I am having a terrible time understand a mathematical derivation in this computer graphics paper: I know this is asking a lot, but if someone could go to page 4, and see if they understand how the constant K is inferred, I would very much appreciate it. I have been trying to figure this out with no success and I have posted on several graphics forums, but no one seems to be able to figure it out. Here are some specific questions regarding the paper: confused for the number 1): Thus the following condition must hold for the decision variable f1 of an adjacent block in order to limit the level differences: (4) f1 < f2 <=> ln1 / (d*d21) < ln2 / [(d/2) * d22] ----------------- Ok, I think he is saying that if we are going to subdivide f2, which is a level lower than f1, then we must also subdivide f1. But, couldn't the relationship also be f1 <= f2, that will still ensure they are both subdivided, right? Also, the original equations of f is (3) f = ln / [d*C*max(c*d2, 1) How is it that he infers (4) from (3). I can see that we can cancel 'C', but I do not follow how he simplified the max. function. He continues: For a point of view falling inside the rectangular region Equation (3) [f < 1] is always satisfied, since ln1 / d is always less than the minimum resolution C. ----------------- Now when he says for a point of view falling inside the region, does he mean the view point (camera position) lies inside the region, or does he mean the region is inside the viewing frustum? I also do not follow how he says ln1 / d < C. He continues: Outside this region the value of the fraction ln1 / (2*ln2) is bounded by 1/2 and the constant K: 1/2 < ln1 / (2*ln2) < K (C > 2) K = L1 / (2*L2) = C / [2*(C - 1)] ---------------- I'm totally lost here. First of all, from the figure in the paper, I cannot make out what ln1 / (2*ln2) means geometrically. Perhaps if I knew where he obtained that relationship I would understand the rest; specifically how it is bounded. Moreover, I do not understand how he obtains the relationship K = L1 / (2*L2) = C / [2*(C - 1)] from that. ==== A brief scan of Dave Rusin's archive and the group didn't seem to turn up any solutions to these two problems that looked much like mine: A2: Proceed by induction. If n=1, then the inequality is clear (in fact, it's an equality...) To get from n-1 to n, apply Holder's inequality with p=n to the two vectors ((a_1a_2...a_{n-1})^{1/n},(b_1b_2...b_{n-1})^{1/n})) and (a_n^{1/n},b_n^{1/n}). A6: Let b(n) be the number of 1's in the base-2 expansion of n; let f(n)=(-1)^b(n). Let A=f^{-1}(1), B=f^{-1}(-1). Then clearly A and B partition the set of nonnegative integers. Also, f(2n)+f(2n+1)=0 for all n. Thus sum_{i=0}^n f(i)=0 if n is odd, and f(n) if n is even. Suppose p+q=n, with p neq q. If p and q are both in A, then f(p)+f(q)=2; if p and q are both in B, then f(p)+f(q)=-2; if one is in each set, then f(p)+f(q)=0. It therefore suffices to show that sum_{p+q=n,p neq q} f(p)+f(q)=0 for all n. If n is odd this sum is just 2sum_{i=0}^n f(i)=0; if n is even, it is 2(sum_{i=0}^n f(i)-f(n/2))=2(f(n)-f(n/2)). By construction, f(n)=f(n/2) for all even n; thus this sum is zero. So we are done. -Micah Smukler ==== Let R denote the real line and let S denote the cartesian product of R with the three element set {-1,0,+1} and give S the order topology for the lexicographical order. I think it is very likely that this space has a name. If you know the name of this space, please let me know. I think it has probably been good for something, such as studying functions f:R->R with left and right hand limits at every point. If you know where that might have been done, and whether in fact that is the reason the space was cooked up, I'd like to know that also. Ignorantly, Allan Adler ara@zurich.ai.mit.edu **************************************************************************** * * * Intelligence Lab. My actions and comments do not reflect * * in any way on MIT. Moreover, I am nowhere near the Boston * * metropolitan area. * * * **************************************************************************** ==== > Let R denote the real line and let S denote the cartesian product of > R with the three element set {-1,0,+1} and give S the order topology for > the lexicographical order. I think it is very likely that this space has a name. If you know the > name of this space, please let me know. I think it has probably been > good for something, such as studying functions f:R->R with left and right > hand limits at every point. If you know where that might have been done, > and whether in fact that is the reason the space was cooked up, I'd like > to know that also. > I have seen R x {0,1} lexicographic. It is sometimes known as the two arrows space. Or the top and bottom of the lexicographic square. I think this is the one you want if you require left and right limits at every point...although it doesn't provide an isolated point in between where you can provide a value different from those two limits. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== > Let R denote the real line and let S denote the cartesian product of > R with the three element set {-1,0,+1} and give S the order topology for > the lexicographical order. > S = lex Rx{-1,0,1} which has isolated pts (r,0) or S = lex {-1,0,1}xR which is disconnected as {-1}xR, {0}xR, {1}xR > I think it is very likely that this space has a name. If you know the > name of this space, please let me know. I think it has probably been > good for something, such as studying functions f:R->R with left and right > hand limits at every point. If you know where that might have been done, > and whether in fact that is the reason the space was cooked up, I'd like > to know that also. > The closest I come to it is the dictonary order unit square. ==== Suppose you have two dictionaries, D1 and D2. Suppose that D1 has a much narrower selection of words than D2 but is generally regarded as being more reliable. Suppose further that D1 is sufficiently long that it would be impractical to go through D1 to see whether D2 handles at least the words in D1 correctly for the most part. It would, however, be practical to sample the words in D1 and compare the treatments those words receive in D1 and D2. How can one design an experiment in which one samples the words in D1, compares the results with the entries of D2, and arrives at a measure of how accurately D2 defines the words that occur in D1? It is very likely that some problem equivalent to this is standard in statistics. I'd be glad to have a reference to where it is treated in detail. Ignorantly, Allan Adler ara@zurich.ai.mit.edu **************************************************************************** * * * Intelligence Lab. My actions and comments do not reflect * * in any way on MIT. Moreover, I am nowhere near the Boston * * metropolitan area. * * * **************************************************************************** ==== > Yes, . means multiplication. I thought that that was a kind of universal > notation. The only universal is that there are no universals. Jon Miller ==== In the context of Innocent's proof number obviously means integer, Doug. Please see my reply to Thomas Bushnell for an improved version of her proof. >prove if x is an irrational number then sqroot of x is irrational?? >>Any rational number has a finite number of digits > You mean, like 1/3 = 0.3333333333333333...? Doug ==== Aren't you unnecessarily confusing this proof rather than improving it? First, why insist on using base 10? Second, fractions seem irrelevant to the original question. Third, pi is already irrational. Here are my improvements on Innocent's proof: Any rational integer plus one is also rational, so that may be taken as a base x such that (x-1) squared is obviously also a rational integer with twice as many digits (in base x) just as she claimed. Thus, since Innocent has shown that any rational integer has a rational square, with the obvious corollary for any rational fraction, it follows that there are no rational candidates for sqroots of irrationals. Q.E.D. It's interesting that for all functions f where root is an integer greater than one, p is prime and i is an irrational, then: f(root p)=i Of course for any integer n not an integral power of an integer > 1 (all primes and composites): f(root n)=i This simple truth has philosophical import in that if reality is fundamentally digital then infinity is rational whereas if it's continuous then infinity is irrational - another way of phrasing that old either/or duality vs. bothand. >prove if x is an irrational number then sqroot of x is irrational?? >>Any rational number has a finite number of digits, so when >>multiplied by itself it would have at most twice a many while any >>irrational number requires infinite digits. > What do you mean any rational number has a finite number of digits? > For example, in base 10, the number 1/3 has infinitely many digits. Or if you mean finitely many different digits (so that 1/3 has one > digit), then pi has only digits. Thomas ==== > Aren't you unnecessarily confusing this proof rather than improving it? My goal was to correct an incorrect proof. You give a perfectly good proof, but solving people's homework for them is not my game. Thomas ==== I apologize, Thomas. Unnecessarily confusing was a poor choice for describing your clarification of loosely defined terms in Innocent's proof. It's better to appreciate that your correction game leads me to look deeper. Emit ==== >>As a matter of fact, the entire universe could be a Swiss cheese of >>giant expanding VDk voids, and we would be...oblivious. For we could >>not observe any of them until one hit us, because of the fact they are >>expanding at light speed. >> [snip - more crap] >> This isn't a relevant subject for either sci.math or rec.puzzles... Whether or not the hypothesis is preposterous, the question asked is >perfectly serious. Given the distributions of radii, velocities and >rates of expansion, he asks a perfectly valid mathematical question. So get off your high horse. You're a SELF-APPOINTED censor How is it that I am a censor exactly? Did I censor something or someone? Do you know what the word censor means? I snipped some of LeRoy's crap out of my own post, but I didn't remove his post from the newsgroup... Unfortunately, I don't have that power... And if LeRoy's question is perfectly valid then why didn't you answer it, smart guy. Ouch! It really hurts me when you say I have contributed nothing to this newsgroup... man, you sure have me pegged right; I'm a high-horse riding censor who has contributed nothing to this newsgroups... usually less quixotic questions. --Ron Bruck > have a nice day, adam ==== > Depends on which multiplication(s) you use. If o is the convolution > operator (sort of multiplication between functions) and * is the > normal multiplication operator, then L(f(x) o g(x)) = L(f(x)) * L(g(x)) > F(f(x) o g(x)) = F(f(x)) * F(g(x)) where L(f(x)) is the Laplace transform and F(f(x)) is the Fourier > transform of f(x). I'm using bad notation here I guess, what if the operator itself is convolution? ie. f(x) o ( S1(x) + S2(x) ) = ( f(x) o S1(x) ) + ( f(x) o S2(x) ) says that convolution is linear given proper boundary conditions, but are there conditions under which this is true: f(x) o ( S1(x) * S2(x) ) = ( f(x) o S1(x) ) * ( f(x) o S2(x) ) I'm coming from a signal processing angle, so the functions S1 & S2 are signals, and f() is a filter kernel. Which i think means that numbers....? ( Have I got the right end [any end] of the stick this? ) S X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBBEAqA31325; ==== Perhaps there's another name for this besides what I'm calling it here. Def. Let X be a set. Call *: X times X -> X an associative if (a*b)*c = a*(b*c) for all a,b,c in X. Let Z be the whole numbers. Then normal multiplication and aĦb = a+b-ab (ab normal multiplication) are associatives. Earlier, I had asked if the set of all associatives of a given set had a particular name (still not sure): Let As(Z) be such a set on Z. Every x in N (naturals) has a unique representation as x =(x_1)(x_2)...(x_n) where x_1, x_2, ...,x_n are prime and x_1 <= x_2 <= ... <= x_n (normal multiplication) If x in Z and x < 0, then x also has a unique prime representation as x =-(x_1)(x_2)...(x_n) where x_1, x_2, ...,x_n are prime and x_1 <= x_2 <= ... <= x_n I would like to show that * in As(Z) where * is defined as: For x,y > 0 x*y = [(-1)^j]xy = [(-1)^j](x_1)(x_2)...(x_n)(y_1)(y_2)...(y_m) = [(-1)^j](z_1)(z_2)...(z_{n+m}) = [(-1)^j]z when (x_1)(x_2)...(x_n) (y_1)(y_2)...(y_m) (z_1)(z_2)...(z_{n+m}) are the (unique) prime representations of x, y and z (=xy) respectively and j is the minimum number of shifts necessary to order (x_1)(x_2)...(x_n)(y_1)(y_2)...(y_m) into (z_1)(z_2)...(z_{n+m}). If x or y = 0, then define x*y = 0. If x < 0 or y < 0, then do the multiplication exactly as if both were positive, but substitute j with (j+k) were k = 1 <-> (x < 0 and y > 0) or (x > 0 and y < 0) k = 2 <-> x,y < 0. Examples: (3*10)*14 = (3*((2)(5))*14 = (-(2)(3)(5))*14 = (-(2)(3)(5))*((2)(7)) = -(2)(2)(3)(5)(7) = -420 Reason: in the 2nd equation, 1 shift was necessary; (2) was shifted once to the left. in the 4th equation, 2 shifts were necessary; (2) was shifted twice to the left. Note: Whether the shift is to the left or the right does not play a role, i.e. j is the same number in either case. 3*(10*14) = 3*((2)(5)*(2)(7)) = 3*(-(2)(2)(5)(7) = -(2)(2)(3)(5)(7) = -420 Reason: in the 2nd equation, (2) was shifted once to the left. alternatively, (5) was shifted once to the right. in the 3rd equation, (3) was shifted twice to the right. C. Dement ==== > Def. > Let X be a set. Call *: X times X -> X an > associative if (a*b)*c = a*(b*c) for all a,b,c in X. > (X,*) is a semi-group. 2x2 matrices is an example of a noncommunative monoid, ie semi-group with identity. > Let Z be the whole numbers. Then normal multiplication > and a°b = a+b-ab (ab normal multiplication) are associatives. > a#b = a+b - ab is know as symmetric difference in set theory. Symmetric differences of a boolean ring produce a boolean algebra. , I'm allergic to permutations. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBBEAmE31262; ==== >Today I landed a pretty idea. Whether it flys or not is unknown as >yet. It came to me whilst thinking of those cylindrical farm bins used >to store grains. Suppose one were to build one not out of steel but >instead from concrete block or brick. Expensive I know but consider >it. And suppose the builder was given a 3/8 joint for mortar. Pretty Idea: then given a standard size brick would dictate the >circumference and diameter of this building. So that given the brick >size would determine the size of the building where no sawing of brick >is allowed. Granted of course the building is not really a circle but >we can consider the midpoint inside each brick once the building is >completed sribes a circle. What is pretty about this idea is the fact that a rectangle determines >a circle. So that the circle is quantized from a rectangle. I have not thought of the reverse where given a circle determines or >quantizes >a rectangle or a square. It maybe even possible that a circle just >does not quantize a square or rectangle and if that is the case gives >deep mystery and deep implications as to why this Cosmos is built >where rectangles can quantize circles but not the reverse. And if that >is the case would go back to the idea that the cosmos is one gigantic >plutonium atom which is cosmically spherical and cylindrical and that >life sees Euclidean geometry not as a reflection of reality but as a >result of the mind trying to make sense of the world. Euclidean Geometry in this sense does not exist in Nature but only as >a figment of the imagination of a mind. Well , This is true . Do lines exist,dots and surfaces ? All this lot is timeless and immaterial. Even third dimention Eucledian/Platonic Forms bear the same qualities. After all this is Pure Geometry. This is the absolute truth that cannot be violated. When matter embodies this backbone structure of Geometry ,naturally deformations occur for its accomodation to form the real world. I, UNDERSTAND THAT EXCHANGE OF FORMS BETWEEN ALL KINDS OF LINEAR ONES (SQUARES,RECTANGLES TRIANGLES)AND CIRCLE ,IF ACCEPTED IN ONE DIRECTION IT IS TRUE FOR THE REVERSE. http://www.stefanides.gr/quadbig.htm Panagiotis Stefanides http://www.stefanides.gr Euclidean Geometry would be >like those many Optical Illusions we see such as the artist Escher. I doubt that Euclidean Geometry is a illusion for zero is a true >existing number and that Euclidean Geometry is the zero curvature >geometry. But, if rectangles and squares can generate circles but the reverse is >not true would have deep implications upon Euclidean Geometry. Archimedes Plutonium >whole entire Universe is just one big atom where dots >of the electron-dot-cloud are galaxies X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBBEApc31314; ==== Dear All, I am trying to solve a system of nonlinear equations which is very complicated. The resulting function is a stochastic variable, which I have to approximate using the Monte Carlo methods. Therefore evaluation of the function is very time consuming (one evaluation is about one minute). My problem is to find an appropriate algorithm which solves the nonlinear system of equations without approximation of the Jacobian/Hessian matrix. Due to the complexity of the stochastic function it does not make sense to use methods like Newton, Newton-Raphson, etc. I was thinking about something like bisections or regula falsi, eventually secant method. They work fine in one dimensional space. But I do not know about higher dimensions. PRzemyslaw X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBBIdL219511; ==== >prove if x is an irrational number then sqroot of x is irrational??? Let y = exp(1) and x = exp(2). x is irrational (For any rational a>0, exp(a) is irrational), and sqrt(x) = exp(1) is irrational :) X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBBG9Lr07886; ==== Perhaps there's another name for this besides what >I'm calling it here. Def. >Let X be a set. Call *: X times X -> X an >associative if (a*b)*c = a*(b*c) for all a,b,c in X. Let Z be the whole numbers. Then normal multiplication >and aĦb = a+b-ab (ab normal multiplication) are associatives. Earlier, I had asked if the set of all associatives of >a given set had a particular name (still not sure): >Let As(Z) be such a set on Z. Every x in N (naturals) has a unique representation as >x =(x_1)(x_2)...(x_n) where x_1, x_2, ...,x_n are prime and x_1 <= x_2 <= ... <= x_n (normal multiplication) If x in Z and x < 0, then x also has a unique prime >representation as x =-(x_1)(x_2)...(x_n) where x_1, x_2, ...,x_n are prime and x_1 <= x_2 <= ... <= x_n I would like to show that * in As(Z) where * is defined as: For x,y > 0 x*y = [(-1)^j]xy = [(-1)^j](x_1)(x_2)...(x_n)(y_1)(y_2)...(y_m) > > = [(-1)^j](z_1)(z_2)...(z_{n+m}) = [(-1)^j]z when (x_1)(x_2)...(x_n) > (y_1)(y_2)...(y_m) > (z_1)(z_2)...(z_{n+m}) are the (unique) prime representations of x, y and z (=xy) respectively and j is the >minimum number of shifts necessary to order >(x_1)(x_2)...(x_n)(y_1)(y_2)...(y_m) >into (z_1)(z_2)...(z_{n+m}). If x or y = 0, then define x*y = 0. >If x < 0 or y < 0, then do the multiplication exactly as if >both were positive, but substitute j with (j+k) were >k = 1 <-> (x < 0 and y > 0) or (x > 0 and y < 0) >k = 2 <-> x,y < 0. Examples: >(3*10)*14 = (3*((2)(5))*14 = (-(2)(3)(5))*14 = (-(2)(3)(5))*((2)(7)) > = -(2)(2)(3)(5)(7) = -420 Reason: in the 2nd equation, 1 shift was necessary; > (2) was shifted once to the left. > in the 4th equation, 2 shifts were necessary; > (2) was shifted twice to the left. > >Note: Whether the shift is to the left or the right does not > play a role, i.e. j is the same number in either case. > >3*(10*14) = 3*((2)(5)*(2)(7)) = 3*(-(2)(2)(5)(7) > = -(2)(2)(3)(5)(7) = -420 Reason: in the 2nd equation, (2) was shifted once to the left. > alternatively, (5) was shifted once to the right. > in the 3rd equation, (3) was shifted twice to the right. > We can now assign each x in N a parity according to: x in P^{+} <-> (x*x)/(xx) = 1 x in P^{-} <-> (x*x)/(xx) = -1 Ex. 10 in P^{-}, 70 in P^{-}, but 560 in P^{+}. Let R be the reverse operator, i.e. if x = (x_1)(x_2)...(x_n)= (x_1)*(x_2)*...*(x_n) is the prime representation of x, then Rx := (x_n)*...*(x_2)*(x_1) A few properties and conjectures (in the order of decreasing obviousness, as I see it now): i) RRx = x ii) x*(Rx) in P^{+} iii) x in P^{+}, y in P^{-} -> x*y in P^{-} x in P^{-}, y in P^{+} -> x*y in P^{-} x in P^{-}, y in P^{-} -> x*y in P^{+} x in P^{+}, y in P^{+} -> x*y in P^{+} iv) x*y = y*x <-> x and y of same parity v) x*(y+z) = x*y + x*z <-> x*y and x*z of same parity >C. Dement > X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBBIdLb19525; ==== >prove if x is an irrational number then sqroot of x is irrational??? What was I thinking !? My previous message was purely NOTHING. Assume that x = irratoinal, and x^(1/2) = (x)^(1/2) = p/q for some positve integers p, q. Then, ln(x) = 2ln(p/q) ln(x)/(ln(p/q)) = 2. If p/q = 2/3, then ln(x) = 2ln(2/3) x = exp(2*ln(2/3)). Sorry if my message was out of present discussion (I would have to do some homework and so there is no time to read the bunch of messages; this is really a bad attitude for a person who wants to be a mathematician). I think the real line is very dense where denotes that the term is used informally. H.S. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBBIdMw19544; ==== >prove if x is an irrational number then sqroot of x is irrational??? PROOF: Assume that (a)x is irrational; (b)x^(1/2) is p/q for some natural numbers p, q. Put x^(1/2) = p/q ln(x) = 2ln(p/q)) x = exp(ln((p/q)^2)) = (p/q)^2. But this contradicts (a). This completes the proof. ==== >prove if x is an irrational number then sqroot of x is irrational??? > PROOF: Assume that > (a)x is irrational; > (b)x^(1/2) is p/q for some natural numbers p, q. Put x^(1/2) = p/q ln(x) = 2ln(p/q)) > x = exp(ln((p/q)^2)) > = (p/q)^2. But this contradicts (a). This completes the proof. > Is there some reason you're using logarithms and exponentials in this problem? I mean, it's not incorrect, but it's truly irrelevant to what you want to show. Why not prove the contrapositive: If sqrt(x) is rational, then x is rational. or more simply put: If y is rational, then y^2 is rational. ??? A word to the wise: If you use things (e.g., concepts, functions, constructions, calculations) that are not required in, or relevant to, an argument, it shows that you don't understand what the argument is about. I suppose that principle may apply more generally than within mathematics. Dale X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBBK93w27283; ==== As part of a software engineering group project to design a math tutor/testing program for middle school pre-algebra students, I have written a C program to generate random angles (to be sent to the GUI) for 3 to 8 sided polygons. Is there a (relatively) simple way of calculating the length of all the sides, assuming that one side is set to an arbitrary length of one Unit? Assume that the first angle would be on the bottom left, that the angles progress clockwise, the first angle splits side 0 and side 1, that side 0 is a horizontal line, that line 1 goes up, and that the polygon is regular?, e.g. describes a more or less circular pattern. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBBK93s27260; ==== Please, send me an order form of ChiWriter word-processor by Horstmann software. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBBK93J27255; ==== I am trying to do some research on alcuin's sequence, but all the references that I can find quote the terms of the sequence as the coefficients in the Maclaurin expansion of [(1-x^2)(1-x^3)(1-x^4)]^- 1. I find this surprising, as I believe the sequence to have been named after Alcuin of York (735-804) who lived many years before Maclaurin expansions were discovered. This sequence also gives the number of different triangles that have integral sides and perimeter n. I would think that it is much more likely that this is what Alcuin discovered, and years later someone found that the above expansion acted as a generating function. That is, of course, if the Alcuin who the sequence is named after is Alcuin of York and not someone else with the same name. I would be most grateful if you could either confirm or refute this, or point me in the direction of any references. Many thanks Tony ==== > This sequence also gives the number of different triangles that have > integral sides and perimeter n. I would think that it is much more > likely that this is what Alcuin discovered, and years later someone > found that the above expansion acted as a generating function. This is the most plausible explanation. Certainly Maclaurin series were unknown to Alcuin! Alcuin did write texts on arithmetic, geometry and astronomy, for the purpose of teaching students. Alcuin is never reported as being a wonderful intellectual innovator, but rather, an exceptional teacher, and of course, calligrapher. (Don't discount that last--his calligraphy was as important in its day as computers are today, and for much the same reasons.) He was also an important liturgist and a strong (but not very creative) theologian. juvenes, with twenty-one simple problems for sharpening the minds of youth. The text of Alcuin's book is available at http://www.thelatinlibrary.com/alcuin.propos.html. There is a and Technology Bulletin, and also an annotated translation by David Singmaster and John Hadley. This is one of his most famous puzzles from the book, number 18 (with my translation): Homo quidem debebat ultra flavium transferre lupum, capram, et fasciculum cauli. Et not potuit aliam navem invenire, nisi quae duos tantum ex ipsis ferre valebat. Praeceptum itaque ei fuerat ut omnia haec ultra illaesa omnino transferret. Dicat, qui potest, quomodo eis illaesis transire potuit. A certain man must bring across a river a wolf, a goat, and a bundle of plants. But he also cannot find any boat, except one that is only able to carry two of them. He was told that he must bring all these across entirely unharmed. Let him say, if he can, how he can go across with them unharmed. And Alcuin's answer is: Solutio. Simili namque tenore ducerem prius capram et dimitterem foris lupem et caulum. Tum deinde venirem, lupumque transferrem: lupoque foris misso capram navi receptam ultra reducerem; capramque foris missam caulum transveherem ultra; atque iterim remigassem, capramque assumptam ultra duxissem. Sicque faciendo facta erit remigatio salubris, absque voragine lacerationis. Solution. Similarly [to the answer to the previous problem], for with a noose I would lead the goat and leave behind the wolf and the plants. Then when I returned, I would carry the wolf over, release the wolf, and return across again leading the goat; and releasing the goat, transfer the plants across; and then again row, and taking the goat, lead it across. And by doing thus, the rowing will be done safely, free from the disaster of tearing teeth. None of the problems are in general form, but all could obviously be adapted to be in such. For example, number XXI: Est campus qui habet in longitudine pedes CC, et in latitudine pedes C. Volo ibidem mittere oves; sic tamen ut unaquaeque ovis habet in longo pedes V, et in lato pedes IV. Dicat, rogo, qui valet, quot oves ibidem locari possint? Solutio. Ispe campus habet in longitudine pedes CC. Et in latitudine pedes C. Duc bis quinquenos de CC, fiunt XL. At deinde C divide per IIII. Quarta pars centenarii XXV. Sive ergo XL vicies quinquies; sive XXV quadragies ducti, millenarium implent numerum. Tot ergo ibidem oves collocari possunt. There is a field which is two hundred feet long and one hundred feet wide. I want to place sheep there; but so that each sheep has five feet long and four feet wide. Let him say, I ask, if he can, how many sheep can be placed there? Solution. The field is two hundred feet long. And in width, one hundred feet. Compute twice: in fifths from two hundred, that makes forty. And then divide one hundred by four. A fourth part of a hundred is twenty-five. Therefore is computed either forty, twenty-five times; or twenty-five, forty times, the full number of a thousand. Therefore so many sheep can be placed there at once. That gives you the flavor. I don't know enough about the sequence to be sure if any of the problems can be easily generalized to something similar. I would guess that the annotated edition by Singmaster and Hadley includes such interesting tidbits. Thomas X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBBL9aa32375; ==== >> Michigan State University grad student Michael Shafer has succeeded in >> identifying the largest known prime number to date, using a distributed >> computer network of more than 200,000 computers located around the world. >> The new number is 6,320,430 digits long and is only the 40th Mersenne >prime >> to have ever been discovered (Mersenne primes are an especially rare breed >> that take the form of 2-to-the-power-of-P, where P is also a prime >number). Sounds like a waste of processor power (and electricity) to me. There is no >reason >in discovering these large numbers (even if you want to use them for >encryption. >I believe the encryption process would need veeeery much time ;-}} ). Karl And everyone now knows what the primes are,,,duh! X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBBLwXp03824; ==== >Dan schrieb: > Strange how (e^2) has no order in its cf but the (n) in my equation >> has. >cf(e^2) = > [ 7, 2, 1, > 1, 3, 18, 5, 1, > 1, 6, 30, 8, 1, > 1, 9, 42, 11, 1, > 1, 12, 54, 14, 1, > 1, 15, 66, 17, 1, > 1, 18, 78, 20, 1, > 1, 21, 90, 23, 1, > 1, 24, 102, 26, 1, > 1, 27, 114, 29, 1, > 1, 30, 126, 32, 1, > ... ] computed by the gnu-program maxima Gottfried Helms I was wrong on that one. I didn't check deep enough. Interesting how the center column progress by 12 and the two columns on either side progress by 3. I guess in some way there is a spinoff after all with my constant (n)'s cf being the most orderly of e and e^2 cf. Dan ==== >Interesting how the center column progress by 12 and the two columns >on either side progress by 3. I'm unable really to read Perron's book on continued fractions because I don't know any German, but Perron appears to make a study of the numbers whose continued fraction is sort of close to periodic in this way: [a1,...,an, b1,...,bn, b1+c1, b2+c2, b3+c3,..., bn+cn, b1+2c1, b1+2c2,..., bn+2cn, ...], with each element of the cycle increasing in an arithmetic progression. Gosper has explained how to compute with continued fraction numbers, and you can calculate the expansion of multiples of e and so on easily enough, and you'll find a bunch of related numbers which have the same type of continued fraction. See for example http://www.inwap.com/pdp10/hbaker/hakmem/cf.html#item101b Keith Ramsay X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBC2ZTG25062; ==== > As noted in the marginalia of the AMS Notices, the cover sheet is intended > to help departments process job applications. As such, it serves a purely > bureaucratic purpose and could perhaps be forgiven on that basis if it were > then not also used to determine how one is to be treated. You seem to be using the word bureaucratic like a curse word. There needs to be *some* way to classify the hundreds of applications that flood into every mathematics department with jobs available. The purpose of the coversheet and the cover letter is to match applicants with faculty members who can evaluate them. > The one category that sounds sufficiently ambiguous for me to feel I'm > not misrepresenting myself is 00, i.e. General. Perhaps by its nature, > after looking at publications with that classification in Math-Sci, I can't > figure out what it means, but that doesn't matter. What does matter is what > assumptions a hiring committee will make about someone who describes his/her > primary interest as 00. And what would they assume? I think what would happen is that the secretaries would put your application in the we're not sure pile. At X University, where I am, they employ someone to go through that pile to check who would be best suited to evaluate the unknowns. Maybe at some other places they don't bother. > If someone can look over my publications on Math-Sci and figure out what > the appropriate codes are for my primary and secondary interests for the > AMS Standard Cover Sheet, I would appreciate the advice. I don't seem to > have been programmed with this capability. Accroding to MathSciNet, at least some of your publications are 14G35 (algebraic geometry of modular varieties). Many of your other publications are in the general area of number theory (the 11-XX hierarchy). So you could put 14G35 and 11-XX as your primary and secondary interests. Maybe you could ask your co-authors for their advice too. The big question is: would you feel happier if a number theorist or (say) a Bayesian statistician read your application? Which would be able to better appreciate your work? X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBC2ZTs25087; ==== >>prove if x is an irrational number then sqroot of x is irrational??? PROOF: Assume that >(a)x is irrational; >(b)x^(1/2) is p/q for some natural numbers p, q. Put x^(1/2) = p/q ln(x) = 2ln(p/q)) >x = exp(ln((p/q)^2)) > = (p/q)^2. But this contradicts (a). This completes the proof. Actually, what is wrong with the proof? Certainly, I did not have to use the natural logarithm :) Put x^(1/2) = p/q. x = (p/q)^2 and so x must have been rational, a contradiction. If x is rational, then of course there exists some x whose square is yet a rational: x := rational -> x^(1/2) is (1) rational, (2) irrational. If x is irrational, then by the proof, x := irrational -> x^(1/2) is (3) irrational. This discussion seems to have been already over, but for making sure. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBC5G6O07453; ==== At 10,000 digits the following {0,1,2,3,4,5,6,7,8,9} occur in the following amounts {987,1015,973,993,1016,1055,996,1022,943,1000} At 100,000 digits they occur at {9983,10178,9947,9926,10035,10186,9929,9845,9938,10033} Seems fairly regular to me. ==== The mathematician comes home early from a conference and finds his wife...eh, you know the routine. Off all the...With a traveling salesman? Well, at least he was P-space hard. -- Hauke Reddmann <:-EX8 For our chemistry workgroup,remove math from the address For spamming, remove anything else ==== Why is it required that Hom(A,B) = Hom(C,D) <=> (A,C) = (C,D) in the definition of a category? Most of the time we're dealing with sets and mappings, so that axiom is trivially verified. Thks, -- J.S ==== |Why is it required that Hom(A,B) = Hom(C,D) <=> (A,C) = (C,D) in the |definition of a category? As someone else pointed out, this implies that each arrow has a unique domain and codomain. Often categories are defined so that each arrow has a designated domain and codomain, and then Hom(A,B) is defined as the set of arrows with A as domain and B as codomain, which makes the above property automatic. |Most of the time we're dealing with sets and mappings, Whether we're dealing with sets and mappings most of the time depends on what kind of category theory we're doing. Lot of categories are subcategories of the category of sets, but this is not always the most appropriate context to put them in. |so that axiom is |trivially verified. It's a very light assumption, basically. If we started out with arrows f defined so that they could be both from A to B and C to D, without A=C and B=D, then we could switch instead to considering our arrows to be triples (f,A,B) where f is one of our original arrows from A to B. Then the axiom would become automatically true. I suppose it's not so obvious why we want arrows to have unique domains and codomains. But I think it amounts to a way to smooth out notation. If category is a subcategory of the category of sets, then it's tempting to think of an f:A->B as equal to f':A->C where B is a subset of C, and f'(a)=f(a) for all a. Yet a preference exists for distinguishing them, perhaps partly because in category theory we don't want to have to say things like B is a subset of C which rely upon equality between elements of one set and elements of another set. Instead we want to say only that there exists an injection g:B->C, so that f' = g o f. If the category is just an arbitrary abstract category, however, permitting f:A->B to be equal to f':C->D is not of much use. Keith Ramsay ==== >Why is it required that Hom(A,B) = Hom(C,D) <=> (A,C) = (C,D) in the >definition of a category? >Most of the time we're dealing with sets and mappings, so that axiom is >trivially verified. So that the domain and codomain are functions? --Dan Grubb ==== A space S is monotoniclly normal when for each x in S and each open U nhood x, there's assigned mx(x,U) another open nhood of x, for which for all x,y, open U,V x in U, y in V, mu(x,U) / mu(y,V) not empty ==> x in V or y in U. Theorem: monotonically normal ==> completely normal ==> normal Problem: to show a linear order space is monotonically ordered Proof with weakness marked ** below. What's to be done about **? Let S be linear order space Let < be order of S for the order-topology Let <_w well-order S define mu(x,U) as follows: let x in (c,d) be interval inside U If (c,x) nonnul, let x_l = <_w-first in (c,x), otherwise let x_l = c If (x,d) nonnul, let x_r = <_w-first in (x,d), otherwise let x_r = d let mu(x,U) = (x_l, x_r) which is an open nhood of x If mu(x,U) / mu(y,V) nonnul then x in (x_l, x_r); y in (y_l, y_r) some z in I = (x_l,x_r) / (y_l,y_r) = (max x_l,y_l, min x_r,y_r) Consider when x_l <= y_l. [(y_l <= x_l) similar] if y_r <= x_r: I = (y_l, y_r); y in I subset U if x_r < y_r: I = (y_l, x_r) assume x,y not in I then we have x <= y_l < z < x_r <= y if x < y_l < z < x_r < y, then comes desired contradiction x_r <=_w y_l; y_l <=_w x_r; y_l = x_r; z in I = nulset ** Otherwise, what to do when x = y_l or x_r = y ? For indeed when x = y_l, then y_l not in nonnul (x,d) so one cannot claim x_r <=_w y_l ---- ==== in message : [...] >> Dear all, >> Trying to figure out the 1-dimensional representations of D_n, the >> dihedral group of order 2n. >> Should be easy... [...] > I guess the last thing I would have to do is analyze the conjugacy classes > of D_n and that is all. (I can do this on my own) In the end you'll find this is a specific instance of a general theorem in character theory which you might also enjoy trying to prove: The number of 1-d complex linear reps of a finite group G is |G/[G,G]|. -- Jim Heckman ==== in message <83f64ab8.0312102356.424eb9c9@posting.google.com>: > im trying to find all groups of order eight. i already know what these > groups are, but i am trying to derive it, mainly from the sylow > theorems. will this approach work? for example, I was able to work > with groups of order 21=3*7, because I was able to find a semidirect > product representation of the group, but 8=2^3 doesnt really lead me > anywhere. is there another method? The approach I've always liked is based on the fact that every non-cyclic group P of order p^n for p prime is generated by 2 subgroups of order p^{n-1}, each necessarily normal in P, intersecting in a subgroup of order p^{n-2}. So look for all possible ways in which 2 groups of order 4 can intersect in a group of order 2 such that the 4-groups normalize each other. (In practice, you need only consider one of the 4-groups normalizing the other.) -- Jim Heckman ==== hi, and thank you for the suggestion. i worked by case analysis on maximal order of an element and it came out quite well, although i ran into quite a few complications. in any case, i feel that trying to describe all the groups of order 8 turned out to be more work than what was actually required for what i am doing. in the end i am trying to show that Aut(Z16) is isomorphic to Z2xZ8, and I dont think I want to use the fact that I know all groups of order 8. I have the curious fact that there is only one sylow subgroup, namely H2 the 2-Sylow subgroup, which of course H2=G. does it seem like any of sylow's theorems could help me in this restricted case? otherwise i suppose i will continue with the rigmarole of describing groups of order 8. thanks again im trying to find all groups of order eight. i already know what these > groups are, but i am trying to derive it, mainly from the sylow > theorems. will this approach work? Trouble is, as you say later, is that 2 is a prime power > so Sylow tells you nowt. for example, I was able to work > with groups of order 21=3*7, because I was able to find a semidirect > product representation of the group, but 8=2^3 doesnt really lead me > anywhere. is there another method? Two possible attacks > (i) the exponent of the group: this is the smallest n such that a^n = 1 > for all a in the group. For a group of order 8 it is 2, 4 or 8. > Divide into these three cases. (ii) Use the centre. A standard theorem states that the centre Z > of a p-group G is nontrivial. Split into cases according > to the structure of the centre. ==== Given a set S = { (1,2),(1,3),(1,4)...(k-1,k)}, is it possible to > partition the above set into 'k' subsets such that the elements of > each subset are pairwise-disjoint? Yes. You can find this in most any introductory textbook on graph > theory. Keywords: edge-coloring, edge-chromatic number, chromatic > index, complete graph. Is this homework? larger problem I was attempting. One of the questions asked by Erd.9as is - Let K_1, K_2,..., K_k be complete graphs, all of size at most k, and assume that for any two of them the intersection is either empty or exactly one vertex. Let G be the union of K_1, K_2,..., K_k. Is G k-colorable? Proof by reverse induction - If P = [0,m] and P(n) is a statement and the following two conditions hold - (1) P(m) is true. (2) for all m <= n <= 0, P(n+1) is true implies P(n) is true then P(n) is true for all 0 <= n <= m. Let us assume that the complete graphs in G are numbered from 1 to k and the intersection point of two or more graphs is represented as a tuple consisting of the number of the intersecting graphs. For instance, if G contains a intersection point between K_1, K_2 and K_3, the intersecting point is represented as (1,2,3). We note that G can be colored in k colors if the intersecting points can be colored in k colors such that each of the complete graph, K_i, is properly (vertex) colored. Basic step [to show P(m) is true] : Let G consist of intersecting points between exactly two complete graphs (i,j) for all i,j <= k. G contains maximum number of intersecting points. G can be colored in k colors by coloring each intersection point with [(i+j) mod k]. We note that for two tuples (i,j) and (m,n) - (i+j) mod k = (m,n) mod k only if i,j,m,n are all different and (i+m) mod k != (i,n) mod k for m != n; which implies, the color assigned to (i,m) is unique for all m <= k; which implies that the coloring of i th complete graph is proper. Hence, the coloring of G is proper. Induction hypothesis - If the k graphs intersect at exactly (n+1) vertices, there is a k-coloring of G. Induction step - To prove that if the k graphs intersect at n vertices, there is a k-coloring of G. I have not been able to complete the proof but I suppose I would have to show that I can create a graph G' from G with (n+1) intersecting points and that, a k-coloring of G' is also a k-coloring of G. I would appreciate any thoughts on this or the whole argument so far. Many thanks for your suggestions. -- Pradip ==== Induction hypothesis - If the k graphs intersect at exactly (n+1) > vertices, there is a k-coloring of G. We note that : m = k(k+1)/2 Induction step - To prove that if the k graphs intersect at n vertices (n < m), there is a k-coloring of G. Consider an intersection point (i,j,...,l) [There is atleast one intersecting point (i,j,...,l) because n < m]. We construct G' from G such that the intersection point (i,j,...,l) in G is split as (i,j,...) and (i,l) in G'. By induction hypothesis, G' has a k-coloring. A k-coloring of G' is also a k-coloring of G by assigning the color of (i,j,...) in G' to (i,j,...,l) in G. -- Pradip ==== Here is my answer: We can say that at least the sequence ...11111111 is not in the list, for example: ...00000000 <--> 1 ...00000001 <--> 2 ...00000010 <--> 3 ...00000011 <--> 4 ...00000100 <--> 5 ...00000101 <--> 6 ...00000110 <--> 7 ...00000111 <--> 8 ... Let us examine the infinite from another point of view. When we have ...11111 AND ...00000 in an ordered combinations list, it means that the list is complete. But this is the whole point, infinitely many objects cannot be completed, otherwise they are finite. Therefore ...11111 AND ...00000 are not in the list of infinity many objects. In other words [...000, ...111) XOR (...000, ...111] . There are 2 possible structural types of infinitely many 01 notations: (?...0] (?...1] We know how some infinitely long combination starts, but its opposite side is unknown (can be 0 XOR 1) and this missing information is essential to the existence of the induction. Therefore we can find a meaningful missing result by Cantor's Diagonalization method, only in a finite combinations list. For more details please look at: http://www.geocities.com/complement...iemannsBall.pdf Because I don’t know how to write my idea in the common formal > way, I am going to do it in a non-formal way, but I will do my best to > write it in the clearest way (PLEASE READ ALL OF IT, OTHERWISE IT > CAN'T BE UNDERSTOOD). So here it is: Let us check these lists. P(2) = {{},{0},{1},{0,1}} = 2^2 = 4 and also can be represented as: 00 > 01 > 10 > 11 > P(3) = {{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}} = 2^3 = 8 and also can be represented as: 000 > 001 > 010 > 011 > 100 > 101 > 110 > 111 Let us call any full 01 list, combinations list. Now, let us use Cantor's Diagonalization method on some finitely long > combinations list, for example, the combinations list of number 3: 000 > 001 > 010 > 011 > 100 > 101 > 110 > 111 We can change the order of the rows, and then use Cantor's > Diagonalization method, for example: 001 > 011 > 010 > 000 > 101 > 100 > 111 > 110 The input for Cantor's Diagonalization method in the first example is > 000 and the output is 111. of course that choice was completely arbitrary. you could have started > with any string. and any ordering of the strings. > The input for Cantor's Diagonalization method in the second example is > 010 and the output is 101. fine, so you're ignoring the bottom 2^n - n rows then? > In both examples we find that the result is already in the > combinations list, and this combination, which is already in the list, > is one of the combinations that Cantor's Diagonal does not cover. The number of the combinations, which are out of the range of Cantor's > diagonal is: 2^n - n > so that's where that number came from. now i understand what it means. > incidentally, you'd (still be wrong but) be better off saying domain. > Every column, which belongs to some combinations list is a sequence of > 01 notations, based on some periodic frequency changes, for example: the right column of number 3 combinations list, is based on 2^0(=1). Therefore the periodic frequency changes are 1, and the result in this > case is: > 01010101. > but a completely arbitrary ordering of the rows wouldn't look so nice > would it? i mean 01110100 is possible under some ordering, so what's > your point? The result of the middle column is based on 2^1(=2), therefore the > sequence is: > 00110011. The result of the left column is based on 2^2(=4), therefore the > sequence is: > 00001111. and we get the full combinations list of number 3: 000 > 001 > 010 > 011 > 100 > 101 > 110 > 111 We can get a combinations list of infinitely many places, by using the > ZF Axiom of infinity induction, on the left side of our combinations > list, by using the induction on the power_value of each column, for > example: 2^0, 2^1, 2^2, 2^3, ... > still passing to infinite constructions without demonstrating it's valid. In this stage we have proven, by induction, that Cantor's diagonal > cannot cover any full 01 combinations list, finite or infinite. > what do you mean cover? at best you'd need transfinite induction wouldn't you to pass from a > finite cardinal to a countably infinite cardinal? otherwise: the sum of 1/r from 1 to n is finite. therefore, using your assumption, > the infinte sum is finite. prove that's a problem if you don't already know Therefore its result is not a new combination (that has to be added to > the list). Because Cantor's diagonal cannot cover the full 01 combinations list, > (of aleph0 places for each combination) we can conclude that 2^aleph0 >> aleph0. But each infinitely long sequence of 01 notations can be mapped with > some natural number, for example: ...000 <--> 1 > ...001 <--> 2 > ...010 <--> 3 > ...011 <--> 4 > ...100 <--> 5 > ...101 <--> 6 > ...110 <--> 7 > ...111 <--> 8 > ... what does the string entirely made of ones get sent to? i can see how that > works for strings with a finite number of ones in them: the string (x_i) i in N gets sent to sum x_i.2^i but where do the ones with infinitely many 1's get sent to? > Therefore we can conclude that 2^aleph0 = aleph0, and we come to > contradiction. (2^aleph0 >= aleph0) = {}, and we have a proof saying that Boolean > Logic cannot deal with infinitely many objects. > Doron > I think that at best you've proven the set of finite subsets of a > countable set is countable. But that's not surprising is it? > You've not got the infinite subsets in your enumeration. if you insist on doing this. note that it is not required to talk about > all those sets and strings. just think about all strings of 0's and 1's, > forget that set stuff, it's just confusing and unnecessary. ==== As it's a friday lunch time and the office party is happening soon, i don't feel too guilty about taking time off to analyse your pdf. Fortunately, it is clearer than the posting to this newsgroup, and therefore it is simple to point out where you are still going wrong. You make some conclusion about finite sets. In fact you are showing there are no bijections from a finite set to its power set. Good. You pass, through an induction, to a set of countably infinitely many elements. The result fails because simple induction cannot do that: I repeat: The product of a finite number of sets with 2 elements in each is finite. The infinite product, if it exists, is not finite. It is not even countable. As CAntor's argument tells us. There is a bijection between the collection of all finite subsets of a countable set and itself. You are using that fact: that's where your sending things to binary expansions comes in. You conclude that cantor's diagonal argument must produce something not in your list because it doesn't in the finite case. That may or may not be true. And isn't important. You argue that as 2^n - n is not zero, this must pass through in the limit. I'm not sure what you want to mean by that. The number of sequences 'in the limit' is not 2^aleph_0, absolutely not. It has cardinality aleph_0 as does the number of entries in each row. There are no missing rows 'not in the range of the diagonalization argument. If there were there would be a first such, where is it? Either the diaginalization produces something on the list, and all is fine, or it doesn't, and so what? I'm almost certain it would be an infinite set, but why does that matter? ==== Thought of an even better way of saying it. Your entire argument depends upon claiming that that list of 0's and 1's completely lists all the power sets of a countable set. How is it produced? It is by taking a (filtered direct) limit over the power sets of an nested increasing collection of finite sets. There are no infinite sets in this limit. Any element of the limit lies in one of the finite power sets over which you are taking your limit (and then in all the succeeding ones), and is thus finite. So you don't have a list of all the elements in the power set. Amazingly you have correctly asserted that 2^aleph_0 > aleph_0 but the proof is not valid. And you've demonstrated that the finite subsets of a countably infinite set are countable. But you've put those things together wrongly and got some odd conclusion. ==== Here is my answer: We can say that at least the sequence ...11111111 is not in the list, > for example: ...00000000 <--> 1 > ...00000001 <--> 2 > ...00000010 <--> 3 > ...00000011 <--> 4 > ...00000100 <--> 5 > ...00000101 <--> 6 > ...00000110 <--> 7 > ...00000111 <--> 8 > ... > So you are saying that there are no infinite sets on the list? And this means what? The finite subsets of a countable set are countable. You can list them. You could do that thing you cantor's diagonalization method. It doesn't follow that what you have is necessarily some infinite set from what you've written. And it doesn't matter. > Let us examine the infinite from another point of view. When we have ...11111 AND ...00000 in an ordered combinations list, it > means that the list is complete. > No, ...00000 is on the list, it is the empty set. And we can simply add ....1111 to the list by placing it in the 0th position. > But this is the whole point, infinitely many objects cannot be > completed, otherwise they are finite. Therefore ...11111 AND ...00000 are not in the list of infinity many > objects. > No, ..00000 is on the list, and ..11111 isn't on the list, but as ...1111 represents some infinite set, and you've only got the finite ones there that is not a contradiction with anything. > In other words [...000, ...111) XOR (...000, ...111] . There are 2 possible structural types of infinitely many 01 notations: (?...0] > (?...1] We know how some infinitely long combination starts, but its opposite > side is > unknown (can be 0 XOR 1) and this missing information is essential to > the existence of the induction. Therefore we can find a meaningful missing result by Cantor's > Diagonalization method, only in a finite combinations list. > For more details please look at: http://www.geocities.com/complement...iemannsBall.pdf So, what are you trying to show? That cantor is wrong? that there is aleph-0 and aleph_1 are the same? By the way, you still cannot pass from a finite cardinal to an infinite cardinal and say results hold by induction on the finite cardinalities alone. You seem want to assume that there are strictly more objects in the natural numbers union a finite set, than in the natural numbers, from all the 2^n - n stuff which I still don't see the relevance of. ==== Give a point-by-point exposition to support your case without insults, > innuendo, or any other attempts at sly or back-handed insults, but > with *mathematics* and logic. > When are you going to get round to answering my questions, James? All in given in an insult free manner > The challenge is to you now, show that you can reason and discuss > mathematics reasonably, if you can. > James Harris still waiting. ==== Hash: SHA1 [Follow-up reset] > Some of you may have noticed these free-wheeling mathematical > discussions I've been having with all these people, and wondered. Well, lots of posters have tried to educate you about math, and you've been freewheeling for everyone. [..] > Now then, use your common sense. Is it simpler that I'm really some crank, Yes, you are a crank. If not, then you're the best fake I've seen outside the circles of professional crank-imitators, but that's a different story. [..] > And don't worry, this post is very unlikely to > change anything. > You mean that the responses are unlkikely to make you change your behaviour? All too true :-( - -- Key ID 0x09723C12, j.tingleff@ieee.org/jens_tingleff@yahoo.com http://www.imaginet.fr/~jensting/ +44 1223 211 585 I told you to write a fake letter, not fake writing a letter 'Cerebus' iD8DBQE/2W28imJs3AlyPBIRAjq0AKCOQ/Z3MPZcwWTDdhnShx0Xiow/bwCgoNvM sfCVf60yBeVZCmUnVXqaAfU= =snCZ ==== I just finished reading an by Manfred Schroeder's excellent 1992 ('93) math book Fractals, Chaos and Power Laws, and he makes the argument that any function is chaotic if its spectral transform is a power law. That means that if P(w) ~ w^x, where w is the frequency omega, P(w) is the power spectrum and x is any small number, typically either rational or convenient round decimals, like 0.5, 1, 1.33, 1.4, 1.45, 1.5, 1.6, 1.75, 1.8, 2, etc. I don't think that x ever exceeded 3 in the many examples he gave. Of course, you can always fit a polynomial of arbitrary order n to any multimodal function if n is the number of peaks in the function, but it makes no sense if n is too large. Besides, power laws and polynomials are not the same thing. If you make the basis function expansion and the result is still extremely noisy, you're on the wrong track. If you use a low-pass filter, you've probably also ruined the math, since all of Schroeder's functions had positive exponents, making them rise monotonically with frequency. In other words, the highest order modes are the most populated states, making them the most important modes. And if you do low pass filtering, you aren't doing wavelets, you're doing sums of polynomials or trigonometrics. I'm still very loyal to the FFT exclusively, since trig functions are the only functions I know that aren't polynomials, and you can get free source codes on-line. I'm just an analyst, I don't do math very well. In fact, I'll bet you can generalize Schroeder's result to the conclusion that a function is chaotic if P(v) ~ v^x, where v is the working parameter of any basis function you choose (Phi(v)), not necessarily frequency. Phi could be a polynomial or your wavelet, and wavelets I've never seen the math for, so I don't know how they fix get past the low-pass problem. And that is out of my reach. > power law ... would you mind to give a little bit > more than you already said? > I think that the idea would then be to simply examine the spectral >> function and see if it obeys a power law. Has that ever been >> observed in the data? Power law dependence would mean that it is aperiodic, that basis >> function expansions are a waste of time (not counting wavelets) and >> that active control wouldn't work. Passive control might still work, (ie. setting interest rates to >> neutral and forgetting about them) but I wouldn't bet on that either. >> I'm not up on the math, but a few years ago there was a breakthrough >> in aerodynamics, where engineers realized when a fighter jet went >> into a tumble (chaotic trajectory) the pilot was causing it by >> fighting the stick. If he let go of the stick, the plane would >> spontaneously come out of the tumble. A sloppy person might be inclined to simply try the same solution for >> stabilizing the business cycle, but that would be a bad idea. It >> might be that there is no stable limit cycle in the available range >> of parameters except the one where all activity stops dead. This would be the same as if the time constant for recovery were so >> long that everybody starved before the economy came back. In the >> absence of actual state equations for the process model, I think >> you'd have to assume that this is the way things are and that the >> business cycle is the best solution allowed by the system. >> Yes, a long time ago. >> The business cycles has a varying period and amplitude. Also cycles >> are not always that clear when you get 'double dips' and the such >> like. > ==== >How about zero? As in There are no prime numbers which are >powers of Mersenne primes. None at all. Ain't gonna happen. >End of story. Next. The poster is wrong of course, but I have discovered powers of Mersenne primes that are neither primes nor composites. I will get my money once the mathematical community acknowledges my discovery or the year 3017, whichever comes first. ==== > Michigan State University grad student Michael Shafer has succeeded in > identifying the largest known prime number to date, using a distributed > computer network of more than 200,000 computers located around the world. > The new number is 6,320,430 digits long and is only the 40th Mersenne prime > to have ever been discovered (Mersenne primes are an especially rare breed > that take the form of 2-to-the-power-of-P, where P is also a prime number). AP is reporting that this genius had 2 gigahertz of memery. I dont get how such a simple story can be mangled by Arts majors so badly. Oh. I just figured it out. Shaun ==== Michigan State University grad student Michael Shafer has succeeded in > identifying the largest known prime number to date, using a distributed > computer network of more than 200,000 computers located around the world. > The new number is 6,320,430 digits long and is only the 40th Mersenne prime > to have ever been discovered (Mersenne primes are an especially rare breed > that take the form of 2-to-the-power-of-P, where P is also a prime number). AP is reporting that this genius had 2 gigahertz of memery. 2 billion cycles per second of memory? That's a rate, not a quantity. I dont get how such a simple story can be mangled by Arts majors so badly. > Oh. I just figured it out. > Shaun ==== > AP is reporting that this genius had 2 gigahertz of memery. 2 billion cycles per second of memory? That's a rate, not a quantity. Welcome to the genius club. -Mike ==== >Michigan State University grad student Michael Shafer has succeeded in >identifying the largest known prime number to date, using a distributed >computer network of more than 200,000 computers located around the world. >The new number is 6,320,430 digits long and is only the 40th Mersenne prime >to have ever been discovered (Mersenne primes are an especially rare breed >that take the form of 2-to-the-power-of-P, where P is also a prime number). > AP is reporting that this genius had 2 gigahertz of memery. ...and a 5 1/4 processor running at 31.2 kps of USB keyboard cache memory. ==== >>Michigan State University grad student Michael Shafer has succeeded in >>identifying the largest known prime number to date, using a distributed >>computer network of more than 200,000 computers located around the world. >>The new number is 6,320,430 digits long and is only the 40th Mersenne prime >>to have ever been discovered (Mersenne primes are an especially rare breed >>that take the form of 2-to-the-power-of-P, where P is also a prime number). >>AP is reporting that this genius had 2 gigahertz of memery. > ...and a 5 1/4 processor running at 31.2 kps of USB keyboard cache memory. I'll have a prime rib, please. Rare. Baked potato. Salad with Italian dressing. And bacon for dessert. --Bill -- The World Wide Web is the hugest vanity press in the history of the human race! http://billwilkinson.home.mindspring.com/index.html ==== > Michigan State University grad student Michael Shafer has succeeded in > identifying the largest known prime number to date, using a distributed > computer network of more than 200,000 computers located around the world. > The new number is 6,320,430 digits long and is only the 40th Mersenne prime > to have ever been discovered (Mersenne primes are an especially rare breed > that take the form of 2-to-the-power-of-P, where P is also a prime number). ==== Every place I look at leaves it as an exercise. I know that the proof involves thinking of A (mxn matrix) as a collection of column vectors, so that [A1 .. An] X B is a subpace of the column space of A1..An. But how do we know that the product is a subspace? How do we know it doesn't increase the dimension? ==== > Every place I look at leaves it as an exercise. I know that the proof involves thinking of A (mxn matrix) as a collection of > column vectors, so that [A1 .. An] X B is a subpace of the column space of A1..An. But how do we know that the product is a subspace? How do we know it > doesn't increase the dimension? of a matrix equals the dimension of the range of the associated linear transformation. ==== > By O(1), I was defining the _computational_ complexity with regard to > the variable 'x' and not defining the size of an error term. Even that sense of O(1) is difficult -- Your forumula contains a number of computations using numbers that are approximately as large as x, so each operation takes at least O(log x) time. I also see 2 logarithms, 2 exponentiations, and 1 cosh. How long does it take to execute those, in the general case? (What degree of accuracy are you demanding?) I think what you really mean is Here is an approximation to pi(x) that is in closed form and gets remarkably close. Dale ==== > The error for 10^23 appears to be 57867510952120449. Why do >> you call this an O(1) approximation? You have some reason to >> think the error is never larger than 57867510952120450 or what? My apologies. By training I am a computer scientist, and so had forgotten that the >O(x) notation is more commonly interpreted as an indication of the size >of difference from some required value. By O(1), I was defining the _computational_ complexity with regard to >the variable 'x' and not defining the size of an error term. Simple formulae are not normally used for calculating, say, the Riemann >Prime Counting function or the Logarithmic integral, as both are usually >calculated by summing an infinite series. Summing an infinite series is computationally O(p) where p is some >chosen limit of precision. Again, sorry for any confusion. No problem. Although of course you must be aware that there are well-known approximations to pi(x) which are O(1) in the same sense as what you gave above, for example x/log(x). >Carl ************************ David C. Ullrich ==== > No problem. Although of course you must be aware that there > are well-known approximations to pi(x) which are O(1) in the > same sense as what you gave above, for example x/log(x). It was that very approximation that led to the one described. Suppose there exists a solution (as a function of x) for N to: pi(x) = x/log_N(x) == log(N)*x/log(x) or even an approximation to such a function, then pi(x) will be calculable to the same degree. The actual value of N tends toward e as x -> oo, but for the values of pi(x) available it appears that N is not less than e for x >= ~100. It was assumed that this would remain true upto infinity. Graphing N over the values of pi(10^n) from the Mathworld page resulted in something resembling that for the function f(x)=x^(1/x), yet converging towards e rather than 1. This idea eventually became: a = L^[2/(3L)] A modified hyperbolic secant was introduced to attempt to eliminate an aberration in the difference between the graphs for N and e*a, peaking at around L = e^2. Why this should be when L is a logarithm to base ten is unclear, but this led to the formula: b = 1 + K1*sech(L-e^2)^K2 ; 0 < K1 < 1 ; 0 < K2 << 1 The 1 + intending to make the curve converge to 1 (as L->oo) so that a/b would converge to 1. After some experimentation, the sech formula became the cosh formula seen in the original post. k is then set to the approximated value of log(N). The 1 + in this case adjusts log(a/b) so that it again converges to 1, allowing N = e^k to converge to e as required. And thus we have a marginal improvement on the standard x/log(x) approximation. Carl ==== Hallo all, Does anybody have a clue how to solve the following definite integral Int( cos x / (1 + 2C^2 - 4C cosh x + 2 C^2 cosh 2x ), x,0..pi ) Karjanto ==== > It is kind of curious to note the observed frequency of the digit 5 > is *always* less than expected. Is there an explanation for this > wierdness? I'd start by asking if the deviations from the expectation values are larger than would be expected from a random variable with that distribution. It may not be so weird. Dale ==== > Suppose x is a real number and 0=0 compute: 1: a[i]:=the largest integer k s.t. floor(k*x[i])=1 >> 2: x[i+1]:=a[i]*x[i]-1 It's not too difficult to see: (1) 2<=a[i]<=5 if i>0, and (2) >> x=1/a[0]+1/(a[0]*a[1])+... So a[i] is reasonably called the ith digit of >x. In a very nice post from several years ago, Rob Johnson showed the expected >> distribution of the digits of a random x is [206,81,30,20]/337 >> (tinyurl.com/ys8t). Thus p(a[i]=5) is 20/337 or about .059347. The exact >> distribution of the first n digits of log(2) is as follows (according to >> Pari/GP): n freq 1000:[587,276,87,50] >> 2000:[1179,537,178,106] >> 3000:[1783,793,267,157] >> 4000:[2404,1035,360,201] >> 5000:[3008,1274,448,270] >> 6000:[3618,1529,528,325] >> 7000:[4240,1753,625,382] >> 8000:[4886,1943,727,444] >> 9000:[5490,2194,912,562] >> 10000:[6113,2413,912,562] >> 20000:[12236,4817,1780,1167] >> 30000:[18337,7266,2678,1719] >> 40000:[24449,9716,3545,2290] >> 50000:[30564,12105,4443,2888] >> 60000:[36692,14483,5368,3457] >> 70000:[42847,16819,6291,4043] It is kind of curious to note the observed frequency of the digit 5 is >> *always* >> less than expected. Is there an explanation for this wierdness? >> Always? Or just when n ends in 000? > Good question. Pari/GP sez for 51<=n<=70000 the observed frequency is less than 20/337. rich ==== > Suppose x is a real number and 0=0 compute: 1: a[i]:=the largest integer k s.t. floor(k*x[i])=1 > 2: x[i+1]:=a[i]*x[i]-1 It's not too difficult to see: (1) 2<=a[i]<=5 if i>0, and (2) > x=1/a[0]+1/(a[0]*a[1])+... So a[i] is reasonably called the ith digit >of >>x. In a very nice post from several years ago, Rob Johnson showed the >expected > distribution of the digits of a random x is [206,81,30,20]/337 > (tinyurl.com/ys8t). Thus p(a[i]=5) is 20/337 or about .059347. The exact > distribution of the first n digits of log(2) is as follows (according to > Pari/GP): n freq 1000:[587,276,87,50] > 2000:[1179,537,178,106] > 3000:[1783,793,267,157] > 4000:[2404,1035,360,201] > 5000:[3008,1274,448,270] > 6000:[3618,1529,528,325] > 7000:[4240,1753,625,382] > 8000:[4886,1943,727,444] > 9000:[5490,2194,912,562] > 10000:[6113,2413,912,562] > 20000:[12236,4817,1780,1167] > 30000:[18337,7266,2678,1719] > 40000:[24449,9716,3545,2290] > 50000:[30564,12105,4443,2888] > 60000:[36692,14483,5368,3457] > 70000:[42847,16819,6291,4043] It is kind of curious to note the observed frequency of the digit 5 is > *always* > less than expected. Is there an explanation for this wierdness? >Always? Or just when n ends in 000? > >Good question. Pari/GP sez for 51<=n<=70000 the observed frequency is less >than 20/337. > However, if n=179569 the observed frequency is greater than 20/337. rich ==== > May I suggest SymbMath for $20: It is an online symbolic math and computer algebra system. > It can perform exact, numeric, symbolic and graphic computation, > e.g. arbitrary-precision calculation, solve equation, plot data and > user-defined functions, linear regression, symbolic differentation > and integration, pattern-match. It is a programming language, in > which you can define conditional, case, piecewise, recursive, > multi-value functions and procedures, derivatives, integrals > and rules. > www.SymbMath.com > www.mathHandbook.com For various obscure reasons I've needed an inexpensive shareware > software which would calculate basic arithmetic functions, ln(n), e^n, > roots, and the basic trig. functions,...all out to pretty high decimal > points, > say 50 or so (or more depending on how playful I get). For a while I've tested the Haxial Calculator which is OK but laborious, > esp. since it has, as yet, no trig. functions. For such trig stuff I've > used > basic formulae to let it calculate actual trig. values, though it kicks up > a > fuss when I stuff certain numbers into it. The Hax. Calc. costs maybe > $15 or 20. Does anyone know of a good shareware, perhaps even a bit more expen- > sive, which will do ALL the functions I described, to fairly large > arbitrary > accuracy?? (There is no way I can afford anything like Mathematica > unless I don a mask and take to the highroad ;- ) Gene ==== > I was wondering whether there be a name for a faithful functor F: C > --> D such that given any isomorphism f': F(a) --> b' in D, there is > unique object b of C such that for some isomorphism f: a --> b, F(f) = > f'? In other words, when F is into the category of sets, given a > structure on a and a bijection f from a into b, there is a unique > structure on b such that f is an isomorphism from a into b. This > seems to be the categorical notion with which you would want to > replace Bourbaki's notion of structure and morphism (as defined in > Theories des Ensembles, Ch. IV) if you want all his results to work > out yet (understandably) want to use a categorical approach. It is > what I prefer to think of when he uses the concepts of morphism and > structure. Such a functor is called uniquely transportable in the terminology of Adamek,Herrlich,Strecker: Abstract and concrete categories (see Def 5.28 for details and a funny picture). More precisely, in the above situation F: C --> D they call C a concrete category over D, when F is faithful. In this setting they call F: C --> D uniquely transportable when the above lifting property holds. Unlike a op-fibration, only _isomorphisms_ are asked to be liftable. So it is easy to produce UT-Functors where you cannot lift arbitrary morphisms, e.g. restrict C by a size-condition on the F(c) (c in C). However these examples look a bit artificial. If you are interested in more of this stuff, the Adamek/Herrlich/Strecker book might be a good starting point. There is also an earlier book Manes: Algebraic Theories, where some of this is discussed. Marc ==== |Such a functor is called uniquely transportable in the terminology of |Adamek,Herrlich,Strecker: Abstract and concrete categories |(see Def 5.28 for details and a funny picture). Subtitled, The Joy of Cats. The illustration at the beginning of each chapter features a cat doing various things. :-) The illustration of unique transportability is of the machine that you should imagine doing the unique transportation, I think. I looked at that book once out of curiosity about the definitions of algebraic functor and topological functor. The forgetful functor from the category of groups to the category of sets is algebraic. The forgetful functor from the category of topological spaces to the category of sets is topological. The forgetful functors from the category of topological groups to the categories of groups and topological spaces are, if I remember correctly, topological and algebraic, respectively. These are concepts of a similar flavor. Keith Ramsay ==== |Such a functor is called uniquely transportable in the terminology of > |Adamek,Herrlich,Strecker: Abstract and concrete categories > |(see Def 5.28 for details and a funny picture). Subtitled, The Joy of Cats. The illustration at the beginning of > each chapter features a cat doing various things. :-) The illustration > of unique transportability is of the machine that you should imagine > doing the unique transportation, I think. > ... > Keith Ramsay > Cat-egory theory--I get it. Or maybe a main inspiration behind the book was a girl named after a cat, which can happen sometimes. So the cat is really altogether quite various in the things it is doing? Meow. ==== post of Tom Leinster (http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8&edition=us&selm =73362839.0308250747.33206a21%40posting.google.com): This is a special case of the notion of discrete fibration, or more accurately, discrete opfibration. A functor Q: J ---> K is called a discrete opfibration if: for any object j of J and any morphism g: Q(j) ---> k in K, there is a unique morphism f: j ---> j' in J such that Q(f) = g. I guess what you are saying is that my diagram condition is equivalent to the discrete opfibration condition holding for the faithful functor obtained by restricting both the domain and range of the faithful functor F I mentioned to the respective subcategories containing as morphisms precisely the isomorphisms. I get what you are saying now. ==== |I see the following definition of discrete opfibration in a |(http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8&edition=us&sel m=73362839.0308250747.33206a21%40posting.google.com): | |This is a special case of the notion of discrete fibration, or more |accurately, discrete opfibration. A functor Q: J ---> K is called a |discrete opfibration if: | | for any object j of J and any morphism g: Q(j) ---> k in K, there | is a unique morphism f: j ---> j' in J such that Q(f) = g. | |I guess what you are saying is that my diagram condition is |equivalent to the discrete opfibration condition holding for the |faithful functor obtained by restricting both the domain and range of |the faithful functor F I mentioned to the respective subcategories |containing as morphisms precisely the isomorphisms. yes. it's quite possible that there's also some special terminology for a functor between (possibly non-groupoid) categories that gives a discrete fibration on the groupoid level, but i don't know what it might be offhand. also, although i don't think that i've read the section of bourbaki that you mentioned, i had gotten the impression from second-hand sources that in that section bourbaki was actually doing something closer to groupoid theory than to category theory. it might have been a false impression though. -- ==== if i'm not misunderstanding what you said too badly, then it sounds > like you really want to consider only the case where c and d are > groupoids rather than arbitrary categories. (in particular when you > take the objects of d to be the sets it seems that you want the > morphisms to be the bijections rather than the arbitrary functions.) > in that case i think the standard name for a functor of the type > you're describing would be something like discrete fibration, and > yes they're used for exactly the purpose that you're describing. No. For instance, F might be the forgetful functor from the category of topological spaces (with morphisms the continuous maps) to the category of sets. Given any bijection f from the underlying set of a topological space U into a set S, there is a unique topology on S making f a homeomorphism. Thus, F would be a functor having the properties I enumerated, notwithstanding neither the category of sets nor the category of topological spaces are groupoids. It is true that Bourbaki (in E.IV) only considers F into the category of sets--in fact he always takes F to be a forgetful functor. But he demands more than that F be a forgetful functor (assuming forgetful is defined as MacLane vaguely defines it)--e.g., to be a legitimate concept of morphism in the Bourbaki sense, the identity map is an isomorphism between two objects having the same underlying set if and only if the structures on these objects are identifical. ==== |> if i'm not misunderstanding what you said too badly, then it sounds |> like you really want to consider only the case where c and d are |> groupoids rather than arbitrary categories. (in particular when |> you take the objects of d to be the sets it seems that you want the |> morphisms to be the bijections rather than the arbitrary |> functions.) in that case i think the standard name for a functor |> of the type you're describing would be something like discrete |> fibration, and yes they're used for exactly the purpose that |> you're describing. | | |No. For instance, F might be the forgetful functor from the category |of topological spaces (with morphisms the continuous maps) to the |category of sets. Given any bijection f from the underlying set of a |topological space U into a set S, there is a unique topology on S |making f a homeomorphism. Thus, F would be a functor having the |properties I enumerated, notwithstanding neither the category of sets |nor the category of topological spaces are groupoids. you might as well just consider u(F) where u is the forgetful functor from the category of categories to the category of groupoids, as far as i can tell from the description you've given. -- ==== >I have the following application: I have a force sensor probe that touches 3 points on a plate. At each >of the points, I can calculate an XYZ point representing the point >that the probe detectes the plate. I can easily calculate the equation of the plane Ax+By+Cz+D = 0. What >I really need is the rotation matrix/Euler angles as represented by >the plane so I can correctly translate positions along the slope of >the plane. What do you mean by rotation as represented by the plane? There is no preferred orientation in 3d, so you have to specify which is the initial orientation relative to which the rotation of the plane is to be calculated. So for example, if the initial plane is the standard xy plane, with normal vector n1=(0,0,1), and the destination plane equation is Ax+By+Cz+D=0 with normal vector n2=(A,B,C) then you can first create the necessary rotation in axis/angle format, then convert to matrix format to extract euler angles. See http://www.geocities.com/scroussette/rotplana.html#rotn Internally, I am storring all my locations and frames of reference as >XYZ points and Euler angles in terms of DZ, DX, DZ rotations. Any thoughts would be appreciated ==== > You've been smoking that stuff for far too many years!!!!!! >... >> -- Two sets A and B are the same size, |A| = |B|, if and only if >> there exists a bijection between the two sets. >If size is a natural number, then you can't take the size of >an infinite set (without additional axioms and definitions). At the moment you say that infinite sets are of the same size >when there exists a bijection, then you already introduce some >of the choices Cantor made. But those are choices, not proofs. > -- If B is a subset of A, then A is at least the same size >> as B, that is, |B| =< |A|. >> (There is more that needs to be shown. For example, for >> all sets A and B, either |A| =< |B| or |B| =< |A|, and >> that |A| =< |B| and |A| =< |B| together imply |A| = |B|. I think >> those two points are all we have to assume, though.) >> We have shown _there does not exist_ a bijection between >> the natural numbers and the reals. Remember we have looked >> at _all_ the potential bijections and _all_ of them failed >> at at least one point. >This is indeed proven. > As we understand the size of infinite sets, that means N >> and R are different sizes. It's easy to show |N| =< |R|, >> since N is a subset of R. Thus |N| < |R|. >This is Cantor's choice, not a proof. You may also conclude that the logical system you are currently >using is not capable of definining all irrational numbers and >leave it there. So, there are at least two choices you can >make: >a. You may refer to R even when there is no logical system > that can list all its elements (this is Cantor's choice). >b. You must be aware that R is never be complete in your > logical system (alternative to Cantor). I don't think choice b is very attractive, but to my opinion >it is a way you can try to go. However, I always questioned if saying that |N| < |R| >has any more meaning than saying that irrational numbers >are green. >Lucas ==== Green's functions often have logarithmic terms in them, so they are often irrational functions. So irrational( function)s can be Green('s fuctions). -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== > ... > -- Two sets A and B are the same size, |A| = |B|, if and only if > there exists a bijection between the two sets. > If size is a natural number, then you can't take the size of > an infinite set (without additional axioms and definitions). At the moment you say that infinite sets are of the same size > when there exists a bijection, then you already introduce some > of the choices Cantor made. But those are choices, not proofs. -- If B is a subset of A, then A is at least the same size > as B, that is, |B| =< |A|. > (There is more that needs to be shown. For example, for > all sets A and B, either |A| =< |B| or |B| =< |A|, and > that |A| =< |B| and |A| =< |B| together imply |A| = |B|. I think > those two points are all we have to assume, though.) We have shown _there does not exist_ a bijection between > the natural numbers and the reals. Remember we have looked > at _all_ the potential bijections and _all_ of them failed > at at least one point. > This is indeed proven. As we understand the size of infinite sets, that means N > and R are different sizes. It's easy to show |N| =< |R|, > since N is a subset of R. Thus |N| < |R|. Although the cardinality of the naturals is smaller than the cardinality of the reals, it is not because the naturals are a subset of the reals. The natural numbers are also a subset of the rational numbers yet these two sets have the same cardinality. In fact, infinite sets are sometimes defined as sets of elements such that there is a one-to-one correspondense between the elements of the set and the elements of a proper subset of that set. Darren > This is Cantor's choice, not a proof. You may also conclude that the logical system you are currently > using is not capable of definining all irrational numbers and > leave it there. So, there are at least two choices you can > make: > a. You may refer to R even when there is no logical system > that can list all its elements (this is Cantor's choice). > b. You must be aware that R is never be complete in your > logical system (alternative to Cantor). I don't think choice b is very attractive, but to my opinion > it is a way you can try to go. However, I always questioned if saying that |N| < |R| > has any more meaning than saying that irrational numbers > are green. > Lucas ==== ... > -- Two sets A and B are the same size, |A| = |B|, if and only if > there exists a bijection between the two sets. If size is a natural number, then you can't take the size of > an infinite set (without additional axioms and definitions). Then size is not a natural number for infinite sets. Is this a problem? > At the moment you say that infinite sets are of the same size > when there exists a bijection, then you already introduce some > of the choices Cantor made. But those are choices, not proofs. OK, those are choices (or definitions, as I would say). They are, however, very good definitions. They allow us to extend the idea of sets size to infinite sets and preserve many of the prperties of size that we are familiar with from finite sets. > -- If B is a subset of A, then A is at least the same size > as B, that is, |B| =< |A|. > (There is more that needs to be shown. For example, for > all sets A and B, either |A| =< |B| or |B| =< |A|, and > that |A| =< |B| and |A| =< |B| together imply |A| = |B|. I think > those two points are all we have to assume, though.) We have shown _there does not exist_ a bijection between > the natural numbers and the reals. Remember we have looked > at _all_ the potential bijections and _all_ of them failed > at at least one point. > This is indeed proven. As we understand the size of infinite sets, that means N > and R are different sizes. It's easy to show |N| =< |R|, > since N is a subset of R. Thus |N| < |R|. > This is Cantor's choice, not a proof. You may also conclude that the logical system you are currently > using is not capable of definining all irrational numbers and > leave it there. So, there are at least two choices you can > make: > a. You may refer to R even when there is no logical system > that can list all its elements (this is Cantor's choice). > b. You must be aware that R is never be complete in your > logical system (alternative to Cantor). If I redefine R using other axioms, it may be that this new R' could be listed, but so what? There are many sets that are countable. (Many is a bit of an understatement.) The fact that R' is countable does not remove the fact that R, described by the original axioms, is not countable. > I don't think choice b is very attractive, but to my opinion > it is a way you can try to go. I find choice b very unattractive, too. It's as though someone decides the consequences of some axioms are wrong or ugly or something. Say they don't like Euler's formula exp(i theta) = cos(theta) + i sin(theta) so they mess around with the axioms for complex analysis until it goes away. It doesn't seem like a very good way to do things. > However, I always questioned if saying that |N| < |R| > has any more meaning than saying that irrational numbers > are green. I know what I mean when I say |N| < |R|. I mean something about the existence of functions between N and R. I also know I can often get away with thinking about |N| and |R| like the sizes of finite sets and when I need to be more careful. We often deal with things in math by pretending we're imagining things we can't really imagine, like infinite- dimensional vector spaces, or the empty set. What can you say about irrationals being green? Jim Burns ==== > However, I always questioned if saying that |N| < |R| > has any more meaning than saying that irrational numbers > are green. thats obviously because they are purple! ==== > However, I always questioned if saying that |N| < |R| > has any more meaning than saying that irrational numbers > are green. thats obviously because they are purple! Dude...take off those colored glasses... I see them clearly as Chartreuse.... with pink spots... ==== > However, I always questioned if saying that |N| < |R| > has any more meaning than saying that irrational numbers > are green. The irrationals are green, and justifiably so IMO, out of envy of the rationals. > thats obviously because they are purple! Dude...take off those colored glasses... I > see them clearly as Chartreuse.... with pink spots... Now somewhat seriously: Besides being a mathematician, I am also a musician. As such, I am aware that some people experience synesthesia. The composer Scriabin was famous for this, experiencing certain color sensations when he heard certain pitches. I would not be greatly surprised if there were some people who experience certain color sensations when contemplating certain numbers. Does anyone know of any instances of such? David ==== Now somewhat seriously: > Besides being a mathematician, I am also a musician. As such, I am aware > that some people experience synesthesia. The composer Scriabin was famous > for this, experiencing certain color sensations when he heard certain > pitches. I would not be greatly surprised if there were some people who > experience certain color sensations when contemplating certain numbers. > Does anyone know of any instances of such? David There is a chapter on this sort of thing in Underwood Dudley's book Mathematical Cranks. That material (number forms I think they are called?) is a bit different than the rest of the book. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== > However, I always questioned if saying that |N| < |R| > has any more meaning than saying that irrational numbers > are green. The irrationals are green, and justifiably so IMO, out of envy of the > rationals. thats obviously because they are purple! Dude...take off those colored glasses... I > see them clearly as Chartreuse.... with pink spots... Now somewhat seriously: > Besides being a mathematician, I am also a musician. As such, I am aware > that some people experience synesthesia. The composer Scriabin was famous > for this, experiencing certain color sensations when he heard certain > pitches. I would not be greatly surprised if there were some people who > experience certain color sensations when contemplating certain numbers. > Does anyone know of any instances of such? David I was only refferancing the subject line of irrational and why I see Chartreuse.... with pink spots... ==== |-|erc says... >You said that it misses non computable numbers, like r = Halt(1) + 1/10 Halt(2) >+ 1/100 Halt(3)... >since Halt(n) is unknown for all n. i.e. r is not computable and will not be listed in UTM(Z). I tried to argue r is just not known, you tried to show r is not computable at >all (standard view), >now I am now suggesting r it not existent! so Computables = Reals. Yes, it is perfectly consistent to believe that the only real numbers that exist are computable reals. However, Cantor's theorem is *still* valid: Under this interpretation, it says that there is no computable listing of all computable reals. -- Daryl McCullough Ithaca, NY ==== > There is only 1 infinity type, the answer is NO, here's a similar exposition of diagonalisation. Not true. There are an infinite number of transfinite cardinals. Bob Kolker ==== > What is an irrational number? Can you count to it? > Can you pin point it? There is no such thing. Not true. The reals can be built from the rationals in several equavalent ways. 1. Dedikind Cuts 2. Equivalence Classes of rational Cauchy Sequences All numbers are the result of computable functions, sqrt(2) is the > result of turing machine(x), where x is some integer, probably > under a million with any crude mapping technique. A Cauchy Sequence converging to sqrt(2) can be generated by a non-stopping turing machine. Bob Kolker <3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de> <1Ff9oJOLyx1$Ew6T@baesystems.com> <2a0cceff.0312101407.2225db91@posting.google.com> ==== In message <2a0cceff.0312101407.2225db91@posting.google.com>, Edward >> In message <3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de>, Bjoern <... > sqrt(2) is the > result of turing machine(x), where x is some integer, probably > under a million with any crude mapping technique. >>That's incomprehensible. Could you rephrase this, please? >> He means that any Turing machine can be represented by a tape fed to a >> universal Turing machine, and the contents of that tape can be >> represented by some number x. Well, let's forget about the Turing machine for a second. Let's just >note that sqrt(2) is computable in the sense that, armed with some >finite algorithm and an indefinite amount of storage, we can compute >an arbitrary number of digits of it. Ok ... that does sound a lot like a Turing machine after all. ;-) This stirs some vague recollection: there are some privileged >irrational numbers which can be specified with a finite amount of >information. Obviously _all_ irrational numbers cannot be so >specified, or they in fact would be countable. So most irrational >numbers are poor lost souls which not only have non-repeating decimal >representations, but can't even be named in any meaningful way -- they >are unknowable. (This was the subject of some (IBM?) news release >within the past decade, possibly one of those over-hyped news releases >which appear regularly, repeating essentially known results as if were >fresh revelation). Can somebody remind me what this concept is called? I think you're thinking of the distinction between algebraic (roots of a polynomial) and transcendental (the rest) ? I wonder if there are any other possibilities - read on... >Do these >unknowable irrational numbers correspond to non-computable functions? Or to unknown unknowns? ;-) That only leaves two types of numbers left that qualify for irrational, > non computable and random numbers. >>Well, I would say that irrational numbers are not computable; why do >>you think otherwise? >> Because he's confused about the halting problem. So am I, evidently -- would you refresh my memory of its significance >here? In essence, you can't devise a universal algorithm which, given _any_ algorithm as input, will tell you whether or not the other halts. That doesn't mean that for any particular algorithm you can't determine what it does. And, repeating the argument above, surely _some_ irrational numbers >are in fact computable? Only the ones you can write down ;-) Don't forget, the complete answer hasn't been computed until the algorithm halts. Non computable numbers is not a proof that irrationals exist, IMO, > that no halting function exists does not clearly define what the halting > number is, it states it is impossible. There is no gap on the number line > from non computable numbers. >>That's incomprehensible again. >> I think he's trying to argue that because there exist TMs for which we >> can't determine to which of the sets halts or does not halt they >> belong, the does not halt set must be empty. Ah yes ... irrational numbers correspond to the output of a machines >which neither halts nor loops. ;-) I have one here that does that - sometimes it just freezes. Perhaps it's running in imaginary time? > As to what he is trying to say, OTOH, >I would feel comfortable leaving it in the incomprehensible bin. Aha! That's the missing category: algebraic, transcendental, or incomprehensible. Ah, knowledge, ah, intellect, ah, thought, I miss you ... you jade. -- Richard Herring ==== Richard Herring most amusingly scrivened in message > In message <2a0cceff.0312101407.2225db91@posting.google.com>, Edward >> In message <3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de>, Bjoern > ... there are some privileged >irrational numbers which can be specified with a finite amount of >information. Obviously _all_ irrational numbers cannot be so >specified, or they in fact would be countable. So most irrational >numbers are poor lost souls which not only have non-repeating decimal >representations, but can't even be named in any meaningful way -- they >are unknowable. ... >Can somebody remind me what this concept is called? I think you're thinking of the distinction between algebraic (roots of a > polynomial) and transcendental (the rest) ? > I wonder if there are any other possibilities - read on... Maybe -- Feuerbacher says the same. But I think I'm thinking of essentially what I am thinking of, if you follow my thought. To put us on the same page, I understand computable to mean produceable as the output of a Turing machine. The term is normally applied to integer functions of the integers, but obviously a decimal representations of numbers can be considered a function on the integers. I presume all algebraic numbers are computable, but at least some of the rest of 'em are too: anything we could write down a finite instruction set for -- given sufficient time and paper -- calculating any needed number of digits of, is computable: a class which includes pi and e. The argument for the existence of the unknowable numbers is that all finite algorithms or specifications form a countable set: so almost all of the irrationals can be neither unambiguously specified nor computed. ... > That only leaves two types of numbers left that qualify for irrational, > non computable and random numbers. I was silent the first time, but this claim is obviously complete nonsense: the definition of irrational is ... well, you know what it is ... not non computable, and a random number is a meaningless idea in this context. The number line contains numbers, not random numbers. But the OP is still to be thanked for giving an occasion to display foils of learnedness. >>Well, I would say that irrational numbers are not computable; why do >>you think otherwise? >> Because he's confused about the halting problem. So am I, evidently -- would you refresh my memory of its significance >here? In essence, you can't devise a universal algorithm which, given _any_ > algorithm as input, will tell you whether or not the other halts. That > doesn't mean that for any particular algorithm you can't determine what > it does. Right ... but -- given the admited silliness of asking for details of a hypothetical nonsensical argument -- what do you _think_ he thinks this has to do with the existence of irrational numbers? >And, repeating the argument above, surely _some_ irrational numbers >are in fact computable? Only the ones you can write down ;-) Exactly -- if write down means give a finite algoritm to compute. > Don't forget, the complete answer hasn't been computed until the > algorithm halts. Hmm... maybe that's right. The Turing machine in general is free to go back and erase parts of the tape, so that you can never be sure any given bit will stand until the machine stops? Is that it? Yet I think there is a concept of computable for infinite strings of digits; I'm just not sure how this fits together. We can imagine a machine with two tapes, one non-erasable, such that the n-th digit is not written on that tape until the algorithm is sure that's the real n-th digit. That's not a Turing machine, but anything any bastardized computing machine can do, a general Turing machine can do better ... right? ... so we ought to be able to emulate this machine with a Turing machine. Maybe this is a red-herring: as you say, the fact that no general halting test exists doesn't prevent us from being able to analyze _some_ algorithms with certainty to know that they halt. Similarly, although a general Turing machine might back up and write over any space in the tape before the algoritm stops, some algorithms might be analyzable to prove that, for example, only the last 5 written positions on the tape are labile, so at any given stage all eariler positions are immutable. Clearly a Turing machine which prints a complete function on N (which is not trivially zero above some range) is never going to stop, yet I don't think computable functions are limited to those with only finite information: this is some semantic problem here. >Ah yes ... irrational numbers correspond to the output of a machines >which neither halts nor loops. ;-) I have one here that does that - sometimes it just freezes. Perhaps > it's running in imaginary time? Pshaw. You know perfectly well that your frozen machine is running in a loop down on the CPU level. > As to what he is trying to say, OTOH, >I would feel comfortable leaving it in the incomprehensible bin. Aha! That's the missing category: algebraic, transcendental, or > incomprehensible. :-))) Similar to a serious classification of statements though: right, wrong, and nonsense. ==== > In message <2a0cceff.0312101407.2225db91@posting.google.com>, Edward ... >This stirs some vague recollection: there are some privileged >irrational numbers which can be specified with a finite amount of >information. Obviously _all_ irrational numbers cannot be so >specified, or they in fact would be countable. So most irrational >numbers are poor lost souls which not only have non-repeating decimal >representations, but can't even be named in any meaningful way -- they >are unknowable. (This was the subject of some (IBM?) news release >within the past decade, possibly one of those over-hyped news releases >which appear regularly, repeating essentially known results as if were >fresh revelation). > >Can somebody remind me what this concept is called? > > I think you're thinking of the distinction between algebraic (roots of a > polynomial) and transcendental (the rest) ? > I wonder if there are any other possibilities - read on... Wrong. Some transcendentals can be calculated. Pi and e to name two. (Where I mean calculated to mean that you can give a finite algorithm to calculate an arbitrary amount of digits.) The concept is computable. And indeed, the set of computable numbers is countable. There is, however, a beautiful problem when you try to apply a Cantor like argument to a (complete) list of computable numbers. This would tell that you get a new computable number (you just computed it) that is not on the list. The problem here is that a complete list of computable numbers is itself not computable. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ <2a0cceff.0312101407.2225db91@posting.google.com> ==== > In message <2a0cceff.0312101407.2225db91@posting.google.com>, Edward >... >This stirs some vague recollection: there are some privileged >irrational numbers which can be specified with a finite amount of >information. Obviously _all_ irrational numbers cannot be so >specified, or they in fact would be countable. So most irrational >numbers are poor lost souls which not only have non-repeating decimal >representations, but can't even be named in any meaningful way -- they >are unknowable. (This was the subject of some (IBM?) news release >within the past decade, possibly one of those over-hyped news releases >which appear regularly, repeating essentially known results as if were >fresh revelation). Can somebody remind me what this concept is called? I think you're thinking of the distinction between algebraic (roots of a > polynomial) and transcendental (the rest) ? > I wonder if there are any other possibilities - read on... Wrong. And season's greetings to you too. _What's_ wrong? Last I heard, those were (near enough) the definitions of algebraic and transcendental. Or are you saying that what I think Ed is thinking is wrong? Unless you are reading his mind, you can't possibly tell. Actually I think you missed the joke. Never mind, it's not worth labouring it. > Some transcendentals can be calculated. Pi and e to name two. >(Where I mean calculated to mean that you can give a finite algorithm >to calculate an arbitrary amount of digits.) To use your own word, Wrong. All you have calculated is a rational approximation to the number. That's not the number itself, though it can approximate it as closely as you would like. >The concept is computable. >And indeed, the set of computable numbers is countable. There is, however, >a beautiful problem when you try to apply a Cantor like argument to a >(complete) list of computable numbers. This would tell that you get a >new computable number (you just computed it) that is not on the list. >The problem here is that a complete list of computable numbers is itself >not computable. So you agree that the computable numbers are incomprehensible. -- Richard Herring ==== >> In message <2a0cceff.0312101407.2225db91@posting.google.com>, Edward >>... >>This stirs some vague recollection: there are some privileged >>irrational numbers which can be specified with a finite amount of >>information. Obviously _all_ irrational numbers cannot be so >>specified, or they in fact would be countable. So most irrational >>numbers are poor lost souls which not only have non-repeating decimal >>representations, but can't even be named in any meaningful way -- they >>are unknowable. (This was the subject of some (IBM?) news release >>within the past decade, possibly one of those over-hyped news releases >>which appear regularly, repeating essentially known results as if were >>fresh revelation). >>Can somebody remind me what this concept is called? >> I think you're thinking of the distinction between algebraic (roots of a >> polynomial) and transcendental (the rest) ? >> I wonder if there are any other possibilities - read on... >>Wrong. And season's greetings to you too. _What's_ wrong? Last I heard, those were (near enough) the definitions > of algebraic and transcendental. Or are you saying that what I think Ed > is thinking is wrong? Unless you are reading his mind, you can't > possibly tell. Actually I think you missed the joke. Never mind, it's not worth > labouring it. > Some transcendentals can be calculated. Pi and e to name two. >>(Where I mean calculated to mean that you can give a finite algorithm >>to calculate an arbitrary amount of digits.) To use your own word, Wrong. All you have calculated is a rational > approximation to the number. That's not the number itself, though it can > approximate it as closely as you would like. He is talking about a perfectly reasonable notion of calculated. Pick a number. It is computable if and only if there is an algorithm A taking input n such that A(n) returns the first n digits of your number. Your point is that for all n, A(n) is merely a rational approximation. And that's certainly correct. in this sense then A is an algorithm that calculates the exact real number. Not just an approximation. The real thing. If you prefer, we could have A() return rational numbers such that the sequence A(i), i = 1 to oo is Cauchy. Or we could have A() take rational numbers as input and return MEMBER or NON-MEMBER so that the set {x: x in Q and A(x) = MEMBER} is an upper Dedekind cut. There are many formalisms that encompass the notion of computable. Deciding whether a particular algorithm calculates a particular real number ought not depend on whether the printer has an infinite supply of ink. John Briggs ==== > In message <2a0cceff.0312101407.2225db91@posting.google.com>, Edward ... > So most irrational >numbers are poor lost souls which not only have non-repeating decimal >representations, but can't even be named in any meaningful way -- they >are unknowable. (This was the subject of some (IBM?) news release >within the past decade, possibly one of those over-hyped news releases >which appear regularly, repeating essentially known results as if were >fresh revelation). > >Can somebody remind me what this concept is called? > > I think you're thinking of the distinction between algebraic (roots of a > polynomial) and transcendental (the rest) ? > I wonder if there are any other possibilities - read on... > >Wrong. > > And season's greetings to you too. > > _What's_ wrong? Last I heard, those were (near enough) the definitions > of algebraic and transcendental. Or are you saying that what I think Ed > is thinking is wrong? Unless you are reading his mind, you can't > possibly tell. Well, this is incorrect. See the word named in a meaningful way? > > Actually I think you missed the joke. Never mind, it's not worth > labouring it. > > Some transcendentals can be calculated. Pi and e to name two. >(Where I mean calculated to mean that you can give a finite algorithm >to calculate an arbitrary amount of digits.) > > To use your own word, Wrong. All you have calculated is a rational > approximation to the number. That's not the number itself, though it can > approximate it as closely as you would like. The same holds for quite a few algebraic numbers. Or do you have some way to calculate (or name) the roots of, say, x^5 + 5x + 3, and to distinguish the roots? finite space you can calculate an arbitrary amount of digits. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > In message <2a0cceff.0312101407.2225db91@posting.google.com>, Edward >... > So most irrational >numbers are poor lost souls which not only have non-repeating decimal >representations, but can't even be named in any meaningful way -- they >are unknowable. (This was the subject of some (IBM?) news release >within the past decade, possibly one of those over-hyped news releases >which appear regularly, repeating essentially known results as if were >fresh revelation). Can somebody remind me what this concept is called? I think you're thinking of the distinction between algebraic (roots of a > polynomial) and transcendental (the rest) ? > I wonder if there are any other possibilities - read on... Wrong. And season's greetings to you too. _What's_ wrong? Last I heard, those were (near enough) the definitions > of algebraic and transcendental. Or are you saying that what I think Ed > is thinking is wrong? Unless you are reading his mind, you can't > possibly tell. Well, this is incorrect. _What_ is incorrect? >See the word named in a meaningful way? Yes, that's in Ed Green's posting, not mine. Take it up with him, don't yell wrong at me. > Actually I think you missed the joke. Never mind, it's not worth > labouring it. Some transcendentals can be calculated. Pi and e to name two. >(Where I mean calculated to mean that you can give a finite algorithm >to calculate an arbitrary amount of digits.) To use your own word, Wrong. All you have calculated is a rational > approximation to the number. That's not the number itself, though it can > approximate it as closely as you would like. The same holds for quite a few algebraic numbers. Of course. >Or do you have some >way to calculate (or name) the roots of, say, x^5 + 5x + 3, and to >distinguish the roots? No. >finite space you can calculate an arbitrary amount of digits. That's not very precise. I think you mean that it is said to be computable if your algorithm can calculate a rational which approximates to it with an arbitrarily small error bound. Note that the thing you _calculate_ is rational. Isn't that what I said above? -- Richard Herring ==== >finite space you can calculate an arbitrary amount of digits. That's not very precise. I think you mean that it is said to be > computable if your algorithm can calculate a rational which > approximates to it with an arbitrarily small error bound. Note that > the thing you _calculate_ is rational. Isn't that what I said above? I think you two are violently agreeing on this point. Still, Dik's is the usual definition, though I think yours is certainly equivalent. Thomas ==== Richard Herring says... >>So most irrational numbers are poor lost souls which not only >>have non-repeating decimal representations, but can't even be >>named in any meaningful way -- they are unknowable... >>Can somebody remind me what this concept is called? Richard Herring responds: >> I think you're thinking of the distinction between algebraic >> (roots of a polynomial) and transcendental (the rest) ? >> I wonder if there are any other possibilities - read on... Dik Winter said Wrong because Ed Green is clearly talking about numbers that are unknowable that can't even be named in any meaningful way. Those descriptions are not true of the numbers pi and e, for example. pi and e are just as knowable as square-root(2). So the algebraic/transcendental distinction doesn't capture what Ed was talking about. >> Some transcendentals can be calculated. Pi and e to name two. >>(Where I mean calculated to mean that you can give a finite algorithm >>to calculate an arbitrary amount of digits.) >> To use your own word, Wrong. All you have calculated is a rational >> approximation to the number. That's not the number itself, though it can >> approximate it as closely as you would like. That's what it *means* to say that a real number is computable. A number like pi or e is computable if and only if we can approximate them as closely as you like. So it really isn't correct to call them unknowable, we can know quite a bit about them. In contrast, there are some numbers that are *not* computable. They cannot be approximated as closely as you like. If a number r is noncomputable, there does not exist an algorithm for computing its digits. So there is a real sense that a noncomputable number is much less knowable than e or pi. >>The same holds for quite a few algebraic numbers. Of course. Then it follows that the algebraic/transcendental distinction is not what Ed was getting at. Algebraic numbers are no less unknowable than computable transcendental numbers. -- Daryl McCullough Ithaca, NY ==== In message , Daryl McCullough >Richard Herring says... So most irrational numbers are poor lost souls which not only >have non-repeating decimal representations, but can't even be >named in any meaningful way -- they are unknowable... >Can somebody remind me what this concept is called? Richard Herring responds: > I think you're thinking of the distinction between algebraic > (roots of a polynomial) and transcendental (the rest) ? > I wonder if there are any other possibilities - read on... Dik Winter said Wrong because Ed Green is clearly talking about >numbers that are unknowable that can't even be named in any >meaningful way. Those descriptions are not true of the numbers >pi and e, for example. pi and e are just as knowable as >square-root(2). So the algebraic/transcendental distinction >doesn't capture what Ed was talking about. clarification. _I_ was referring to Ed's earlier words in the same paragraph: >>This stirs some vague recollection: there are some privileged >>irrational numbers which can be specified with a finite amount of >>information. where I took the finite amount of information to mean coefficients of a [finite] polynomial. If you take the name of a number to be the set of polynomial coefficients which defines it, then clearly the transcendentals, even pi and e, don't have a name in that sense. > Some transcendentals can be calculated. Pi and e to name two. >(Where I mean calculated to mean that you can give a finite algorithm >to calculate an arbitrary amount of digits.) To use your own word, Wrong. All you have calculated is a rational > approximation to the number. That's not the number itself, though it can > approximate it as closely as you would like. That's what it *means* to say that a real number is computable. >A number like pi or e is computable if and only if we can >approximate them as closely as you like. So it really isn't correct >to call them unknowable, we can know quite a bit about them. In contrast, there are some numbers that are *not* computable. They >cannot be approximated as closely as you like. If a number r is >noncomputable, there does not exist an algorithm for computing its >digits. So there is a real sense that a noncomputable number is >much less knowable than e or pi. The same holds for quite a few algebraic numbers. >>Of course. Then it follows that the algebraic/transcendental distinction >is not what Ed was getting at. Algebraic numbers are no >less unknowable than computable transcendental numbers. > But they are less unnameable. -- Richard Herring ==== So you agree that the computable numbers are incomprehensible. Nonsense. Computability can be clearly defined hence the concept is comprehensible, as is the concept of a computable number. The square root of two is computable. I know an algorithm that will give you the n-th digit of the decimal expansion for any n. The sqare root of two is totally comprehnsible and easily constructed with a compass and straightedge. Bob Kolker <2a0cceff.0312101407.2225db91@posting.google.com> ==== In message , Robert J. And indeed, the set of computable numbers is countable. There is, however, >a beautiful problem when you try to apply a Cantor like argument to a >(complete) list of computable numbers. This would tell that you get a >new computable number (you just computed it) that is not on the list. >The problem here is that a complete list of computable numbers is itself >not computable. >> So you agree that the computable numbers are incomprehensible. Nonsense. Computability can be clearly defined hence the concept is >comprehensible, as is the concept of a computable number. Comprehensible has more than one meaning: http://www.ccel.org/creeds/athanasian.creed.html The computable numbers cannot be _comprehended_ within a computable list. >The square root of two is computable. I know an algorithm that will >give you the n-th digit of the decimal expansion for any n. That's some puny rational approximation. > The sqare root of two is totally comprehnsible and easily constructed >with a compass and straightedge. > Ah. That's more like it. -- Richard Herring ==== > In message , Robert J. >>The square root of two is computable. I know an algorithm that will >>give you the n-th digit of the decimal expansion for any n. That's some puny rational approximation. No. That's a set of puny rational approximations, one for each n. Take a course in real analysis someday and see what a real number is defined to be in the Cauchy model. John Briggs <3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de> <1Ff9oJOLyx1$Ew6T@baesystems.com> <3FD83D0A.1BF67E01@ix.urz.uni-heidelberg.de> ==== In message <3FD83D0A.1BF67E01@ix.urz.uni-heidelberg.de>, Bjoern >> In message <3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de>, Bjoern [snip] > sqrt(2) is the > result of turing machine(x), where x is some integer, probably > under a million with any crude mapping technique. >>That's incomprehensible. Could you rephrase this, please? >> He means that any Turing machine can be represented by a tape fed to a >> universal Turing machine, and the contents of that tape can be >> represented by some number x. How can the content of the whole tape be represented with a single >number x? John Briggs answered this. And where does he get the probably under a million from? Heaven knows. [snip] > Non computable numbers is not a proof that irrationals exist, IMO, > that no halting function exists does not clearly define what the halting > number is, it states it is impossible. There is no gap on the number > line from non computable numbers. >>That's incomprehensible again. >> I think he's trying to argue that because there exist TMs for which we >> can't determine to which of the sets halts or does not halt they >> belong, the does not halt set must be empty. *scratches head* Does anyone understand this logic? > That's my point. Logic it is not. -- Richard Herring ==== that is self referential, can anyone give me an actual number that isn't computable? Herc > You would think that it was impossible to specify a number that wasn't computable - the act of specifying the number would seem to require specifying the algorithmic process which generates it. Well, you would be wrong. Here is exactly what you asked for: http://mathworld.wolfram.com/ChaitinsConstant.html ==== > It seems odd Information Theory is wiped out of existence from a single > 'proof' > that is self referential, can anyone give me an actual number that isn't > computable? Herc > You would think that it was impossible to specify a number that wasn't > computable - the act of specifying the number would seem to require > specifying the algorithmic process which generates it. Well, you would be wrong. Here is exactly what you asked for: http://mathworld.wolfram.com/ChaitinsConstant.html Is this completely fair? If there is a Turing Halting problem that is undecidable within our own system of reasoning? Is it then still a proper defined number? Lucas ==== > You would think that it was impossible to specify a number that wasn't > computable - the act of specifying the number would seem to require > specifying the algorithmic process which generates it. Well, you would be wrong. Here is exactly what you asked for: http://mathworld.wolfram.com/ChaitinsConstant.html Beautiful! ==== In sci.physics, Uncle Al <3FD74A11.E8852E20@hate.spam.net>: What is an irrational number? Can you count to it? >> Can you pin point it? There is no such thing. sqrt(2) http://mathforum.org/library/drmath/view/57117.html > http://medialab.dyndns.org/bignum/ > All numbers are the result of computable functions, sqrt(2) is the >> result of turing machine(x), where x is some integer, probably >> under a million with any crude mapping technique. The concepts irrational and transcendental are rigorousy defined > and rigorously provable. Do some reading. If you think sqrt(2) is > not irrational, supply integers for the ratio. Note that sqrt(2) has > been taken to ridiculous lengths - though not to the 1.241 trillion > places of pi by Yasumasa Kanada - to do statistical correlations. Of course, any such pair of integers has a minor problem. Assume sqrt(2) = p/q, then 2 = p^2/q^2. Assume gcd(p,q) = 1, p must be even. Writing p^2/q^2 = 4 r^2/q^2 and flipping, we find 2 = q^2/r^2, and therefore q must be even as well. Whoops. Or one could pull a James Harris and let p and q be algebraic integers. Since sqrt(2) is an algebraic integer (x^2 - 2 = 0 is irreducible and monic), ta-daah! Except that that doesn't do much for sqrt(2)'s (ir)rationality. ftp://metalab.unc.edu/pub/docs/books/gutenberg/etext94/2sqrt10a.zip > sqrt(2) to 5 million decimal places > http://antwrp.gsfc.nasa.gov//htmltest/gifcity/sqrt2.10mil > sqrt(2) to 10 million decimal places > http://www.sciencenews.org/20021214/mathtrek.asp sqrt(A): 1) Choose a rough approximation G of sqrt(A). 2) Divide A by G and then average the quotient with G, G* = ((A/G)+G)/2 3) If G* is sufficiently accurate, stop. Otherwise, let G = G* and > return to step 2. The number of correct decimal places roughly doubles with each > repetition of step 2. That's a cute one. I'll have to remember that technique. :-) If G = sqrt(A) + e, then G* = ((A/(sqrt(A) + e)+sqrt(A)+e))/2 = sqrt(A)/2 - e/2 + e^2/(sqrt(A)) + ... + sqrt(A)/2 + e/2 = sqrt(A) + e^2/(sqrt(A)) + ... and Al's right again, damnit. :-) > That only leaves two types of numbers left that qualify for irrational, >> non computable and random numbers. Jesus you do whine. The digit occurances of sqrt(2) are demonstrated > to be random (with one exception) with a huge battery of statistical > qualifiers including string poker hands (no flushes), but fractionally > less so than those of pi (also with the obvious excpetion). [snip] > -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > That's a cute one. I'll have to remember that technique. :-) It's a special case of something more general: Newton's Method. If G is an approximate solution of f(x)=0, you can often get a better approximation with G* = G - f(G)/f'(G) In this case, f(x) = x^2 - A and f'(x) = 2x, so it reduces to the given formula. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W ==== > For example, I might not know whether the least upper bound > of some complicated set is computable, but I do know that, > if the set is bounded, the LUB exists. Under your scheme, > I might not know whether the LUB exists, but I would know > that, if it existed, it would be computable. You might > prefer to know the second and not the first, but generally > speaking, knowing the first and not the second is more > useful in real analysis. > Its not more useful, its a superimposed mutually sustaining > impression that mathematics lies outside the realMs of > computability. You say it's not more useful; I say it is more useful. How should we decide this point? I could give you many examples where the existence of the LUB is important to the conclusion drawn _in real analysis_. Can you do the same with the computability property? The second sentence here is difficult to understand. I think you're saying the LUB property is part of the whole complex that gives the (false) impression that mathematics extends beyond the computable. Why is this impression false? You haven't gone into your reasons for asserting this yet, unless you're counting That would mean the reals are countable as a reason. I'm going to assume you agree with me that this would be a bad reason. > Your proof jumps to the fact of a contradiction, what is this > contradiction? At worst, my proof _derives_ a contradiction. It does not pull it out of thin air. It's true that I glossed over a some the details, but they were details you've been over many times. I'm willing to fill them in, if you really want me to. However, several people have pointed out that you do not need a contradiction for this result. In fact, Cantor did not use contradiction. In the latest version, the one you most recently snipped, there is no contradiction, just a demonstration that every list misses at least one real number. This is a contradiction only to the assumption that |N| = |R|. > Is it self referencing? YES > Does the formulation of extra numbers occur purely depending on > the original formulation of all numbers? YES I don't know what you mean by self-referencing, but I suspect that if I did, I would disagree. There is no spooky Goedel-style mixing of levels going on here. OK, the not-found number Cantor(L) is a function of the list, L, it is not found in. I don't see how it could be otherwise. Every real is in some list or other; there is no real which is missing from every list. > Does this contradiction rely on specifying the new number is > *different* from the original forumlation at finite set points? > YES > Does the extra number reinsert into the original list to > demontrate the imcompatibility? YES NO. The extra number is not re-inserted into the original list. If you think I did this, show me where. > The extra number is simply this : > Given a number on a list, change it and put it back in the list > without changing the list. The most generous interpretation of this I can make is that you meant to say something else. (1) The extra number is not one of the numbers on the list (hence extra). (2) Changing a number on the list and putting it back in the list must change the list. (3) I haven't done anything like this. I suggest you try again. This time say what you mean. > Does the diagonal transform form a well defined number? > NO -> computables = reals > YES -> computables C reals Well, it would be handy for me to agree with your implication here. The Cantor diagonal number is very well defined, for a well-defined list, so it would follow that the computables are a proper subset of the reals. However, you have not made any argument for the pair of implications. I think I saw you _assert_ in a separate post that uncomputable nnumbers are not reals, but you don't give a reason for that beyond allowing you to conclude what you want to conclude, that |N| = |R|. > There is only 1 infinity type, the answer is NO, here's a similar > exposition of diagonalisation. Apart from anything else in this thread, there is more than one infinity type. You would have to re-work set theory to get rid of that conclusion, if even that would do it. Questions of computability just do not enter into it. For any set A, |A| < |P(A)|, where P(A) is the powerset of A. Original Format > alt.paranormal, alt.sci.physics > I don't see anything new here. Jim Burns ==== ---------------------------------------------------------------------------- ------ > For example, I might not know whether the least upper bound > of some complicated set is computable, but I do know that, > if the set is bounded, the LUB exists. Under your scheme, > I might not know whether the LUB exists, but I would know > that, if it existed, it would be computable. You might > prefer to know the second and not the first, but generally > speaking, knowing the first and not the second is more > useful in real analysis. > Its not more useful, its a superimposed mutually sustaining > impression that mathematics lies outside the realMs of > computability. You say it's not more useful; I say it is more useful. How > should we decide this point? I could give you many examples > where the existence of the LUB is important to the conclusion > drawn _in real analysis_. Can you do the same with the > computability property? do any examples apply outside of number theory? what is the contribution to technology of uncountability? anything? The second sentence here is difficult to understand. I think > you're saying the LUB property is part of the whole complex > that gives the (false) impression that mathematics extends > beyond the computable. Why is this impression false? You haven't gone into your reasons > for asserting this yet, unless you're counting That would > mean the reals are countable as a reason. I'm going to > assume you agree with me that this would be a bad reason. Your proof jumps to the fact of a contradiction, what is this > contradiction? At worst, my proof _derives_ a contradiction. It does not pull > it out of thin air. It's true that I glossed over a some > the details, but they were details you've been over many times. > I'm willing to fill them in, if you really want me to. However, several people have pointed out that you do not need a > contradiction for this result. In fact, Cantor did not use > contradiction. In the latest version, the one you most recently > snipped, there is no contradiction, just a demonstration that > every list misses at least one real number. This is a contradiction > only to the assumption that |N| = |R|. misses an infinitely long number, at a digit that it specifies itself must be contradictory. the number is defined in terms of itself. you treat an infinitely long sequence like a simple object you can manipulate. where is the contradictory bit? can you point it out? why cannot the number be computed to any precision specified? its a simple trick of enumeration, the number is part of the list and poorly defined at its own referential digit, it doesn't lie outside the list. why does logic confine itself to non self reference yet mathematics yields it. there are simple limitations to computer programs when they self reference, its a practical base of limitations. natural language and set theory both get around self referential paradoxes by specifying before hand their use in the domains. limitations to computers are very limited in scope, all other fields of science focus on them, they are trivial. Is it self referencing? YES > Does the formulation of extra numbers occur purely depending on > the original formulation of all numbers? YES I don't know what you mean by self-referencing, but I suspect that > if I did, I would disagree. There is no spooky Goedel-style > mixing of levels going on here. OK, the not-found number Cantor(L) is a function of the list, L, > it is not found in. I don't see how it could be otherwise. > Every real is in some list or other; there is no real which > is missing from every list. Does this contradiction rely on specifying the new number is > *different* from the original forumlation at finite set points? > YES > Does the extra number reinsert into the original list to > demontrate the imcompatibility? YES NO. The extra number is not re-inserted into the original list. > If you think I did this, show me where. right! depends on how you conclude the proof as an assertion or contradiction. The extra number is simply this : > Given a number on a list, change it and put it back in the list > without changing the list. The most generous interpretation of this I can make is that > you meant to say something else. (1) The extra number is not one of > the numbers on the list (hence extra). (2) Changing a number > on the list and putting it back in the list must change the > list. (3) I haven't done anything like this. I suggest you try again. This time say what you mean. Would it help to tally the list of reals as a single number? DIGIT 1 2 3 4 5 6 __________________________ UTM(1) 4 3 6 4 2 4 UTM(2) 7 4 3 4 3 2 UTM(3) 0 1 0 1 1 1 UTM(4) 1 2 2 2 2 2 UTM(5) 7 7 7 7 7 7 Working along the diagonal in this fashion : 1 2 4 7 3 5 6 that would give the single real number : 437640.... Does the diagonal transform form a well defined number? > NO -> computables = reals > YES -> computables C reals Well, it would be handy for me to agree with your implication > here. The Cantor diagonal number is very well defined, for a > well-defined list, so it would follow that the computables are > a proper subset of the reals. However, you have not made any argument for the pair > of implications. I think I saw you _assert_ in a separate post > that uncomputable nnumbers are not reals, but you don't give > a reason for that beyond allowing you to conclude what you want to > conclude, that |N| = |R|. There is only 1 infinity type, the answer is NO, here's a similar > exposition of diagonalisation. Apart from anything else in this thread, there is more than > one infinity type. You would have to re-work set theory to get > rid of that conclusion, if even that would do it. Questions > of computability just do not enter into it. For any set A, |A| < |P(A)|, where P(A) is the powerset of A. > Original Format > alt.paranormal, alt.sci.physics > I don't see anything new here. > Its a diagonalisation proof of Godels theorom by Roger Penrose, Steven Hawkings assistant. Probably accepted as a correct proof by mainstream mathematicians to this day, yet the definition of the diagonal function is obviously flawed and not well defined. Read the capitalisation part of the post and tell me if its a valid proof. When a proposed theorem passes certain credulity test it is automatically given the label of well formed. A second examination will bring down the whole encapsulation to naive eyes, rather than expose the single theorem. Herc creáduáliáty ( P ) Pronunciation Key (kr-dl-t, -dy-) n. A disposition to believe too readily. ==== |-|erc says... >misses an infinitely long number, at a digit that it specifies itself must >be contradictory. the number is defined in terms of itself. No, that's not true. Cantor's method works like this: if you give me a list of real numbers, then I can give you a real number that does not appear anywhere on the list. It always works, which implies that there cannot be a list that contains every real number. For example. Consider the following list of real numbers (in decimal notation): 0.1 0.01 0.001 0.0001 0.00001 ... Cantor's method produces a real number that is not on the list, namely 0.222... = 2/9 This is a perfectly ordinary real number. It isn't defined in terms of itself---it is defined in terms of the above list of real numbers. Given *any* list of numbers, Cantor's method will allow you to construct a real number that doesn't appear anywhere on the list. The meaning of construct a real number is to give a method for computing its decimal places. Which number Cantor's method returns depends on which list you provide. Give me a different listing of reals, and I will give you a different real that is not on that list. So *every* list misses some real number, although different lists miss different reals. So what Cantor's method does is to construct a function Diagonal from lists of reals to reals with the property: forall f: list of reals, Diagonal(f) is not on the list f which logically implies forall f: list of reals, exists r: real, r is not on list f which logically implies there does not exist a list containing all reals Notice that if we replace real by rational, the result is not true. There *does* exist a list of all possible rational numbers: 0/1, -1/1, 0/2, 1/1, -2/1, -1/2, 0/3, 1/2, 2/1, -3/1, -2/2, -1/3, 0/4, 1/3, 2/2, 3/1, -4/1, -3/2, -2/3, -1/4, 0/5, 1/4, 2/3, 3/2, 4/1, ... This list contains every possible rational number. -- Daryl McCullough Ithaca, NY ==== -- www.StealthHostiing.com You rule Truman. http://tinyurl.com/iky4 Hey Trueman...love the show. YOU ARE the Truman I heard him. Very spooky! >Is the truman living in Townsville? I've been hearing stuff, yeah. Webmasters help the TRUEman by joining www.theBanner.net Current:1 Goal:1000 ---------------------------------------------------------------------------- ------ > |-|erc says... misses an infinitely long number, at a digit that it specifies itself must >be contradictory. the number is defined in terms of itself. No, that's not true. Cantor's method works like this: > if you give me a ***infinite*** list of real numbers, then I can give note **** > you a real number that does not appear anywhere on the list. > It always works, which implies that there cannot be a list > that contains every real number. It doesn't work on every list. It wont work on a list that is defined to be complete because then it self references. For example. Consider the following list of > real numbers (in decimal notation): 0.1 > 0.01 > 0.001 > 0.0001 > 0.00001 > ... Cantor's method produces a real number that is not on the list, > namely 0.222... = 2/9 This is a perfectly ordinary real number. It isn't defined > in terms of itself---it is defined in terms of the above > list of real numbers. Given *any* list of numbers, Cantor's > method will allow you to construct a real number that doesn't > appear anywhere on the list. The meaning of construct a real > number is to give a method for computing its decimal places. Which number Cantor's method returns depends on which list > you provide. Give me a different listing of reals, and I > will give you a different real that is not on that list. > So *every* list misses some real number, although > different lists miss different reals. So what Cantor's method does is to construct > a function Diagonal from lists of reals to reals with the > property: forall f: list of reals, Diagonal(f) is not on the list f which logically implies forall f: list of reals, > exists r: real, > r is not on list f which logically implies there does not exist a list containing all reals Notice that if we replace real by rational, the > result is not true. There *does* exist a list of all > possible rational numbers: 0/1, > -1/1, 0/2, 1/1, > -2/1, -1/2, 0/3, 1/2, 2/1, > -3/1, -2/2, -1/3, 0/4, 1/3, 2/2, 3/1, > -4/1, -3/2, -2/3, -1/4, 0/5, 1/4, 2/3, 3/2, 4/1, > ... This list contains every possible rational number. > DIGIT 1 2 3 4 5 6 __________________________ UTM(1) 4 3 6 4 2 4 UTM(2) 7 4 3 4 3 2 UTM(3) 0 1 0 1 1 1 UTM(4) 1 2 2 2 2 2 UTM(5) 7 7 7 7 7 7 Here is the procedure : 1/ Construct a UTM and an array of processors to evaluate each applied to increasing integers. 2/ Eventually you will construct a TM ~ UTM(7777) say ~ that performs the diagonal transform calculation. The output of UTM(7777) starts by emulating UTM(1), get the 1st digit 4, adds 1 = 5. The next output of UTM(7777) is UTM(2), gets the 2nd digit 4, adds 1 = 5, Then the next output of UTM(7777) is UTM(3), gets the 3rd digit 0, adds 1 = 1. So far UTM(7777) = 5 5 1 UTM(7777) seems clearly defined, but when it reaches row 7777 it has to emulate that function just like it emulated UTM(1) and UTM(2). UTM(7777) emulates UTM(7777), which again emulates UTM(7777). The digit at (7777,7777) never gets calculated, it is not a properly defined TM or a properly defined number. When you define a number you are defining the algorithm to exploit it. You *know* that that algorithm will appear on the list of algorithms, all permutations of computer programs are there, so you *know* you are defining the number to reference itself. There are 2 approaches to diagonalisation : 1 The number does not appear on the list. -> The diagonal forms a new number -> The new number does not appear on the list (nothing new here) 2 The number appears on the list -> The diagonal forms a self referencing number -> The contradiciton refutes A The diagonal was on the list -> the list is not complete -> sci.math thinks Cantors proof works B The diagonal is well formed -> the list is complete for well formed numbers -> single oo Herc ==== |-|erc says... >> |-|erc says... >>misses an infinitely long number, at a digit that it specifies itself must >>be contradictory. the number is defined in terms of itself. >> No, that's not true. Cantor's method works like this: >> if you give me a ***infinite*** list of real numbers, then I can give note **** >> you a real number that does not appear anywhere on the list. >> It always works, which implies that there cannot be a list >> that contains every real number. >It doesn't work on every list. Yes, it does. >It wont work on a list that is defined to >be complete because then it self references. How do you create a list that is defined to be complete? A list of decimal numbers is by definition a function which, given a natural number n, returns a decimal expansion for a real. In turn, a decimal expansion is a function which given a natural number i, returns an integer between 0 and 9. So, how can there be a list of decimal numbers that is defined to be complete? > DIGIT 1 2 3 4 5 6 >__________________________ >UTM(1) 4 3 6 4 2 4 >UTM(2) 7 4 3 4 3 2 >UTM(3) 0 1 0 1 1 1 >UTM(4) 1 2 2 2 2 2 >UTM(5) 7 7 7 7 7 7 >Here is the procedure : 1/ Construct a UTM and an array of processors to evaluate each applied to >increasing integers. >2/ Eventually you will construct a TM ~ UTM(7777) say ~ that performs >the diagonal transform calculation. No, there is no such TM. If your list includes only *total* TMs (that is, they halt on all possible inputs), then there cannot be a UTM that performs the diagonal calculation. >The output of UTM(7777) starts by emulating UTM(1), get the 1st >digit 4, adds 1 = 5. >The next output of UTM(7777) is UTM(2), gets the 2nd digit 4, adds 1 = 5, >Then the next output of UTM(7777) is UTM(3), gets the 3rd digit 0, adds 1 = 1. So far UTM(7777) = 5 5 1 >UTM(7777) seems clearly defined, but when it reaches row 7777 it >has to emulate that function just like it emulated UTM(1) and UTM(2). >UTM(7777) emulates UTM(7777), which >again emulates UTM(7777). The digit at (7777,7777) never gets calculated, >it is not a properly defined TM or a properly defined number. That's right. If you broaden your collection of Turing Machines to include both total functions and partial functions (functions that may fail to halt on some input), then you *can* enumerate all possible such Turing Machines. Then, as you say, there will be a number (say 7777) such that digit 7777 if UTM(7777) is undefined. >When you define a number you are defining the algorithm to exploit it. >You *know* that that algorithm will appear on the list of algorithms, >all permutations of computer programs are there, so you *know* you >are defining the number to reference itself. If you want to include nonhalting Turing Machines in your list, then you have to modify the diagonalization procedure slightly. Given an infinite list UTM(i) of turing machines, you define a decimal expansion r according to the rule: If digit number n produced by UTM(n) is undefined, then make digit n of r be 0. If digit number n produced by UTM(n) is 9, then make digit n of r be 0. Otherwise, make digit n of r be equal to 1 + digit number n of UTM(n). It is still the case that r cannot be equal to the output of any Turing machine on the list. >There are 2 approaches to diagonalisation : 1 The number does not appear on the list. >-> The diagonal forms a new number >-> The new number does not appear on the list >(nothing new here) 2 The number appears on the list >-> The diagonal forms a self referencing number That isn't possible. -- Daryl McCullough Ithaca, NY ==== > where is the contradictory bit? can you point it out? why cannot the number > be computed to any precision specified? its a simple trick of enumeration, > the number is part of the list and poorly defined at its own referential digit, > it doesn't lie outside the list. We're talking about a list of computable numbers. For every number on the list [at least] _one_ Turing Machine that can correctly compute _every_ digit in the number. That's what it means for a number to be computable. There are, of course, other equivalent definitions. If you choose to relax your definition of computability to say that a number is computable if no matter how many digits you want there is at least one Turing Machine that computes those digits correctly then you get a different set. It's obvious that all real numbers are computable under that definition. Heck, under that definition, all real numbers are computable by finite state machines. You want 100 digits, pick a machine that spits out 100 digits and halts. You want 200 digits, pick a machine that spits out 200 digits and halts. No matter how many digits you want, there's always a machine that will give them to you. But under that definition of computability the fact that there are only countably many finite state machines no longer implies that there are only countably many real numbers. Because now the correspondence is between real numbers and equivalence classes of [Cauchy-like] sequences of finite state machines and not between real numbers and Turing machines. The cardinality of the set of equivalence classes of sequences of finite state machines is greater than the cardinality of the set of all finite state machines. John Briggs ==== > What is an irrational number? Can you count to it? > Can you pin point it? There is no such thing. Really? Hmmm...let's try creating an isosceles right triangle with each side 1. What do you do with the length of the hypotenuse if irrationals don't exist. In fact, for *ANY* right triangle, if any two sides are of integer length, the remaining length must be either another integer or else (usually) an irrational number! Good Luck with your rewriting of Geometry. Jonathan Hoyle Older and Wiser than Age 11 ==== -- www.StealthHostiing.com You rule Truman. http://tinyurl.com/iky4 Hey Trueman...love the show. YOU ARE the Truman I heard him. Very spooky! >Is the truman living in Townsville? I've been hearing stuff, yeah. Webmasters help the TRUEman by joining www.theBanner.net Current:1 Goal:1000 ---------------------------------------------------------------------------- ------ > What is an irrational number? Can you count to it? > Can you pin point it? There is no such thing. Really? Hmmm...let's try creating an isosceles right triangle with > each side 1. What do you do with the length of the hypotenuse if > irrationals don't exist. In fact, for *ANY* right triangle, if any two sides are of integer > length, the remaining length must be either another integer or else > (usually) an irrational number! Good Luck with your rewriting of Geometry. The answer would be sqrt(1^2 + 1^2) = sqrt(2). That is a terminating, consise and usable representation of the solution. Herc well you can keep rationals separate to functions if you like, a division for the division_is_special_computation crowd. ==== > The answer would be sqrt(1^2 + 1^2) = sqrt(2). That is a terminating, consise and usable representation of the solution. Huh??? How does that change the fact that sqrt(2) is irrational? You claimed that irrationals did not exist...now you seem to be agreeing that sqrt(2) exists. So, are you saying that sqrt(2) is rational? Or are you admitting that your original statement was wrong? Jonathan Hoyle Educating adults with 11 year old child-like minds ==== Jonathan Hoyle > Educating adults with 11 year old child-like minds You time would be better spent in vertically urinating up an elongated fibrerous fastening device. Bob Kolker ==== > You time would be better spent in vertically urinating up an elongated > fibrerous fastening device. No doubt. It reminds me of the adage to never argue with a fool, as people might not recognize which one the fool is. Ah well, my futile attempts to bring intelligence to our adolescent impersonator. ==== You are confused because you believe the fact that a TM can enumerate all the algebraic numbers contradicts the uncountability of the reals. Algebraic numbers are countable, hence there is no contradiction. James ==== ---------------------------------------------------------------------------- ------ > ----------------------------------------------------------------------------- ----- >To extend, basically I'm reopening a 75 post thread from several months ago >with Daryl McCollough (I recognise Jim Burns and BruceS my skeptic >counterparts too). Sorry, I don't recall what it was about. I think the Cantor diagonalisation across UTM(Z) is more a concern anyway, > but the thread is here : http://tinyurl.com/yo9w I said reals are countable by UTM(Z) You said that it misses non computable numbers, like r = Halt(1) + 1/10 Halt(2) + 1/100 Halt(3)... > since Halt(n) is unknown for all n. i.e. r is not computable and will not be listed in UTM(Z). I tried to argue r is just not known, you tried to show r is not computable at all (standard view), > now I am now suggesting r it not existent! so Computables = Reals. > If r is the sequence of Halt values along all indexed functions, then parts of r are unknown. We can find out more values of r as we go along but it will never be complete, some computer programs are impossible to test! This make r a loci of numbers not a number itself. Herc ==== > I found this on a web search, the result was stated without the > details of derivation: gravitational field around a central mass m, the eccentricity (e) is > given by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the total > energy (kinetic plus potential), h = rv_t is the angular momentum, v > is the total speed, v_t is the tangential component of the speed, and > r is the radial distance from the center of the mass. (Note v_t is v > subscript t). > I'd suggest you reference Becker's book Introduction to Theoretical > Mechanics, Section 10-5 in which he addresses Equation of the Orbit > by the Energy Method. While the equations of the derivation are a > bit too much to post on Usenet, essentially Becker begins with > expressions for the total energy and angular-momentum : T + V = W > mr^2*Theta(dot) = J After a bit of integration and routine substitutions, Becker imports a > previously derived relationship derived earlier in chapter 10: r = 1/((1/ep) - (1/p)Cos(Theta)) and obtains: e = sqrt((2WJ^2)/mk^2) + 1) This is a remarkably similar to form as the equation that you posted, > and by expanding to equivalent terms as that you posted, I suspect > that they will or should turn out to be equivalent. I wasn't > sufficiently ambitious to perform this final step. Harry C. p.s. My copy of Becker's text is dated 1954 (Lord, how time flies!), > so you may have a difficult time locating a copy. If so, I'd be happy ==== > I found this on a web search, the result was stated without the > details of derivation: gravitational field around a central mass m, the eccentricity (e) is > given by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the total > energy (kinetic plus potential), h = rv_t is the angular momentum, v > is the total speed, v_t is the tangential component of the speed, and > r is the radial distance from the center of the mass. (Note v_t is v > subscript t). > Solve Newton's Equations of motion for such orbits and you will see that it comes out of the wash. This is a standard first year classical mechanics problem, dealt with in any worth while text book on the subject Franz Heymann ==== >I found this on a web search, the result was stated without the >details of derivation: gravitational field around a central mass m, the eccentricity (e) is >given by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the total >energy (kinetic plus potential), h = rv_t is the angular momentum, v >is the total speed, v_t is the tangential component of the speed, and >r is the radial distance from the center of the mass. (Note v_t is v >subscript t). > formulas, which with a small bit of work, verify this formula. In that the initial position, p, and velocity, v, of the satellite along with the gravitational constant, G, and the mass of the primary, M. 1 k = - |p x v| [3a] 1 2 1 2 GM k = - |v| - --- [6a] 2 2 |p| k_1 is the rate of area swept out per unit time; it is also your h/2. k_2 is the energy per unit mass of the satellite; it is also your E. derive the formula of the orbit as 1 r = ---------------- [10] b + c cos(@-k_3) where b = GM/(4 k_1^2) and c^2 = b^2 + k_2/(2 k_1^2). I match that formula with the formula in terms of the semimajor axis, a, and eccentricity, e. a(1-e^2) r = ------------ [10a] 1 + e cos(@) Looking at these formulas, c e = - b k_2 = sqrt( 1 + ----------- ) 2 k_1^2 b^2 8 k_2 k_1^2 = sqrt( 1 + ----------- ) (GM)^2 2 E h^2 = sqrt( 1 + ------- ) (GM)^2 which is your equation using m = GM. This implies that if E < 0, the orbit is elliptical and if E > 0, the orbit is hyperbolic. Rob Johnson take out the trash before replying ==== > I found this on a web search, the result was stated without the > details of derivation: gravitational field around a central mass m, the eccentricity (e) is > given by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the total > energy (kinetic plus potential), h = rv_t is the angular momentum, v > is the total speed, v_t is the tangential component of the speed, and > r is the radial distance from the center of the mass. (Note v_t is v > subscript t). > This is quite adequately derived in most theoretical mechanics textbooks (at either the undergrad or grad level). The one I recommend is Herbert Goldstein. It's the most thorough treatment of orbital motion I'm aware of. ==== > Epilog The Interesting Question has already been satisfactorily answered now. > But I thought that some people might enjoy a nice, pertinent passage > (which I only just now discovered) in the excellent _Concrete Mathematics_ > by Graham, Knuth, and Patashnik. On page 481 (2nd ed.), where m is the number of terms of Stirling's > asymptotic expansion of ln(n!) to be used, they say ... if n is fixed and m increases, > the error bound |B_(2m+2)|/((2m+2)(2m+1)n^(2m+1)) decreases to a certain > point and then begins to increase. Therefore the approximation reaches a > point beyond which a sort of uncertainty principle limits the amount by > which n! can be approximated. David You can find an asymptotic series for n! rather than that for ln(n!) in A new derivation of Stirling's Approximation to n!, G. and J. Marsaglia, Am. Math. Monthly, 97,No. 9, 1990. Mixing TeX and * for multiplication: n! = n^n e{-n}sqrt{2pi n}* [1 + 3a_3/n + 3*5a_3/n^2 + 3*5*7a_5/n^3 + 3*5*7*9a_9/n^4+...] where the a's satisfy the recursion, (n+1)a_n = a_{n-1} - 2a_2a_{n-1} - 3a_3a_n-2- ... - (n-1)a_{n-1}a_2 initialized with a_1=1, a_2=1/3. The first few a's are 1,1/3,1/36,-1/270,1/4320,1/17010,-139/5443200 Thus n! = n^n e^{ 1+ c_1/n + c_2/n^2 + c_3/n^3 + ... with the first nine c's: 1/12, 1/288, -139/51840, -571/2488320, 163879/209018880, 5246819/75246796800, -534703531/902961561600, -4483131259/86684309913600, 432261921612371/514904800886784000, (sign pattern +,+,-,-), If S_k = n^n exp{-n}sqrt(2pi n}*[1+c_1/n+...+c_k/n^k] sums the series through k terms, then these examples show the ultimately chaotic behaviour of the asymptotic series, but for which early terms may be quite useful: With n=4, 4!= 24 S_1 = 23.995888... S_2 = 24.000988281... S_3 = 24.000003472778576 S_4 = 23.999982402422244949 ... S_24= 23.99999999997607404482055067828621313524 S_25= 24.00000000002184951216601742396435065363 S_26= 24.00000000002280566142889597320458957977 S_27= 23.99999999997569894109331075218544444326 ... S_65= 24.07902081946226925298335074817255083162 ... S_70= 28.02611844348535334766765116652722021875 S_71= -6.281830892605180810481922429588177923873 ... S_75= -2098.109030254598135895004889287527652778 S_76= -2147.591820869521698236087744269317078347 S_77= 19280.21714886293068959493792076383534356 ... With n=10, 10! = 3628800 S_1 = 3628684.748897... S_2 = 3628809.703606... S_3 = 3628800.054325... S_4 = 3628799.971745... ... S_10 = 3628800.0000000672569... ... S_23 = 3628799.99999999999999936190019... ... S_40 = 3628799.999999999999999999987543840408047... ... S_70 = 3628800.000000000000000000000130930739823... ... S_130= 3628800.000000000038873676252536002626173... ... S_150= 3628800.00029462746487456259981407627620... ... S_160= 3628797.730095008848234664534843871313008... S_161= 3628814.537100312802960926738475176189011... ... S_166= 3629460.850717572983759390565123042386732... ... S_170= 3661581.5010228672... ... S_174= 5416236.0003953219... S_175= -9909389.3801847535217... S_176= -10037108.898098820886747... S_177= 109529312.221996667266697... You may find this interesting, or, if you have Maple or Mathematika or other such package for rational arithmetic, do some examples yourself. George Marsaglia ==== > Suppose I want to studdy a function f : G -> H > where H is the quaternion division algebra, hypercomplex > numbers of dimension 4 and G is a given (commutative) group, > for example a finite one e.g. Z/nZ. Is there a kind of 'representation theory' with values on H ? > What can be said for the 'caracter' of Z/nZ with walue in > H ? I think they are of the form > phi_k : l -> e^{ j*2*pi*k*l/n } for j a unit pure quaternion. > Once you have fixed j, you get a complete set > of caracter, and you can compute Fourier-like decomposition, > with an inversion formula. Is it that simple ? You can certainly define representations of G in a vector space V over H. Such a representation can also be considered as a representation over C of twice the dimension/degree. There are fairly straightforward criteria for determining if a representation over C arises in this way. (To start with, the dimension must be even -- if finite -- and the character must be real.) I don't think one can usefully define the character of a representation over H as a function G -> H, but I could be mistaken. Normally one considers the character of the corresponding representation over C, I think. (I'm not clear what k,l are in your formula above.) -- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 ==== I think that 2^n = sum (nCk) (k ranges from 0 to n) (by nCk I mean n!/(k!(n-k)!)) but can't prove it. does anyone have any hints? also, is there any standard way to represent sigma notation in ascii? the above expression is a little vague as I've written it. cheers, Gareth ==== > I think that 2^n = sum (nCk) (k ranges from 0 to n) (by nCk I mean n!/(k!(n-k)!)) but can't prove it. does anyone have any hints? binomial theorem? > also, is there any standard way to represent sigma notation in ascii? > the above expression is a little vague as I've written it. I would write that in pseudoTeX as 2^n = sum_{k=0}^n {n choose k}. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) ==== Oh please, you NPD afflicted loser! You are such a LIAR and CHEAT and you > are a super-troll and a super-crank! Just look at some of the quotes from you below! This has been going on for > 8 years now and you are about as close to a proof as you were before you > started, you ignoramus! > Whereas James makes valuable contributions to sci.math and math world the detractors and their negativity add nothing, you might try not posting for a month and instead really read and try to understand James's posts. ==== Pauvre andouille ==== > If you do a Google Search on my name James Harris and math, like you > put in James Harris math, you'll currently see a link to > www.crank.net coming up third, and I don't know how many of you know > that site is run by a guy named Erik Max Francis who used to be a > regular sci.math poster. I want you to do that search, look at what he has on me, and please > explain to me exactly why this person is calling me names, like, what > makes me a crackpot. Can *any* of you go there to what he has and tell me what's supposed > to be so bad about what I'm doing? The evidence lies in the history of your posting. That makes it clear that you are a minor, self important, cranklet. Franz Heymann ==== > That makes it clear that you are a minor, self important, cranklet. I am not an expert on cranks, but I think that JSH deserves better than minor cranklet. I think he is a primary figure of modern Usenet crankery. He's no founder of Lawsonomy, but his work is not without a certain kind of merit. I think that his name ranks up there as a pioneer of sorts (but not one without peers, even on Usenet -- still, I've always found him more engaging than the competition). (The follow .sig was chosen merely coincidentally.) -- Jesse F. Hughes My proofs are out there. -- James S. Harris ==== > If you do a Google Search on my name James Harris and math, like you > put in James Harris math, you'll currently see a link to > www.crank.net coming up third, and I don't know how many of you know > that site is run by a guy named Erik Max Francis who used to be a > regular sci.math poster. I want you to do that search, look at what he has on me, and please > explain to me exactly why this person is calling me names, like, what > makes me a crackpot. Can *any* of you go there to what he has and tell me what's supposed > to be so bad about what I'm doing? The evidence lies in the history of your posting. > That makes it clear that you are a minor, self important, cranklet. Franz Heymann The question has to do with the website http://www.crank.net/harris.html and what Erik Max Francis presents there to defend his insult. I'm curious about whether or not anyone can find justification in what he presents. Why don't you try again given that further explanation of what is requested? James Harris <3c65f87.0312120708.bb9dace@posting.google.com> ==== The evidence lies in the history of your posting. >> That makes it clear that you are a minor, self important, cranklet. Franz Heymann > The question has to do with the website http://www.crank.net/harris.html and what Erik Max Francis presents there to defend his insult. That's curious. You are *not* alleging that the total evidence available is insufficient to prove you're a crank? You are only claiming that Francis failed to provide enough evidence that you're a crank? I guess you could check Google, then. Maybe that would shed light on whether Francis's claim is justified regardless of the evidence his site includes. -- Jesse Hughes Surround sound is going to be increasingly important in future offices. -- Microsoft marketing manager displays his keen insight ==== As can be seen by the number of posts in this thread, > and the references to his web site in thousands of other posts, > a computer programmer, who took some data processing classes > at a third rate California college, has become a highly regarded expert > in math, physics, and other science disciplines, and > many people, who pretend to be rational, intelligent, open-minded > scientists (Or at least, pretend to have a scientific mind.), > frequently use this programmer as a major reference. So what-if it is true? What is your point? My point is, that only an fool, would use a web site like that as a reference. And only a sociopath would put up such a web site. There are many web sites that reference non-conventional ideas, in an informative, entertaining way, and the web site owner does not try to get an ego trip by demeaning other folks. Better questions would be: What's the point of posters in science newsgroups using that web site as a reference? What's the point in that web site? (Ego tripping, hate, neurosis, sociopathy, revenge, etc?) -- Tom Potter http://tompotter.us =============== WHO instigates conflict and war for power and wealth? WHO instigated the class wars of the 1900's? WHO is instigating the religious wars of the 2000's? WHO has a well organized propaganda machine? WHO gang attacks all who expose their agenda and methods? Visit my web site, and download the world's best physics tutorial! =============== ==== > As can be seen by the number of posts in this thread, > and the references to his web site in thousands of other posts, > a computer programmer, who took some data processing classes > at a third rate California college, has become a highly regarded expert > in math, physics, and other science disciplines, and > many people, who pretend to be rational, intelligent, open-minded > scientists (Or at least, pretend to have a scientific mind.), > frequently use this programmer as a major reference. So what-if it is true? What is your point? It is true. The point is that Erik Max Francis tells a lot of people what they want to hear, and few seem to care that he lacks the expertise to do a real evaluation of the work he pans. So they ignore his lack of credentials and support him. It's fascinating behavior from a sociological viewpoint. James Harris ==== > It's fascinating behavior from a sociological viewpoint. > James Harris If any behavior in this NG is fascinating from that viewpoint, it is JSH's sociopathy. ==== It's fascinating behavior from a sociological viewpoint. > James Harris If any behavior in this NG is fascinating from that viewpoint, it is > JSH's sociopathy. Everyone has there own diagnosis for him. :) ==== > As can be seen by the number of posts in this thread, > and the references to his web site in thousands of other posts, > a computer programmer, who took some data processing classes > at a third rate California college, has become a highly regarded expert > in math, physics, and other science disciplines, and > many people, who pretend to be rational, intelligent, open-minded > scientists (Or at least, pretend to have a scientific mind.), > frequently use this programmer as a major reference. So what-if it is true? What is your point? It is true. The point is that Erik Max Francis tells a lot of people > what they want to hear, and few seem to care that he lacks the > expertise to do a real evaluation of the work he pans. One does not have to be a chicken to identify a bad egg -- Thomas A. Edison How would you know? Did a scientist or mathematician tell you that? Or, more likely, did you just make that up to make *yourself* feel better? > So they ignore his lack of credentials and support him. Predictable explanation, except for your cryptic reference to 'support'. But here's a better one. Namely, that you are a pompous, arrogant, unsufferable idiot and this observable fact is obvious to everyone but you. > It's fascinating behavior from a sociological viewpoint. There's nothing fascinating about it -- unless you find it fascinating that people run from a deluded paranoiac who flings manure. > James Harris Wacky, isn't it? But, hey, it's just basic math. Yup, yup, yup. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== at 01:38 PM, Bruce Harvey said: >I like the definition of a tensor as the set of all possible >representations of a vector in all possible co-ordinate systems. That's not a definition. Further, it seems to be an alternate way of looking at vectors and to say nothing about tensors of higher rank. >Lets face it, we are dealing with maths developed by guys who could >not define the length of a vector in the complex plane becaue they >missed the lectures on complex numbers. WTF is that supposed to refer to? >If you just assume that space is Euclidian and the universe knows >only the moment of now, then its rulers that shrink and clocks that >slow and the Schwarzchild metric pops out in a few lines. Please show us thos few lines. >All this tensor stuff is just a way of hidding the fudges. ROTF,LMAO! This tensor stuff is just as important to Classical Mechanics as it is to Relativity. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== at 06:58 PM, Boris Schaefer said: > A multilinear function T : V^k --> R is called a k-tensor on V and > the set of all k-tensors, denoted J^k(V), becomes a vector space > (over R), if for S,T in J^k(V) and a in R we define: I assume that he gives a more general definition latter, for a mixed tensor T: V^kxV*^l ->R. >Now my problem is that I cannot reconcile this with my previous >notion of tensors. That's because you have not encountered tensors before, but rather tensor fields on a manifold, and your texts defined them in terms of components in particular coordinate systems. That doesn't really make much sense once you start working with local coordinate systems. >Also, these quantities have to transform in a certain way. What does that mean when you don't have global coordinate systems? >I thought a second-rank tensor was a matrix No. If you pick a particular basis for a vector space then you can express the components of a second rank tensor with respect to that basis. As a Physicist, you should understand that neither 0 nor 32 is a temperature, but that you can express the temperature as either number by picking an appropriate scale, e.g., C versus F. >and an (m x n)-matrix is a linear mapping from V^n --> V^m. Likewise, the matrix just groups component in terms of a particular basis. Change the basis and you need a different matrix. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== >Yes, very much. This got me interested. Do you know any good books >and/or webpages that explain these two different viewpoints on >tensors and their relationship. Unfortunately, there are not many good sites for such a comparison. Wald's book on General Relativity is better than most. Another thing to point out is that most physics texts do tensors on manifolds and the change of basis for the vector space comes from a change of coordinate system at the point of the manifold. That's why all those partial derivatives come up. However, it is quite possible (and useful) to consider bases of the tangent space that are not derived from some coordinate system at the point. If this is done, those partial derivatives are not relevant, even for tensors ona manifold. --Dan Grubb X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== at 11:09 PM, Matthew Senn said: >I've always been strong in math, ave you had any exposure to Mathematics, or only what passes for it in the public stool system? >but matrices and vectors spaces have got me entirely stumped. Matrices? That sounds like you're taking a computation course. Try a strictly Mathematical text like Halmos's Finite Dimensional Vector Space. If you can get through that, it might help to put the computational material in context. OTOH, if your problem is grasping abstractions, then you might have trouble with it. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== >I need help on the following problem. >such that >Kerv={h in H| v(h)=0}={0} and v(H)is dense in H but does not coincide >with it. Is it true that v(H)=H space the answer is no, because you specified but does not coincide with it. Drop one or the other and the answer is that such an -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org <3ik3tvosm69hgkat84jr0dpuur49s8gatf@4ax.com> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== at 02:01 PM, Jonathan Miller said: >> All degrees, credentials, etc., should be by comprehensive >> examinations ALONE. Anything else is anti-educational. >Written by someone with a degree which is a result of a long >paper. But none the less valid. Multiple-choice questions are almost useless in testing comprehension. And *this* is written by someone with a degree that is *not* the result of a long paper ;-) -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== >> All degrees, credentials, etc., should be by comprehensive >> examinations ALONE. Anything else is anti-educational. Perhaps the appropriate method depends on the degree. For example, one could imagine that a degree in music composition could have a final hurdle where the candidate composes a musical composition... A Ph.D. in mathematics could have a final hurdle where the candidate produces original theorems. And so on. X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== >You'll need to convince the graduate faculty that you really do >have the stuff, not such an easy task, unfortunately. Why is that unfortunate? Certainly it should not be oppressively difficult, but neither should it be so easy as to allow unqualified applicants to get in. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== |>You'll need to convince the graduate faculty that you really do |>have the stuff, not such an easy task, unfortunately. | |Why is that unfortunate? Certainly it should not be oppressively |difficult, but neither should it be so easy as to allow unqualified |applicants to get in. Indeed. Some people here may not be aware of how much of a problem it can be when someone enters a graduate program without really being ready to benefit from it the way they want to. I've heard of students who scraped their way through several years of unpromising work and lost a lot of income without getting what they were after. It really isn't just bureaucratic or just for the convenience of the department that the people deciding admissions are cautious. A BA is not an absolute requirement. I know one very well respected mathematician who never got his (although he did attend college). It's also not enough-- some people with one are not ready enough. Keith Ramsay ==== Been too busy this week to read sci.logic, but here's a belated response. >the category of all sets IS AMBIGUOUS. Absolute unambiguity is probably an unattainable ideal in any context. What one needs is to be unambiguous enough for the purposes at hand. It is perfectly reasonable to assume that a professional mathematician will understand the category of all sets accurately enough in this context---i.e., he will not mistakenly latch on to some other thing, or inquire with puzzlement as to what you are talking about. >You may NOT just gloss over these >in the name of professional mathematical context IF the reason WHY you >are alleging that all categories are imitative of the category of sets >is to HELP people achieve a BASIC understanding! Depends on what basic understanding means...more on this below. > : No. I use the word naturally here in the sense that it is found, > : implicitly at least, Nice dodge. If we had to deal with explicit mentions of the word >natural, especially as it relates to category theory, you'd >just be lying here. Ah, the word it in my sentence has an ambiguous antecedent! I meant it to refer to category of sets, not the word `naturally.' That is, I meant to say: When I said that the category of sets is found naturally, I meant that the category of sets is found, implicitly at least, in the mathematical literature. I did not mean to say: When I used the word naturally, I was using the word in the same way that the word naturally is used in the mathematical literature. >Once you say all sets, one >becomes tempted to think of a function as a set of ordered pairs! No, I don't think so. It's true that if one says, everything is a set, then we are tempted to think of a function as a set. But if one does not already think of a function as a set, then the phrase all sets is not going to start making you think that way. >GETTING him familiar with it would REQUIRE >him to notice the difference between a function-set as an object >and a non-function-set being isomorphic to the identity-function-arrow >on the same set. It INVOLVES DETAILS that he'd RIGHTLY refuse to be >BOTHERED with! Familiarity is again a relative thing. I don't have to understand every last detail and every theorem in number theory to have some familiarity with the natural numbers. Similarly, these details about the category of sets aren't really needed to grasp james dolan's point---a superficial familiarity is enough. > : Open up Jacobson's Basic Algebra II and you will see that the category > : of sets is one of the first examples he gives. >There are a LOT of misguided textbooks in the world. In my opinion, this isn't one of them. In any case, whether or not Basic Algebra II is misguided doesn't matter so much, since I'm citing it not as an example of stellar pedagogy, but to illustrate how working mathematicians talk with each other. They may talk with each other in misguided ways, but even if that's the case, you still need to listen if you want to understand how they think. >But in general, if one wants to understand category theory, >understanding why it is a good alternative foundation is >simply indispensable. And here again, we hit the fundamental issue. For a working mathematician, a basic understanding of category theory means an understanding that is sufficient for him to follow the way it is used in topology or algebraic geometry or other subjects that he is more interested in. He can easily spend his entire career using the language of basic category theory and its simplest theorems, without ever even realizing that it can be used as an alternative foundation. I daresay many mathematicians do. I repeat, working mathematicians study category theory *not* because it can be used as a foundation for mathematics, but because its language and basic results are useful for proving/formulating theorems and solving problems that they are interested in. That's the same reason they study calculus or group theory or any other core topic in mathematics. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences ==== tchow@lsa.umich.edu says... Been too busy this week to read sci.logic, but here's a belated response. >the category of all sets IS AMBIGUOUS. Absolute unambiguity is probably an unattainable ideal in any context. >What one needs is to be unambiguous enough for the purposes at hand. Actually, when people talk about some sort of collection of *all* whatevers, it is usually (always?) good enough that it satisfy certain closure properties. We just want to make sure that any whatever that is going to come up will be included. Saying all sets is just a way to avoid saying precisely *which* closure properties. It's closed under whatever operations you're going to be interested in. -- Daryl McCullough Ithaca, NY ==== > A category has two sorts, objects and arrows. (In a pinch, you can do > without the objects.) It has operations of domain and codomain and a > partial operation of composition of arrows, whose domain is definable > in terms of the domain and codomain operations. > That takes about 6 axioms. Actually, only one structure is required: the product, defined as a > partial operation. Most categories DO NOT HAVE products. The product you MEANT to be referring to here is composition. In the context of category theory, product ALREADY means something ELSE. : Maybe around 4 axioms are required. Classical FOL doesn't do partial function application. And if you are going to count axioms then you'll be needing to outlaw conjunction except in special cases. I will be happy to post some actual axiom-sets if you persist in tilting at this windmill. > Among the categories > we single out some very special ones called toposes. > That takes at least 3 more axioms, because [...] You have to be careful about reasoning this way. Axioms > don't add. For example, non-Abelian groups are characterized by 1 > operation (/) and 3 axioms: > (a/c)/(b/c) = a/b > a/(b/b) = a > (a/a)/(b/b) = b/b > or something like that. I don't recall the exact forms, but > it was posted it to sci.math.research several years ago and > you can probably find it there. Abelian groups, however, are characterized by 1 operation > (-) and only 2 axioms: > a-(a-b) = b > a-(b-c) = c-(b-a). More stuff, here, means fewer axioms. You can characterize groups by a single axiom if you are willing to make it complicated enough. More importantly, though, for purposes of THIS discussion, you ALSO CAN *make* the axioms add IF you make them SIMPLE enough. If you'll excuse me for overloading (0-ary identity and 1-ary inverse) i, you can define the group axioms by i*a = a i(a)*a = i (a*b)*c = a*(b*c) Then, you CAN add, to THAT. In particular, for Abelian groups, you can add a*b = b*a . ==== > Consider the events A_n = {X_n > a n log n} where a is a constant. > This is where my problem is....how do I calculate P(A_n)? I understand Show for your distribution that P(X_n > t ) geq (const)/t for all t. ==== >> For the first part, see Feller's An Introduction to Probability >> Theory and Its Applications: Volume 1. In my 3rd edition (1968), >> it's in section X.4 starting on p. 251. Where can I see your 3rd edition? What is your textbook called? > (more text below) Sorry, I meant my 3rd edition copy of Feller's An Introduction to > Probability Theory and Its Applications: Volume 1. Now, is the following proposition true? For any sequence of independent random variables {Z_k}, if > Summation (from k = 1 to infinity) of P( |Z_k| > epsilon) = infinity for > all epsilon > 0, > then Z_k converges almost surely to infinity. (Maybe I can prove this with > borel-cantelli lemma of independent events?) No, it's not true. Constant random variables (i.e., random > variables that are always equal to a specific constant) are > independent of anything, so any deterministic sequence is a sequence > of independent random variables, too, but your proposition fails for > Z_k=-k (where the limit is -infinity instead of infinity) or Z_k=k for > k even and =0 for k odd (where there is no limit). It *is* true that a sequence of nonnegative, independent random > variables {Z_k} satisfying sum_{k=1}^infinity P(Z_k > epsilon) = infinity for all epsilon > 0 will have limsup Z_k = infinity, and this can be > proved by Borel-Cantelli. (Note that we have added the hypothesis of > nonnegativity and changed the conclusion to specify the limsup rather > than the limit.) If true, then I'd like to use Markov's inequality, by saying that > P((X_1 + X_2 + .... + X_n)/nlog(n) > epsilon) <= (1/epsilon) * E((X_1 + > X_2 + .... + X_n)/nlog(n))) or some variation > of markov's inequality, and then sum up these values to possibly get > infinity and our proof would be done. The problem is that the inequality goes the wrong way. To apply the > above result, you need to show that the sum of the probabilities is > infinite, but the inequality will only let you conclude that this sum > is less than or equal to something, and that won't help. If this doesn't work, how can I view the events {X_n > nlog(n)}? I agree > this would be helpful. Well, if n=6, say, then nlog(n) is about 10.75. So, how would you go > about calculating P(X_n > nlog(n)) = P(X_6 > 10.75)? You should be > able to generalize this to an arbitrary n. > To make sure we're on the same page, $P(X_1 > 1 $log(1))$ = 1$ $P(X_2 > 2 $log(2))$ = 1$ $P(X_3 > 3 $log(3))$ = 1/2$ But the results do not follow a simple pattern...to write out some more terms.... $P(X_4 > 4 $log(4))$ = 1/4$ $P(X_5 > 5 $log(5))$ = 1/8$ $P(X_6 > 6 $log(6))$ = 1/8$ $P(X_7 > 7 $log(7))$ = 1/8$ $P(X_8 > 8 $log(8))$ = 1/16$ So I would like to say the sum of these probabilities is infinite as we take the infinite series...but it seems to be a bit messy. In particular, P(X_n > nlog(n)) = P(X_n = 2^a_1) + P(X_n = 2^a_2) + ..., where a_1 is the first integer that is greater than [log(nlog(n))]/log(2) since we want 2^a_1 > nlog(n) where a_1 is the first integer that satisifies that inequality. So this gets a bit messy it seems... > I don't see how I can tell if they occur infinitely often. > Furthermore, if this were true, how would this imply that the limsup > is infinite? Well, okay, I spoke too glibly. It still wouldn't be *quite* enough, > but it would tell you that, infinitely often (i.e., whenever X_n nlog(n)), you'd have to have: (X_1 + ... + X_n)/nlog(n) > X_1/nlog(n) > 1 Now, if you could do the same thing with {X_n > a n log(n)} for every > a > 0 (instead of just for a=1), *then* that would be enough. Do you > see why? > I see why this would be enough. The trouble is proving this now. But if I haven't proved it for a = 1, there's no chance I can get it for arbitrary a. So you must be thinking of an easier method to show that given the events A_n = {X_n > n log(n)}, that the sum_{n=1}^{infty} P(A_n) = infty. > And which Borel-Cantelli lemma would you use? Which do you have? The two I know about are the one that applies when > the sum of the probabilities is infinite and the other that applies > when the sum of the probabilities is finite. So, once you've > determined whether the sum of the probabilities is infinite or finite, > you'll know. > (I had figured this part out before you posted) > -- > Kevin Cancel-Lock: sha1:P2d0jr69ANVq7W3pvQWQpG5jfcY= ==== In particular, P(X_n > nlog(n)) = P(X_n = 2^a_1) + P(X_n = 2^a_2) + ..., > where > a_1 is the first integer that is greater than [log(nlog(n))]/log(2) since > we want 2^a_1 > nlog(n) where a_1 is the first integer that satisifies that > inequality. > So this gets a bit messy it seems... assume you meant a_2=a_1+1, and so on. Let's introduce some standard notation here: for a real number c, the quantity [c] is the greatest integer that's less than or equal to c. So, [3]=3 and [3.6]=3. So, for any real constant c, you have: P(X_n > c) = P(X_n = 2^{[log(c)/log(2)]+1}) + P(X_n = 2^{[log(c)/log(2)]+2}) + ... c=nlog(n). If you work out that right-hand side, it should be a geometric series that sums to a pretty nice function of [log(c)/log(2)]. Now, since you know that log(c)/log(2) <= [log(c)/log(2)] < log(c)/log(2)+1, you can bound that nice function of [log(c)/log(2)] above and below by even nicer (after simplification) functions of c that don't involve this new [...] notation. Finally, if you substitute in c=n log n, you get a value for P(X_n > n log n) that's a pretty nice function of [log(nlog(n))/log(2)] and is bounded above and below by very nice functions of n. If you look at the infinite sum of those probabilities over all n, you'll be able to bound it above and below by infinite sums that involve only n (without the [...] notation), and you can use the convergence or divergence of those sums to show whether or not your probabilities sum to infinity. -- Kevin ==== In particular, P(X_n > nlog(n)) = P(X_n = 2^a_1) + P(X_n = 2^a_2) + ..., > where > a_1 is the first integer that is greater than [log(nlog(n))]/log(2) since > we want 2^a_1 > nlog(n) where a_1 is the first integer that satisifies that > inequality. > So this gets a bit messy it seems... assume you meant a_2=a_1+1, and so on. Let's introduce some standard notation here: for a real number c, the > quantity [c] is the greatest integer that's less than or equal to c. > So, [3]=3 and [3.6]=3. So, for any real constant c, you have: P(X_n > c) = P(X_n = 2^{[log(c)/log(2)]+1}) > + P(X_n = 2^{[log(c)/log(2)]+2}) + ... c=nlog(n). If you work out that right-hand side, it should be a geometric series > that sums to a pretty nice function of [log(c)/log(2)]. Now, since > you know that log(c)/log(2) <= [log(c)/log(2)] < log(c)/log(2)+1, you > can bound that nice function of [log(c)/log(2)] above and below by > even nicer (after simplification) functions of c that don't involve > this new [...] notation. Finally, if you substitute in c=n log n, you get a value for P(X_n > n > log n) that's a pretty nice function of [log(nlog(n))/log(2)] and is > bounded above and below by very nice functions of n. If you look at > the infinite sum of those probabilities over all n, you'll be able to > bound it above and below by infinite sums that involve only n (without > the [...] notation), and you can use the convergence or divergence > of those sums to show whether or not your probabilities sum to > infinity. > Dear Kevin, Ok, I am getting closer and closer... By the way in Feller where you mention X.4 for the first original problem that we haven't discussed, it is proven that P{ |S_n/nlog(n) - 1| > epsilon } ---> 0 which is equivalent to saying S_n/nlog(n) converges in probability to 1. However, in Feller, the log is base 2. Is it henceforth equivalent to say that S_n/nlog(n) converges in probability to log(2) where log is base e? My problem originally was to Show that (S_n)/(n*log(n)) converges in probability to log(2), which is different from what Feller says in X.4 edition 3. (I have edition 2, but I believe edition 3 is just a reprint). Robert > -- > Kevin ==== > In particular, P(X_n > nlog(n)) = P(X_n = 2^a_1) + P(X_n = 2^a_2) + ..., > where > a_1 is the first integer that is greater than [log(nlog(n))]/log(2) > since > we want 2^a_1 > nlog(n) where a_1 is the first integer that satisifies > that > inequality. > So this gets a bit messy it seems... assume you meant a_2=a_1+1, and so on. Let's introduce some standard notation here: for a real number c, the > quantity [c] is the greatest integer that's less than or equal to c. > So, [3]=3 and [3.6]=3. So, for any real constant c, you have: P(X_n > c) = P(X_n = 2^{[log(c)/log(2)]+1}) > + P(X_n = 2^{[log(c)/log(2)]+2}) + ... c=nlog(n). If you work out that right-hand side, it should be a geometric series > that sums to a pretty nice function of [log(c)/log(2)]. Now, since > you know that log(c)/log(2) <= [log(c)/log(2)] < log(c)/log(2)+1, you > can bound that nice function of [log(c)/log(2)] above and below by > even nicer (after simplification) functions of c that don't involve > this new [...] notation. Finally, if you substitute in c=n log n, you get a value for P(X_n > n > log n) that's a pretty nice function of [log(nlog(n))/log(2)] and is > bounded above and below by very nice functions of n. If you look at > the infinite sum of those probabilities over all n, you'll be able to > bound it above and below by infinite sums that involve only n (without > the [...] notation), and you can use the convergence or divergence > of those sums to show whether or not your probabilities sum to > infinity. > Dear Kevin, Ok, I am getting closer and closer... By the way in Feller where you mention X.4 for the first original problem > that we haven't discussed, it is proven that > P{ |S_n/nlog(n) - 1| > epsilon } ---> 0 > which is equivalent to saying > S_n/nlog(n) converges in probability to 1. However, in Feller, the log is base 2. > Is it henceforth equivalent to say that S_n/nlog(n) converges in probability to log(2) > where log is base e? My problem originally was to > Show that (S_n)/(n*log(n)) converges in probability to > log(2), which is different from what Feller says in X.4 edition 3. > (I have edition 2, but I believe edition 3 is just a reprint). Robert I believe what I have just said above is true, but please affirm this if you would. Robert > -- > Kevin ==== n distinct parallel lines are given in the euclidean plane (n>2). construct a regular n-gon having each of its vertices on a different parallel line. any idea? m ==== >n distinct parallel lines are given in the euclidean plane (n>2). construct a regular n-gon having each of its vertices on a different >parallel line. any idea? m > It's clear that there must be constraints on the set of parallel lines for this to be possible. For example, for n = 4, draw a unit square (the problem is invariant under similarity transformations), then consider the space of quadruples of parallel lines, one line passing through each corner of the square. The first line determines the remaining lines, so the set of such quadruples is smoothly parametrized by the angle of the first line. The set of all quadruples of parallel lines, modulo similarities, is a two-dimensional space (defined by ratios of distances between the lines). The set of quadruples that arise from a square form a curve in this space, so there is no solution for most such quadruples. The case n = 3 is special. The space of triplets of parallel lines, modulo similarities, is one-dimensional, and it's easy to see that all such triplets arise from some equilateral triangle. John Mitchell ==== John Mitchell ha scritto nel n distinct parallel lines are given in the euclidean plane (n>2). construct a regular n-gon having each of its vertices on a different >parallel line. any idea? m > It's clear that there must be constraints on the set of parallel lines > for this to be possible. For example, for n = 4, draw a unit square > (the problem is invariant under similarity transformations), then > consider the space of quadruples of parallel lines, one line passing > through each corner of the square. The first line determines the > remaining lines, so the set of such quadruples is smoothly > parametrized by the angle of the first line. The set of all quadruples > of parallel lines, modulo similarities, is a two-dimensional space > (defined by ratios of distances between the lines). The set of > quadruples that arise from a square form a curve in this space, so > there is no solution for most such quadruples. The case n = 3 is special. The space of triplets of parallel lines, > modulo similarities, is one-dimensional, and it's easy to see that all > such triplets arise from some equilateral triangle. very good. actually i found a very simple way to construct the triangle (case n=3), but couldn't find a way for the square; now you convinced me it was somehow impossible... ==== > n distinct parallel lines are given in the euclidean plane (n>2). construct a regular n-gon having each of its vertices on a different > parallel line. any idea? Yeah, I have the idea that this is impossible for the general case. Only a few special cases can be solved. A triangle will always fit; but for instance imagine four parallel lines with successive distances d1, d2 and d3 with d1 <> d3, then I think that no square will fit. Unless I read this quastion in the wrong way...(?) -- M.vr.gr. Dave (d-dot-langers-at-wxs-dot-nl) ==== Dave Langers ha scritto nel messaggio > n distinct parallel lines are given in the euclidean plane (n>2). construct a regular n-gon having each of its vertices on a different > parallel line. any idea? Yeah, I have the idea that this is impossible for the general case. Only > a few special cases can be solved. A triangle will always fit; but for instance imagine four parallel lines > with successive distances d1, d2 and d3 with d1 <> d3, then I think that > no square will fit. Unless I read this quastion in the wrong way...(?) no you read it right! thanks for your hint! ==== I haven't tried to prove it, but it occurred to me after posting that if one does the same construction with the closed interval [a,b], but leaves out (a,-1) and (b,+1), then one ought to get the maximal ideal space of the algebra of functions on [a,b] with right and left hand limits at every point, which ought to be a Banach algebra in the sup norm. Of course, even if that is correct, it isn't a name... Ignorantly, Allan Adler ara@zurich.ai.mit.edu **************************************************************************** * * * Intelligence Lab. My actions and comments do not reflect * * in any way on MIT. Moreover, I am nowhere near the Boston * * metropolitan area. * * * **************************************************************************** ==== Friday April 26th 2002 116/249 16504 R I Z Z U T O 18 9 26 26 21 20 15 = 135 I am posting Vito Rizzuto's stats again because Shriner/Freemason Don Ocean and/or his friend James Takayama is repeatedly calling me a pedophile. I have never sexually assaulted anybody, I have never been charged with sexual assault, I have never been arrested for sexual assault, and I am not sexually attracted to children. In 1988 I criticized the phallic worship in the Protestant and Catholic churches and was repeatedly arrested and tortured at the U of S for years. Ruby would have me arrested for failing to kiss her God-damned ass and wear the clothes she was always trying to force upon me (and because Protestants and Catholics were upset with my words and lobbied her to shut me up), Ruby would have me arrested and chemically lobotomized, and then she would come to the psychiatric ward and force her choice of clothes upon me there. Eventually I found myself internet access at the U of S and posted Collecting Mail For The Coming Anti-Christ, in the essay I spoke in defense of people to wear their own choice of clothes, and this included defending the rights of women and girls to go topless if they wanted. Now Don Ocean calls me a pedophile, and as a result of Don Ocean, James Takayama in Hawaii has begun doing the same. First the fellow libels me in the WaxyOrg website using the name of Nospam, then using the name of Thomas (aka MauiCop), James Takayama quotes his own material posted as Nospam. Now another website has begun to quote the libel as being truthful, saying that I have been repeatedly arrested for sexual assault and that I was a pedophile (and it is possible that James Takayama is responsible for this website and may start others where he will claim I have been repeatedly arrested for sexual assault). Now as a result, people in Saskatoon are calling me a pedophile and are threatening to beat me up and kill me. The Saskatoon police are not in the habit of charging people for assault when they give me beatings (in fact they arrested me after James De Witt brutally assaulted me at the Seventh Day Adventist Church, I was brutally assaulted and then the police arrested me and took me to the U of S relievers when these people crack and break my ribs. The situation is quite unfair, and now Vito Rizzuto will have his stats posted daily until the situation is resolved. I get tortured year after year and begged people in futility for assistance to flee the country, now watch Vito spend huge piles of cash again this year turning trees into decorated idols, for his compassion is limited to traditions. 203 Nicolo Nicolo 68 Rizzuto 135 225 Libertina Libertina 90 Rizzuto 135 201 Vito 21 2 46 52/313 +4014 Vito 66 Rizzuto 135 177 Maria Maria 42 Rizzuto 135 Mom's first name adds to 90 (66th non-prime), Exodus contains 1213 verses (66+66+66th prime) and terminates at chapter 90 (66th non-prime). Vito adds to 66, mom and the little sister have first names averaging 66. The kids have first names adding together for 108 (the first 6 primes in prime positions). The kids have first names differing in value by the 24 chapters of Bible Books 6 and 10 (6th non-prime). Dad has a 6 lettered first name, there are 24 letters in all the first names, the number of chapters in Books 6 and 10 (6th non-prime). All first names add together for 266. Dad and Vito have first names adding together for 134, corresponding to Numbers 17 with 13 verses (the 6th prime). Mom and Vito have first names averaging 78 (6 times the 6th prime). Dad and Vito have full names adding together for the 404 verses of Bible Book 66, Revelation. Vito was born on the 21st, corresponding to Ecclesiastes (chapters 660 to 671). Vito has consonants adding to 132 (66+66). Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 --- 108 Vito Rizzuto (135) was hot for Cammalleni (83), the names differ in value by 52, pretty as Vito was born on the 52nd day of the year. Giovanna (83) Cammalleni (83) was born with names adding to the 23rd prime and to the 23rd prime, together for 46, and she gets a husband that was born in 46. 166 Giovanna 48 Giovanna 83 Cammalleni 83 Vito married Giovanna (83) Cammalleni (83). She was born with 18 (6+6+6) letters adding to 166. Both of her names added to the 83 verses of Second Timothy (the 16th Book of the New Testament). Her names added to 83 and 83, corresponding to Exodus 33 and Exodus 33, together for 66. At birth, Vito and Giovanna had 29 letters in their names, or 6 plus the 6th prime (13) plus the 6th non-prime (13), there are 29 (6+6p+6np) chapters in Bible Book 13 (the 6th prime), there are 29 (6+6p+6np) verses in chapter 666 (Ecclesiastes 7). At birth Vito and Giovanna had 29 letters adding together for 367 (First Chronicles 29). Vito's sister's first name adds to 42 (the 29th non-prime), Bible Book 29 is Joel (42). Now Giovanna's name adds to 218 (twice the 29th prime). Nicolo's name adds to 203 (7x29). Leonardo's name adds to 219, or 3 times the 73 verses of Bible Book 29. Libertina was born in 73 (the length of Book 29 and is the Lucas numbers up to 29). Leonardo and Libertina have first names adding together for 174 (6x29). Dad and the first two kids have first names adding together for 218 (twice the 29th prime). The males were born in years adding to 182 (Deuteronomy 29 with 29 verses). 201 Vito 21 2 46 52/313 +4014 Vito 66 Rizzuto 135 218 Giovanna 48 Giovanna 83 Rizzuto 135 203 Nicolo 67 Nicolo 68 Rizzuto 135 219 Leonardo 69 Leonardo 84 Rizzuto 135 225 Libertina 73 Libertina 90 Rizzuto 135 Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime They have three kids, the first and last of the kids bear Vito's parent's names, so the daughter's first name adds to 90 (66th non-prime). The kids have first names adding together for 242 (First Samuel 6), all names in the family add together for 1066. The kids were born in 67, 69 and 73, these are the 19th prime, 50th non-prime and the 21st prime, together for 90 (66th non-prime and the value of the daughter's first name). Mom and her sons have first names adding together for 235... the 184th prime (1097) and the 184th non-prime (235) averages 666. The brothers have names averaging 211, it is approximately 66.6% of the 66th prime (317) and is the terminating chapter of Bible Book 6. Book 6 chapter 6 (193) plus the terminating chapter of Book 6 (211) adds together for the 404 verses of Bible Book 66. Daughter's first name not only adds to 90 (66th non-prime), but her first name adds to 66.666...% of her last name. This is a family of 5, the 4014 days dad is older than me is the 959 verses of Bible Book 5 short of the 666th prime (4973). Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 26 61 27 67 28 71 30 73 32 79 33 83 34 89 35 97 36 101 38 103 39 107 40 109 42 113 44 127 45 131 46 137 48 139 49 149 50 151 51 157 52 163 54 167 55 173 56 179 57 181 58 191 60 193 62 197 63 199 64 211 65 223 <-48th-> 66 227 68 229 69 233 70 239 72 241 74 251 75 257 76 263 77 269 78 271 80 277 81 281 82 283 84 293 85 307 86 311 87 313 88 317 <-66th-> 90 Vito adds to 66 (48th non-prime). Vito's first name adds to 48.888...% of his last name, God gives him a wife that was born in 48. Vito and his wife have first names adding together for the 149 verses of Bible Book 48, Galatians. Vito's 201 valued name exceeds his 52nd day of birth by the 149 verses of Bible Book 48. Rizzuto (135) was hot for Cammalleni (83), the 83 value of mom's maiden name is 61.48% of the 135 value of Rizzuto. The males were born in years adding to 182, the females in years adding to 121 (66.48%). The kids are missing 15 letters from their first names, these missing letters add to 248. The 4014 days dad is older than me is 18 times the 48th prime (223), prettier as I was born on the 48th day of the year and with 317 days remaining in the year (66th prime). 1-50 - Genesis 51-90 - Exodus 91-117 - Leviticus 118-153 - Numbers 154-187 - Deuteronomy 188-211 - Joshua 930-957 - Matthew 958-973 - Mark 974-997 - Luke 998-1018 - John 1019-1046 - Acts 1047-1062 - Romans 123 <-Numbers 6, it is three times the 13th prime (41+41+41), keeping in mind that 13 is the 6th prime 188 <-the opening chapter of Book 6 is 6x6x6 short of the 404 verses of Bible Book 66, it is the 6th prime squared (13x13) short of the 357 verses of Daniel (also in part about 666) 193 <-Book 6 chapter 6 is the 44th prime, while 44 is in turn 66.666...% of 66 211 <-the terminating chapter of Book 6 is approximately 66.6% of the 66th prime (317) 357 <-the opening chapter of Book 6 plus the 6th prime squared is the 357 verses of Daniel (in part about 666) 404 <-the 6th prime squared (13x13) plus the 6th prime squared (13x13) plus 66 adds to the 404 verses of Bible Book 66 1062 <-666 plus 6x66 is a combination of the 658 verses of Bible Book 6 plus the 404 verses of Bible Book 66, and is the terminating chapter of New Testament Book 6 1070 <-666 plus the 404 verses of Book 66 is the 1070 verses of Job (Book 6+6+6) 1213 <-Exodus terminates at chapter 90 (66th non- prime) with 1213 verses (the 198th or the 66+66+66th prime) 1292 <-the 658 verses of Book 6 plus twice the 66th prime (317) is the 1292 verses of Isaiah (the Book contains 66 chapters) Vito is 4014 days older than me, his name adds to 201 (Joshua 14), his mom and daughter share the same 225 valued name (Judges 14). The kids have odd valued letters in their first names adding together for 140. The letters that are neither prime nor square in the kids' given names add to 196 (14x14). Vito and his daughter have names differing in value by 14. All first names add to 391 (314th non-prime). Dad's name adds to 67+67+67. The first of the kids was born in 67 and has names differing in value by 67. These first three family members have first names adding to 66, 83 and 68, corresponding to Exodus 16, 33 and 18, together for 67. This 67 is the 19th prime, the second kid was born in 69 (Exodus 19) and has a name adding to 219. The brothers were born in years adding to 136 (Numbers 19). Vito and the last of the kids were born in years adding to 119. Perhaps mom was 19 years old when she gave birth in 67 (the 19th prime). Mom's name adds to twice 109 (Leviticus 19) while dad's name adds to 201 (3 times the 19th prime). The parents have names adding together for 419 (Nehemiah 6 with 19 verses). Vito and his kids were born in years adding to 255 (First Samuel 19). Vito and his kids have first names averaging 77 (the primes up to 19). The kids were born in years adding to 5909 (19x311). Libertina's last name adds to 150% of her first name (150 chapters in Bible Book 19). Vito was born a multiple of 19 days into the century (16853=19x887). Vito's vowels add to 69 (Exodus 19). The 2460 verses of Bible Book 19 is 19x19+19x19+19x19+19x19+19x19+19x19+19x19 minus 67 (the 19th prime). Primes Non-Primes 2 1 3 4 5 6 7 <-4th-> 8 <-Bible Book 8 contains 11 4 chapters, pretty as 13 8 is the 4th non-prime 17 19 -- 77 <-the first 8 primes plus 8 more adds to the 85 verses of Bible Book 8 Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 18 61 19 67 <- 67 <-the 8th prime in prime position R U T H <-Book 8 18 21 20 8 = 67 Vito's names differ in value by 69 and his second kid was born in 69. Vito was born on the 21st and his third kid was born in 73 (21st prime). Vito was born on the 21st and his name adds to 201, and he was likely 21 years old when the first of the kids was born. Vito was born in 46, his 201 valued name exceeds it's 155th non-prime position by 46. The kids have prime and square valued letters in their first names adding together for 46. Vito was born on day 52, it's the number of chapters in Bible Book 24, while his daughter gets a name that exceeds his by 24. Vito and I were together born 530 days closer to the beginning of our years than to the end of our years (Psalm 52). This 530 is a combination of the first three perfect numbers (6, 28 and 496), they have factors that add to form themselves: Perfects 6 - 1, 2, 3 28 - 1, 2, 4, 7, 14 496 - 1, 2, 4, 8, 16, 31, 62, 124, 248 The males have first names adding together for 218, and mom's name adds to 218. Mom was born with 18 letters and took a last name that begins with the 18th letter of the alphabet, it is a last name that adds to 135 (Numbers 18). Vito and his sister have first names adding together for 108 (Leviticus 18). Rizzuto adds to 135 (the 103rd non-prime), the 11 different letters utilized in the construction of the first names for the kids add together for 103. 3x3x3x3x3x3 3x3x3 103 <- the 3x3x3rd prime ----------- 859 <- the number of verses in Bible Book 3 The parents were born in 46 and 48, Bible Books 46 and 48 contain 437 and 149 verses. These Books contain an average of 293 verses while the 437 is 293.28% of the 149. The 437 and 149 are the 353rd non-prime and 35th prime, and so it is pretty that there would be 35 letters in the family first names, and pretty that the last name would add to 135. Mom and her sons have first names adding together for 235. The brothers have first names adding together for 152 (Numbers 35). Libertina Kabatoff adds to 152 (Numbers 35), prettier as 293 is the 62nd prime while Kabatoff adds to 62. Rizzuto adds to 135, or 57 plus the 57th non-prime (78). The main Books of end-times prophecy are Daniel with 357 verses and Revelation with 404 verses, or 57 plus the 57th prime (269) plus the 57th non-prime (78). Primes Non-Primes 2 1 3 4 5 6 7 8 11 <-5th-> 9 -- -- 28 28 Vito and his sister Maria have first names adding to 66 and 42, together these Bible Books contain 1555 verses. Vito has 11 letters and a first name adding to a multiple of 11, his kids have first names adding together for 11x11+11x11 (11 is the 5th prime). Then Vito marries Giovanna (83) Cammalleni (83), there are 83 verses in Bible Book 55, and she soon finds herself in a family of 5. The 5th of 5 family members has a name adding to 225, prettier as 25 is not only 5x5 but is 5 plus the 5th prime (11) plus the 5th non-prime (9), keeping in mind that Bible Book 5 contains 959 verses. Perhaps mom was 25 (5x5 or 5+5p+5np) years old when she gave birth to the 5th of 5 family members. Libertina's first name adds to a multiple of 5 and also a multiple of 9 (5th non-prime). Libertina's (the 5th of 5 family members) first 5 letters add together for dad's 46th year of birth. My name adds to 187 (the terminating chapter of Bible Book 5), it is 5x5x5 plus twice the 5th prime in prime position (31). If Libertina took my 62 (twice the 5th prime in prime position) valued last name, then her names would have an average value of 76 (55th non-prime). Lamentations is Bible Book 25 with 154 verses, the 154th prime is 887 while Vito was born on the 19x887th day of the century, pretty because if Libertina married me then he would be lamenting. Vito and his kids already have first names adding together for 308, or twice the 154 verses of Lamentations. Note that Bible Books 5 and 5x5 differ in length by 805 verses (the 666th non-prime). And Old Testament Book 9 (the 5th non-prime) and New Testament Book 9 (the 5th non-prime) together contain 959 verses (the number of verses of Book 5). Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 --- 167 Esther Book 17 Vito (66) was born on day 52, it is an average of 59 (the 17th prime). have names averaging 189 (the first 17 primes minus the first 17 non-primes). Vito and Giovanna have names differing in value by 17 (7th prime, the primes up to 7 add to 17). Giovanna's name adds to 218 (Book 7 chapter 7). Vito and Giovanna were born with last names adding together for 218 (Book 7 chapter 7). The first of the kids gets a first name adding to a multiple of 17 and he was born in 67 (Exodus 17). The second gets a first name adding to 7 times the 7th non-prime (12). The brothers have first names adding to 68 and 84 (a span of 17). The daughter gets a name adding to 225 (Book 7 chapter 7+7). The kids were born in 67, 69 and 73, corresponding to Exodus 17, 19 and 23, together for 59 (the 17th prime). Libertina's names differ in value by 45 (45 chapters contain the length of 17 verses), she might take my name and end up with 17 letters. Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 <-17th-> 26 --- --- 440 251 I wanted 17 French fry girls at my weddink, they would throw French fries high up into the air, the french fries would fall to the ground and get dirty, and nobody would ever eat french fries again. And I wanted 59 big nosed Greeks with their big noses throwing hamburglers at us from across the street. And I wanted American Noel Nibblett blowing his bugle and leading a marching regiment of cadets up and down the street while American Don Ocean and his Shriner friends ride circles around them on their little motorscooters. Scientists have recently discovered that tomatoes contain properties that help to prevent prostate cancer, in men, and since A&W allows one to take as much ketchup as they wanted, the Great A&W Rootbear was on the fast track to becoming an international symbol of health and fertility... so of course I wanted the Great A&W Rootbear to be the best man at my weddink. My weddink was soon approaching and it was goink to be a glorious affair, probababbly. 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 207 Libertina 73 Libertina 90 Acevedo-Kabatoff 55-62 Anyway, if you people think that you have the right to use my abusive parents as tools and arrest and torture me, then I think that I should have the right to ask women to marry me, or to marry Marcia and me, our last names add together for the 117 verses of Song of Solomon, it's the Bible's Book of Love. The nubile sweety was born on the 6th and has a 6 lettered first name). Isaiah is the Book with 66 chapters, pretty as it is Book 23, or the 6th prime plus the 6th non-prime (13+10=23). Isaiah 4, 12 and 20 (adds to 6x6) each contains 6 verses. Isaiah 4:1 is about Marcia (and me), and 6 other women who are capable of feeling shame rather than pride, greed or lust, or who limit their love for traditions and for people who abide by their traditions. You people have Egyptian penises on the roofs of your churches and lined city streets with representations of penises, and had me tortured for years for saying so, others just sat back in silence while they were doing this to me, and similarly you remain silent and compassionless now that the arrests and torture have ceased. You people spent millions of dollars having me tortured, and then annually you spend billions on your decorated trees, I begged and begged for assistance to flee the country (they tortured me for years at the U of S) and you people are so cheap that you can't even offer to buy me a cookie when I bust my ass to show you evidence that your very name is a gift from God!!! Should Libertina marry me, great, but if Marcia marries me and then Libertina marries Marcia and me, then Nicolo and Leonardo are both goink to win themselves a shiny new Cadillac!!! And if Libertina turns out to be some sort of sexual acrobat then Vito and Giovanna are both goink to win themselves a shiny new Cadillac too. Or if Melinda Jarocki sleeps with me then Vito will win himself a used Pontiac. I have Scripture to support taking seven brides (Isaiah 4:1) and I have Scripture to support sleeping with Melinda Jarocki outside of wedlock (First Kings 1:1-5), while you people have a vast multitude of Scriptures condemning your decorated trees, phallic-capped churches and your violence against me for daring to point out your pagan traditions. Good luck and may God bless you!!! Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! What a wonderful weddink there will be What a wonderful day for you and me Church bells will chime You will be mine In apple blossom time... Note that every day Freemason Don Ocean (or his friend James Takayama) libels me and calls me a pedophile (or such) on the usenet, then I will repost the stats for the following people: Daniel Bruner, Angel Cadwell, Brittanie Cecil, Carl Koopang, Jay Larson, Brian Lott, Adam Millikan, Cody Milliken, Christopher Ridsdale, Vito Rizzuto, Melissa Schultz, Erin Sorenson, Ann Wigdahl, Michael Winkler, or maybe more (more or less). ==== I am working with Calculus of Variations by Gelfand and Fomin. I am stuck on canonical transformations and Hamilton-Jacobi equations. Do you guys have any pointers (web site, book) I would like to have sort of the intuition and the proof. Tony ==== I am working with Calculus of Variations by Gelfand and Fomin. I am stuck >on canonical transformations and Hamilton-Jacobi equations. Do you guys have >any pointers (web site, book) I would like to have sort of the intuition and the proof. Tony You can try Mathematical Methods of Classical Mechanics by V. I. Arnold, Graduate Texts in Mathematics, Springer-Verlag, 1978. I don't know if it will be any clearer than Gelfan and Fomin, but it can't hurt to take a look. John Mitchell ==== I'd like to get some opinions about this problem, please. The dependable capacity C available in an electric system in a month of the future, say Dec 2008, is a random variable with density function f defined on [0, Cmax]. We can admit f is continuous on this interval. On the cited month you have to meet a known and deterministic demand r<=Cmax. Define a random variable deficit, D, by D= r-C if C=r. Supposing the density function f is independent of r, the expected value of D corresponding to a demand r is given by E(r) = Integral [0,r] (r-c) f(c) dc = r*Integral [0,r] f(c) dc - Integral [0,r] c f(c) dc. Since f is continuous, E(r) exists for r<=Cmax and E is differentiable wrt r. Applying the F.T. Of Int. Calculus, we get E'(r) = r* f(r) + Integral [0,r] f(c) dc - r*f(r) = Integral [0,r] f(c) dc = Probability (C<=r) = Probability (D>=0). Though I don't have f in a closed form, I can estimate Probability (D>=0) by means of simulation models, using a process similar to Monte Carlo's. Then, for variations on r of about 5%, I can make the estimate Delta E(r) = (Delta r) * Probability (D>=0). All I need to know is that f exists and is continuous on [0, Cmax]. I don't need to know how exactly f sends c into f(c). In some situations this is reasonable, but there are cases when it's not admittable at all to suppose f is independent of r. In such cases, for each r there's a particular density function f_r. You can still suppose r is known, but it affects the distribuition of C on [0, Cmax] (Cmax is always known and independent of r]. Then I think I have something like E(r) = Integral [0,r] (r-c) g(r,c) dc where g is defined on R^2 with values on R and, for a given r, g(r,c) = f_r(c). Supposing g is continuous, is that OK if I use Leibiniz formula to compute E'(r)? Anyway, if f_r depends on r than that beautiful conclusion I came to before is no longer true, right? I'm a bit confused here. Artur ==== The surface area of the base of a non-regular tetrahedron is unknown, but the surface areas of the three other faces are known (A1, A2, A3 respectively). Why is the surface area of the base equal to: Ai/{cosine(angle between the normal of the base and the normal to face i)} where i=1,2 or 3? ==== The surface area of the base of a non-regular tetrahedron is unknown, > but the surface areas of the three other faces are known (A1, A2, A3 > respectively). Why is the surface area of the base equal to: Ai/{cosine(angle between the normal of the base and the normal to face > i)} where i=1,2 or 3? > Let vectors a,b,c from one vertex define three edges. Then the oriented surface vector of the adjoining faces is one half of C == a X b , A == b X c, B == c X a and the oriented surface vector of the fourth face is ( c - a ) X ( b - a ) = c X b - a X b - c X a = - b X c - a X b - c X a == D ( a form of Stokes' theorem ) Now the magnitude of this vector is twice the area of the face, and is given by the sqrt ( D dot D ), but |D|^2 = D dot D = - D dot A - D dot C - D dot B and dividing both sides by |D| gives you your answer. ... Well, almost, you wanted the sum of the terms you described. Note that if you define the normal vectors as all pointing outwards the normal to the base points down. Lew Mammel, Jr. ==== >The surface area of the base of a non-regular tetrahedron is unknown, >but the surface areas of the three other faces are known (A1, A2, A3 >respectively). Why is the surface area of the base equal to: Ai/{cosine(angle between the normal of the base and the normal to face >i)} where i=1,2 or 3? > Any one of the three faces F1, F2, F3 projects orthogonally onto the base face. Think about how orthogonal projection from one plane to another affects area. If you imagine viewing the planes in the direction of the planes' intersection line (if the planes are parallel, the answer is trivial), this reduces to thinking about how projection from one line to another affects lengths, which should be easy to figure out. John Mitchell ==== Can anyone solve this? Use the implicit function theorem to show that as a subspace of R^{n+1} an n-surface M is locally homeomorphic to an open subset of R^n. That is, for each p in M, there exists a neighborhood of p (call it O, which is a subset of M) and an open set U (subset of R^n) and a homeomorphism f : U --> O Note : An n-surface M (which is a subset of R^{n+1}) is a non-empty subset such that M = f^{-1}(c) where f : U ----> R (U is a subset of R^{n+1}) , f is C-infinity, and such that for all p in M, p is a regular point of f (i.e. it's not a critical point of f). That's all I am given. Steven Rossi ==== >Can anyone solve this? Use the implicit function theorem to show that as a subspace of >R^{n+1} an n-surface M is locally homeomorphic to an open subset of R^n. >That is, for each p in M, there exists a neighborhood of p >(call it O, which is a subset of M) and an open set U (subset of R^n) >and a homeomorphism f : U --> O Note : An n-surface M (which is a subset of R^{n+1}) is a non-empty subset >such >that M = f^{-1}(c) where >f : U ----> R (U is a subset of R^{n+1}) , f is C-infinity, and such >that for all p in M, p >is a regular point of f (i.e. it's not a critical point of f). That's all I >am given. >Steven Rossi > You got an answer when you posted this question yesterday. Why start a new post on the same question? John Mitchell ==== Can anyone solve this? Use the implicit function theorem to show that as a subspace of >R^{n+1} an n-surface M is locally homeomorphic to an open subset of R^n. >That is, for each p in M, there exists a neighborhood of p >(call it O, which is a subset of M) and an open set U (subset of R^n) >and a homeomorphism f : U --> O Note : An n-surface M (which is a subset of R^{n+1}) is a non-empty subset >such >that M = f^{-1}(c) where >f : U ----> R (U is a subset of R^{n+1}) , f is C-infinity, and such >that for all p in M, p >is a regular point of f (i.e. it's not a critical point of f). That's all I >am given. >Steven Rossi > You got an answer when you posted this question yesterday. Why start a > new post on the same question? > My apologies. I wouldn't have posted again if I had seen your last response in the previous thread...I was still confused. > John Mitchell ==== > Can anyone solve this? Use the implicit function theorem to show that as a subspace of > R^{n+1} an n-surface M is locally homeomorphic to an open subset of R^n. > That is, for each p in M, there exists a neighborhood of p > (call it O, which is a subset of M) and an open set U (subset of R^n) > and a homeomorphism f : U --> O Note : An n-surface M (which is a subset of R^{n+1}) is a non-empty subset > such > that M = f^{-1}(c) where > f : U ----> R (U is a subset of R^{n+1}) , f is C-infinity, and such > that for all p in M, p > is a regular point of f (i.e. it's not a critical point of f). That's all I > am given. We have grad(f)(p) nonzero. Write p = (p1, p2, ..., p(n+1)). WLOG, the partial derivative of f with respect to the last variable is nonzero. The IFT then says that near p, M is the graph of a C^oo function defined near (p1, p2, ..., pn) in R^n. The graph function will give you your homeorphism. ==== Can anyone solve this? Use the implicit function theorem to show that as a subspace of > R^{n+1} an n-surface M is locally homeomorphic to an open subset of R^n. > That is, for each p in M, there exists a neighborhood of p > (call it O, which is a subset of M) and an open set U (subset of R^n) > and a homeomorphism f : U --> O Note : An n-surface M (which is a subset of R^{n+1}) is a non-empty subset > such > that M = f^{-1}(c) where > f : U ----> R (U is a subset of R^{n+1}) , f is C-infinity, and such > that for all p in M, p > is a regular point of f (i.e. it's not a critical point of f). That's all I > am given. We have grad(f)(p) nonzero. Write p = (p1, p2, ..., p(n+1)). WLOG, the > partial derivative of f with respect to the last variable is nonzero. The > IFT then says that near p, M is the graph of a C^oo function defined near > (p1, p2, ..., pn) in R^n. The graph function will give you your homeorphism. When you say M is the graph, what do you mean by graph? And later when you say graph function, what do you mean by graph function? Steven Rossi ==== I have a problem that I can't seem to get...can anyone help please? Consider : X_1, X_2, ... is a sequence of independent random variables with finite variances and a common distribution F such that F(0) = 0. What is the relationship between 2*E[(X_n)^2] and E[1 / ( (X_{n-1})^2 + (X_{n-2})^2 ) ]? Is one always at least as large as the other? (E denotes expectation) If I replaced 2*E[(X_n)^2] by C*E[(X_n)^2], what value of C would still make this true? (C is a real number) Henrique ==== > Consider : X_1, X_2, ... is a sequence of independent random variables with > finite variances and a common distribution F such that F(0) = 0. What is the relationship between > 2*E[(X_n)^2] > and > E[1 / ( (X_{n-1})^2 + (X_{n-2})^2 ) ]? Is one always at least as large as the other? If the random variables take values in [a,b], where, 0 < a < b < oo, then your first expression lies between 2a^2 and 2b^2, and your second expression lies between 1/(2b^2) and 1/(2a^2). If the first expression is <= the second, then we have 2a^2 <= 1/(2a^2) for all a > 0, nonsense. Same for the other way around. ==== Consider : X_1, X_2, ... is a sequence of independent random variables with > finite variances and a common distribution F such that F(0) = 0. What is the relationship between > 2*E[(X_n)^2] > and > E[1 / ( (X_{n-1})^2 + (X_{n-2})^2 ) ]? Is one always at least as large as the other? If the random variables take values in [a,b], where, 0 < a < b < oo, then > your first expression lies between 2a^2 and 2b^2, and your second > expression lies between 1/(2b^2) and 1/(2a^2). If the first expression is > <= the second, then we have 2a^2 <= 1/(2a^2) for all a > 0, nonsense. Same > for the other way around. Dear World Wide Wade, So you are saying that I can't say anything about the relationship between these expectations? What if I replace the 2 in front of 2*E[(X_n)^2] by an arbitrary constant C? What I am trying to get at eventually is to look at : Z_n = product (from k = 1 to n) of { (C(X_k)^2) / ( (X_{k-1})^2 + (X_{k-2})^2 ) } and to determine which values of C would guarantee that Z_n converges with probability one to a limit W which may be a constant or a random variable. What do you think? Henrique ==== I want to ask how i can proof this theorm: If R is a semi-prime ring with maximum condition on annihilators then evrey annihilator ideal K contain a finit generated ideal N such that annihilator K equal annihilator N.. or send to me where can i found the useful thing suspened with goldie ring. thanks ==== >Problem : >>Use the implicit function theorem to show that as a subspace of >>R^{n+1} an n-surface M is locally homeomorphic to an open subset of R^n. >>That is, for each p in M, there exists a neighborhood of p >>(call it O, which is a subset of M) and an open set U (subset of R^n) >>and a homeomorphism >>f : U --> O >>Steven Rossi >> The details depend on how your surface is defined. If it's defined (at >> least locally) by an equation of the form F(x) = 0, with x in R^{n+1} >> and F: R^{n+1} -> R being non-singular on the zero-set M, the implicit >> function theorem says that F locally looks like a linear function L: >> R^{n+1} -> R. Clearly, the zero-set of a linear function is globally >> (and therefore locally) diffeomorphic to an open subset of R^{n+1}, so >> the same is true for F. The phrase looks like means that there are >> locally diffeomorphisms of the domain and range space that transform F >> to L. My surface is defined as follows : An n-surface M (which is a subset of R^{n+1}) is a non-empty subset such >that M = f^{-1}(c) where >f : U ----> R (U is a subset of R^{n+1}) , f is C-infinity, and such >that for all p in M, p >is a regular point of f (i.e. it's not a critical point of f). That's all I >am given. > John Mitchell > John Mitchell ==== Prinzip: Ich habe einen Spielautomaten entwickelt, das ein Geschicklichkeitsspiel anbietet, so wie z.B. ein J.8ager geschickt sein muss sein Wild zu fangen - das Wild wiederum wiederum manchmal einfach und manchmal schwieriger zu kriegen ist, was nicht eindeutig voraussehbar ist f.9fr den J.8ager, einmal wird die Geschicklichkeit mehr gefordert, manchmal weniger, im Wesentlichen abh.8angig von den Erfolgen/Misserfolgen der anderen J.8ager (Konkurrenten), die durch Dezimieren des Wildes den Wildbestand mitbestimmen, d.h. die Schwierigkeit zu jagen. Zum Ger.8at selber: F.9fr die Spieler ist es m.9aglich mit dem Einsatz von einer Einheit Geld max 20x der Einheiten (pro Spiel) zu gewinnen - mit Geschicklichkeit/Strategie/etc. Die Spieler k.9annen jederzeit und Intensit.8at teilnehmen wie sie wollen. Das heisst es ist eine offene Konkurrenz. Das Ger.8at funktioniert wie ein Messger.8at, das die Geschicklichkeit der Spieler in ***angepasster Schwierigkeit wiedergibt. Wir haben folgende Kriterien die mathematisch relevant w.8aren, meiner Ansicht nach: Einsatz: 1 Einheit Auszahlung: mal 20 Einheiten Geschicklichkeit der Spieler (GamblerSkill): Anfangs unbekannt, das Ger.8at startet bei 50% der technisch m.9aglichen Schwierigkeit/Aufgabenstellung, lernt aber hinzu und berechnet einen Durchschnitt bei jeder Aktion. Schwierigkeit des Ger.8ats (MachineSkill): ***120% des Durchschnittes (kommerziellen Umsetzung 20%) des GamblerSkills. Erkl.8arung: Die Bandbreite der m.9aglichen Schwierigkeit ist bekannt und rein technisch vorgegeben. Dies wird in Prozenschritten am Ger.8at angezeigt. Von 0% (leichtestet Stufe) bis 100% (schwierigste Stufe) - intern aber im Promille-Bereich berechnet. Bandbreite der Kaskadierungsgeschwindigkeit in welcher sich die Schwierigkeit dem Durchschnitt anpasst ist gleich 50fache der m.9aglichen Hochstauszahlung (hier 20 Einheiten) will heissen 1000. Das w.9frde sich dann folgendermassen auswirken: ein Ger.8at hat eine Schwierigkeitsstufe von 52%, ein Anderes (Lokal) 56%. Sind die Spieler (Spielergemeinschaft/(Versicherungsgemeinschaft)) an einem Ort konstant geschickter, pendelt sich das Ger.8at auf 56% ein. Die (durchschnittliche) Geschicklichkeit l.8asst sich direkt am erziehlten Erfolg der gewonnen oder verlorenen Spiele ablesen. Trainingsspiele (mit weniger Einsatz --.20) basieren auf dem ***aktuellen Stand des Ger.8ates ohne diesen zu ver.8andern - weil auch keine Gewinne m.9aglich sind. Die Herausforderung ist - wen wunderts - ein Spiel zu entwickeln, das f.9fr m.9aglichst viele Spieler attraktiv ist, eine interessante Safari zu veranstalten - um auf die J.8ager zur.9fckzukommen. In der Praxis funktioniert es - mit empirischen Versuchen ermittelt. Mein Wunsch w.8are es dies mittels einer mathematischen Formel oder/und Modell-Theorie darzustellen. Im Grunde empfinde ich dies als Marktwirtschaft in spielerischer Weise umgesetzt - auf dem Markt der Geschicklichkeit/Strategie - Der Spielautomat als Spiegel des Geschicklichkeits-Marktes. Wobei ich mich selber wiederum in diesem Markt selber bew.8ahren muss (kommerzielle Umsetzung). Es geht mir um die Darstellung des Systems (Spieler/Automat) (Geschicklichkeit/Schwierigkeit) mittels einer math.-formel und entsprechender Umschreibung damit auch ein Jurist sie verstehen kann, inkl. mich selber. Juristen behaupten, dass es keine math. Formel gibt die funktioniert, denen w.9frde ich gerne das Gegenteil beweisen, mangels math. Talent ich keine Chance habe dies zu tun. Wer kann mir dabei weiterhelfen? Karl Fay CH-Z.9frich info@fay.ch ==== I tried to solve another task on this issue today and failed. {/-( First of all, I wanted to know if there is a general approach to solve tasks of the form: You have given a fraction of the form P(x)/Q(x). There P(x) and Q(x) are polynoms. You have to express this fraction in the form sum(a_k*(m*x+n)^k, k, 0, infinity). ? [m, n are constants and a_k represents k constants.] If somebody knows a general approach or a website (understandable for a beginner) there this is explained then please tell me. (I somehow must understand this matter.) Anyway, this was my task today: ### task: ^^^^^^ Express 14 / ( 8*(x^2-1) ) as an infinite sequence of the form: sum(a_k*(2-x)^k, k, 0, infinity). ____________ |my attempt(s): ^^^^^^^^^^^^^^^^^^ 14 / ( 8*(x^2-1) ) = (7/4) * ( 1 / ( (x-1)(x+1) ) ) partial fraction decomposition (Do I name it right?): ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1 = A*(x+1) + B(x-1) if x = 1 => A = 1/2 if x = -1 => B = -1/2 ==> (7/4) * ( 1 / ( (x-1)(x+1) ) ) = (7/4) * ( 1/( 2*(x-1) ) - 1/( 2*(x+1) ) ) = (7/8) * ( 1/(x-1) - 1/(x+1) ) *** Attempt NĦ 1: ^^^^^^^^^^^^^^^^ *** = (7/8) * ( -1/(1-x) - 1 / ( 1-(-x) ) ) = (7/8) * ( -sum( x^k, k, 0, infinity) - sum( (-x)^k, k, 0, infinity) ) = (7/8) * ( -sum( x^k - (-1)^k * x^k, k, 0, infinity) ) = (7/8) * ( -sum( ( 1 - (-1)^k ) * x^k, k, 0, infinity) ) = ? I didn't know how to transform this to the demanded form. Attempt NĦ 2: ^^^^^^^^^^^^^^^^ *** = (7/8) * ( 1/(1-2+x) + 1/(1-2-x) ) = (7/8) * ( 1 / ( 1 - (2-x) ) + 1 / ( 1-(2+x) ) ) = (7/8) * ( sum( (2-x)^k, k, 0, infinity ) + sum( (2+x)^k, k, 0, infinity ) ) = ? I didn't know how to simplify (2-x)^k + (2+x)^k, so that I get (..)*(2-x)^k {the demanded form}. ### That's it. It would be very nice if you told me a general approach for these tasks. Karl ==== [snip] > Express 14 / ( 8*(x^2-1) ) as an infinite sequence of the form: > sum(a_k*(2-x)^k, k, 0, infinity). > ____________ > |my attempt(s): > ^^^^^^^^^^^^^^^^^^ > 14 / ( 8*(x^2-1) ) = (7/4) * ( 1 / ( (x-1)(x+1) ) ) partial fraction decomposition (Do I name it right?): > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > 1 = A*(x+1) + B(x-1) if x = 1 > => A = 1/2 if x = -1 > => B = -1/2 ==> (7/4) * ( 1 / ( (x-1)(x+1) ) ) > = (7/4) * ( 1/( 2*(x-1) ) - 1/( 2*(x+1) ) ) > = (7/8) * ( 1/(x-1) - 1/(x+1) ) *** Partial fractions is the right name. 1/(x+1) = 1/(3-(2-x)) = (1/3) / (1 - (2-x)/3 ) Jim Buddenhagen -- To reply copy jbuddenh@REMOVEtexas.net to address bar and edit out REMOVE Call an infinite subset X of the real numbers recognizable if and only if there is a bijection f:R->R such that for all x in X there is an integer k (possibly dependent on x) with f^k(x)=0. [f^2(x)=f(f(x)), ect.] Say a real number x is recognizable iff x in X for some recognizable X. What numbers are recognizable? rich ==== Call an infinite subset X of the real numbers recognizable if and only > if there is a bijection f:R->R such that for all x in X there is an > integer k (possibly dependent on x) with f^k(x)=0. [f^2(x)=f(f(x)), > ect.] Say a real number x is recognizable iff x in X for some recognizable > X. What numbers are recognizable? If b is a positive real number, let X be the set of all positive integer multiples of b, and let f:R->R be f(x) = x-b. Then X is recognizable, and so is b. The negative reals are handled similarly. 0 is of course recognizable. So the answer is all of them. ==== Call an infinite subset X of the real numbers recognizable if and only >> if there is a bijection f:R->R such that for all x in X there is an >> integer k (possibly dependent on x) with f^k(x)=0. [f^2(x)=f(f(x)), >> ect.] Say a real number x is recognizable iff x in X for some recognizable >> X. What numbers are recognizable? If b is a positive real number, let X be the set of all positive >integer multiples of b, and let f:R->R be f(x) = x-b. Then X is >recognizable, and so is b. The negative reals are handled similarly. >0 is of course recognizable. So the answer is all of them. > Yes. Kind of a dumb definition, I suppose. Wait a minute, what if X is required to be a *dense* subset of R? The rationals are recognizable with f:x->1/x-floor(1/x). Is sqrt(2) recognizable? How about that awful number log(2)? thanks, rich ==== Call an infinite subset X of the real numbers recognizable if and only > if there is a bijection f:R->R such that for all x in X there is an > integer k (possibly dependent on x) with f^k(x)=0. [f^2(x)=f(f(x)), > ect.] Say a real number x is recognizable iff x in X for some recognizable > X. What numbers are recognizable? >>If b is a positive real number, let X be the set of all positive >>integer multiples of b, and let f:R->R be f(x) = x-b. Then X is >>recognizable, and so is b. The negative reals are handled similarly. >>0 is of course recognizable. So the answer is all of them. > >Yes. Kind of a dumb definition, I suppose. Wait a minute, what if X is >required to be a *dense* subset of R? The rationals are recognizable with >f:x->1/x-floor(1/x). No, they are not, at least, not with this function: your f is not injective: both 3 and 3/4 map to 1/3: 3|-> (1/3) - floor(1/3) = (1/3)-0 = 1/3. (3/4) |-> (4/3) - floor(4/3) = 4/3 - 1 = 1/3. -- ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > Call an infinite subset X of the real numbers recognizable if and only >> if there is a bijection f:R->R such that for all x in X there is an >> integer k (possibly dependent on x) with f^k(x)=0. [f^2(x)=f(f(x)), >> ect.] >> Say a real number x is recognizable iff x in X for some recognizable >> X. >> What numbers are recognizable? If b is a positive real number, let X be the set of all positive >integer multiples of b, and let f:R->R be f(x) = x-b. Then X is >recognizable, and so is b. The negative reals are handled similarly. >0 is of course recognizable. So the answer is all of them. >Yes. Kind of a dumb definition, I suppose. Wait a minute, what if X is >>required to be a *dense* subset of R? The rationals are recognizable with >>f:x->1/x-floor(1/x). No, they are not, at least, not with this function: your f is not >injective: both 3 and 3/4 map to 1/3: 3|-> (1/3) - floor(1/3) = >(1/3)-0 = 1/3. (3/4) |-> (4/3) - floor(4/3) = 4/3 - 1 = 1/3. You are, of course, correct! Ok, forget about all of R and just consider the interval [0,1). Then f:x->1/x-floor(1/x), f(0)=0 recognizes the rationals in [0,1). Are there any more recognizable numbers? [X a dense subset of [0,1); positive k; bijective f:[0,1)->[0,1)] thanks, rich ==== Call an infinite subset X of the real numbers recognizable if and only if there >is a bijection f:R->R such that for all x in X there is an integer k (possibly >dependent on x) with f^k(x)=0. [f^2(x)=f(f(x)), ect.] And I assume f^{-2}(x) means f^{-1}(f^{-1}(x)), etc... >Say a real number x is recognizable iff x in X for some recognizable X. What numbers are recognizable? THEOREM. All real numbers are recognizable. Proof. Given an arbitrary real number r, define g_r:R->R to be the function g_r(x) = x+r, and let X_r be: X_r = {kr: k an integer}. g_r is a bijection; and (g_r)^k(x) = x+kr for every integer k. In particular, for every x in X_r, x=kr for some k, we have (g_r)^{-k}(x)= x -kr = 0. Since r is in X_r, it follows that any real number is in a recognizable set, and therefore, that every real number is recognizable. QED -- ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >Call an infinite subset X of the real numbers recognizable if and only if >there >>is a bijection f:R->R such that for all x in X there is an integer k >(possibly >>dependent on x) with f^k(x)=0. [f^2(x)=f(f(x)), ect.] And I assume f^{-2}(x) means f^{-1}(f^{-1}(x)), etc... Fair assumption. I was thinking k>0, but didn't include that restriction. >Say a real number x is recognizable iff x in X for some recognizable X. >>What numbers are recognizable? THEOREM. All real numbers are recognizable. Proof. Given an arbitrary real number r, define g_r:R->R to be the >function g_r(x) = x+r, and let X_r be: X_r = {kr: k an integer}. g_r is a bijection; and (g_r)^k(x) = x+kr for every integer k. In >particular, for every x in X_r, x=kr for some k, we have >(g_r)^{-k}(x)= x -kr = 0. Since r is in X_r, it follows that any real number is in a >recognizable set, and therefore, that every real number is >recognizable. QED Yes, thanks. Are there any simple restrictions on k and/or f that will make the recognizable numbers a proper subset of R that includes the algebraic numbers? rich ==== >Call an infinite subset X of the real numbers recognizable if and only if >>there >is a bijection f:R->R such that for all x in X there is an integer k >>(possibly >dependent on x) with f^k(x)=0. [f^2(x)=f(f(x)), ect.] > >>And I assume f^{-2}(x) means f^{-1}(f^{-1}(x)), etc... Fair assumption. I was thinking k>0, but didn't include that restriction. Even so all numbers are recognizable: given an arbitrary real number r, the set X_r would become {-kr : k a positive integer}, and then X_r is the set recognized by g_r(x) = x+r. Then -r is in X_r, so every real number is in some X_r, hence every real number is recognizable. >Are there any simple restrictions on k and/or f that will make the >recognizable numbers a proper subset of R that includes the algebraic >numbers? I doubt it. The algebraic numbers are zeros of polynomials with rational coefficients, but polynomials are not usually one-to-one anyway... -- ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== The question was whether there are > necessarily arbitrary choices involved in constructing the > Riemann surface for a function... Then as a technique analytical continuation >is only practical with the aid of a computer program. >If cuts are >allowed I can treat some algebraic functions (I do not claim all) >without using it. Huh? Nobody's said anything about cuts not being allowed. > The question was [see above]. So I am allowed to use cuts after all. Perhaps now it can be agreed that there are two divergent theories, one in analysis, one into geometry. Then the question which relates to functions was answered in the first paragraph of my first post. ==== >The question was whether there are >> necessarily arbitrary choices involved in constructing the >> Riemann surface for a function... >>Then as a technique analytical continuation >>is only practical with the aid of a computer program. >If cuts are >>allowed I can treat some algebraic functions (I do not claim all) >>without using it. Huh? Nobody's said anything about cuts not being allowed. >> The question was [see above]. So I am allowed to use cuts after all. Perhaps now it can be agreed >that there are two divergent theories, one in analysis, one into >geometry. Then the question which relates to functions was answered >in the first paragraph of my first post. You're not making any sense. ************************ David C. Ullrich ==== > James Takayama posts under the names of Thomas, Nospam and Maui Cop. He > posted libel against me using the name of Nospam in the WaxyOrg web site and << The following (courtesy of Waxy.org) is sort of an unofficial FAQ explaining the psychotic nonsense posted to Usenet by Shawn Daryl Kabatoff AKA Dar, AKA Probababbilities. And now AKA marcia and me. WARNING: Read below before even thinking about responding to this twit. http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml#000643 Usenet has the tendency to provide a public forum for those who would normally be scribbling in a closet. For example, take Daryl Shawn Kabatoff. For the last few years, he's methodically gathered statistics from various sources, ranging from local newspaper obituary pages to the food court of the Saskatoon Midtown Plaza mall. With all the raw data he's collected, he's attempting to prove daily that our full names are in mathematical harmony with our birthdays. about, starting with calculations related to their birthdate and full names, blending in whatever other personal information about their family members, spouses, birthplace, and career he's been able to zealotry, and personal torment. I've never seen anything like it. With all the prime numbers, Fibonacci sequences and biblical references, it's like reading the notebooks of Maximillian Cohen and John Nash combined. Unsurprisingly, several posts unfold to reveal a history of painful mental illness. If you have some time, take a look. I've detailed his posting history and a several sample posts below. January 27, 1999 to July 5, 2000 as Catsco@home.com December 9, 2000 to May 4, 2001 as s.kabatoff@sk.sympatico.ca Oct 30, 2001 to Oct 31, 2001 as kabatoff@the.link.ca posts have been removed from Google Groups archive) Selected Posts: Tessa Lynne Smith Dastageer Sakhizai and Helen Smith Brett David Maki Andrew Meredith Cotton Amanda Dawn Newton Mona Marie Etcheverry Tony Peter Nuspl Lisa Charlene McMillan Grant Allyn Wood Comments scarier still is that saskatoon is my hometown, though not my current residence. and every single place he's mentioned in his posts (most notably nervous harold's and the roastary) were either places i've been (as it's a small city of 200K) or hangouts, ie. the two places out if they know of him, they (my friends that is) being of the broadway-centred slacker ilk. myself, too, until i got out of there. eh, anyways. thought it odd to see all this. midtown mall. i ate my meals there, whilst waiting several days in line for star wars episode one, at the theatre across the street. posted by andy raad on May 22, 2002 06:20 PM Fascinating. It's like he's trying to take chaos and bind it into whatever rules he can find, religious, logical and otherwise. Numbers and math have a reliable pattern, something that can always be proven to true or false. People and religion do not. It reminds me of Darren Aronofsky's movie Pi. It's the story of an paraniod genius who is trying to find a pattern in Pi. A group that takes interest in his work is convinced that the existence of Pi, a number whose existence can be proven but no quantified, is proof of the existence of God. Kabatoff's hunt for patterns in something as random as name selection is a way to reconcile his deeply logical thought process with his conflicting religious views. posted by matt on May 23, 2002 11:19 AM asking him if he'd be willing to create a numerological analysis for me. I also asked him if he had seen either Pi or A Beautiful Mind, and what he thought of them. If he replies, I'll be sure to post it. posted by Andy Baio on May 23, 2002 11:24 AM I baked many pumpkin pies for Shawn (he likes pumpkin pies). I rubbed pumpkin pie all over my breasts for him, and my breasts turned orange. I am a pumpkin for Shawn. posted by Trisha Blondie on July 24, 2002 10:41 PM Um, that's swell. So, you're in love with him? posted by Andy Baio on July 25, 2002 07:10 AM Shawn once went to a funeral for a Jehovah Witness that shot himself and the lemon tarts were very bad, they were not only sour but were rubbery as well. Shawn said that the guy was some kind of Jehovah Witness prophet, he saw in advance that the lemon tarts at his funeral were to be very very bad, and so he shot himself. Shawn said that he never ate pumpkin pie at a funeral but would like to some day. Shawn likes pumpkin pie and so I have been practicing to make very good pumpkin pies. posted by Trisha Blondie on July 25, 2002 02:49 PM Shawn said that the lemon tarts were sour, bitter and rubbery. posted by Trisha Blondie on July 30, 2002 12:32 AM I don't think this guy takes notes. I think he has Total Recall, and it has driven him insane... posted by Todd Smith on December 26, 2002 11:00 AM Oh... I almost forgot... I didnt spend thousands of dollars a day tormenting Daryl... We got a deal on tormenting that fiscal year, it only came to about 37cents a day.... posted by Dr Claw on December 30, 2002 01:56 AM Mr. Kabatoff attempts to portray himself as a victim, but in fact he is a violent predatory pedophile who is well known to his local law enforcement. In his post to multiple newsgroups with the subject Collecting Mail For The Coming Anti-Christ, he encourages mothers to send him photos of their naked daughters. Mr Kabatoff explains, I Ant-Christ) that were of underage children unless the parent was signing consent. He is banned from virtually all the shopping malls in his community because he stalks young people and sexually harasses them. He has an extensive arrest record which includes sexual molestation charges. He's been hospitalized in mental institutions about his contact with young girls in many posts. Search newsgroup archives for posts by him containing the word nubile. As part of his harrassment, he provides personal details in a public forum, such as the real names of real children, in these and other posts. About one wanted her and her sister dead. http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+dead+or+in+my+bed&hl= en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%40ID-136124.news.dfncis.de&rnu He not only curses children and prays for their death in his posts, he also enjoys attending the funerals of young people: And so, since nubile sweeties are found in greatest abundance at the funerals of high school students, then it is the funerals of high school students that make the very very best funerals, especially if there is food... I stuff my face (and my pockets) with all the good food and look at all the pretty nubile sweeties and have the time of my life.. .http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+nubile+sex&hl=en&l r=&ie=UTF-8&scoring=d&selm=LfXN8.63042%24R53.25142039%40twister.socal.rr. com&rnum=1 Many of his posts are sent to alt.teens.advice. However, he liberally spams, floods and crossposts his off-topic threatening and offensive missives to countless newsgroups. Some people HAVE problems and some folks ARE problems. Don't dismiss Mr. Kabatoff as a harmless nut. When he sends these posts to any newgroup, please help by reporting him to I knew of him when I was attending the University of Saskatchewan. He'd hang out in the Arts computer lab and all you'd see is screens of numbers racing by on his laptop. I have an original copy of his Collecting Mail for the Coming Anti-Christ pamphlet, and have seen him be hauled away by campus security on more than one occasion. My friends and I refer to him as Crazy Number Man. I've been posting to (and about) Shawn for over two years with big gaps in between. He has seen Pi and didn't like it and didn't think it resembled him at all. (Wrong, it fits him to a tee) He doesn't have total recall and has stated that he travels with a lap top to notate items. Also, he uses cut n' paste a lot if you read all the way through his ramblings. He is anti-social as shown by his angry statements towards those who, by his own admission, have been kind (but not kind enough) to him. Still, he's intelligent and seems to be able to take a joke on occassion. That's where I came in. ALOHA if it comes from anyone not already in my addressbook. I'll never even see it) ==== It was a dark and stormy night, and Marcia and Me managed to scribble: > James Takayama posts under the names of Thomas, Nospam and Maui Cop. He > posted libel against me using the name of Nospam in the WaxyOrg web site > and then quotes the material using the name Thomas, quoting your libel and > attributing it to someone else is just more libel. He has repeatedly cited > my interest in Callie to support his unfounded allegations that I am a > pedophile but Callie is not a child but is instead a woman. I have never > been hospitalized in mental institutions for accosting women (nor do I > accost women as he purports), but only for daring to criticize Protestant > and Catholic churches and daring to try to show people that God provides > them with their names (and for failing to abide by my parent's good > advice). -Daryl S. Kabatoff HALOHA!!! Well I only quote Daryl Shawn Kabatoff but year after year after year I tell people of my plight and of their filthy penises, The homosexuals will say that I am embarrased about my penis At some appeal panel hearings I am drugged so heavily that I do not know that I am even there http://groups.google.com/groups?selm=fPID6.299%24pp.2818127%40tomcat.sk.symp atico.ca gtoomey ==== What with the war on terror, and Canada's puny military and corrupt > police force, the only way to insure peace and safety for that > country, is for it to be annexed by America. It is the only way foward for Canada aka Canuckistan. >>You seem to forget that the Canadians kicked our ass in two conflicts.... >> Really? Which two? > 1775 >http://www.civilization.ca/cwm/chrono/1000invasion_e.html There was no country then called Canada. Ergo there were no Canadians. The people that lived there were simply British subjects just like their neighbors to the south pre-revolution. BTW the so-called canadians kicked so much ass the their neighbors to the south became truly independent while the Canadians had to wait another 80+ years. Well even longer considering Britain ran their foreign affairs well into the twentieth century. 1812 >http://www.civilization.ca/cwm/chrono/1774invasion_repelled_e.html > Your fantasizing again (see above explanation). Even if it was true it just shows how pathetic you are that you have to go back almost two hundred years to find something. dkp ==== What with the war on terror, and Canada's puny military and corrupt > police force, the only way to insure peace and safety for that > country, is for it to be annexed by America. It is the only way foward for Canada aka Canuckistan. >>You seem to forget that the Canadians kicked our ass in two conflicts.... >> Really? Which two? > 1775 >http://www.civilization.ca/cwm/chrono/1000invasion_e.html you apparently didn't read it very well. PLUS Canada didn't exist then, it was just another British puppet and the forces that worked their way north went after the British, not the so called Canadians. The area then was commonly called New France and the name Kanata is an indian word for cluster of villages or houses. Canada was formed in 1867 with the Constitution Act of 1867, and in Canada's original constitution the name of the new dominion would be . . . Canada. 1812 >http://www.civilization.ca/cwm/chrono/1774invasion_repelled_e.html that was a war with the British. Canada did not exist then. THOM ==== > What with the war on terror, and Canada's puny military and corrupt > police force, the only way to insure peace and safety for that > country, is for it to be annexed by America. It is the only way foward for Canada aka Canuckistan. >>You seem to forget that the Canadians kicked our ass in two conflicts.... >> Really? Which two? > 1775 >http://www.civilization.ca/cwm/chrono/1000invasion_e.html you apparently didn't read it very well. PLUS Canada didn't exist > then, it was just another British puppet and the forces that worked > their way north went after the British, not the so called Canadians. By that logic, those that were repelled weren't american, either. > The area then was commonly called New France and the name Kanata > is an indian word for cluster of villages or houses. And...? Canada was formed in 1867 with the Constitution Act of 1867, and in > Canada's original constitution the name of the new dominion would be . > . . Canada. And...? >1812 >http://www.civilization.ca/cwm/chrono/1774invasion_repelled_e.html that was a war with the British. Canada did not exist then. > THOM Wow, you really are a nation in denial... ==== Thom a .8ecrit dans le message de > you apparently didn't read it very well. PLUS Canada didn't exist > then, it was just another British puppet and the forces that worked > their way north went after the British, not the so called Canadians. > The area then was commonly called New France and the name Kanata > is an indian word for cluster of villages or houses. Canada was formed in 1867 with the Constitution Act of 1867, and in > Canada's original constitution the name of the new dominion would be . > . . Canada. 1812 >http://www.civilization.ca/cwm/chrono/1774invasion_repelled_e.html that was a war with the British. Canada did not exist then. THOM --------------------------------------------------------------------------- ---------- Jacques Cartier was the first Frenchman to write the word Canada, in 1535. -- http://www.civilisations.ca/vmnf/reper/glossair/r-ge-02.htm Anemic Combatant ==== How do I use the sinh, cosh functions on my calculator? I'm not sure what buttons to use. I have a Casio fx-991MS. ==== > How do I use the sinh, cosh functions on my calculator? I'm not sure what > buttons to use. I have a Casio fx-991MS. Should I really encourage to hyp(e) before sin? :-) I never used a casio but... Raymond ==== > How do I use the sinh, cosh functions on my calculator? I'm not sure what > buttons to use. I have a Casio fx-991MS. Is there an instruction booklet that came with it? ==== How do I use the sinh, cosh functions on my calculator? I'm not sure what > buttons to use. I have a Casio fx-991MS. Is there an instruction booklet that came with it? http://ftp.casio.co.jp/pub/world_manual/edu/en/fx570MS_991MS_E.pdf ==== How do I use the sinh, cosh functions on my calculator? I'm not sure what buttons to use. I have a Casio fx-991MS. >>Is there an instruction booklet that came with it? > http://ftp.casio.co.jp/pub/world_manual/edu/en/fx570MS_991MS_E.pdf For simpler operations try rather this one : http://ftp.casio.co.jp/pub/world_manual/edu/en/fx115MS_991MS_E.pdf For sinh do hyp sin x = For sinh^(-1) do hyp shift sin-1 x = and so on... Raymond ==== >> I bet you already knew that cos pi = -1 and such. But I was shocked to find, >> just now, that there appears to be a _vegetable_ in the domain of cos x. >> (Yes, you read that correctly. Obviously this is a silly pseudo-mathematical >> post, but then you could tell that from the thread's title.) What is that vegetable? What is the value of cos x there? (I guess I should wait for other people to try to answer first. But this is >> far too silly to be drawn out that way, and so I've given the answers below.) David The vegetable is lettuce. According to my dictionary: cos lettuce = romaine >You're wrong, lettuce is NOT in the domain of cos. It would be if it >was cos(lettuce) = romaine, which is not the case. I really like puns, which is why i'm feeling annoyed now - i'm not -- -john February 28 1997: Last day libraries could order catalogue cards from the Library of Congress. ==== >> I bet you already knew that cos pi = -1 and such. But I was shocked to >> find, just now, that there appears to be a _vegetable_ in the domain >> of cos x. (Yes, you read that correctly. Obviously this is a silly >> pseudo-mathematical post, but then you could tell that from the >> thread's title.) >> What is that vegetable? What is the value of cos x there? >> (I guess I should wait for other people to try to answer first. But >> this is far too silly to be drawn out that way, and so I've given the >> answers below.) >> The vegetable is lettuce. >> According to my dictionary: >> cos lettuce = romaine I really like puns, which is why i'm feeling annoyed now - i'm not Sorry. I didn't want anyone to be annoyed. There's nothing deep here. _The American Heritage Dictionary of the English Language_ says that cos lettuce is the same as romaine. [The reason for the name cos lettuce is that that type of lettuce was originally exported from the island of Cos (or Kos).] And so we have the well known cos pi = -1, etc., together with the silly (and perhaps not well known) cos lettuce = romaine, which appears to place lettuce in the domain of the cosine function. David ==== Does anyone know why for example my calculate can produce the argument and modulus for 1 - 3i no problems at all, but if I enter (1-3i)^4 I get a Math Error?