any clarifying explanations of significant problems and mysteries, e.g the UFO. They are not bogus physics simply wrong physics in my humble opinion. A few specifics: 1. On the zero point origin of inertia - it is a mistake to look only at the virtual photons. It is the virtual electron-positrons that is most important. 2. There is no vacuum coherence in their idea set. That throws the baby out with the bathwater because guv Einstein's metric field for curved spacetime emerges from that non-perturbative vacuum coherence of the virtual electron-positron pairs primarily. So does the dark energy/matter that is 96% of the universe that does not appear anywhere in their models. They have been working on this for almost 20 years with little to show really. 3. There is no PV (i.e. no quantum polarized fluctuations) in Puthoff's PV math. 4. Puthoff and Ibison mis-interpret the physical meaning of their isotropic radial r coordinate in their toy model K = e^2GM/c^r Puthoff is very interested in UFOs and that is a primary motivation for this zero point gravity work. Nowhere do Puthoff and Haisch et-al squarely face the number String Tension = c^4/8piG = 10^19 Gev per 10^-33 cm which prevents any plausible explanation of UFO metric engineering with their brute force approach. Space-time geometry is simply too stiff to bend with the energy schemes they have in their paradigm. They are missing some very essential new concepts. Cook is a very approachable fellow; have you cross-checked with him also? -Andrew him about the flaky stuff he was pushing so he erased me from his book. That told me he was not intellectually honest, but had some hidden non-scientific agenda. I find this disturbing since he is associated with Jane's Defence Weekly in UK. === Subject: Re: UFO Bogus Physics what use do you get from these postings, when you never reply to your critics?... do you think that we're going to drop our gaurd, letting go of Eternal Vigilance? Ode to the Reichsaucer?... or can it all be ascribed to Puthoff's remote viewing research? I cannot imagine the repartee to follow your address to the Austin Powers Society for Science and Lit.! > http://www.salon.com/books/review/2002/08/05/zero_gravity/ > I did some investigating and interviewed retired senior USG > executives > who laid the Nazi flying saucer (disguised as UFOs) claim (a la > Cook's book) to rest - it didn't exist and it didn't fly. > erased me from his book. That told me he was not intellectually honest, > but had some hidden non-scientific agenda. I > find this disturbing since he is associated with Jane's Defence Weekly --ils duces d'Enron! === Subject: Re: UFO Bogus Physics >what use do you get from these postings, >when you never reply to your critics?... do you think that >we're going to drop our gaurd, >letting go of Eternal Vigilance? >Ode to the Reichsaucer?... or can it all be ascribed >to Puthoff's remote viewing research? > I cannot imagine the repartee to follow your address >to the Austin Powers Society for Science and Lit.! >> http://www.salon.com/books/review/2002/08/05/zero_gravity/ >> I did some investigating and interviewed retired senior USG >> executives >> who laid the Nazi flying saucer (disguised as UFOs) claim (a la >> Cook's book) to rest - it didn't exist and it didn't fly. >> erased me from his book. That told me he was not intellectually honest, >> but had some hidden non-scientific agenda. I >> find this disturbing since he is associated with Jane's Defence Weekly I think that there is enough evidence to support the claims that these ET craft exist, and have an exotic power supply. My educated guess is that they are using em waves to create a low pressure em area and then they fall into it. The vast difference between their physics and ours, IMHO is that we look at gravity as a field, and a separate force from electromagnetism, whereas they may be viewing gravity as a side effect, the basic pressure of the universe, fueled by expansion. radiation is a clue here. These waves combine, and cause gravity. Below the level of a pressurized zone, the basic pressure of the universe. Examine th eloaf of bread expanding, from all points within. The expansion as a ray from all points, equals the force of gravity. One pole. By using em waves, which permeate all of space anyway, they are able to affect even the zone which lies under our level of detection, and can project a low pressure area, or gravity well, whereupon, the vehicle then falls into that well. See the Hutchinson videos, Hutchison effect, and the broom which takes off like a rocket, while under the influence of em waves from powerful RF transmitters, as proof. === Subject: Re: UFO Bogus Physics QncyMI@netscape.net says... >what use do you get from these postings, >when you never reply to your critics?... do you think that >we're going to drop our gaurd, >letting go of Eternal Vigilance? >Ode to the Reichsaucer?... or can it all be ascribed >to Puthoff's remote viewing research? > I cannot imagine the repartee to follow your address >to the Austin Powers Society for Science and Lit.! >> http://www.salon.com/books/review/2002/08/05/zero_gravity/ >> I did some investigating and interviewed retired senior USG >> executives >> who laid the Nazi flying saucer (disguised as UFOs) claim (a la >> Cook's book) to rest - it didn't exist and it didn't fly. >> erased me from his book. That told me he was not intellectually honest, >> but had some hidden non-scientific agenda. I >> find this disturbing since he is associated with Jane's Defence Weekly > I think that there is enough evidence to support the claims that > these ET craft exist, and have an exotic power supply. > My educated guess is that they are using em waves to create > a low pressure em area and then they fall into it. > The vast difference between their physics and ours, IMHO is that > we look at gravity as a field, and a separate force from > electromagnetism, whereas they may be viewing gravity as > a side effect, the basic pressure of the universe, > fueled by expansion. > radiation is a clue here. > These waves combine, and cause gravity. Below the level of > a pressurized zone, the basic pressure of the universe. > Examine th eloaf of bread expanding, from all points within. > The expansion as a ray from all points, equals the force > of gravity. One pole. > By using em waves, which permeate all of space anyway, > they are able to affect even the zone which lies under > our level of detection, and can project a low pressure > area, or gravity well, whereupon, the vehicle then falls > into that well. > See the Hutchinson videos, Hutchison effect, and the broom > which takes off like a rocket, while under the influence > of em waves from powerful RF transmitters, as proof. No silly that's not how they do it... This is: Http://Paul.Mays.Com/device.html Note the predicted flight characteristics.... Non Newtonian flight, undetectable while under power, faster than light velocity, no thruster audio just the sounds of the power system when not fully charged/stressed, materials would be required to be strong but extremely low mass (ala Roswell) , ability to very quickly accelerate and decelerate with out inertia impingements, no impact with massive objects at trans light speed, very low power requirements since the Quantum State is modified by a 0 current, high voltage static charge once the field is charged it requires no current to maintain and directed motion is only varying the plate voltage relationships so only current is required when making a change in direction or velocity to recharge any plate that was reduced in voltage to induce the change..... Once the craft/device is in the gravitational well of a large massive object the device /craft must continuously vary its plate charge relationships to match the local variance in local gravitational anomalies giving the device/craft a bouncy, jittery motion when in slow velocity low stress field state which would seem smoother as velocity increased... Seem to match observed flight characteristics quite well.... And here's a working demo you can build to see the concept in action.. Http://ripkaboroski.com/device5b.jpg Science is fun for all ages..... Now I got a customer calling for a bridge I have up for sale on EBay.... so gota run... Ta Ta.... Paul R. Mays ---------------------------------------------------------------------------- - Some where within the Quantum State Http://Paul.Mays.Com/story.html http://paul.mays.com/mayday.html http://paul.mays.com/rainy.html We must be clear that, when it comes to atoms, language can be used only as in poetry. Niels Bohr === Subject: A Double-Sum Congruence Here is a trite little theorem, yet possessing some beauty. (And so I post...) Let p = prime, and let m be any positive integer <= p^p -1. Then: m --- --- | 1/p | | m | / / |------| is congruent to --- --- |_ j _| k=1 j|k j<= k^(1/p) 1/p [m ] --- --- | m | / / |-----| (mod p) --- --- |_ j _| k=1 j|k Linear-mode: sum{k=1 to m} sum{j|k, j<= k^(1/p)} floor(m^(1/p)/j) is congruent to sum{k=1 to floor(m^(1/p))} sum{j|k} floor(m/j) (mod p) (I think.) Leroy Quet === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days >> Let's be serious for once. >> Consider an object being accelerated by a idealistic jet of water or a >> continuous 'stream of elastic ping pong balls'. What is its subsequent velocity >> pattern? >> accelerated beyond the operating speed of the accelerating fields, ie at 'c'. I >> hope it might also produce a relationship that is equivalent to mass >> 'appearing' to increase with velocity by gamma.) >Please define what you mean by the operating speed of an >accelerating field. I've been in physics research for over 20 >years, and I've never heard that term. >Do you mean phase velocity, group velocity, how fast the >operators turn the knobs, what? >*static* fields, while most RF acclerating structures use >standing waves, so your point is dead on arrival. >Why are you blathering on about jets of water and ping pong >balls? They are nothing like the acceleration of an EM >field. Even if they were, what you're describing is a rocket, >and a rocket can easily exceed the velocity of it's propellant. >If you'd ever taken a physics class, you'd know that the >final speed of a rocket starting at rest is (in the >absense of gravity): > v = v_0 * ln(m_initial/m_final) Here the jet speed is constant wrt the rocket not the ground. It is NOT the same problem. >where v_0 is the propellant (or water or ping pong ball) velocity, >so as long as that mass ratio is greater than e, you're point- >even though it was flawed to begin with - is *still* dead on arrival. >Maxwell's Equations and the Lorentz Force Equation have >been around for a very long time, and no one's ever found >a flaw with them. If you're going to talk about accelerating >fields, please do so in the proper language - or is it >that you *can't*? >Just because you think the Earth is flat doesn't stop >other from sailing around it. Find a hobby you can understand. Oh what an obvious genius we have here! If the x/t curve for my 'object' turns out to be about the same as the one for a charge in a constant field, your conceited smirk might fade away. Henri Wilson. See why relativity is wrong: http://www.users.bigpond.com/HeWn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) >> Let's be serious for once. >> >> Consider an object being accelerated by a idealistic jet of water or a >> continuous 'stream of elastic ping pong balls'. What is its subsequent velocity >> pattern? >> >> accelerated beyond the operating speed of the accelerating fields, ie at 'c'. I >> hope it might also produce a relationship that is equivalent to mass >> 'appearing' to increase with velocity by gamma.) >> >Please define what you mean by the operating speed of an >accelerating field. I've been in physics research for over 20 >years, and I've never heard that term. >Do you mean phase velocity, group velocity, how fast the >operators turn the knobs, what? >*static* fields, while most RF acclerating structures use >standing waves, so your point is dead on arrival. >Why are you blathering on about jets of water and ping pong >balls? They are nothing like the acceleration of an EM >field. Even if they were, what you're describing is a rocket, >and a rocket can easily exceed the velocity of it's propellant. >If you'd ever taken a physics class, you'd know that the >final speed of a rocket starting at rest is (in the >absense of gravity): > v = v_0 * ln(m_initial/m_final) > Here the jet speed is constant wrt the rocket not the ground. It is NOT the > same problem. OK, so you have something which is neither a model for a rocket NOR an EM field. Of what use is that supposed to be, exactly? >where v_0 is the propellant (or water or ping pong ball) velocity, >so as long as that mass ratio is greater than e, you're point- >even though it was flawed to begin with - is *still* dead on arrival. >Maxwell's Equations and the Lorentz Force Equation have >been around for a very long time, and no one's ever found >a flaw with them. If you're going to talk about accelerating >fields, please do so in the proper language - or is it >that you *can't*? >Just because you think the Earth is flat doesn't stop >other from sailing around it. Find a hobby you can understand. > Oh what an obvious genius we have here! > If the x/t curve for my 'object' turns out to be about the same as the one for > a charge in a constant field, your conceited smirk might fade away. But of course, since you can't even do *this* simple math, you'll never know, will you? Grown-ups deal with it all the time, and it's nothing involving ping pong balls or water or any such nonsense. If you can't handle math, you can't handle math, but don't pretend there's anything profound about that except your ignorance. -E > Henri Wilson. > See why relativity is wrong: > http://www.users.bigpond.com/HeWn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days >> same problem. >OK, so you have something which is neither a model for >a rocket NOR an EM field. Of what use is that supposed to >be, exactly? I want to drive a boat across a river by hosing it with water from the bank. I want to know how fast it will travel. I want to know exactly how it reaches the maximum speed. By the way the river water has no viscosity. >> >>where v_0 is the propellant (or water or ping pong ball) velocity, >>so as long as that mass ratio is greater than e, you're point- >>even though it was flawed to begin with - is *still* dead on arrival. >> >>Maxwell's Equations and the Lorentz Force Equation have >>been around for a very long time, and no one's ever found >>a flaw with them. If you're going to talk about accelerating >>fields, please do so in the proper language - or is it >>that you *can't*? >> >>Just because you think the Earth is flat doesn't stop >>other from sailing around it. Find a hobby you can understand. >> Oh what an obvious genius we have here! >> If the x/t curve for my 'object' turns out to be about the same as the one >for >> a charge in a constant field, your conceited smirk might fade away. >But of course, since you can't even do *this* simple math, >you'll never know, will you? >Grown-ups deal with it all the time, and it's nothing involving ping >pong balls or water or any such nonsense. If you can't handle >math, you can't handle math, but don't pretend there's anything >profound about that except your ignorance. I'm sure the x/t equation for moving charges has been calculated many times, including the so-called 'relativistic mass increase'. I just want to compare that curve with a couple af mechanical models. Anything wrong wth that? Henri Wilson. See why relativity is wrong: http://www.users.bigpond.com/HeWn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) >> >> same problem. >> >OK, so you have something which is neither a model for >a rocket NOR an EM field. Of what use is that supposed to >be, exactly? > I want to drive a boat across a river by hosing it with water from the bank. > I want to know how fast it will travel. > I want to know exactly how it reaches the maximum speed. > By the way the river water has no viscosity. For this you need to specify three things: - Water stream velocity - Water flow - Boat mass Hopefully, the computation itself should be obvious by applying the concepts of impulse and momentum. Harry C. === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days >> Let's be serious for once. >> Consider an object being accelerated by a idealistic jet of water or a >> continuous 'stream of elastic ping pong balls'. What is its subsequent velocity >> pattern? >> accelerated beyond the operating speed of the accelerating fields, ie at 'c'. I >> hope it might also produce a relationship that is equivalent to mass >> 'appearing' to increase with velocity by gamma.) >> We have: >> S= -------m/s->Vo--------->M->v >> A stream of 'matter' emerges from source S at velocity Vo and m 'mass units' >> per second. It strikes unconstrained object (mass M), which subsequently >> accelerates away. Both energy and momentum must be conserved during this >> operation. >> The problem is to find the x/t relationship for three different situations. >> 1) nonelastic case where the jet ends up having the same velocity as the object >> after collision. >> 2) where the jet bounces back perfectly elastically. >> 3) (maybe) where the jet is absorbed into the object. >> WRT the source frame, the equation describing the nonelastic jet case is: >> M.dv/dt=m(Vo-v) or: >v = Vo*(1 - exp(-m*t/M)) That's roughly what I got - but I thought there was more to it? Are you sure that's the full story? >> (d2x/dt2)=K(Vo-dx/dt) >> Here, the water ends up traveling at M's velocity. >> In the elastic case, the water (or stream of ping pong balls) ends up moving >> backwards at Vo-2v, so the change in momentum per second is 2m(Vo-v) >v = Vo*(1 - exp(-2*m*t/M)) OK here's a harder problem. When the jet strikes the object, it is collected inside the object so that it increases the weight of the object. What is the x/t equation? Note, this is not the reverse of the 'rocket engine' problem because in that case the jet velocity is constant wrt the ship not the ground. >> Do you all agree? >> I now wish to solve this equation. My maths is a bit rusty but I can probably >> work it out eventually. I get something starting with a factor e^-K (which is >> good). I'm not sure about the rest. Any offers? >> What are mathematicians for anyway? >There is no reason for making this more complicated than it is, Henry. >The mass M will in all cases end up moving with the speed Vo. >Or rather, v will approach Vo asymptotically. >But you knew this, didn't you? >And this have absolutely nothing to do with what happens >Which you also knew, didn't you? Not at all Paul. I have to compare the curve given by the above solution with that predicted by SR for a charge accelerated in a constant field. Can you tell me what that might be? >Paul Henri Wilson. See why relativity is wrong: http://www.users.bigpond.com/HeWn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days > Let's be serious for once. > Consider an object being accelerated by a idealistic jet of water or a > continuous 'stream of elastic ping pong balls'. What is its subsequent velocity > pattern? > accelerated beyond the operating speed of the accelerating fields, ie at 'c'. I > hope it might also produce a relationship that is equivalent to mass > 'appearing' to increase with velocity by gamma.) > We have: > S= -------m/s->Vo--------->M->v > A stream of 'matter' emerges from source S at velocity Vo and m 'mass units' > per second. It strikes unconstrained object (mass M), which subsequently > accelerates away. Both energy and momentum must be conserved during this > operation. > The problem is to find the x/t relationship for three different situations. > 1) nonelastic case where the jet ends up having the same velocity as the object > after collision. > 2) where the jet bounces back perfectly elastically. > 3) (maybe) where the jet is absorbed into the object. > WRT the source frame, the equation describing the nonelastic jet case is: > M.dv/dt=m(Vo-v) or: >>v = Vo*(1 - exp(-m*t/M)) >That's roughly what I got - but I thought there was more to it? >Are you sure that's the full story? > (d2x/dt2)=K(Vo-dx/dt) > Here, the water ends up traveling at M's velocity. > In the elastic case, the water (or stream of ping pong balls) ends up moving > backwards at Vo-2v, so the change in momentum per second is 2m(Vo-v) >>v = Vo*(1 - exp(-2*m*t/M)) >OK here's a harder problem. >When the jet strikes the object, it is collected inside the object so that it >increases the weight of the object. What is the x/t equation? >Note, this is not the reverse of the 'rocket engine' problem because in that >case the jet velocity is constant wrt the ship not the ground. > Do you all agree? > I now wish to solve this equation. My maths is a bit rusty but I can probably > work it out eventually. I get something starting with a factor e^-K (which is > good). I'm not sure about the rest. Any offers? > What are mathematicians for anyway? >>There is no reason for making this more complicated than it is, Henry. >>The mass M will in all cases end up moving with the speed Vo. >>Or rather, v will approach Vo asymptotically. >>But you knew this, didn't you? >>And this have absolutely nothing to do with what happens >>Which you also knew, didn't you? >Not at all Paul. >I have to compare the curve given by the above solution with that predicted by >SR for a charge accelerated in a constant field. >Can you tell me what that might be? >>Paul Incidentally, particularly in the case of the elastic 'ping-pong ball drive', momentum is balanced but does the (kinetic) energy equation match? Henri Wilson. See why relativity is wrong: http://www.users.bigpond.com/HeWn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days >> accelerated beyond the operating speed of the accelerating fields, ie at 'c'. >Explain that phrase, please. >As you know, the accelerating electic field in an accelerator >So what do you mean by: >the operating speed of the accelerationg field? You don't know what happens in close proximity to the accelerating charge. Back radiation? >Paul Henri Wilson. See why relativity is wrong: http://www.users.bigpond.com/HeWn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days >> Let's be serious for once. >> >> Consider an object being accelerated by a idealistic jet of water or a >> continuous 'stream of elastic ping pong balls'. What is its subsequent velocity >> pattern? >> Explain to me how this model applies to an attractive >> electric field. >> What are the ping-pong balls and where are they coming >> from? >I will seriously sit back and seriously enjoy the reply. >Good one :-)) >Dirk Vdm Same applies to you. Henri Wilson. See why relativity is wrong: http://www.users.bigpond.com/HeWn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days >> Let's be serious for once. >> Consider an object being accelerated by a idealistic jet of water or a >> continuous 'stream of elastic ping pong balls'. What is its subsequent velocity >> pattern? >Explain to me how this model applies to an attractive >electric field. >What are the ping-pong balls and where are they coming >from? Don't worry about it Randy. I doubt if you know what a differential equation is. > - Randy Henri Wilson. See why relativity is wrong: http://www.users.bigpond.com/HeWn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) > Let's be serious for once. > > Consider an object being accelerated by a idealistic jet of water or a > continuous 'stream of elastic ping pong balls'. What is its subsequent velocity > pattern? >>Explain to me how this model applies to an attractive >>electric field. >>What are the ping-pong balls and where are they coming >>from? >Don't worry about it Randy. I doubt if you know what a differential equation >is. Well, you could write one and we could see. But you don't write equations, do you? Meanwhile, why don't you explain to me how this model applies to an attractive electric field. What are the ping-pong balls and where are they coming from? Or do you really think nobody noticed that you avoided answering any questions about your model? - Randy === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days >> Let's be serious for once. >> >> Consider an object being accelerated by a idealistic jet of water or a >> continuous 'stream of elastic ping pong balls'. What is its subsequent velocity >> pattern? >Explain to me how this model applies to an attractive >electric field. >What are the ping-pong balls and where are they coming >from? >>Don't worry about it Randy. I doubt if you know what a differential equation >>is. >Well, you could write one and we could see. But you don't write >equations, do you? >Meanwhile, why don't you explain to me how this model applies to an >attractive electric field. What are the ping-pong balls and where are >they coming from? This is called theoretical research Randy. The aim is to try to match a hypothetical physical process with the prediction of an equally hypothetical maths theory, namely relativity. >Or do you really think nobody noticed that you avoided answering any >questions about your model? Paul Anderson understood perfectly. Is he sarter than you?. > - Randy Henri Wilson. See why relativity is wrong: http://www.users.bigpond.com/HeWn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) >> >> Let's be serious for once. >> >> Consider an object being accelerated by a idealistic jet of water or a >> continuous 'stream of elastic ping pong balls'. What is its subsequent velocity >> pattern? > >Explain to me how this model applies to an attractive >electric field. > >What are the ping-pong balls and where are they coming >from? >> >>Don't worry about it Randy. I doubt if you know what a differential equation >>is. >Well, you could write one and we could see. But you don't write >equations, do you? >Meanwhile, why don't you explain to me how this model applies to an >attractive electric field. What are the ping-pong balls and where are >they coming from? > This is called theoretical research Randy. > The aim is to try to match a hypothetical physical process with the prediction > of an equally hypothetical maths theory, namely relativity. It's just a theory? Are we a creationist now? Of course the theory is just a theory. The point is, it makes predictions about such things like the maximum speed You have an alternate theory. are accelerated by receiving momentum from little objects moving at c. Fine. If this is your explanation for the limiting speed, please tell me what corresponds to the little ping-pong balls in the case of an attractive force, and where they are coming from. You have now alluded to theoretical research without a theory, differential equations without an equation, and in general spun a lot of words while refusing to answer the question. Could it be that you know that your model is not a model >Or do you really think nobody noticed that you avoided answering any >questions about your model? > Paul Anderson understood perfectly. Is he sarter than you?. If Paul Anderson has given you an answer about your model which you can crib and offer as your answers, by all means feel free to do so. Do you really think nobody has noticed that you've avoided answering any questions about your model? Explain to me how this model applies to an attractive electric field. What are the ping-pong balls and where are they coming from? - Randy === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) >> Let's be serious for once. >> >> Consider an object being accelerated by a idealistic jet of water or a >> continuous 'stream of elastic ping pong balls'. What is its subsequent velocity >> pattern? >Explain to me how this model applies to an attractive >electric field. >What are the ping-pong balls and where are they coming >from? > Don't worry about it Randy. I doubt if you know what a differential equation > is. Brilliant :-) Dirk Vdm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) > Let's be serious for once. > > Consider an object being accelerated by a idealistic jet of water or a > continuous 'stream of elastic ping pong balls'. What is its subsequent velocity > pattern? >> >>Explain to me how this model applies to an attractive >>electric field. >> >>What are the ping-pong balls and where are they coming >>from? >> Don't worry about it Randy. I doubt if you know what a differential equation >> is. >Brilliant :-) All right. Now tell me which diff. eq. is on the number 2 ball in the Mass. Millions lottery. /BAH Subtract a hundred and four for e-mail. === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days >> Let's be serious for once. >> >> Consider an object being accelerated by a idealistic jet of water or >> continuous 'stream of elastic ping pong balls'. What is its >subsequent velocity >> pattern? > >Explain to me how this model applies to an attractive >electric field. > >What are the ping-pong balls and where are they coming >from? > Don't worry about it Randy. I doubt if you know what a differential >equation > is. >>Brilliant :-) >All right. Now tell me which diff. eq. is on the number 2 ball >in the Mass. Millions lottery. Ask an SRian. He claims to know everything. >/BAH >Subtract a hundred and four for e-mail. Henri Wilson. See why relativity is wrong: http://www.users.bigpond.com/HeWn/index.htm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Expires: 28 days >>Let's be serious for once. >>Consider an object being accelerated by a idealistic jet of water or a >>continuous 'stream of elastic ping pong balls'. What is its subsequent velocity >>pattern? >>accelerated beyond the operating speed of the accelerating fields, ie at 'c'. I >>hope it might also produce a relationship that is equivalent to mass >>'appearing' to increase with velocity by gamma.) > What is being ignored is the fact that no matter what >in the same direction will still be going faster by precisely c ! > Why don't you find a useful agenda, you are doing >nothing but costing a lot of people a lot of money on a >worthless and futile effort to show that henry is smarter >than Einstein and all physicists since. What I proposed is a simple and interesting mechanical problem. I gather you cannot solve the rather elementary differential equation. OR - Maybe you fear that I am right and this will give the same relationship that SR give for fictitious 'mass increase' >Joe Fischer Henri Wilson. See why relativity is wrong: http://www.users.bigpond.com/HeWn/index.htm === Subject: Help teaching stat 11/4/03 I could sure use your help. I was just asked to speak next January at the National Institute for the Teaching of Psychology and would like to ask you for a few minutes of your time. My presentation at NITOP is titled, What's difficult to teach in introductory statistics and how to do it. Can you please take ten minutes and help me prepare for this presentation by answering the following questions? 1. In your intro stat class, what three topics do you find most difficult to teach your students? 2. How do you teach these topics? What techniques, strategies, ideas, and tools help you? 3. What's the one coolest, most fun thing you do in your intro stat class to help students learn? Your help is tremendously appreciated and I will send you a copy of the final presentation which should be ready sometime in January. Can you please send your responses via email to njs@ku.edu and please place the words intro stat in the subject line? Please be sure to include your email address so I may send you a copy of the final presentation. Also, can you please forward this same letter to two other colleagues of yours that teach intro stat? Neil Salkind, Ph.D. University of Kansas Lawrence, KS 66045 === Subject: Sum{k|m, k<= sqrt(m)} mu(k) If mu(k) is the Mobius(Moebius) function, then sum{k|m, k<= sqrt(m)} mu(k) gives us the sequence: 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, -1, 1, 0, 0, 0,... (Sequence A068101 in the EIS: http://www.research.att.com/projects/OEIS?Anum=A068101 ) I may have asked this already a long time ago, but what can be said about this sequence, such as its asymptotics, the expected number of 0's, +-1's, +-2's, etc below any given value, and (perhaps) even what its closed-form might be?? All I can deduce right now is that, unspectacularly, sum{m=1 to n} sum{k|m, k<= sqrt(m)} mu(k) = sum{1<=k<=sqrt(n)} mu(k) (floor(n/k) +1 -k). Leroy Quet electron-dot-cloud are galaxies === Subject: VonNeumann Gametheory Optimal Strategy for playing Stockmarket Portfolio of PAF as of 4NOV03 50 BCE 22.42 $1,121.00 50 BLS 25.87 $1,293.50 100 BMY 24.81 $2,481.00 50 DT 15.51 $775.50 51,050 Q 3.52 $179,696.00 11,600 SBC 23.90 $277,240.00 50 VZ 32.92 $1,646.00 50 WYE 43.35 $2,167.50 realestate land 3APR03 of 3 lots $19,000 art of science-lithographs & porcelain JAN-JUN03 for $12,000 realestate land 30JUL03 another lot $11,500 Last time I reported on PAF was 26Sept03 where I simply took some dividend cash sitting around in the portfolio saved up and bought 100 shares of SBC and 1,000 more shares of Qwest. Today I had some new dividend cash and with the proceeds went ahead and bought 150 more shares of SBC. The Optimal Strategy for playing the StockMarket should have long lull periods where nothing is in action except for dividend re-investment. Although there are signs of action stirring in the telecom sector with BellSouth recently breaking off with AT&T as a possible merger where BLS cited that the price for AT&T was too high considering it is a company of shrinking assets. The obvious match for BLS is Qwest since the baby bells were the prize assets for the past several decades. The drug industry will stay in the doldrums until human-cloning lifts it back as the premier industry. That means we need a president in the WhiteHouse who is not anti-science and antitechnology. In the past 3 years, the Bush administration could have spurred the economy without its tax cuts and ballooning deficits by simply legislating a new Antitrust Law which says that almost any company can buy out any other company when the economy is in a recession and during expansions the government breaks apart large companies. In that way, society smooths out and evens out expansions so that they do not become crazy like the 1990s and that the economy does not fall steeply into recessions. So this new Antitrust law would balance out the economy instead of booms and busts. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: simple variational/control problem? >The task is to choose h(t) that maximizes the integral: >int_a^b e^(-rt) * (p(t) - c(t)) * h(t) dt, Let's call this int_a^b f(t) h(t) dt >subject to: >int_a^b h(t) dt = W (some given constant) >and where p(t)-c(t) is concave down (ie, upside-down-parabola-like). >My book says that it is evident, due to the simple form of the >integrand in the objective, that the required h(t) is given by: >h(t) = W * delta (t - t_m), I hope the book doesn't call this a function... >where t_m that time giving the maximum value of e^(-rt) * (p(t) - >c(t)). >I don't dispute the truth of the claim, I just dont see what's so >evident about it - can anyone out there explain the obviousness of the >Dirac delta as the solution to this problem? Is it required that h >= 0? Otherwise (except in the trivial case where f is constant) the claim is false and there is no such maximum for the integral. Namely, taking points c,d in the interval where f(c) > f(d), you can get an arbitrary value v by taking h(t) = (W + x) delta(t-c) - x delta(t-d) where x = (v - W f(c))/(f(c) - f(d)). Suppose M = max(f(t): a <= t <= b). Then (if h >= 0) int_a^b h(t) f(t) dt <= M int_a^b h(t) dt = M W = int_a^b f(t) delta(t-t_m) dt Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: simple variational/control problem? >My book says that it is evident, due to the simple form of the >integrand in the objective, that the required h(t) is given by: >h(t) = W * delta (t - t_m), > I hope the book doesn't call this a function... lol - no worries - the book explicitly points out that this is cheating, since the variational apparatus developed there was for PWS functions - that's what immediately preceded the intuitive analysis I asked about, in fact. > Is it required that h >= 0? ack! I was writing from memory - that one slipped my mind - my bad - yes, h is nonnegative. ( h is a harvest rate, and the rest of the integrand is present-value-profit.) sorry about the omission, cdj === Subject: Magnetism: compare magnetic fields Looking for a direct mathmatical comparison between the magnetic field produced by an electronic current (i) and that of a fixed magnet of a given strength. === Subject: Re: Magnetism: compare magnetic fields > Looking for a direct mathmatical comparison between the magnetic field > produced by an electronic current (i) and that of a fixed magnet of a > given strength. See: http://scienceworld.wolfram.com/physics/Magnetism.html Note the differences in o diamagnetism o Paramagnetism Ferromagnetism http://scienceworld.wolfram.com/physics/Ferromagnetism.html The development of extremely strong magnetic properties in certain materials which occurs when magnetic domains (regions at most 1 mm in dimension) become aligned in the absence of an applied field, below a temperature known as the Curie temperature. The net magnetization depends on the magnetic history (the hysteresis effect). Above the Curie temperature, these materials become paramagnetic. Iron, nickel, cobalt, and gadolinium are ferromagnetic at room temperature. Ferromagnetism is believed to be caused by magnetic fields generated by the electrons' spins in combination with a mechanism known as exchange coupling, which aligns all the spins in each magnetic domain. === Subject: Re: Magnetism: compare magnetic fields > Looking for a direct mathmatical comparison between the magnetic field > produced by an electronic current (i) and that of a fixed magnet of a > given strength. > See: http://scienceworld.wolfram.com/physics/Magnetism.html > Note the differences in > o diamagnetism > o Paramagnetism > Ferromagnetism > http://scienceworld.wolfram.com/physics/Ferromagnetism.html > The development of extremely strong magnetic properties in certain > materials which occurs when magnetic domains (regions at most 1 mm in > dimension) become aligned in the absence of an applied field, below a > temperature known as the Curie temperature. The net magnetization > depends on the magnetic history (the hysteresis effect). Above the > Curie temperature, these materials become paramagnetic. Iron, nickel, > cobalt, and gadolinium are ferromagnetic at room temperature. > Ferromagnetism is believed to be caused by magnetic fields generated > by the electrons' spins in combination with a mechanism known as > exchange coupling, which aligns all the spins in each magnetic > domain. === Subject: Re: Magnetism: compare magnetic fields | Looking for a direct mathmatical comparison between the magnetic field | produced by an electronic current (i) and that of a fixed magnet of a | given strength. Keyword: solenoid FrediFizzx === Subject: Re: Magnetism: compare magnetic fields > Looking for a direct mathmatical comparison between the magnetic field > produced by an electronic current (i) and that of a fixed magnet of a > given strength. See Biot-Savart Law. Also look up solenoids. A tightly wound solenoid has a magnetic field that is exactly like that of a natural bar magnet. The workings of a magnetic are due to the spin of electrons in the atoms of the magnet and the fact that the spins are aligned over large regions of the magnet. All magnetism, as far as anybody knows, is do to electric charges in motion. No one has ever seen a magnetic monopole. Bob Kolker === Subject: Re: [JSH] On the Rewriting of a Polynomial > To reprise James Harris' proof, or at least the first two steps > (I've included '*' so that GP/Pari works): > [begin excerpt] > 1. Let P(x) = 14706125 * x^3 - 900375 * x^2 - 17640 * x + 1078, where x is > in the ring of algebraic integers, notice that P(x) has a constant > term that is 1078. > 2. It can be shown that > P(x)= 7^2*(2401*x^3 - 147*x^2 + 3*x)*(5^3) - 3*(-1 + 49*x)*(5)*(7^2)+7^3 > where the *same* polynomial has been put in a form which allows a > factorization into non-polynomial factors so that I have > P(x) = (5*a_1(x) + 7)(5*a_2(x)+ 7)(5*a_3(x) + 7) > where the a's are roots of > Q(a) = a^3 + 3*(-1 + 49*x)*a^2 - 49*(2401*x^3 - 147*x^2 + 3*x). > [end excerpt] > Now, here's where I have a problem. How did Mr. Harris get to > this point? You can get there quickly by noticing that if you let a = -7/5 then -5^3Q(a=-7/5) = P(x) Similarly in the factored from Q(a)=(a-a_1)(a-a_2)(a-a_3) let a = -7/5 then -5^3Q(a) = (5a_1 + 7)(5a_2 + 7)(5a_3 + 7) = P(x) This confirms that the P(x) is a correct expression for the a_1, a_2 and a_3 and that they are indeed algebraic integers. The problem starts when James divides them by 7 and declares that because a_1/7 and a_2/7 are not necessarily algebraic integers, something is wrong. I can't work out what he means. For example P(x) = x(x + 1) is a multiple of 2 for all x integer therefore P(x)/2 is integer for all x integer P(x)/2 = x/2*(x + 1) (the only way it can go when you set x = 0, or something) But x/2 is not an integer for all x integer. Presumably there is something wrong with integers as well. HTH, Phil. === Subject: Re: [JSH] On the Rewriting of a Polynomial In sci.math, Phil : >> To reprise James Harris' proof, or at least the first two steps >> (I've included '*' so that GP/Pari works): >> [begin excerpt] >> 1. Let P(x) = 14706125 * x^3 - 900375 * x^2 - 17640 * x + 1078, where x > is >> in the ring of algebraic integers, notice that P(x) has a constant >> term that is 1078. >> 2. It can be shown that >> P(x)= 7^2*(2401*x^3 - 147*x^2 + 3*x)*(5^3) - 3*(-1 + 49*x)*(5)*(7^2)+7^3 >> where the *same* polynomial has been put in a form which allows a >> factorization into non-polynomial factors so that I have >> P(x) = (5*a_1(x) + 7)(5*a_2(x)+ 7)(5*a_3(x) + 7) >> where the a's are roots of >> Q(a) = a^3 + 3*(-1 + 49*x)*a^2 - 49*(2401*x^3 - 147*x^2 + 3*x). >> [end excerpt] >> Now, here's where I have a problem. How did Mr. Harris get to >> this point? > > You can get there quickly by noticing that if you let a = -7/5 then > -5^3Q(a=-7/5) = P(x) > Similarly in the factored from Q(a)=(a-a_1)(a-a_2)(a-a_3) > let a = -7/5 then > -5^3Q(a) = (5a_1 + 7)(5a_2 + 7)(5a_3 + 7) = P(x) Hm...yes, that looks fairly simple. :-) I'll admit what confused me was the requirement that Q(a,x) have two independent variables. > This confirms that the P(x) is a correct expression for the a_1, a_2 and a_3 > and that they > are indeed algebraic integers. > The problem starts when James divides them by 7 and declares that because > a_1/7 and a_2/7 are not necessarily algebraic integers, something is wrong. > I can't work out what he means. Join the club. :-/ > For example > P(x) = x(x + 1) is a multiple of 2 for all x integer > therefore P(x)/2 is integer for all x integer > P(x)/2 = x/2*(x + 1) (the only way it can go when you set x = 0, or > something) > But x/2 is not an integer for all x integer. > Presumably there is something wrong with integers as well. Well, James has also tried to argue that Z[1/2] contains all reals. :-) (It doesn't even contain 1/3.) > HTH, Phil. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: [JSH] On the Rewriting of a Polynomial Most readers will be able to see that Dik W. and Ghost ITM are cheating, by using more than a handheld calculator. Well, I've never been wrong, yet! > Actually it is quite simple. Q(a) is one of the many possible polynomials. > And it works the other way around, when you have a1(x), a2(x) and a3(x) > roots of Q(a), then P(x) = (5 a1(x)+7)(5 a2(x)+7)(5 a3(x)+7). > Q(a) gives the values for (a1+a2+a3), (a1 a2+a1 a3+a2 a3) and (a1 a2 a3). > Writing out the factorisation of P(x) as a single expression and filling > in these three values gives the original form of P(x). --ils duces d'Enron! http://tarpley.net/bush8.htm http://www.wlym.com/PDF-SpReps/SPRP13.pdf === Subject: Re: [JSH] On the Rewriting of a Polynomial > Most readers will be able to see that Dik W. and Ghost ITM are cheating, > by using more than a handheld calculator. Well, I've never been wrong, > yet! Ah, you are also cheating. You are using more than a handheld calculator -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Derivatives II I'm reposting this because it didn't show up on my newsreader, only > Phil Holman grava .88 la saucisse et au marteau: > f(p) = q > where f(140) = 15000 > and f'(140) = -100 > R = pq > What is dR/dp (p=140) > I get (f'*p) + (f+f')*dp > = (-100*140) + (14900*1) > = -14000 + 14900 > = +900 > R = pf(p) > dR/dp = f(p) + pf'(p) > So dR/dp |p =140 = f(140) + 140*f'(140) > = 15000 - 14000 > = 1000 If we apply this in a practical example where p=price, q=quantity sold, R=revenue. If we increase the price from 140 to 141, we decrease the quantity sold from 15000 to 14900. This will increase the revenue by 900 and not 1000. I guess the problem here is with a changing derivative between a p=140 and p=141. This gave me a real problem and hence my hacked solution. With no definition of f(p) except for a single point the function could be defined in an infinite number of ways. Does anyone have any other comments about this. Phil Holman === Subject: Re: Future of mathematics >which do you claim? That mere humans really ARE that predictable that you can tell what a mehum is going to do or say even a decade in advance. === Subject: c-number Howdy everybody, I'm reading Peskin and Schroeder's Intro to QFT and in this book the author constantly refers to c-numbers... without saying what a c-number is. I couldn't find c-number on mathworld.wolfram.com or physicsworld.wolfram.com. Can someone please tell me what the hell a c-number is? adam === Subject: Re: c-number : : Howdy everybody, : : I'm reading Peskin and Schroeder's Intro to QFT and in this : book the author constantly refers to c-numbers... without : saying what a c-number is. I couldn't find c-number : on mathworld.wolfram.com or physicsworld.wolfram.com. : : Can someone please tell me what the hell a c-number is? I think it means a classical number, as opposed to a q-number (quantum number). : : : adam : === Subject: Re: c-number > Howdy everybody, > I'm reading Peskin and Schroeder's Intro to QFT and in this > book the author constantly refers to c-numbers... without > saying what a c-number is. I couldn't find c-number > on mathworld.wolfram.com or physicsworld.wolfram.com. > Can someone please tell me what the hell a c-number is? > adam I haven't read Peskin and Schroeder but if their usage of the term c-number is the same as everyone else's usage then they are referring to commuting variables. For example, Bryce DeWitt introduces the concepts of c-numbers (commuting variables) and a-numbers (anticommuting variables) in his book Supermanifolds. If a,b are c-numbers then a*b = b*a while if x,y are a-numbers we have x*y = -y*x. More generally, denoting the parity of an element x of a Z2-graded algebra by |x| we have x*y = (-1)^{|x|.|y|}*y*x, where |x| = 0 (resp. 1) if x is a c-number (resp. a-number). davidoff === Subject: Re: c-number >> Howdy everybody, >> I'm reading Peskin and Schroeder's Intro to QFT and in this >> book the author constantly refers to c-numbers... without >> saying what a c-number is. I couldn't find c-number >> on mathworld.wolfram.com or physicsworld.wolfram.com. >> Can someone please tell me what the hell a c-number is? >> adam >I haven't read Peskin and Schroeder but if their usage of the term >c-number is the same as everyone else's usage then they are referring >to commuting variables. For example, Bryce DeWitt introduces the >concepts of c-numbers (commuting variables) and a-numbers (anticommuting >variables) in his book Supermanifolds. If a,b are c-numbers then Eh? I thought it was just a complex number. -- Let us learn to dream, gentlemen, then perhaps we shall find the truth... But let us beware of publishing our dreams before they have been put to the proof by the waking understanding. -- Friedrich August Kekul.8e === Subject: Re: c-number >Howdy everybody, >I'm reading Peskin and Schroeder's Intro to QFT and in this >book the author constantly refers to c-numbers... without >saying what a c-number is. I couldn't find c-number >on mathworld.wolfram.com or physicsworld.wolfram.com. >Can someone please tell me what the hell a c-number is? >adam >>I haven't read Peskin and Schroeder but if their usage of the term >>c-number is the same as everyone else's usage then they are referring >>to commuting variables. For example, Bryce DeWitt introduces the >>concepts of c-numbers (commuting variables) and a-numbers (anticommuting >>variables) in his book Supermanifolds. If a,b are c-numbers then > Eh? I thought it was just a complex number. Again, I can't say for sure since I don't have a copy of Peskin and Schroeder to hand. However, in the literature on supersymmetry and supergravity c-numbers and a-numbers always refer to elements of a Grassmann algebra (the so-called Grassmann variables). Anticommuting variables rock. Non-commutative geometry rocks. Berezin rocks. DeWitt rocks. Gelfand and Likhtman rock. Supergravity, however, most certainly /does not/ rock. davidoff === Subject: Re: c-number Originator: grubb@lola >> I'm reading Peskin and Schroeder's Intro to QFT and in this >> book the author constantly refers to c-numbers... without >> saying what a c-number is. I couldn't find c-number >> on mathworld.wolfram.com or physicsworld.wolfram.com. >> Can someone please tell me what the hell a c-number is? >> adam >I haven't read Peskin and Schroeder but if their usage of the term >c-number is the same as everyone else's usage then they are referring >to commuting variables. For example, Bryce DeWitt introduces the >concepts of c-numbers (commuting variables) and a-numbers (anticommuting >variables) in his book Supermanifolds. If a,b are c-numbers then >a*b = b*a This is correct. The problem is that Peskin and Schroeder don't go into 'grassman numbers' in any detail at all. DeWitt's book is a good intro. --Dan Grubb === Subject: Re: Math dependency logic permission for an emailed response. > Ultimately my argument relies on numbers like 7 being NUMBERS, not > variables dependent on x, and on the distributive property. 7 is a number, but so is x. Neither can change (in a given context); each refers to exactly one number at a time. === Subject: Re: Math dependency logic > Ultimately my argument relies on numbers like 7 being NUMBERS, not > variables dependent on x, and on the distributive property. > 7 is a number, but so is x. Neither can change (in a given context); > each refers to exactly one number at a time. What does time have to do with it? How do you define time with mathematical argument Thomas Bushnell, BSG? James Harris === Subject: Re: Math dependency logic permission for an emailed response. > > Ultimately my argument relies on numbers like 7 being NUMBERS, not > variables dependent on x, and on the distributive property. > > 7 is a number, but so is x. Neither can change (in a given context); > each refers to exactly one number at a time. > What does time have to do with it? > How do you define time with mathematical argument Thomas Bushnell, BSG? Good grief. Time in that sentence refers to scope, it's a repetition of the in a given context paretheses. This is English usage. I refer you to In any given context, x refers to exactly one number (that is, x refers to the number x), never more than one. It certainly doesn't change. It doesn't vary. For example, in the sentence x = 2, x refers to 2. In the sentence x - 2 = 0, the symbol x again refers to 2. In both sentences, x refers to exactly one number, and doesn't vary or change at all. In the sentence for all x, if x is an integer, then either x is even, or x is odd--in that sentence, x doesn't appear at all! That's just how quantification works. In the phrase x + 2, we have two ways of understanding that phrase. We might mean x + 2 to be a function f, such that f:R->R, with the definition that for all x, f(x) = x+2. In that case, the variable x is bound by the quantifier, and so again it doesn't appear. This can be confusing, because for all x, f(x) = x+2 gets abbreviated as f(x) = x+2, and the function f:R->R such that f(x) = x+2 gets abbreviated as x+2. So when x+2 is taken to refer to a function, then the variable x doesn't occur in it, because x+2 would just be shorthand for the function f such that for all x, f(x) = x+2, and neither f nor x are free in that specification thank to the rules for quantifiers. Or, we might mean by x+2 a number, rather than a function. In that case, it refers to exactly one number, and x also refers to exactly one number (though we don't know which absent further information). Thomas === Subject: Re: Math dependency logic permission for an emailed response. > > That's stupid. Given that 7 is a NUMBER it can't just change, and > given that dividing P(x) by 49 *does* change it to 1, then 1 can't > change as a variable dependent on x. > > But x is a number too. And it can't change. > Well what is it then? I told you, it's a number. > Obviously x is a symbol that can stand in for LOTS of different > numbers, which is what makes algebra so powerful. x is a symbol that stands for whatever you like. You can't tell without more information; it might be an arbitrary set, a number, an ultrafilter, a topology, or whatnot. x might stand for a number in some context. In the context here, x stands for a number. Which means that x is a number. > On the other hand, 2 is just a number. 2 stands for a number: the number 2. Unlike x, 2 always stands for the same thing. We don't have to have that convention; it's purely a matter of convenience. We might use the symbol 2 for something completely different. Thomas === Subject: One algebra question Hi I am learning algebra by myself I am trying to do some problems but I can't solve this one could you please do me a favor to give me some help here is the question: Let f be a homomorphism defined on a finite group G and let H subset G a) show that |f(G):f(H)| divides |G:H| b) show that |f(H)| divides |H| === Subject: Re: One algebra question > Hi I am learning algebra by myself > I am trying to do some problems > but I can't solve this one > could you please do me a favor to give me some help > here is the question: > Let f be a homomorphism defined on a finite group G and let > H subset G > a) show that |f(G):f(H)| divides |G:H| > b) show that |f(H)| divides |H| The claims are false unless H is actually a subgroup. If that is the case, let N be the kernel of f. Then |f(G)| = |G:N| and |f(H)| = |H:Ncap H| so that |f(G):f(H)| = |G:N|/|H:Ncap H| = |G:H|/|N:Ncap H| which the makes a) obvious and b) already is. === Subject: Re: One algebra question > Let f be a homomorphism defined on a finite group G and let > H subset G Do you mean that H is a subgroup of G? > a) show that |f(G):f(H)| divides |G:H| > b) show that |f(H)| divides |H| Let K be the kernel of f. Note that |f(H)| = |H|/|H intersect K| .... -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: new way of describing ellipse for math > If you are looking for a unique elipse defined only by an inscribed > rectangle, you are out of luck. Four points arranged as the corners of > a rectangle define a whole family of elipses including one circle. > You have not read yesterday's suggestion of 2 inscribed rectangles. > I am not sure if 2 inscribed rectangles defines a unique ellipse or if it > even creates an ellipse. > Anyway, if it does then 9 points uniquely define a ellipse where 8 are the > corners of 2 rectangles and 1 point is what I call the *center of the ellipse*. > And in keeping with this method then 5 points uniquely define a circle > where 4 corners and the center of the circle which coincides with the center > of the square inscribed. Arch, It only takes two points to define a circle if one is the center, three if they lie on the circle. > An elipse is not a 'squashed` circle, it is a rotated one. > (Conic Sections - remember?) > In some meaningful sense a rectangle is a squashed square. Likewise, > a ellipse is a squashed circle and now I have to engineer and create the > mathematics to conform to that idea. > This is important for physics in that the Unified Force is one force > where all the other forces are broken symmetry (the squashing). In three > dimensions the circle becomes a sphere and the ellipse becomes ellipsoids > or lobes. Thus the hydrogen atom is the nearest to the unified force and > the uranium atom far away in broken symmetry. > Whether you like it or not, I believe that the relationship you are > looking for > does involve the rotation posted earlier. > I have not thrown out the rotation. I simply am saying that I can get at it > in a simplier method than rotation. An equivalent but much more simple > method. > Consider the circle with its inscribed square. (You didn't say you > had trouble with that.) If you rotate this figure about one side of > the square, and project it, you get a rectangle, (the square with one > set of opposite sides forshortened), and a unique circumscribed > elipse, (the circle rotated). > This is Projective Geometry. I think a simplier method, but equivalent > will be inscribed and circumscribed. I studied Projective Geometry for > one year in college and know that points can be equivalent to rotations. Didn't do too well eh? > Hint: if you wanted to graph this elipse, you could multiply one set > of ordinates from the graph of the circle with its center at the > projected intersection of the diagonals of the square by the cosine of > the angle of rotation. > This is High School stuff. If you have trouble with these concepts, > you are wasting your time in attempting to understand, much less > modify, string theory. > your mind is too restless with the creation of new ideas or the assembling of > a new method of looking at things and as a result, your mind bulks and then > tosses out these ad hominems. If you cannot stand new ideas or new approaches > then just do not ever respond to any of my posts and stick to your teaching > in schools, for you do not belong in any of my adventures which invariably seek > new ideas, new methods. > I have made this detour of ellipse,sphere,spheroid, ellipsoid because the original > intent of String theory in the 1960s was that the Euler Gamma Function seemed > to clarify some rules of the StrongNuclear Force, or the nucleus of atoms. And since nuclei are > symmetrical in spherical or ellipsoid directly implies that the mathematics of the geometry of nuclei > involve distorting spheres into ellipsoids. > Why the s orbital makes precisely p orbitals or d orbitals or f orbitals. So if we link up Schrodinger > Equation with Euler Gamma Function that broken symmetry > becomes squashing. > If the StrongNuclear force is really NuclearElectrons and that the StrongNuclear > force is paired to the WeakNuclear Force as a Coulomb Unification then this > ellipse defined by 2 rectangles where one rectangle is the StrongNuclear force and the other rectangle > is the WeakNuclear Force. > When all is said and done, I expect to link and tie Schrodinger Equation with > Euler Gamma Function with circle/sphere/ellipse/ellipsoid for the StrongNuclear > paired to the WeakNuclear Force. > Coulomb Unification of Forces in Physics: > Coulomb force, perfect and spheriod, region of existence is the > nucleus protons to the electrons > / > / > StrongNuclear to WeakNuclear, broken coulomb symmetry, and > region of existence is the nucleus of atoms > / > / > Gravity to Antigravity, again a coulomb force when combined and > region of existence is the electronic-space of atoms > Trouble with Stringtheory is that they never really applied it to the nuclear > region of atoms and to fully garner the understanding of the strongnuclear force, > and instead they tried to overblow stringtheory as some theory of everything. > Archimedes Plutonium, a_plutonium@hotmail.com > whole entire Universe is just one big atom where dots > of the electron-dot-cloud are galaxies === Subject: Re: new way of describing ellipse for math > (snipped) > > If you use two rectangles, I believe you can generate a both a unique > ellipse or circle by connecting their verticies, so long as the > rectangles do not have any identical verticies. I suppose if you were > to take a limit case, you might be able to generate some equation that > would give the result you're looking for. > > Eg: > |-/----------------/-| > | / / | > |-/----------------/-| > > Connect the eight outer verticies, and you get an ellipse. As the > verticies converge on each other, you maintain the ellipse. If you > took the limit as V1 -> V1', V2 -> V2' etc, you might be able to > generate a single equation for the ellipse in terms of the verticies > of the rectangle(s). > > Also, I wonder Aaron, if your suggestion of 2 rectangles to make a > unique > ellipse. I wonder why the 4 corners of a solo rectangle is unable to > make > it unique. I do not see the mechanism for why 1 rectangle was not > sufficient. > Because the differences between the first and second rectangle provide the slopes at the points of contact. To put it another way - the relationship between the aspect ratios of the two rectangles determines the aspect ratio of the elipse. > Perhaps the answer is that with 2 rectangles the one rectangle > provides the > major axis and the other is the minor axis. So I really do not get > away from > axis. I suppose there is a Projective Geometry theorem then that would > say something to the effect that 8 points of the 8 corners of the 2 > rectangles > is the equivalency of a major axis and minor axis. > > Question: since a unique circle is formed from a rectangle that > circumscribes > the rectangle. Then a unique ellipse formed from 2 rectangles has 2 > unique > circles. What relationship do those 2 unique circles have with the > ellipse??? > > A.P. > My guess would be that the diameters of the circles correspond to the > major/minor axis. I think there will always be a major/minor axis in > some form for your problem, since they are part of what defines the > ellipse. You got it. Did Arch? === Subject: Re: new way of describing ellipse for math it takes five points to determine a conic section, not four ... I'm not sure if that's correct, but an infinity of conics will pass through the vertices of a ractangle, as they're not in general position. so, the use of two rectangles on the same center (?) is overdetermined, in a way, but it seems to work. obviously, any 3 of the corners of a rectangle determine the circle. > |-/----------------/-| > | / / | > |-/----------------/-| > ellipse. I wonder why the 4 corners of a solo rectangle is unable to > make > it unique. I do not see the mechanism for why 1 rectangle was not --ils duces d'Enron! http://tarpley.net/bush8.htm http://www.wlym.com/PDF-SpReps/SPRP13.pdf === Subject: Citation for a theorem in topology? I seem to have shown the following. It seems too fundamental not to be known. Can anyone supply a citation? Theorem: Suppose X is a closed subspace of the normal space Y and X is a subspace of the completely regular space Z. Then the amalgamated sum Y +_X Z is completely regular. Moreover both Y and Z embed as subspaces of it. === Subject: Re: A 'basic' topology question about interiors <3FA725C2.6040506@rutcor.rutgers.edu> NNTP-Posting-User: bbscott >>Of interest perhaps are these formulas: >> >>When U subset A subspace S >>cl_A (U) = A / cl_S (U) >>int_A (U) = A / int_S (SA / U) >> >> >> >Ben Scott > And thank *you* for using my preferred plural of formula! > -- > Stephen J. Herschkorn herschko@rutcor.rutgers.edu Of course. But y'know, 'you say formulas, I say formulae...' Ben Scott === Subject: Re: A 'basic' topology question about interiors > Of course. But y'know, 'you say formulas, I say formulae...' formulae formulae formulas formularum formulis formulis ?? Woohoo! > Ben Scott === Subject: The Mystery Of Mass (was: The need of geodesics in physics) >I don't know where you got that notion from. QT says nothing about m >changing discretely. Mass is quantized on a discrete spectrum in QFT, but the origin of the mass matrix and its spectrum is unknown. What *is* known only furthers the mystery: the mass operator doesn't commute with the SU(2) symmetry generators corresponding to the W and anti-W. The mystery only deepens, considering that [anti-]W can only interact with and right helicity states is relative. But interacting vs. not interacting can't be relative too! Therefore, the helicity states Which means, in turn, that their mass is not intrinsic, but dynamic in origin, arising from an interaction with an otherwise unknown or unseen/undiscovered field. The mass matrix, then, just describes Were it not for this extra kink, all the charge states of all the completely indistinguishable from one another. The mass matrix mixes between these groups, and the mass eigenstates for each charge state are actually mixtures across the 3 groups with 3 separate eigenvalues. The mystery deepens yet further when you step back and loop at the charge states. They form a pattern: they reside at the points of a 6 dimensional Cartesian lattice forming within it a 5-dimensional hypercube! Indeed, you can write down an orthonormal basis (X,A,B,C,D,E), and the 32 vertices are precisely those for the 32 combinations of vectors xX + (x-a)A + (x-b)B + (x-c)C + (x-d)D + (x-e)E for which a,b,c,d,e = +/- 1/2 with x undetermined (possibly with one set for each of the 3 groups). And the mystery deepens yet further when you consider what the W and anti-W actually do; and what the 6 (charged) gluons in SU(3) do. The W switches a (+a,-b) to a (-a,+b); anti-W the other way. The gluons switch a (+c,-d) to a (-c,+d); and the other 5 combinations involving c, d and e. Left and right are distinguished solely by one of the a or b, say, by +b vs. -b, with left being (+,-) and (-,+) and right being (+,+) or (-,-); so (+,-) & (+,+) are the left & right another. === Subject: Re: How did Euler determine Euler's Constant? > I know that : > Euler's constant = Lim (n-> infinity) [ Sum(i/j) from j=1 to n - ln(n)] > but, I can't see how one actually evaluates this relation to get the value > of Euler's constant. > MB Details are given in H. Goldstine, A History of Numerical Analysis from the 16th through the 19th Century, Springer, 1977, page 129. He used the Euler-Maclaurin formula (he had just derived it using a generating function approach) and found the value to 16 places. The 16th one was wrong. Not bad for a blind man. === Subject: Re: How did Euler determine Euler's Constant? > I know that : > Euler's constant = Lim (n-> infinity) [ Sum(i/j) from j=1 to n - ln(n)] > but, I can't see how one actually evaluates this relation to get the value > of Euler's constant. Luis Rodriguez answer: Calling g the Euler's constant. First he deducted the formula g = Sum [j=1 to n] 1/j - Ln(n)-1/2n + 1/2n^2 - 1/(120n^4),utilizing the Euler-Maclaurin algorithm Then he calculates the sum of inverses of first 100 integers with pencil and paper. So he obtained g with 13 decimal digits,replacing n = 100 . === Subject: Re: JSH: My victories get lost Yes it is true that every one of your victories has gotten lost. It is also true that every one of your victories has not gotten lost. It is also true that every one of your victories is the pope. It is also true that every one of your victories has any predicate you want including four wheel drive, but not as it happens, existence. - William Hughes (with a bit of help from Tom Stoppard) === Subject: Re: JSH: My victories get lost >There are quite a few posters who reply in my threads. Now I've >noticed that I trounce one poster and then another pops up and starts >yapping. Later some poster who got his ass kicked is back trotting >out the *same* crap. > That's your impression. In _fact_ you have never trounced > anyone here. David Ullrich is a ing piece of dog. I think it's funny that I can call a professor at Oklahoma State University a ing piece of dog knowing that he'll keep replying in my threads. You see, he has to keep replying pushing the same old lies. He's stuck. He's trapped in something that he can't get out of, so it doesn't matter what I call him, or what I say about him, he has to come back. You see I'm the person who has the correct math argument, so posters like David Ullrich or Arturo Magidin are *compelled* to reply out of fear that if they go away, then I'll get some people who'll pay attention to the truth. So David Ullrich, the math professor at Oklahoma State University, is demeaned by me as the piece of ing dog he is, and he *has* to keep coming back. James Harris === Subject: Re: JSH: My victories get lost >>There are quite a few posters who reply in my threads. Now I've >>noticed that I trounce one poster and then another pops up and starts >>yapping. Later some poster who got his ass kicked is back trotting >>out the *same* crap. >> That's your impression. In _fact_ you have never trounced >> anyone here. >David Ullrich is a ing piece of dog. You know, the _next_ time you complain to my employer about me being mean to you it's going to be even funnier than the last time... >I think it's funny that I can call a professor at Oklahoma State >University a ing piece of dog knowing that he'll keep replying >in my threads. >You see, he has to keep replying pushing the same old lies. >He's stuck. He's trapped in something that he can't get out of, so it >doesn't matter what I call him, or what I say about him, he has to >come back. >You see I'm the person who has the correct math argument, so posters >like David Ullrich or Arturo Magidin are *compelled* to reply out of >fear that if they go away, then I'll get some people who'll pay >attention to the truth. >So David Ullrich, the math professor at Oklahoma State University, is >demeaned by me as the piece of ing dog he is, and he *has* to >keep coming back. Yeah, you got all that right. Doesn't change the fact that you've never trounced anyone here, except in your imagination. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: My victories get lost > David Ullrich is a ing piece of dog. > I think it's funny that I can call a professor at Oklahoma State > University a ing piece of dog knowing that he'll keep replying > in my threads. Yes, it is funny that a ing piece of dog can call a professor at Oklahoma State University a ing piece of dog. I think it is more than five years since I posted that there is a difference of at least 30 in our IQs, and I know which direction, and you know it as well. === Subject: Re: JSH: My victories get lost >There are quite a few posters who reply in my threads. Now I've >noticed that I trounce one poster and then another pops up and starts >yapping. Later some poster who got his ass kicked is back trotting >out the *same* crap. > That's your impression. In _fact_ you have never trounced > anyone here. > David Ullrich is a ing piece of dog. > I think it's funny that I can call a professor at Oklahoma State > University a ing piece of dog knowing that he'll keep replying > in my threads. Mr. Harris, is there no level to which you will stoop to propagate your fierce lies? You are pathetic and you can continue repeating your lies and conspiracy theories, but the truth has and will always be against you. Go see a doctor! === Subject: Re: Determinant of a special matrix > Adj(Adj(A))=det(A)^(n-2) A >Or references if it is already proven by someone (which will > probably be the case, though I cannot find it). Google gives three hits on Adj(Adj(A))=det(A)^(n-2) A. Judging from your email address ( I assume nl stands for Netherlands?) you should be able to read this one easily :-) [PDF] D:DocsOnderwijsMatrixtheorieTent 02-05tent 02-05.dvi File Format: PDF/Adobe Acrobat - View as HTML ... c) Als n 3 en ?(A ) < n, dan geldt adj(adj(A )) = 0 (d) Als n = 2, dan geldt adj(adj(A )) = A (e) Voor n 3 geldt adj(adj(A )) = (det A ) n?2 A. Normering 1 ... www.win.tue.nl/~wscomalo/math-onderwijs-2F400/ tent-02-05.pdf - Similar pages === Subject: Re: Key Factorization Proof [...] Why did you post this to sci.math.num-analysis with not even a cross-post to sci.math? V. -- email: lastname at cs utk edu homepage: cs utk edu tilde lastname === Subject: Re: Conversion of red phosphorus to white phosphorus Nice, but quasi green bob suffers from Plumbum paranoia of legendary dimensions which forced him to hallucinate that his NG comments will enhance mankind's survival against the doomsday prospects of lead in message which he delivered under great inner torment and psychotic fear, because he says... > I wonder if you are a danger to others at times. His comment here is embedded in a master piece of a post, that teeters on the intellectual niveau of a petrock, the belletristic values of an ameba and his understanding of the field with the depth comparable to that of a flatlander's. His efforts nevertheless elicit a faint chuckle, at best, but, me being magnanimous will bless him with a roaring AHAhahahhaha.......ahahahahanson === Subject: Re: Conversion of red phosphorus to white phosphorus hanson spewed: > Nice, but quasi green bob suffers from > Plumbum paranoia of legendary dimensions which forced him > to hallucinate that his NG comments will enhance mankind's > survival I'm betting on marsupials. > against the doomsday prospects of lead in message > which he delivered under great inner torment and psychotic fear, > because he says... > I wonder if you are a danger to others at times. Well, are you? Do you take antipsychotics? If not, has a doctor ever suggested that you take them? > His comment here is embedded in a master piece of a post, > that teeters on the intellectual niveau of a petrock, the belletristic > values of an ameba and his understanding of the field with > the depth comparable to that of a flatlander's. That didn't flow very well Hanson. You get a C-. > His efforts > nevertheless elicit a faint chuckle, at best, but, me being > magnanimous will bless him with a roaring > AHAhahahhaha.......ahahahahanson A twitter when you say it aloud. === Subject: Re: Repeat: White Noise Dilemma >>I find it odd (2n+1) that no one has responded to this. I'm guessing >>that either a) everyone thinks it is a homework problem (it is not), >>or b) no one knows the answer. Come on, all you big strong mathematicians - >>certainly you can answer a little ol' question like this, can't you? ... >>Here is a question that has had me in a quandary for several >>years now. >>Let X(t) be a zero-mean IID process with variance of >>sigma^2. Now we know that since this function is >>IID, it has a white PSD and therefore its autocorrelation >>function at lag 0, R_XX(0), should be b*delta(tau), where >>delta(tau) is the usual Dirac delta function and b is some >>constant. > What are you using as a definition of PSD (power spectrum density)? The Wiener-Khinchine theorem states that the PSD of a wide-sense stationary random process, Sxx(w), is the Fourier transform of its autocorrelation function Rxx(tau) = E[X(t)X(t+tau)]. > This would seem to be a problem for a process such as X(t) in which > t--> X(t) is not measurable, or continuous in L^2. What do you mean by t-->X(t)? -- % Randy Yates % ...the answer lies within your soul %% Fuquay-Varina, NC % 'cause no one knows which side %%% 919-577-9882 % the coin will fall. %%%% % 'Big Wheels', *Out of the Blue*, ELO http://home.earthlink.net/~yatescr === Subject: Re: Repeat: White Noise Dilemma > > >>I find it odd (2n+1) that no one has responded to this. I'm guessing >>that either a) everyone thinks it is a homework problem (it is not), >>or b) no one knows the answer. Come on, all you big strong mathematicians - >>certainly you can answer a little ol' question like this, can't you? ... >> >>Here is a question that has had me in a quandary for several >>years now. >> >>Let X(t) be a zero-mean IID process with variance of >>sigma^2. Now we know that since this function is >>IID, it has a white PSD and therefore its autocorrelation >>function at lag 0, R_XX(0), should be b*delta(tau), where >>delta(tau) is the usual Dirac delta function and b is some >>constant. > > > What are you using as a definition of PSD (power spectrum density)? > The Wiener-Khinchine theorem states that the PSD of a wide-sense stationary > random process, Sxx(w), is the Fourier transform of its autocorrelation > function > Rxx(tau) = E[X(t)X(t+tau)]. But what is the *definition* of the term PSD? > This would seem to be a problem for a process such as X(t) in which > t--> X(t) is not measurable, or continuous in L^2. > What do you mean by t-->X(t)? The mapping of R into L^2 that sends time t to r.v. X(t). -- A. === Subject: Re: Repeat: White Noise Dilemma > Incidentally, I had a thought on the applied side. In physics, for > example, if we apply a strong force over a short period time, like > when two billiard balls collide, we can describe that by a function > f(t) which equals the strength of the force at time t. (Or something > like that. Sorry, I'm no physicist.) I think it's called an > impulse(?). Anyway, it's the integral of f that determines the overall > effect of the impulse. If we want to assume the entire impulse is > applied at a single instant in time, so that f is suuported at the > origin, then f *must* be a delta function, or there will be no effect > at all. If, for example, we wanted to apply the impulse f(t) which > is zero everywhere except f(0)=1, then we're saying that we apply some > finite amount of force over an arbitrarily small period of time, which > in the limit amounts to applying no force at all. Maybe your process > is analogous to this. Maybe a noise which satisfies the properties > you prescribe would amount to no noise at all. I think this is almost right. I have some heuristics and some mathematics to support this. Those processes X_r(t)=[B(t+r)-B(t)]/r are converging to white noise. Formally, white noise is all that abstract business about distributions and what not. But what's really happening as r-->0? Well, X_r becomes wilder and wilder. It's oscillations become more frequent, with roughly the same number of ups and downs. Normally, this would be bad because it would cause things to cancel out. What I mean is that, given some function f(t), if we look at int{X_r(t)f(t)dt}, then the fact that there are as many ups as downs of X_r(t) and the fact that these ups and downs are so close together would cause this integral to tend to zero. In particular, all the Fourier coefficients of X_r(t) would go to zero. What prevents this from happening in the case of white noise is that the magnitudes of the ups and downs grow in their variance as r-->0. So some are very large, some are very small, and they are unable to cancel each other out, leaving something random and nontrivial in the limit. In your case, you would want to look at the processes Y_r(t)=[B(t+r)-B(t)]/sqrt{r}. The difference here is that you've bounded the variance. So as r-->0, the ups and downs will have roughly the same magnitude and all the Fourier coefficients will go to zero. So it's not that your process results in no noise at all; it results in a noise that too regular. It's like looking at sin(nt) as n-->infty. It's not converging to zero, but if you look at int{sin(nt)f(t)dt} for any reasonable function f, this will tend to zero. Here's some sketchy computations. See the Oksendal SPDE reference for more details on the notation and concepts. Let W(x)=int{f(t-x)dB(t)}, where f is a Schwartz function. (If f is sufficiently close to a delta function, then W(x) is close to white noise.) Let g(x) be some reasonable function, say of compact support. Then int{W(x)g(x)dx} = int{int{f(t-x)dB(t)}g(x)dx} = int{-int{B(t)f'(t-x)dt}g(x)dx} = -int{B(t)int{f'(t-x)g(x)dx}dt}. Let F(t)=int{f(t-x)g(x)dx}=int{f(x)g(t-x)dx}, so that int{W(x)g(x)dx} = -int{B(t)F'(t)dt} = int{F(t)dB(t)}. Hence, the variance of int{W(x)g(x)dx} is int{[F(t)]^2 dt}. For the process you're trying to construct, we would want to take a sequence of positive f's that converge to zero everywhere except the origin, where they converge to 1 (still assuming sigma=1). Then, unlike a sequence which coverges to the delta function, int{f(x)dx} would go to zero. Now, note that |F(t)| <= (sup|g(x)|)*int{f(x)dx}. So we see that int{W(x)g(x)dx} would actually converge to zero as the f's converged to the desired function and we have a noise that behaves the way the heuristics expected it would. === Subject: Re: elliptic integral of first kind > let n(r') = constant =1. and lets evaluate this in the plane i.e. z=0 > V(r,0) = 1/2 Integrate(0,oo) J(rk)*J(r'k) dk Integrate(0,a) r'dr' > the integral over the bessel functions (i think) is > V(r,0) = 1/(pi*r) * Integrate(0,a) E((r'/r)^2) r' dr' > Up to here, you are correct (as far as I can tell). The integral > formula however is only correct if r' must be symmetric in r and r' it is the other way round if r Integrate(0,oo) J(rk) J(r'k) dk = 2/(pi r) K((r'/r)^2) if r' Integrate(0,oo) J(rk) J(r'k) dk = 2/(pi r') K((r/r')^2) if r'>r dr' but i meant Integrate(0,a) K((r'/r)^2) r' dr' I have to slow down and double check everything when i post on these groups...i seem to rush a bit. i appologise. i did actually use K not E in all my calculations from that point on and the rest would seem correct. i have gone over it all again and i do get what i originaly posted as... V(r,0) = 1/(pi*r)* [ r^2 *E((a/r)^2) + (a-r)*(a+r)K((a/r)^2)] electric field is = -1/r*dV/dr e(r) = [ K((a/r)^2) - E((a/r)^2) ] / pi The next thing is to find out what happens to E for the interior of the disc! I dont think its as simple as swapping r' and r. I *think* we have to break it up into two integrals over two domains: i.e. we break up the disc into two domains: 1) points r' that belong to [0,r) 2) points r' that belong to [r,a] if this is the case we have V(r,0) = Integrate(0,r) K((r/r')^2) dr' + 1 Integrate(r,a) K((r'/r)^2) r' dr' ok but the first integral does not converge over the domain of integration...so lets offset it by a small positive constant b to get V(r,0) =Integrate(0,r-b) K((r/r')^2) dr' + Integrate(r+b,a) K((r'/r)^2) r' dr' this avoids the singularity, while still maintaining the applicability of both forms under the integrals for the given domains. at the end we let b->0 Integrating this we have V(r,0) = r^2 * E( (b-r)^2 / r^2 ) + a*E((r/a)^2) - (b+r)*E( r^2/(b+r)^2) + b*(b-2*r)*K((b-r)^2/r^2) letting b->0 im not sure! i dont know what happens in the limit of z*K(z -1) for z->0 i am assuming we cannot discount it...and i am assuming that it probably blows up!. difficult. im not sure this is the right way to get at potential for the domain of the disc. cheers moth === Subject: Re: elliptic integral of first kind > Up to here, you are correct (as far as I can tell). The integral > formula however is only correct if r' must be symmetric in r and r' it is the other way round if r Integrate(0,oo) J(rk) J(r'k) dk = 2/(pi r) K((r'/r)^2) if r' Integrate(0,oo) J(rk) J(r'k) dk = 2/(pi r') K((r/r')^2) if r'>r > Sorry, for beeing so late with this remark, but I only just found > time for it. Also forget about my comment about E(-4 r/...), it > was just a stupid error on my part. Hello Dr. Ulm, assuming that i have botched the integration, and the above integrals are in fact correct....the electric field still comes out as looking like e_r ~ K(a/r) for r>a e_r ~ K(r/a) for ra heading to r=a make sense to me intuitively. but the above also suggests that there is a singularity at on the rim when approaching it from any direction. why would it be the case for r grava .88 la saucisse et au marteau: > f(p) = q > where f(140) = 15000 > and f'(140) = -100 > > R = pq > If we apply this practically. p=price, q=quantity sold and R=revenue. > So if we increase the price from 140 to 141, we decrease the quantity > sold from 15000 to 14900. This will increase the revenue by 900 and > not 1000. I guess the problem here is with a changing derivative > between a p=140 and p=141. This gave me a real problem and hence my > hacked solution. Do you have any other comments about this. Actually, you're applying a discrete derivative, which is not the same thing. For you, dR/dp = R(p+1) - R(p) where f(p+1) = f(p)+ 1*f'(p). In this case, your formula is right and we have dR/dp = (p+1)(f(p)+f'(p)) - pf(p) = f(p) + (p+1)f'(p) = 15000 + 141*(-100) = 900 But you have to keep in mind that the discrete derivative isn't the same thing as the true derivative. Furthermore, in your case, you're forced to use an approximation of f(p+1) using the first order at p (I don't know if this is the way to say it in English, but I think it is understandable). -- Nicolas === Subject: Re: Derivatives > Phil Holman grava .88 la saucisse et au marteau: > f(p) = q > where f(140) = 15000 > and f'(140) = -100 > > R = pq > If we apply this practically. p=price, q=quantity sold and R=revenue. > So if we increase the price from 140 to 141, we decrease the quantity > sold from 15000 to 14900. This will increase the revenue by 900 and > not 1000. I guess the problem here is with a changing derivative > between a p=140 and p=141. This gave me a real problem and hence my > hacked solution. Do you have any other comments about this. > Actually, you're applying a discrete derivative, which is not the same > thing. For you, dR/dp = R(p+1) - R(p) where f(p+1) = f(p)+ 1*f'(p). > In this case, your formula is right and we have dR/dp = (p+1)(f(p)+f'(p)) - > pf(p) = f(p) + (p+1)f'(p) = 15000 + 141*(-100) = 900 > But you have to keep in mind that the discrete derivative isn't the same > thing as the true derivative. Furthermore, in your case, you're forced > to use an approximation of f(p+1) using the first order at p (I don't > know if this is the way to say it in English, but I think it is > understandable). before. The charge dissipation of a capacitor is seen as a continuous function of time but if you are a physicist, an electrical charge is made up of electrons with a discrete charge. It may be ok to assume a continuous discharge function when the number of electrons is large but when we get down to only a few, it is not the case. Your English is excellent by the way. Phil Holman === Subject: Re: Unit interval homemorphic to Circle help >If I = [0,1], and I / ~ is the quotient space of I obtained by identifying 0 >and 1, then the circle S^1 is homeomorphic to I / ~. >This is intuitively clear, but how can I prove this? >I think I want to show that p : I ---> S^1 given by p(x) = e^(2ix*pi) is a >quotient map. > That's exactly what you must show. > In this case it is quite easy (fortunately). > The map p is already continuous, as I and S^1 are compact metric the map > p is > automatically closed (i.e., if A is closed then so is p[A]). Why does a continuous map from a compact space to a compact space necessarily have to be a closed map? Is this what you are implying? > This implies that p is a quotient map. > KP > -- > E-MAIL: K.P.Hart@EWI.TUDelft.NL PAPER: Faculty EWI > PHONE: +31-15-2784572 TU Delft > FAX: +31-15-2786178 Postbus 5031 > URL: http://aw.twi.tudelft.nl/~hart 2600 GA Delft > the Netherlands === Subject: Re: Unit interval homemorphic to Circle help >If I = [0,1], and I / ~ is the quotient space of I obtained by > identifying 0 >and 1, then the circle S^1 is homeomorphic to I / ~. >This is intuitively clear, but how can I prove this? > >I think I want to show that p : I ---> S^1 given by p(x) = e^(2ix*pi) is > a >quotient map. > > > That's exactly what you must show. > In this case it is quite easy (fortunately). > The map p is already continuous, as I and S^1 are compact metric the map > p is > automatically closed (i.e., if A is closed then so is p[A]). > Why does a continuous map from a compact space to a compact space > necessarily > have to be a closed map? Is this what you are implying? If X is a compact space and Y is a Hausdorff space, then any continuous map f: X --> Y is closed. This follows from: (1) If X is compact and A subset X is closed, then A is compact. (2) If Y is Hausdorff and C subset Y is compact, then C is closed since continuous maps preserve compactness. > This implies that p is a quotient map. > KP > -- > E-MAIL: K.P.Hart@EWI.TUDelft.NL PAPER: Faculty EWI > PHONE: +31-15-2784572 TU Delft > FAX: +31-15-2786178 Postbus 5031 > URL: http://aw.twi.tudelft.nl/~hart 2600 GA Delft > the Netherlands === Subject: Re: Unit interval homemorphic to Circle help >>The map p is already continuous, as I and S^1 are compact metric the map >>p is >>automatically closed (i.e., if A is closed then so is p[A]). >> >Why does a continuous map from a compact space to a compact space >necessarily >have to be a closed map? Is this what you are implying? Yes, as A is closed it is compact, as p is continuous p[A] is compact, as p[A] is compact it is closed in the metric space S^1. KP -- E-MAIL: K.P.Hart@EWI.TUDelft.NL PAPER: Faculty EWI PHONE: +31-15-2784572 TU Delft FAX: +31-15-2786178 Postbus 5031 URL: http://aw.twi.tudelft.nl/~hart 2600 GA Delft the Netherlands === Subject: Re: Unit interval homemorphic to Circle help >>The map p is already continuous, as I and S^1 are compact metric the map >>p is >>automatically closed (i.e., if A is closed then so is p[A]). >> >> >Why does a continuous map from a compact space to a compact space >necessarily >have to be a closed map? Is this what you are implying? > Yes, as A is closed it is compact, as p is continuous p[A] is compact, > as p[A] is compact it is closed > in the metric space S^1. > KP > -- > E-MAIL: K.P.Hart@EWI.TUDelft.NL PAPER: Faculty EWI > PHONE: +31-15-2784572 TU Delft > FAX: +31-15-2786178 Postbus 5031 > URL: http://aw.twi.tudelft.nl/~hart 2600 GA Delft > the Netherlands === Subject: Ex(~x=x), counterpart theory, and contingent identity In another post I suggested that John Correy's ideas about non-reflexive identity might have a rational formulation with respect to something called counterpart theory. (I think categoreal would be the right term here--in my humble opinion, the apparent self-contradictions associated with John's intuitions arise from vagueness and ambiguity associated with standard presuppositions rather than irrational error on his part. For the record, Langholm's investigation of determinability and indeterminability in first-order contexts includes incoherent formulas. Exclusion negation is not informationally well-behaved.) The link, http://www.sussex.ac.uk//Users/muralir/kct_final.pdf <>((x=x) & ~(x=x)) is discussed as well as what is actually done in counterpart theory to exclude such an incoherent result. :-) mitch === Subject: how to identify data from different distributions? Hi sorry to bother you guys. This problem suddently comes into my mind. Given two data sets from two different distributions, if I mixed them together, i.e., you don't know which point comes from which distribution. Is it possible to use some learning or statistical algorithms to separate the two data set out? For example, X=(x1,x2,...,xn) ~ Gaussia Y=(y1,y2,...,yn) ~ Uniform Given (x1, y1, x2,x3,y2,x4,y3,...), can you separate out X and Y? === Subject: Re: how to identify data from different distributions? >Hi sorry to bother you guys. >This problem suddently comes into my mind. >Given two data sets from two different distributions, >if I mixed them together, i.e., you don't know which point comes from >which >distribution. Is it possible to use some learning or statistical >algorithms to separate the two data set out? >For example, >X=(x1,x2,...,xn) ~ Gaussia >Y=(y1,y2,...,yn) ~ Uniform >Given (x1, y1, x2,x3,y2,x4,y3,...), can you separate out X and Y? It can be partially used to separate them. The most that one can get by learning or statistical algorithms is the number from each group, and the precise distributions. If an individual has a prior probability h of coming from the first group, the posterior probability is given by Bayes' Theorem. More than this cannot be done for one individual. If it is known exactly how many are from each group, one can get the posterior probability that a particular partition is the correct partition. -- 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: how to identify data from different distributions? roy grava .88 la saucisse et au marteau: > For example, > X=(x1,x2,...,xn) ~ Gaussia > Y=(y1,y2,...,yn) ~ Uniform > Given (x1, y1, x2,x3,y2,x4,y3,...), can you separate out X and Y? Try looking for Kernel PCA on Google. Basically, you learn the eigenfunctions of the covariance matrix with respect to a kernel K. Every eigenfunction will correspond to a distribution (or am I mistaken?) -- Nicolas === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) > So in ANY ring that contains the integers, 7 and 22 are coprime > (under either definition). >> That's true in any ring R since R contains a homomorphic image of Z, >> Z -> Z*1_R, i.e. any ring is a Z-algebra. But any ring homomorphism >> must preserve the relation 22 - 3(7) = 1. > Hmmm... Only if you assume that ring morphisms map 1 to 1, which > is not necessarily a given either. Even assuming rings have a 1, > the zero map is usually considered a valid homomorphism, > and your conclusion would be incorrect there. For Rings (with 1, as I assume above) ring morphisms must preserve 1, so the zero map is not a morphism of Rings with 1. If, as you claim, one considered the zero map as a valid homomorphism of Rings with 1 then basic theorems on rings would fail, e.g. the image of the zero morphism would fail to be a subring (except if the target ring is 0). Therefore my above quoted statement is in fact correct as written. >> By the way, as I mentioned in a prior post [1], coprime is >> a highly overloaded term whose meaning depends upon context. >> JSH is using one of the most common definitions and it is >> incorrect to criticize him for that. > I will quibble that what is most common depends on context as well. > Most ring theorists I know would object to using the definition > depending on common divisors, since to them 'prime' refers to ideals, > almost never to elements; and most number theorists would certainly > disagree that the definition via common divisors is 'the most common' > (for the latter the definition ->is<- invariably related to ideals, > never to elements). This issue is not what _should_ be the proper definition but, rather, what _is_ past and current usage. The fact of the matter is that due to the way this and related ideas evolved, the denotation of coprime was - and still is - highly overloaded. Thus it is best to explicitly specify its meaning in any context where possible ambiguity exists. > To me, and particularly given that JSH's work is taking place in > subrings of the ring of all algebraic integers, it seems that the > most common usage from algebraic number theory should prevail. But often such subrings are of infinite degree over Q and are not Dedekind domains. Therefore they lie outside the scope of classical algebraic number theory. Hence common usage may not even apply. -Bill Dubuque === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) Adjunct Assistant Professor at the University of Montana. >> >> So in ANY ring that contains the integers, 7 and 22 are coprime >> (under either definition). > That's true in any ring R since R contains a homomorphic image of Z, > Z -> Z*1_R, i.e. any ring is a Z-algebra. But any ring homomorphism > must preserve the relation 22 - 3(7) = 1. >> Hmmm... Only if you assume that ring morphisms map 1 to 1, which >> is not necessarily a given either. Even assuming rings have a 1, >> the zero map is usually considered a valid homomorphism, >> and your conclusion would be incorrect there. >For Rings (with 1, as I assume above) ring morphisms must preserve 1, >so the zero map is not a morphism of Rings with 1. Granted; like I said, if you assume that ring morphisms map 1 to 1. On the other hand, you lose some things by assuming that: you don't get isomorphic copies of the rings in direct products. Some conventions are better than others, depending on the situation. > If, as you claim, >one considered the zero map as a valid homomorphism of Rings with 1 >then basic theorems on rings would fail, e.g. the image of the zero >morphism would fail to be a subring (except if the target ring is 0). >Therefore my above quoted statement is in fact correct as written. If by Ring you meant with 1 and by homomorphism you meant preserve 1, then ->of course<- they were correct. But you must grant that not ->everyone<- means both things, and that there was at least some sense in my stating so explicitly... [.snip.] 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: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) > ... > > > > My point is that you have to focus on the *factorization* and its > > > > validity in particular rings. > > > > > > > > Some factorizations will be valid in one ring, but not another. > > > > > > Yes, and your factorisation > > > P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22) > > > is in general not valid in the ring of algebraic integers. So what > > > are you trying to show? > > > > > > That's the point. I *prove* that if you have coprimeness between 7 > > > and 22 in the ring in which the factorization is valid, where 7 is NOT > > > a unit (and neither is 22), then the constant terms of the factors > > > that result from dividing P(x) by 49 *MUST* be coprime to 7. > > > > If you are talking here about the factorisation > > P(x)/49 = (5 a1(x)/7 + 1)(5 a2(x)/7 + 1)(5 b3(x) + 22) > > being valid in that ring (for all x), that is vacuously true. No > > proof needed, 1 is coprime to 7 and 22, and 22 is coprime to 7 in > > any ring that contains 1, 7 and 22. Even if in that ring 7 and/or 22 > > are units. So we need no proof in that case. > > > > Oh yeah, so ignore my statement that 7 is NOT a unit Dik Winter as > > you're showing your lack of reasonableness. > > You are ignoring my statement that in *every* ring that contains 1, 7 and 22, > they are coprime. Whether 7 is a unit or not. > Are you just totally stupid? That's irrelevant Dik Winter as the > point is that 7 is NOT a factor of 22, which my saying that it's not a > unit points out. Why do you not read what I write, rather than what you think I write? > I'm getting sick of stupid games from stupid people who apparently > have nothing better to do with *their* time. > Now are you or are you not intelligent enough to understand what it > means for 7 NOT to be a factor of 22? Yup. That means that you think your factorisation of P(x)/49 is valid in some ring where 7 is not a unit. I am not convinced. I *know* it is not valid in the algebraic integers, because there is ample proof that for varying x the factors of 49 distribute differently amongst the three factors of the polynomial when you wish to stay in the algebraic integers. Furthermore, I *know* that in the algebraic integers (5 b3(x) + 22) is *not* coprime (either the standard definition or your definition) to 7 for most values of x. So you should construct your ring such that both (5 a1(x) + 7) and (5 a2(x) + 7) are divisible by 7 and that (5 b3(x) + 22) is coprime to 7 *for all x*, and that 7 is not a unit. I am not satisfied that such a ring does exist. Moreover, you wish to divide all three factors by a constant. Why do you assume it must be a constant? Given w1(x), w2(x) and w3(x) such that their product is 49, and w1(0) = 7, w2(0) = 7 and w3(0) = 1. Wouldn't a distribution like: P(x)/49 = (5 a1(x)/w1(x)+7/w1(x))(5 a2(x)/w2(x)+7/w2(x))(5 b3(x)/w3(x)+22/w3(x)) be feasable? If I calculate correctly, the constant terms in this factorisation are now 1, 1 and 22, just as you wish. So why can this not satisfy your requirements? Assuming the w's are constant is just plain stupid if you wish to stay in a particular ring. The functions w depend on divisibility considerations within the ring, and so are not continuous in any sense. -- 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: Factorization P(x) = 2(x(x+1)/2 + 1) > ... > > > > > My point is that you have to focus on the *factorization* and its > > > > > validity in particular rings. > > > > > > > > > > Some factorizations will be valid in one ring, but not another. > > > > > > > > Yes, and your factorisation > > > > P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22) > > > > is in general not valid in the ring of algebraic integers. So what > > > > are you trying to show? > > > > > > > > That's the point. I *prove* that if you have coprimeness between 7 > > > > and 22 in the ring in which the factorization is valid, where 7 is NOT > > > > a unit (and neither is 22), then the constant terms of the factors > > > > that result from dividing P(x) by 49 *MUST* be coprime to 7. > > > > > > If you are talking here about the factorisation > > > P(x)/49 = (5 a1(x)/7 + 1)(5 a2(x)/7 + 1)(5 b3(x) + 22) > > > being valid in that ring (for all x), that is vacuously true. No > > > proof needed, 1 is coprime to 7 and 22, and 22 is coprime to 7 in > > > any ring that contains 1, 7 and 22. Even if in that ring 7 and/or 22 > > > are units. So we need no proof in that case. > > > > > > Oh yeah, so ignore my statement that 7 is NOT a unit Dik Winter as > > > you're showing your lack of reasonableness. > > > > You are ignoring my statement that in *every* ring that contains 1, 7 and 22, > > they are coprime. Whether 7 is a unit or not. > > > > Are you just totally stupid? That's irrelevant Dik Winter as the > > point is that 7 is NOT a factor of 22, which my saying that it's not a > > unit points out. > Why do you not read what I write, rather than what you think I write? > > I'm getting sick of stupid games from stupid people who apparently > > have nothing better to do with *their* time. > > > > Now are you or are you not intelligent enough to understand what it > > means for 7 NOT to be a factor of 22? > Yup. That means that you think your factorisation of P(x)/49 is valid > in some ring where 7 is not a unit. I am not convinced. I *know* it That's because you're *refusing* to be rational!!! Now I've found another way your hacks go away from my argument as the polynomial I use, with v=-1+49x is simply P(x) = (v^3+1)5^3 - 3v(5)7^2 + 7^3 which shouldn't look totally unfamiliar to you, but you're a hack, apparently dedicated to convincing newsgroup readers of what you believe, without concern about the actual mathematical truth. Just my problem you and your cabal have been successful enough that I have to break you. > is not valid in the algebraic integers, because there is ample proof that > for varying x the factors of 49 distribute differently amongst the three > factors of the polynomial when you wish to stay in the algebraic integers. And my point remains that I've found an error in core so your claim of a proof using core is stupid in context as I've *repeatedly* explained it to you. Why do you refuse to be logical Dik Winter? Why do you refuse to follow a stepped out mathematical proof? What's wrong with you? > Furthermore, I *know* that in the algebraic integers (5 b3(x) + 22) is > *not* coprime (either the standard definition or your definition) to 7 > for most values of x. You're irrational Dik Winter, desperate to hold on to what you think you know, as if mathematics need care about your beliefs. You've been refuted. I've shown yet again how your hack diverges from my work. Tell the truth, or go away. James Harris === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) >>If 7 and 22 are *not* units, I'm pretty sure they won't have non-unit >>divisors, but I'm not sure how to go about showing that. Maybe when >>I've got some time later I'll look at it. > That's why the usual definition is better: > LEMMA. If R is a ring, and x and y in R are coprime in the sense that > there exist a and b in R such that ax+by = 1, then for any ring that > contains R, x and y are also coprime (in the same sense). > Proof: a and b serve as witnesses in both R and the larger ring. QED > REMARK. If R is a ring, x and y in R are coprime in the sense that > any common divisor in R of x and y is a unit in R, then it is possible > for there to be a larger ring S, containing R, where x and y are no > longer coprime (in that sense). > Example: R= Z[sqrt(-5)]; x = 2, y = (1+sqrt(-5)), S= > Z[sqrt(-5),sqrt(2)]; note that y is a multiple of sqrt(2), since > (1+sqrt(-5)) = sqrt(2)*sqrt(3+sqrt(-5)). > PROP. Let R be a ring. If x and y are coprime in R in the sense that > there exists a and b in R such that ax+by=1, then x and y are coprime > in R in the sense that any common divisor in R is a unit in R. > Proof. Let u be a common divisor of x and y. Then it is a divisor of > ax, and it is a divisor of by, so it is a divisor of ax+by=1. Divisors > of 1 are units. So u is a unit. QED > So in ANY ring that contains the integers, 7 and 22 are coprime (under > either definition). -- Will Twentyman email: wtwentyman at copper dot net === Subject: any info on a^n+a^(n-1)b+a^(n-2)b^2+..+ab^(n-1)+b^n Hi anybody could please give me some info on a^n+a^(n-1)b+a^(n-2)b^2+..+ab^(n-1)+b^n it looks like geometric and/or binomial series but nothing is clear (to me). === Subject: Re: any info on a^n+a^(n-1)b+a^(n-2)b^2+..+ab^(n-1)+b^n > Hi anybody could please give me some info on > a^n+a^(n-1)b+a^(n-2)b^2+..+ab^(n-1)+b^n > it looks like geometric and/or binomial series but > nothing is clear (to me). For a = b, it is n*a^n = n*b^n. For a and b distinct, it is [a^(n+1) - b^(n+1)]/[a-b] === Subject: Re: any info on a^n+a^(n-1)b+a^(n-2)b^2+..+ab^(n-1)+b^n > Hi anybody could please give me some info on > a^n+a^(n-1)b+a^(n-2)b^2+..+ab^(n-1)+b^n > it looks like geometric and/or binomial series but > nothing is clear (to me). Multiply it by a-b and see what you get. === Subject: Re: any info on a^n+a^(n-1)b+a^(n-2)b^2+..+ab^(n-1)+b^n GR56 > a^n+a^(n-1)b+a^(n-2)b^2+..+ab^(n-1)+b^n > it looks like geometric and/or binomial series but > nothing is clear (to me). The ratio of each term to the next is a/b, which is independant of n, so it's a geometric series. If the sum is s, we get (b-a)s = b^(n+1) - a ^(n+1) and therefore a formula for s. Equally good is (1 - a/b)s = etc. LH === Subject: Re: Gaussian Continued Fractions > Conjugates of alpha have the same norm; is there always a conjugate of > alpha with purely positive coefficients? If so, then one could take > the algebraic integer nearest zero in the cell alpha lies in. This > guarantees that the coefficients remain positive Oops! The coefficients are all positive on the remainder, but certainly not its inverse. I haven't worked out if there's always some integer that gives all positive coefficients on the inverse of the remainder while keeping the magnitude of the remainder within some bound. Another definition of positive would be to consider those numbers with positive real part; I think this is preserved with the definition of greatest integer above. Mike Stay === Subject: Re: RFI: Anybody know if this is true/known? > yet] If X is odd, X^2 = 1 mod 4; If X is even, X^2 = 0 mod 4 If all 3 of a,b,c are odd then d^2 = 3 mod 4 (impossible) If 2 of a,b,c are odd, then d^2 = 2 mod 4 (impossible) -Tralfaz > There exist no solutions to: > a^2+b^2+c^2=d^2 with > gcd(a,b)=gcd(b,c)=gcd(c,a)=1 > If a, b, c, d are integers and > a^2 + b^2 + c^2 = d^2 > then at least two of a, b, c are even. > -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > Needless to say, I had the last laugh. > Alan Partridge, _Bouncing Back_ (14 times) === Subject: A NOTy problem. G'day G'day Folks, I have been puzzling over what ought to be a simple logic problem but am at a loss to come up with a method. A crime is committed by one of the following; Alan, Bob, Chris or Dave. Each makes a statement to the police but three of them lie. Alan says, I didn't do it. Bob says, Alan is lying. Chris says,Bob is lying. Dave says, Bob did it. Usually one can solve such problems with a grid L. N. I. G. A. B. C. D. Where A. B. C. D represent their names and L. = lying, N. = not_lying, I. = innocent, G. = guilty. Somehow that doesn't seem to get anywhere is this case. Maybe it is because N = NOT(lying) and G = NOT(innocent) and one has to use another method say Boolean logic or Karnaugh maps. I am less interested in knowing the answer to the puzzle than understanding why it is different from the routine logic puzzles. Best wishes, -- Quentin Grady ^ ^ / New Zealand, >#,#< [ / / ... and the blind dog was leading. http://homepages.paradise.net.nz/quentin === Subject: Re: A NOTy problem. Adjunct Assistant Professor at the University of Montana. >G'day G'day Folks, > I have been puzzling over what ought to be a simple logic problem >but am at a loss to come up with a method. These sort of things can be solved by brute force, by assigning all possible values of truth or false to each and seeing which yield contradictory information. On the other hand, sometimes there are shortcuts based on noticing certain things. >A crime is committed by one of the following; Alan, Bob, Chris or >Dave. >Each makes a statement to the police but three of them lie. >Alan says, I didn't do it. >Bob says, Alan is lying. >Chris says,Bob is lying. >Dave says, Bob did it. There's two easy observations here. Note that Dave and Allan's statements are compatible. If Dave is telling the truth, then so is Alan. Since only one person is telling the truth, Dave must be lying. Note also that it is impossible for BOTH Bob and Chris to be lying. That means that the person telling the truth is either Bob or Chris; which means that Alan must be one of the liars. Which solves the problem: Alan committed the robbery, and is lying; Bob is telling the truth, and both Chris and Dave are lying. 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: A NOTy problem. This post not CC'd by email >>G'day G'day Folks, >> I have been puzzling over what ought to be a simple logic problem >>but am at a loss to come up with a method. >These sort of things can be solved by brute force, by assigning all >possible values of truth or false to each and seeing which yield >contradictory information. On the other hand, sometimes there are >shortcuts based on noticing certain things. >>A crime is committed by one of the following; Alan, Bob, Chris or >>Dave. >>Each makes a statement to the police but three of them lie. >>Alan says, I didn't do it. >>Bob says, Alan is lying. >>Chris says,Bob is lying. >>Dave says, Bob did it. >There's two easy observations here. Note that Dave and Allan's >statements are compatible. G'day G'day Arturo, Yes. That is neat. Dave and Alan are talking about who did the crime. Bob and Chris are talking about who is lying. This forms natural groupings to test for contradiction. >There's two easy observations here. Note that Dave and Allan's >statements are compatible. Compatible? I can see that the statements are not inconsistent. > If Dave is telling the truth, then so is Alan. >Since only one person is telling the truth, Dave must be lying. You lost me there. What if I had a let's pick on Alan day? If Alan is telling the truth can we say that so it Dave. Since only one person is telling the truth, Alan must be lying. Wish I could see it as clearly as you do but I don't. >Note also that it is impossible for BOTH Bob and Chris to be >lying. That means that the person telling the truth is either Bob or >Chris; which means that Alan must be one of the liars. Which solves >the problem: Alan committed the robbery, and is lying; Bob is telling >the truth, and both Chris and Dave are lying. >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 -- Quentin Grady ^ ^ / New Zealand, >#,#< [ / / ... and the blind dog was leading. http://homepages.paradise.net.nz/quentin === Subject: Re: A NOTy problem. > G'day G'day Folks, > I have been puzzling over what ought to be a simple logic problem > but am at a loss to come up with a method. > A crime is committed by one of the following; Alan, Bob, Chris or > Dave. > Each makes a statement to the police but three of them lie. > Alan says, I didn't do it. > Bob says, Alan is lying. > Chris says,Bob is lying. > Dave says, Bob did it. > Somehow that doesn't seem to get anywhere is this case. > Maybe it is because N = NOT(lying) and G = NOT(innocent) and one has > to use another method say Boolean logic or Karnaugh maps. > I am less interested in knowing the answer to the puzzle than > understanding why it is different from the routine logic puzzles. It depends on the situation as to what method is most useful. I would attempt a trial&error approach. There is exactly one truth teller. If it's Alan, then Bob is lying which means Chris is telling the truth. Contradiction. If it's Bob, the other three are lying and Alan did it. If it's Chris, Bob is lying so Alan told the truth. Contradiction. If it's Dave, then Alan told the truth. Contradiction. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: A NOTy problem. This post not CC'd by email >> A crime is committed by one of the following; Alan, Bob, Chris or >> Dave. >> Each makes a statement to the police but three of them lie. >> Alan says, I didn't do it. >> Bob says, Alan is lying. >> Chris says,Bob is lying. >> Dave says, Bob did it. >It depends on the situation as to what method is most useful. >I would attempt a trial&error approach. >There is exactly one truth teller. >If it's Alan, then Bob is lying which means Chris is telling the truth. > Contradiction. >If it's Bob, the other three are lying and Alan did it. >If it's Chris, Bob is lying so Alan told the truth. Contradiction. >If it's Dave, then Alan told the truth. Contradiction. G'day G'day Will, This is method certainly has the merit of simplicity. It is like the brute force methods for finding integer values that make 1/A + 1/B = 1/C true. Try a value and see if it works. If it doesn't try the next one. As I understand it computer programmers like such procedures because there are so few procedural decision points. Basically the same procedure is followed throughout. Best wishes, -- Quentin Grady ^ ^ / New Zealand, >#,#< [ / / ... and the blind dog was leading. http://homepages.paradise.net.nz/quentin === Subject: Re: A NOTy problem. > G'day G'day Folks, > I have been puzzling over what ought to be a simple logic problem > but am at a loss to come up with a method. > A crime is committed by one of the following; Alan, Bob, Chris or > Dave. > Each makes a statement to the police but three of them lie. > Alan says, I didn't do it. > Bob says, Alan is lying. > Chris says,Bob is lying. > Dave says, Bob did it. > Usually one can solve such problems with a grid I find such grids helpful, but not perfect. It's hard to track some statements with them, and to keep track of what ifs. > Where A. B. C. D represent their names and > L. = lying, > N. = not_lying, > I. = innocent, > G. = guilty. > Somehow that doesn't seem to get anywhere is this case. > Maybe it is because N = NOT(lying) and G = NOT(innocent) Well because of that, an easier grid to use would be T G A B C D (T for telling truth) Then you just need a method to indicate whether the statement is true or false for each one. For instance, an X means its false and a dot means it's true. Thus, each column will have one dot and three X's when you're done. If you put a dot, you have to put three corresponding X's. > I am less interested in knowing the answer to the puzzle than > understanding why it is different from the routine logic puzzles. It's not. But you do need to be able to backtrack and reset your grid after a false hypothesis. For instance, let's start with the first statement. > Alan says, I didn't do it. Suppose that's true. Then Alan gets an O in column one (he's telling the truth) and an X in column two (he's not guilty). T G A O X B X C X D X Once we put an O, everybody else gets an X. Now Bob says Alan is lying. We assumed Alan is a truth teller, so Bob is lying. We already marked that. Chris says Bob is lying. This is true. But we already marked Chris as a liar. Inconsistency. Alan is not the truth teller. Restart the grid with somebody else as truth teller and X under Alan for column one. Of course, if Alan is lying, then this statement is false: Alan says, I didn't do it. So no matter who the truth teller is, Alan did it. T G A X O B X C X D X Now we just have to see if putting an O under B, C, or D works out consistently. But actually, I see that Bob's statement is now true. T G A X O B O X C X X D X X Is this consistent with the other two statements? Better check (if not, there is no solution). > Chris says,Bob is lying. If Chris is a liar, then this statement is wrong. Which it is. It checks. > Dave says, Bob did it. If Dave is a liar, then Bob didn't do it. And he didn't. The grid system works fine, you just set up an improper grid. - Randy === Subject: Re: A NOTy problem. This post not CC'd by email >> G'day G'day Folks, >> I have been puzzling over what ought to be a simple logic problem >> but am at a loss to come up with a method. >> A crime is committed by one of the following; Alan, Bob, Chris or >> Dave. >> Each makes a statement to the police but three of them lie. >> Alan says, I didn't do it. >> Bob says, Alan is lying. >> Chris says,Bob is lying. >> Dave says, Bob did it. >> Usually one can solve such problems with a grid >I find such grids helpful, but not perfect. It's hard to >track some statements with them, and to keep track of >what ifs. >> Where A. B. C. D represent their names and >> L. = lying, >> N. = not_lying, >> I. = innocent, >> G. = guilty. >> Somehow that doesn't seem to get anywhere is this case. >> Maybe it is because N = NOT(lying) and G = NOT(innocent) >Well because of that, an easier grid to use would be > T G >(T for telling truth) G'day G'day Randy, That is much simpler and it handles the negation aspect. >Then you just need a method to indicate whether the >statement is true or false for each one. For instance, >an X means its false and a dot means it's true. >Thus, each column will have one dot and three X's >when you're done. If you put a dot, you have to put >three corresponding X's. >> I am less interested in knowing the answer to the puzzle than >> understanding why it is different from the routine logic puzzles. >It's not. But you do need to be able to backtrack and >reset your grid after a false hypothesis. I thought there had to be a way to avoid backtracking ie a method of doing the grid WITHOUT using an eraser. I imagined solving a Boolean must arrive at the answer using reductio absurdum ie it can't be this answer or that one so it must be whatever is left. >For instance, let's start with the first statement. > > Alan says, I didn't do it. >Suppose that's true. Then Alan gets an O in column >one (he's telling the truth) and an X in column >two (he's not guilty). > T G >A O X >B X >C X >D X >Once we put an O, everybody else gets an X. >Now Bob says Alan is lying. We assumed Alan is a >truth teller, so Bob is lying. We already marked that. >Chris says Bob is lying. This is true. But we already >marked Chris as a liar. Inconsistency. Alan is not >the truth teller. Restart the grid with somebody >else as truth teller and X under Alan for column one. >Of course, if Alan is lying, then this statement >is false: > Alan says, I didn't do it. >So no matter who the truth teller is, Alan did it. > T G >A X O >B X >C X >D X >Now we just have to see if putting an O under >B, C, or D works out consistently. >But actually, I see that Bob's statement is now >true. > T G >A X O >B O X >C X X >D X X >Is this consistent with the other two statements? Better >check (if not, there is no solution). >> Chris says,Bob is lying. >If Chris is a liar, then this statement is wrong. Which >it is. It checks. >> Dave says, Bob did it. >If Dave is a liar, then Bob didn't do it. And he didn't. >The grid system works fine, you just set up an improper grid. I love it. Personally I find I need a visual grid of some sort to much. > - Randy -- Quentin Grady ^ ^ / New Zealand, >#,#< [ / / ... and the blind dog was leading. http://homepages.paradise.net.nz/quentin === Subject: Re: A NOTy problem. === Subject: A NOTy problem. > I have been puzzling over what ought to be a simple logic problem >A crime is committed by one of the following; >Alan, Bob, Chris or Dave. >Each makes a statement to the police but three of them lie. >Alan says, I didn't do it. >Bob says, Alan is lying. >Chris says,Bob is lying. >Dave says, Bob did it. If Chris told truth: Alan told truth; two soothe sayers. Thus Chris lied; Bob told truth; Alan did it. Alan lied; Dave is wrong ---- === Subject: Re: A NOTy problem. This post not CC'd by email === >Subject: A NOTy problem. > I have been puzzling over what ought to be a simple logic problem >A crime is committed by one of the following; >Alan, Bob, Chris or Dave. >Each makes a statement to the police but three of them lie. >Alan says, I didn't do it. >Bob says, Alan is lying. >Chris says,Bob is lying. >Dave says, Bob did it. >If Chris told truth: Alan told truth; two soothe sayers. >Thus Chris lied; Bob told truth; Alan did it. >Alan lied; Dave is wrong G'day G'day Alan, Well that is brilliant but I am not. Somehow you must have eliminated other entry points to the problem. That is quite a skill backtrack because of further reductio absurdum. Best wishes, -- Quentin Grady ^ ^ / New Zealand, >#,#< [ / / ... and the blind dog was leading. http://homepages.paradise.net.nz/quentin === Subject: Re: A NOTy problem. Hello > I have been puzzling over what ought to be a simple logic problem > but am at a loss to come up with a method. > A crime is committed by one of the following; Alan, Bob, Chris or > Dave. > Each makes a statement to the police but three of them lie. > Alan says, I didn't do it. > Bob says, Alan is lying. > Chris says,Bob is lying. > Dave says, Bob did it. Sounds like Dave plays no role here. Either Bob or Alan is telling the truth (if not, than Bob lies and Alan is not lieing, which is absurd). Either Chris or Bob is telling the truth (same proof) But there is only one truth-teller, hence Bob tells the truth. Alan did commit the crime! -hopefully, he's living in New Zealand so he won't be grilled like chicken-. === Subject: Re: A NOTy problem. This post not CC'd by email >Hello >> I have been puzzling over what ought to be a simple logic problem >> but am at a loss to come up with a method. >> A crime is committed by one of the following; Alan, Bob, Chris or >> Dave. >> Each makes a statement to the police but three of them lie. >> Alan says, I didn't do it. >> Bob says, Alan is lying. >> Chris says,Bob is lying. >> Dave says, Bob did it. >Sounds like Dave plays no role here. G'day G'day Julian, I like the approach. Reducing a four possibility problem to a three possibility problem. My question is how did you do that? Are you working that out from what Dave said or because of the statements that follow? >Either Bob or Alan is telling the truth (if not, than Bob lies and Alan is >not lieing, which is absurd). >Either Chris or Bob is telling the truth (same proof) >But there is only one truth-teller, hence Bob tells the truth. That is fantastic. As I see it you aren't forced to do an eventual back track. Your description fits finding the intersection of two sets. The reductio absurdum is confined to establishing the _initial_ is that the backtracks are confined to a single stage and not to multiple stages in the logic. (Hey, it saves on erasers.) >Alan did commit the crime! -hopefully, he's living in New Zealand so he >won't be grilled like chicken-. Indeed. Best wishes, -- Quentin Grady ^ ^ / New Zealand, >#,#< [ / / ... and the blind dog was leading. http://homepages.paradise.net.nz/quentin === Subject: Re: A NOTy problem. > I am less interested in knowing the answer to the puzzle than > understanding why it is different from the routine logic puzzles. I am not familiar with the routine logic puzzles, so I don't know why its solution would be different. However, it seems to be straigtforward to solve. You did not make use of the information that exactly three are lying. I would guess that the routine puzzle would be that exactly three are telling the truth. The same method can be used to solved both types of problems. This method must be a routine method to solve puzzles like this. -- Bill Hale === Subject: Re: A NOTy problem. This post not CC'd by email >> I am less interested in knowing the answer to the puzzle than >> understanding why it is different from the routine logic puzzles. >I am not familiar with the routine logic puzzles, >so I don't know why its solution would be different. >However, it seems to be straigtforward to solve. >You did not make use of the information that exactly three are lying. >I would guess that the routine puzzle would be that exactly >three are telling the truth. G'day G'day William, In the routine logic puzzles the person setting the puzzles provides the grid. This was my inexpert attempt at creating a grid and solving a logic puzzle. >The same method can be used to solved both types of problems. >This method must be a routine method to solve puzzles like this. >-- Bill Hale -- Quentin Grady ^ ^ / New Zealand, >#,#< [ / / ... and the blind dog was leading. http://homepages.paradise.net.nz/quentin === Subject: Re: Classic wave equation vs. Klein-Gordon >>The ordinary one-dimensional wave equation may be read to say that the >>tension creates a restorative force proportional to the curvature >>(second derivative), which in turn produces an acceleration. >>The one-dimensional Klein-Gordon equation can be read the same way, >>except that we have added a force term proportional direction to the >>displacement -- the string is joined to its equilibrium position by >>springs. >>Comment. >The way the Klein-Gordon equation is usually derived (in quantum >theory) is by taking the relativistic formula E^2 = (pc)^2 + >(mc^2)^2 and converting it to an operator equation by replacing E with >i hbar d/dt and p with -i hbar d/dx, so you end up with a second order >diff eq in x and t in the form of a classical wave equation with an >additional term in the form of a mass-related constant times the wave >function. Write down the equations of motion for a bunch of point masses joined by springs... then take the continuum limit and you end up with the wave equation (the nearest neighbor terms due to the springs turn into spacial dirivatives) Write down the equations of motion for a bunch of point masses joined by springs and also held in place by zero length springs .. then take the continuum limit and you end up with the K-G equation (the nearest neighbor terms due to the springs turn into spacial dirivaties and the zero length spring terms turn into the mass term.) If you need more info you should find a book on classical field theory... or the first chapter of a book on quantum field theory. adam === Subject: Re: How to find probability mass function of this? >>Let U and V be independent random variables each with Uniform(0,1) >>distribution. Determine the joint density of X = maj(U,V) and Y= U+V. >> >I get this: >P(X= (1/2)(ab)^2 And that answer is wrong. >since: >P(UP(U1? Care to try again before we provide a solution? Do you know about the use of conditional probability? >How do you like this solution? Please feel more than free to point out >any eventuall flaws. I think this part of mathematics is hilariously >interesting. I really enjoy going into it. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Skeptickal Inquirer UFO > I had mentioned yesterday that I believed that since major > contact appeared to take place just after Hiroshima, > that these sorts of things were a beacon, inviting > investigation. > Taking this further, we have every reason to believe that > some sort of galactic government then intervened on earth, > and essentially took control of the earth. Covertly, > and not pushing any overt agenda, other than to > slowly and methodically assist mankind in dealing with > such tools of destruction. To bring man to the point > where he was capable of handling the technology without > destroying the earth. > That would be the best case scenario. > It would appear however, that their mandate was quite > strict in that they could not necessarily interfere in > the affairs of men, otherwise we would certainly not have > seen subsequent wars and our civilization would not be where > it is today. We would expect a more utopian society based > on intelligence, rather than the continued power struggles > we see. > However, people, are just people, and even under threat > of death, or sworn to secrecy cannot keep a secret and things > do leak out. You're making a lot of assumptions here. First, you're assuming that the ET's exist; second you're making the assumption that they would care one way or another about what happens to humans (no reason why they should) third, and much more profoundly, you're making the assumption that they ET's are CAPABLE of doing something about it even if they wanted to. I don't mean in the sense of technology, necessarily, but it is naive to think that a species that evolved on another planet (assuming that evolution is a universal law) would be able to speak, see, listen, think, smell, feel or taste just because we can. Who says the ET's are even in a dimension that we can see (they could be two dimensional, restricted only to moving on a plane), they would be completely incapable of seeing us. They could be only visible at a different frequency... X-ray, infrared, and we would be completely incapable of seeing them. And quite frankly, if the alien gov't was controlling all world gov'ts simultaneously, they've done a pretty wretched job of keeping us from killing ourselves off (we're not dead yet, but we're in a much more likely position to do it after the arms race in the Cold War) You're assuming that SOMEHOW the ET's are able to detect a very small nuclear explosion from a considerable distance, despite the fact the amount of radiation it caused would be negligible at any distance outside the Earth's atmosphere, and then are able to travel here in 2 Earth years or less and convince world leaders to let them rule the planet. The whole thing seems a little far-fetched to me... === Subject: Re: Skeptickal Inquirer UFO Lt.Col. Corso, may he rest in peace, used his onetime position at the Roswell bomber base to illucidate his silly story, We dumb Americans found this little pile of ****, and I sent it out to the assembled brethren of the MIC, and we reverse-engineered RADAR, fiber optics, transistors ... um, le'see ... balsa wood and mylar and latex balloons, to boot! do you know of the cache that Roswell has with WW2?... really, the whole thing could be denial, denial, denial. Corso's essential thesis was: human being cannot create ideas; it's got to be cargo-cult from upstairs! his coauthor exposed him on the booktour, which Corso had to sit out, as he was dying, but I don't know if it's actually in the book, as well. >history to now, i.e. about 7 billion years ago. This dark energy is the >same as Kip Thorne's exotic matter in his 1986 Star Gate paper and it >is the same stuff needed for Alcubierre's weightless warp drive metric >of the early 1990's and it is the same negative matter needed for >exotic propulsion that British Ministry of Defense Chief Scientist >Herman Bondi told us Cornell students about in ~ 1960 that Stalin's top > physics Spy Master Y. Terletskii was also very keen on. > http://www.rense.com/general44/nmxx.htm > one of the few middle of the road outlets for information of > the strange bizarre and unusual. > such tools of destruction. To bring man to the point > where he was capable of handling the technology without > destroying the earth. > That would be the best case scenario. > I have no reason to doubt the credibility of this man. > Mr. Phillips further states that there are records and filmed > documentation of meetings in California in 1954 between ETs > and leaders of the USA. He lists a few of the technologies > we were able to develop because of the ETs: computer chips, > lasers, night vision, bulletproof vests, and concludes, > Are these ET people hostile? Well, if they were hostile, > with their weaponry they could have destroyed us a long time > ago or could have done some damage. Mr. Phillips now develops > technologies that can help eliminate environmental pollutants > and reduce the need for fossil fuels: energy generation > systems that use natural energies from planet Earth. > And speaking of debunkers, I look at the people on the above > page, not to mention a person like Corso, and I examine > their credibility. Then I look at the Amazing Randi, > and I want to laugh out loud. > We are not alone in the universe, that I am sure of. --ils duces d'Enron! http://tarpley.net/bush8.htm http://www.wlym.com/PDF-SpReps/SPRP13.pdf === Subject: NOVA strings and branes === Subject: NOVA Look at Elegant Universe NOVA tonight. http://www.pbs.org/wgbh/nova/elegant/ Brian Greene telephoned an ET Gray on brane universe next door a millimeter away using presumably Ray Chiao's gravity radio transducer of off-brane world gravity waves to EM waves. ;-) Also long part on Star Gate time and space-travel. Ed Witten & Co http://superstringtheory.com/people/witten.html would be considered real kooks by the SI CSICOPS if their actual words were given to the CSICOPS without identifying who they were in a blind fold Turing Test. The tidy classical universe of James Oberg & Co is not the Universe of Ed Witten & Co. BTW Q to Ed Witten: How can the cosmological constant be so close to zero but not zero? Ed Witten's answer: I really don't know. It's very perplexing that astronomical observations seem to show that there is a cosmological constant. It's definitely the most troublesome, for my interests, definitely the most troublesome, observation in physics in my lifetime. In my career that is. My answer to the same question is at http://qedcorp.com/APS/StarGate1.mov For the record, I do not think that the intelligence agencies of the the major powers, US in particular, have the ET UFO anti-gravity technology on the shelf and are hiding it. In other words I disagree with Nick Cook's book Hunt for the Zero Point and with Stephen Greer's thesis in the Disclosure Project. It is quite obvious that no human physicist today understands how such stuff would work. We are only now beginning to get to that level of understanding of how metric engineering would work by manipulating dark energy. It may be possible that USG et-al has retrieved damaged craft and alien creatures from the Universe Next Door perhaps only a millimeter away across the extra bosonic space dimensions like Robert Bigelow's NIDS people reported on his Utah Ranch, or that may have happened at Roswell in 1947 etc. Even if that were true, and I am not saying it is true, the fact is that none of the intelligence people who might have that stuff have the slightest understanding how that hypothetical and/or alleged alien technology actually works. It's The Sorceror's Apprentice. Hi Jim, OK, I believe you. But I do suspect that you may have been subtly conditioned by your NASA(?) experience to accept uncritically (despite your overt commitment to critical thinking) the Government's long-standing public position re the phenomenon in question, that is, that UFOs (and aliens) are all either misidentifications of natural phenomena, hoaxes, or hallucinations, mass and otherwise. This, I suspect-taking you at your word that you're not cynically mouthing a party line that you know to be false-is reinforced by the logically flawed assumption that they shouldn't exist, therefore they don't exist. Clearly, it's very difficult for most mainstream members of the scientific and engineering establishment-that is, those not privy to (or part of) any Government-sponsored disinformation campaign-to accept even the remote possibility that we human beings are, in effect, Wogs vis-a-vis what I've come to think of as an Alien Raj, an extraterrestrial (or possibly extra-dimensional) colonial establishment that has maintained clandestine hegemony over this planet for at least the last eleven or twelve millennia, if I read the late Upper Paleolithic evidence correctly. No, of course I can't as yet prove any of this definitively. But the vast amount of observational evidence, as gathered by people like Richard Dolan, et al., is extremely persuasive, as is the comparative mythological and folkloric evidence, which you may not be aware of (this is my anthropological specialty), and which, after stripping away the pre-modern glosses and metaphors, closely resembles what has been reported by contemporary abductees and other experiencers. And now that physicists like Jack Sarfatti are beginning to give us a handle on these remarkable craft appear to work, the probability of their presence, both today and in the past, is significantly reinforced. Again, once we get over the ego problem, and can accept-and live with-at least the possibility that we've long since been trumped by others when it comes to the technology of space exploration, we can begin to approach this supremely important phenomenon objectively, and from a variety of perspectives, including that of the cultural anthropologist, free from the hysterical sneering, ridicule, and marginalizing that all too frequently tarnishes the discussion of UFOs, whether by mainstream scientists or by the media. (That much of this sneering and ridicule is the result of a conscious policy on the part of the Government, or some elements thereof, to discredit anyone who takes the subject seriously so as to keep the truth from becoming general knowledge-perhaps due to a legitimate fear of massive and socially disruptive nativistic movements that would put the New Guinean Cargo Cults to shame-is a can of peas which I won't re-open any further in this message.) Scott C. SCOTT LITTLETON President, Phi Beta Kappa Alumni in Southern California Professor of Anthropology, Emeritus Occidental College Los Angeles, CA 90041 TEL (323) 255-5477 FAX (323) 982-0264 http://www.oxy.edu/~yokatta/home.htm Any sufficiently advanced technology is indistinguishable from magic --Sir Arthur C. Clarke I think we're property. . . --Charles Fort -----Original Message----- === Subject: Fw: Skeptical Inquirer UFO fyi ----- Original Message ----- === Subject: Re: Skeptical Inquirer UFO I can solve this one. The answer, for me at least, is 'No'. My agenda is to find out about mysterious aerospace reports and events (see www.jamesoberg.com) and share what I've found, for discussion, with no directives, constraints, or other external controls over my activities whatsover. You may stop wondering, and start thinking, now. Jim Oberg Phi Beta Kappa, Ohio Wesleyan University, 1966 Or do debunkers like Sheaffer, Oberg, Klass, and the rest have another, more devious agenda, that is, to further the Government's long-standing policy of keeping the lid on by systematically ridiculing those of us who suspect the truth about this phenomenon, all the while being privy to above-top-secret knowledge that would prove us right? One wonders.... C. SCOTT LITTLETON President, Phi Beta Kappa Alumni in Southern California === Subject: Re: Geometry And Newtonian Mechanics of Action Device > > OK, let us see Pythagores in Action.. > ______________________________________________________ > > Pythagores Theorem: > In a right angled triangle, the square of hypotenuse is equal to the > sum of the squares of the other two sides. > > c^2 = a^2 + b^2 > ______________________________________________________ > You would be more credible if you started by spelling his name > correctly. The Pythagoran theorem is trivial special case of the > general triangle theorem. Look it up to learn why the right angle is > so convenient. If I spell your name as Yncal Al, will it change your persona, actions, understanding of Geometry and Physics... I am trying to explain step by step Geometry, Newtonian Mechanics of this action device. Please, Uncle Al, concentrate on meaning, not on words. I don't have much time left. -Abhi. === Subject: Re: Geometry And Newtonian Mechanics of Action Device >OK, let us see Pythagores in Action.. >______________________________________________________ >Pythagores Theorem: >In a right angled triangle, the square of hypotenuse is equal to the >sum of the squares of the other two sides. >c^2 = a^2 + b^2 >______________________________________________________ >Now draw X, Y axis on paper. We refer origin as B instead of >conventional O. >Line AC of length c is on X axis and initially point C of line AC >coincides with origin B. Let us move or slide line AC along X axis by >distance d(where d<slides along Y axis in vertical direction. Now, we have a right angle >traingle ABC. >In this triangle AC = c, AB = a = c-d, BC = b >According to Pyathagores theorem, in this right angle triangle >c^2 = a^2 + b^2 >We replace a by c-d as AB = a = c-d >Hence c^2 = (c-d)^2 + b^2 >c^2 = c^2-2cd+d^2 + b^2 >We get, b^2 = 2cd-d^2 >****************** >b = sqrt[d(2c-d)] >****************** >This is the basic equation which controls this action device. Please >note that when, initially, we move line AC along X axis by very small >distance d, point C slides along Y axis and BC = b > d >Please tell me whether you have understood this or not. Yes I understand Abhi, you are speaking about this. http://dbarkertv.com/UPDATE.htm So why do you think your life is in danger? Why not just come forward. You can talk here. People here are just pretending they don't know what goes on in this world. We are all alright. Just like the song says. Cheaptrick-Surrender.mp3 Although I know at times, it seems like just being alright is not enough. At least that is what Tatu says in this song. Tatu - All The Things She Said.mp3 If you'd like to hear those songs now, you need but visit this site and download Kazaa Lite. Here is the page where you can choose the version of Kazaa Lite you want to download. http://home.hccnet.nl/h.edskes/mirror.htm I use this version http://le4.edskes.com/klitekpp210e.exe found on that page. But then you might want the french version. =*= Rick === Subject: Re: Geometry And Newtonian Mechanics of Action Device > OK, let us see Pythagores in Action.. > ______________________________________________________ > Pythagores Theorem: > In a right angled triangle, the square of hypotenuse is equal to the > sum of the squares of the other two sides. > c^2 = a^2 + b^2 > ______________________________________________________ > Now draw X, Y axis on paper. We refer origin as B instead of > conventional O. > Line AC of length c is on X axis and initially point C of line AC > coincides with origin B. Let us move or slide line AC along X axis by > distance d(where d< slides along Y axis in vertical direction. Now, we have a right angle > traingle ABC. > In this triangle AC = c, AB = a = c-d, BC = b > According to Pyathagores theorem, in this right angle triangle > c^2 = a^2 + b^2 > We replace a by c-d as AB = a = c-d > Hence c^2 = (c-d)^2 + b^2 > c^2 = c^2-2cd+d^2 + b^2 > We get, b^2 = 2cd-d^2 > ****************** > b = sqrt[d(2c-d)] > ****************** > This is the basic equation which controls this action device. Please > note that when, initially, we move line AC along X axis by very small > distance d, point C slides along Y axis and BC = b > d > Please tell me whether you have understood this or not. > -Abhi. Understood, yes. What's the point? This tells us nothing about the mysterious Action Device. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Geometry And Newtonian Mechanics of Action Device > And by the way, Al, did you ever think that perhaps > Abhi's language might be French, for instance? > Abhi speaks fluent crackpot. Then where should you be placed on a list if you admit to understanding Crackpoteze..... For if you did not understand it you could not realize he was fluent in it..... > -- > Pyriform Paul R. Mays ---------------------------------------------------------------------------- - Some where within the Quantum State Http://Paul.Mays.Com/story.html http://paul.mays.com/mayday.html http://paul.mays.com/rainy.html Science is facts. Just as houses are made of stones, so is science made of facts. But a pile of stones is not a house and a collection of facts is not necessarily science. - Jules Henri Poincare (1854~1912) === Subject: Re: Geometry And Newtonian Mechanics of Action Device >> Abhi speaks fluent crackpot. > Then where should you be placed on a list if > you admit to understanding Crackpoteze..... > For if you did not understand it you could not > realize he was fluent in it..... Not so. I know enough French and German to recognise fluency in others, even though I am not fluent myself. Same with Crackpot. -- Pyriform === Subject: DESIGN OF A PRIMITIVE NANOFACTORY Design of a Primitive Nanofactory Chris Phoenix Director of Research, Center for Responsible Nanotechnology http://CRNano.org Abstract: Molecular manufacturing requires more than mechanochemistry. A single nanoscale fabricator cannot build macro-scale products. This paper describes the mechanisms, structures, and processes of a prototypical macro-scale, programmable nanofactory composed of many small fabricators. Power requirements, control of mechanochemistry, reliability in the face of radiation damage, convergent assembly processes and joint mechanisms, and product design are discussed in detail, establishing that the design should be capable of duplicating itself. Nanofactory parameters are derived from plausible fabricator parameters. The pre-design of a nanofactory and many products appears to be within today's capabilities. Bootstrapping issues are discussed briefly, indicating that nanofactory development might occur quite soon after fabricator development. Given an assembler, a nanofactory appears feasible and worthwhile, and should be accounted for in assembler policy discussions. . . . More here: http://www.jetpress.org/volume13/Nanofactory.htm Jai Maharaj http://www.mantra.com/jai Om Shanti Shubhanu Nama Samvatsare Dakshinaya Jeevan Ritau Tula Mase Shukl Pakshe Mangal Vasara Yuktayam Poorvaprostapad-Uttaraprostapad Nakshatr Vyaghat-Harshan Yog Vishti-Bav Karan Ekadashi-Dvadashi Yam Tithau Hindu Holocaust Museum http://www.mantra.com/holocaust Hindu life, principles, spirituality and philosophy http://www.hindu.org http://www.hindunet.org The truth about Islam and Muslims http://www.flex.com/~jai/satyamevajayate o Not for commercial use. Solely to be fairly used for the educational purposes of research and open discussion. The contents of this post may not have been authored by, and do not necessarily represent the opinion of the poster. The contents are protected by copyright law and the exemption for fair use of copyrighted works. o If you send private e-mail to me, it will likely not be read, considered or answered if it does not contain your full legal name, current e-mail and postal addresses, and live-voice telephone number. are not necessarily those of the poster. === Subject: Re: DESIGN OF A PRIMITIVE NANOFACTORY > Design Shut the up. === Subject: Re: DESIGN OF A PRIMITIVE NANOFACTORY > Design of a Primitive Nanofactory http://www.geocities.com/drjosemariachi/jay_faq.html#bb Troll FAQ for Jai Maharaj (Hindi for cracked athletic cup) -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: c-number >: Howdy everybody, >: I'm reading Peskin and Schroeder's Intro to QFT and in this >: book the author constantly refers to c-numbers... without >: saying what a c-number is. I couldn't find c-number >: on mathworld.wolfram.com or physicsworld.wolfram.com. >: Can someone please tell me what the hell a c-number is? >I think it means a classical number, as opposed to a q-number (quantum >number). Okay, then what is a classical number? adam === Subject: Re: Covering a rectangle > Is it possible to cover a 1 x sqrt(2) rectangle with a finite number of > disjointed squares (with eventually different measures)? No. Covering a rectangle with squares is possible iff the ratio of the sides is rational. This follows from the relation between square coverings and electrical networks: see for instance, Bollobas's _Graph Theory_. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Key Core Error Argument The following proof steps through a rather basic argument which is key in proving an over one hundred year old error resulting from previously unexpected consequences resulting from the definition of the ring of algebraic integers. Note that ultimately the proof relies on 22 NOT having 7 as a factor, and constant terms like 7 and 22, being constant, and not variables dependent on x, which may seem like odd things to emphasize, but I've faced posters who've gotten away with challenging those truths because people seem unaware that's what they're doing. 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is in the ring of algebraic integers, notice that P(x) has a constant term that is 1078. 2. It can be shown that P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 where you should note that using v = -1 + 49x, gives P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 where the *same* polynomial has been put in a form which allows a factorization into non-polynomial factors so that I have P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) where the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). 3. Now let x=0, so P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic defining the a's at x=0 is a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and a_2(0) to equal 0, which leaves a_3(0) with a value of 3. 4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I have P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). 5. Now P(x) has a factor of 49 as P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 which means that (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) has a factor of 49. 6. However, the constant term of P(x)/49 is 22, which is verified by again setting x=0, which gives P(0)/49 = 22. But for two of the factors of P(x), the constant terms is 7, which is NOT a factor of 22. Therefore, *none* of the constant terms of P(x)/49 as they multiply to give 22 can have 7 as a factor. (By saying that 7 is NOT a factor of 22, I'm making a choice as to where the proof is going. Since I've been talking about algebraic integers, where 7 is NOT a factor of 22, it's natural to go with a choice where 7 is NOT a factor of 22.) Given that the constant terms are independent of x's value, it must be the case that dividing P(x) by 49 divides the two constant terms equal to 7, by 7. 7. But to divide 7 from those constant terms requires dividing through two of the factors, so (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 from reverse use of the distributive property, which gives constant terms that don't have 7 as a factor, as required. Notice that it's a rather short and direct argument, where if you accept that 22 does not have 7 as a factor, it's obvious enough what the constant terms of the factors must be as you go from 7, 7 and 22, necessarily to 1, 1, and 22, when you divide P(x) by 49. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: Key Core Error Argument > The following proof steps through a rather basic argument which is key > in proving an over one hundred year old error resulting from > previously unexpected consequences resulting from the definition of > the ring of algebraic integers. > Note that ultimately the proof relies on 22 NOT having 7 as a factor, > and constant terms like 7 and 22, being constant, and not variables > dependent on x, which may seem like odd things to emphasize, but I've > faced posters who've gotten away with challenging those truths because > people seem unaware that's what they're doing. > 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is > in the ring of algebraic integers, notice that P(x) has a constant > term that is 1078. > 2. It can be shown that > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > where you should note that using v = -1 + 49x, gives > P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 > where the *same* polynomial has been put in a form which allows a > factorization into non-polynomial factors so that I have > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > where the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > 3. Now let x=0, so > P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) > as the cubic defining the a's at x=0 is > a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and > a_2(0) to equal 0, which leaves a_3(0) with a value of 3. > 4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I > have > P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) > P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). > 5. Now P(x) has a factor of 49 as > P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 > which means that > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) > has a factor of 49. > 6. However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. > But for two of the factors of P(x), the constant terms is 7, which is > NOT a factor of 22. Therefore, *none* of the constant terms of > P(x)/49 as they multiply to give 22 can have 7 as a factor. > (By saying that 7 is NOT a factor of 22, I'm making a choice as to > where the proof is going. Since I've been talking about algebraic > integers, where 7 is NOT a factor of 22, it's natural to go with a > choice where 7 is NOT a factor of 22.) > Given that the constant terms are independent of x's value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7. > 7. But to divide 7 from those constant terms requires dividing > through two of the factors, so > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > from reverse use of the distributive property, which gives constant > terms that don't have 7 as a factor, as required. > Notice that it's a rather short and direct argument, where if you > accept that 22 does not have 7 as a factor, it's obvious enough what > the constant terms of the factors must be as you go from 7, 7 and 22, > necessarily to 1, 1, and 22, when you divide P(x) by 49. > James Harris > http://mathforprofit.blogspot.com/ Start with g(x) = ( a(x) + f ) where we know nothing about a(x). If we know that f divides g(0) we can conclude that f divides a(0). However, as a(x) is just a general function we don't know if f divides a(1). Indeed, we can see immediately that f divides a(1) if and only if f divides g(1). More generally, f divides a(x) for all x, if and only if, f divides g(x) for all x. Lemma 1: Let P(x) be a polynomial divisible by p, and let g1(x) and g2(x) be polynomials such that P(x) = g1(x) * g2(x) Them p divides at least one of g1 and g2. Now suppose we have P(x) divisible by p, polynomials g1 and g2 such that P(x) = g1(x) * g2(x) p divides g1(0) and p does not divide g2(0). Then using Lemma 1 we can conclude that p divides g1(x) for all x. So can we do the following? P(x) = (a1(x)+p) (a2(x) +p) with a1(x)=0, a2(x)=1. Thus p divides (a1(0) +p) and p does not divide (a2(0) + p). We conclude that p divides (a1(x) +p) for all x. Thus p divides a1(x) for all x. No. The problem is that (a1(x)+p) and (a2(x) +p) are not polynomials. Lemma 1 is only true for polynomial factorization. Counterexample. Let P(x) = 15-3x. Then P(x) is divisible by 3. Let g1(x)=4-sqrt(1+3x) and g2(x)=4+sqrt(1+3x). Then we have a non-polynomial factorization 15-3x = (4-sqrt(1+3x)) (4+sqrt(1+3x)) list a few values P( 0) = 15, g1( 0) = 3, g2( 0) = 5 P( 1) = 12, g1( 1) = 2, g2( 1) = 6 P( 5) = 0, g1( 5) = 0. g2( 5) = 8 P( 8) = -9, g1( 8) = -1. g2( 8) = 9 P(16) = -33, g1(16) = -3. g2(16) = 11 Note how the factor of 3 switches back and forth between g1 and g2 depending on the value of x. Thus neither g1(x) nor g2(x) is divisible by 3 for all x. In particular the fact that g1(0) is divisible by 3 does not mean that g1(1) is divisible by 3. Application to your polynomial: You have the non-polynomial factorization (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) with a_1(0)=a_2(0)=b_3(0)=0 You conclude that 7 divides (5 a_1(0)+ 7) 7 divides (5 a_2(0) + 7) 22 divides (5 b_3(0) + 22) However because you have a non-polynomial factorization you cannot conclude that 7 divides (5 a_1(x)+ 7) 7 divides (5 a_2(x) + 7) 22 divides (5 b_3(x) + 22) for all x. And thus you cannot conclude that 5a_1(x)/7 or 5a_2(x)/7 is an algebraic integer for all x. Thus your factorization of step 7 (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) is not only not the only possible factorization, but not even guaranteed to be a factorization into algebraic integers. -William Hughes === Subject: Re: Key Core Error Argument > The following proof steps through a rather basic argument which is key > in proving an over one hundred year old error resulting from > previously unexpected consequences resulting from the definition of > the ring of algebraic integers. > Note that ultimately the proof relies on 22 NOT having 7 as a factor, > and constant terms like 7 and 22, being constant, and not variables > dependent on x, which may seem like odd things to emphasize, but I've > faced posters who've gotten away with challenging those truths because > people seem unaware that's what they're doing. > 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is > in the ring of algebraic integers, notice that P(x) has a constant > term that is 1078. > 2. It can be shown that > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > where you should note that using v = -1 + 49x, gives > P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 > where the *same* polynomial has been put in a form which allows a > factorization into non-polynomial factors so that I have > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > where the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > 3. Now let x=0, so > P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) > as the cubic defining the a's at x=0 is > a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and > a_2(0) to equal 0, which leaves a_3(0) with a value of 3. > 4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I > have > P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) > P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). > 5. Now P(x) has a factor of 49 as > P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 > which means that > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) > has a factor of 49. > 6. However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. > But for two of the factors of P(x), the constant terms is 7, which is > NOT a factor of 22. Therefore, *none* of the constant terms of > P(x)/49 as they multiply to give 22 can have 7 as a factor. > (By saying that 7 is NOT a factor of 22, I'm making a choice as to > where the proof is going. Since I've been talking about algebraic > integers, where 7 is NOT a factor of 22, it's natural to go with a > choice where 7 is NOT a factor of 22.) > Given that the constant terms are independent of x's value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7. You are still making the mistake of thinking the constant terms of a factor determine the divisibility of the factor. This has been repeatedly demonstrated to be incorrect. Your conclusion in step 6 is wrong. > 7. But to divide 7 from those constant terms requires dividing > through two of the factors, so > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > from reverse use of the distributive property, which gives constant > terms that don't have 7 as a factor, as required. > Notice that it's a rather short and direct argument, where if you > accept that 22 does not have 7 as a factor, it's obvious enough what > the constant terms of the factors must be as you go from 7, 7 and 22, > necessarily to 1, 1, and 22, when you divide P(x) by 49. > James Harris > http://mathforprofit.blogspot.com/ -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Key Core Error Argument > The following proof steps through a rather basic argument which is key > in proving an over one hundred year old error resulting from > previously unexpected consequences resulting from the definition of > the ring of algebraic integers. > Note that ultimately the proof relies on 22 NOT having 7 as a factor, > and constant terms like 7 and 22, being constant, and not variables > dependent on x, which may seem like odd things to emphasize, but I've > faced posters who've gotten away with challenging those truths because > people seem unaware that's what they're doing. > 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is > in the ring of algebraic integers, notice that P(x) has a constant > term that is 1078. > 2. It can be shown that > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > where you should note that using v = -1 + 49x, gives > P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 > where the *same* polynomial has been put in a form which allows a > factorization into non-polynomial factors so that I have > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > where the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). I have previously pointed out that, when x = 1, *another* solution for the a's is a1 = 1, a2 = 1562, a3 = 28. This set of numbers satisfies all your requirements for x = 1, because P(1) = (5*1 + 7)*(5*1562 + 7)*(5*28 + 7) = 13789188. Maybe you should explain why you don't want this solution, but instead prefer the solutions which satisfy a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > 3. Now let x=0, so > P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) > as the cubic defining the a's at x=0 is > a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and > a_2(0) to equal 0, which leaves a_3(0) with a value of 3. > 4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I > have > P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) > P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). > 5. Now P(x) has a factor of 49 as > P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 > which means that > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) > has a factor of 49. > 6. However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. Yes. This is OK. > But for two of the factors of P(x), the constant terms is 7, which is > NOT a factor of 22. Therefore, *none* of the constant terms of > P(x)/49 as they multiply to give 22 can have 7 as a factor. Also true. > (By saying that 7 is NOT a factor of 22, I'm making a choice as to > where the proof is going. Since I've been talking about algebraic > integers, where 7 is NOT a factor of 22, it's natural to go with a > choice where 7 is NOT a factor of 22.) > Given that the constant terms are independent of x's value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7. No, this is your central error. You factor your function P(x) into three parts, g1, g2, and g3. Each part is a function of x: thus g1(x), g2(x), g3(x), where g1(x)*g2(x)*g3(x) = P(x). Let's focus on g3(x). It is g3(x) = 5*b3(x) + 22. The constant term is indeed 22, and 22 is indeed coprime to 7. Now suppose I pick a particular value of x. Say, x = 1. I have g3(1) = 5*b3(1) + 22. You say 22 is coprime to 7. I accept that. We all do. But why would that imply that 5*b3(1) + 22 is also coprime to 7? After all, you can have two numbers, each of which is coprime to 7, but their sum is NOT coprime to 7. For example, 20 + 22. Right? How does 22 being coprime to 7 prove that 5*b3(1) + 22 is also coprime to 7 ? Give me a straightforward, honest answer to this simple question, and I will start thinking you have something. Nora B. > 7. But to divide 7 from those constant terms requires dividing > through two of the factors, so > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > from reverse use of the distributive property, which gives constant > terms that don't have 7 as a factor, as required. > Notice that it's a rather short and direct argument, where if you > accept that 22 does not have 7 as a factor, it's obvious enough what > the constant terms of the factors must be as you go from 7, 7 and 22, > necessarily to 1, 1, and 22, when you divide P(x) by 49. > James Harris > http://mathforprofit.blogspot.com/ === Subject: Re: Key Core Error Argument Your proof is hard to follow. Is this a good rewrite of your proof? 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is in the ring of algebraic integers. 2. P(0) = 1078 3. P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 4. Let v = -1 + 49x, then P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 5. Let F(a(x)) = a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) 6. Let the roots of F(a(x)) be a_1(x), a_2(x) and a_3(x). 7. P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) 8. P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) = 1078 9. F(a(0)) = a^3 - 3a^2, which has roots 0, 0, and 3. 10. Let a_1(0) = 0, a_2(0) = 0, and a_3(0) = 3. 11. Let a_3(x) = b_3(x) + 3 12. P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) 13. P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). 14. P(x) has a factor of 49 as P(x)/49 = 300125 x^3 - 18375 x^2 - 360x + 22 15. Then (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) has a factor of 49. 16. The constant term of P(x)/49 is P(0)/49 = 22. 17. But for two of the factors of P(x), the constant terms is 7, which is NOT a factor of 22. Therefore, *none* of the constant terms of P(x)/49 as they multiply to give 22 can have 7 as a factor. Note: (By saying that 7 is NOT a factor of 22, I'm making a choice as to where the proof is going. Since I've been talking about algebraic integers, where 7 is NOT a factor of 22, it's natural to go with a choice where 7 is NOT a factor of 22.) 18. Given that the constant terms are independent of x's value, it must be the case that dividing P(x) by 49 divides the two constant terms equal to 7, by 7. 19. But to divide 7 from those constant terms requires dividing through two of the factors, so (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 from reverse use of the distributive property, which gives constant terms that don't have 7 as a factor, as required. 20. Notice that it's a rather short and direct argument, where if you accept that 22 does not have 7 as a factor, it's obvious enough what the constant terms of the factors must be as you go from 7, 7 and 22, necessarily to 1, 1, and 22, when you divide P(x) by 49. > The following proof steps through a rather basic argument which is key > in proving an over one hundred year old error resulting from > previously unexpected consequences resulting from the definition of > the ring of algebraic integers. > Note that ultimately the proof relies on 22 NOT having 7 as a factor, > and constant terms like 7 and 22, being constant, and not variables > dependent on x, which may seem like odd things to emphasize, but I've > faced posters who've gotten away with challenging those truths because > people seem unaware that's what they're doing. > 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is > in the ring of algebraic integers, notice that P(x) has a constant > term that is 1078. > 2. It can be shown that > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > where you should note that using v = -1 + 49x, gives > P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 > where the *same* polynomial has been put in a form which allows a > factorization into non-polynomial factors so that I have > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > where the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > 3. Now let x=0, so > P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) > as the cubic defining the a's at x=0 is > a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and > a_2(0) to equal 0, which leaves a_3(0) with a value of 3. > 4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I > have > P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) > P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). > 5. Now P(x) has a factor of 49 as > P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 > which means that > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) > has a factor of 49. > 6. However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. > But for two of the factors of P(x), the constant terms is 7, which is > NOT a factor of 22. Therefore, *none* of the constant terms of > P(x)/49 as they multiply to give 22 can have 7 as a factor. > (By saying that 7 is NOT a factor of 22, I'm making a choice as to > where the proof is going. Since I've been talking about algebraic > integers, where 7 is NOT a factor of 22, it's natural to go with a > choice where 7 is NOT a factor of 22.) > Given that the constant terms are independent of x's value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7. > 7. But to divide 7 from those constant terms requires dividing > through two of the factors, so > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > from reverse use of the distributive property, which gives constant > terms that don't have 7 as a factor, as required. > Notice that it's a rather short and direct argument, where if you > accept that 22 does not have 7 as a factor, it's obvious enough what > the constant terms of the factors must be as you go from 7, 7 and 22, > necessarily to 1, 1, and 22, when you divide P(x) by 49. > James Harris > http://mathforprofit.blogspot.com/ === Subject: Re: Key Core Error Argument [snip umpteenth repeat of previous flawed argument] I thought you were going to just post your proof elsewhere and stop cluttering up this newsgroup. You've given a URL, now go away. -- 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: Key Core Error Argument > [snip umpteenth repeat of previous flawed argument] > I thought you were going to just post your proof elsewhere and stop > cluttering up this newsgroup. You've given a URL, now go away. Such a remarkably silly posting. What's the definition of insanity again? How many YEARS have I posted, and how many posters through the years have ordered me to go away? Yet here you are C. Bond doing the same silly thing, as if it matters. You're funny though in your irrationality. > -- > There are two things you must never attempt to prove: the unprovable -- > and the obvious. > -- > Democracy: The triumph of popularity over principle. But you see there's talk and there's substance. My proof is substance, and in fact, there is no error in it. If you disagree, then point out an error in what follows. The following proof steps through a rather basic argument which is key in proving an over one hundred year old error resulting from previously unexpected consequences resulting from the definition of the ring of algebraic integers. Note that ultimately the proof relies on 22 NOT having 7 as a factor, and constant terms like 7 and 22, being constant, and not variables dependent on x, which may seem like odd things to emphasize, but I've faced posters who've gotten away with challenging those truths because people seem unaware that's what they're doing. 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is in the ring of algebraic integers, notice that P(x) has a constant term that is 1078. 2. It can be shown that P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 where you should note that using v = -1 + 49x, gives P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 where the *same* polynomial has been put in a form which allows a factorization into non-polynomial factors so that I have P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) where the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). 3. Now let x=0, so P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic defining the a's at x=0 is a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and a_2(0) to equal 0, which leaves a_3(0) with a value of 3. 4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I have P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). 5. Now P(x) has a factor of 49 as P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 which means that (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) has a factor of 49. 6. However, the constant term of P(x)/49 is 22, which is verified by again setting x=0, which gives P(0)/49 = 22. But for two of the factors of P(x), the constant terms is 7, which is NOT a factor of 22. Therefore, *none* of the constant terms of P(x)/49 as they multiply to give 22 can have 7 as a factor. (By saying that 7 is NOT a factor of 22, I'm making a choice as to where the proof is going. Since I've been talking about algebraic integers, where 7 is NOT a factor of 22, it's natural to go with a choice where 7 is NOT a factor of 22.) Given that the constant terms are independent of x's value, it must be the case that dividing P(x) by 49 divides the two constant terms equal to 7, by 7. 7. But to divide 7 from those constant terms requires dividing through two of the factors, so (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 from reverse use of the distributive property, which gives constant terms that don't have 7 as a factor, as required. Notice that it's a rather short and direct argument, where if you accept that 22 does not have 7 as a factor, it's obvious enough what the constant terms of the factors must be as you go from 7, 7 and 22, necessarily to 1, 1, and 22, when you divide P(x) by 49. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: Key Core Error Argument > [snip umpteenth repeat of previous flawed argument] > I thought you were going to just post your proof elsewhere and stop > cluttering up this newsgroup. You've given a URL, now go away. > Such a remarkably silly posting. Oh? And I suppose incessantly repeating the same flawed argument is *not* silly? [snip umpteenth+1 repeat of same flawed argument] When you finally get back to your sandbox, don't be surprised if the other kids have grown up and moved on to more productive lives. There is still the possibility that you can secure a job as a crash dummy, however. -- 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: Key Core Error Argument > > [snip umpteenth repeat of previous flawed argument] > > I thought you were going to just post your proof elsewhere and stop > cluttering up this newsgroup. You've given a URL, now go away. > Such a remarkably silly posting. Yes! After all our experience of JSH's behavior, expecting him to do anything rqtions is quite silly. > What's the definition of insanity again? It is synonymous with JSH. You cannot reason with JSH, as he is immune to it. > How many YEARS have I posted, and how many posters through the years > have ordered me to go away? > Yet here you are C. Bond doing the same silly thing, as if it > matters. Certainly what JSH posts here doesn't matter, except as an irritant. This continued irritation by JSH has created some pearls, but only in refutation of his inanity. > But you see there's talk and there's substance. My proof is > substance, and in fact, there is no error in it. The only substance in JSH's proofs is one not referred to in polite society. [snipped the remaining oongah] === Subject: Re: [JSH: Nonsense Continues: Key Core Error Argument > [snip umpteenth repeat of previous flawed argument] > I thought you were going to just post your proof elsewhere and stop > cluttering up this newsgroup. You've given a URL, now go away. > Such a remarkably silly posting. > What's the definition of insanity again? > How many YEARS have I posted, and how many posters through the years > have ordered me to go away? > Yet here you are C. Bond doing the same silly thing, as if it > matters. > You're funny though in your irrationality. > -- > There are two things you must never attempt to prove: the unprovable -- > and the obvious. > -- > Democracy: The triumph of popularity over principle. > But you see there's talk and there's substance. My proof is > substance, and in fact, there is no error in it. Of the 6 billion plus people on the Earth, only you are capable of determining just how great and beyond reproach your arguments are. Listen to yourself, you irrational megalomaniac! Your proof? You can't even understand terminology let alone errors that are repeatedly brought to your attention and you have the gall to say that your proof is substance. Perhaps I agree that it is substance, but it is the sort of substance that most use to fertilize fields, but yours stinks ten times worse! You DO NOT have a proof and wouldn't know proof if you hit in the ass and knocked sense into your brains (and I think they are in your ass). Take your medication! === Subject: Re: Key Core Error Argument > The following proof steps through a rather basic argument which is key > in proving an over one hundred year old error resulting from > previously unexpected consequences resulting from the definition of > the ring of algebraic integers. > Note that ultimately the proof relies on 22 NOT having 7 as a factor, > and constant terms like 7 and 22, being constant, and not variables > dependent on x, which may seem like odd things to emphasize, but I've > faced posters who've gotten away with challenging those truths because > people seem unaware that's what they're doing. > 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is > in the ring of algebraic integers, notice that P(x) has a constant > term that is 1078. > 2. It can be shown that > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > where you should note that using v = -1 + 49x, gives > P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 > where the *same* polynomial has been put in a form which allows a > factorization into non-polynomial factors so that I have > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > where the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > 3. Now let x=0, so > P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) > as the cubic defining the a's at x=0 is > a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and > a_2(0) to equal 0, which leaves a_3(0) with a value of 3. > 4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I > have > P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) > P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). > 5. Now P(x) has a factor of 49 as > P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 > which means that > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) > has a factor of 49. > 6. However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. > But for two of the factors of P(x), the constant terms is 7, which is > NOT a factor of 22. Therefore, *none* of the constant terms of > P(x)/49 as they multiply to give 22 can have 7 as a factor. > (By saying that 7 is NOT a factor of 22, I'm making a choice as to > where the proof is going. Since I've been talking about algebraic > integers, where 7 is NOT a factor of 22, it's natural to go with a > choice where 7 is NOT a factor of 22.) > Given that the constant terms are independent of x's value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7. > 7. But to divide 7 from those constant terms requires dividing > through two of the factors, so > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > from reverse use of the distributive property, which gives constant > terms that don't have 7 as a factor, as required. Define three functions w1(x), w2(x), w3(x), such that w1(x).w2(x).w3(x) = 49 for all x, and w1(0) = w2(0) = 7, w3(0) = 1. Now (5 a1(x)/w1(x)+7/w1(x))(5 a2(x)/w2(x)+7/w2(x))(5 b3(x)/w3(x)+22/w3(x)) = 300125 x^3 - 10375 x^2 - 360 x + 22 So why is that *not* a possibility? If I understand your terminology correctly (you still have *not* defined the concept constant term as you use it), the constant terms of the three factors are 1, 1 and 22. -- 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: Key Core Error Argument > Define three functions w1(x), w2(x), w3(x), such that w1(x).w2(x).w3(x) = 49 > for all x, and w1(0) = w2(0) = 7, w3(0) = 1. Now > (5 a1(x)/w1(x)+7/w1(x))(5 a2(x)/w2(x)+7/w2(x))(5 b3(x)/w3(x)+22/w3(x)) = > 300125 x^3 - 10375 x^2 - 360 x + 22 > So why is that *not* a possibility? If I understand your terminology > correctly (you still have *not* defined the concept constant term > as you use it), the constant terms of the three factors are 1, 1 and 22. Let's say you have f(x)/g(x) in the ring of algebraic integers. That is, both f(x), and g(x) give algebraic integers for algebraic integer x. Then necessarily, you have *another* algebraic integer function I'll call h(x). Understand? Now then, if you have h(x), then you can set x=0 to get the constant term. Otherwise you'd have a backdoor to making numbers like 7, that are constants, into variables, which would be mathematically inconsistent. So if you have 7/w_1(x), and it gives algebraic integers for any algebraic integer x, then you have some function h(x) = 7/w_1(x), which STILL has to have a constant term of 1, found by checking at x=0. It's a simple but powerful technique Dik Winter, and no matter how much you try, you can't find a way to invalidate its results or make them inconsistent. So despite all your efforts Dik Winter, you end up with the same result: constant terms for the factors that are 1, 1, and 22. Apparently, you think writing the functions as ratios changes things, but *if* they're in the ring of algebraic integers, you can flatten them out. For other readers who don't understand what my original post is, it's a math proof. Yup, an actual real math proof, which means that it begins with a truth and proceeds by logical steps to a conclusion which MUST be true. Attacking a math proof, as you may know, is silly, as ultimately it relies on rather basic facts, which are simple enough that everyone accepts them, and in my case the proof relies on some basic algebra, the constancy of 7 and 22, and 22 not having 7 as a factor. I mention that because when it comes to knowing that you have found a proof, simplicity rules. Now if any of you have decided that algebra might be bunk, or if you think that 7 and 22 are variables, or wish to claim that 7 is a factor of 22, in the appropriate rings, then that's another matter, of course, as in you're no longer a part of the intellectual community, but are then a crank. James Harris === Subject: Re: Key Core Error Argument > Define three functions w1(x), w2(x), w3(x), such that w1(x).w2(x).w3(x) = 49 > for all x, and w1(0) = w2(0) = 7, w3(0) = 1. Now > > (5 a1(x)/w1(x)+7/w1(x))(5 a2(x)/w2(x)+7/w2(x))(5 b3(x)/w3(x)+22/w3(x)) = > 300125 x^3 - 10375 x^2 - 360 x + 22 > So why is that *not* a possibility? If I understand your terminology > correctly (you still have *not* defined the concept constant term > as you use it), the constant terms of the three factors are 1, 1 and 22. > Let's say you have f(x)/g(x) in the ring of algebraic integers. That > is, both f(x), and g(x) give algebraic integers for algebraic integer > x. Then necessarily, you have *another* algebraic integer function > I'll call h(x). > Understand? So, h(x) = f(x)/g(x)? > Now then, if you have h(x), then you can set x=0 to get the constant > term. Ok. So you define the constant term of a function as the value of the function when the argument is 0. True? That is what I expected. 1. f(x) = (5 a1(x) + 7), g(x) = w1(x), h(x) = f(x)/g(x). h(0) = f(0)/g(0) = 7/7 = 1 2. f(x) = (5 a2(x) + 7), g(x) = w2(x), h(x) = f(x)/g(x). h(0) = f(0)/g(0) = 7/7 = 1 3. f(x) = (5 b3(x) + 22), g(x) = w3(x), h(x) = f(x)/g(x). h(0) = f(0)/g(0) = 22/1 = 22 So the constant terms of the functions work out as 1, 1 and 22. This is even independent of whether the h's are algebraic integers or not. > Otherwise you'd have a backdoor to making numbers like 7, that are > constants, into variables, which would be mathematically inconsistent. > So if you have 7/w_1(x), and it gives algebraic integers for any > algebraic integer x, then you have some function h(x) = 7/w_1(x), > which STILL has to have a constant term of 1, found by checking at > x=0. Note that I maybe do not have that 7/w1(x) is an algebraic integer, I do have that (5 a1(x)/w1(x) + 7/w1(x)) is an algebraic integer, and as shown above, the constant term of it is 1. But in the actual case w1(x) is a divisor of 7 in the algebraic integers. On the other hand, w3(x) is *not* a divisor of 22 in general, but (5 b3(x)/w3(x) + 22/w3(x)), *is* an algebraic integer, although 22/w3(x) is not. But what you tell is true, but trivial. It only means that w1(0) = 7. > It's a simple but powerful technique Dik Winter, and no matter > how much you try, you can't find a way to invalidate its results or > make them inconsistent. Not so very powerful. A lot of words to say that w1(0) must be 7. > So despite all your efforts Dik Winter, you end up with the same > result: constant terms for the factors that are 1, 1, and 22. Yup, just as I suspected, also when you use functions w1(x), w2(x) and w3(x) with the properties I give above. > Apparently, you think writing the functions as ratios changes things, > but *if* they're in the ring of algebraic integers, you can flatten > them out. The meaning of this sentence escapes me completely. So, *why* can 49 not be the product of three functions w1(x), w2(x) and w3(x) where each of the w's gives an algebraic integer when x is an integer (or probably even an algebraic integer), such that also for every x, (5a1(x) + 7)/w1(x), (5a2(x) + 7)/w2(x) and (5b3(x) + 22)/w3(x) are also algebraic integers? The constant terms for these three are 1, 1 and 22, as shown above, so that does *not* violate your principle. And as one of my polynomial example shows (in that case it was easy) the three w's acted as follows (for x = 0, 1 and 2): x = 0: w1(0) = w2(0) = 7, w3(0) = 1 x = 1: w1(1) = w2(1) = w3(1) = cbrt(49) x = 2: w1(2) = 7, w2(2) = [-sqrt(5)+3i]/sqrt(2), w3(2) = [-sqrt(5)-3i]/sqrt(2). Here also the constant terms remained 1, 1 and 22. It has been shown that for your polynomial there are also functions w1 to w3 that act similarly. They are pretty wild functions, yes, but what is the problem with that? We can actually define them (as we can use gcd in the algebraic integers, we can give it a sensible meaning there): w1(x) = gcd( (5 a1(x) + 7), 49) w2(x) = gcd( (5 a2(x) + 7), 49) w3(x) = gcd( (5 b3(x) + 22), 49) When x (which is an integer in your case) is not equal to 7y + 5 (for any integer y), the above equations define the functions w. The product is 49 upto unit factors (as the remaining polynomial is coprime to 7 when x is not equal to 7y + 5). If x equals 7y + 5 for some integer y, we have to adjust the values, as in that case the polynomial is not only divisible by 49, but also by 343, so some of the w's may have too many factors of 7. So we have to adjust some of the w's. But that is trivial. See also that w1(0) = 7, w2(0) = 7 and w3(0) = 1, as required (the constant terms are 1, 1 and 22). Yup, the gcd is a very wild function. Try it in the integers with gcd(x, 210). Expect it to go wilder in the algebraic integers, as there every number has an unlimited amount of factors... > Now if any of you have decided that algebra might be bunk, or if you > think that 7 and 22 are variables, or wish to claim that 7 is a factor > of 22, in the appropriate rings, then that's another matter, of > course, as in you're no longer a part of the intellectual community, > but are then a crank. You are only claiming that other people are claiming this... However, in the ring of rational numbers, 7 *is* a factor of 22... Also in the ring of integers mod 29, 7 *is* a factor of 22 (indeed, 7 = -22)... So there are rings where 7 is a factor of 22. -- 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: Key Core Error Argument >>Define three functions w1(x), w2(x), w3(x), such that w1(x).w2(x).w3(x) = 49 >>for all x, and w1(0) = w2(0) = 7, w3(0) = 1. Now >> (5 a1(x)/w1(x)+7/w1(x))(5 a2(x)/w2(x)+7/w2(x))(5 b3(x)/w3(x)+22/w3(x)) = >> 300125 x^3 - 10375 x^2 - 360 x + 22 >>So why is that *not* a possibility? If I understand your terminology >>correctly (you still have *not* defined the concept constant term >>as you use it), the constant terms of the three factors are 1, 1 and 22. > Let's say you have f(x)/g(x) in the ring of algebraic integers. That > is, both f(x), and g(x) give algebraic integers for algebraic integer > x. Then necessarily, you have *another* algebraic integer function > I'll call h(x). > Understand? > Now then, if you have h(x), then you can set x=0 to get the constant > term. > Otherwise you'd have a backdoor to making numbers like 7, that are > constants, into variables, which would be mathematically inconsistent. > So if you have 7/w_1(x), and it gives algebraic integers for any > algebraic integer x, then you have some function h(x) = 7/w_1(x), > which STILL has to have a constant term of 1, found by checking at > x=0. It's a simple but powerful technique Dik Winter, and no matter > how much you try, you can't find a way to invalidate its results or > make them inconsistent. All this means is that w_1(0) = 7. This would satisfy your observations of what happens at x=0. But that doesn't guarantee that w_1(1) = 7. In fact, it's not. > So despite all your efforts Dik Winter, you end up with the same > result: constant terms for the factors that are 1, 1, and 22. But we also get *more*: we get a result different from yours that doesn't take us out of the algebraic integers, despite what your result claims must happen. > Apparently, you think writing the functions as ratios changes things, > but *if* they're in the ring of algebraic integers, you can flatten > them out. Flatten them out? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Key Core Error Argument > Apparently, you think writing the functions as ratios changes things, > but *if* they're in the ring of algebraic integers, you can flatten > them out. I don't get it. I flunked flattening out. Can you explain? === Subject: teaching descriptive statistics / proposal When teaching descriptive statistics, it is usual to interpret the moments of a distribution in R as follows: moment of order 1 (mean) : location centered moment of order 2 (variance): dispersion reduced centered moment of order 3 (skewness): degree of asymmetry reduced centered moment of order 4 (kurtosis): degree of flatness Now let us look a bit more to the skewness (goes back to 1895). It is known that some non symmetric distributions have a null skewness. Nevertheless, owing to its simplicity, skewness is still used. Here comes my proposal: Replace the skewness by the chiral index Chi. Having a sample and a pocket calculator, the chiral index is simpler to compute than the skewness, and it is such that: Chi = 0 <=> symmetric distribution (i.e. indirect symmetry) The chiral index exists also for multivariate distributions. (http://www.mdpi.net/entropy ; This journal is free for readers) Sections 2.9 and 4.2; chiral index for univariate distributions: equations 2.9.3 to 2.9.5. Michel Petitjean, Email: petitjean@itodys.jussieu.fr ITODYS (CNRS, UMR 7086) ptitjean@ccr.jussieu.fr http://petitjeanmichel.free.fr/itoweb.petitjean.html === Subject: Asymtotic normal distribution Hello everybody, I'm working on a project and I want to prove the following. Let Ggr ~ BIN( m/2 , pgr) Gro ~ BIN( m/2 , pro) Zgr ~ BIN( n/2 , pgr) Zro ~ BIN( n/2 , pro) I've already proved that S is a unbiased estimator for the ratio n/m is: S = Sqrt{ (Zgr Zro)( Ggr Gro) } What I want to prove now is that LOG(S) ~ Normal distributed. I already have: LOG(S)=1/2LOG(Zgr) +1/2LOG(Zro) - 1/2LOG(Ggr) - 1/2LOG(Gro) = .. Can somebody help me with this prove? === Subject: Re: Asymtotic normal distribution >Hello everybody, >I'm working on a project and I want to prove the following. >Let >Ggr ~ BIN( m/2 , pgr) >Gro ~ BIN( m/2 , pro) >Zgr ~ BIN( n/2 , pgr) >Zro ~ BIN( n/2 , pro) >I've already proved that S is a unbiased estimator for the ratio n/m is: > S = Sqrt{ (Zgr Zro)( Ggr Gro) } > What I want to prove now is that LOG(S) ~ Normal distributed. >I already have: > LOG(S)=1/2LOG(Zgr) +1/2LOG(Zro) - 1/2LOG(Ggr) - 1/2LOG(Gro) = .. > Can somebody help me with this prove? I do not see how one can say that S (or LOG(S)) is unbiased, as the numerator or denominator in the square root can be 0. That said, the variable Ggr/(m/2) is asymptotically (as m goes to infinity) normally distributed with mean pgr and variance 2/(m*pgr*(1-pgr)), etc., and as LOG(S) is a differentiable function of its arguments, it is asymptotically (for m and n large) normally distributed with mean LOG(n/m) and variance given by linearizing the logarithms. This is the usual delta-method. -- 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: Differential Equation Hi there. Does the differential equation (d^2/dt^2-grad^2+m^2)f=0 for m real have a solution in the space of Schwartz functions? I don't believe so, but would feel much better, if one of you could gie me a definite answer. Benjamin Bahr === Subject: Re: Usenet Posting Guide? > I'm considering writing up a guide to posting on newsgroups. Why? Why would your guide be better than any of the very many guides available... > I wouldn't put in a lot of technical things, nor would I talk about > other news readers or means of posting besides Google Groups ...especially when your guide is obviously going to be so limited? > Or I could talk about the real Nettiquette, > what rules you can bend, Just out of interest, which 'rules' can you bend? > and which ones you not only can break but you *must* break and which 'rules' *must* you break? > if you're trying to break through. === Subject: Re: Usenet Posting Guide? > I'm considering writing up a guide to posting on newsgroups. > Why? Why would your guide be better than any of the very many guides > available... Actually there are several already in existence. I will have to check with my friend for the URL, but there is a good one. Unfortunately people take advantage of ignoring it too often. Norma > I wouldn't put in a lot of technical things, nor would I talk about > other news readers or means of posting besides Google Groups > ...especially when your guide is obviously going to be so limited? > Or I could talk about the real Nettiquette, > > what rules you can bend, > Just out of interest, which 'rules' can you bend? > and which ones you not only can break but you *must* break > and which 'rules' *must* you break? > if you're trying to break through. === Subject: Re: Usenet Posting Guide? > Sometime I miss the days when getting to USENET required configuring > UUCP, compiling, installing and configuring B News or C News, finding > and negotiating with an admin somewhere for a news feed... or, if you > were a student, getting an account from your university news admin. I am still using 'rn'. An improvement over 'readnews', but until now I have seen nothing better. (And, no, no graphical interfaces for me, please. I have problems when I use the mouse too much.) -- 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: Usenet Posting Guide? > I am still using 'rn'. An improvement over 'readnews', but until now I > have seen nothing better. (And, no, no graphical interfaces for me, > please. I have problems when I use the mouse too much.) I prefer 'tin' but remember 'rn' (and 'readnews') fondly and could use it if necessary. Graphical interfaces do have their uses; for instance, the X Window System lets me open multiple xterm windows to run command-line clients like tin (or rn). -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Usenet Posting Guide? >I prefer 'tin' but remember 'rn' (and 'readnews') fondly and could use it >if necessary. Graphical interfaces do have their uses; for instance, the >X Window System lets me open multiple xterm windows to run command-line >clients like tin (or rn). What benefit do you get by having two newsreader windows open at once? === Subject: Re: Solving definite integral problem - newbie question > I am studying for an exam and don't have access to my instructor > today. > Can someone help me with the following? I have a graph of the right > half of a bell curve, a portion of which I must solve. > The portion is from 0 to 1 (that is x=0 to x=1) and the equation is > y=e^(-x^2/2). > That is, e raised to (negative x squared, divided by two). > I just need some help in the proper approach. I have been given the > answer, but so far nothing that I have tried has helped me to get the > answer. > Pam Pam === Subject: Re: Cantor's diagonal number: why do we just ASSUME such a number exists? >I am of the firm belief that you and the author of this thread are one and >the same (and possibility just an alias of JSH, which would explain his >immaturity). Both of you sign your name in the same way and right your age >afterwards. Both of you refer to proofs but do not provide links to them >(Charlie, neither you nor Justin have websites with your name). You both >sign with different names than the email address states. Lastly, both of >you are wrong. You cannot state that the natural numbers are uncountable. >By definition, a countable set is that of the same cardinality as the >naturals. By the way, Nathan how have you been through University at age >11, and Charlie, how have you finished law school by age 9? You're thinking linearly. Nathan's age does not progress linearly. He's been 11 many times. Gets up to about 13 or 14 and recycles. - Randy === Subject: Re: Cantor's diagonal number: why do we just ASSUME such a number exists? A8EwTYfhf*u~,Eu,tf6$HN*MY&)u0G =N' x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|'1{w70ZY=ih,=)kMY_}?{%)x0)];K~@J6m5.EN?>Zh Xh;Y V|',x(js'Jfq02joVpj|#x linux) > > For anyone who hadn't noticed yet, this post contains ALL the 35 items of > the crackpot index IN ORDER. didn't realize it was a parody with method. -- Jesse Hughes I often told you of the dangers of hubris, and most importantly of all, I TOLD you that I wanted to change the institution of mathematics worldwide. -- James Harris, on the evils of pride === Subject: Re: Cantor's diagonal number: why do we just ASSUME such a number exists? > I am of the firm belief that you and the author of this thread are one and > the same (and possibility just an alias of JSH, which would explain his > immaturity). Both of you sign your name in the same way and right your age > afterwards. Both of you refer to proofs but do not provide links to them > (Charlie, neither you nor Justin have websites with your name). You both > sign with different names than the email address states. Lastly, both of > you are wrong. You cannot state that the natural numbers are uncountable. > By definition, a countable set is that of the same cardinality as the > naturals. By the way, Nathan how have you been through University at age > 11, and Charlie, how have you finished law school by age 9? > Steven Hey, I just assumed it was a big joke. Did you notice how the OP scored absolutely top marks in the crackpot test, by satisfying *all* criteria? Not a coincidence, me thinks.... -Michael. === Subject: Re: Cantor's diagonal number: why do we just ASSUME such a number exists? permission for an emailed response. > On my webpage I have posted a proof of the 1-1 mapping, N -> R. However, as > a corrolary to my proof, I demonstrate that N itself is uncountable. It's > there for anyone to view, and unless someone can find a flaw in my proof, we > will have to agree that it is correct. The definition of countable is that a set can be in a 1-1 mapping with N. A set is uncountable if it is infinite and not countable. Since the identity mapping is a 1-1 mapping from N->N, N is countable. Thomas === Subject: Re: Cantor's diagonal number: why do we just ASSUME such a number exists? > Mike Deeth says... >I have obtained empirical evidence that Cantor's diagonal number >does not exist, through a long thought experiment. > The diagonal number isn't just one number. You failed to see that the original post was an attempt to get into the Guinness book of records: highest crank index of all times. I have not actually counted the score, but it must be very high. -- 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: Cantor's diagonal number: why do we just ASSUME such a number exists? > Mike Deeth says... > >I have obtained empirical evidence that Cantor's diagonal number >does not exist, through a long thought experiment. > > The diagonal number isn't just one number. >You failed to see that the original post was an attempt to get into the >Guinness book of records: highest crank index of all times. I have >not actually counted the score, but it must be very high. Someone else said that, but I didn't know that he meant it literally. But yes, he goes right down the list of things not to do in the crackpot index. It's a significant accomplishment to be the world's foremost crackpot. -- Daryl McCullough Ithaca, NY === Subject: Re: Jaynes' book on probability - thoughts? As a matter of interest, didn't Jayne's book used to be available on-line? Is it still? Or am I mistaken? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Jaynes' book on probability - thoughts? >As a matter of interest, didn't Jayne's book used to be available on-line? >Is it still? Or am I mistaken? http://omega.albany.edu:8008/JaynesBook John Bailey http://home.rochester.rr.com/jbxroads/mailto.html === Subject: Re: Equation Object converting? > Probably can do so with a quick macro. > To what version of Word are you converting? That is MS Word 2002. === Subject: Re: Product of Reals >> Is it possible to determine what the uncountable product of all real >> numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1, >> but is this concept studied in general anywhere (e.g. the concept of >> uncountably infinite products or sums of real numbers)? >> Lookup the term summable family. This is a concept which allows definition >> of convergence for any numbers a_i with i in I where I is an arbitrary >> index set. By applying exp-log-formalism (transform your product into a >> sum) you get the analogue theory for products. >> A necessary condition for such a family to converge is that its support (the >> largest set J subset I such that a_j <> 0 forall j in J) is countable - >> this means that your product cannot converge. >> Because equally well than your reasoning I could argue that it converges to >> any other real number <> 1. Cf. Riemann's rearrangement theorem. > This is something that I have wondered about. Occasionally in > previous threads the idea of summing an uncountable set of numbers has > come up. It is usually commented this is not possible unless all but > a countable number are zero. However my reaction has always been: is > there any definition for infinite sums of any sort other than the > common series. Suppose {a_i} is a family of real numbers, where the subscripts i range over some (possibly uncountable) index set I. If S is a finite subset of I, we can consider the finite sum A_S = sum_{s in S} a_s. We say the family {a_i} for i in I is summable (and the sum = L) if, for every epsilon > 0, there exists a finite subset S of I (depending on epsilon) such that | A_S' - L | < epsilon for every finite set S' such that S subseteq S' subseteq I. This can be stated more concisely by saying that the net of finite sums converges to L. > I wondered about summing the values of a function f from a set to R or > C. I wondered what sort of structure the set would have to have and > what class of functions could be handled. Even when the set is > countable, the answer is not obvious since the sequence can affect the > result. I mentioned in a recent thread, the distinction between > absolutely and conditionally convergent series. In the definition above there is no structure imposed on the index set I. It does not even have an order. It does turn out that with this definition, a summable family must necessarily be zero except for an countable number of terms. If uncountably many terms are nonzero, then for some n we must have | a_i | > 1/n for uncountably many i, implying that the sum diverges. > This is the first reference that I have noticed to a generalisation of > infinite sums. Unfortunately I could not find much on Summable > families on the net. A search of the groups revealed little but this > thread. A search of the web found more but many seemed to mention it > only in passing or were in formats that I could not read. Mathworld > did not seem to have anything. > Can you point me somewhere or give a bit more detail? > For non-countable sets, could measure theory be considered a > generalisation of these sums? Yes. In fact, counting measure is a particular type of measure, and the net of finite sums approach amounts to integration with respect to counting measure. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Naive Q: Set theory, logic - which comes first? ....................... >> Natural numbers pose a special problem since they may be viewed in two >> different ways. We may view them either as something defined by >> induction, or as the smallest algebraic structure which is closed >> under the operations 0, 1 , +, and * (theese being subject to the >> usual laws). >> Isn't the field of two elements a smaller algebraic structure which is >> closed under the operations 0, 1 , +, and * (theese (sic) being subject to >> the usual laws) than the natural numbers. >You are right! Of course, Z_2 is the smallest such structure. One >needs to add some axiom that forces there to be infinitely many >elements. This is not enough. The algebraic closure of Z_p has this property, and it is not a smallest. The closure of Z_p under solutions of quadratics is (I believe) one of the smallest, although this takes proof. I do not know how to define characteristic 0 without first having the integers. Alternatively, one could add < to the list of primitives; this will get the natural numbers, even without *. -- 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: Naive Q: Set theory, logic - which comes first? ..................... > The problem is exclusively related to the phrase defined by induction. > just as it has been for the last several thoushand years. > Since you have to know something about geometry rather than > natural numbers to define inductive logic. This is totally false. In fact, one can do geometry without using inductive logic at any point, and a key part of the Peano Postulates is the induction postulate. BTW, inductive logic is usually used for inference from observations, and as such, has nothing to do with mathematical induction. Geometry is unnecessary for this, but properties of the real numbers as such is of great importance. -- 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: Naive Q: Set theory, logic - which comes first? >Two questions whose answers I am seeking are: >1) What exactly might we mean by representation? >2) How do we define a mathematical object if we do not want to >identify it with some representation of it? > This is done quite often. The Peano Postulates CHARACTERIZE > the positive (or non-negative) integers. There are many other > ways to characterize them. So we say that 0 is a natural number, that the successor operation takes us from a natural number to another natural number, and then list properties that natural numbers satisfy. But what we say is consistent with their being sets, functions in lamda calculus, or something else. We need to exclude such interpretations, but how? Mattias === Subject: Re: Naive Q: Set theory, logic - which comes first? >>Two questions whose answers I am seeking are: >>1) What exactly might we mean by representation? >>2) How do we define a mathematical object if we do not want to >>identify it with some representation of it? >> This is done quite often. The Peano Postulates CHARACTERIZE >> the positive (or non-negative) integers. There are many other >> ways to characterize them. >So we say that 0 is a natural number, that the successor operation >takes us from a natural number to another natural number, and then >list properties that natural numbers satisfy. But what we say is >consistent with their being sets, functions in lamda calculus, or >something else. We need to exclude such interpretations, but how? Why? We only characterize what it means for a particular object to be considered as a model for the integers. It is a mistake to attempt to say what an integer is. -- 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