mm-2059 === Subject: Re: Jacobson's theorem! > And sorry lastly I found another theorem which is Wedderburn's > famous theorem about finite division rings being commutative, it > says that it has many applications and extensions but only lists one > of projective geometry. Could anyone point me in the way of any other > applications of this theorem. If you have a simple representation of a finite group over a finite field then the endomorphism ring is a division algebra, hence a finite field. Ps I didn't understand your formulation of Jacobson's theorem. -- 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: Complex Question #2 >infinite product [(n^3-z)/(n^3+z)] >from n = 1 to infinity >defines a bounded analytic function on G = {z : Re(z) > 0} vanishing on {n^3 >: n = 1,2,3,..} and nowhere else on G. >Well, I have done everything except prove it vanishes on {n^3 : n = >1,2,3,...}. Is there anything difficult here? It is obvious that it >vanishes if and only if one of the terms [one of the _factors_] >vanish, which is if and only if >n^3 - z vanishes for some n, which is if and only if z = n^3 for n = >1,2,3,.. >Is there anything I'm missing here? Hard to say whether you're missing anything since you don't tell us how you proved the things you proved. It seems possible you're missing something, because it's not that obvious that it vanishes if and only if one of the factors vanishes, that really does need to be proved. (Consider g(z) = product_1^infinity (1/2). That vanishes everywhere, even though none of the factors has a zero anywhere. How do you know that that sort of thing doesn't happen here? Also how do you know the product converges, and that it converges to something analytic? That's the part where we can't tell whether you're missing something...) >Tony ************************ David C. Ullrich === Subject: Re: Complex Question #2 >>infinite product [(n^3-z)/(n^3+z)] >>from n = 1 to infinity >>defines a bounded analytic function on G = {z : Re(z) > 0} vanishing on >>{n^3 >>: n = 1,2,3,..} and nowhere else on G. >>Well, I have done everything except prove it vanishes on {n^3 : n = >>1,2,3,...}. Is there anything difficult here? It is obvious that it >>vanishes if and only if one of the terms > [one of the _factors_] >>vanish, which is if and only if >>n^3 - z vanishes for some n, which is if and only if z = n^3 for n = >>1,2,3,.. >>Is there anything I'm missing here? > Hard to say whether you're missing anything since you don't tell > us how you proved the things you proved. It seems possible you're > missing something, because it's not that obvious that it vanishes > if and only if one of the factors vanishes, that really does need > to be proved. > (Consider g(z) = product_1^infinity (1/2). That vanishes everywhere, > even though none of the factors has a zero anywhere. How do you > know that that sort of thing doesn't happen here? > Also how do you know the product converges, and that it converges > to something analytic? That's the part where we can't tell whether > you're missing something...) I have shown that the sum (from 1 to infinity) of [(n^3-z)/(n^3+z) - 1] converges absolutely and uniformly on compact subsets of the right half plane, and thus it converges in H(G) where G = {z : Re(z) > 0}. I see your point about the infinite producct of 1/2. How do I know that sort of thing doesn't happen here? Certainly it vanishes on z = n^3 n =1,2,3,... But anywhere else? >>Tony > ************************ > David C. Ullrich === Subject: Re: Complex Question #2 >infinite product [(n^3-z)/(n^3+z)] >from n = 1 to infinity >defines a bounded analytic function on G = {z : Re(z) > 0} vanishing on >{n^3 >: n = 1,2,3,..} and nowhere else on G. >Well, I have done everything except prove it vanishes on {n^3 : n = >1,2,3,...}. Is there anything difficult here? It is obvious that it >vanishes if and only if one of the terms >> [one of the _factors_] >vanish, which is if and only if >n^3 - z vanishes for some n, which is if and only if z = n^3 for n = >1,2,3,.. >Is there anything I'm missing here? >> Hard to say whether you're missing anything since you don't tell >> us how you proved the things you proved. It seems possible you're >> missing something, because it's not that obvious that it vanishes >> if and only if one of the factors vanishes, that really does need >> to be proved. >> (Consider g(z) = product_1^infinity (1/2). That vanishes everywhere, >> even though none of the factors has a zero anywhere. How do you >> know that that sort of thing doesn't happen here? >> Also how do you know the product converges, and that it converges >> to something analytic? That's the part where we can't tell whether >> you're missing something...) >I have shown that the sum (from 1 to infinity) of [(n^3-z)/(n^3+z) - 1] >converges absolutely and uniformly on compact subsets of the right half >plane, and thus >it converges in H(G) where G = {z : Re(z) > 0}. Right, because a theorem says that this implies that. Now the same theorem says (or _should_ say) that it follows that the product cannot vanish except where one of the factors vanishes - is that not included in whatever book you're using? >I see your point about the >infinite producct of 1/2. >How do I know that sort of thing doesn't happen here? Certainly it vanishes >on z = n^3 n =1,2,3,... >But anywhere else? >Tony >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: Re: Complex Question #2 >>infinite product [(n^3-z)/(n^3+z)] >>from n = 1 to infinity >>defines a bounded analytic function on G = {z : Re(z) > 0} vanishing on >>{n^3 >>: n = 1,2,3,..} and nowhere else on G. >>Well, I have done everything except prove it vanishes on {n^3 : n = >>1,2,3,...}. Is there anything difficult here? It is obvious that it >>vanishes if and only if one of the terms > [one of the _factors_] >>vanish, which is if and only if >>n^3 - z vanishes for some n, which is if and only if z = n^3 for n = >>1,2,3,.. >>Is there anything I'm missing here? > Hard to say whether you're missing anything since you don't tell > us how you proved the things you proved. It seems possible you're > missing something, because it's not that obvious that it vanishes > if and only if one of the factors vanishes, that really does need > to be proved. > (Consider g(z) = product_1^infinity (1/2). That vanishes everywhere, > even though none of the factors has a zero anywhere. How do you > know that that sort of thing doesn't happen here? > Also how do you know the product converges, and that it converges > to something analytic? That's the part where we can't tell whether > you're missing something...) >>I have shown that the sum (from 1 to infinity) of [(n^3-z)/(n^3+z) - 1] >>converges absolutely and uniformly on compact subsets of the right half >>plane, and thus >>it converges in H(G) where G = {z : Re(z) > 0}. > Right, because a theorem says that this implies that. > Now the same theorem says (or _should_ say) that it > follows that the product cannot vanish except where > one of the factors vanishes - is that not included > in whatever book you're using? >>I see your point about the >>infinite producct of 1/2. >>How do I know that sort of thing doesn't happen here? Certainly it >>vanishes >>on z = n^3 n =1,2,3,... >>But anywhere else? >>Tony > ************************ > David C. Ullrich > ************************ > David C. Ullrich === Subject: Re: condition for convergence of a series >> Hi all >> I'd like some hints to prove that, if (a_n) is a sequence of positive >> numbers, then Sum(a_n) converges if, and only if, Product((1- a_n)) >> converges. So far, all the solutions I tried got quite messy. >> Artur >1. You need some additional condition to rule out the case a_n = 1 for >some n or a_n->1. >2. With the appropriate additional condition, you will see that both >conditions imply a_n->0. Show that for sufficiently small x>0 > x < -log(1-x) < 2x >Or if you want to avoid logarithms, I'm in favor of avoiding logarithms here, because although there's no problem really in the complex case they always make me nervous. >you could try using the 3rd from >last displayed equation of >http://www.math.missouri.edu/~stephen/preprints/disttail11/node2.html Or Lemma 6.4 from my complex notes: If $z_1,dots,z_ninC$ and $sum_{j=1}^{n-1}|1-z_j|<1/2$ then $$left|1-prod_{j=1}^nz_jright|le2sum_{j=1}^n|1-z_j|.$$ (Proof by induction.) That must be where you got the idea - you'll be hearing from my lawyer. >Stephen ************************ David C. Ullrich === Subject: Re: condition for convergence of a series >> I'd like some hints to prove that, if (a_n) is a sequence of positive >> numbers, then Sum(a_n) converges if, and only if, Product((1- a_n)) >> converges. So far, all the solutions I tried got quite messy. >> Artur >1. You need some additional condition to rule out the case a_n = 1 for >some n or a_n->1. >2. With the appropriate additional condition, you will see that both >conditions imply a_n->0. Show that for sufficiently small x>0 > x < -log(1-x) < 2x >Or if you want to avoid logarithms, > I'm in favor of avoiding logarithms here, because although there's no > problem really in the complex case they always make me nervous. Why?? I found Stephen suggestion really interesting and could finish up the proof! It seems logs simplify things. For instance, if you know that the sequence of arithmetics means of a convergent sequence converges to the same limit of the original sequence, then, by using logs, you can immediately extend this result to the sequence of geometric means, if its terms are positive. Actually, I think we have to rule out the case a_n = 0 for some n, at least according to the usual definition of convergent products. But I really thought the product (1/2) * (1/2) * (1/2)*.... converged to 0. The result we are discussing was suggested to me to show that the series Sum((1/p_n^a)), where p_n is the n_th positive prime, diverges for a=1 and converges for a>1. But analysing the convergence of Product (1 - (1/p_n)^a) doesn't look any easier. At least so far. Artur >you could try using the 3rd from >last displayed equation of >http://www.math.missouri.edu/~stephen/preprints/disttail11/node2.html > Or Lemma 6.4 from my complex notes: > If $z_1,dots,z_ninC$ and > $sum_{j=1}^{n-1}|1-z_j|<1/2$ then > $$left|1-prod_{j=1}^nz_jright|le2sum_{j=1}^n|1-z_j|.$$ > (Proof by induction.) > That must be where you got the idea - you'll be hearing from my > lawyer. >Stephen > ************************ > David C. Ullrich === Subject: Re: condition for convergence of a series > 1. You need some additional condition to rule out the case a_n = 1 for > some n or a_n->1. This is covered in the definition of convergence for an infinite product. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: condition for convergence of a series >> 1. You need some additional condition to rule out the case a_n = 1 for >> some n or a_n->1. >This is covered in the definition of convergence for an infinite >product. Is it? Yes, that definition is not what people often assume it is, for example the product (1/2)(1/2)... is said to diverge. But I thought that a product where one of the factors equals 0 is said to _converge_ to 0 regardless of the other factors (if so then Stephen is right, there is an additional condition needed in what the OP said - the case of a_n <> 1 but a_n -> 1 is covered but the case where a_n = 1 for some n is not.) At least I thought that's what the definition was - don't have any suitable references at hand right now (Rudin [gasp] gets the definition wrong, saying a product converges if the sequence of partial products converges, period.) ************************ David C. Ullrich === Subject: Re: condition for convergence of a series > 1. You need some additional condition to rule out the case a_n = 1 for > some n or a_n->1. >>This is covered in the definition of convergence for an infinite >>product. >Is it? Never mind - it seems that definition is not what I thought it was, and yes it does cover all this. >Yes, that definition is not what people often assume it is, >for example the product (1/2)(1/2)... is said to diverge. But >I thought that a product where one of the factors equals 0 is >said to _converge_ to 0 regardless of the other factors (if so >then Stephen is right, there is an additional condition needed >in what the OP said - the case of a_n <> 1 but a_n -> 1 is >covered but the case where a_n = 1 for some n is not.) >At least I thought that's what the definition was - don't >have any suitable references at hand right now (Rudin [gasp] >gets the definition wrong, saying a product converges if >the sequence of partial products converges, period.) >************************ >David C. Ullrich ************************ David C. Ullrich === Subject: Re: The state-of-the-art in mathematics >[...] >To avoid repetition, Uh, it's a little late for that. ************************ David C. Ullrich === Subject: Number Factorization Machine by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j05DEnI12915; Hi all I am not sure how far the subject I am discussing interests this group. But anyway please go through this http://techtrix.batcave.net/ esearch paper. I would like to share few commends on this. === Subject: Re: Not relevant to mathematics but, > when buying textbooks for mathematics course, a huge price difference exists > between the book being sold in U.S and in South Korea, even though they are > exactly same books. > For example > Real Analysis 3rd ed. by Royden is about $100 at Amazon, but the same > book(same edition, same publisher), which is imported from U.S., is about > $30 at a bookstore in South Korea. > Do any of you guys know why this happens? I don't know about this particular case, but one can find copies of US texts produced in Asia where copyright laws are lax or ignored. Probably the same way you can find the latest Hollywood films on sale there for $5 ... === Subject: Re: Not relevant to mathematics but, > when buying textbooks for mathematics course, a huge price difference exists > between the book being sold in U.S and in South Korea, even though they are > exactly same books. > For example > Real Analysis 3rd ed. by Royden is about $100 at Amazon, but the same > book(same edition, same publisher), which is imported from U.S., is about > $30 at a bookstore in South Korea. > Do any of you guys know why this happens? Because corporations are all greedy and corrupt. The same is happening with prescription drugs. Hey, why don't you make some money selling text books from S.Korea to US students? === Subject: Re: Present value of a growing perpetuity C/(r-g) Where did you get this formula? Isn't it absurd? It could involve division by zero (r=g) or give the investment a negative value (g>r). > I'm currently taking a finance course and hope someone can help me with > a query regarding growing perpetuities. > I know the formula for calculating the present value of a growing > perpetuity is > C/(r-g) > where C is the cash flow each period, r is the interest rate, and g is > the rate of growth. But this formula assumes that the first payment in > the perpetuity occurs in one year's time. What happens if the first > payment won't occur for a few years. > For example, say that the first payment at the end of year 1 will be > $100, the annual rate of interest is 10%, and the perpetuity needs to > grow at an annual rate of 6%. In this case the present value will be > 150/(0.10-0.06)=$3750 (I think). > However, what would be the case if the first payment didn't happen > until the end of year 5? Do you simply work out the present value of > 150 and then plug that into the equation? > Hope you can help === Subject: Re: Present value of a growing perpetuity This formula is, as I understand it, industry standard but only works for values where r > g for the reasons you have indicated. > C/(r-g) > Where did you get this formula? Isn't it absurd? It could involve division > by zero (r=g) or give the investment a negative value (g>r). > I'm currently taking a finance course and hope someone can help me with > a query regarding growing perpetuities. > I know the formula for calculating the present value of a growing > perpetuity is > C/(r-g) > where C is the cash flow each period, r is the interest rate, and g is > the rate of growth. But this formula assumes that the first payment in > the perpetuity occurs in one year's time. What happens if the first > payment won't occur for a few years. > For example, say that the first payment at the end of year 1 will be > $100, the annual rate of interest is 10%, and the perpetuity needs to > grow at an annual rate of 6%. In this case the present value will be > 150/(0.10-0.06)=$3750 (I think). > However, what would be the case if the first payment didn't happen > until the end of year 5? Do you simply work out the present value of > 150 and then plug that into the equation? > Hope you can help === Subject: Re: tetrahedron problem > Let T1, T2, T3, T4 be the vertices of the regular tetrahedron and > suppose > the distance between any two vertices is 1. > Let R(T1) be the closed ball centered at T1 of radius 1 > , R(T1) = { X in R^3 : d(X,T1) <= 1 }, where > d(X,Y) = (the distance between the points X and Y). > Let S be the intersection of R(T1), R(T2), R(T3), and R(T4). > Let A, B be arbitrary two points in S. > then, can anyone prove or disprove one or both of the following? > (i) d(A,B) <= 1. > (ii) if d(A,B) = 1, then either A or B is one of T1, T2, T3, T4. > My guess is that the two statements are true. > Dae-jung Yoo > BTW, why could you not have simply posed: > If distance between any two points in the intersection volume of 4 > spheres, each centered at a vertex of regular unit edge tetrahedron > passing through other 3 vertices..<=1 or ..etc. ? I would prefer the > problem to be posed this way verbally. Is this way you posed more > general as symbolic or modern? You are suggesting a simple and natural way to rewrite my original question. I am not much fluent in English. so I tend to write symbolically. -- Dae-jung Yoo (IGNJSA YOO) === Subject: Re: Complex Question #4 : Harmonic function >>message > f is real on the open square S, ie, f(S) is a subset of R. >>I understand that. But then what? Why can't f(S) be open? Am I missing >>something here? > A subset of R, open in C? > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Re: Complex Question #4 : Harmonic function question about a problem whose solution is actually given : > Suppose that f is analytic on G = {z : |z| < 3} and real-valued on the > boundary of the square {z : |Re(z)| <= 1, |Im(z)| <= 1}. Prove that f is > constant on {z : |z| < 3}. > Proof: Since f is analytic on G, f = u + iv on G, but v = 0 on the > boundary > of the square. Applying maximum/minimum principles of harmonic functions > to > v, we get v = 0 on the entire closed square. Now consider f on the open > line. Then it goes on to say that the open mapping theorem implies that > f > is constant on the open square. Why does the Open Mapping theorem imply > that f is constant on the open square? I know if f is non-constant and > analytic, then the image of an open set is open. > But what is the image of the open square in this problem? >> f is real on the open square S, ie, f(S) is a subset of R. >I understand that. But then what? Why can't f(S) be open? Am I missing >something here? You understand that f(S) is a subset of R but don't see why it can't be open? What's the definition of open set? >Why is f(S) constant? ************************ David C. Ullrich === Subject: Re: Complex Question #4 : Harmonic function >> I have a question about a problem whose solution is actually given : >> Suppose that f is analytic on G = {z : |z| < 3} and real-valued on the >> boundary of the square {z : |Re(z)| <= 1, |Im(z)| <= 1}. Prove that f >> is >> constant on {z : |z| < 3}. >> Proof: Since f is analytic on G, f = u + iv on G, but v = 0 on the >> boundary >> of the square. Applying maximum/minimum principles of harmonic >> functions >> to >> v, we get v = 0 on the entire closed square. Now consider f on the >> open >> real >> line. Then it goes on to say that the open mapping theorem implies >> that >> f >> is constant on the open square. Why does the Open Mapping theorem >> imply >> that f is constant on the open square? I know if f is non-constant and >> analytic, then the image of an open set is open. >> But what is the image of the open square in this problem? > f is real on the open square S, ie, f(S) is a subset of R. >>I understand that. But then what? Why can't f(S) be open? Am I missing >>something here? > You understand that f(S) is a subset of R but don't see why it > can't be open? What's the definition of open set? >>Why is f(S) constant? > ************************ > David C. Ullrich === Subject: Re: Complex Question #5 : Integral, Rouche >> Evaluate >> integral (over Gamma) of dz/(e^z-1-z) >> where Gamma(t) = e^(it) for 0 <= t <= 2pi >> Tony > [I have removed the part already done by Tony] > You were on the right track! > The residue at 0 is obtained by noticing that e^z-1-z=z^2/2+z^3/6+... > Thus 1/(e^z-1-z)=2z^(-2)-2z^(-1)/3+... How did you get this inverse? Did you just match up term by term in (z^2/2+z^3/6+...) * (something) = 1 ? Is every power series invertible if we allow Laurent series to be the inverses? (If we only allow power series to be the inverses, I think power series are invertible if and only if the constant term is invertible) But I never learned a similar thing if we allow Laurent series to be the inverses... >and the residue is -2/3. > Now observe that e^z-1-z=z^2(1/2+z/6+...+z^(n-2)/n!+...) > the part inside the parentheses has its modulus at least > 1/2 - |z|/6 -|z|^2/24- ... - |z|^(n-2)/n!-... > that is at least > 1/2 - 1/6 -1/24 ... -1/n! - ... > where 1/2 - ( e - 5/2) =3-e >0 is to be recognized. > Thus 0 is the only pole of 1/(e^z-1-z) inside the unit circle > and we get the value > -4 pi i/3 > for the integral > -- > Charles Delorme tous les m.8egalomanes > LRI ont une signature > cd@lri.fr .88 .8etages === Subject: Re: Complex Question #5 : Integral, Rouche >>The residue at 0 is obtained by noticing that e^z-1-z=z^2/2+z^3/6+... >>Thus 1/(e^z-1-z)=2z^(-2)-2z^(-1)/3+... > How did you get this inverse? Did you just match up term by term in > (z^2/2+z^3/6+...) * (something) = 1 I do not know how he did it, but that's how I would have done it. > Is every power series invertible if we allow Laurent series to be > the inverses? (If we only allow power series to be the inverses, I think > power series are invertible if and only if the constant term is invertible) > But I never > learned a similar thing if we allow Laurent series to be the inverses... Yes, it is true, with one obvious exception: the null series. It's rather easy to prove. Try it! Jose Carlos Santos === Subject: Re: Focussing Lens without Spherical Aberration > eccentricity of hyperboloid lens = refractive index You have finally succeeded in getting me to go look at my old copy of Optics by Hecht & Zajac, and sure enough they have two pages devoted to refraction by aspheric surfaces. There they give the same result for hyperbolic surfaces, eccentricity = refractive index. The section of interest is in the chapter on Geometrical Optics -- Paraxial Theory. -- Mark === Subject: Re: Ayn Rand's greatest idea >> What kind of moron would think that experience is subjective, hence >> unimportant? > The same kind of moron who believes that life is the battle of strong > against the weak, or who believes that people make their own reality > (and everyone who experiences atrocity and misfortune is responsible > for it), or who believe that good things in their lives are God's > reward for their righteousness. You mean, like most so-called libertarians? http://geolib.pair.com/essays/sullivan.dan/royallib.html Actually, the person that most sounds like is George W. Bush. > Basically, bullies. === Subject: Re: Ayn Rand's greatest idea > Actually, the person that most sounds like is George W. Bush. Dubya is most certainly NOT a libertarian. He is as much a Statist as any of the crazy-left Democrat liberals. Bob Kolker === Subject: Re: Cantor's diagonal proof wrong? <41aa5b47$13$fuzhry+tra$mr2ice@news.patriot.net> <41ad5139$15$fuzhry+tra$mr2ice@news.patriot.net> <41b2427f$15$fuzhry+tra$mr2ice@news.patriot.net> <41c22df7$7$fuzhry+tra$mr2ice@news.patriot.net> <41C25DB8.EF7C5F2F@tiki-lounge.com> > Can you prove, whether my number Integer(pi*10^10^100) + > Integer(sqrt(2)*10^10^100) is prime? It would at least ascertain a > small part of its reality. It has a definite number of digits (10^100+1). Does that count as a small part of its reality? Can you say that a thing whose length I know doesn't exist? - Randy === Subject: Re: Solving inequalities that involve abs value functions > and anyway you will have to > wrestle with the issue of the ANDS and ORS and THEREFORES in your > statements. > I know it seems like I've taken far too many bytes just to say > yadda yadda yadda, but trust me, if you can't turn this kind of > problem into a _logical analysis_ the way I just did, you'll never get > past the level of symbol pushing; Not quite true. Ivan didn't seem to have any intuition about the problem. I pointed him to the number line because that is the easiest way to settle the AND vs. OR question. If he gains a good feel for what's going on in his inequality, then he'll have gotten past symbol pushing, without turning the problem into a rigourous proof. Bart === Subject: Re: #2: New_Year's Quiz > So what do you think the answer of my original Quiz ? I mean What do you think the answer of my original Quiz is ? === Subject: Re: Stone Topology: Metrizable? > Does any one know if the Stone Topology, define on > the collection of n-types over a set M is metrizable? Stone Topology is a space of ultrafilters with any set of ultrafilters containing a common point, being an open base set? Do you a different definition? Have you checked to see if it's first countable? === Subject: Re: Giant Quantum Tsunami Wave from Earthquake inside Dirac Sea toHiggs Ocean The Prophecy scribed: >> The Prophecy scribed: >>huh? usenet and morons don't mix. >> You prove how wrong that is, in an entertaining way. >>stupid morons like you, made popular by putting >>their persecution disorders into practice, learn to articulate their inane ramblings. >>I don't want to hear it. >> Hearing voices now? >when the boys are out to play Dr Postman is never far behind. >why don't you explain to the good people of sci.military, sci.maths, sci.physics, sci.astro The people on those groups are not laughing with you, jHerk. >how you conclusively proved there is no such person as The Truman! Simply because Hollywood movies are NOT reality. There is no Batman, or Spiderman, as much as that may disappoint you. Your paranoid delusions are simply that, and nothing more. >It didn't say based on a true story in the moofie now wasn't it? That's because it wasn't, kook. >avoidance expected with tangential frothing, let me guess idiot moron retard kook stoopid, >kos it said so in the moofie Aren't you supposed to be on some sort of heavy meds? -- DrPostman USPS, MBMC, BsD; Disgruntled, But Unarmed Member,Board of Directors, afa-b, SKEP-TI-CULTĀ #15-51506-253. You can email me at: DrPostman(at)gmail.com Nothing compares to the complicated futility of ignorance. -Kurt Vonnegut === Subject: Re: Giant Quantum Tsunami Wave from Earthquake inside Dirac Sea toHiggs Ocean alt.fan.art-bell in message <341genF45oaibU1@individual.net>: >no, still amazed it can be public in a 100,000 population town >and no one on usenet will check a Truman claim. SPACE-TIME ANOMOLY DETECTED 300 METERS OFF THE PORT BOW, CAPTAIN! BRACE FOR IMPACT! SET LOGIC SHIELDS TO FULL POWER! -- V.G. Change pobox dot alaska to gci. Actually the Law of Conservation of Energy proves there's free energy. - Professor Alexa connects some brand new dots. Sarcasm is my sword, Apathy is my shield. === Subject: Re: Giant Quantum Tsunami Wave from Earthquake inside Dirac Sea toHiggs Ocean alt.fan.art-bell in message <3422ifF459lg4U1@individual.net>: >It didn't say based on a true story in the moofie now wasn't it? You're too smart for me! -- V.G. Change pobox dot alaska to gci. Actually the Law of Conservation of Energy proves there's free energy. - Professor Alexa connects some brand new dots. Sarcasm is my sword, Apathy is my shield. === Subject: Re: Giant Quantum Tsunami Wave from Earthquake inside Dirac Sea toHiggs Ocean alt.fan.art-bell in message <341eajF442pcbU1@individual.net>: >jab jab jab quote, Im a critical thinker. BWAHAHAHAHAHAHAHAH!!!!!!!!!!!!!!! -- V.G. Change pobox dot alaska to gci. Actually the Law of Conservation of Energy proves there's free energy. - Professor Alexa connects some brand new dots. Sarcasm is my sword, Apathy is my shield. === Subject: Re: Giant Quantum Tsunami Wave from Earthquake inside Dirac Sea toHiggs Ocean alt.fan.art-bell in message <3419mvF452fq1U1@individual.net>: >this is a limited opportunity There's the understatement of the week. -- V.G. Change pobox dot alaska to gci. Actually the Law of Conservation of Energy proves there's free energy. - Professor Alexa connects some brand new dots. Sarcasm is my sword, Apathy is my shield. === Subject: Re: Giant Quantum Tsunami Wave from Earthquake inside Dirac Sea toHiggs Ocean > why don't you explain to the good people of sci.military, sci.maths, > sci.physics, sci.astro > how you conclusively proved there is no such person as The Truman! > Perhaps you can demonstrate how one proves a negative. next you'll be asking me to calculate the square root of -1!! oh you've done that already, the rest of that thread involved some lessons in complex geometry that might have been a tad unsuitable for alt.usenet.kooks, and the rest of the thread I'd rather forget. Herc === Subject: Re: Giant Quantum Tsunami Wave from Earthquake inside Dirac Sea toHiggs Ocean The Prophecy scribed: >> why don't you explain to the good people of sci.military, sci.maths, >> sci.physics, sci.astro >> how you conclusively proved there is no such person as The Truman! >> Perhaps you can demonstrate how one proves a negative. >next you'll be asking me to calculate the square root of -1!! >oh you've done that already, the rest of that thread involved some >lessons in complex geometry that might have been a tad unsuitable >for alt.usenet.kooks, and the rest of the thread I'd rather forget. Start off by explaining how you prove a negative. That's all. -- DrPostman USPS, MBMC, BsD; Disgruntled, But Unarmed Member,Board of Directors, afa-b, SKEP-TI-CULTĀ #15-51506-253. You can email me at: DrPostman(at)gmail.com Nothing compares to the complicated futility of ignorance. -Kurt Vonnegut === Subject: Re: Giant Quantum Tsunami Wave from Earthquake inside Dirac Sea toHiggs Ocean alt.fan.art-bell in message <34270mF3hrmv2U1@individual.net>: >rest of the thread I'd rather forget Looks like you already have. YOU are the one who made the extraordinary claim, and then asked Postman to show that he had disproved it. -- V.G. Change pobox dot alaska to gci. Actually the Law of Conservation of Energy proves there's free energy. - Professor Alexa connects some brand new dots. Sarcasm is my sword, Apathy is my shield. === Subject: Re: Giant Quantum Tsunami Wave from Earthquake inside Dirac Sea toHiggs Ocean proposed: >> why don't you explain to the good people of sci.military, sci.maths, >> sci.physics, sci.astro >> how you conclusively proved there is no such person as The Truman! >> Perhaps you can demonstrate how one proves a negative. >next you'll be asking me to calculate the square root of -1!! >oh you've done that already, No, mindless one, that was not CS. > the rest of that thread involved some >lessons in complex geometry that might have been a tad unsuitable >for alt.usenet.kooks, and the rest of the thread I'd rather forget. As usual, Hork, casually packs his bags and runs for the CAVE OF DOOM! >Herc === Subject: Students - Attend AAAS Annual Meeting for Free! Student Session Aide volunteers receive a free meeting registration to the AAAS Annual Meeting in Washington, DC. The meeting will be held February 17 -21, 2005. Graduate and undergraduate students at the sophomore level or higher may apply. You must be able to work at least 8 hours and attend the orientation session. A one-year membership to AAAS is included for those who volunteer for 16 hours. For more information or to apply as a session aide please go to: http://www.aaas.org/meetings/Annual_Meeting/02_PE/Student_01.shtml Please e-mail aaasmeeting@aaas.org with any questions. === Subject: Re: [OT] I hate being American >>30% of the employees in my company are foreign-born, >I suppose you mean 'ethnically-distinct' rather than just 'foreign- >born'. No, I meant foreign-born. Foreign, in this context, means not USA. John === Subject: Re: [OT] I hate being American >> Salads make you fart and have very little by way of protein. > Add tofu. Problem solved. > Bob Kolker I didn't fight my way to the top of the food chain to eat tofu! === Subject: Pole of order 3? Quick question about a pole : The function 1 / [(z^2)(e^z - e^(-z))] has a pole of order 3 at z = 0, correct? At least a pole of order 2. But then I investigate e^z - e^(-z) = 2z + ..., so we need a z^3 in the numerator of 1 / [(z^2)(e^z - e^(-z))] to get an analytic function about 0 (right?) Tony === Subject: Re: Pole of order 3? > The function > 1 / [(z^2)(e^z - e^(-z))] has a pole of order 3 at z = 0, correct? > At least a pole of order 2. But then I investigate e^z - e^(-z) = 2z + ..., > so we need a z^3 in the numerator of 1 / [(z^2)(e^z - e^(-z))] > to get an analytic function about 0 (right?) You're right, the order is 3. I would put it like this: it is 3 because the limit lim_{z -> 0} z^3/(z^2(e^z - e^(-z))) exists in C and it is not 0. Jose Carlos Santos === Subject: Re: metamath scope mathematics = theory, ... > theory = axioms, theorems > axioms = axiom, ... > theorems = theorem, ... > theorem = statement, its-proof How would you feel about a definition of architecture that was also devoid of ideas, esthetics, humanity, ... for example defining architecture as architecture= building, ... building = construction material, ... construction material = brick, mortar, steel, concrete, ... ... concrete = cement, sand, gravel steel = .... You would, I think, be missing the essence of architecture, which I think includes relating buildings to their uses, their environment, their occupants, their historical context, etc. One can easily come up with theorems, say 1+1=2 1+2=3 1+3=4 .... which are quite dull, in my opinion. Mathematics as you define it sounds like a hollow pointless shell. While this DOES describe some mathematics, it would be most distressing to learn that a PhD student in advanced mathematics was, for example, studying 1+23423234242= ? It would be (is?) distressing to see computer scientists studying systems to do mathematics that are, in that sense, simulating the creation of mathematics in the hollow pointless and (to humans) dull sense. On the other hand, I think training a computer to reflect upon its own program (e.g. this is a dull theorem) would be an interesting accomplishment. RJF === Subject: Help with set equality proof Let Dm = UNION (k=0,3^(m-1) - 1) ((3k+1)/3^m) , (3k+2)/3^m)) and Em = UNION (j=0, 0.5*(3^m - 3)) ((2j+1)/3^m , (2j+2)/3^m)) How does one show that UNION (m=1,n) Dm = UNION (m=1,n) Em All help appreciated. === Subject: Re: MATHEMATICS AND REALITY listed above. > MATHEMATICS AND REALITY > Copyright 1984 to 2005 Allen C. Goodrich > Newton's equation for gravitational force > F = Gm^2/L^2 is only true for particular values of L > , m and G. > And it provides the value of a force that does not exist > in the real world. Force of gravity is an illusion which > is created by the equation and which Newton assumed > to exist. So the force of gravity didn't exist before the mid-seventeenth century? Interesting. Was everything just kind of floating about randomly until then? === Subject: Re: A measure on sets of integers: Why p(even)=1/2 makes sense > Suppose A is a set of integers. Let C be any countable collection of > arithmetic progressions which cover A. So C=I_1+I_2+..., where > I_n={a_n*k+b_n:a_n,b_n in Z+, k in Z}. Let f(C)=1/a_1+1/a_2+... > Define: > m*A= inf {f(C):C as above} > Then m*(B) = 0 for all subsets B of Z. > .snip Yes. My bad. So suppose A is a infinite set of positive integers, each I_n={a_n*k+b_n:k in Z+} for some a_n,b_n in Z+, m*({})=0 and m*(any other finite set) is undefined. Does m*({2*k))=1/2? > If m is a measure on a countable set A, and m({x}) = 0 for all x in B, > then m(B) = 0. This is rather a problematic property for trying to > define any sort of uniform probability measure on the integers (or > any other countable set). Rich === Subject: Re: A measure on sets of integers: Why p(even)=1/2 makes sense >Suppose A is a set of integers. Let C be any countable collection of >arithmetic progressions which cover A. So C=I_1+I_2+..., where >I_n={a_n*k+b_n:a_n,b_n in Z+, k in Z}. About the notation: Saying I_n={a_n*k+b_n:a_n,b_n in Z+, k in Z} implies I_n = Z, which is not what you meant. You meant where I_n={a_n*k+b_n : k in Z} for some a_n,b_n in Z+. >Let f(C)=1/a_1+1/a_2+... >Define: >m*A= inf {f(C):C as above} >Now consider those sets E such that, for all A: >m*(A)=m*(A.intersect.E)+m*(A.intersect.E~) >Does the collection of all such E form a sigma-algebra of subsets of >Z? Is the set of even numbers, say EVEN={2*k}, such a set? Is >m(EVEN)=1/2? >Oh yeah, one more silly question: Why is it unreasonable to call the >nonmeasurable sets *random*? >Rich >*-----------------------* > www.GroupSrv.com >*-----------------------* ************************ David C. Ullrich === Subject: Analytic mappings, normal family I'm having trouble with the following question : Let F be the famiy of all analytic mappings of {z : Re(z) > 0} into {w : |w| < 1}. Let a = sup |f ''(1)| over f in F. (i) Find an upper bound for a. The solution that is given is like this : Let Gamma(t) = 1 + .5e^(it) 0 <= t <= 2pi If f is in F, the Cauchy integral formula gives f ''(1) = (2/(2pi*i)) * integral over Gamma of [ f(w) / (w-1)^3 ] dw So |f ''(1)| <= 1/pi * (1*2pi*.5)/(.5^3) = 8 My question is : Why did we select Gamma(t) to have radius .5? We could have easily made the radius .7, and then our bound would be even better since w - 1 in the denominator would be .7. .9 would have been better as well. I must be missing something here. (ii) Now show that there exists a g in F such that |g ''(1)| = a. The solution starts out by saying Consider f_n in F such that |f_n ''(1)| ---> a. Why must such a sequence exist? (Then it talks about normal family, and more stuff that I think I understand.) Tony === Subject: Re: Complex Question #1 >> (i) Find a 1-1 analytic mapping f of {z : |z| < 1} onto C {x : x >= 0} >> such that f(0) = -1. >> (ii) Determine with proof all analytic maps f of {z : |z| < 1} into C >> {x : x >= 0} such that f(0) = -1 and f(1/2) = -1/9. >> For (i), I mapped the disk to the upper half plane by (i+iz)/(1-z) and >> then to C {x : x >= 0} by squaring, so the final map is h(z) = >> [(i+iz)/(1-z)]^2. >> This also maps 0 to -1. >> But I'm not sure how to do part (ii). I want to use Schwarz Lemma. So >> now look at a general h that maps disk D to C {x : x >= 0}. Now I can >> map C {x : x >= 0} to D by >> g(z) = (sqrt(z) - i)/(sqrt(z) + i), or even g(z) = (sqrt(z) - w)/(sqrt(z) > And what do you mean with sqrt(z)? sqrt(z) is defined on C {x : x >= 0}, sending it to the upper half plane. >> +w) where Im(w) > 0. Now, the composition f = g o h maps the disk to the >> disk and f(0) = 0. (Note also that if w is anything else, I don't >> believe we have f(0) = 0, so we wouldn't be able to apply Schwarz.) So, >> |f(z)| <= |z|. But now what? I think I'm pretty close but I can't see >> something here... > Put F(z) = 1/f(z); What is f(z) ? >then F is a 1-1 analytic mapping from {z | |z| < 1} > onto C {x | x >= 0}, F(0) = -1 and F(1/2) = -1/9. Suppose that there's > another one; call it g and put h = g^(-1) o F. Then h is a 1-1 analytic > mapping from the open unit circle into itself, h(0) = 0, and h(1/2) = > = 1/2. It follows from Schwarz's lemma that h is the icdentity function > and therefore g = F. > Jose Carlos Santos === Subject: Re: Complex Question #1 >>Put F(z) = 1/f(z); > What is f(z) ? Sorry, I was talking here about your function h, so I should have written h(z). Jose Carlos Santos === Subject: Re: Epistemology 102 >***Please disregard the previous post. It was send in error.*** > Unfortunately, my reply to the previous post has already been send. > How about if I just disregard you instead? You normally disregard facts, history and logic. It makes perfect sense that you disregard anybody who points that out. Milan === Subject: Re: Epistemology 102 <64fCd.3834$7n1.247601@news20.bellglobal.com> <41d998fd.31630860@netnews.att.net> <41dacf32.41796668@netnews.att.net> > No, I meant effects in the sense that the empirical result of mental > effects is human behavior. Look at the problem this way. We have > circumstantial contingencies and we have human behavior. I maintain > that mentation, the mind, or mental effects, M, mediate contingencies > and behavior B=MC the way inertial mass mediates acceleration and > force, f=ma. And this is true regardless of what M may turn out to be. Crank alert > No one understands mass in any irreducible, fundamental sense anymore > than we understand the mind. We keep f=ma around on purely utilitarian > grounds because its use enables us to explain acceleration in terms of > force consistently if not absolutely for palpable bodies. The concept of mass leads to confirmed experimental prediction and it is therefore well-founded. Crank alert > But we don't keep the concept of mind around on utilitarian grounds > the way we do mass. We keep it around on reductionist grounds because > it regresses finitely even if we don't understand all of the mechanics > involved. Thus M is definable in the sense of finite reduction whereas > m is not. Crank alert > It's an empirical observation nonetheless. Crank alert > Which I observe is to be explained in terms of contingencies through > the mediation of M. As I understand behaviorism - what little I > understand of behaviorism, I should add - that philosophy maintains > that B=C+P (private behavior) or perhaps B=PC. The problem with this > explanation is that behaviorism admits it doesn't understand what P > is. So there is no reduction and hence no explanation of B in terms of > C. Crank alert > Taken on faith the way we take mass on faith. In the case of mass > there is a demonstrable consistency of reduction between f and a > which is lacking in M. But in the case of M there is a demonstrably > finite reduction in mediation lacking in m. Crank alert at maximum > Well, there is more than experience I can check on. I can also check > on whether there is a finite reduction in the mediation provided by M. > If so we have a definable M regardless of whether we understand M on > utilitarian grounds the way we understand m. Crank alert above all known levels > At first pass I > suggest that the term mechanical just means relations between > empirical observations drawn through tautologies. Mechanical in the philosophy of science is a term used to ground causality. The other alternative is conterfactuals. It is usually not relations between empirical observations but the use of term mechanical suggests a causal connection between some empirical quantity and some unobservable quantity. This is used mainly in Realist's grounds whereas anti-realists argue that the mechanical hypothesis is not well-founded and therefore adhere to an acausal world. Crank alert has hit level unkown in this galaxy, Mike === Subject: Re: Epistemology 102 >> [. . .] >LOL. Empiricism can be traced to Locke and Hume. Positivism emerges in >the >1920s with the Vienna Circle etc. >>Gee, that online reference to Pierre Auguste Comte and positivism >>that I read must have been erroneous. Not to mention contributions >>of Aristotle. I think you really need to be correcting the history of >>philosophy on such points and not merely lil ole me. Various others >>have been referring to logical positivism and not myself except in >>reply to them. You might really consider learning to read what is >>written before you reply. >Comte was a guy who made the Norman Vincent Peale kind of positivism >popular, and gave it a name. Has nothing whatver to do with logical; >positivism. Having never suggested otherwise I can agree. === Subject: Re: .9999... = 1 proof > And I guess you still don't seem to understand. No. YOU don't understand. Once more look at the -series- Sum (n >= 1) [3*/10^n]. Now look at the -sequence- of partial sums a_n = {Sum (1 <= k <=n) 3/10^k). What does the sequence {a_n} converge to? Get our head our of your ass and think for a change. Bob Kolker === Subject: Re: countability of reals > Theorem. A one-to-one correspondence can be set up between the set IQ > of all rational numbers of the interval (0,1) and the set IX of all > irrational numbers of the interval (0,1) Wrong. Bob Kolker === Subject: Re: countability of reals > Theorem. A one-to-one correspondence can be set up between the set IQ > of all rational numbers of the interval (0,1) and the set IX of all > irrational numbers of the interval (0,1) > Clue: when something has been proved to be impossible, and > the proof is simple enough to be understood by anyone with > a passing score in ninth grade algebra, posting a muddled > mess claiming to have done that impossible thing anyway to > a newsgroup frequented by folks with earned PhDs in math > is a fast track to ridicule and Usenet kookdom nominations. > FYI YOU'RE AN IDIOT XANTHIAN. there IS a problem with the proof, and the fact your HYPERINFINITY rubbish *relies* on the definitions of infinite streams that are comprehensible by 10 years olds is the problem. grow up a bit. 0xxxxxxx.. x0xxxxxx.. xx0xxxxx.. xxx0xxxx.. xxxx0xxx.. .. SEE THE DIAGONAL? Its NOT A NUMBER. You don't prove anything by proving its not a number on the list, we know that. its an illusion of a unique sequence. If you treat it like a number then you get contradictions. OBVIOUSLY (n,i) =/= (i,i) has an inbuilt contradiction, that means you can't use it as a definition and an argument over N. hththththht.. hththththhth.. hththththth.. hhhhhhhhhh.. ttttttttttttttt.. hththththth.. hhhhhhhhhh.. ttttththththth.. .. Here is an infinite binary list. Infnite people flipping infinite random coins. For some reason people here can't comprehend that the list contains every sequence of H and T to infinite length. The word random makes sci.math squeal out literal excuses, they think everyone will toss Heads. That's OK, Use a computer instead. UTM(1,1)mod2 UTM(1,2)mod2 UTM(1,3)mod2 UTM(1,4)mod2.. UTM(2,1)mod2 UTM(2,2)mod2 UTM(2,3)mod2 UTM(2,4)mod2.. UTM(3,1)mod2 UTM(3,2)mod2 UTM(3,3)mod2 UTM(3,4)mod2.. UTM(4,1)mod2 UTM(4,2)mod2 UTM(4,3)mod2 UTM(4,4)mod2.. UTM(5,1)mod2 UTM(5,2)mod2 UTM(5,3)mod2 UTM(5,4)mod2.. .. No the HALT proof is not the end of the world, diagonalistion works while some processes are still ticking over. Get over the halt proof sci.math, stop trashing computer science with your idiot math jargon just because another discipline can set limitations. That UTM(x,y) mod 2 is just the same list as the random flippers. HTHTHTH.. THTHTHTH.. THTHTHH.. HHHHHHH.. THTHTHTH.. .. See that final .. ! NONE OF YOU ARE INTERPRETING WHAT IT MEANS. Assume a finite segment is missing from the initial segments of the list. It has some length L. There are 2^L possible segments. That means with any significant number of sequences larger than 2^L, the Prob(initial segment of lenght L on list) -> 1. POSSIBLE..........OF ANY LENGTH. Now THINK what that means. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. IF YOU LEARN ONE THING IN 2005 MAKE IT THIS Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. Every coin sequence is computable to infinite length. 3 months I've been saying this, its refuted every post by smart alec and dumb alec morons alike. Its not refutable, the alternative is some finite unique initial segment. The conclusion of a unique sequence missing from the list is a contradiction. Mathematics doesn't prove SQUAT, you find *allowable branches* in the theory and CALL IT TRUE, until more constraints on the formula are discovered. Herc === Subject: Re: countability of reals <3428nfF46vp3aU1@individual.net> SEE THE DIAGONAL? Its NOT A NUMBER. You don't prove anything by > proving its not a number on the list, we know that. its an illusion of a unique sequence. > If you treat it like a number then you get contradictions. Two things (it's been a long time since I've thought about this stuff): 1. pi: 3.1415926....., is a number exactly like that diagonal, because we can permissibly say it has an infinite number of digits - that is, no matter how large a number you bring forth, we can find a digit of pi that is further along than that number's number of digits. There's no inherent contradiction in claiming pi is a number - however it's not a RATIONAL number. 2. I thought that was the purpose of the infamous diagonalization argument - that the diagonal was not a RATIONAL number, thus proving the existence of IRRATIONALS - which are all 'infinite' in length IN THE STANDARD WAY OF EXPRESSING THEM - that is, base-n-I-forget-the-term-strings-of-digits. In fact, there are infinite numbers that aren't irrational - 1/3 = .3333333.... There's no inherent problem with having an infinite number of digits, but there are certain kinds of numbers that can't be represented as a RATIONAL number - quotient of two integers. c. Could someone pls. post here the TEXT of the diagonalization argument, the one I've described above, so we can be sure we're all on the same page? (I could stand the refresher). And if I've said anything above that is not right, pls correct. Ken === Subject: Re: countability of reals <3428nfF46vp3aU1@individual.net> |> SEE THE DIAGONAL? Its NOT A NUMBER. You don't prove anything by > proving its not a number on the list, we know that. its an illusion of a unique sequence. > If you treat it like a number then you get contradictions. Two things (it's been a long time since I've thought about this stuff): 1. pi: 3.1415926....., is a number exactly like that diagonal, because we can permissibly say it has an infinite number of digits - that is, no matter how large a number you bring forth, we can find a digit of pi that is further along than that number's number of digits. There's no inherent contradiction in claiming pi is a number - however it's not a RATIONAL number. 2. I thought that was the purpose of the infamous diagonalization argument - that the diagonal was not a RATIONAL number, thus proving the existence of IRRATIONALS - which are all 'infinite' in length IN THE STANDARD WAY OF EXPRESSING THEM - that is, base-n-I-forget-the-term-strings-of-digits. In fact, there are infinite numbers that aren't irrational - 1/3 = .3333333.... There's no inherent problem with having an infinite number of digits, but there are certain kinds of numbers that can't be represented as a RATIONAL number - quotient of two integers. c. Could someone pls. post here the TEXT of the diagonalization argument, the one I've described above, so we can be sure we're all on the same page? (I could stand the refresher). And if I've said anything above that is not right, pls correct. Ken c. I forget the diagonalization argument I am embarrassed to say. === Subject: Re: countability of reals > In other words: Everybody is crazy except me and this invisible purple elephant that is standing beside me. === Subject: Re: countability of reals > Clue: when something has been proved to be impossible, and > the proof is simple enough to be understood by anyone with > a passing score in ninth grade algebra, posting a muddled > mess claiming to have done that impossible thing anyway to > a newsgroup frequented by folks with earned PhDs in math > is a fast track to ridicule and Usenet kookdom nominations. In my view it is just possible (though very unlikely) that a proof simple enough to be understood by anyone with a passing score in ninth grade algebra might contain a flaw so subtle that it has not yet been spotted by folks with earned PhDs in math. Do you really consider this to be an absolute impossibility? -- Alec McKenzie === Subject: Re: countability of reals http://mygate.mailgate.org/mynews/talk/talk.bizarre/eadb3c5add68a9466026a936 7 9df62e0.48257%40mygate.mailgate.org > In my view it is just possible (though very > unlikely) that a proof simple enough to be > understood by anyone with a passing score in ninth > grade algebra might contain a flaw so subtle that > it has not yet been spotted by folks with earned > PhDs in math. > Do you really consider this to be an absolute > impossibility? Considering that the finding that the reals are uncountable has been the Kook-of-the-month-club target in sci.math for as long as I've been on Usenet (roughly two decades), and has yet to even receive a competent rebuttal, much less a correct one, I put the likelihood of the next candidate kook succeeding right up there with the chance of all the air molecules in this room ending up by chance on the other side and me exploding into the resulting vacuum: i.e., it doesn't dominate my lifetime list of things I worry much about or lose sleep over. HTH xanthian. -- === Subject: Re: countability of reals >> Clue: when something has been proved to be impossible, and >> the proof is simple enough to be understood by anyone with >> a passing score in ninth grade algebra, posting a muddled >> mess claiming to have done that impossible thing anyway to >> a newsgroup frequented by folks with earned PhDs in math >> is a fast track to ridicule and Usenet kookdom nominations. >In my view it is just possible (though very unlikely) that a proof >simple enough to be understood by anyone with a passing >score in ninth grade algebra might contain a flaw so subtle that >it has not yet been spotted by folks with earned PhDs in math. >Do you really consider this to be an absolute impossibility? Well, personally, I would be less surprised if I were to be unexepctedly teleported to Mars (which, according to the laws of quantum mechanics, is not absolutely impossible) than I would be if there were to be a flaw in the proof that there is no bijection between the rational and irrational numbers. Derek Holt. === Subject: Re: countability of reals >> Clue: when something has been proved to be impossible, and >> the proof is simple enough to be understood by anyone with >> a passing score in ninth grade algebra, posting a muddled >> mess claiming to have done that impossible thing anyway to >> a newsgroup frequented by folks with earned PhDs in math >> is a fast track to ridicule and Usenet kookdom nominations. >In my view it is just possible (though very unlikely) that a proof >simple enough to be understood by anyone with a passing >score in ninth grade algebra might contain a flaw so subtle that >it has not yet been spotted by folks with earned PhDs in math. >Do you really consider this to be an absolute impossibility? > Well, personally, I would be less surprised if I were to be unexepctedly > teleported to Mars (which, according to the laws of quantum mechanics, > is not absolutely impossible) than I would be if there were to be a flaw > in the proof that there is no bijection between the rational and irrational > numbers. > Derek Holt. funny, are you sure you're not quoting a mob of text book authors there. 100s of you online posted the same thing about the halt proof just months ago, then after 20,000 posts here over 4 months, I disproved Halt and its been silent since. you all stop talking. the sci.math law, no one can make new mathematics online. === Subject: R3COGNITION OF THE HALTING SOLN > the Halting Problem recently when you were feeling well. Of all > those who talked about that in real-world context, you had the > best idea in my opinion. I think that was a major contribution, > and I am sure I will be able to use that idea someday. > Larry the solution for those who missed it among the heated discussion. halt(f, a) -> pHalt2(n) -> P(UTM(n, f(a))='halt') = 0.5 !halt(f, a) -> pHalt2(n) -> UTM(n, f(a))='don't know' pHalt2 is an (infinite) set of functions that satisfy pHalt below, with the added contraint P(UTM(n, f(a))='halt') = 0.5. Therefore a randomly selected pHalt function has 50% odds of answering the halt program for some given input. (different set notation) halt(f, a) -> En, n e pHalt, UTM(n, f(a))=true !halt(f, a) -> An, n e pHalt, UTM(n, f(a))='don't know' For example, pHalt could be an infinite set of Universal Turing Machines, but they timeout after various amounts of time. If the parameter f(a) halted in that time then it outputs TRUE, otherwise it timed out and outputs DONTKNOW. Since the number of these partial Halt functions is infinite, we should come across one with enough test cycles to determine if any function halts. By itself pHalt is still useless as it only gives one output value, HALTS. But if each pHalt function has 50% chance of being correct (something more powerful than the UTM), then determining NOTHALT is a simple probabilistic procedure. e.g. pHalt2() = {2, 6, 33, 655, ..} i.e. pHalt(2) = true pHalt2(6) = true pHalt2(33) = true .. pHalt2 is an infinite set of godel numbers whos functions can be parsed by a UTM. each of these functions has independant probability of 50% of answering if the input function halts, otherwise it will return DONTKNOW. As an example. the program with godel number 999 is 10 goto 10 UTM(2, 999) = DONTKNOW (ignoring the parameter of function 999) UTM(6, 999) = DONTKNOW UTM(33, 999) = DONTKNOW UTM(655, 999) = DONTKNOW ... Since : !halt(f, a) -> pHalt2(n) -> UTM(n, f(a))='don't know' the program with godel number 1000 is 10 print 10 UTM(2, 1000) = HALT UTM(6, 1000) = DONTKNOW UTM(33, 1000) = HALT UTM(655, 1000) = DONTKNOW Given : halt(f, a) -> pHalt2(n) -> P(UTM(n, f(a))='halt') = 0.5 1/ program 999 does not halt with probability of error 1/16 2/ program 1000 halts A *solution* to the halting problem is apparent, but it cannot be fully represented as an *algorithm*, so the halting proof will fail here to surface a contradiction. Herc === Subject: Re: countability of reals http://mygate.mailgate.org/mynews/talk/talk.bizarre/e947e9cc327496c2bc055d50 c 05463a6.48257%40mygate.mailgate.org > 100s of you online posted the same thing about the > halt proof just months ago, then after 20,000 > posts here over 4 months, I disproved Halt and its > been silent since. Well, no, your proof was worthless, and, after I pointed folks toward your online posting record of kookery, your reputation didn't warrant anyone wasting further time to respond, as you have amply demonstrated your inability to participate in Usenet's life of reason. Quit working so hard, Herc, you've already earned the Lifetime Kook Ribbon with Gold Clusters, and your math proofs, numerology beliefs, and repeated claims of deity status have been retired to the Kooknet Hall of Fame there to be the subject of awe that anyone could possibly be that stupid, and continuous 24x7 ridicule by passing throngs of newbies to whom you are shown as an object lesson in ways not to behave on Usenet. You're in no danger of being taken seriously on any subject anywhere on Usenet in your remaining lifetime, or in the remaining lifetime of the universe down to the last surviving baryon, come to that. You've got your well earned laurel, rest on it. In fact, since it is one of the hardier varieties, you can safely sit and spin on it. HTH xanthian. -- === Subject: Re: tards Please do not respond to this message, thank you. === Subject: Re: Limit of sequence >Dwayne escribi.97: >> Well, this gets interesting... >> The answer in the book (e^(3/2) comes from the author applying >> L'hopitals' rule to: >> [ (n^2+2)/(2n^2+1) ]^(n^2) = e^ [ n^2 * Ln( (n^2+2)/(2n^2+1) ) >> ], or >> = e^ [ Ln( (n^2+2)/(2n^2+1) ) / (1/n^2) ]. >> Then as n goes to infinity, both numerator and denominator of the >> exponential funciton go to zero. >As n goes to infinity, Ln((n^2+2)/(2n^2+1)) goes to Ln(1/2) = - Ln(2), then >you can`t apply L'H.99pital's rule. >The limit is, as it is stated, clerly 0. But I guess that it is wrongly >transcripted. Nope, the book answer was right. You misapplied L'Hopital's Rule. ans. ~.5 >-- >Ignacio Larrosa Ca.96estro >A Coru.96a (Espa.96a) >ilarrosaQUITARMAYUSCULAS@mundo-r.com >> Thus, L'hopital's Rule was applied >> and that gave 3/2 as the limit, and hence the limit of the original >> function would be e^(3/2). >> Now, is this the limit or was L'hopital's Rule misapplied? I leave >> that as your exercise. >> Dwayne Hickman Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: Limit of sequence >Dwayne escribi.97: >> Well, this gets interesting... >> The answer in the book (e^(3/2) comes from the author applying >> L'hopitals' rule to: >> [ (n^2+2)/(2n^2+1) ]^(n^2) = e^ [ n^2 * Ln( (n^2+2)/(2n^2+1) ) >> ], or >> = e^ [ Ln( (n^2+2)/(2n^2+1) ) / (1/n^2) ]. >> Then as n goes to infinity, both numerator and denominator of the >> exponential funciton go to zero. >As n goes to infinity, Ln((n^2+2)/(2n^2+1)) goes to Ln(1/2) = - Ln(2), then >you can`t apply L'H.99pital's rule. >The limit is, as it is stated, clerly 0. But I guess that it is wrongly >transcripted. That's not how you apply L'hopitals Rule. lim f(n)/g(n) = f'(n)/g'(n) --> to convergence n-->oo I get. ans. ~ .5 >-- >Ignacio Larrosa Ca.96estro >A Coru.96a (Espa.96a) >ilarrosaQUITARMAYUSCULAS@mundo-r.com >> Thus, L'hopital's Rule was applied >> and that gave 3/2 as the limit, and hence the limit of the original >> function would be e^(3/2). >> Now, is this the limit or was L'hopital's Rule misapplied? I leave >> that as your exercise. >> Dwayne Hickman Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/