mm-989 >I am trying to improve my perception of imaginary numbers. Think of them as Mathematic objects that happen to have utility in other Þelds. >Can anybody point me to any resource describing their TRUE physical >meaning, There is none. What is the true physical meaning of 5? Is 5 degrees Celsius the same as 5 parsecs? Is either the same as 5 cc? >I just can¹t believe that mathematicians came up with them just >because they couldn¹t solve x^2+1=0... Why not? That is an imanently respectable reason all by itself. -- 872 === Subject: Re: Rudin¹s proof of the open mapping theorem >> The fact that y_n is in the closure of L(V_n) shows that y_n -> 0. >I wish I knew why... >> where I waved my hands is exactly what goes wrong in the >> not-quite-counterexample below. We¹re set if we can show that >> in any TVS, if O is a neighborhood of 0 then there exists another >> neighborhood of 0, V, such that the closure of V is contained >> in O, right? This ustbe true... >> Aha. It follows from the (joint!) continuity of the operations >> that the origin has a neighborhood V such that V - V >> = {x - y : x, y in V} is a subset of O. Suppose that x is in >> the complement of O. Then x + V is a neighborhood of x, >> and V intersect (x + V) is empty, hence x is not in the >> closure of V. >Well, I already knew that, for topological groups, T_0 <=> T_3. Figured it must be in Rudin, I don¹t have the book here. This morning I thought that this was the only problem. Let¹s see: We want to show that y_n -> 0. Let O be a neighborhood of the origin. Choose W, a neighborhood of the origin such that the closure of W is contained in O. Now since L is continuous there exists delta > 0 so that d(x, 0) < delta implies Lx is in W. Hence there exists N so that L(V_n) is a subset of W for all n > N. So the closure of L(V_n) is contained in O for all n > N. So y_n is in O for all n > N. So for every O there exists N such that y_n is in O for all n > N. Maybe I¹m missing something here... >But I do >not see how to deduce from that fact that lim_n y_n = 0. Besides, to >which neighborhoods of 0 do you apply this? The closures of L(V_n) are >already closed neighborhoods. I suppose that you apply it to the >interiors of the closures of L(V_n), but in my fake counter-example >these are equal to the whole space. >Jose Carlos Santos ************************ David C. Ullrich === Subject: Re: limit of sequence of functions >I got this problem from an archive of functional analysis qualifying exam >problems... Well ok then, revise the question: Do you know anything about functional analysis? About Banach spaces in particular? There are like three big results that you learn at the start of the study of Banach spaces, one of which is directly applicable here. Hmm. Do you know any results about Banach spaces where the conclusion is hen the norms of ___ are bounded ************************ David C. Ullrich === Subject: Re: JSH: GOT IT!!! Algebraic integers check proof >Well I kept Þddling at it, and now have on my blog the complete proof >that there¹s a problem with the ring of algebraic integers by checking >what happens with the roots of x^2 - x + 42 and y^2 + by - 7, with an >algebraic integer b. You got it? But you¹ve been saying for some time that you had it, haven¹t you? Does the fact that you say you¹ve Þnally found a correct proof mean that the previous ones were wrong? Huh. >It¹s all at >http://mathforproÞt.blogspot.com/ >which is easier to keep up with than posting on Usenet having to do >updates and corrections in a thread. >Now I can write another paper (other one is STILL at the math journal >where it¹s been since Augus, but that¹s not necessarily strange as it >can take months). >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Splitting Þeld, algebraic integer factors > > That¹s a nifty and powerful result. Why am I the one who had to > discover it? > > > James Harris > > Because you¹re so special in God¹s eyes. > > The math isn¹t that complicated. Consider that I¹m using > > x^2 - x + 42 = 0, which has (1 + sqrt(-167))/2 as on of its roots, and > > y^2 - by - 7 = 0, which has as one of its roots > > (b + sqrt(b^2 + 28))/2. > > So I can simply introduce z, where > > (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2. > > Now it doesn¹t take a math genius to see that you can¹t multiply by an > algebraic integer z, and manage that sqrt(-167) on the left side > without having b^2 + 28 = -167n^2, which is why I brought up the > splitting Þeld. > > But to see replies in this thread you¹d think I brought up the Tooth > Fairy. > You brought up the equation > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0, derived from > (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2, after several false > starts, and claime that that equation > . > But for b = 6, which was an integer last time I looked, > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 > = 7z^4 + 6z^3 + 299z^2 - 36z + 252 > = (z^2 + z + 42)(7z^2 - z + 6). > which certianly Þts my deÞnition of being reducible. Which brings up > teh question of what your deÞnition of irreducible versus reducible > must look like. AARGGHHH!!! and YEAH!!!! Why did it have to be you Virgil? Is Virgil your real name or a pseudonym? Rick Decker and ora BaronFINALLY concede? (If you I¹m crazy here then you haven¹t looked carefully at Virgil¹s result. Think about it. I¹ll add more later if necessary.) James Harris === Subject: Re: JSH: Splitting Þeld, algebraic integer factors > > > That¹s a nifty and powerful result. Why am I the one who had to > discover it? > > > James Harris > > Because you¹re so special in God¹s eyes. > > The math isn¹t that complicated. Consider that I¹m using > > x^2 - x + 42 = 0, which has (1 + sqrt(-167))/2 as on of its roots, and > > y^2 - by - 7 = 0, which has as one of its roots > > (b + sqrt(b^2 + 28))/2. > > So I can simply introduce z, where > > (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2. > > Now it doesn¹t take a math genius to see that you can¹t multiply by an > algebraic integer z, and manage that sqrt(-167) on the left side > without having b^2 + 28 = -167n^2, which is why I brought up the > splitting Þeld. > > But to see replies in this thread you¹d think I brought up the Tooth > Fairy. > > You brought up the equation > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0, derived from > (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2, after several false > starts, and claime that that equation > > . > > > But for b = 6, which was an integer last time I looked, > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 > = 7z^4 + 6z^3 + 299z^2 - 36z + 252 > = (z^2 + z + 42)(7z^2 - z + 6). > > which certianly Þts my deÞnition of being reducible. Which brings up > the question of what your deÞnition of irreducible versus reducible > must look like. > AARGGHHH!!! and YEAH!!!! > Why did it have to be you Virgil? > Is Virgil your real name or a pseudonym? How is my name relevant to the correctness of my methematics? > Rick Decker and ora BaronFINALLY concede? They have nothing to concede. You do. Yoou have claimed that there was no raational integer value of b for which 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 was reducible, and no altgebraic integer value of b for which the equation in zm 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0, had an algebraic integer solution. Both of these claims are now trivially false. > (If you I¹m crazy here then you haven¹t looked carefully at Virgil¹s > result. Think about it. I¹ll add more later if necessary.) > James Harris JSH is now taking a new tack: that counterexamplees to his case prove his case. It is not a tack that many of the mathematically apt will follow. === Subject: A question on inÞnite integral domains Hi Has any one seen a proof that showed that an inÞnite integral domain cannot be a Þeld? All the books I have prove that a Þnite intergal domain is a Þeld but don¹t give any indication about why inÞnite domains can¹t be. For instance, Z is not a Þeld; I know why it is not; but does the same argument hold for any inÞnite domain. A hint about the proof would be welcome; for I am sure it is trivial. Navin Kadambi === Subject: Re: A question on inÞnite integral domains > Has any one seen a proof that showed that an inÞnite integral domain cannot > be a Þeld? All the books I have prove that a Þnite intergal domain is a > Þeld but don¹t give any indication about why inÞnite domains can¹t be. Who says they can¹t be? >For > instance, Z is not a Þeld; I know why it is not; but does the same argument > hold for any inÞnite domain. Z is not a Þeld because 2 is not a unit in Z. > A hint about the proof would be welcome; for I am sure it is trivial. Is 2 a unit in Q? Can you generalize? -- Dave Seaman Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling. === Subject: branch of log z >You can picture log z as being like a ramp in a parking > garage, where each each time you go around you wind up at the next level > of the garage. So You proceed from a plane to rotation on the plane, to a movement? The difference of two directions is an angle, which never exceeds 360 degrees, but a rotation is unlimited. The local time is the difference of direction to the dateline plus the time of the dateline. ( Our ancestors also got worried about this, when the shadow on the sundial grew longer and longer. The erohour of the new day started in twilight with bats þuttering around.) What is Your picture of z ? Usually an arrow or a position on a plane. Try to picture ln z on the same plane (where You picture z) - You¹ll get something spooky like a Riemann surface. You have one plane for the domain, for z and another one for the range, for ln z. Can You combine these into one in 3D or in 3D plus time? The ramp in the parking-garage spiralling up over the plane is only a partial picture, as the height over ground (corresponding to the angle) is only the second component of ln z, the Þrst is ln |z| or ln (length of z-arrow). Solve this and You solved the next item on the agenda after ln too: the visualization of raising a base z to the power of an exponent w, of z^w, z to the power of w. You can look at it as a pair of two vectors.The result of the calculation w*ln z and e^(w*ln z) reduces to the 2D-domain or to the 2D-range, so reducing the information, but giving some knowledge to you. What i really like in Your letters, as well in Lynn Kurtz letter to is the treatment of complex numbers as pairs of two reals, no differennce between i-axis and y-axis. Is this only for insiders ?! Why not propagate this? My attempt: http://i-is-no-longer-imaginary.gmxhome.de or, the same: http://www.i-z.eu.tt . Have fun Hero === Subject: Re: Hausdorff dimension of graph of Weierstrass function. >> The lower bound of the Hausdorff dimension of the graph of >> f(x)=sum^{infty}_{i=1}t^{(s-2)i}sin(t^ix) For 11. >> is s. Is it equal to s? >If anyone is still interested > s >= HausDim(graph f) >= s - c/log(t) > Huh. GIve us a hint/sketch of how you prove that. It is stated, and a proof method merely hinted at, on page 151 of Kenneth Falconer ractal Geometry Mathematical Foundations and Applications I¹ll quote: The known lower bounds come from mass distribution methods depending on estimates for L{t: (t, f(t)) in B} where B is a disc and L is Lebesgue measure. The rapid small-scale oscillation of f ensures that the graph is inside B relatively rarely, so that this measure is small. In this way it is possible to show that there is a constant c such that s >= dim_H graph y >= s - c/log(lambda) so when lambda is large the Hausdorff dimension cannot be much less than the conjectured value. -- G.C. === Subject: Re: How fast is the ConÞnued Fraction factorization algorithm? > I¹m factoring numbers n using the Continued Fraction algorithm, Þnding > pairs (x,y) such that x^2==y mod n. The y values are factorized as y = > product(p_i^e_i) * p, where p_i are primes<=307, p<307^2 is another > prime, and e_i>=0. > Isn¹t this usually called the Quadratic Sieve? > Where do continued fractions come into it? Well, the y have the property that |y| < 2 sqrt(n) and hence are comparatively easy to factor. This is what the Continued Fraction algorithm ensures. This is because if p/q is close to sqrt(n) then p^2 is close to n*q^2 and hence p^2 mod n is much smaller than n. > Regarding the speed, there is a standard trick using logarithms > to determine if a number is smooth (has small factors), > using the fact that if it is not smooth > there must be quite a large iscrepancy > Do you use this? No, and I¹m not sure what that trick is. Could you explain it in a few more words. First of all, are we talking about discrete logarithm or þoating point logarithm? -Michael. === Subject: Re: Knuth Toom-Cook, exercise 4.3.3-(4) > Has anyone implemented the n-way Toom-Cook multiplication algorithm from > Knuth 4.3.3-T with the improvements from exercise 4.3.3-(4)? I want pointers > to how to implement the Toom-Cook algorithm with evaluation points at > $-r,..0..r$ or at $1^2,2^2,4^2,8^2,..,(2^r)^2$. You could have a look at GNU MP (http://www.swox.com/gmp/). -Michael. === Subject: Re: Checking ring of algebraic integers > > It turns out that there¹s an incredibly simple way to check the ring > of algebraic integers and see that there¹s a problem with how it¹s > currently taught. > > > Is your problem now with the TEACHING of the subject? Can you be more > speciÞc? > My assertion is that mathematicians teach that given > x2 - x + 42 = 0 > the roots, which are algebraic integers, have non-unit algebraic > integer factors in common with 7 in the ring of algebraic integers. > It turns out they do not, and I¹m working out the argument--drafts > Þrst and then reÞning--as I¹ve done before. > This thread is now the key one, as I take the step of eliminating > sqrt(-167) without multiplying it out which goes to the splitting > Þeld of > x2 - x + 42 = 0 > for those who wonder what splitting Þeld I¹m discussing. > Does that make sense to you? > James Harris Since you fail to state what it is that you are eliminating sqrt(-167) from, no it does not make sense. === Subject: Re: e is transcendental > Since e^[iPi]=cosPi+isinPi > or , e^[iPi]=-1+i[0] > then there are two solutions here, to the given equatio: > A) e^[ipi]=-1 the real part solution and > B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio. No, the conclusion from e^[iPi]=-1+i[0] is Re(e^[iPi]) = Re(-1+i[0]) = -1 AND Im(e^[iPi]) = Im(-1+i[0]) = 0 -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: expectation problem === >Subject: Re: expectation problem >Message-id: === >Subject: Re: expectation problem >Message-id: === >Subject: Re: expectation problem >Message-id: In a nutshell: >>To keep total prize money paid to a minimum, how fast do you want >the >>voice guessed? >Never >But this would be a promotional gimmick the cost of witch is Þxed. >the comp would stop when the budget is use. >> In this schedule, there is clearly a point that minimizes >> the liability to the BBC. They would want the questions >> hard enough such that the audience needs 16 tries to >> get it right, but no more than 16. >the goal of the BBC is promotion the cost is FIXED Not the way the situation was described. The cost is Þxed only if the average prize/contestant is 30 quid. That is possible but unlikely (it would only happen if every voice was correctly guessed on the Þrst try). The reality is that the cost to the BBC is not Þxed but varies depending on how difÞcult it is to recognize the voice. If they simply increased the prize by 30 quid every time an incorrect guess was made, then the cost would be Þxed and there would be no motivation to moderate the difÞculty of guessing the voices. If the BBC wanted to minimize the cost, they would want the voices as difÞcult as possible. Higher difÞculty means lower cost. >> With the original schedule, there is no motivation to >> moderate the difÞculty. >Carl >-- >If its Monday then I am a fool but not ignorant. >> -- >> Mensanator >> Ace of Clubs >This kind of thing is promotion not something to minmize, the cost is >in a sense Þxed ie no more than 60 per day. the promoter has to balance >reward against excitment, But this intangible was not part of the OP¹s question. It may be that running a contest for three weeks without a winner creates boredom and subsequent loss of ratings, but we weren¹t asked to factor that in. And no one would budget the worse case scenario of 60 quid/day. They would look at the average number of trials to get the answer, say 6, and set the budget accordingly. The game planners task is then to see that the contest remains popular by giving away the amount they have budgeted for, but not give away too much so as to go over budget and get their bosses mad at them. Without altering the prize schedule, the proper way to frame the original question would be: Given a budget of x quid/week, how fast do you want the voice guessed? >-- >If its Monday then I am a fool but not ignorant. -- Mensanator Ace of Clubs === Subject: Re: Probability of Þnding a Prime Number I think maybe what he is trying to say is that PROD[p/(p-1)] is the expected distance from n to the next prime after n. So let n=11. We expect to Þnd the next prime near n + (3/2)(5/4)(7/6) = 11 + 35/16 = 13.1875. Still, it¹s not really a probabilistic event at all. You deÞnitely have NOT stated things correctly, but maybe it¹s not entirely bunk.... > I do think that I have stated things correctly. Yes, the probability of > Þnding primegreater than p is 1. However, the probability of Þnding > the ext primeis not 1. You can think of it by asking how far past p do > you have to go before you get to the next prime? If the probability is 1/10 > then you can expect to Þnd a prime somewhere between p and p+10. If the > probability if 1/100 then you can expect to Þnd a prime somewhere between p > and p+100. Exactly where can¹t be determined. > What I am fairly certain I have discovered is that this probability depends > only on the primes from 3 to p and equals the product((p-1)/p). > David > I have placed a paper on my web site at http://www.guffy.net/primes.pdf. > In it I prove the following: > Given the prime numbers 3 to p, the probability of Þnding the next prime > is > exactly: > Product((p-1)/p) > The paper is 3 pages and contains 4 equations. I would appreciate any > comments. > David === Subject: Re: Probability of Þnding a Prime Number > I found this in one of your posts (they¹re easy to Þnd given that you sign > everything with that horse¹s ass comment): > he probability that a large integer N is prime is the probability that > it is not divisible by 2,3,5,7,11...sqrt(N). This > probability is (1-1/2) * (1-1/3) * (1-1/5) .... And this was in response to: > If I (or anyone else) has a list of all the primes > from 3 to p, the probability that I (or anyone else) > can Þnd the next time is 1. I can ALWAYS > Þnd the next prime. David, the point that Bob Silverman is making is that your claim that he probability of Þnding the next primeis PROD((p-1)/p) is a very incorrect statement. But you have since corrected this statement by saying that your probability applies to a speciÞc N and the product is over all prime less than N... as opposed to saying anything about a Œnext prime.¹ J === Subject: Re: Probability of Þnding a Prime Number >So, the conclusion I came to is nothing new. However I think the approach I >describe in the paper is relatively novel -- at least I enjoyed discovering >it. Good! Discovery of even simple curiosities can be quite enjoyable. >On a personal note, I am an amateur who enjoys reading about math and trying >to make some serious attempts at solving some problems. Your response (as >well as some others) to my oringinal post was very discouraging. Mathematics, as taught and learned, is often designed to be discouraging. Rich === Subject: Re: Rationals are Uncountable > 1. N = smallest inductive set (the deÞnition we have been using). > a. DeÞne inductive set. > b. Show that the intersection of all inductive sets is a nonempty > set. > c. DeÞne this set to be N. > d. Show that N is a Peano set. This is the method we were discussing. > Would you consider this set to be inductive? > [0,1,2,...,w,w+1,w+2,...] > Yes. That¹s the same set that I called w+w in a previous post. It¹s the > next limit ordinal after w. The AoI guarantees there is at least one set that is closed under the successor function. Let¹s call this set A. Obviously, A can¹t have a successor or it wouldn¹t be closed under the successor function. For the sake of argument let¹s say: A = (0,1,2,...,w,w+1,w+2,...) 0 is not the successor of any member of A, so we could remove it and still have a set closed under the successor function. Similarly, we could remove (0,1), (0,1,2), etc. None of these sets would be inductive sets because they wouldn¹t include 0. We might be able to remove w from A and still have a set that is closed. We know w is not the successor of any natural number. But, we also know that A contains members that are not natural numbers. w could be the successor of one of these other members. I don¹t see how we can show there exists an inductive set that is closed under the successor function that is smaller than A. A obviously contains members that are not natural numbers (in our example). Russell - 2 many 2 count === Subject: Strings, Loops & Weaves as Opera Paul, you say: ... Hilbert arrived at a set of Þeld equations by a variational procedure from a Lagrangian that was not supported by detailed physical arguments. .... JS: Hardly unusual in physics today. :-) TS: I think that in fact the Hilbert Lagrangian IS supported by physical arguments, although it may be true that Hilbert did not present or know the physical arguments. PZ: Yes, this is what I meant -- Hilbert did not in his paper justify all the details of his Lagrangian by physical arguments. Of course, others later provided a deeper foundation for Hilbert¹s original assumptions. As far as I know, Einstein did not object to Hilbert¹s formulation.. My point here is that Hilbert was clearly guided by the physical ideas and arguments that were presented by Einstein at Goettingen in formulating his theory, and yet did not justify all the details of his derivation by physical arguments. TS: This is important, not because of any personal credit Þght between some followers of Hilbert and Einstein, but because the physical basis of the Hilbert approach is quite different from the physical basis of the Einstein approach, and the physical basis of the Hilbert approach is quite similar to the Lagrangian formulation of the Standard Model and path-integral quantum theory, while the physical basis of the Einstein approach is (in my view) the source of the widely held attitude that General Relativity and Standard Model Quantum Theory have little in common and are hard to unify. PZ: Are you saying that the two models are completely independent? Or that they are simply different perspectives on the same physics? JS: My position on this is that the standard model, i.e. local gauge invariance in globally þat space-time with conventional special relativistic quantum Þeld theory ---> supersymmetric string theory is the UNSTABLE substratum even at T = 0 degK. The resulting vacuum phase transition is essentially the vacuum coherent inþation and the phase of the inþation Þeld gives Einstein¹s ZERO TORSION limiting case of Einstein¹s 1915 GR (basically from locally gauging ONLY the 4-parameter translation group in my current toy model, i.e. compensating gauge force Þeld is the ODLRO lattice distortion world hologram (t¹Hooft-Susskind) Þeld: du(x) ~ (Loop Quantum of Area)(Phase of Vacuum Coherent Inþation Field),u (1) u = 0, 1, 2, 3 *The primacy of World Hologram Quantum of Area is obvious, indeed trivial, from my eq. (1). Einstein¹s c-number curved spacetime is ODLRO metric elasticity emergent guv(Curved) = Globally Flat + Strain Tensor of World Crystal Lattice from du(x). U(1) local gauge transforms on Phase of Inþation Vacuum Coherence --> generalized Diff(4) coordinate transformations on guv. Eq (1) is to the 4D metric elasticity (Sakharov) world crystal lattice (H. Kleinert) as Bohm¹s 3D constraint v(IT) = (h/m)Grad(Phase of BIT pilot wave of active information) is to the linear unitary probabilistic nonlocal micro-quantum liquid --> localMACRO-QUANTUM superþuid in More is different (PW Anderson) sense. (ODLRO of Oliver Penrose, BCS non-perturbative transition, Lambda transition ...) where generalized phase rigidity makes the fragile nonlocal micro-quantum potential with signal locality into a nonlinear nonunitary non-probabilisitc robust local MACRO-QUANTUM potential with signal nonlocality. Note that Tegmark¹s environmental decoherence only works on fragile micro-quantum nonlocal phases! Local MACRO-QUANTUM phase Þelds are IMMUNE to decoherence as seen, for example, in the enormous space-time stiffness where the Witten string tension is huge: Witten String Tension ~ hc/(Loop Quantum of Area) Loop Quantum of Area ~ (QED coupling)^-1(Planck Area) Stable Loop Quantum Gravity with dynamical background-independent curved spacetime responding, albeit weakly, to mostly zero point þuctuation dark stress-energy density engineerable exotic vacuum currents, is dual to Witten string theory with unstable absolute non-dynamical globally þat space-time and string world sheet Þbers weaving the fabric of reality for our Elegant Tailors Greene & Co of Saville Row. :-) http://math.boisestate.edu/gas/patience/html/index.html 38. When I go out of the door ... 39. I¹m a Waterloo House Young Man Note 3D quantum computer Penrose spin network 0-Dim point is a 3D volume. Spin network 1D (string) line is a 2D area. Therefore, the quBIT string is dual to the IT loop Area Quantum! Discrete string BITs of the Cosmic Computer Program -> Quantized Area of the World Crystal Lattice. The scale of the unit cell is related to the inverse string tension. All 3D lattice tilings are spin networks, but not all spin networks are lattice tilings. Go to spin foams for 4D World Crystal. TS: It is widely known that the Einstein approach is based on taking curved spacetime and the equivalence of inertial and gravitational mass/energy as fundamental. PZ: Depending on what you mean by curved spacetime. My information is that Einstein did not view *Riemann* spacetime curvature, but rather the *connection Þeld*, as fundamental. JS: Splitting hairs, I mean strings :-) PZ: As you know, non-vanishing connection Þelds do not necessarily imply non-vanishing Riemann curvature. TS: It is less widely known that the Hilbert Lagrangian can be obtained by gauging (very similarly to the way Standard Model forces such as SU(3) color force are gauged) a pre-gravity force that is either an anti-deSitter force with Spin(2,3) = Sp(2) gauge group or a conformal force with Spin(2,4) = SU(2,2) gauge group. PZ: OK. But exactly how does this provide a physical motivation for Hilbert¹s choice of Lagrangian? JS: The constant curvature in homogeneous dS or Ads is Einstein¹s cosmological constant that I make into a local inhomogeneous zero point stress-energy current density exotic vacuum w = -1 uniÞed dark energy/mattter Þeld killing the Killing tangent vector Þeld symmetries I suppose? :-) TS: The anti-deSitter force is the one that was initially used back in the 1970s by MacDowell and Mansouri (and others) as a way to get the Hilbert Lagrangian, and it was soon after realized that the Conformal force could also produce the Hilbert Lagrangian. PZ: But aren¹t these pre-gravity Þelds as hypothetical as Hilbert¹s Lagrangian? Just a naive question. TS: The motivation for the anti-deSitter / Conformal / Hilbert approach to gravity was to produce a gravity model that would Þt naturally into supergravity theories which were based on Lie Superalgebras such as OSp(N|2). OSp(N|2) supergravity is sort of an amalgam of two gauge groups: O(N) (or SO(N)) from which people tried to get the Standard Model and Sp(2) = Spin(2,3) = anti-deSitter group from which they did get gravity, in the Hilbert Lagrangian formulation. PZ: OK, this is interesting. I see that this is motivation, but from my POV it is also purely hypothetical and formal in character. JS: There is very little motivation in Physics Today. Physics Today is Physics without physics done by Prancing Mathematicians in Physicist¹s G-Strings. :-) TS: The supergravity program was seen to have failed in the early 1980s, not because of problems with anti-deSitter / Conformal / Hilbert Lagrangian gravity, but because of problems with getting SO(8) to produce the Standard Model (in my opinion, that failure was due to their misuse of SO(8), but their work motivated me to use it in a different way to produce my physics model which in my opinion PZ: OK. TS: On the other hand, the Einstein physical view of General Relativity, taking curved spacetime and equivalence as fundamental, has not led to a successful uniÞcation of gravity and the Standard Model, although Wheeler did succeed in formulating a Geometrodynamics that uniÞed gravity with classical electromagnetism. PZ: Yes -- and I have a theory under construction as to why this might be. TS: Einstein¹s efforts toward a uniÞed Þeld theory led him to work with antisymmetric parts of his geometric tensors, a general approach that was also explored by very smart people including Schroedinger, Pauli, and Heisenberg without success. In my opinion, the physical errors of the Einstein approach are: 1 - taking a continuum structure (curved spacetime) as fundamental, rather than a discrete structure; and PZ: Maybe. Are you saying you don¹t think the logjam in quantum gravity can be broken without going to discrete spacetime at the Planck scale? JS: Wake up and smell the coffee that¹s Loop Quantum Gravity from Ashtekar, Smolin & Co. However I am not sure if they have derived GR yet in a kosher way the way I claim to have done! Problem for Elegant Tailors and Hoola Loopers: Get Guv = -8pi(G/c^4)Tuv from Loop Quantum of Area only! Well with Vacuum Coherence it¹s easy! TS: 2 - using the equivalence principle when it has been shown clearly that the equivalence principle is violated by quantum Þeld theories at Þnite temperature (see for example the work of Donaghue and of Kulikov). PZ: *What* equivalence principle? JS: Who¹s on Þrst? http://www.baseball-almanac.com/humor4.shtml PZ: Einstein¹s classic equivalence principle is simply invalid even at the macro- level IMHO. There is in fact no actual Einstein equivalence, either global or local, macro- or micro-scopic. Wheeler¹s so-called equivalence principle (Wheeler and Ciufolini 1995) is not actually an equivalence principle -- and even as a stripped-down correspondence principle doesn¹t work as advertised, as pointed out by a number of authors. So I see plenty of scope for internal criticism here without even going to discrete spacetime or even quantum Þeld theory. TS: As I see it, the Einstein physical view of gravity is a useful simpliÞed approximation that works pretty well at large distances where spacetime looks effectively smooth and low temperatures where equivalence works pretty well, but it is the Hilbert Lagrangian that lets you see physically how gravity is uniÞed with the Standard Model. PZ: OK. IMHO the Einstein equivalence model works only as a heuristic tool -- a disposable guide for the development of a tensor-Þeld theory of gravity. IMO, as a serious physical model, Einstein¹s classic model is at this point standing in the way of deeper physical insight into the true nature of the gravitational Þeld. Z. Tony PS - You also say ... I know about Mach and Poincare. I know about Voight. I know about Fitzgerald and Lorentz. I know about Minkowski. I know about Gauss, Riemann, Ricci, and Levi-Civita. And I know about Hilbert. .... To that list I would add Clifford, who in 1870 showed that energy and matter can be considered as different types of curvature of space. It is interesting that in 1876, Clifford, who had been working on Riemann surfaces, discovered Clifford Algebras, thus anticipating Dirac¹s rediscovery of Clifford Algebras when he formulated the Dirac Equation of Quantum ElectroDynamics. PS PZ: Yes, I should have included him. This is not exactly an academic discussion. The list was not intended to be exhaustive! My point here was that I am certainly not unaware of the many contributors to the development of physical relativity, but that I still regard Einstein as the author of his own theories. Van Flanderen was trying to argue that Einstein plagiarized Hilbert¹s work on general relativity, which is sheer nonsense from a historical POV, IMO. JS: No that is Koen Vlaenderen not Tom Van Flanderen, though both hold, IMHO, crankish views on Einstein¹s theory of relativity. Is this, perhaps, some kind of Dutch Elm disease that jumped from leaves of grass to humans via Mad Cows? :-) TS: It is interesting that in 1876, Clifford, who had been working on Riemann surfaces, discovered Clifford Algebras, thus anticipating Dirac¹s rediscovery of Clifford Algebras when he formulated the Dirac Equation of Quantum ElectroDynamics. PZ: Yes. Actually, I was aware of Clifford¹s work, and I completely agree with you that he is worthy of mention in this context. JS: http://math.boisestate.edu/gas/yeomen/web_opera/yeomen_15. html :-) === Subject: International Journal of Mathematics - Vol 14 No 10 International Journal of Mathematics View table-of-contents and abstracts at http://www.worldscinet.com/ijm.html Contents: On Weakly Hyperbolic Spaces And A Convergence-Extension Theorem In Weakly Hyperbolic Spaces Pham Viet Duc On Branched Coverings Of Pn By Products Of Discs A. Muhammed Uludag Coordinates for the Teichm.9þler Space of Planar Surface NEC Groups B. Estrada and E. Martinez Rationality Properties Of Manifolds Containing Quasi-Lines Paltin Ionescu And Daniel Naie Reþexivity Of The Translation-Dilation Algebras On L2(R) R. H. Levene and S. C. Power A Remark On A Question Of LempertöHenkin Viet Anh Nguyen Stable Rank-2 Bundles On CalabiöYau Manifolds Wei-Ping Li And Zhenbo Qin For more information, go to http://www.worldscinet.com/ijm.html === Subject: Re: Rationals are Uncountable >> 1. N = smallest inductive set (the deÞnition we have been using). >> a. DeÞne inductive set. >> b. Show that the intersection of all inductive sets is a nonempty >> set. >> c. DeÞne this set to be N. >> d. Show that N is a Peano set. > This is the method we were discussing. >> Would you consider this set to be inductive? >> [0,1,2,...,w,w+1,w+2,...] >> Yes. That¹s the same set that I called w+w in a previous post. It¹s the >> next limit ordinal after w. > The AoI guarantees there is at least one set that is closed > under the successor function. Let¹s call this set A. > Obviously, A can¹t have a successor or it wouldn¹t > be closed under the successor function. Closure means s(x) is deÞned for every x in A. That is, A is the domain of s. Assuming regularity, A itself is not in the domain. > For the sake of argument let¹s say: > A = (0,1,2,...,w,w+1,w+2,...) > 0 is not the successor of any member of A, so we could > remove it and still have a set closed under the successor > function. Similarly, we could remove (0,1), (0,1,2), etc. > None of these sets would be inductive sets because > they wouldn¹t include 0. Correct. > We might be able to remove w from A and still > have a set that is closed. We know w is not > the successor of any natural number. > But, we also know that A contains members > that are not natural numbers. w could be the > successor of one of these other members. Not with the successor function s(x) = x U {x }. > I don¹t see how we can show there exists > an inductive set that is closed under the > successor function that is smaller than A. > A obviously contains members that are > not natural numbers (in our example). If, in our hypothetical model, there is no inductive set that doesn¹t include w, then w is a natural number according to our model. That doesn¹t happen if the successor operation is deÞned as for the standard ordinals. -- Dave Seaman Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: What is the meaning of imaginary numbers? .... > I was always going to get around and give a talk along these lines to > the local high school algebra classes titled There¹s nothing > imaginary about complex numbers, but I never got around to it. High time the notation imaginary been changed, is n¹t it? __ as these are just as real and concrete, if only a different type, becoming complex by mixing (union) with real. === Subject: Re: What is the meaning of imaginary numbers? > Apologies if this question has been answered before... > I am trying to improve my perception of imaginary numbers. I already > have done an extensive search in the web but I keep Þnding > information about how to manipulate them (add, multiply etc), about > their representation on a plane with two axes and other things like > that, that I am already familiar with. > Can anybody point me to any resource describing their TRUE physical > meaning, Their true physical meaning is that they make it easier to describe certain phenomena. Complex exponentials are easier to manipulate than trigonometric functions, so you tend to add an imaginary part that you can later throw away. So they don¹t have a TRUE physical meaning, but neither have real numbers or integers. > I just can¹t believe that mathematicians came up with them just > because they couldn¹t solve x^2+1=0... Oh, but that is as good a reason as any. Certainly a much more pressing reason than making harmonic oscillations easier to describe. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Equation bounds >I¹ve been looking at this for a while. I can¹t seem to Þgure it out. >I don¹t think it should be that complicated :( >i have the following equation: >V31 = (V32 + V21) / (1 + V32*V21) >How do I verify that if -1 < V21, V32 < 1, then -1 < V31 < 1??? Hint: factor 1+V31 and 1-V31. 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: Some problem in simulation with Matlab I have a question that has to do with a problem encountered while using Matlab to verify a cumulative distribution function (CDF) that I worked out. Hope someone on this forum may give me some help. Anyway, here it goes. I have derived a certain CDF expression for a random variable t, and it is obtained from a double-integral expression. t is a function of two variables, v and phi, and it should be able to take on all validity. What I got was a curve that looks like the CDF when t is greater than a certain value (t_a). But when t0) instead of having to treat the cases below and above t_a separately. Has anybody ever encountered this kind of problem in his/her career? I certainly would appreciate some pointer on this. -Ed === Subject: Re: What is the meaning of imaginary numbers? >Apologies if this question has been answered before... >I am trying to improve my perception of imaginary numbers. I already have >done an extensive search in the web but I keep Þnding information about how >to manipulate them (add, multiply etc), about their representation on a >plane with two axes and other things like that, that I am already familiar >with. >Can anybody point me to any resource describing their TRUE physical meaning, >how the need for their introduction into mathematics rose and generally the >whole concept behind them? You might Þnd Paul Nahin¹s AN IMAGINARY TALE useful. More info at http://www.amazon.com/exec/obidos/ASIN/0691027951/. --gregbo gds at best dot com === Subject: Re: What is the meaning of imaginary numbers? > Apologies if this question has been answered before... > I am trying to improve my perception of imaginary numbers. I already have > done an extensive search in the web but I keep Þnding information about how > to manipulate them (add, multiply etc), about their representation on a > plane with two axes and other things like that, that I am already familiar > with. > Can anybody point me to any resource describing their TRUE physical meaning, > how the need for their introduction into mathematics rose and generally the > whole concept behind them? > I just can¹t believe that mathematicians came up with them just because they > couldn¹t solve x^2+1=0... Take a look at the book _Imagining Numbers (particularly the square It has a well written historical discussion and other cultural stuff that might help you. On a different note, you could say that in a sense there are no imaginary numbers since everything can be done with ordered pairs of reals. === Subject: Re: wooooooooooosssssssssshhhhhhhhhhh............. > silver object in my hand and communicate, was something that, as little > as 200 years ago, would have had one burned at the stake. Now it¹s > commonplace and easily understood. Could you explain what you mean? === Subject: Re: rank and eigenvalue >Given the eigenvalues, can we know the rank of a matrix? > I assume it¹s a square matrix... > Given the eigenvalues _with their algebraic multiplicities_, we can > (of course we only need the multiplicity of 0). Oops: I meant geometric multiplicities. 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: Rationals are Uncountable > 1. N = smallest inductive set (the deÞnition we have been using). > a. DeÞne inductive set. > b. Show that the intersection of all inductive sets is a nonempty > set. > c. DeÞne this set to be N. > d. Show that N is a Peano set. > This is the method we were discussing. > Would you consider this set to be inductive? > [0,1,2,...,w,w+1,w+2,...] > Yes. That¹s the same set that I called w+w in a previous post. It¹s the > next limit ordinal after w. > The AoI guarantees there is at least one set that is closed > under the successor function. Let¹s call this set A. > Obviously, A can¹t have a successor or it wouldn¹t > be closed under the successor function. The successor function does not work on the set A, but on the members of the set A, so your last comment makes no sense. > For the sake of argument let¹s say: > A = (0,1,2,...,w,w+1,w+2,...) Let¹s not. > 0 is not the successor of any member of A, so we could > remove it and still have a set closed under the successor > function. Similarly, we could remove (0,1), (0,1,2), etc. > None of these sets would be inductive sets because > they wouldn¹t include 0. > We might be able to remove w from A and still > have a set that is closed. We know w is not > the successor of any natural number. > But, we also know that A contains members > that are not natural numbers. w could be the > successor of one of these other members. > I don¹t see how we can show there exists > an inductive set that is closed under the > successor function that is smaller than A. > A obviously contains members that are > not natural numbers (in our example). But B = (0,1,2,3,...) is a proper subset of A, B contians 0 and is inductive, so A is not minimal in that respect, and, therefore, is not the set of natural numbers. You alwauys seem to need to leave something out. Is your brain so limited that it cannot hold more than one small idea at a time? === Subject: Re: Monotonic function > Situation: > Let y = a^n + b^n (1) where (a, b) = 1, b > a > 1 and odd n 3. > Let a1, a2 are the two values of a and b1,b2 are the two values of b This, then, doesn¹t make much sense to me. If I¹m reading you incorrectly and they aren¹t integers, what are they? > Here a2 > a1 and b2 > b1 > y1 = a1^n + b1^n (2) and y2 = a2^n + b2^n (3) > Therefore, y2 > y1 > Statement: > y in (1) also increases monotonically for a given odd n. This is another problem with viewing a and b as integers. How does their sum increase linearly? > Any comment upon the correctness of the above statement will be > appreciated. === Subject: envelope stufÞng probability problem (question) i read the following problem and solution in a text. given n letters matched to n envelopes - letter 1 goes into envelope 1, etc. given that the nephew of the boss randomly stuffs the envelops, what is the probability that at least one letter is in the correct envelop? notation: p(Ei + Ej) probability that (letter i is in envelop i OR letter j is in envelop j) p(Ei Ej) probability that (letter i is in envelop i AND letter j is in envelop j) p(Ei¹) probability that envelop i does NOT contain letter 1 p(Ei/Ej) probability that letter i is in envelop i given that letter j is in envelop j S( ) summation (sigma notation). the problem is solved by starting with p(at least one letter in correct envelope) = p(E1 + E2 +...+ En). This expands into p(E1) + .. + p(En) - S(p(Ei Ej)) + S(p(Ei Ej Ek)) - .... this is used to Þnd that the probability is 1 - 1/2! + 1/3! - 1/4! .... 1/n! which tends to 1 - e**-1 as n -> inÞnity. the text derivation seems correct (though i have omitted some details). now, i believe that p(at least one correct match) = 1 - p(there are no correct matches). because, 1 = p(0 matches) + p(1 match) + p(2 matches) ... + p(n matches). and p(at least one match) = p(1 match) + p(2 matches) + p(3 matches) + ... + p(n matches) so, p(there are no correct matches) = p(E1¹ E2¹ ... En¹) = p(E1¹) * p(E2¹/E1¹) * p(E3¹/E2¹ E1¹) ... p(En¹/E1¹...E(n-1)¹) now (i think), p(E1¹) = (n-1)/n , p(E2¹/E1¹) = (n-2)/(n-1) ... thus, the product collapses to 1/n so, p(at least one correct match) = 1 - 1/n, which does not match the original solution (and doesn¹t make sense, as n-> oo p(0 matches) = 0. what am i doing wrong?? (please be gentle. i haven¹t done probability problems in 20 years, which is why i am reading a probability text). === Subject: Re: What is the meaning of imaginary numbers? |Can anybody point me to any resource describing their TRUE physical meaning, |how the need for their introduction into mathematics rose and generally the |whole concept behind them? which is supposed to help you visualize imaginary numbers, if you¹re interested. For the origin, you can look in the chapter on the Renaissance in Boyer and Merzbach¹s history of mathematics, in the section on the cubic equation and complex numbers. People working on methods to solve cubic equations found that they could solve them using formal methods that sometimes manipulated square roots of negative numbers along the way. In particular, square roots of negative numbers would appear in the case where the cubic has three distinct real roots. They were still reluctant at that time to consider negative numbers valid, by the way, and complex numbers must have seemed all the more Þctional to them. Eventually, of course, all of it was justiÞed by the fact that instead of using complex numbers a+bi, one can always use pairs (a,b) of real numbers. It simpliÞes things a lot to treat them as single entities, however. In hindsight, a lot of the useful properties of the complex numbers come from the fact that they contain the real numbers, are a Þeld (in the sense used in abstract algebra, so that many key algebraic manipulations work equally well for complex numbers as for real numbers), and that they are algebraically closed, meaning each nonzero polynomial with complex numbers as coefÞcients can be factored as a product of linear terms. They are in a sense relatively simple, since each can be written as a+bi for real numbers a and b. Later on, one is also glad to Þnd that there are guarantees that standard differential equations like f¹=f have local solutions in the complex numbers, so that familiar functions like e^x, log x, and trig functions can be deÞned for complex numbers too. I don¹t really know what you mean by physical meaning. Long ago we stopped thinking that a mathematical structure should somehow correspond to some one physical thing, like space or time. The issue is what kinds of thing are suitably described with the use of the mathematical structure. Lots of physical applications of complex numbers basically use them as a combination of a magnitude and a phase angle. Keith Ramsay === Subject: Re: Proof that v.a < 0 = deceleration (Attn: David A Smith) John Schoenfeld: >> John Schoenfeld: >> >> An obvious counter-example is v(t) = v_0 sin(wt). In the >> >> fourth quadrant, the velocity is obviously increasing, since >> >> it¹s going from more negative to less negative values because >> >> the acceleration is positive. Since v is negative and a is >> >> positive, v.a is obviously negative. >> > >> >[PREVIOUS POST HAD TYPO] velocity _magnitude_ is decreasing, so it is >> >decelerating, so your counter-example is false. >> >> Magnitude is not sufÞciemt to deÞne a vector. Vector functions have a >> direction, too. Why do you think I told you to take the derivative of v^2 >> and avoid the silliness of your assertion about vector functions? Just >I didn¹t realize I was dealing with either, >1. A kid >2. A spastic >3. A crackpot In otherwords, you think a magnitude is sufÞcient to deÞne a vector. === Subject: Re: Proof that v.a < 0 = deceleration (Attn: David A Smith) John Schoenfeld: >dubious@radioactivex.lebesque-al.net (Bilge): >> John Schoenfeld: >> >> >Here is the proof that v.a < 0 = deceleration. >> >> >> >> An obvious counter-example is v(t) = v_0 sin(wt). In the >> >> fourth quadrant, the velocity is obviously increasing, since >> >> it¹s going from more negative to less negative values because >> >> the acceleration is positive. Since v is negative and a is >> >> positive, v.a is obviously negative. >> > >> >I am shocked to see you make this blunder, Bilge. >> > >> >If the velocity is going from more negative numbers to less negative >> >numbers then the velocity is decreasing as it is approaching 0. >> >> Look, you¹re the one who chose to say vector function rather than >> use v^2. A velocity which goes from a larger negative value to a smaller >> one is an increasing velocity. Just draw a line for such a velocity and >> you¹ll see it has a positive slope. >I Þnd it unbelievable that you persist with your ridiculous position >that >v(t)=sin(t) is somehow a counter-example to v.a<0 being deceleration. >Listen idiot, something decelerates when d|v|/dt < 0 I Þnd it unbelievable that you are attempting to use an absolute value to deÞne a vector quantity, especially after I told how to deÞne it as thescalar v^2 avoid that problem. === Subject: Re: C^n proper inclusions problem The function f(x) = |x|*x^n is in C^n(R) but not in C^(n+1)(R). If a is in U, an open subset of R^k, then the function x -> f(x1 - a1) is in C^n(U) but not in C^(n+1)(U). === Subject: Learning measure theory I am taking a course on functional analysis and the Þrst chapter of the book is on measure theory(Fubini¹s theorem, Radon¹s theorem) The book by Barry Simon kinda expects you to know measure theory already. I was wondering if anyone had any ideas on how to learn measure http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: Final Rout of Synchronization Clocks in Relativity Germaine to this Œdicussion¹ is the fact that contrary to all claims that experiments have been performed which showed Œactually¹ that light was independent of source motion, apparently only one has been conÞrmed, AND TURNS OUT TO BE A CROCK! admitted, and they WERE travelling at c...........sounds good to DHR¹s! What they forgot to mention was that the SOURCE, at the precise moment of emission of photons, STOPPED!! The experiment is a fraud (perhaps unknowingly)----the source was NOT moving when the light was emitted. (just as a photon is not moving at the surface of a mirror ....and that is ALL DHR¹s have to cling to. All the rest involve tweaked clocks and wavelength. Jim G === Subject: Re: I thought it was the tuna can > At some point in my educational training, I thought I was taught that > a metal can with height=radius (similar to a tuna can) was special, > like the greatest volume for the least surface or something > counterintuitive like that. Alas, it really is the expected tomato > sauce can (height=diameter) that has the greatest volume/surface > ratio. So . . . is there anything special about the tuna can or do I > just need to move on with my life? Standard maxima/minima application problem. For a given surface area(cylindrical surface area+ bottom+ top lid areas ) = 2 Pi r^2 + 2 Pi r h , Volume = Pi r^2 h ; eliminate either r or h to bring to a single variable equation, differentiate and equate to zero, getting h = 2 r. === Subject: Mathematics forums Are there any other mathematics forums? I just want to be a part of a couple and this is the only one I can Þnd. I am interested in tricky competion style problems or advancements in pure mathematics. http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: What is the meaning of imaginary numbers? >>Physicists like nice mathematical >>operations, so they adopted them. > Welllllll...... > Complex numbers make the description of waves and vibrations, including > electrical phenomena, a lot easier. That sounds like a better reason to > adopt them, and it wouldn¹t surprise me if physicists had actually > invented something like complex numbers for this purpose. > V. Why would they need to invent something that had already been used for over a hundred years before Newton¹s Principia was published? Cardano¹s formulas, published in Ars Magna (1545), made use of complex numbers, and Bombelli (1572) initiated the notation i for the imaginary unit, and published rules for the manipulation of complex quantities. The Principia was only published in 1687. I am continually amazed at the almost romantic yearning the many express, for physicists to have invented mathematics. Dale === Subject: History of stochastic calculus? I¹m slowly teaching myself stochastic calculus. First Einstein¹s paper on Brownian Motion, then The Wiener process and gaussian white noise: dX=a(t)Xdt+b(t)dW and hopefully then on to Ito¹s formula and Black Scholes. My question is historically where does the subject begin? Was Einstein the Þrst to posit a randomly þuctuating force as the driving function in a 1st order linear differential equation? Did he consider it simply a extremely fast þuctuating yet still smooth differentiable function? I¹m looking for a good textbook on the subject. === Subject: Re: envelope stufÞng probability problem (question) >i read the following problem and solution in a text. > given n letters matched to n envelopes - letter 1 goes into > envelope 1, etc. > given that the nephew of the boss randomly stuffs the envelops, > what is the probability that at least one letter is in the > correct envelop? > notation: p(Ei + Ej) probability that > (letter i is in envelop i OR letter j is in envelop j) > p(Ei Ej) probability that > (letter i is in envelop i AND letter j is in envelop j) > p(Ei¹) probability that envelop i does NOT contain letter 1 > p(Ei/Ej) probability that letter i is in envelop i given > that letter j is in envelop j >[book¹s correct solutin ommitted] > now, i believe that > p(at least one correct match) = 1 - p(there are no correct matches). > because, > 1 = p(0 matches) + p(1 match) + p(2 matches) ... + p(n matches). > and > p(at least one match) = p(1 match) + p(2 matches) + p(3 matches) + ... > + p(n matches) > so, > p(there are no correct matches) = p(E1¹ E2¹ ... En¹) = > p(E1¹) * p(E2¹/E1¹) * p(E3¹/E2¹ E1¹) ... p(En¹/E1¹...E(n-1)¹) > now (i think), > p(E1¹) = (n-1)/n , p(E2¹/E1¹) = (n-2)/(n-1) ... There¹s your problem: For example, for P(E2¹ | E1¹), you need to consider whether envelope 1 has letter 2 or not. P(E2¹ | E1 Œ) = P(E1¹ E2¹) / PE1¹ For E1¹ E2¹, either envelope 1 has letter 2 and envelope 2 has any of n-1 letters, or envelope 1 has neither letter 1 nor 2 and envelope 2 has any of n-2 letters. Thus, P(E1¹ E2¹) = [n-1 + (n-2) (n-2)] / [n(n-1)] = (n^2 - 3n + 3) / [n(n-1)] PE1¹ = (n-1) / n P(E1¹ | E2¹) = (n^2 - 3n + 3) / (n-1)^2 -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Learning measure theory >I am taking a course on functional analysis and the Þrst chapter of >the book is on measure theory(Fubini¹s theorem, Radon¹s theorem) The >book by Barry Simon kinda expects you to know measure theory already. > I was wondering if anyone had any ideas on how to learn measure By far, the most clearly written introduction I have seen is Bartle¹s slim volume, The Elements of Integration. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Unatural Numbers <87k73s8fsk.fsf@phiwumbda.org> <-PKdnRWH4KtqKprdRVn-iQ@comcast.com> <871xq0rs04.fsf@phiwumbda.org> <8765f9lih3.fsf@phiwumbda.org> <87zncljrjk.fsf@phiwumbda.org> Discussion, linux) >> These are Peano¹s Axioms as given at Mathworld: >> http://mathworld.wolfram.com/PeanosAxioms.html >> 1. Zero is a number. >> 2. If a is a number, the successor of a is a number. >> 3. zero is not the successor of a number. >> 4. Two numbers of which the successors are equal are themselves equal. >> 5. (induction axiom.) If a set S of numbers contains zero and also the >> successor of every number in S, then every number is in S. >> Stupid questions about PA. >> 1) What prevents a natnum from being its own successor? >> Proof by induction. Let S be the set of all natural numbers n such >> that s(n) != n. Then 0 is in S. Suppose that n is in S, and we must >> show that s(n) is in S, i.e., that s(s(n)) != s(n). But, by (4), is >> s(s(n)) = s(n), then s(n) = n, contradicting our supposition that n is >> in S. Hence, S is inductive, i.e., S = N. > At Þrst I thought (4) prevented this, but it doesn¹t. > If there exists an n that is its own successor then (4) > proves that every natnum is equal to zero. > This may be boring, but I don¹t see an axiom > that prevents it. Using induction we prove N=(0). No. This doesn¹t follow, since if N = {0}, then s(0) must be 0, contradicting (3). Unless you were simply pointing out that (4) and (5) together do not sufÞce to prove for all n, n != s(n). If so, you¹re right: you have to have (3) for the induction to get started. >> 3) What prevents circular successors (a successor of b, b successor >> of a)? >> A proof by induction similar to (1). > Again, I think all induction would prove is that every natural > number is equal to zero. And again, this would necessitate that s(0) = 0, contradicting (3). It cannot be the case that N = {0}. >> 4) What prevents a natnum from not having a successor (like 0)? >> Successor is a function by deÞnition. > I meant to say what prevents a natnum from not being > a successor, like zero. (3) says zero is not the successor > of any natural number, but it doesn¹t say that zero is > the ONLY such natural number. Oh. Again, induction. Let S = {n in N | n = 0 or exists m in N . s(m) = n}. So, a natural number n is a non-zero, non-successor iff n is *not* in S. If we prove that S = N, then we have answered your (corrected) question (4). Clearly 0 is in S. Suppose that n is in S. Then s(n) is also in S (obviously). Hence, by (5), S = N. >> 5) Why do we need axiom 5? >> To give proofs like that in (1). To ensure that every natural number >> is given by a Þnite iteration of successor applied to 0. > So, all of this falls apart without induction. Sure. Of course. One way to regard induction is that it says exactly this: Every natural number is reached by a Þnite number (perhaps 0) of iterations of s applied to 0. Now, this is only motivational, since that former statement isn¹t really a Þrst order statement as is. But, if you *don¹t* have induction, then you don¹t have a least <0,s> structure and you don¹t have much control over what you have. For instance, remove induction and pick any cardinal k and the k-fold disjoint union of N would satisfy (1) - (4) (where one of those copies of N would contain the real distinguished 0). Also, I suppose Z satisÞes (1) - (4). Induction is every bit as central to the deÞnition of N as, say, Cauchy completeness is to R. >> It doesn¹t answer any of these questions. >> One would think that something as important as the deÞnition >> of successor would be part of the axioms. >> No, it¹s entirely inconsequential. A model for PA is any set X >> together with a distinguished element 0 in X and a *function* s:X -> X >> satisfying (1) - (5). > Any function satisfying (1)-(5) is a successor function? > Sounds like a challenge. I can¹t imagine what it would prove. If you pick a set X and suitable 0 and s, then the resulting structure is isomorphic to your favorite N. -- Jesse F. Hughes I have written many words to sci.math, some of them are not even meaningless. --Ross Finlayson === Subject: Now for a card problem (Followup to: Yet another dice game problem.) I always liked this exercise: Thoroughly shufþe a standard deck of playing cards. Turn over cards one by one. Find, in an efÞcient manner, the expected numbers of cards up to and including a) the ace of spades and b) the Þrst ace. [Bonus points: determine the variances as well.] -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: JSH: Focusing on counting prime numbers Discussion, linux) > Isn¹t replying to your own posts considered trolling? Not when you are correcting errors, no. In fact, replying to one¹s own posts in general is just not trolling. (Still, there are trolls that *do* reply to their own posts more than normal/necessary/healthy/etc.) On the other hand, slinging accusations of trolling about willy-nilly, well, it may not be a sign of trolling, but it¹s a pretty good sign of, well, something... (Again, not to say that JSH isn¹t a troll, but David¹s reasoning here is silly.) -- Sure, [my Usenet presence is] like Shaq playing against you in your backyard, but that has its perks, as I Þnd ways to have my fun *and* I can send messages to certain people in the United States Government without concern that the rest of you understand them. -- James Harris === Subject: Re: Data analysis software > It plots and analyses any x-y data for peak location, peak height, > peak > width, semi-derivative, derivative, integral, semi-integral, > convolution, > deconvolution, curve Þtting, and separating overlapped peaks and > background. www.chemSoftware.com === Subject: Re: SymbMath.com: web-based computer algebra system > Would you please show these results? Hey I tried some of the functions and they gave results inconsistent > with matlab. Do you think I should tell them to Þx their program? > www.SymbMath.com === Subject: Re: Now for a card problem (Followup to: Yet another dice game problem.) <4009C9CF.9000705@rutcor.rutgers.edu> playing cards. Turn over cards one by one. Find, in an efÞcient > manner, the expected numbers of cards up to and including > a) the ace of spades and 1 + 51/2 = 26.5; each of the other 51 cards has a 50-50 chance of turning up before the ace of spades. > b) the Þrst ace. 1 + 48/5 = 10.6; each of the 48 non-aces has one chance in Þve of beating out all of the aces. > [Bonus points: determine the variances as well.] Too hard, I¹d need pencil and paper for that one. === Subject: Re: Rationals are Uncountable <87k73s8fsk.fsf@phiwumbda.org> <-PKdnRWH4KtqKprdRVn-iQ@comcast.com> <871xq0rs04.fsf@phiwumbda.org> Discussion, linux) >> The AoI guarantees there is at least one set that is closed >> under the successor function. Let¹s call this set A. >> Obviously, A can¹t have a successor or it wouldn¹t >> be closed under the successor function. > Closure means s(x) is deÞned for every x in A. That is, A is the > domain of s. Assuming regularity, A itself is not in the domain. This is a possible point of confusion for the usual successor operators on Set. When we use the axiom of inÞnity stating that there exists a set A such that 0 in A and for every x in A, s(x) in A, clearly s(A) *does* make sense, since A is a set. And clearly A *does* have a successor. But just as clearly, A¹s successor has nothing at all to do with whether A is closed or not, since A is not in A (given regularity/foundation). What you *meant* to say is: Closure means that the operator s induces a (total) function s:A -> A. And also that, since A is not in A, it doesn¹t matter whether s(A) is in A or not (although, again, by regularity, it is not in A for the usual successors). -- Sale or rental of this disc is ILLEGAL. If you have rented or purchased this disc, please call the MPAA at 1-800-NO-COPYS. -- The MPAA begins a new anti-piracy program, found on a DVD purchased in China === Subject: Re: Now for a card problem (Followup to: Yet another dice game problem.) >I always liked this exercise: Thoroughly shufþe a standard deck of >playing cards. Turn over cards one by one. Find, in an efÞcient >manner, the expected numbers of cards up to and including >a) the ace of spades and >b) the Þrst ace. >[Bonus points: determine the variances as well.] Wouldn¹t the expected number of cards to the ace of spades be 25.5, or half of the 51 cards in the deck? -- dgates@spamfreelinkline.com === Subject: Re: Proposal for responding to idiots > Daniel McLaury > Many newsgroup readers show the threads which have had the most recent > responses at the top of the list. This leads to a problem when some > idiot comes on this board and says that .9999.. != 1, or that he¹s > squared the circle, or whatever, and like a million people post > replies to messages like this. > ... > But that¹s the problem -- such threads will get responses. As I¹m sure the > late C. N. Parkinson would say, the length of a thread is inversely > proportional to its value. > LH Indeed, this is the problem; that¹s why I¹m trying to help Þx it. If there¹s no problem, there¹s no need to come up with a solution. I¹m requesting that such threads do NOT get responses. If enough people vow not to answer such threads in a way that takes up the bandwidth of the rest of us, we can at least ameliorate the problem to some extent. === Subject: Re: Proposal for responding to idiots >I¹m requesting that such threads do NOT get responses. While you¹re at it, why not request a house in the Bahamas? === Subject: Re: Axiom exploring software? |Surely there are general programs for the purpose of |shortening strings with some given simple axioms |(with the usual problem that some strings must be |heavily lengthened before they can be shortened)? Well, for equational theories there¹s the Knuth-Bendix completion procedure, which does this for some cases. I don¹t know where one would look for an implementation. Keith Ramsay === Subject: Re: What is the meaning of imaginary numbers? |> I just can¹t believe that mathematicians came up with them just |> because they couldn¹t solve x^2+1=0... | |Oh, but that is as good a reason as any. Certainly a much more |pressing reason than making harmonic oscillations easier to describe. I don¹t know about that. Now and then we get people who¹d like to have a solution to 0*x=1 included in their number system, just for the sake of having a solution to that. I don¹t think that¹s much of a reason. Logicians give the theory of algebraically closed Þelds (such as the the complex numbers) as an example of a more general idea called model completion. Indeed it does have something to do with including as new elements ideal elements solving systems of formulas which would not necessarily have solutions in an arbtrary Þeld. But the fact that one is working with a well-behaved theory like the theory of Þelds to begin with gives the enterprise more discipline. Otherwise it can be like building castles in the air. Keith Ramsay === Subject: Re: What is the meaning of imaginary numbers? |Another use is when Einstein introduced the unit imaginary in the |interval (x,y,z,ict), which, when squared produces | |(x,y,z,ict)^2 = x^2 + y^2 + z^2 - c^2t^2 | |Einstein¹s trick of introducing the unit imnaginary here is not |favored much these days. Something similar is used in Euclidean gravity, though. They analyze gravitation pretending as it were that the signature is x^2+y^2+z^2+t^2, as if it were Euclidean 4-space, and then try to relate it back to a more realistic model using analytic continuation. Hawking likes this approach. See <46i2v4$iif@galaxy.ucr.edu>. Keith Ramsay === Subject: Re: JSH: My quick math quide | Yes, I completely agree on that. In fact it seems like |incredible luck that gcd¹s can be found at all in the algebraic |integers. I wonder if something even deeper, not yet discovered, |might be another explanation. Perhaps. The fact that the ring of algebraic integers is Bezout[*] is equivalent to every Þnitely generated ideal being principal. That¹s in turn equivalent to every ideal in the ring of integers of a number Þeld becoming principal in the ring of integers of some number Þeld containing it. Those equivalences seem reasonably direct and straightforward. We¹ve dug up what seems to be the standard proof of that, using the Þniteness of the class group of the ring of integers in a number Þeld. [*] Bezoutness is stronger than the existence of GCDs, but it seems like Bezoutness is the essential reason why one has GCDs in this case. If there¹s some deeper reason, I¹d say it should still be in terms of what it is that enables ideals to become principalized. In _Introduction to the Construction of Class Fields_ Harvey Cohn motivations for the construction of class Þelds. Theorem 7.4.14 says that the ideals in the ring of integers of a number Þeld become principal in the (weak or strong) Hilbert class Þeld. So perhaps one could come up with a more fundamental reason in terms of class Þeld theory. As far as just proving the result goes, it seems like serious overkill! All we need is to Þnd an extension where one particular ideal becomes principal, not necessarily where *all* the ideals of the smaller number Þeld become principal. There are times, however, when it seems like the explanation given by the easiest proof of a result is not quite as fundamental as one provided by deriving it from some deeper results which might be hard to prove. If I wanted to look for a more essential or fundamental reason for the existence of GCDs in this case, I¹d look around in the vicinity of this result on principalization to see whether there¹s a related explanation for why we are able to succeed in making the one particular ideal principal in an extension. Keith Ramsay === Subject: Re: Final Rout of Synchronization Clocks in Relativity Expires: 28 days >Germaine to this Œdicussion¹ is the fact that contrary to all claims >that experiments have been performed which showed Œactually¹ that >light was independent of source motion, apparently only one has been >conÞrmed, AND TURNS OUT TO BE A CROCK! >admitted, and they WERE travelling at c...........sounds good to >DHR¹s! What they forgot to mention was that the SOURCE, at the precise >moment of emission of photons, STOPPED!! >The experiment is a fraud (perhaps unknowingly)----the source was NOT >moving when the light was emitted. (just as a photon is not moving at >the surface of a mirror >....and that is ALL DHR¹s have to cling to. All the rest involve >tweaked clocks and wavelength. >Jim G It¹s the same with the pions. They stop too, before decaying. Henri Wilson. www.users.bigpond.com/hewn/index.htm === Subject: Re: Monotonic function > Situation: > Let y = a^n + b^n (1) where (a, b) = 1, b > a > 1 and odd n 3. > Let a1, a2 are the two values of a and b1,b2 are the two values of b > Here a2 > a1 and b2 > b1 > y1 = a1^n + b1^n (2) and y2 = a2^n + b2^n (3) > Therefore, y2 > y1 This much is deÞnitely true. > Statement: > y in (1) also increases monotonically for a given odd n. I¹m not exactly sure what you mean by a + b increases LINEARLY. Linearly in what variable? Clearly it¹s always linear in a and b; do you mean one of them is a constant? Do you mean that a(x) + b(x) is linear in x? What exactly do you mean/ === Subject: Re: I thought it was the tuna can > Consider the perimeter of the side view (2*(height+diameter)). > The tuna can has maximal volume for Þxed perimeter. > I expect this has nothing whatever to do with the reason they > pack tuna in cans of this shape. The wide/shallow can make it easier to get th tuna out? === Subject: Re: One-sheeted cone, regular surface > I am reading an example in my diff. geom. book, and I don¹t really > understand why the one-sheeted cone isn¹t a regular surface. >>It goes funny at (0,0,0). >>I like to see it like this. Each point on a regular surface >>has a neighbourhood in the surface homeomorphic to an open disc. >>But the point (0,0,0) doesn¹t. Each connected neighbourhood of (0,0,0) >>in the cone, can be disconnected by removing (0,0,0). The open disc >>hasn¹t the property of being disconnectable by removing one point. > I suspect you¹re thinking of the surface deÞned by z^2 = x^2 + y^2; > the problem asked about the surface z = sqrt(x^2 + y^2), which is > very different. (In particular it _is_ a continuous manifold, just not > a differentiable one.) OK so he¹ll have to do some analysis, like think about the tangent space of the cone at (0,0,0). :-0 -- 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: Final Rout of Synchronization Clocks in Relativity >Germaine to this Œdicussion¹ is the fact that contrary to all claims >that experiments have been performed which showed Œactually¹ that >light was independent of source motion, apparently only one has been >conÞrmed, AND TURNS OUT TO BE A CROCK! >admitted, and they WERE travelling at c...........sounds good to >DHR¹s! What they forgot to mention was that the SOURCE, at the precise >moment of emission of photons, STOPPED!! >The experiment is a fraud (perhaps unknowingly)----the source was NOT >moving when the light was emitted. (just as a photon is not moving at >the surface of a mirror >....and that is ALL DHR¹s have to cling to. All the rest involve >tweaked clocks and wavelength. >Jim G > It¹s the same with the pions. They stop too, before decaying. > Henri Wilson. > www.users.bigpond.com/hewn/index.htm Not really, Henri. What the relativist¹s claim is that pions are created when a material is bombarded with a proton beam. Because they happen to have extremely short life span before they decay, what they actually do (in lab experiments) is decay whilst still inside the material. Then any radiation they emit is emitted from within the material. Every relativist knows that the material is really a vacuum, and that all the pions are really moving in one direction only at nearly c inside the vacuum of the material because that conÞrms relativity. Androcles === Subject: Re: Would you recommend a school of math for a soon-to-be college student > If you think you have the qualiÞcations, you should go for the > absolute top schools. Don¹t necessarily agree. U.C. Berkeley has the one of the top-rated math departments in the world, but freshman and sophomore classes are still taught as huge lectures. If you went to a community college for two years and transferred in as a junior you¹d be just as well off. Being a freshperson at Berkeley has its pleasures and beneÞts, but academically you¹re just another mad cow in a real big herd. Same goes for any other large state university. === Subject: Re: Mathematics forums > Are there any other mathematics forums? I just want to be a part of a > couple and this is the only one I can Þnd. I am interested in > tricky competion style problems or advancements in pure mathematics. > News==---- > http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 > Newsgroups > ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption > =--- When you sign up for a newsgroup access, your access provider is supposed to provide your newsgroup software with a list of accessable newsgroups, though how this is done, and how to search the list, will depend on the software you use to do the accessing. You should look for newsgroups with math in their names. There are quite a few of them. I Þnd sci.math the one most to my taste, though I monitor several others as well. You can also post your questions about newsnetting to news.newusers.questions. === Subject: Re: Equation bounds > i have the following equation: > V31 = (V32 + V21) / (1 + V32*V21) > How do I verify that if -1 < V21, V32 < 1, then -1 < V31 < 1??? Alternate hint: look up tanh(x + y) === Subject: Re: Now for a card problem (Followup to: Yet another dice game problem.) <4009C9CF.9000705@rutcor.rutgers.edu> <872n00d5tuik30plv3nmf4ht1a7rg0sabr@4ax.comI always liked this exercise: Thoroughly shufþe a standard deck of >playing cards. Turn over cards one by one. Find, in an efÞcient >manner, the expected numbers of cards up to and including >a) the ace of spades and >b) the Þrst ace. >[Bonus points: determine the variances as well.] > Wouldn¹t the expected number of cards to the ace of spades be 25.5, or > half of the 51 cards in the deck? Right, except he said up to and including, so add one for the ace of spades itself, making 26.5. === Subject: Re: Bound for roots of a polynomial > Hi all, > Let P(x) ba a polynomial with real coefÞcients. Is there > a simple way of Þnding a number R > 0 such that every real > root of P(x) is in [-R,R]? I¹m thinking in terms of R being > obtained as a function of the coefÞcients of P(x). > Jose Carlos Santos Look at: http://sepwww.stanford.edu/oldsep/stew/descartes.pdf Theorem 1 (Lagrange) states: If in p(x) the Þrst negative coefÞcient is preceded by k coefÞcients which are positive or zero, and if G denotes the greatest of the magnitudes of the negative coefÞcients, then p(x) is always positive for x > 1+ {G / p_n}^{1/k}, and so all real roots are less than that value. Use it for your P and for Q(x):=P(-x) (to get negative bound). -- Rafal Kucharski. === Subject: Re: Now for a card problem (Followup to: Yet another dice game problem.) * dgates@nospamlinkline.com > Wouldn¹t the expected number of cards to the ace of spades be 25.5, or > half of the 51 cards in the deck? Or at 26,5 as it asked for including. Using an anologous reasoning concerning the expected number of cards the Þrst ace would reside: (a) divide the 48 cards without the aces into Þve equal heaps, i.e. 9.6 cards in each heap. (b) Put an ace on top of the Þrst four heaps. (c) Put each heap on top of each other so that the heap without an ace is on top. (d) You now have the card deck with the aces spread out, i.e. put on their expected place. (e) The answer thus is that the Þrst ace is expected to appear on place 10.6. This is not rigorous reasoning, so we use formalism: (a) The probability that the Þrst ace is at place n (1 <= n <= 49) is PRODUKT(i=0 to n-2) (48-i)/(52-i) * 4/(53-n), i.e. the probability that no ace is found at the Þrst n-1 cards times the probability that the ace is at card no n. (b) This can be simpliÞed to: aceAt(n) = [ (52-n)(51-n)(50-n)] / {13*49*50*51] (c) The expected value is by deÞnition: E = SUM(i=1 to 49) i*aceAt(i) (d) Unfortenately, I am not able to simplify this expression by the means of manipulation. However, I do own a computer, and a simple Lisp expression or two made E=10.6. (e) Amazing! :-) -- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@iÞ.uio.no http://www.iÞ.uio.no/~jonhaug/, Phone: +47 22 95 21 52 === Subject: Re: Nothing¹s Zero (0) > In sci.math, Garry Denke > Consider the equation: x - 3 = 0 >> The solution for x - 3 is the number 0 >> >> No, it¹s x = 3. > > Are you having a problem with the equals sign > or 0 the number to the right of the equals sign? > > The equation > > x - 3 = 0 > > cannot be solved by declaring that the solution is Œx - 3 = 0¹ > or just Œ0¹. That¹s just stupid. > So you are having a problem with the = 0. The answer is not stupid, > the answer is 0. If you want to solve for x, feel free to do so, but > the fact of the matter is that the answer is 0. It¹s not my fault, it > just is. That¹s the problem with mathematicians (and computer > programmers), they do not stand behind their statements. They hmmm, > and haaa, and rant away... Heh, heh, hu, heh heh, heh hah heh. > Of course for such a simple equation it¹s fairly obvious anyway > that the solution is x = 3; that sort of thing is probably > learned in the third grade -- although I¹m not entirely sure > when simple algebra is Þrst taught anymore. (*I* learned it > in third grade, but that¹s me...and I¹m not entirely sure if > the taught material in the third grade is where I learned it, > as I have an old algebra book which now I can¹t Þnd.) > Yep. The old personal attack manuever. Second grade? You¹re a riot. This is one of the best trolls I¹ve seen. >> For example: >> >> Consider the equation x^3 - 3x - 4 = 0 >> The solution for x^3 - 3x - 4 is the number 0 >> >> I¹d have to work it out, but x = 0 won¹t work here (0^3 - 3*0 - 4 = -4). > > Are you having a problem with the equals sign > or 0 the number to the right of the equals sign? > > Are *you* having a problem with the notion of solving > an arbitrary equation f(x) = 0, f(x) = K, or f(x) = g(x), > for the variable x? > No, that¹s boring to me. No, my study is 0, and 0/0. >> For example: >> >> Consider the equation x - log x = 0 >> The solution for x - log x is the number 0 >> >> That has no solution at all in the reals. I¹m not sure regarding >> the complex plane minus the origin. > > Are you having a problem with the equals sign > or 0 the number to the right of the equals sign? > > Polly wanna cracker? > Yea. The old personal attack manuever. First grade? No, I think he said he learned this stuff in the _third_ grade. Didn¹t you? :p >> For example: >> >> Consider the equation y - 4 x + 1 = 0 >> The solution for y - 4 x + 1 is the number 0 >> >> That solution is an inÞnite line (x,y) which passes through >> the points (0, -1), (1/4, 0) and (1, 3), among uncountably >> inÞnite others. Or, if you prefer, the slope is 4 and the >> y-intercept is -1. > > Are you having a problem with the equals sign > or 0 the number to the right of the equals sign? > > Polly wanna cracker? > Yes. The old personal attack manuever. Playground? > Consider the equation x^2 + 1 = 0 > The solution for x^2 + 1 is the number 0 > > Polly wanna cracker? > Oh no, not again. > ObSheesh: Sheesh. > Pre-school? > Write again if you need help. > > I got a degree in this stuff, Rocks-In-Head-Boy. :-P > I would hope to have at least half a clue. > Good for you, really. > Yep, 0 is the big rock of 0123456789, 8 a couple of smaller rocks, and > 6¹s and 9¹s swinging rocks. But whenever you¹re Þnished playing with > your x¹s, y¹s, and z¹s, write again, and I will help you with the > solution to every algebraic equation, namely, = 0, and 0/0. Wait, so does 0 = 0/0? I thought 0/0 = 1, since the 0¹s cancel. (heh) Œcid Œooh === Subject: Re: category/topos theory and OOP >I recently came across a book which recasts music theory using categories >as a foundation--different topologies for rhythm, harmony, melody and so >forth are given. It¹s called The Topos of Music by Guerino Mazzola, and >easily). It¹s all very interesting though much of it is way over my head. >The author develops some computer-implemented analytic methods that >represent data structures of music as objects. It struck me at times that >the language used to discuss manipulating these data structures was very >similar to the language of category theory itself. (Not that I¹m an expert >in either of those Þelds.) Does anyone else think that categories and >object-oriented programming are a natural pair? Is the charge ever >levelled that some people only use category theory because it Þts the >data structures selected? (Not that this is a bad reason; just that if >this is how it is, then for me at least it removes some of the deepness >about the applicability of categories in this case.) >Curious to see your comments, about this question and about the work >mentioned, if anyone has had a look at it. --Henri. > I shared this with a good friend who is well-versed in both music and > computers and databases. His reply: > I suspect those people think too much. I¹ve not yet seen an answer to > the question why does a certain chord progression make me cry?. Tell him that¹s Idiosyncratic and it makes him wierd. :) Œcid Œooh === Subject: sinusoidal 2d equation Somebody would be kind enough to remind me the equation of a sinusoidal surface in 2dimension (aim to do plot a wavy surface) in cartesian system. Actually in polar coordinates would appreciated as well. Thierry === Subject: re:Generators? was not actually for any of my classes, it was actually some extra work that i will not be learning in school. Tthe actual context was: > I want to know if anyone knows how to prove this: > (Sum of Generators in p)(Mod p) = 0, 1 or -1 > (Note p is a prime) Although the question was given to me by a fellow student, so there could be errors.[/quote] http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: Analicity in the complexes >Good morning >I was studying Complex Analysis in Ahlfors book and would like help >with a question. Suppose a function f is deÞned on an open domain D >of the complex plane C and has values in C. If z= x+ i*y, then f can >be given by f(z)= v(x,y) + i w(x,y), where v and w are functions from >R^2 to R. If f is analytic, then it¹s easy to show that v and w >satisfy the Cauchy-Riemann differential equations. Then Ahlfors says >they also satisfy Laplace equation and are harmonic conjugate, because >they have continuous partial derivatives with respect to x and y. But >he doesn¹t prove it. >Next, Ahlfors shows that, if v and w have continuous partial >derivatives, then f¹ exists in the same domain of f. > ??? Surely you mean if v and w have continuous partial > derivatives and satisfy the Cauchy-Riemann equations. Yes, sure! >This is not hard >to understand, the proof takes into account that, for having >continuous partial derivatives, v and w are differentiable in D. But >then Ahlfors just claims the converse is true. I¹m confused here. If f >is analytic in D, then we se all directional derivatives of v and w >and so are v and w. But are these conditions enough strong to ensure >continuity of the Þrst partial derivatives, or at least >differentiability? It¹s possible that all directional derivatives of a >function exist without the function being differentiable. >I¹m confused here. > I¹m not sure exactly what the question is, but I suspect that > this may be an answer: It is not obvious, but it¹s true, that > if f has a derivative at every point of D then f (and hence > v and w) are inÞnitely differentiable. This follows from the > Cauchy-Goursat theorem. I¹m actually not familiar with > Ahlfors, but I Þnd it hard to believe that there¹s no proof > of this fact in there. (Hmm, actually although I¹m not > actually familiar with the book I¹m almost certain there¹s > a proof of the Cauchy-Goursat theorem, because if I > recall correctly I was talking to someone the other month > about the fact that the proof in Ahlfors uses subdivision > of a rectangle into smaller rectangles, while the One > True Proof uses subdivision of a triangle into smaller > triangles.) > Are you certain that there¹s no proof of these things in > the book, or could it be that you just haven¹t got there > yet? Look for Cauchy-Goursat - seems likely to me > that there will be a proof of the fact that one derivative > implies inÞnite differentiability a little later... Artur === Subject: Showing the following function is onto? Hi to prove f(x)= { (2x-1)/x , x in (0,1/2) and (2x-1)/(1-x) x in [1/2, 1) ] , f:(0,1)->R is onto you have to split into cases ? i.e take b in R , either b=0 , b>0 or b<0 and show that for any b the x must lie in (0,1) ? any shorter way? Actually I want to prove f is a bijection, but I already proved f is 1-1 so i¹m stuck in showing surjectivity, how do you proceed? === Subject: Re: Generators? >> I want to know if anyone knows how to prove this: >> (Sum of Generators in p)(Mod p) = 0, 1 or -1 >> (Note p is a prime) Assuming this means generators in Z_p, then if p is prime then all elements of Z_p except the neutral element are generators. The sum of all generators is then: 1 + 2 + ... + (p-1) = (p-1)p/2 = p (p/2 - 1/2) If p = 2 we get simply p (p/2 - 1/2) == 1 (mod 2). If p > 2 then p is a factor in p (p/2 - 1/2) so p (p/2 - 1/2) == 0 (mod p). I don¹t immediately see where p (p/2 - 1/2) == p-1 (mod p). === Subject: Re: What is the meaning of imaginary numbers? >On a different note, you could say that in a sense there are no >imaginary numbers since everything can be done with ordered pairs of >reals. There are no reals since everything can be done with Cauchy sequences of rationals. There are no rationals since everything can be done with equivalence classes of ordered pairs of integers. There are no integers since everything can be done with elementary sets. I¹m not sure there are sets, either. === Subject: Re: Now for a card problem (Followup to: Yet another dice game problem.) <4009C9CF.9000705@rutcor.rutgers.edu> <872n00d5tuik30plv3nmf4ht1a7rg0sabr@4ax.com> Wouldn¹t the expected number of cards to the ace of spades be 25.5, or > half of the 51 cards in the deck? > Or at 26,5 as it asked for including. Using an anologous reasoning > concerning the expected number of cards the Þrst ace would reside: > (a) divide the 48 cards without the aces into Þve equal heaps, > i.e. 9.6 cards in each heap. > (b) Put an ace on top of the Þrst four heaps. > (c) Put each heap on top of each other so that the heap without an > ace is on top. > (d) You now have the card deck with the aces spread out, i.e. put on > their expected place. > (e) The answer thus is that the Þrst ace is expected to appear on > place 10.6. > This is not rigorous reasoning, so we use formalism: Consider any non-ace, say the queen of hearts. The probability that the queen of hearts precedes all four of the aces is clearly 1/5; ignoring the other 47 cards, one of those 5 cards must come Þrst, and they are all equally likely. So the expected number of queens of hearts turned over before the Þrst ace is 1/5. It is the same for any other non-ace. Summing over all 48 of them, the expected number of cards turned over before the Þrst ace is 48/5 = 9.6. Add one to that because we are supposed to count the Þrst ace. This is more or less the same as your argument above. === Subject: Re: Biomathematics - > [...] > [...] > [...] > Nobody can usefully model turbulence > through space and over time. > [...] True, but only because of an implicit Falsehood. No such thing as Œtime¹ exists within physical reality. What¹s been referred to as time is an Erroneous partial conceptualization of the one-way þow of energy from order to dis-order that is what¹s =described= by 2nd Thermo [WDB2T]. When this Falsehood is eliminated, turbulence is easy. In the limit of 3-D space -> 0, energy will þow toward decreasing order. Period. K. P. Collins === Subject: Re: Now for a card problem (Followup to: Yet another dice game problem.) * Fred Galvin >> [Graphical fairness argument deleted} > [Semiformal argument deleted] > This is more or less the same as your argument above. Nah, yours is more convincing. -- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@iÞ.uio.no http://www.iÞ.uio.no/~jonhaug/, Phone: +47 22 95 21 52 === Subject: Re: Generators? > I want to know if anyone knows how to prove this: > > (Sum of Generators in p)(Mod p) = 0, 1 or -1 > > (Note p is a prime) > Assuming this means generators in Z_p, then if p is prime then all > elements of Z_p except the neutral element are generators. The sum of > all generators is then: > 1 + 2 + ... + (p-1) = (p-1)p/2 = p (p/2 - 1/2) > If p = 2 we get simply p (p/2 - 1/2) == 1 (mod 2). If p > 2 then p is > a factor in p (p/2 - 1/2) so p (p/2 - 1/2) == 0 (mod p). > I don¹t immediately see where p (p/2 - 1/2) == p-1 (mod p). By generators he probably means primitive roots. It is the case that the sum of the primitive roots modulo p is 0, 1 or -1. This can be seen by considering the coefÞcients of the (p-1)-th cyclotomic polynomial. -- 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: groups and permutations Let G be a group, S the symmetric group on G, and H the holomorph of G. We say that G is strong if the normalizer of H in S is H. Do these groups have a standard name? Is there any non-trivial necessary and sufÞcient condition to characterize these groups? If not, are there some sufÞcient conditions for a group to be strong (or at least examples of famous strong groups)? If G is Þnite, are these groups classiÞed? I thank any help. C. === Subject: Re: Gloves in the dark > But can it be proved that 29 is the maximum? > Yes. I have programmatically generated the 317 partitions of 30 into at > least three parts, each part being at least 3. Programmatically solving the > problem for each partition shows that 21 is never the least number of gloves > to be picked. And thus, clearly, it cannot be the least number of gloves for > any number greater than 30. > I do not know of any more elegant proof. -- Alec McKenzie mckenzie@despammed.com === Subject: Re: One-sheeted cone, regular surface >> I am reading an example in my diff. geom. book, and I don¹t really >> understand why the one-sheeted cone isn¹t a regular surface. >It goes funny at (0,0,0). >I like to see it like this. Each point on a regular surface >has a neighbourhood in the surface homeomorphic to an open disc. >But the point (0,0,0) doesn¹t. Each connected neighbourhood of (0,0,0) >in the cone, can be disconnected by removing (0,0,0). The open disc >hasn¹t the property of being disconnectable by removing one point. >> I suspect you¹re thinking of the surface deÞned by z^2 = x^2 + y^2; >> the problem asked about the surface z = sqrt(x^2 + y^2), which is >> very different. (In particular it _is_ a continuous manifold, just not >> a differentiable one.) >OK so he¹ll have to do some analysis, like think about the tangent >space of the cone at (0,0,0). Yup. Not an easy thing to think about, though... >:-0 ************************ David C. Ullrich === Subject: Re: Proof that v.a < 0 = deceleration (Attn: David A Smith) > John Schoenfeld: > >> John Schoenfeld: > >> >> An obvious counter-example is v(t) = v_0 sin(wt). In the > >> >> fourth quadrant, the velocity is obviously increasing, since > >> >> it¹s going from more negative to less negative values because > >> >> the acceleration is positive. Since v is negative and a is > >> >> positive, v.a is obviously negative. > >> >> >[PREVIOUS POST HAD TYPO] velocity _magnitude_ is decreasing, so it is > >> >decelerating, so your counter-example is false. > >> > >> Magnitude is not sufÞciemt to deÞne a vector. Vector functions have a > >> direction, too. Why do you think I told you to take the derivative of v^2 > >> and avoid the silliness of your assertion about vector functions? Just > >I didn¹t realize I was dealing with either, > >1. A kid > >2. A spastic > >3. A crackpot > In otherwords, you think a magnitude is sufÞcient to deÞne a vector. My friend Bilge, d|v|/dt is not a vector, it is a scalar - it is the rate of change of speed. d|v|/dt = (v.a) / |v| The rate of change of speed in the direction of the velocity is d|v| / dt = (v.a) / |v| = (v.a) / |v|^2 v JS === Subject: Re: Showing the following function is onto? >to prove f(x)= { (2x-1)/x , x in (0,1/2) and (2x-1)/(1-x) x in [1/2, >1) ] , >f:(0,1)->R is onto you have to split into cases ? Yes. >i.e take b in R , >either b=0 , b>0 or b<0 and show that for any b the x must lie in >(0,1) ? Not really - there is no the x which must lie in (0,1). What you need to show (probably what you meant to say?) is that for any b in R there _evists_ an x in (0,1) such that f(x) = b. >any shorter way? >Actually I want to prove f is a bijection, but I already proved f is >1-1 >so i¹m stuck in showing surjectivity, how do you proceed? ************************ David C. Ullrich === Subject: Re: Proof that v.a < 0 = deceleration (Attn: David A Smith) > >I Þnd it unbelievable that you persist with your ridiculous position > >that > >v(t)=sin(t) is somehow a counter-example to v.a<0 being deceleration. > >Listen idiot, something decelerates when d|v|/dt < 0 > I Þnd it unbelievable that you are attempting to use an absolute > value to deÞne a vector quantity, especially after I told how > to deÞne it as thescalar v^2 avoid that problem. The absolute quantity that you are refering to is |v|, which is the speed. v = velocity |v| = speed dv/dt = acceleration d|v|/dt = rate of change of speed Something is decelerating when d|v(t)|/dt < 0, because the speed is decreasing. Decreasing velocity (i.e. dv/dt < 0) does not mean decreasing speed (i.e. d|v|/dt < 0). I posted the theorem and it¹s proof over 20 times, and you still forgot to disprove it (or even quote it). Let me try again. Point out the mathematical error: THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? THEOREM: v(t).a(t) < 0 denotes deceleration. PROOF: Deceleration is when d|v|/dt < 0 d|v|/dt = (v(t).a(t)) / |v(t)| So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 Do you understand this blindingly simple logic yet? If this is nonsense as you claim, then show the error. JS === Subject: Re: Proof that v.a < 0 = deceleration (Attn: David A Smith) John Schoenfeld: >My friend Bilge, d|v|/dt is not a vector, So, what you¹re saying is that all the nonsense about vector functions was just that - nonsense and you are simply taking the the derivative of v^2 = v.v like I said. Sheeesh.... >it is a scalar - it is the rate of change of speed. So much for your argument about vector functions. === Subject: Re: Proof that v.a < 0 = deceleration (Attn: David A Smith) John Schoenfeld: >> >I Þnd it unbelievable that you persist with your ridiculous position >> >that >> >v(t)=sin(t) is somehow a counter-example to v.a<0 being deceleration. >> > >> >Listen idiot, something decelerates when d|v|/dt < 0 >> >> I Þnd it unbelievable that you are attempting to use an absolute >> value to deÞne a vector quantity, especially after I told how >> to deÞne it as thescalar v^2 avoid that problem. >The absolute quantity that you are refering to is |v|, which is the >speed. You are clearly a complete crackpot. >v = velocity >|v| = speed >dv/dt = acceleration >d|v|/dt = rate of change of speed >Something is decelerating when d|v(t)|/dt < 0, because the speed is >decreasing. >Decreasing velocity (i.e. dv/dt < 0) does not mean decreasing speed >(i.e. d|v|/dt < 0). >I posted the theorem and it¹s proof over 20 times, and you still >forgot to disprove it (or even quote it). Let me try again. >Point out the mathematical error: >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >THEOREM: v(t).a(t) < 0 denotes deceleration. >PROOF: >Deceleration is when d|v|/dt < 0 >d|v|/dt = (v(t).a(t)) / |v(t)| >So clearly, IF d|v|/dt < 0 THEN (v(t).a(t)) < 0 >Do you understand this blindingly simple logic yet? >If this is nonsense as you claim, then show the error. >JS === Subject: Explicit prime counting formula, complexity The way I calculate explicit prime formulas is straightforward, as I started with dS(N,3) = þoor(N-4)/6. with even N and iterated. So N/2 - þoor((N-4)/6) - þoor((N-16)/10) + þoor((N-16)/30) - þoor((N-36)/14) + þoor((N-22)/42) + þoor((N-106)/70) - þoor((N-106)/210) + 2. is the result of two iterations covering 5 and 7. I¹m curious now about whether or not there are smaller expressions that do what it does out to inÞnity as I sat down and Þgured out how terms get added. In general, with x=sqrt(N) there are approximately 2^{x/ln x}/ln N terms. James Harris === Subject: Re: Mathematics forums > Are there any other mathematics forums? I just want to be a part of a > couple and this is the only one I can Þnd. I am interested in > tricky competion style problems or advancements in pure mathematics. If you want Web-based forums (not Usenet like this one) there are many to choose from. I don¹t know if there is a good list of them, though. http://archives.math.utk.edu/ http://mathforum.org/library/ http://www.sciencegroups.com/ http://mathforum.org/discussions/ http://www.cut-the-knot.org/content.shtml http://www.bham.ac.uk/ctimath/gateway/news.htm http://www.math-net.org/ http://e-math.ams.org/mathweb -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Rationals are Uncountable > The AoI guarantees there is at least one set that is closed > under the successor function. Let¹s call this set A. > Obviously, A can¹t have a successor or it wouldn¹t > be closed under the successor function. >> Closure means s(x) is deÞned for every x in A. That is, A is the >> domain of s. Assuming regularity, A itself is not in the domain. > This is a possible point of confusion for the usual successor > operators on Set. When we use the axiom of inÞnity stating that > there exists a set A such that 0 in A and for every x in A, s(x) in A, > clearly s(A) *does* make sense, since A is a set. Right. I forgot that since we were talking about AofI, a speciÞc successor operator was implied. > And clearly A *does* have a successor. But just as clearly, A¹s > successor has nothing at all to do with whether A is closed or not, > since A is not in A (given regularity/foundation). > What you *meant* to say is: Closure means that the operator s induces > a (total) function s:A -> A. And also that, since A is not in A, it > doesn¹t matter whether s(A) is in A or not (although, again, by > regularity, it is not in A for the usual successors). Yes. -- Dave Seaman Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: USE LINUX OR PERISH! > Oy, I better get to work! Fortunately I have my ofÞshal Alien Detector > which glows reddish in the presence of AYLEENNNS. Ayleens? Once I knew a girl named Ayleen (maybe not), but I do not recall your detector glowing, or maybeee? Was I stoned? That Ayleen was a crazy bitch. === Subject: Action Device Tragedy http://www.geocities.com/inertial_propulsion 17 years ago, in 1986, when I was 17 year old, I ran away from one of the most prestigious engineering college in India, VJTI Bombay. Reason was, I don¹t want to be Engineer. I want to become Pilot. through which we can hang the things in air. We can build air vehicles and any one can become Pilot. But I am back to square 1. I will have to become Engineer to build this action device. There are no mechanical engineers in any workshop in this town. All you people in world are forcing me to build this device myself even though you are aware that at least one person has built one of the parts of this device and it works. Shame! Shame! Shame! I don¹t want to be Engineer. I want to be Pilot, you Morons! I am not going to build this action device. That spring in hardware shop struck to my Þnger very hard. -Abhi. === Subject: Dishonesty I note that dishonesty has become the normal thing for ŒBilge¹. He makes the return address something other than sci.physics.relativity, such as alt.troll or alt.moron. Obviously such a cheap trick can be considered amusing, but wears thin when repeated. He claims to be able to Œderive¹ sqrt(1-v^2/c^2) as though it were something special that he alone can do, using the trigonometric functions of a conic section. Obviously this is the full extent of his mathematical capabilities, which, when exhausted, causes him to resort to personal abuse and cheap practical jokes as his only counter argument when his absurd claims are challenged. We note that in ŒMichelson¹s Interference Experiment¹, H.A.Lorentz (p. 7, The Principle of Relativity¹ Dover publications, SBN 486-60081-5) uses this in a supposed Œ...shortening in the direction of motion in the proportion of 1 to sqrt(1-v^2/c^2)...¹ by which method Lorentz assumes a spherical Earth becomes an ellipsoidal Earth pressed upon by aether. That this ratio is such is hardly surprising, since the ratio of the minor axis of an ellipse to the major axis is given by sqrt(1-e^2):1, e being the eccentricity. Being unable to correspond with a fool that indulges in such childish behaviour and dishonest claims, I merely ignore any further communiqu.8es he may post. Androcles === Subject: Re: Final Rout of Synchronization Clocks in Relativity > Germaine to this Œdicussion¹ is the fact that contrary to all claims > that experiments have been performed which showed Œactually¹ that > light was independent of source motion, apparently only one has been > conÞrmed, AND TURNS OUT TO BE A CROCK! > admitted What on earth are you talking about? How does one run a partcle accelerator up to speed? , and they WERE travelling at c...........sounds good to > DHR¹s! What does that mean? Could you please elaborate? > What they forgot to mention was that the SOURCE, at the precise > moment of emission of photons, STOPPED!! the pi0? If not, please elucidate what you are talking about. Please do so to reassure us that you are not just regurgitating some garbled piece of half a story you heard somewhere. If it is the pi0 decay experiment of which you speak, the following apply: They were most certainly not travelling at zero speed when they decayed. They decayed in þight. This is borne out by the fact that the forward going photons were shifted up in energy in full accordance with the relativistic Doppler effect and the backward going photons were shifted downwards in energy by the expected amount, for the same reason. > The experiment is a fraud (perhaps unknowingly)----the source was NOT > moving when the light was emitted. You have said this twice now. I have showed you why you are ill-informed. (just as a photon is not moving at > the surface of a mirror There is no analogy of any description whatsoever between the experiment which I saw being conducted and a photon moving or not moving at the surface of a mirror. Perhaps I have the wrong experiment in mind. That could be cleared up easily by you telling us what the actual experiment is of which y ou are speaking, and where it was done. > ....and that is ALL DHR¹s have to cling to. On the contrary, that experiment is but a tiny weapon in the armoury of physicists. The really abundant evidence in favour of SR comes from the designed in accordance with the principles of relativistic mechanics and kinematics as accurately as can be measured. > All the rest involve > tweaked clocks and wavelength. Twaddle. Franz Heymann === Subject: Number of Cuts in a non-complete graph im am intersted in Þnding the number of cuts (not minimum cuts) of a graph who ist not complete. For complete graphs i found: Number of Cuts = 2 ^(Number of Vertices) - 1 First question: What ist the name or author of this Theorem. Does anybody now an formula to calculate the number of cuts for a (given) non complete graph? === Subject: Re: Equation bounds Not sure, but I think I found an instance where this is not true? FindInstance[{ V31 == (V32 + V21)/(1 + V32*V21), -1 < V21, V32 < 1, V31 >= 2}, {V31, V32, V21} ] V31 -> 4, V32 -> -2, V21 -> 2/3 With your two constraints implying that -1 < V31 < 1, this found an example of V31 that was outside your implied range. (V31>=2). Hope I said this right. -- Dana > I¹ve been looking at this for a while. I can¹t seem to Þgure it out. > I don¹t think it should be that complicated :( > i have the following equation: > V31 = (V32 + V21) / (1 + V32*V21) > How do I verify that if -1 < V21, V32 < 1, then -1 < V31 < 1??? > This is what I was thinking: > Well we know the extreme values are > -2 < V32 + V21 < 2 and 0 < 1 + V32*V21 < 2 > But we have a problem when both V32+V21 and 1+ V32*V21 are close to 0. > Maybe I don¹t know how to manipulate the equation to make things > clearer. > Any help is appreciated. === Subject: Re: Data Þtting problem with chebyshev polynomials Here¹s a suggestion. Split y into real and imaginary values, and Þt them separately: yr = p0 T0(z) +p1 T1(z)+...+ pn Tn(z) yi = q0 T0(z) + q1 T1(z) +...+ qn Tn(z). Here z = x/i is a real number. All numbers are now real. -Michael. === Subject: Re: Biomathematics - > > (Latin American Spanish):(literate Spanish)::(Ebonics):(literate > > hey Uncle Al, aren¹t we getting caustic about something that doesn¹t > even exist? There is no such thing as Latin American Spanish. You¹re > acting like > Thiotimoline ;P > I live in Southern California. Spanish as spoken by literate > Spaniards is a lyrical Romance language. Spanish as spoken by local > swine is an agrammatical slurred patois that sounds like a gutteral > chicken cackling while Þring a machine gun. > New World wetback women are astoundingly ugly as a class. OOh, Uncle Al, you are SO cute when you¹re being outrageous! === Subject: Re: Now for a card problem (Followup to: Yet another dice game problem.) >I always liked this exercise: Thoroughly shufþe a standard deck of >playing cards. Turn over cards one by one. Find, in an efÞcient >manner, the expected numbers of cards up to and including >a) the ace of spades and >b) the Þrst ace. *sigh*. 1. I can never resist the temptation to do a trick shufþe. Seriously, off the top of my head (very efÞcient, not very accurate), I¹d guess a) is 26.5 and b) is 11 and change. -- Matthew T. Russotto mrussotto@speakeasy.net Extremism in defense of liberty is no vice, and moderation in pursuit of justice is no virtue. But extreme restriction of liberty in pursuit of a modicum of security is a very expensive vice. === Subject: Re: Hope You Haven¹t Sold Your Soul Yet! On a supercalifragilisticexpialidocious day, after dancing about singing Bibbety bobbety boo!, reotpreeoj ishkabibbled: ^ ^> I yet again forgot the command to block messages by a certain person, as ^> rarely do people annoy me. This guy obviously understands a few interesting ^> features of quantum mechanics and string theory, as well as concepts of a ^> sinularity-based collective conciousness, but this juvenile continued ^> obsessing on trivial details is closer to psychosis than enlightenment, in ^> my opinion. ^ ^there is only trivial. 1=0. your ego destroys insight. ^ ^0=1 ^1=8 ^8=0 ^0=totality. ^ ^i annoy you because you speak from a viewpoint of individualism. if you ^had payed attention you would know truth. it is right in front of you; a ^part of everything - all energy, all words, all that exists. ^ Aw shoot, I sold mine... only got 10 bucks for it, too. Ya think I got ripped off?? Should I try to buy it back, or is it even worth foolin with?? -- The Queen of DXers, as well as Queen of the Commonwealth of Virginia, as well as The Ruler of A.D.P., as well as Saint Debbe, as well as Our Lady of the Black Hole Exploratory Input Services as OhFishAlly Appointed by the Psychedelic Pope, a/k/a Saint Isidore of Seville An Ointed Minister of the Universal Life Church Reverant of the Church of the SubGenius, UnOrthodox Superior Mutha Superior of the Little Sistahs of the Politically Incorrect Worshipper of Eris, Goddess of Discord I WON¹T grow up!! -- Peter Pan === Subject: Re: Monotonic function > Situation: > Let y = a^n + b^n (1) where (a, b) = 1, b > a > 1 and odd n 3. > Let a1, a2 are the two values of a and b1,b2 are the two values of b > Here a2 > a1 and b2 > b1 > y1 = a1^n + b1^n (2) and y2 = a2^n + b2^n (3) > Therefore, y2 > y1 > This much is deÞnitely true. > Statement: > y in (1) also increases monotonically for a given odd n. > I¹m not exactly sure what you mean by a + b increases LINEARLY. > Linearly in what variable? Clearly it¹s always linear in a and b; do > you mean one of them is a constant? Do you mean that a(x) + b(x) is > linear in x? What exactly do you mean/ Yes, a(x) + b(x) is linear in x. Even if this condition is relaxed the statement is still true. Kindly comment === Subject: any good solution/model for this modiÞed knapsack problem? The knapsack problem in its original form, has a bunch of goods, each with certain value, and each has certain size, there is a knapsack, limited in size; now if you are asked to collect goods into the knapsack, what is the maximum value under the constraint of knapsack size... ModiÞed Problem 1: What if the size and value of each goods are interdependent? that¹s to say, after item 1 is picked, item 2 Œs value may be devalued, and its size maybe inþated, etc. ModiÞed Problem 2: What if the items are continuous, such as þour, etc., the value is propotional to how much is picked into the knapsack... If you have seen solutions to similar problems, please point out for me. Maybe they are simple problem... any comments are appreciated! -Walala === Subject: Re: Showing the following function is onto? > Hi > to prove f(x)= { (2x-1)/x , x in (0,1/2) and (2x-1)/(1-x) x in [1/2, > 1) ] , > f:(0,1)->R is onto you have to split into cases ? i.e take b in R , > either b=0 , b>0 or b<0 and show that for any b the x must lie in > (0,1) ? any shorter way? > Actually I want to prove f is a bijection, but I already proved f is > 1-1 > so i¹m stuck in showing surjectivity, how do you proceed? Note that (2x-1)/x <= 0 for x in (0,1/2), (2x-1)/x --> 0 as x --> 1/2 and has a vertical asymptote with the negative y axis. Note that (2x-1)/(1-x) is 0 at x=1/2 and is >=0 for x in [1/2,1) and has a vertical asymptote with x=1. === Subject: Re: History of stochastic calculus? Good question about its origins. Ask the question here: http://mathforum.org/epigone/historia_matematica/all > I¹m slowly teaching myself stochastic calculus. First Einstein¹s paper on > Brownian Motion, then The Wiener process and gaussian white noise: > dX=a(t)Xdt+b(t)dW and hopefully then on to Ito¹s formula and Black Scholes. > My question is historically where does the subject begin? Was Einstein > the Þrst to posit a randomly þuctuating force as the driving function in a > 1st order linear differential equation? Did he consider it simply a > extremely fast þuctuating yet still smooth differentiable function? I¹m > looking for a good textbook on the subject. === Subject: Re: What is the meaning of imaginary numbers? > I am continually amazed at the almost romantic yearning the many > express, for physicists to have invented mathematics. I guess the idea of thinking about stuff that has physical reality but no proÞt potential is overly abstract for many people. When you go beyond physical reality, it becomes totally incomprehensible. Jon Miller === Subject: Re: Action Device Tragedy >http://www.geocities.com/inertial_propulsion >17 years ago, in 1986, when I was 17 year old, I ran away from one of >the most prestigious engineering college in India, VJTI Bombay. Reason >was, I don¹t want to be Engineer. I want to become Pilot. >through which we can hang the things in air. We can build air vehicles >and any one can become Pilot. >But I am back to square 1. I will have to become Engineer to build >this action device. There are no mechanical engineers in any workshop >in this town. All you people in world are forcing me to build this >device myself even though you are aware that at least one person has >built one of the parts of this device and it works. Shame! Shame! >Shame! >I don¹t want to be Engineer. I want to be Pilot, you Morons! >I am not going to build this action device. That spring in hardware >shop struck to my Þnger very hard. >-Abhi. What little is left of your mind is missing again. -- Dr.Postman USPS, MBMC, BsD; Disgruntled, But Unarmed Member,Board of Directors of afa-b, SKEP-TI-CULT¨ member #15-51506-253. You can email me at: TuriFake(at)hotmail.com Shake it like a polaroid picture. - Andre 3000 of Outkast === Subject: area of a spherical right triangle following formula for the area of a spherical right triangle is derived: tan(E/2) = tan(a/2) tan(b/2) Where E is the area (spherical excess) and a and b are the legs of the right spherical triangle. I have not seen this formula before and I have been unable to Þnd it on the web or in the books that I have searched. Does anyone have a reference for it? I can¹t imagine that such a simple formula has not been recorded before. Rob Johnson take out the trash before replying === Subject: Re: Halting problem and Cantor¹s Diagonal > Can someone tell me how Turing proof that for compute the Cantor¹ s > number we must execute the programs that output the numbers in the > diagonal. > Why we must execute the programs and go in the halting problem? > Why is not there another way to know the result of a program without > execute it? Hi Denis, The diagonal in Turing¹s proof is not a program that is (or MUST be) executed if we are ever to solve the Halting Problem. Rather, it is a program that CAN be constructed and executed, if the Halting Problem were solvable, that produces a contradiction. In particular, it is a program that, by deÞnition of its construction, halts iff it doesn¹t halt. Since this is impossible, by reductio ad absurdum we conclude that no solution to the Halting Problem can exist. (This is reminiscent of the Liar Paradox, This is false. that is true iff it is false. In the case of the Liar Paradox, the sentence exists, so it is neither true nor false. In the case of Turing Machines, the program constructed must halt or not halt, so the program that solves the Halting Problem cannot exist. I have posted elsewhere how we can derive both of these results, and others, from a single general incompleteness theorem.) Thus we can take any purported solution to the Halting Problem and transform (map) it into this program that produces a self contradiction. This is done using 2 principles: A. For any predicate P(x,y) that is decidable, the predicate P(x,x) is decidable. That is, if we have a program that calculates (solves) the two-place predicate P(x,y), then we can write a program that takes one input, x, uses x for both x and y in the Þrst program, and calculates predicate P(x,x). (This is actually a special case of Kleene¹s s-m-n theorem.) B. Given a program that computes predicate P, we can construct a program that behaves as follows: If P then the program loops. If ~P then the program halts. Then the proof proceeds as follows: 1. Assume that the Halting Problem is solvable. Then we have a program that computes the predicate Program x halts on input y. 2. By principle A, we can construct a program that computes Program x halts on input x. 3. By principle B, we can construct a program (call it P3) that loops if its input x halts on an input of x, and halts if its input x loops (does not halt) on an input of x. 4. We now run program P3 on itself, P3(P3), that is, x=P3. Then by its construction described in step 3, P3(P3) loops if P3(P3) halts, and halts if P3(P3) loops. Since every program (plus input), including P3(P3), must halt or loop, and no program can do both, P3(P3) cannot exist. 5. Thus the assumption in step 1 is false, and no program can solve the Halting Problem. We can express this rigorously if we generalize the notion of an axiomatic system. In Logic, an axiom is a proposition (sentence) that is accepted without proof. Originally the intent was that the axioms be true statements (e.g. Euclid¹s axioms of geometry, aka Euclid¹s Postulates.) However, over the years, Logicians studying the nature of deduction focused on the deductive process itself, independent of whether the expressions being generated were true statements, or even statements at all. We can take this a step further and replace the notion of a true statement or axiom with a statement for which we have a computer program (Turing Machine) that halts yes iff the statement is true, and halts no iff it is false. In the case of a predicate with degree > 1, we say that the predicate is recursive (decidable). For a 0-place predicate, i.e., sentence, we have the curiosity that every 0-place predicate, by deÞnition, is recursive, since there is a program that always halts yes and there is one that always halts no. People sometimes debate the signiÞcance of applying the concept of a recursive predicate to a 0-place predicate (sentence.) For example, see the current Google Groups thread Halting of a Given Machine at: However, what I do is to associate with any predicate, whether it be of degree 0 or >0, a program, and call that an axiom. Then the special case of an axiom in classical Logic is simply a statement whose program is Halt yes. Then the Rules of Inference are functions that map programs which calculate known predicates (or functions), the axioms (and theorems), this we can develop a very general Program Synthesis system. By diagonalizing, we can directly show that the set of programs that do not halt yes on themselves is not recursively enumerable. This I call the Axiom of Incompleteness. Then, working backwards from this one axiom (a formalization of Reduction) we can conclude that various other programs cannot exist, including a solution to the Halting Problem. (See my paper cited below.) Various researchers, including Marvin Minsky and Hartley Rogers, have written over the years that there is no practical use for the Theory of Computation. However, we have just seen that: Program Synthesis + 1 Axiom = Theory of Computation Program Synthesis is arguably the most practical problem there is in all of Computer Science. (It makes program analysis and program veriÞcation obsolete!) The very same principles that show us how to (automatically) write programs can be used to show us what cannot be programmed. Charlie Volkstorf Cambridge, MA http://www.arxiv.org/html/cs.lo/0003071 PS Don¹t worry if the Halting Problem is a little mysterious. In Computer, Paradoxes and the Foundations of Mathematics (http://www.cs.auckland.ac.nz/CDMTCS/chaitin/amsci.pdf) Gregory Suppose you could write a computer program that checks whether any given computer program eventually halts. Create a second program. If the program under investigation terminates, go into an inÞnite loop. Here comes the subtle part: Feed your new program a copy of itself. What does it do? The answer is, Nothing., because you just fed it a program without also feeding it that program¹s input. It can¹t test a program without having that program¹s input. If you instead feed it itself plus itself as input, then you¹re not feeding it a copy of itself anymore. He missed the real subtle part: how to input a copy of yourself to yourself, where yourself depends on the input. This is the trick that Turing accomplished in 1937. Chaitin is confusing a program (which needs input, even if that input is ignored) with a complete program + input. Any purported solution to the Halting Problem is a program (which needs input), while its input is a complete program + input. The self-contradictory program that is constructed doesn¹t take in a complete program + input and see if it halts or not. Rather, it takes in a program (without its input) and sees if that program would halt if it had itself as its input. What Chaitin needs to say is that If the program under investigation, when run with itself as input, terminates, then go into an inÞnite loop, and terminate otherwise. Chaitin got the self-reference part wrong. So don¹t worry. After writing about the unsolvability of the Halting Problem for over 30 years, Gregory Chaitin still hasn¹t gotten it right! === Subject: re:Mathematics forums Maybe I am just looking in the wrong places, but it appears to me that none of the webpages you provided led to post-reply style forums like this one. If anyone has a link to a forum where you can post your math problems i would appreciate if you provided it here for me. looking for please tell me quickly how to post something there. http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: groups and permutations >Let G be a group, S the symmetric group on G, >and H the holomorph of G. We say that G is strong >if the normalizer of H in S is H. Do these groups have a >standard name? Is there any non-trivial necessary and >sufÞcient condition to characterize these groups? >If not, are there some sufÞcient conditions for a group to be strong > (or at least examples of famous strong groups)? >If G is Þnite, are these groups classiÞed? Well, the Þrst thing to note is that G strong implies G is abelian. If G is nonabelian, then the permutation of G that interchanges g and g^-1 for all g in G lies in the normalizer of H in S, but not in H. If G is abelian things are more complicated, but I am sure it is possible to determine exactly which abelian groups G are strong, at least when G is Þnite. After a little thought and experimentation I have: 1. Elementary abelian p-groups are strong. 2. C4 (cyclic group) is strong, but C2^n is not strong for n>=3. 3. C2 x C4, C4 x C8, C2 X C8 are not strong, but C4 x C4, C2 x C2 x C4 and C2 x C4 x C4 are. 4. C9, C27, C3 x C9 are all strong. Make what you can of that! I have not thought at all about inÞnite abelian groups, except that the inÞnite cyclic group if obviously strong, since all elements in H - G have order 2 in that case. Derek Holt. === Subject: Re: area of a spherical right triangle > following formula for the area of a spherical right triangle is derived: > tan(E/2) = tan(a/2) tan(b/2) > Where E is the area (spherical excess) and a and b are the legs of the > right spherical triangle. > I have not seen this formula before and I have been unable to Þnd it > on the web or in the books that I have searched. Does anyone have a > reference for it? I can¹t imagine that such a simple formula has not > been recorded before. It¹s a nice formula, but it can be derived from the other well-known formulas in spherical trigonometry: The half-side formulas are tan(a/2) = K * cos(S-A) and tan(b/2) = K * cos(S-B), where K^2 = - cos(S) / [cos(S-A) * cos(S-B) * cos(S-C)], and 2S = A+B+C = E + pi. Multiplying the two equations and inserting C = pi/2 leads to the above result. Hope that helps. -Michael. === Subject: Re: Action Device Tragedy >17 years ago, in 1986, when I was 17 year old, I ran away from one of >the most prestigious engineering college in India, VJTI Bombay. Reason >was, I don¹t want to be Engineer. I want to become Pilot. Indian Air Force? Bombay Flying Club? >through which we can hang the things in air. We can build air vehicles >and any one can become Pilot. That¹s been true for a century. Most of them do it in Cessnas. -- Richard Herring === Subject: Re: Equation bounds Oooops. You said... >> if -1 < V21, V32 < 1, then ... I think you meant for constraints: -1 < V21 < 1, -1 < V32 < 1 (Both variable between -1 & 1) With this added constraint, then I think -1 < V31 < 1 would be true. So, please disregard my other idea. Anyway, I too am trying to Þgure this out with the hints given. === Subject: Re: Action Device Tragedy > I am not going to build this action device. That spring in hardware > shop struck to my Þnger very hard. I predict it won¹t stop him from posting about it. -=-=-=-=- === Subject: Re: The most linear sigmoid What is wrong with using a numerical approach? You want a exact answer? I¹m not sure I understand your problem. You have a linear function from [0, 1] to [1/2, 1] and you want to approximate this with a sigmoid function with some unspeciÞed slope. Is that right? Do you require that the approximating function maps 0 to 1/2 and 1 to 1. In that case your sigmoid does not work. -Michael. > I¹m looking for solving analytically this equation : > f1(x) = (x+1) / 2 linear > f2(x) = (1/(1+exp(-ax)) sigmoid > d ( int( (f1(x)-f2(x))^2, x=0..1) ) / da = 0 > I Þnd a numerical solution : a = 2.46 > The problem is to adjusting the sigmoid slope (a) for having the most closed sigmoid to the linear function. > Thierry Hoinville. === Subject: Help in Computing Lie algebras I have this question that is bugging me to death: I¹m reading about Lie algebras, and in particular, the lie algebras of the orthogonal groups and the heisenberg group. Try to compute the lie algebra of SO(n) : Well an n x n real matrix R is orthogonal if and only if R^T = R^{-1}. Thus, given a n x n real matrix X, e^{tX} is orthogonal if and only if (e^{tX})^T = (e^{tX})^{-1} or *** e^{t(X^T)} = e^{-tX}. My text goes on to say that clearly a sufÞcient condition for this to hold is that X^T = -X. Then the next line is what confuses me : If *** holds for all t, then by differentiating at t = 0, we must have X^T = -X. Why? Isn¹t the derivative of e^{tX} at t = 0 equal to X? Then what is the point of differentiating? I would probably understand a lot better if I understood the case for the Heisenberg group, which explicitly says that things change once you look at the derivative (but I don¹t understand) : In trying to compute the lie algebra of the heisenberg group H which is the group of all 3 x 3 real matrices of the form 1 a b 0 1 c 0 0 1 we know that the exponential of the matrix of the form 0 d f 0 0 e 0 0 0 is in the heisenberg group H. The next line says On the other hand, if X is any matrix such that e^{tX} is of the form 1 a b 0 1 c 0 0 1 (which is of course what we want), then all of the entries of X = d/dt e^{tX} | t = 0 which are on or below the diagonal must be zero, so that X is of the form 0 d f 0 0 e 0 0 0 . How can you tell this from taking the derivative? Moshe Adrian === Subject: Re: Dishonesty > I note that dishonesty has become the normal thing for ŒBilge¹. > such as alt.troll or alt.moron. This activity is called setting follow-ups and there is nothing dishonest about it. http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ VectorSpace.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ NegTime.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ TheLiar.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ CrapHuh.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Loadcrap.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Vision.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ HelpPretend.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ ProveProof.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ AndroMMX.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Equation.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Relativist.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Humour.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Chuckle.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Gibberish.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ AboutTheories.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ ConArtist.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ ProvePostulate.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Abstraction.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ EnergyConservation.htm l http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ SpeedInvariant.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Androrgasm.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ AndersenLogic.ht.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ SqrtAnswers.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ PartialDiff.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ AndArg.html Dirk Vdm === Subject: Re: Action Device Tragedy > 17 years ago, in 1986, when I was 17 year old, I ran away from one of > the most prestigious engineering college in India, VJTI Bombay. You misspelled expelled for incompetence and mental disorder. For a crazy person to be ostracized rather than worshipped in India is saying something truly vile. Why are you trolling your disease through professional newsgroups? -- Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: Action Device Tragedy > http://www.geocities.com/inertial_propulsion > But I am back to square 1. I will have to become Engineer to build > this action device. Why? It¹s not that complex. Three joints, four magnets, Þve sticks. Even I could build it if I saw any proÞt in doing so. > There are no mechanical engineers in any workshop > in this town. All you people in world are forcing me to build this > device myself even though you are aware that at least one person has > built one of the parts of this device and it works. Shame! Shame! > Shame! > I don¹t want to be Engineer. I want to be Pilot, you Morons! You. Pilot. Ok, start by learning to fold paper airplanes. > I am not going to build this action device. That spring in hardware > shop struck to my Þnger very hard. And that stops you from making a device that one day could conquer the universe? Well, it¹s been nice to deal with you. sci.astro will be less entertaining now. Volker === Subject: Re: Number of Cuts in a non-complete graph >im am intersted in Þnding the number of cuts (not minimum cuts) of a graph >who ist not complete. >For complete graphs i found: Number of Cuts = 2 ^(Number of Vertices) - 1 Should be 2^(n-1) - 1. >First question: What ist the name or author of this Theorem. It hardly deserves a name. Just observe that cuts are in 1-1 correspondence with ways to partition the set of vertices into two nonempty sets. >Does anybody now an formula to calculate the number of cuts for a (given) >non complete graph? The set C*(G) consisting of the cuts and the empty set forms a vector space over the 2-element Þeld Z_2. For a connected graph with n vertices, this has dimension n-1 (see e.g. R. Diestel, Graph Theory, Springer 1997, Theorem 1.9.6). For any graph, C*(G) is the direct sum of the corresponding spaces for each connected component, so it has dimension n-k if there are n vertices and k connected components. And thus the number of cuts is 2^(n-k) - 1. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Dishonesty > I note that dishonesty has become the normal thing for ŒBilge¹. Having a hard time with objectively demonstrated competence, are you? Why don¹t you slit open your abdomen and augur the future from your splayed guts? > We note that in ŒMichelson¹s Interference Experiment¹, H.A.Lorentz Phys. Rev. Lett. 88(1) 010401 (2002) Phys. Rev. Lett. 42(9) 549 (1979) Phys. Bull. 21 255 (1970) Europhysics Lett. 56(2) 170 (2001) Gen. Rel. Grav. 34(9) 1371 (2002) Try education rather than psychosis and personal snit. Try emulating adult behavior. Here¹s a hint: God doesn¹t speak to you. -- Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: any good solution/model for this modiÞed knapsack problem? >The knapsack problem in its original form, has a bunch of goods, each with >certain value, and each has certain size, there is a knapsack, limited in >size; now if you are asked to collect goods into the knapsack, what is the >maximum value under the constraint of knapsack size... >ModiÞed Problem 1: >What if the size and value of each goods are interdependent? that¹s to say, >after item 1 is picked, item 2 Œs value may be devalued, and its size maybe >inþated, etc. This is likely to complicate things considerably, depending on the details of the interdependence. It may be still be possible to write it as an integer linear programming problem, but you may need variables such as x_{ij} which is to be 1 if both items i and j are picked, 0 otherwise. >ModiÞed Problem 2: >What if the items are continuous, such as þour, etc., the value is >propotional to how much is picked into the knapsack... That makes the problem trivial (assuming you didn¹t do the Þrst modiÞcation): just take as much as possible of the good with the highest value::size ratio. 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: sinusoidal 2d equation >Somebody would be kind enough to remind me the equation of a >sinusoidal surface in 2dimension (aim to do plot a wavy surface) in >cartesian system. Actually in polar coordinates would appreciated as >well. >Thierry If you want ripples like after you throw a rock into a pond try: z = a*cos(b*r) in cylindrical (polar) coordinates with suitable a, b. Or if you want something like a corrugated roof: z = a*sin(b*x) as a function of x and y will do. --Lynn === Subject: Re: USE LINUX OR PERISH! ,n a supercalifragilisticexpialidocious day, after dancing about singing Bibbety bobbety boo!, MiRo ishkabibbled: ^ ^ ^> Oy, I better get to work! Fortunately I have my ofÞshal Alien Detector ^> which glows reddish in the presence of AYLEENNNS. ^ ^Ayleens? Once I knew a girl named Ayleen (maybe not), but I do not ^recall your detector glowing, or maybeee? Was I stoned? That Ayleen was ^a crazy bitch. ^ ^ Coulda been... I don¹t remember MY detector glowing, but maybe she was wearing her protective aluminum foil beanie to divert the signals. Plus I¹m known LOTSA crazy biotches who didn¹t set off the detector, which just proves that you might be able to get the alien outta tha bitch, but not the bitch outta the alien. My 1st hubby¹s exwoof (as he called her) was BOTH. I¹m hoping for the mothership to pick her up soon- -though they might not want HER back either. -- The Queen of DXers, as well as Queen of the Commonwealth of Virginia, as well as The Ruler of A.D.P., as well as Saint Debbe, as well as Our Lady of the Black Hole Exploratory Input Services as OhFishAlly Appointed by the Psychedelic Pope, a/k/a Saint Isidore of Seville An Ointed Minister of the Universal Life Church Reverant of the Church of the SubGenius, UnOrthodox Superior Mutha Superior of the Little Sistahs of the Politically Incorrect Worshipper of Eris, Goddess of Discord I WON¹T grow up!! -- Peter Pan === Subject: Re: USE LINUX OR PERISH! On a supercalifragilisticexpialidocious day, after dancing about singing Bibbety bobbety boo!, John GrifÞn ishkabibbled: ^ ^> On a supercalifragilisticexpialidocious day, after dancing ^> about singing Bibbety bobbety boo!, ClueStickMan ^> ishkabibbled: ^ ^> [ snipped...not the point ] ^ ^>- ^>The Queen of DXers, as well as ^>Queen of the Commonwealth of Virginia, as well as ^>The Ruler of A.D.P., as well as ^>Saint Debbe, as well as ^>Our Lady of the Black Hole Exploratory Input Services ^> as OhFishAlly Appointed by the Psychedelic Pope, a/k/a ^> Saint Isidore of Seville ^>An Ointed Minister of the Universal Life Church ^>Reverant of the Church of the SubGenius, UnOrthodox ^>Superior Mutha Superior of the Little Sistahs of the ^> Politically Incorrect ^>Worshipper of Eris, Goddess of Discord ^>I WON¹T grow up!! -- Peter Pan ^ ^Just when it begins to seem that maybe lame has been ^completely deÞned, you come along and extend it. ^ ^ ^ ^ My dear, the only time I extend my lameness is when circumstances (illness or injury) force me to use various appliances to move about in the world. And then I endeavor to retrieve the lameness at the Þrst possible chance. Or is this just a case of a wannabe lamer jealous of a TRUE lamer??? Don¹t fret, I¹m sure you¹re receive your certiÞcation of Absolute Lameness soon; you know how the holiday mail deliveries can be delayed. -- The Queen of DXers, as well as Queen of the Commonwealth of Virginia, as well as The Ruler of A.D.P., as well as Saint Debbe, as well as Our Lady of the Black Hole Exploratory Input Services as OhFishAlly Appointed by the Psychedelic Pope, a/k/a Saint Isidore of Seville An Ointed Minister of the Universal Life Church Reverant of the Church of the SubGenius, UnOrthodox Superior Mutha Superior of the Little Sistahs of the Politically Incorrect Worshipper of Eris, Goddess of Discord I WON¹T grow up!! -- Peter Pan === Subject: BandMatrix elements.. I have the dimensions of bandmatrix of size N*M, where N is the number of rows, and M is the number of columns in the matrix. I also have the Upper Bandwidth U, and the Lower Bandwidth L. Does there exist a formula to calculate the number of elements in the band? I would like to construct a function function(N,M,U,L) Which returns the number of elements in the band. Any Help appreciated, Pat === Subject: Re: any good solution/model for this modiÞed knapsack problem? > The knapsack problem in its original form, has a bunch of goods, each with > certain value, and each has certain size, there is a knapsack, limited in > size; now if you are asked to collect goods into the knapsack, what is the > maximum value under the constraint of knapsack size... The knapsack problem as stated at http://mathworld.wolfram.com/KnapsackProblem.html is slightly different. > ModiÞed Problem 1: > What if the size and value of each goods are interdependent? that¹s to say, > after item 1 is picked, item 2 Œs value may be devalued, and its size maybe > inþated, etc. Sounds a lot more difÞcult than the unmodiÞed problem posed above! > ModiÞed Problem 2: > What if the items are continuous, such as þour, etc., the value is > propotional to how much is picked into the knapsack... I¹m not sure if I understand the problem posed. As stated, surely you calculate which item contributes most value per unit volume of the knapsack and take as much as possible, then take as much as possible of the next most valuable item, and so on. Mark Atherton === Subject: Re: I thought it was the tuna can > At some point in my educational training, I thought I was taught that > a metal can with height=radius (similar to a tuna can) was special, > like the greatest volume for the least surface or something > counterintuitive like that. Alas, it really is the expected tomato > sauce can (height=diameter) that has the greatest volume/surface > ratio. So . . . is there anything special about the tuna can or do I > just need to move on with my life? It could be that tuna cans were once twice as tall, in which case their volume/surface ratio would have been the same as those of tomato sauce cans. Then perhaps some marketing guy decided to sell novelty half cans, for little old ladies to feed their cats or something, and over time these proved so much more popular than the original can size that the latter stopped being used. Obviously, to minimize changes in canning machinery and make boxing and stacking easy, the diameter would stay the same rather than the entire can being scaled down accordingly, and hence the anomaly. BTW in the UK one can buy full-sized cans of tuna. I¹m amazed you can¹t in the US (assuming that is where you are based), considering the vast profusion of food brands and variations I recall seeing in US stores when I was last there! John Ramsden === Subject: Re: Dishonesty > I note that dishonesty has become the normal thing for ŒBilge¹. [unsnip] He makes the return address something other than sci.physics.relativity, such as alt.troll or alt.moron. Obviously such a cheap trick can be considered amusing, but wears thin when repeated. He claims to be able to Œderive¹ sqrt(1-v^2/c^2) as though it were something special that he alone can do, using the trigonometric functions of a conic section. Obviously this is the full extent of his mathematical capabilities, which, when exhausted, causes him to resort to personal abuse and cheap practical jokes as his only counter argument when his absurd claims are challenged. We note that in ŒMichelson¹s Interference Experiment¹, H.A.Lorentz (p. 7, The Principle of Relativity¹ Dover publications, SBN 486-60081-5) uses this in a supposed Œ...shortening in the direction of motion in the proportion of 1 to sqrt(1-v^2/c^2)...¹ by which method Lorentz assumes a spherical Earth becomes an ellipsoidal Earth pressed upon by aether. That this ratio is such is hardly surprising, since the ratio of the minor axis of an ellipse to the major axis is given by sqrt(1-e^2):1, e being the eccentricity. Being unable to correspond with a fool that indulges in such childish behaviour and dishonest claims, I merely ignore any further communiqu.8es he may post. > Having a hard time with objectively demonstrated competence, are you? > Why don¹t you slit open your abdomen and augur the future from your > splayed guts? Arsehole, I¹d rather slit open yours, you fucking ignorant cunt! > We note that in ŒMichelson¹s Interference Experiment¹, H.A.Lorentz > Phys. Rev. Lett. 88(1) 010401 (2002) > Phys. Rev. Lett. 42(9) 549 (1979) > Phys. Bull. 21 255 (1970) > Europhysics Lett. 56(2) 170 (2001) > Gen. Rel. Grav. 34(9) 1371 (2002) > Try education rather than psychosis and personal snit. Try emulating > adult behavior. Example of adult behaviour from Auntie Alice:- quote > Hell, Dumb Donny shitHead, you would be off by more than 1000 mph at > the Earth¹s equator anyway - or short by the same amount, or anywhere > in-between. Yer stooopid. unquote. Yeah, that¹s really adult, you fucking imbecile! > Here¹s a hint: God doesn¹t speak to you. I don¹t happen to accept your stupid gods, moron, so here¹s a hint. I don¹t accept you or your stupid relativity either. Try using a little logic instead of you pathetic attempts at intimidation, I can piss all over you at that game, stoopid! http://www.androc1es.pwp.blueyonder.co.uk/Fundamental_rev_2.1 .htm Androcles === Subject: Re: Mathematics forums >Maybe I am just looking in the wrong places, but it appears to me that >none of the webpages you provided led to post-reply style forums like >this one. If anyone has a link to a forum where you can post your >math problems i would appreciate if you provided it here for me. >looking for please tell me quickly how to post something there. Well, this is probably going to turn out to be a stupid question, but why not post whatever it is _here_? I mean it sounds like this is an emergency, and evidently you¹ve already Þgured out how to post and even how to read replies to your posts... >http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups >---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- ************************ David C. Ullrich === Subject: Re: Dishonesty [snot] > I don¹t accept you or your stupid relativity either. Try using a little > logic like http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Gibberish.html Dirk Vdm > instead of you pathetic attempts at intimidation, I can piss all over you at > that game, stoopid! > http://www.androc1es.pwp.blueyonder.co.uk/Fundamental_rev_2.1 .htm > Androcles === Subject: Re: Help in Computing Lie algebras >I have this question that is bugging me to death: >I¹m reading about Lie algebras, and in particular, the lie algebras of the >orthogonal groups and the heisenberg group. >Try to compute the lie algebra of SO(n) : Well an n x n real matrix R is >orthogonal if and only if R^T = R^{-1}. Thus, given a n x n real matrix X, >e^{tX} is orthogonal if and only if (e^{tX})^T = (e^{tX})^{-1} or >*** e^{t(X^T)} = e^{-tX}. >My text goes on to say that clearly a sufÞcient condition for this to hold >is that X^T = -X. Then the next line is what confuses me : If *** holds >for all t, then by differentiating at t = 0, we must have X^T = -X. Why? >Isn¹t the derivative of e^{tX} at t = 0 equal to X? Yes. So for example the derivative of e^{-tX} at t = 0 is -X, and the derivative of e^{t(X^T)} at t = 0 is X^T. >Then what is the point >of differentiating? To show that X^T = -X, as claimed. > I would probably understand a lot better if I >understood the case for the Heisenberg group, which explicitly says that >things change once you look at the derivative (but I don¹t understand) : >In trying to compute the lie algebra of the heisenberg group H which is the >group of all 3 x 3 real matrices of the form >1 a b >0 1 c >0 0 1 >we know that the exponential of the matrix of the form >0 d f >0 0 e >0 0 0 >is in the heisenberg group H. The next line says On the other hand, if X >is any matrix such that e^{tX} is of the form >1 a b >0 1 c >0 0 1 >(which is of course what we want), then all of the entries of X = d/dt >e^{tX} | t = 0 which are on or below the diagonal must be zero, so that X >is of the form >0 d f >0 0 e >0 0 0 >How can you tell this from taking the derivative? If e^(tX) = [[1, a(t), b(t)], [0, 1, c(t)], [0, 0, 1]] then when you differentiate both sides and set t = 0 you get X = [[0, a¹(0), b¹(0)], [0, 0, c¹(0)], [0, 0, 0]]. The fact that the derivative of the left side is X is mentioned above, while the fact that the derivative of the right side is what I say follows from the fact that the derivative of 1 is 0. (I¹m having a hard time Þguring out what you¹re missing about all this, so as to give a better answer. It _seems_ like what you¹re missing is the fact that if f(t) = g(t) for all t then f¹(0) = g¹(0); that¹s the key to both the questions you ask. But you can¹t be missing _that_...) >Moshe Adrian ************************ David C. Ullrich === Subject: Re: Help in Computing Lie algebras > I have this question that is bugging me to death: > I¹m reading about Lie algebras, and in particular, the lie algebras of the > orthogonal groups and the heisenberg group. The right way to see what te Lie algebra of a subgroup G of GL(n,R) is, it yo consider the tangent space at the identity. More concretely, consider all smooth functions f:R -> G with f(0) = I. Then Lie(G) consists of all f¹(0). > Try to compute the lie algebra of SO(n) : Well an n x n real matrix R is > orthogonal if and only if R^T = R^{-1}. Or A is in SO(n) iff A A^T = I an det(A) = I. Let f:R -> SO(n) with f(0) = I. Then f(t)f(t)^T = I. Differentiating, f¹(t) f(t)^T + f(t) f¹(t)^T = 0. Evalute at 0: f¹(0) + f¹(0)^T = 0, so f¹(0) is skew-symmetric. So Lie(SO(n)) is contained in the space of skew-symmetric matrices. On the other hand, if B is skew-symmetric, then g(t) = exp(tB) is in O(n) for all t and g(0) = I and g¹(0) = B. So the space of skew-symmetric matrices is contained in Lie(SO(n)). -- 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 base for R? I am studying KOLMOGOROV & FORMIN. Introductory Real Analysis. Dover In section 9.3, page 81, they say that the set of all open intervals with rational end points is a base on the real line, since any open interval can be represented as the union of such intervals. According to the above stated, how could I represent the interval (-pi,pi) ??? Diego Andr.8es === Subject: Re: a base for R? > I am studying > KOLMOGOROV & FORMIN. Introductory Real Analysis. Dover > In section 9.3, page 81, they say that the set of all open intervals > with rational end points is a base on the real line, since any open > interval can be represented as the union of such intervals. > According to the above stated, how could I represent the interval > (-pi,pi) ??? For instance, as the union of all (a,b) where -pi < a < b < pi. -- 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: Dishonesty > I note that dishonesty has become the normal thing for ŒBilge¹. > such as alt.troll or alt.moron. He¹s always done that. Nothing new for bilge. It just means that he either doesn¹t want to discuss it or he¹s unable to Þgure something out so he doesn¹t want to be anymore embarassed than he alread is. === Subject: Polynomials and the AC Method I am moving forward with my math studies. I am working on a text called Intermediate Algebra. I recently was working on systems of three equations. I encountered for the Þrst time matrix operations. The operations themselves are pretty easy. I think you just have to memorize them. The hard part is understanding why they work (which I don¹t really intend to do). Otherwise, I have a math question: There is a method to solve polynomials called the ac method. If you have a polynomial of the form ax^2+bx+c, you can factor this by substituting m and n for b, such that m+n=b and mn=ac. Here is my question: the method is pretty easy, and works. But I can¹t Þgure out why it works. I see why m+n must equal b. This is obvious. But where does the requirement that mn=ac come in? There is no obvious reason that I can see. === Subject: Re: Now for a card problem (Followup to: Yet another dice game problem.) Aw, Fred - I was hoping we geezers would let the youngsters play with this one. Guess I should have said that. >>I always liked this exercise: Thoroughly shufþe a standard deck of >>playing cards. Turn over cards one by one. Find, in an efÞcient >>manner, the expected numbers of cards up to and including >>a) the ace of spades and >> >1 + 51/2 = 26.5; each of the other 51 cards has a 50-50 chance of >turning up before the ace of spades. Easier: The position of the Þrst ace is uniformly distributed on {1, 2,..., 52}. I included this part because too many students tend to make a difÞcult problem out of this one. >>b) the Þrst ace. >> >1 + 48/5 = 10.6; each of the 48 non-aces has one chance in Þve of >beating out all of the aces. Since Jon Haugsand seemed to want a formalization of this, here it is: Label the 4 aces a1, a2, a3, and a4, the 48 non-aces o1, o2, ..., o48. Let I_i be the indicator for o_i turning up before any ace. Then the total number of cards N = 1 + sum(i=1..48, I_i), whence EN = 1 + sum(i=1..48, EI_i) = 1 +48 EI_1 by the linearity of expectation and symmetry. EI_1 = P{o1 turns up Þrst amongst o1, a1, a2, a3, a4} = 1/5 by symmetry. >>[Bonus points: determine the variances as well.] >> >Too hard, I¹d need pencil and paper for that one. (a) should be easy now. (b) is more tedious but uses the same ideas as in the expectaion. Here¹s another: Find the covariance of the two counts. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: need help in understanding Torkel¹s ZFC comment >I¹d love to see a couple of speciÞc examples of formal proofs of >consistency of subsets of ZFC in ZFC. (Actually, I just want to test >out my simpliÞcation of ZFC and formal proofs within ZFC are so hard >to Þnd.) > Actually ZFC has inÞnitely many axioms. Those inÞnitely many > axioms can be described by Þnitely many axiom _schemata_. > (Schema? Oh hell, schemes?) >Charlie Volkstorf > ************************ > David C. Ullrich Ok. I have always seen them listed as 10 (or less) axioms, but I agree with the interpretation given by you and others - to a degree. I have always felt that axiom schemas are just rules of inference. Each instance of the axiom schema is actually a theorem that the actual rule of inference maps to what is called an axiom. So, for example, Separation is a rule that says that if a given collection is expressible by some formula, and another collection is expressible by some set, then the intersection is expressible by a set. For those (if anyone) who have followed my notation, that¹s: Separation: P[TW] , Q[SE] => P^Q[SE] Replacement: P(x)[SE] , Q(I,x)[TW] => (eA)P(A)^Q(A,x)[SE] where P(x)[Q(a,b)] means there is an N such that N # P(x)[Q(a,b)], and N # P(x)[Q(a,b)] means P(a) <=> Q(N,a), and P[Q] means P[Q(a,b)]. Relation SE(a,b) is Set a contains element b. and TW(a,b) means Formula a with b substituted for its free variable is true. Anyway, I think that many (such as Peano and Zermello) seem to feel that everything must be an axiom, when in fact their systems are really a combination of axioms, rules of inference, deÞnitions and theorems, e.g. ZFC is really: DeÞnition: Extensionality Axiom: Power Set, InÞnity, Empty Set, Foundation Rule of Inference: Separation, Sum Set, Replacement, Choice Theorem: Pairing Actually, the truth isn¹t that clear-cut. In some cases, a ZFC axiom is a theorem that is being used as a Rule of Inference. When structured correctly, ZFC is much simpler, smaller and easier to use. You see, I for one am not someone who believes everything he reads. (I believe everything that I program.) I¹m still waiting for my formal (or at least detailed) proofs. Charlie Volkstorf === Subject: Automobile Speeds. Does anyone know why automobile speeds are measures in miles per hour rather than feet/sec or perhaps meters/sec? === Subject: Re: Dishonesty Gauge: >> I note that dishonesty has become the normal thing for ŒBilge¹. >> >> He makes the return address something other than sci.physics.relativity, >> such as alt.troll or alt.moron. >He¹s always done that. Nothing new for bilge. It just means that he >either doesn¹t want to discuss it or he¹s unable to Þgure something >out so he doesn¹t want to be anymore embarassed than he alread is. Or your ranting is off-topic so follow-ups are set appropriately. You always semm to overlook the obvious. === Subject: Re: Automobile Speeds. > Does anyone know why automobile speeds are measures in miles per hour > rather than feet/sec or perhaps meters/sec? They are not measured in miles per hour. They are measured in kilometers per hour. But YMMV. Dirk Vdm === Subject: Re: Explicit prime counting formula, complexity Isn¹t it remarkable how populart prime counting has become with JSH since he was show to be so spectacularly wrong about properties of the ring of algrabraic integers? === Subject: Re: Action Device Tragedy > http://www.geocities.com/inertial_propulsion > 17 years ago, in 1986, when I was 17 year old, I ran away from one of > the most prestigious engineering college in India, VJTI Bombay. Reason > was, I don¹t want to be Engineer. I want to become Pilot. > through which we can hang the things in air. We can build air vehicles > and any one can become Pilot. > But I am back to square 1. I will have to become Engineer to build > this action device. There are no mechanical engineers in any workshop > in this town. All you people in world are forcing me to build this > device myself even though you are aware that at least one person has > built one of the parts of this device and it works. Shame! Shame! > Shame! > I don¹t want to be Engineer. I want to be Pilot, you Morons! > I am not going to build this action device. That spring in hardware > shop struck to my Þnger very hard. Awww, Abhi got a boo-boo? :-) > -Abhi. === Subject: Re: Explicit prime counting formula, complexity ---snip--- > I¹m curious now about whether or not there are smaller expressions > that do what it does out to inÞnity as I sat down and Þgured out how > terms get added. > In general, with x=sqrt(N) there are approximately 2^{x/ln x}/ln N > terms. > James Harris I thought I had an answer for you. Alas, I do not. My primary mathematics resource, a sci.math post by a highly regarded member of the sci.math community, is called a Quick Math Guide. Surprisingly, it offered no guidance to your particular question. Possibly you can e-mail the author in private (and keep it private). Less surprisingly, I am being sarcastic here. The quick math guide won¹t help anyone, ever. Have a nice day, Jay === Subject: Re: Automobile Speeds. > Does anyone know why automobile speeds are measures in miles per hour > rather than feet/sec or perhaps meters/sec? Probably because the average US driver knows approximately how many miles he is from his destination and wants to know approximately how many hours it will take to get there, without needing to do any units conversion. Let¹s see, it¹s about 30 miles so if I do 55 I can make it in a little over half an hour. That¹s as much mental math as most drivers want to do. -- Wayne Brown (HPCC #1104) | When your tail¹s in a crack, you improvise fwbrown@bellsouth.net | if you¹re good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Help in Computing Lie algebras >I have this question that is bugging me to death: >I¹m reading about Lie algebras, and in particular, the lie algebras of the >orthogonal groups and the heisenberg group. >Try to compute the lie algebra of SO(n) : Well an n x n real matrix R is >orthogonal if and only if R^T = R^{-1}. Thus, given a n x n real matrix X, >e^{tX} is orthogonal if and only if (e^{tX})^T = (e^{tX})^{-1} or >*** e^{t(X^T)} = e^{-tX}. >My text goes on to say that clearly a sufÞcient condition for this to hold >is that X^T = -X. Then the next line is what confuses me : If *** holds >for all t, then by differentiating at t = 0, we must have X^T = -X. Why? >Isn¹t the derivative of e^{tX} at t = 0 equal to X? > Yes. So for example the derivative of e^{-tX} at t = 0 is -X, and > the derivative of e^{t(X^T)} at t = 0 is X^T. >Then what is the point >of differentiating? > To show that X^T = -X, as claimed. > I would probably understand a lot better if I >understood the case for the Heisenberg group, which explicitly says that >things change once you look at the derivative (but I don¹t understand) : >In trying to compute the lie algebra of the heisenberg group H which is the >group of all 3 x 3 real matrices of the form >1 a b >0 1 c >0 0 1 >we know that the exponential of the matrix of the form >0 d f >0 0 e >0 0 0 >is in the heisenberg group H. The next line says On the other hand, if X >is any matrix such that e^{tX} is of the form >1 a b >0 1 c >0 0 1 >(which is of course what we want), then all of the entries of X = d/dt >e^{tX} | t = 0 which are on or below the diagonal must be zero, so that X >is of the form >0 d f >0 0 e >0 0 0 >. >How can you tell this from taking the derivative? > If e^(tX) = [[1, a(t), b(t)], [0, 1, c(t)], [0, 0, 1]] then when > you differentiate both sides and set t = 0 you get > X = [[0, a¹(0), b¹(0)], [0, 0, c¹(0)], [0, 0, 0]]. The fact > that the derivative of the left side is X is mentioned > above, while the fact that the derivative of the right > side is what I say follows from the fact that the > derivative of 1 is 0. > (I¹m having a hard time Þguring out what you¹re missing > about all this, so as to give a better answer. It _seems_ > like what you¹re missing is the fact that if f(t) = g(t) for > all t then f¹(0) = g¹(0); that¹s the key to both the questions > you ask. But you can¹t be missing _that_...) THanks for the help I understand now. For one thing I didn¹t realize that a(t),b(t),c(t) were indeed supposed to be functions as opposed to constants (which of course they are functions now that I think about it), so that was a(t),b(t),c(t)...I should have Þgured this out. >Moshe Adrian > ************************ > David C. Ullrich === Subject: Re: Dishonesty > Being unable to correspond with a fool that indulges in such childish > behaviour and dishonest claims, I merely ignore any further communiqu.8es he > may post. Promise? Socks === Subject: For those with aspirations to higher mathematics NNTP-Posting-User: mckay Follow Higher Mathematics in http://www.renyi.hu/ Jm -- But leave the wise to wrangle, and with me the quarrel of the universe let be; and, in some corner of the hubbub couched, make game of that which makes as much of thee. === Subject: Re: Explicit prime counting formula, complexity > The way I calculate explicit prime formulas is straightforward, as I > started with > dS(N,3) = þoor(N-4)/6. with even N > and iterated. > So > N/2 - þoor((N-4)/6) - þoor((N-16)/10) + þoor((N-16)/30) - > þoor((N-36)/14) + þoor((N-22)/42) + þoor((N-106)/70) - > þoor((N-106)/210) + 2. > is the result of two iterations covering 5 and 7. > I¹m curious now about whether or not there are smaller expressions > that do what it does out to inÞnity as I sat down and Þgured out how > terms get added. > In general, with x=sqrt(N) there are approximately 2^{x/ln x}/ln N > terms. Should be that with x=sqrt(N), there are approximately 2^{x/ln x - 1) terms, which I think is a lot forcing me to rethink my assertions before that the formula is the smallest possible. James Harris === Subject: Re: Equation bounds > Oooops. You said... >> if -1 < V21, V32 < 1, then ... > I think you meant for constraints: > -1 < V21 < 1, -1 < V32 < 1 > (Both variable between -1 & 1) > With this added constraint, then I think -1 < V31 < 1 would be true. So, > please disregard my other idea. Anyway, I too am trying to Þgure this out > with the hints given. To simplify the notation a bit, let f(x,y) = (x + y)/(1 + x*y) for |x| < 1 and |y| < 1. Find one-sided limits, lim_{x -> -1+} f(x,y) and lim_{x -> +1-} f(x,y) and show that df(x,y)/dx > 0 when |x| < 1 and |y| < 1 === Subject: Re: The Liar Paradox as a Nonfactual Statement > Just because someone calls it a Quantum Turing Machine doesn¹t mean > it¹s a Turing Machine. You¹re not allowed to change Turing¹s > deÞnition. A QTM is another Base of Computing, not another Turing > Machine. > > I see. Still, there is the following view: > The original Turing machine was deterministic (DTM): the head would be > always in a single state, which would uniquely determine which direction > it would go into and how far. There is a variant of the Turing machine, > which is not deterministic. . . . This began with my statement that an English sentence can be both true and false no more than a Turing Machine can both halt yes and halt no. You countered with the assertion that a Turing Machine can both halt yes and halt no, and a reference to a Quantum Turing Machine which is a model of computing that halts yes and halts no at the same time. Allow me to straighten this mess out. In my original statement I was referring to what Turing deÞned in 1937. Given that deÞnition, do you agree that no Turing Machine can both halt yes and halt no? (debate # 1) (In this case you¹re allowed to answer Indeed. because you DO know that already.) Thus your assertion really is that my choice of words was bad, and that Turing Machine refers to a whole host of models that people have proposed. (debate # 2) In that case, the notion that no Turing Machine can both halt yes and halt no means that None of the models of computing called ÎTuring Machines¹ can both halt yes and halt no. Thus you gave a counter-example: a model of computing called a Turing Machine that both halts yes and halt no. At this point, the question is whether or not that is a reasonable interpretation of the assertion that no Turing Machine can both halt yes and halt no (as opposed to the interpretation that it refers to Turing 1937 only.) Do you really believe (and can justify) that this is a reasonable interpretation of that assertion? If that were the case, then the assertion would mean that nobody in the history of mankind has ever proposed a model of computing in which (a) it was called a Turing Machine, and (b) it had the ability to published in refereed journals of mathematics and computer science? All posts to Google Groups? All notes left in bottles for milkmen or by people shipwrecked on an island? Is that interpretation even well-deÞned?? > An axiom is not a true statement. > Ah? > Let¹s see... > ...every [mathematical] discipline begins with a list of a small number > of sentences, called axioms or primitive sentences, which seem to be > intuitively evident and which are recognized as true without any further > justiÞcation. > (A. Tarski, Truth and Proof) He is talking about axioms developed with the intent of deriving the true assertions concerning a well-deÞned branch of mathematics. Didn¹t you recently say something about some Set Theories having an axiom (or theorem?) that a Quine Atom does (or can?) exist and others that have the opposite? Note that I am talking about true in reality, not the limited space within some particular set of axioms and rules. > It is quite arbitrary. > Nonsense. > ...we use the notion of truth as a guide; for we do not wish to add a > new axiom or a new rule of proof if we have reasons to believe that the > new axiom is not a true sentence, or that the new rule of proof when > applied to true sentences may yield a false sentence. > (A. Tarski, Truth and Proof) Then all this talk of inconsistent (unsound, actually) systems is a waste of time? > That seems to be related to the idea that if something appears in print > then it must be true. > Only if the name of the author is G.9adel or Tarski. :-) How about Hilbert? Charlie Volkstorf > Remember: Easier said than done. - Talk is cheap. (AT & T > advertisement?) > Accepted. ;-) > F. === Subject: Re: need help in understanding Torkel¹s ZFC comment > I¹m still waiting for my formal (or at least detailed) proofs. Why are you waiting? Why not read up on the subject? === Subject: DC Proof: Rationale for a more User-Friendly Notation (Þxed typo) Last month here, I announced my new proof-writing program, DC Proof. Some users have commented on its use of nonstandard notation. Following is a rationale for the notation I have adopted (also posted at my homepage). Dan Visit DC Proof Online at http://www.dcproof.com -- FREE Download (new release yesterday) DC Proof: Rationale for a more User-Friendly Notation While it may not be the standard notation of logicians, in most cases, the notation used by DC Proof does not vary signiÞcantly from it. It was designed primarily for the non-specialist mathematics student in Þrst or second year university or in advanced senior high school courses. To give just one example, most university algebra and calculus (analysis) textbooks, even the advanced ones it seems, do not make use of universal and existential quantiÞers (the logician¹s inverted A¹s and backwards E¹s respectively). They usually spell out for all x or there exists a y such that etc. Students should be used to working with such notation. It was felt that the use of ALL and EXIST would be a more natural and suggestive notation. By the way, users need not type out these words or commit the notation to memory. To minimize keystrokes and to help students learn the notation, there is a convenient, pull-down menu of all symbols and notation used in DC Proof (click Notation on the Premise form, etc.). Also, in a effort to make variable and predicate names more meaningful, DC Proof allows names of any length. This, of course, necessitates the use of some extra delimiters (round and square brackets) for readability. DC Proof automatically minimizes bracketing, however, with its built-in, order of precedence (or operations). This makes it easier for the user to recognize the occurrence of structures like double negations, for example. Generally speaking, the DC Proof notation was designed to be as familiar as possible to students, and to make it easy to learn and master. Return === Subject: Re: 0¹s the solution > Hi Garry, > > I have one number, zero, that I like better than others. > 0¹s the solution to every algebraic equation. That¹s why 0¹s my favorite > number. > What¹s your favorite? > > My favorite is 17. > > The third Fermat number, or the Þrst to end in 7. > And why? > > It¹s the Þrst random number. > > Frank > What makes it random? God === Subject: Re: 0¹s the solution > I like 137 as well, > since it is > 3*4*5*(1 +1/2 +1/3 +1/4 +1/5). > But even better is the related 1764, > which is > 1*2*3*4*5*6*(1 +1/2 +1/3 +1/4 +1/5 +1/6), > and, amazingly, > 1764 also = .... > 42 *42 (!)... > Leroy > Quet 12 = 3 x 4 56 = 7 x 8 etc. etc. Charlie Volkstorf Cambridge, MA === Subject: Re: Explicit prime counting formula, complexity >>The way I calculate explicit prime formulas is straightforward, as I >>started with >>dS(N,3) = þoor(N-4)/6. with even N >>and iterated. >>So >>N/2 - þoor((N-4)/6) - þoor((N-16)/10) + þoor((N-16)/30) - >>þoor((N-36)/14) + þoor((N-22)/42) + þoor((N-106)/70) - >>þoor((N-106)/210) + 2. >>is the result of two iterations covering 5 and 7. >>I¹m curious now about whether or not there are smaller expressions >>that do what it does out to inÞnity as I sat down and Þgured out how >>terms get added. >>In general, with x=sqrt(N) there are approximately 2^{x/ln x}/ln N >>terms. > Should be that with x=sqrt(N), there are approximately 2^{x/ln x - 1) > terms, which I think is a lot forcing me to rethink my assertions > before that the formula is the smallest possible. Take your time. Gib === Subject: Re: What is the meaning of imaginary numbers? > .... > I was always going to get around and give a talk along these lines to > the local high school algebra classes titled There¹s nothing > imaginary about complex numbers, but I never got around to it. > High time the notation imaginary been changed, is n¹t it? __ as > these are just as real and concrete, if only a different type, > becoming complex by mixing (union) with real. My line when introducing complex numbers to my electronics students, If you think the AC voltage across that capacitor is imaginary, just touch the two leads! Then I explain that they¹d get blasted the same as if they¹d touched the resistor leads (for appropriate R, C, V, and omega), but a quarter cycle later... === Subject: Re: Final Rout of Synchronization Clocks in Relativity >>It is actually quite simple. >>There are many types of what loosely can be called >>laser gyros or ring gyros, the two main types >>are the Sagnac ring gyro and the ring laser. >>Common to them all is that the light source >>is rotating with the ring (in ring lasers the sources >>are the atoms in the lasing gas), and that the speed >>of light is isotropic (c or c/n) in the non rotating (inertial) frame. >>This is as predicted by SR and ether theories, but >>blatantly obvious as opposed to the source dependent >>light theory, where the speed of light should be isotropic >>in the rotating frame (the ring frame) where the source >>is stationary. >> Ah! Paul, rotating frames are very dangerous things. They can lead one up a >> blind alley. >> Ring gyros are like 4-mirror sagnac¹s but with an inÞnite number of mirrors. >Did you have a point? >Thousands of ring lasers are in operation at any time, you will >Þnd several of them in every commercial and military aircraft. >If the speed of light were dependent on the velocity of the source, >they wouldn¹t work. They do. > They work perfectly well under source depedency. See my reply to George. Wishful thinking! You can twist anything if you have enough faith. >>The Sagnac experiment conÞrms Michelson¹s ether theory >>The Sagnac experiment falsiÞes source dependent light theory. >> Every experiment MUST support reality, ie, source dependency. >Now, THAT¹S a beauty. :-) >And very illustrative of Henry Wilson¹s way of reasoning: > I, Henry Wilson, deÞnes reality. My fantasy world IS reality. > All experiments MUST support my idea of reality. > If they don¹t, they are faked. Or Œmisinterpreted¹. Or þawed. > Every experiment ever perfomed conÞrming SR falls in > this category. > As a wise man stated recently (that Greek philosopher, Androcles, IIRC) if you > have enough faith in something, you will be able to see evidence of it > everywhere. My very point! >>The MMX conÞrms SR. >> Well, indirectly, yes. SR relies on source dependency. It clearly requires that >> light travels at c from any source. Any observer at rest with that source will >> receive the same light at c. >> If the light is returned to the source by a mirror at rest wrt the source that >> the return travel time of that light will be constant, irespective of the speed >> of the system. >It is an indisputable fact that the MMX conÞrms SR. >That¹s why it isn¹t disputed. > The MMX proves source dependency, fair and square. No doubt about it. Amen. >>The MMX conÞrms source dependent light theory >> Absolutely correct. No doubt about that. >That¹s why it isn¹t disputed. >>The MMX falsiÞes Michelson¹s ether theory. >> Not exactly. >> As I have previously explained, it is impossible to prove the non-existence of >> anything. >> Similarly, if something doesn¹t exist, it is impossible to assign to it >> properties that can be tested experimentally. >> However, my alternative explanation refutes the conventional argument for the >> null result. Mine takes into account the angular departure of the splitting >> mirror from 45 degrees due to aetherian length contraction and shows how the >> cross beam would actually be deþected backwards even though the return travel >> time of the beam elements remains constant. >Mindless babble. >It is an indisputable fact that the MMX falsiÞes Michelsons ether theory. >That¹s why it isn¹t disputed. > The reason it isn¹t disputed is that its NULL result was easily manipulable to > give the answer that the physics maÞa wanted. So in 1886 Michelson was part of a physics maÞa that wanted to falsify the ether to prove that the establishment were wrong. So that¹s why they manipulated the highly unwanted NULL result to give the answer they didn¹t want, but which the maÞa wanted. Right? > Null results prove very litle if anything. > They usually reveal a þaw in the experiment. > My demo shows that þaw. IF the michelson aether existed, then the 45 mirror > would not remain at 45 after its length was contracted. The cross beam would be > bent backwards as shown, thus exactly compensating for the presumed Œdiagonal > effect¹.. AH! Length contraction in Michelson¹s ether!. So THAT¹s what your demo now shows. :-) You are SO funny, Henry! :-) >>So which of the three mentioned theories is not >>falsiÞed by one of these two experiments? >> ŒSource dependency¹ of course. >Every experiment MUST support [my idea of] reality, ie, source dependency. >Thus even the experiments falsifying source dependency MUST support it! >So there! > There are NO experiments that falsify source dependency. De Sitter was wrong > and there are no others. Every experiment MUST support reality, ie, source dependency So the experiments that falisÞes source dependency simply CANNOT exist. So there! Paul === Subject: Re: 0¹s the solution > > I have one number, zero, that I like better than others. > > 0¹s the solution to every algebraic equation. > > That¹s why 0¹s my favorite number. > > What¹s your favorite? > > And why? > > Actually, I like 26 a lot. Can anyone guess why? Hint: Think of > > peanut butter and jelly. > Thinking about it, the closest number I get is 57, but I do not even > know whether that is valid. > 26 = number of required dimensions for string theory to work. > l8r, Mike N. Christoff Wrong! What¹s that got to do with peanut butter and jelly? Charlie Volkstorf Cambridge, MA === Subject: Re: Making of Action Device > http://www.geocities.com/actiondevice > http://www.geocities.com/v_device > So ultimate war between Mind Vs Matter is over. > Today I woke up in evening at 05.00 PM. At about 07.00 PM, I went to > bicycle shop and purchased two springs and two spokes for Rs.9/- > I decided to use only one spring Þrst because in V-shaped two > stretched springs, magnitude of force at vertex point B is given by c > = ab*sin(theta). So if theta = 180 degree, c = ab*0 = 0. But if these > two springs make 180-degree angle to each other, it resembles one > stretched spring. So I decided to use only one spring. > But this spring is very stiff. I just can¹t pull it apart. I will have > to purchase spring of less stiffness. > But, Hey, WAIT..... > spoke and placed a stone at the center of bended spoke. Mass of stone > is far greater than mass of spoke. And I released both ends of spoke. > I am Þnding that spoke and stone are propelled in only one direction. > No reaction mass is expelled or propellant is used. We have inertial > propulsion. Will some body out there repeat this small nice > experiment? > and Arrows. > And Welcome To The Universe! > -Abhi. Did it come back down? Did the spring compress? Was it in contact with anything else if the spring did compress? As we have not seen any photos/video, it is hard to know what exactly happened. -- Will Twentyman email: wtwentyman at copper dot net === Subject: problem from gallian¹s contemporary abstract algebra (this is not homework by the way.) ch. 16 problem 38, let R be a commutative ring with unity, if I is a prime ideal in R then I[x] is a prime ideal in R[x]. This seems to be incredibly hard. i have tried showing a(x)*b(x) in I[x] => a(x) in I[x] or b(x) in I[x], i could not make this work so i tried using the contrapositive, if a(x) and b(x) are NOT in I[x] => a(x)*b(x) is NOT in I[x]. This seems easier, but I can¹t seem to isolate the coefÞcient of a*b that is not in I. If the Þrst or last coeff. of a and b are not in I then it is easy, but when the not in I coeff is lost in the middle of a and b, I have a hard time determining where the not in I coeff is in a*b, since so many things can happen when you add terms not in I. ie, it is possible for x or y to NOT be in I, but x+y is in I. I Þnd this very frustrating, thank you for any hints/help you can give. === Subject: Re: Final Rout of Synchronization Clocks in Relativity >Henry, here is the statement of mine you responded to: >| The P-code (military) pseudo random pattern is transmitted >| with 10.23Mbps, but this bit rate isn¹t normally measured >| from the ground. >| The important point is that the 10.23MHz signal is >| the frequency standard of the satellite clocks. >| That is, the satellite clock counts 10 230 000 cycles for >| each second it advances. The clock time is transmitted. >| (Coded as a bit pattern modulated on two carriers.) >| We know the frequency must be right since the satellite >| clocks stay in synch with the GPS time, not because >| the frequency is measured from the ground. >To this you have responded: >| It is all explained by the fact that lioght speed is source dependent. >| Unless a GPS clock is directly overhead, it will have a velocity component wrt >| the receiver. >| To cut a long story short, the doppler method used by the system confuses this >| effect with another that it tries to explain using relativity. >| The error due to c+v may be rather small when the clocks are near vertical but >| generally they are not. The transverse doppler correction assumes light speed >| is c and not c+(3770.cos theta) >Of course anybody understands that IF you from the ground >measure the frequency of the carrier or anything modulated onto it, >it will be Doppler shifted, and the Doppler shift will change all the time >as the satellite moves relative to the receiver. Of course anybody >will understand that it therefore blazingly obvious is practically impossible >to precisely measure the frequency emitted by the satellite with a receiver >on the ground. >THAT¹S WHY IT ISN¹T DONE. > The travel time of the signal from the OC to the receiver is c+v and not c. > Therefore the GC reading when the signal arrives is affected, which causes an > small error in the Œtransverse doppler correction¹ incorporated into the > positioning process. It turns out that, on average, this error is of the order > of 4x10^-10. Ah! the transverse doppler correction incoporated into the process! You do live in your own fantasy world, don¹t you? :-) >So what the heck are you babbling about, and what is the relevance to >the posting of mine you responded to? >You are not seriously claiming that source dependency can explain >why an uncorrected clock in GPS orbit gains 38 us a day every day >compared to the GPS coordinated time, or that this have anything >whatsoever with Doppler shift to do, are you? >If you are, you must have lost your mind completely. >Which, BTW, is a possibility I will not exclude. >It¹s up to you. > Do the calculation yourself if you don¹t believe me. Well. It WAS up to you. Sad. Paul === Subject: Re: Proposal for responding to idiots