mm-1030 === Subject: Re: e is transcendental >O.k. then , as long there is no conßict with : >Re(e^[iPi]) = Re(-1+i[0]) = -1 That has never been in doubt. Neither has the observation that Im(e^[i pi]) = Im(-1+i[0]) = 0. The statement that caused all the problem was the statement that e^[i pi] = 0. === Subject: Re: Clean coordinates on a unit sphere >> Hi - IÕm a computer science student working in the field of vision and >> graphics. IÕm looking for a clean way to represent a _direction_ in 3d >> space (this can also be imagined as a position on a unit sphere). >> I donÕt like the traditional methods of using a 3d vector or spherical >> coordinates for the following reasons: >> - 3d Vector e.g. (x, y, z): This is of course very common - can be >> rotated with matrices, quaternions... but it also describes length, >> which I donÕt need. >> - Spherical coordinates e.g. (theta, phi): This is more appropriate as >> it does not describe length. But it still appears ugly to me since >> multiple coordinate pairs can describe the same point e.g. (x, pi/2). >> Is there a better way? IÕm sure IÕm not the first who asked this >> question. It seems similar to the problem of creating a 2d map of earth, >> so itÕs probably been thought about before... And since there doesnÕt >> seem to be a _popular_ solution, itÕs probably hard - but thatÕs what >> makes it FUN !!! >> My ideal solution would map a unit sphere with a uniform density (if >> that makes any sense to any one). Could a person use some type of >> tesselation of a sphere with an arbitrary density??? >> Any ideas are good ideas! Please keep the language simple - IÕm not as >> smart as you. >> Nathan > How about using unit vectors (x,y,z), i.e., satisfying x^2+y^2+z^2=1? > This seems clean, unless by clean you really mean efficient, since > (x,y,z) does contain redundant information. For efficiency, you could > just store (x,y,sign(z)), since then z is uniquely determined by z = > sign(z)*sqrt(1-x^2-y^2). But then youÕre back some reduncency, since > on the equator x^2+y^2=1, the sign of z is automatically zero. > You say My ideal solution would map a unit sphere with a uniform > density. Do you mean that you want a bijective map f:U-->S from a > subset of the Euclidean plane to the unit sphere that preserves area, > or at least so that for (nice) subsets V of U, f satisfies > Area(f(V))=c*Area(V) for a constant c that does not depend on V? IÕm > not quite sure how otherwise to interpret your word density. If you circumscribe a cylinder with radius r and height 2*r about the sphere, then the horizontal projection from the cylinderÕs lateral surface onto the sphere preserves area. Hence there is an area-preserving map from a rectangle with width 2*pi*r and height 2*r onto the surface of the sphere with radius r. There is still the objection that the mapping is not a bijection, but it does at least preserve area, if not cardinality. -- Dave Seaman Judge YohnÕs mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Interesting problem Trivially not ! take the element ab where a in A and b in B, then it is neither in A nor in B on account of the mult. closure > I have the following problem: > A and B are two disjoint sets whose union is |R+. Both A and B are closed > under sum and multiplication. Is it possible that neither A nor B is the > VOID set? > {V} === Subject: Re: Interesting problem >Trivially not ! >take the element ab where a in A and b in B, >then it is neither in A nor in B on account >of the mult. closure perhaps that isnÕt what is meant by multiplicative closure - that would seem to be ideal closure, but I believe it can be extended to show the answer. If ab is in B, and 1/b is in B, then so is a# so 1/b is in A. Now wlog 1 is in B, and hence so is 2, thus 1/2 is in A, hence 1/2+1/2=1 is in A# >> I have the following problem: >> A and B are two disjoint sets whose union is |R+. Both A and B are closed >> under sum and multiplication. Is it possible that neither A nor B is the >> VOID set? >> {V} === Subject: Re: Interesting problem >perhaps that isnÕt what is meant by multiplicative closure - that >would seem to be ideal closure, but I believe it can be extended to >show the answer. >If ab is in B, and 1/b is in B, then so is a# contradiction assuming a in A and ab in B, that is... >so 1/b is in A. Now wlog 1 is in B, and hence so is 2, thus 1/2 is in >A, hence 1/2+1/2=1 is in A# No, you could only say thus 1/2 is in A if there was some a in A such that 2a is in B. Of course that canÕt happen. Why canÕt 1/2, 1, 2 all be in B? In fact if 1 is in B, then all positive rationals are in B. Moreover, all rational powers of rationals are in B. But there are lots of other numbers, e.g. perhaps some transcendentals, that could be in A. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Interesting problem Robert Israel ha scritto nel messaggio > No, you could only say thus 1/2 is in A if there was some a in A such > that 2a is in B. Of course that canÕt happen. Why canÕt 1/2, 1, 2 all > be in B? In fact if 1 is in B, then all positive rationals are in B. > Moreover, all rational powers of rationals are in B. > But there are lots of other numbers, e.g. perhaps some transcendentals, > that could be in A. I can show that both A and B have to be dense in |R+. The problem i found in this problem are trascendental numbers.... i have no idea about a solution... === Subject: Re: Interesting problem <1mZ_b.18593$gk.818633@news3.tin.it> sci.math, Interesting problem >> A and B are two disjoint sets whose union is |R+. >> Both A and B are closed under sum and multiplication. >> Is it possible that neither A nor B is the VOID set? >In fact if 1 is in B, then all positive rationals are in B. >Moreover, all rational powers of rationals are in B. >But there are lots of other numbers, e.g. perhaps some >transcendentals, that could be in A. Generalization: b in B ==> bQ+ = { bq | q in Q, q > 0 } subset B b in B ==> b^Q+ = { b^q | q in Q, q > 0 } subset B b in B ==> [b^Q+ * Q+]^Q+ = b^Q+ * (Q+)^Q+ subset B and similar for A. Thus as Q+ subset B, B is dense subset R+ and if a in A, aQ+ subset A shows A is dense subset R+ ---- === Subject: Re: RFI:WanlessÕ Fourth Conjecture/DirichletÕs Geometric Theorem >For every pair of integers a, b with a > 1 and ab <> 2 (thus leaving >out the case of the Mersenne numbers), there are infinitely many >primes p such that a^p + b is composite. Sorry, thereÕs a typo here: it should say ab <> -2. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: measure of stuck-togetherness So the earthÕs surface is about one-quarter land. That land could be distributed across the surface of the globe in different ways - I can imagine at one extreme, all the land stuck together in one roundish continent, and at the other extreme, all the land spread out evenly, in equally-spaced tiny islands. Is there a measure, preferably just one number, that captures these differences? If so, how is it calculated? === Subject: Re: measure of stuck-togetherness >So the earthÕs surface is about one-quarter land. That land could be >distributed across the surface of the globe in different ways - I can >imagine at one extreme, all the land stuck together in one roundish >continent, and at the other extreme, all the land spread out evenly, in >equally-spaced tiny islands. Is there a measure, preferably just one >number, that captures these differences? If so, how is it calculated? I donÕt know if thereÕs an accepted measure, but hereÕs one for you: Partition the surface into many equal-sized regions. In each region, measure the percentage of land. Take the standard deviation across all the regions. ThatÕs your number. -- Matthew T. Russotto mrussotto@speakeasy.net Extremism in defense of liberty is no vice, and moderation in pursuit of justice is no virtue. But extreme restriction of liberty in pursuit of a modicum of security is a very expensive vice. === Subject: Re: measure of stuck-togetherness >So the earthÕs surface is about one-quarter land. That land could be >distributed across the surface of the globe in different ways - I can >imagine at one extreme, all the land stuck together in one roundish >continent, and at the other extreme, all the land spread out evenly, in >equally-spaced tiny islands. Is there a measure, preferably just one >number, that captures these differences? If so, how is it calculated? > I donÕt know if thereÕs an accepted measure, but hereÕs one for you: > Partition the surface into many equal-sized regions. In each region, > measure the percentage of land. Take the standard deviation across > all the regions. ThatÕs your number. I could be wrong, but that doesnÕt seem to work. Call each of the equal-sized regions a cell, and suppose there are n cells. Let p_k be the proportion of the kth cell-area thatÕs land. Then the number you describe is m = s.d.{p_1,...,p_n}. If P is the overall proportion of the global area thatÕs land, then avg{p_1,...,p_n} = P, and m**2 = (1/n) sum( (p_k - P)**2, k=1..n ). There seem to be two cases: n is either large enough to give a stable value for m (so making n even larger wonÕt matter much), or n is too small (and m will be reßecting n rather what itÕs supposed to be measuring). In the acceptable, stable case, it seems that n would eliminate most boundary effects, leaving a sum that comes mainly from two contributions: m**2 ~ (1/n) [ (#cells on land)(1-P)**2 + (#cells off land)(0-P)**2 ] ~ (1/n) [ n P (1-P)**2 + n (1-P) (0-P)**2 ] ~ P(1-P) In other words, either n is too small for m to measure what itÕs supposed to, or n is large enough but then m measures only the global property P(1-P). --r.e.s. === Subject: re:measure of stuck-togetherness Suppose that only the land on the surface has mass, and the rest of the earth (the oceans and the interior of the earth) is massless. We still suppose the earth holds together. In each point X in the interior of the earth measure the absolute value of the gravitaional pull F(X) and gravitational potential E(X). The following are possible measures of dispersion: 1) max F(X) 2) mean F(X) or mean F(X)^2 etc. 3) max (E(X) - E(Y)) 4) mean (E(X) - E(Y)) or mean (E(X) - E(Y))^2 etc. For uniform distribution those are all about zero. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: measure of stuck-togetherness > So the earthÕs surface is about one-quarter land. That land could be > distributed across the surface of the globe in different ways - I can > imagine at one extreme, all the land stuck together in one roundish > continent, and at the other extreme, all the land spread out evenly, in > equally-spaced tiny islands. Is there a measure, preferably just one > number, that captures these differences? If so, how is it calculated? YouÕve received lots of suggestions, but let me just add the fractal dimension of the boundary (coastline). - Risto - === Subject: Re: measure of stuck-togetherness > So the earthÕs surface is about one-quarter land. That land could be > distributed across the surface of the globe in different ways - I can > imagine at one extreme, all the land stuck together in one roundish > continent, and at the other extreme, all the land spread out evenly, in > equally-spaced tiny islands. Is there a measure, preferably just one > number, that captures these differences? If so, how is it calculated? Various measures of dispersion could work. For points x and y on the surface, let dist(x,y) be the length of the shortest great-circle arc connecting x and y. Then two such measures are E dist(x,y) and sqrt(E dist(x,y)**2), where x and y are iid uniform on the land-region. As approximations, you could replace the expectations by averages with x and y suitably discretized. --r.e.s. === Subject: Re: measure of stuck-togetherness Standard deviation of Statistical mathematics, measures the difference. === Subject: Re: measure of stuck-togetherness > So the earthÕs surface is about one-quarter land. That land could be > distributed across the surface of the globe in different ways - I can > imagine at one extreme, all the land stuck together in one roundish > continent, and at the other extreme, all the land spread out evenly, in > equally-spaced tiny islands. Is there a measure, preferably just one > number, that captures these differences? If so, how is it calculated? How about how many islands there are or the area of the largest one. Depends on what you mean by differences. Bill === Subject: Re: measure of stuck-togetherness > So the earthÕs surface is about one-quarter land. That land could be > distributed across the surface of the globe in different ways - I can > imagine at one extreme, all the land stuck together in one roundish > continent, and at the other extreme, all the land spread out evenly, in > equally-spaced tiny islands. Is there a measure, preferably just one > number, that captures these differences? If so, how is it calculated? The perimeter is one such measure. === Subject: Re: The Dot and Line--A Movie They played it this morning. Unfortunately, they cannot foretell much in advance when certain shorts will be shown. I guess IÕll just have to have my VHS recorder ready for action. >>The animator Chuck Jones directed a animated feature in 1965 called The Dot >>and Line, >>which was nominated for an Academy Award. I happened to see part of it on the >>TMC >>channel this morning. Does anyone know if itÕs on video, VHS or DVD? I went >>to Amazon >>and looked and found mention of it on one Chuck Jones feature, but a reviewer >>noted >>that it was incomplete. IÕm told it sometimes appears on The Cartoon Channel. > There is a book, from which I think the movie was made, called > The Dot and the Line, A Romance in Lower Mathematics by Norton Juster > See for exaxmple > http://www.amazon.com/exec/obidos/tg/detail/-/1587170663/103- 8735420-9207 > 022?v=glance -- Wayne T. Watson (121.015 Deg. W, 39.262 Deg. N, 2,701 feet, Nevada City, CA) DonÕt hassle me with your signs, Chuck -- Peppermint Patty, Peanuts Web Page: sierra_mtnview -at- earthlink -dot- net Imaginarium Museum: === Subject: Re: Bounds on Moments > Suppose x is a non-negative random variable. Given that the first > moment of x, E(x^1) and E(x^2) are m1 and m2, respectively. What are > the bounds upon m3=E(x^3)? Why should we expect such a bound? For example, let f(x) = c/(1+x^4), where c is chosen so that the integral of f over (0,oo) is 1. Then for the probability density function f we have E(x^1) and E(x^2) < oo, but E(x^3) = oo. === Subject: Re: More transcendentals > The other transcendentals than e and pi got my quriosity perched. > So besides the obvious generalization to LousvilleÕs construction, where one > has all the digits after the decimal point equal to k except the digits at > positions n! which might equal m=/=k, are there any famous decimals that are > constructed by using digits taken from a particular sequence a_n? The factorial base allows you to create a lot of transcendentals from e. The allowed coefficients at position n are (0 through n). Each position represents 1/n! (or n! left of radix). Every finite representation in base ! is rational. So we only care about infinite representations. Let c = e-2. c = 0.111... base ! (1/2 + 1/6 + 1/24 + ...) Let q be a rational number. q * c = transcendental. I think this means that any infinite base ! representation that has a repeating period can be shown to be of the form q * c, q rational. I donÕt have a proof yet. (multiplication rules for base ! are complicated.) 1.12 * 0.111... (base !) is transcendental. 1.12 * .1 = 1+1/2+2/6 (11/6) * 1/2 = 11/12 = .122 (base !) 1.12 * .11 = 11/6 * 2/3 = 11/9 = 1.01114 (base !) It would be interesting if PI has a repeating base ! representation. number. I have no idea what this means. Russell - 2 many 2 count === Subject: Re: More transcendentals Ì Russell Easterly ó.8d.98.87.8b.8c .97.99.95 .92.86.94.9d.92.87 > The factorial base allows you to create a lot of transcendentals from e. > The allowed coefficients at position n are (0 through n). > Each position represents 1/n! (or n! left of radix). > Every finite representation in base ! is rational. > So we only care about infinite representations. > Let c = e-2. [snip actual constructions for brevity] have been saved and will be analyzed by noodle at school. > Russell > - 2 many 2 count -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === Subject: Re: More transcendentals > You might be interested in one or more of the following. -snip- > MR0965057 (90a:11082) > Loxton, J. H.(5-MCQR); van der Poorten, A. J.(5-MCQR) > Arithmetic properties of automata: regular sequences. > J. Reine Angew. Math. 392 (1988), 57--69. > 11J85 (11K16 11R04 68Q70) > Not very much is known about the decimal expansions of algebraic > irrational numbers. One might conjecture that these expansions are > random, or even normal. In this paper it is shown that they cannot be > generated by a finite automaton. The result appears as a corollary of a > theorem on algebraic independence. Therefore, heavy Mahler-type > machinery is used. > Reviewed by F. Schweiger I thought it might be worth mentioning that the reviewer apparently overstated what was proved in the paper when he said In this paper it is shown that they cannot be generated by a finite automaton. According to Allouche & Shallit (in their book Automatic Sequences, paper by Cobham ... ... which was a step towards proving that all numbers in L(k,b) are either rational or transcendental. This last assertion was approached by Cobham [1968b] (who incorrectly claimed a proof) and Loxton and van der Poorten [1982, 1988], but the general result has not been fully proved yet. Here L(k,b) is the set of all (k,b)-automatic reals. --r.e.s. === Subject: Re: BETA-BINOMIAL EXTENSION FORMULA ??? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1P2cGO15493; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1P2UBi14992 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1P2UAk02756; >I am a programmer working on a project that performs calculations on TV >ratings data. The beta-binomial extension formula is used to express reach >and frequency information (reach and frequency refer to how many people view >a schedule of commercials and how often they view them, respectively). >However, I cannot get my hands on the formula itself. Does anyone know what >PS: I have a mathematics background, but it is based more on theory than >statistics. Could this formula have something to do with a beta >distribution? > === Subject: Calculus help please! hi, iÕm having troubles trying to understand section 7.5 out of StewartÕs single variable calculus 5e - does anyone know of any good sites ( or could possibly explain to me in a nutshell) inverse trig === Subject: ArcCos vs arcCos I recall that one of ArcCos and arcCos which is a function. I think that ArcCos is the function but I was hoping that someone can confirm this for === Subject: Re: ArcCos vs arcCos > I recall that one of ArcCos and arcCos which is a function. I think that > ArcCos is the function but I was hoping that someone can confirm this for ThereÕs no way to confirm this because there are two competing, diametrically opposite conventions. See, below, a copy of a portion of an old post of mine for details. (But in any event I canÕt see any reason to capitalize the second c in Arccos -- unless youÕre using Mathematica, that is.) David Cantrell ---------------------------------------- There is a convention according to which, if a multivalued relation is denoted in lower case, then capitalization is used to distinguish the corresponding principal-valued function. [Referring again to the entry for principal value in the _HarperCollins Dictionary of Mathematics_, The function that has as its values the principal values of a many-valued function is conventionally indicated by writing it with a capital letter; thus, Cotan^(-I) is the principal value of the inverse cotangent, and, especially in complex functions, Ln is the principal value of the natural logarithm.] ... Unfortunately, another convention is the precise reverse of that described above! [For example, in _An Atlas of Functions_, arcsin denotes the inverse sine function while Arcsin denotes the multivalued relation.] And yet another practice is to make no notational distinction whatsoever between a multivalued relation and the corresponding principal- valued function! [See, for example, the treatment of inverse trigonometric functions in the CRC Standard Mathematical Tables and Formulae, 30th ed.] === Subject: Re: ArcCos vs arcCos >> I recall that one of ArcCos and arcCos which is a function. I think that >> ArcCos is the function but I was hoping that someone can confirm this for >ThereÕs no way to confirm this because there are two competing, >diametrically opposite conventions. See, below, a copy of a portion of >an old post of mine for details. (But in any event I canÕt see any >reason to capitalize the second c in Arccos -- unless youÕre using >Mathematica, that is.) ThereÕs a fairly common convention in calculus books (when defining inverse trig functions - it almost never shows up outside that section) to write Cos for the restriction of cos to the interval [0,pi]. So (although I donÕt recall seeing anybody actually do it), itÕs possible the function Cos, not of cos. I donÕt know why youÕd want to capitalize the A as well, unless you just donÕt want it to look like a Java method. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: ArcCos vs arcCos >> I recall that one of ArcCos and arcCos which is a function. I think >> that ArcCos is the function but I was hoping that someone can confirm >ThereÕs no way to confirm this because there are two competing, >diametrically opposite conventions. See, below, a copy of a portion of >an old post of mine for details. (But in any event I canÕt see any >reason to capitalize the second c in Arccos -- unless youÕre using >Mathematica, that is.) > ThereÕs a fairly common convention in calculus books (when defining > inverse trig functions - it almost never shows up outside that section) > to write Cos for the restriction of cos to the interval [0,pi]. So > (although I donÕt recall seeing anybody actually do it), itÕs possible > the function Cos, not of cos. you, I canÕt recall having ever seen anyone actually use arcCos). I might also add that one does sometimes see Cos^{-1} used to denote the principal-valued inverse cosine function. David David > I donÕt know why youÕd want to capitalize > the A as well, unless you just donÕt want it to look like a Java method. === Subject: Re: Do Prime Algebraic Numbers even exist? I am reminded of one error and one deficiency in my original post, so some additions: > In sci.math, Dik T. Winter > > : > Indeed, but there are *no* primes in the algebraic integers. Now you > could define class 1 primes to be primes in the integers, but I fail > to see how you could define class 2 primes or anything else. Not > all primes in number fields are of the form n-th root of p with p > a prime. For instance, one of the primes in Q(sqrt(2)) is 1 + sqrt(2) > (I think). I should have thought further, as I was reminded, 1 + sqrt(2) is a unit, so no prime. To get a prime we have to know something about norms in quadratic fields. In Q(sqrt(m)) the norm of a number a + b.sqrt(m) is a^2 - m.b^2. So the norm of 1 + sqrt(2) = -1, and hence it is a unit. You may verify that the norm so defined is multiplicative, also that the norm is integer if and only if a and b are both integers, or are both half of an odd integer when m = 1 mod 4. And these are precisely the algebraic integers in the field. To get a prime you look for a number with (integer) prime norm. In this case 3 + sqrt(2), with norm 7. > The number of units in quadratic fields (Q(sqrt(m)) ) is as follows: > m > 0: infinitely many > m = -1: 4 > m = -3: 6 > m < 0: 2 in all other cases. > (again, assuming m to be square free.) > However, when m > 0 it is not easy to always find a unit. What I meant here is that it may take a lot of work. There are deterministic methods to get a unit (or the fundamental unit), but the computational complexity becomes pretty big. I have done extensive calculations with discriminant < 10000. (The discriminant is equal to m when m = 1 mod 4, otherwise it is 4m.) The first one I did not complete was D = 409 (m = 409) where in the fundamental unit x + y.sqrt(m), x exceeds 10^11. And there are 19 more in the range D < 1000. (Yes, I could have gone to some bignum package, but did not feel inclined to do so...) This is similar to the case of showing the gcd of two algebraic integers, even if they are both from a quadratic field. You may have reasons to know that there is a non-unit common factor, but showing its value is something else. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: antisymmetric relation = ~symmetric? > Recall that antisymmetric mean aRb ^ bRa => a = b, > or for all a not = b, aRb => ~bRa. > In other words this relation is never symmetric! No! Equality is both symmetric and assymmetric. Assymmetric is what you call antisymmetric. Antisymmetric is not(aRb & bRa). === Subject: Exortic Coordinate Systems. Hello All, I need to create the direction cosines for an orthogonal coordinate system for the following two geometries: 1. A cylynder of varying elliptical cross section. 2. A torid of varying elliptical cross section. I know how do derive the direction cosines for the three orthogonal directions (radial, circumferential, and longitudinal) at every point, but it is fairly complex. Before I embark on this adventure, I want to make sure this informaton is not readily available elsewhere. Since my geometry is not particularly complex (just modified cylynders and toroids), I suspect that this has been done before. I am looking for references. If you know of any, plese let me know. Nanri. Ashok.R === Subject: Re: Bounds on Moments by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1P4mTh27257; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1P4iBi27094 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1P4iBd15034; >> Suppose x is a non-negative random variable. Given that the first >> moment of x, E(x^1) and E(x^2) are m1 and m2, respectively. What are >> the bounds upon m3=E(x^3)? >Why should we expect such a bound? For example, let f(x) = c/(1+x^4), where >c is chosen so that the integral of f over (0,oo) is 1. Then for the >probability density function f we have E(x^1) and E(x^2) < oo, but E(x^3) = >oo. No upper bound, but there is a lower bound. For example, for any p >= 0 we have Q(t) = t (t-p)^2 >= 0 for all t >= 0, so E[Q(x)] = m3 - 2 p m2 + p^2 m1 >= 0, i.e. m3 >= 2 p m2 - p^2 m1. In particular, take p = m2/m1 to get m3 >= m2^2/m1. Conversely, consider a distribution with mass m1^2/m2 at m2/m1 and 1-m1^2/m2 at 0 (noting that 0 < m1^2/m2 <= 1): we have E[x] = m1, E[x^2] = m2 and E[x^3] = m2^2/m1, so that bound is best possible. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: the anticlassicalist }{ ii: the spectre continues sci.lang,sci.logic,sci.math: [...] >:> Do you disagree with any of the points I have made in the towards a >:> constructive education or more focus... posts? Or do you believe >:> that I have in some other way violated the constraints of the groupsÕ >:> topicalities? >: You obviously have in respect of sci.lang. You also >: obviously either violated one of the first principles of >: netiquette, namely, that one should become familiar with the >: actual content and customary practices of a newsgroup before >: posting to it, or deliberately posted to sci.lang something >: that you should have known was inappropriate. Quite clearly >: you *do* feel entitled to grab attention at othersÕ expense. > Expense? Yes, expense. You are, for example, directly responsible for cluttering sci.lang with off-topic mathematics and complaints about the lack of physical content in your posts from sci.physics. > I took the time to write several long posts concerning a topic > relevant to all newsgroups posted to. In your opinion. The relevance is not apparent to many of those better qualified than you to judge. [...] > As for sci.lang, obviously there are problems in understanding reasoning on > topological trees (and linguistic phylogeny) that have been evidenced > several times recently in that forum. Jacques erred in calling his three trees topologically equivalent (though in fact two of them are -- as unlabelled trees), but I doubt that he was trying very hard: youÕve offered nothing to suggest that you should be taken any more seriously on the subject of linguistic phylogeny than, say, Cavalli-Sforza, who is a linguistic ignoramus. > Plus the whole cognitive origins of > language, semiotics and natural language models, etc. features of my > exposition that IÕve been willing to discuss in depth make your statement > patently false. Bollocks. IÕve not seen you offer anything substantive on any of those topics, which with the possible exception of the first are in any case not topics of great interest to most of the regulars here, judging by the eight or so years that IÕve been reading the group. > Oh, did you miss the thread on modality in language as well? What, Andrew PattersonÕs nonsense? That was (a) independent of your posturing and (b) obviously received largely with indifference. > Yeah, IÕm reading... > Are you? Everything? Of course not. I read the threads in which the professional linguists and knowledgeable sci.lang regulars participate. IÕve also read some of the stuff dragged in from sci.physics. I read this bit because from the days when I used to read sci.math I recognized Keith Ramsay as someone likely to have something sensible to say. >: [...] >:> I have seen many pleas against the cross posting. >: Which is prima facie evidence that it was inappropriate. It >: is ultimately the members of a newsgroup who determine what >: is appropriate, not some document. If I want to read >: physics, IÕll go to sci.physics; I donÕt want it cluttering >: up sci.lang. If I want to read mathematics, IÕll go to >: sci.math. If I want to read logic, IÕll go to sci.logic. >: If I want to read philosophy, IÕm ill. > No, itÕs evidence of nothing of the kind! None of those pleas has ever > described in any way how I have violated the topicality of their newsgroup. You still donÕt get it, do you? WhatÕs on topic is determined by the group itself, especially when it has a core of regulars. If youÕd paid the slightest attention, youÕd have recognized that sci.lang runs to historical linguistics (about which youÕve demonstrated considerable ignorance) and sociolinguistics, with occasional forays into syntax. (Oh, and food.) > Even you didnÕt make any such description above. You just accuse and fight > your alpha games like you are the arbiter of truth and justice. Not at all, though I fear that you are far too self-centred and convinced of your own virtue to recognize that what you call an accusation was in fact (1) a partial explanation of the reception youÕve had, (2) implicit advice on how to get a better reception, and (3) a fairly mild expression of exasperation. In point of fact I have for the most part treated you as you deserve: apart from a brief observation and followup when you first appeared, I have ignored you. But your question to Keith was too inviting an opportunity to be ignored. > Your desire to want to avoid certain discussions of an interdisciplinary > nature is easily avoided by ignoring threads you find distasteful. I can > walk you through that procedure if you are having any difficulties. The interdisciplinary nature of these ÔdiscussionsÕ is mostly in your own mind. WhatÕs been posted here has mostly been mathematics and handwaving. >:> None of them have been >:> very convincing in my opinion, >: Which is largely irrelevant. > Unfortunately, that is quite relevant. Convincing me is the only way > someone is going to get me to stop posting. I quite believe it. I was discussing the facts of the matter, to which your opinion is largely irrelevant, and not your future behavior. their own >:> newsgroups, not asking for intelligent discussion, just >:> insults and the like. >: Which again is a very good indication that your content was >: widely considered inappropriate. Like it or not, many >: newsgroups are communities. Outsiders are not necessarily >: unwelcome, but outsiders who barge in and presume to lecture >: from a pedestal are likely to get the rough reception that >: theyÕve earned. > No, its an indication that there are quite a lot of jerks out there who, > when faced with a topic they do not understand and do not want to > understand, find solace in insults. I do like the fact that newsgroups are > communities, particularly that they are communities of wide ranges of views > about the topics they discuss. There are certainly members, such as > yourself mister Scott, who dislike the fact that others may begin a > discussion confident of the knowledge that they have such a right, but > unfortunately you are in the wrong and I am in the right. And playing your > power games, with their complete absence of any rational points, just > illustrates to me that you recognise your complete lack of power in this > circumstance. IÕm afraid that itÕs you who are playing power games. You are the one thumbing his nose at the rest of us and going ÔNyaa, you canÕt make me leaveÕ. In this you are quite correct: Usenet is an open forum, and I wouldnÕt have it any other way. But just as you are free to shove your id.8ee fixe in front of peopleÕs faces at preposterous (and singularly ineffective) length, so am I free to point out that you are doing so. >: Bluntly, youÕre a rude, arrogant bastard with the social >: intelligence of a pet rock. On top of that you write some >: of the ßabbiest, most turgid prose that itÕs been my >: misfortune to read anywhere, let alone on Usenet, and >: exhibit several of the familiar stigmata of the Usenet crank >: or monomaniac. If you donÕt like your reception, mend your >: manners. > I have never been rude to anyone who was not first rude to me, and then only > enough to play the alpha game they initiated to its proper closure. You are mistaken, owing to your inability to recognize the rudeness of your behavior thus far. > I donÕt > seek contentless arguments; it is others who feel inclined to provide me > with such. I am a bastard; that is true. I was born with an unmarried > mother. Irrelevant, since I use the term in its figurative sense, and certainly no concern of mine in any case. > I can be quite humble when speaking to others who engage in > rational critique or otherwise educate me of my errors. I see you borrowed No, I did not. I have no idea even to whom you refer. But I am hardly surprised that someone else used so obviously apt a description. [...] If you are serious about getting together those who are genuinely interested in your views, I suggest that you set up a web-based bulletin board; IÕm given to understand that this is very easy to do these days. You can then announce it in the newsgroups in which you think it might be of interest. Reply or not, as you wish; IÕll not be responding again outside of threads with genuine linguistic content, if I even bother to read. === Subject: Re: the anticlassicalist }{ ii: the spectre continues > Jacques erred in calling his three trees topologically > equivalent (though in fact two of them are -- as unlabelled > trees) Well, in fact, I would not mind arguing that they are. But since I left the vital bit out, arguing so would smack of a post-hoc argument. > but I doubt that he was trying very hard: youÕve > offered nothing to suggest that you should be taken any more > seriously on the subject of linguistic phylogeny than, say, > Cavalli-Sforza, who is a linguistic ignoramus. Just right. IÕll just give a hint of what I left out, as to me it went without saying for any comparative linguist with a smattering of topology and graph theory. The arcs are not one-dimensional lines, they are (sorry, donÕt know the proper jargon) they are ribbons, paths, roads with a _width_. > Unfortunately, that is quite relevant. Convincing me is the only way > someone is going to get me to stop posting. Post, post, post, post. And post. On sÕen bat lÕoeil, Ducon. (Ack, argh, help! Ad hominem, ad hominem! --No, ad mentulam, dickhead--shit, whereÕs the editor of Maledicta when we need him?) their own >:> newsgroups, not asking for intelligent discussion, just >:> insults and the like. At least, insults, in these circumstances, have the merit of being honest. JÕappelle un chat un chat et Rollet un fripon. As for you, gaga-lact.8ee, itÕs all mealy-mouthed innuendos and the arguments statements ex cathedra perc.8ee. > No, its an indication that there are quite a lot of jerks out there who, > when faced with a topic they do not understand and do not want to > understand, find solace in insults. Well, well, well... like people who, faced with matters of comparative linguistics, find solace in calling comparative linguists jerks eh? === Subject: Re: the anticlassicalist }{ ii: the spectre continues sci.lang,sci.logic,sci.math: [...] >:> Do you disagree with any of the points I have made in the towards a >:> constructive education or more focus... posts? Or do you believe >:> that I have in some other way violated the constraints of the groupsÕ >:> topicalities? >: You obviously have in respect of sci.lang. You also >: obviously either violated one of the first principles of >: netiquette, namely, that one should become familiar with the >: actual content and customary practices of a newsgroup before >: posting to it, or deliberately posted to sci.lang something >: that you should have known was inappropriate. Quite clearly >: you *do* feel entitled to grab attention at othersÕ expense. > Expense? Yes, expense. You are, for example, directly responsible for cluttering sci.lang with off-topic mathematics and complaints about the lack of physical content in your posts from sci.physics. > I took the time to write several long posts concerning a topic > relevant to all newsgroups posted to. In your opinion. The relevance is not apparent to many of those better qualified than you to judge. [...] > As for sci.lang, obviously there are problems in understanding reasoning on > topological trees (and linguistic phylogeny) that have been evidenced > several times recently in that forum. Jacques erred in calling his three trees topologically equivalent (though in fact two of them are -- as unlabelled trees), but I doubt that he was trying very hard: youÕve offered nothing to suggest that you should be taken any more seriously on the subject of linguistic phylogeny than, say, Cavalli-Sforza, who is a linguistic ignoramus. > Plus the whole cognitive origins of > language, semiotics and natural language models, etc. features of my > exposition that IÕve been willing to discuss in depth make your statement > patently false. Bollocks. IÕve not seen you offer anything substantive on any of those topics, which with the possible exception of the first are in any case not topics of great interest to most of the regulars here, judging by the eight or so years that IÕve been reading the group. > Oh, did you miss the thread on modality in language as well? What, Andrew PattersonÕs nonsense? That was (a) independent of your posturing and (b) obviously received largely with indifference. > Yeah, IÕm reading... > Are you? Everything? Of course not. I read the threads in which the professional linguists and knowledgeable sci.lang regulars participate. IÕve also read some of the stuff dragged in from sci.physics. I read this bit because from the days when I used to read sci.math I recognized Keith Ramsay as someone likely to have something sensible to say. >: [...] >:> I have seen many pleas against the cross posting. >: Which is prima facie evidence that it was inappropriate. It >: is ultimately the members of a newsgroup who determine what >: is appropriate, not some document. If I want to read >: physics, IÕll go to sci.physics; I donÕt want it cluttering >: up sci.lang. If I want to read mathematics, IÕll go to >: sci.math. If I want to read logic, IÕll go to sci.logic. >: If I want to read philosophy, IÕm ill. > No, itÕs evidence of nothing of the kind! None of those pleas has ever > described in any way how I have violated the topicality of their newsgroup. You still donÕt get it, do you? WhatÕs on topic is determined by the group itself, especially when it has a core of regulars. If youÕd paid the slightest attention, youÕd have recognized that sci.lang runs to historical linguistics (about which youÕve demonstrated considerable ignorance) and sociolinguistics, with occasional forays into syntax. (Oh, and food.) > Even you didnÕt make any such description above. You just accuse and fight > your alpha games like you are the arbiter of truth and justice. Not at all, though I fear that you are far too self-centred and convinced of your own virtue to recognize that what you call an accusation was in fact (1) a partial explanation of the reception youÕve had, (2) implicit advice on how to get a better reception, and (3) a fairly mild expression of exasperation. In point of fact I have for the most part treated you as you deserve: apart from a brief observation and followup when you first appeared, I have ignored you. But your question to Keith was too inviting an opportunity to be ignored. > Your desire to want to avoid certain discussions of an interdisciplinary > nature is easily avoided by ignoring threads you find distasteful. I can > walk you through that procedure if you are having any difficulties. The interdisciplinary nature of these ÔdiscussionsÕ is mostly in your own mind. WhatÕs been posted here has mostly been mathematics and handwaving. >:> None of them have been >:> very convincing in my opinion, >: Which is largely irrelevant. > Unfortunately, that is quite relevant. Convincing me is the only way > someone is going to get me to stop posting. I quite believe it. I was discussing the facts of the matter, to which your opinion is largely irrelevant, and not your future behavior. their own >:> newsgroups, not asking for intelligent discussion, just >:> insults and the like. >: Which again is a very good indication that your content was >: widely considered inappropriate. Like it or not, many >: newsgroups are communities. Outsiders are not necessarily >: unwelcome, but outsiders who barge in and presume to lecture >: from a pedestal are likely to get the rough reception that >: theyÕve earned. > No, its an indication that there are quite a lot of jerks out there who, > when faced with a topic they do not understand and do not want to > understand, find solace in insults. I do like the fact that newsgroups are > communities, particularly that they are communities of wide ranges of views > about the topics they discuss. There are certainly members, such as > yourself mister Scott, who dislike the fact that others may begin a > discussion confident of the knowledge that they have such a right, but > unfortunately you are in the wrong and I am in the right. And playing your > power games, with their complete absence of any rational points, just > illustrates to me that you recognise your complete lack of power in this > circumstance. IÕm afraid that itÕs you who are playing power games. You are the one thumbing his nose at the rest of us and going ÔNyaa, you canÕt make me leaveÕ. In this you are quite correct: Usenet is an open forum, and I wouldnÕt have it any other way. But just as you are free to shove your id.8ee fixe in front of peopleÕs faces at preposterous (and singularly ineffective) length, so am I free to point out that you are doing so. >: Bluntly, youÕre a rude, arrogant bastard with the social >: intelligence of a pet rock. On top of that you write some >: of the ßabbiest, most turgid prose that itÕs been my >: misfortune to read anywhere, let alone on Usenet, and >: exhibit several of the familiar stigmata of the Usenet crank >: or monomaniac. If you donÕt like your reception, mend your >: manners. > I have never been rude to anyone who was not first rude to me, and then only > enough to play the alpha game they initiated to its proper closure. You are mistaken, owing to your inability to recognize the rudeness of your behavior thus far. > I donÕt > seek contentless arguments; it is others who feel inclined to provide me > with such. I am a bastard; that is true. I was born with an unmarried > mother. Irrelevant, since I use the term in its figurative sense, and certainly no concern of mine in any case. > I can be quite humble when speaking to others who engage in > rational critique or otherwise educate me of my errors. I see you borrowed No, I did not. I have no idea even to whom you refer. But I am hardly surprised that someone else used so obviously apt a description. [...] If you are serious about getting together those who are genuinely interested in your views, I suggest that you set up a web-based bulletin board; IÕm given to understand that this is very easy to do these days. You can then announce it in the newsgroups in which you think it might be of interest. Reply or not, as you wish; IÕll not be responding again outside of threads with genuine linguistic content, if I even bother to read. === Subject: Re: convolution of measures Since weÕre talking about convolution of measures, is there any Fourier stuff we can do (since this would transform convolution into multiplication in the Fourier transform space)...? >Suppose M and N are distributions (measures), and itÕs given that >M * N = d_1, >where * indicates convolution and d_1 indicates the delta measure >concentrated at 1. How do you show that either M = d_c or N = d_c for >some constant c? This should probably follow from some really simple >manipulations with convolutions, but I lamentably donÕt see how to do >it. > Here is a solution which works for finite measures; perhaps it can be > modified by approximation to all measures? > The problem is equivalent to showing that if X and Y are independent > random variables and X + Y is almost surely 1, then either X or > Y is almost surely constant. Actually, they both are, since > Var(X + Y) = Var(X) + Var(Y) = 0. > -- > Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: convolution of measures >Since weÕre talking about convolution of measures, is there any Fourier >stuff we can do (since this would transform convolution into multiplication >in the Fourier transform space)...? Yes, you could look at it via the Fourier transform. I was going to mention this, decided not to bother, since working out all the details would be more work than, say, working out the details in what Edgar suggested. But since you ask: First we need to clarify what sort of measures weÕre talking about; itÕs been pointed out that the assertion becomes false with some sorts of measures. LetÕs assume that theyÕre probability measures (this _must_ be what the OP meant, else M = 2 d_0, N = d_0/2 would be a trivial counterexample...) Now we have (M^)(N^) = 1 and |M^| <= 1, |N^| <= 1. It follows that |M^| = 1, and it follows from _that_ that M = d_c So how does that last bit follow? One could give a slightly fussy direct argument... ah: thereÕs a simple elegant argument from a not-quite-trivial fact: If M is a complex measure a result of Wiener shows that sum_{x in R} |M({x})|^2 = lim_{T->infinity} int_{-T}^T |M^(t)|^2 dt. (I leave it to you to figure out which ^Õs denote the Fourier transform and which ones denote exponentiation. I think this result is in Stein&Weiss Fourier Analysis on Euclidean Spaces or Stein Singular Integrals; the corresponding result for the circle is in Katznelson An Introduction to Harmonic Analysis, p. 42.) Now if M is as above then we have sum_{x in R} M({x}) <= 1 sum{_x in R} M({x})^2 = 1; the only way this can happen is if there exists c with M({x}) = 1 for x = c, 0 for x <> c. Then since M is a probability measure it follows that M = d_c. >>Suppose M and N are distributions (measures), and itÕs given that >>M * N = d_1, >>where * indicates convolution and d_1 indicates the delta measure >>concentrated at 1. How do you show that either M = d_c or N = d_c for >>some constant c? This should probably follow from some really simple >>manipulations with convolutions, but I lamentably donÕt see how to do >>it. >> Here is a solution which works for finite measures; perhaps it can be >> modified by approximation to all measures? >> The problem is equivalent to showing that if X and Y are independent >> random variables and X + Y is almost surely 1, then either X or >> Y is almost surely constant. Actually, they both are, since >> Var(X + Y) = Var(X) + Var(Y) = 0. >> -- >> Stephen J. Herschkorn herschko@rutcor.rutgers.edu ************************ David C. Ullrich === Subject: linear operator Let T be a linear operator on an inner product space V and suppose |T(x)| = |x| for all x. Prove that T is 1-1. I started off with suppose T(x) = T(y) (and want to conclude that x=y) then |T(x)| = |T(y)|. Then |x| = |y|. The thing is that just because |x| = |y| we canÕt conclude x=y (consider x=(1,0,0) and y= (0,0,1) for example). There must be something else I can use. I guess it is the fact that T preserves the norm but I do not see how to use this. Any hints will be appreciated. === Subject: Re: linear operator > Let T be a linear operator on an inner product space V and suppose |T(x)| = > |x| for all x. Prove that T is 1-1. > I started off with suppose T(x) = T(y) Then T(x-y) = 0 ... === Subject: Re: linear operator > Let T be a linear operator on an inner product space V and suppose |T(x)| = > |x| for all x. Prove that T is 1-1. > I started off with suppose T(x) = T(y) (and want to conclude that x=y) > then |T(x)| = |T(y)|. Then |x| = |y|. The thing is that just because |x| = > |y| we canÕt conclude x=y (consider x=(1,0,0) and y= (0,0,1) for example). > There must be something else I can use. I guess it is the fact that T > preserves the norm but I do not see how to use this. Any hints will be > appreciated. Since T preserves norm, when is |Tx|=0? What does this imply about the null space of T? === Subject: Re: projective geometry axioms > Why not combine the first two axioms of projective geometry? It seems > like any proof that involved either two would be upheld with an =1 . > IÕll list them all here from mathworld: > 1. If A and B are distinct points on a plane, there is at least one > line containing both A and B. > 2. If A and B are distinct points on a plane, there is not more than > one line containing both A and B. > 3. Any two lines in a plane have at least one point of the plane > (which may be the point at infinity) in common. > 4. There is at least one line on a plane. > 5. Every line contains at least three points of the plane. > 6. All the points of the plane do not belong to the same line That sort of mess is a long way out of date. In the early 20th century various such axioms were proposed, usually something like: (1) Any two points both lie on a unique line, (2) Any two lines both lie on a unique point, (3), (4), ... Non-triviality conditions such as your 4, 5, 6 above. But in 1943 Marshall Hall produced a much simpler non-triviality condition: (3) There exist four points of which no three are collinear, i.e. there exists a quadrangle. This is a purely existential statement about finitely many points, as against the older universal-existential things like your 5. (1), (2), (3) above are now widely regarded as the basic axioms for plane projective geometry. An important special case satisfies also the Desargues condition. An important special case of _that_ satisfies also the Pappus condition (which implies Desargues, by HessenbergÕs Theorem). But you may not be interested in those refinements just yet. Ken Pledger. === Subject: Re: Goading Godel > Most of the above passage is utter nonsense. Your concept of > infinities needs some *serious* refining as itÕs so vague that itÕs > basically meaningless. For example, if you want to show that an > arbitrary infinite set canÕt be mapped to a finite set, just consider > the set of all real numbers, which cannot be put into correspondence > with the set of integers. Since every finite set can be considered as > a subset of the integers (there exists a bijective mapping), this > shows that you canÕt map ALL infinite sets to finite representations. Sure you can. Map every infinite set to 1. Or 2. Or even 3. I suspect you had something else in mind. Perhaps *into* a finite set? You definitely didnÕt mean *onto,* as that is incredibly easy to do. Ôcid Ôooh === Subject: Re: e is transcendental by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1P6o2U04174; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1P6fvi03850 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1P6fvp20699; >>O.k. then , as long there is no conßict with : >>Re(e^[iPi]) = Re(-1+i[0]) = -1 >That has never been in doubt. Neither has the observation that >Im(e^[i pi]) = Im(-1+i[0]) = 0. >The statement that caused all the problem was the statement that >e^[i pi] = 0. Actualy stated e^[ipi]=i0=0 [REF: A) e^[ipi]=-1 the real part solution and B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio.] Panagiotis Stefanides >David McAnally > Despite anything you may have heard to the contrary, > the rain in Spain stays almost invariably in the hills. === Subject: Re: e is transcendental >O.k. then , as long there is no conßict with : >Re(e^[iPi]) = Re(-1+i[0]) = -1 >>That has never been in doubt. Neither has the observation that >>Im(e^[i pi]) = Im(-1+i[0]) = 0. >>The statement that caused all the problem was the statement that >>e^[i pi] = 0. >Actualy stated e^[ipi]=i0=0 Which is exactly the same statement in the long run. And I have not yet seen you retract the statement, as you should have by now. >[REF: > A) e^[ipi]=-1 the real part solution and > B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio.] Arguments which you have now acknowledged as invalid (they are either both valid or both invalid, and the false conclusion in B from the true premise that exp[i pi] = -1+i[0] demonstrates that the argument used in B is invalid). David McAnally At the moment, they (the Time Lords) are far from being all-powerful. ThatÕs why itÕs been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Does Euclidean geometry exist as a physical entity, or does only Riemannian and Lobachevskian > I am going to have to revamp File 103 on FLT, and File 120 of 3 and > only 3 geometries and File 125 of two proofs of the Riemann > Hypothesis in my website of www.iw.net/~a_plutonium/ > I did not do much mathematics after 1997 and recently when I reviewed > my Riemann Hypothesis proof I realized that it is the p-adics that are > on the 1/2 Real line which means that lines are curved when out at > infinity. There are no straightlines. I have to change and revise my > Poincare Conjecture proof also. Physics is duality and not triality. If I go by that presumption then I have to concede that there are only really 2 geometries and not 3. Of the three known geometries of Riem, Loba, and Eucl, I would bet that Euclidean is the nonexistant one. The P-adics create a nice point by point Riemannian geometry and they naturally curve back around. But I have trouble forming Lobachevskian geometry with a one to one correspondence with algebraic numbers. Perhaps the Doubly-Infinites? Perhaps the negative Reals. But then I have trouble with the Reals for they seem to be Euclidean geometry. But are they really? Euclidean geometry is zero curvature. But the number 0 exists in p-adics and doubly-infinites. The complex-numbers I can rule out as just a gimmick that gives added dimensions. So I am faced with 3 number sets of P-adics, Reals, and Doubly-Infinites. If physics is the final word on this that duality exists but triality is nonsense, then I am going to have to find out what 2 and only 2 geometries exist and that one of them is a mental illusion for human minds. Just as Newtonian absolute space and absolute time was just a mental illusion. The most perfect match is P-adics to Riemannian Geometry because the P-adics are all positive numbers which Riem geometry deals with only positive quantities and they naturally curve back around such as in 10-adics the number ....99998 is equivalent of -2 and then ....99999 is equivalent to -1. So it is a beautiful matchup. But I have trouble in finding a number set to matchup with Lobachevskian Geometry. It cannot be the Reals because they have both negative and positive but Loba requires only negative. So are the Doubly-Infinites intrinsically negative quantities similar to the fact that P-adics are intrinsically positive quantities? Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: cracks in Euclidean Geometry and why Reals are fake Re: Does Euclidean geometry exist as a physical entity, or does only Riemannian and Lobachevskian > I am going to have to revamp File 103 on FLT, and File 120 of 3 and > only 3 geometries and File 125 of two proofs of the Riemann > Hypothesis in my website of www.iw.net/~a_plutonium/ > > I did not do much mathematics after 1997 and recently when I reviewed > my Riemann Hypothesis proof I realized that it is the p-adics that are > on the 1/2 Real line which means that lines are curved when out at > infinity. There are no straightlines. I have to change and revise my > Poincare Conjecture proof also. > > Physics is duality and not triality. If I go by that presumption then > I have to concede that there are only really 2 geometries and not 3. > Of the three known geometries of Riem, Loba, and Eucl, I would bet > that Euclidean is the nonexistant one. The P-adics create a nice point > by point Riemannian geometry and they naturally curve back around. But > I have trouble forming Lobachevskian geometry with a one to one > correspondence with algebraic numbers. Perhaps the Doubly-Infinites? > Perhaps the negative Reals. > But then I have trouble with the Reals for they seem to be Euclidean > geometry. But are they really? Euclidean geometry is zero curvature. > But the number 0 exists in p-adics and doubly-infinites. The > complex-numbers I can rule out as just a gimmick that gives added > dimensions. > So I am faced with 3 number sets of P-adics, Reals, and > Doubly-Infinites. If physics is the final word on this that duality > exists but triality is nonsense, then I am going to have to find out > what 2 and only 2 geometries exist and that one of them is a mental > illusion for human minds. Just as Newtonian absolute space and > absolute time was just a mental illusion. > The most perfect match is P-adics to Riemannian Geometry because the > P-adics are all positive numbers which Riem geometry deals with only > positive quantities and they naturally curve back around such as in > 10-adics the number ....99998 is equivalent of -2 and then ....99999 > is equivalent to -1. So it is a beautiful matchup. > But I have trouble in finding a number set to matchup with > Lobachevskian Geometry. It cannot be the Reals because they have both > negative and positive but Loba requires only negative. So are the > Doubly-Infinites intrinsically negative quantities similar to the fact > that P-adics are intrinsically positive quantities? Another major crack in Euclidean Geometry is that Real Numbers really do not coincide with Euclidean Geometry, do they? Perhaps I am making a statement or perhaps I am asking a question, for I am only at the beginning of this inquiry. If we look at the P-adics, they are all positive numbers, even the zero point can be said to be positive and so they are ideal for coinciding with Riemannian Geometry with its *positive curvature*. P-adics are ideal for representing Riemannian Geometry, or, making a 1-to-1-correspondence. So one can say that P-adics are the points of Riemannian Geometry and ideally such because the p-adics have a natural curvature to them for as we start in the 10-adics with 0 and then next is .000001 and next is .000002 and going way out it comes back to ...99997 which we can conceive of as -3 then ....999998 which is -2 and excitedly ....999 which is -1 and finally back to our starting point of 0. So the P-adics can be said to be the actual algebraic points of Riemannian Geometry. But now inspecting Reals with the geometry of Euclid, it just simply does not fit together does it. Because Euclid geometry is 0 curvature and the only number in the Reals that obeys curvature is a single point which is zero itself. The positive Reals disobey Euclidean Geometry because they are positive signifying Riem geom and not Euclidean and likewise the reverse for negative Reals for they signify Lobachevskian geom. So one is left with the conclusion that the Reals never represented the geometry entailed by Euclidean Geometry. Do the Real Number system represent any geometry??? I suspect not. I suspect the Reals are as mythical or imaginary as was Newtonian Mechanics of absolute space and absolute time. Humans have minds that can dream up things which really have no physical reality such as ghosts, witches, and Newtonian absolute space and absolute time. Are the Real Numbers another dream-up thing which has no physics reality? I suspect so. But I am troubled with what numbers correspond to Lobachevskian geometry. I would like to think that the negative REals suit the Loba geometry, but that leaves the nasty question of the positive Reals. I think I can draw some clues as what the numbers that make up Lobachevskian Geometry from the P-adics making up Riemannian Geometry. If I start with this claim: P-adics == Riemannian Geometry and accept it as fully true, then there are some other numbers called Doubly-Infinites. Doubly-Infinites are what the name implies. You see, p-adics are infinite leftward strings. Doubly-Infinites would then be numbers that are both infinite leftward but also infinite rightwards. In some sense, the Real Numbers should be doubly-infinite. Perhaps that is the reason the Reals are fake and nonphysical just as the NaturalNumbers = finite-integers was a fake and dreamed-up illusions. TEST: the test of the above would be to show that the Doubly-Infinites are numbers that are all negative in sign value. What I mean is that the P-adics are all positive (even ....9999 is positive but it can represent -1). So, if I can show that Doubly-Infinites are all negative, then I will have shown a vast amount of knowledge and understanding. Because if I can show the Doubly- Infinites are all negative in sign value then these numbers are what compose Lobachevskian geometry. And that where I have: All-P-adics == Riemannian Geometry I will also have All-Doubly-Infinites == Lobachevskian Geometry This would then conclude that REals as a number system were a fake entity and would imply that the centuries of gaps and holes found in Reals such as the myriad types of differentiation and integration Lebesgue integral to name one is because the Reals are a hodge-podge-mess just as Newtonian absolute space and absolute time was a hodge-podge-mess that saddled Quantum Mechanics. Driver Motivation for the above: what drives me to many of these conclusions is that Quantum Mechanics is duality based and not threesome. Because Physics is twosome, then geometries should be twosome and not threesome. Therefore, geometry should have 2 and only 2 indepedent geometries. So, I have a choice of 2 of these geometries that really exist (1) Euclid (2) Riemannian (3) Lobachevskian. My choice is Riem and Loba. And since there are only two entails that Algebraically in Mathematics there exists only two real Number Systems. I am fully confident that P-adics are one true number system. I suspect Doubly-Infinites is the second truly existing number-system. Hence I suspect the Reals to be a fake system just as Newtonian Absolute Space and Absolute Time was a fake concept. Research: All I need is hard-core evidence that Doubly-Infinites are all negative numbers. If I can get that evidence, then all of my above thoughts would be confirmed. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies (www.iw.net/~a_plutonium) website of the science of AP under revision what used to be my old science website www.newphys.se/elektromagnum/physics/LudwigPlutonium from years 1993 === Subject: Re: as usuall , help is needed!!! > Use the chain rule to get dy/dt in terms of dy/dx. > -Michael. > dear Michael > if your means is [dy/dx]=[dy/dt]*[dt/dx] > i can not solve with it if you can then so0lve Yep, but it helps to reverse it: [dy/dt] = [dy/dx]*[dx/dt] You need the second derivative too: [d^2y/dt^2] = [dx/dt]^2 * [d^2y/dx^2] + [d^2x/dt^2] * [dy/dx] Now, for x=tan(t) you have [dx/dt] = 1 + x^2. I hope you can do the rest yourself. -Michael. === Subject: Re: inequality Jerome Davies > I trying to establish the inequality: > (x^a - 1)/(x^b - 1) <= a/b If x>1 and 00. Since a/b = ka/kb and k>0, we can assume k=1. Rearranging it, we want (e^a - 1)/a <= (e^b -1)/b. This is clear from the power series for e^x, but to see it another way, compare the slope of the line from (0,1) to (a,e^a) to the slope of the line from (0,1) to (b,e^b) and remark that all the derivatives of e^x are positive everywhere. The hypothesis 0Let S be the real numbers (0,1) >Since it is a set of real numbers it is partially ordered and every chain is >obviously bounded by 1 >Yet (0,1) does not have a maximal element. >What am I missing here? The chain (0,1), for example, does *not* have a least upper bound *in S.* -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: ZornÕs Lemma Question >>Let S be the real numbers (0,1) >>Since it is a set of real numbers it is partially ordered and every chain is >>obviously bounded by 1 >>Yet (0,1) does not have a maximal element. >>What am I missing here? >The chain (0,1), for example, does *not* have a least upper bound *in S.* More relevant to a question about ZornÕs lemma is the fact that (0,1) does not have an upper bound in S. ************************ David C. Ullrich === Subject: Q11 format What is Q11 Format ? === Subject: a linear algebra thought All, I have a question on linear algebra. Let V be a finite dimensianl vector space, S be a noninvertable linear transformation on V. I wonder if the Ker(S) is perpendicular to Range(S), or ker(S)+Range(S) is direct sum and Ker(S)+Range(S)=V. === Subject: Re: a linear algebra thought > All, I have a question on linear algebra. Let V be a finite dimensianl > vector space, S be a noninvertable linear transformation on V. Let V be 2 dimensional with basis b_1, b_2. Let S(b_1) = 0 and S(b_2) = b_1, then Ker(S) = Range(S). >I wonder if the Ker(S) is perpendicular to Range(S), or >ker(S)+Range(S) is direct sum and Ker(S)+Range(S)=V. === Subject: Re: a linear algebra thought > All, I have a question on linear algebra. Let V be a finite dimensianl > vector space, S be a noninvertable linear transformation on V. I > wonder if the Ker(S) is perpendicular to Range(S), or ker(S)+Range(S) > is direct sum and Ker(S)+Range(S)=V. Perhaps that I misunderstood your questions, but consider V = R^2 and S(x,y) = (y,0). Then ker(S) = Range(S). I suppose that this answers your questions. Jose Carlos Santos === Subject: Re: Genetics and Math-Ability > > > I think most people would agree that human reasoning is far from > perfect. How can an inconsistent reasoning ability give a survival > advantage? Evolution proves that some inconsistent systems are > better than others. > > Restricting to only those conclusions that are 100% certain > and inescapable would lead to a pityful small number > of conclusions. > The requirement wasnÕt for being that could reason to a correct > conclusion, but that the resoning was consistent. > Both abilities (drawing conclusions beyond the trivial and > drawing them fast) will account for the survival advantage. > Definately. There was an AI research team (I forget where or when) who > developed a robot. This robot had cameras, wheels and a very > sofisticated AI that should allow it to navigate around a room. When it > was switched on, it simply sat still and did nothing. Examining the > robots datapath they found it was sitting there quietly analizing > everything before making a descision as to what to do. They added more > code which should enable it to selectively ignore the unimportant parts, > when it was switched on it simply sat still again. Examining the robots > datapath again showed it was quietly analysing everything so it could > decide what to ignore. > The moral, if human reasoning was forced to be consistent weÕd be stuck > in same situation as the robot. Well, they are stuck in the same situation as robots. WeÕve always, throughout all of human history, restricted both Mathematicians and Lawyers to the study of simple machines. Since the only statement thatÕs ever been true about machines is: Simple is, as simple does. === Subject: re:Alternative ways to solve a quadratic equation HereÕs a simple method which I came up with before I learnt quadratic equations: Write the equation in the form (x + b)x = c Find A and B so that A + B = x + b and A - B = x Obviously A = x + b/2, B = b/2 The equation becomes A^2 - B^2 = c, or A^2 = c + b^2 / 4 ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: Alternative ways to solve a quadratic equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1P5KUi29951 === Subject: Re: Alternative ways to solve a quadratic equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1P02Ri02705 Originator: fishbowl@conservatory.com (james) >But other methods do exist. >I will give a method based on Galois theory. William, youÕve been folderized. === Subject: Re: Alternative ways to solve a quadratic equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1P00Ki02172 Originator: fishbowl@conservatory.com (james) >required us to find a totally new method to solve a quadratic equation >with. Totally new? That would be quite a feat. Oh, not totally new as in, hasnÕt been discovered before. Your chances of finding a method that the Greeks didnÕt know are pretty slim :-) >It can involve the quadratic formula but has to be different. IÕd look for a quadratic identity using trig substitution, or maybe find roots with the Newton Method. Sounds like your assignment is supposed to be an exercise in finding identities for the quadratic formula. If you come up with something totally new here *DONÕT* give it to your professor, submit a paper to James watched Good Will Hunting one time too many M. === Subject: Re: Alternative ways to solve a quadratic equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1ONkTi00513 > required us to find a totally new method to solve a quadratic equation > with. It can involve the quadratic formula but has to be different. I assume you mean totally new to you as opposed to totally new to mathematics. You might look for some techniques for solving higher order equations. Or here is one very simple and inefficient one called the half interval search - once you have established that a root lies between point A and B, because of a sign change. Evaluate the function at (A + B)/2. If it is the same sign as A this becomes the new A otherwise the new B. Keep going until you are tired or happy. Or you might try to find some kind of analog computer. That is, instead of using mathematics to represent a physical phenomenon (the usual way of doing things) find some sort of physical representation of the math. and go measure it. Bill === Subject: re:2 and only 2 geometries where Euclidean is like NewtonÕs abs No offense, but you sound like a crank. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: A probability question I have a probability question that looks deceptively easy but somewhat I just canÕt prove it. There are three events A, B and C. Events A and C are independent. Events B and C are independent. Prove or disprove P(A intersect B | C) = P(A intersect B). Yeoster === Subject: Re: A probability question >There are three events A, B and C. Events A and C are independent. >Events B and C are independent. Prove or disprove P(A intersect B | C) >= P(A intersect B). Consider rolling a 4-sided die with the following events: A: rolling 1 or 2 B: rolling 1 or 3 C: rolling 2 or 3 Rob Johnson take out the trash before replying === Subject: Re: A probability question >I have a probability question that looks deceptively easy but somewhat >I just canÕt prove it. >There are three events A, B and C. Events A and C are independent. >Events B and C are independent. Prove or disprove P(A intersect B | C) >= P(A intersect B). Big hint: Consider a pair of dice and the event of a seven being rolled. Cite sources of assistance on submitted assignments. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: plotting hyperelliptic curves!! could someone tell how to plot hyperelliptic curves using MAPLE, Mathematica, Matlab..anything for example..something like v^2=u^5 - 5*(u^3) + 4*u + 3, over real numbers thanx in advance for all the help.. OP. === Subject: Re: plotting hyperelliptic curves!! > could someone tell how to plot hyperelliptic curves using MAPLE, > Mathematica, Matlab..anything > for example..something like > v^2=u^5 - 5*(u^3) + 4*u + 3, over real numbers > thanx in advance for all the help.. > OP. Another way in maple that works well with many algebraic curves is: with(algcurves): f:=v^2-(u^5-5*u^3+4*u+3); plot_real_curve(f,u,v); Jim Buddenhagen === Subject: Re: plotting hyperelliptic curves!! > could someone tell how to plot hyperelliptic curves using MAPLE, > Mathematica, Matlab..anything > for example..something like > v^2=u^5 - 5*(u^3) + 4*u + 3, over real numbers > thanx in advance for all the help.. With Maple, load the plots package and then do: implicitplot(v^2=u^5-5*u^3+4*u+3,u=-1..3,v=-4..4); Mathematica also has an ImplicitPlot command. Jose Carlos Santos === Subject: Re: branch of log z > |What exactly is to choose a branch of log z for complex z? IÕd > |appreciate if someone help me picture this. > The basic problem is ... log or ln changes an arrow z of length 1 of a plane into a rotational vector ln z of length @ perpendicular to the plane - this is the height in the parking garage, but as an axial vector-arrow it can be translated freely in R3. @ can not exceed 360 degrees or 2*pi, if it is created this way. Rotating an arrow w in a plane at any height is different from this - itÕs first changing an rotational arrow, picture them as screws, of length beta perpendicular to the plane into an arrow of this plane. This arrow is showing the difference of direction of the grove on top of the screw, by e^(i*beta)=exp(i*beta). Second is multiplying this with w. Now an approach to branch of log z or ln z is, that it is the set or manifold of parallel axial-arrows, differing in length less than 2*pi - if we retrict to length z=1 for a start. Feel like a snail, reaching the first level in the parking garage. Did it get it ? Hero by the way, links to f(z) and planegraph, my two favourite programms for visualisation, are on my website. === Subject: Re: Trying to unify axioms. > Nothing. > > You forever spewing fucking imbecile, an axiom by definition is > irreducible and unprovable. > > False. > > First off, an axiom isnÕt necessarily irreducable, unless you want to > claim that reducing it makes the axiom not reducable and therefore not > an axiom. Sometimes we have an axiom that we find out can be reduced > into other axioms. Does that therefore disprove the axiom, or destroy > the usefulness of the axiom, or of taking it as such? No it doesnÕt. > > Second, axioms are not unprovable. They CAN be proven. For an example > of this, consider the law of identity. Can you prove it? If not, then > how do we even know itÕs true? We do know itÕs true, and it IS an > axiom, so that just proves that axioms are not unprovable. > > (...Starblade Riven Darksquall...) > An axiom is unprovable by definition. Usually, if an axiom can be > shown to be provable based on other axioms or postulates, it no longer > remains an axiom and is technically a theorem, which by definition is > provable. And of course, nothing is ever provable in and of itself, > which is why a single isolated statement can only be true by > definition. Then explain the human mind. Which axioms did we start with? If I can explain it to you using reason, then does it no longer become an axiom? (...Starblade Riven Darksquall...) === Subject: Re: Trying to unify axioms. >> An axiom is unprovable by definition. Usually, if an axiom can be >> shown to be provable based on other axioms or postulates, it no longer >> remains an axiom and is technically a theorem, which by definition is >> provable. And of course, nothing is ever provable in and of itself, >> which is why a single isolated statement can only be true by >> definition. >Then explain the human mind. Which axioms did we start with? If I can >explain it to you using reason, then does it no longer become an >axiom? Poincare has a lot to say about that in his book Science and Hypothesis, $4 at Amazon. Basically, you canÕt explain it using reason. You canÕt explain things like width to someone that hasnÕt shared some of your experiences. -- Very well, he replied, I allow you cowÕs dung in place of human excrement; bake your bread on that. -- Ezekiel 4:15 === Subject: Re: Trying to unify axioms. >> An axiom is unprovable by definition. In logic, an axiom is provable by definition. === Subject: Re: Trying to unify axioms. > An axiom is unprovable by definition. > In logic, an axiom is provable by definition. A self-evident principle or one that is accepted as true without proof as the basis for argument; a postulate. If an axiom was provable, what would you call the statements from which the axiom is deduced? -- DonÕt try to teach a pig how to sing. YouÕll waste your time and annoy the pig. === Subject: Re: Trying to unify axioms. > If an axiom was provable, what would you call the statements from which > the axiom is deduced? In logic, an axiom is provable because it is deducible from itself. === Subject: Re: Trying to unify axioms. > If an axiom was provable, what would you call the statements from which > the axiom is deduced? > In logic, an axiom is provable because it is deducible from itself. Only one line can be drawn parallel to a given line through an exterior point. Everybody knows that from high school geometry. It is obviously true as EuclidÕs Fifth Postulate (here restated as PlayfairÕs Axiom). There is only one glitch: It is empirically wrong. There are no lines parallel to a given line on the EarthÕ surface. You cannot accurately navigate or survey with Euclid. A mile square on the EarthÕs surface bounds more than a square mile. All triangles on EarthÕs surface have their interior angles sum to more than 180 degrees (as much as 540 degrees!). Given a circle drawn on the surface of the Earth, the ratio of the circumference to the diameter is always less than pi. By trivial demonstration, you are full of shit. A axiom is a stated unprovable assumption, It is indefensible for being an axiom - and can be falsified by a single reproducible counterdemonstration. A refrigerator is not a refrigerator if its motor is run backwards to create an oven. All acceptible theories of gravitation must give the same predictions to the extreme limits of experimental error because they all describe the same unique reality. There are only two exceptions: 1) They can disagree about observations that have not been made (e.g., Planck energy regimes), and 2) They can disagree about the Equivalence Principle - that all local bodies fall identically in vacuum. An unmade observation cannot be a constraint. The Equivalence Principle has not been exhaustively tested for violation. Like Euclid, the most elegant and comprehensive internally self-consistent and exhaustively empirically confirmed axiomatic system is only as strong as its weakest axiom. Given just one reproducible counterdemonstration, it all comes crashing down. Philosophy is crap. Humanity languished in pestilence, disease, poverty, famine, filth, and ignorance given 5000 years of continuously refined philosophies. If you want to ßush away the crap, you need an engineer. If you wish to wash your hands of it, you need a chemist. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Trying to unify axioms. >> If an axiom was provable, what would you call the statements from which >> the axiom is deduced? >> In logic, an axiom is provable because it is deducible from itself. >Only one line can be drawn parallel to a given line through an >exterior point. Everybody knows that from high school geometry. It >is obviously true as EuclidÕs Fifth Postulate (here restated as >PlayfairÕs Axiom). There is only one glitch: It is empirically wrong. Nonsense. >There are no lines parallel to a given line on the EarthÕ surface. Again, nonsense. What you call lines on Earthj surface are *not* straight lines. TheyÕre curved. The Euclidean postulate is about *straight lines. The claim that Euclidean geometry cannot deal with circles would sure get an amused reaction from the Greek geometers. But you do *not* deal with circles by pretending that theyÕre straight lines. If I take a square and calculate its area assuming that it is a triangle (base times height divided by two), IÕll be off by a factor of two. The failing will be mine, *not* EuclidÕs. You do have good posts, at times. This was *not* one of them. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Trying to unify axioms. > If an axiom was provable, what would you call the statements from which > the axiom is deduced? >> In logic, an axiom is provable because it is deducible from itself. >>Only one line can be drawn parallel to a given line through an >>exterior point. Everybody knows that from high school geometry. It >>is obviously true as EuclidÕs Fifth Postulate (here restated as >>PlayfairÕs Axiom). There is only one glitch: It is empirically wrong. >> Nonsense. >>There are no lines parallel to a given line on the EarthÕ surface. >> Again, nonsense. What you call lines on Earthj surface are *not* >> straight lines. TheyÕre curved. The Euclidean postulate is about >> *straight lines. The claim that Euclidean geometry cannot deal with >> circles would sure get an amused reaction from the Greek geometers. >> But you do *not* deal with circles by pretending that theyÕre straight >> lines. >> If I take a square and calculate its area assuming that it is a >> triangle (base times height divided by two), IÕll be off by a factor of >> two. The failing will be mine, *not* EuclidÕs. >> You do have good posts, at times. This was *not* one of them. >A straight line (geodesic path) on the surface of a sphere is a Great >Circle. Nope. A geodesic path on the surface of a sphere is a Great circle. This is not, (repeat, ***not***) a Euclidean straight line. You can use a geometry in which these lines are defined as straight, but this is a different geometry. The straight line joining the North and South pole is the line going from one to the other *through* the center of the Earth. YouÕll say but this is not on the surface? There is ***no*** definition in Euclidean geometry to the effect that the straight line joining to points on an arbitrary surface must be imbedded in the surface. A geodesic line, yes, a straight line, no. Elliptic (Bolyai-Lobechevsky, positive curvature) geometry >then proceeds to spec. Euclid (except those parts independent of the >Fifth Postulate) doesnÕt work on the surface of a sphere, or an >ellipsoid in general. You progress from the idiotic to the moronic. Again, did it occur to you that the sphere is described by x^2 + y^2 + z^2 = R^2 in Euclidean geometry. >Hyperbolic (Riemann, negative curvature) surfaces similarly fail >Euclid. It doesnÕt fail anything, it is just a different geometry. A mile square encompasses than a square mile, a triangleÕs >interior angles always sum to less than 180 degrees, and the ratio of >circumference to diameter of all circles is greater than pi. >In both cases, a straight line is defined as the path taken by a >light ray in vacuum in that 2-D space. YouÕre totally confused. ThatÕs whatÕs coming from learning geometry from coffee table books. A geometry is a mathematical system. It is a set of axioms and definitions plus the conclusions that can be drawn from these. Get a different set of axioms and definitions and youÕve a different geometry. You can have as many as you wish. And the fact that geometry A differs in its axioms and results from geometry B *does not* mean that either A or B have been falsified. This concept *does not* apply here. There is no empirical verification of mathematical theories. Euclid is fine, and Lobatchevsky is fine, and Bolyi is fine. These are just different geometris, thatÕs all. Now, if you want to use a specific geometry (or, in fact, any mathematical theory) to model some observable aspect of reality, then it is a different story. What you need to do then is to establish some mapping between concepts of the theory and elements of said observable aspect of reality. And then, once youÕve done it, you can check whether the relationships between the observables follow the relationships between the theoretical entities they map into. If this is true, fine. And if it isnÕt true? If it isnÕt true, all it means is that *for the specific mapping* you selected, the mathematical theory youÕve chosen does not apply as a physical model (of whatever it is youÕre modeling). It says *nothing* about the validity of the mathematical theory per se (the very notion is ridiculous) only on its applicability as physical model, for the phenomena required, and using the selected mapping. This is *all*. Am I getting through to you? Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Trying to unify axioms. >> If an axiom was provable, what would you call the statements from which >> the axiom is deduced? >> In logic, an axiom is provable because it is deducible from itself. >Only one line can be drawn parallel to a given line through an >exterior point. Everybody knows that from high school geometry. It >is obviously true as EuclidÕs Fifth Postulate (here restated as >PlayfairÕs Axiom). There is only one glitch: It is empirically wrong. > Nonsense. >There are no lines parallel to a given line on the EarthÕ surface. > Again, nonsense. What you call lines on Earthj surface are *not* > straight lines. TheyÕre curved. The Euclidean postulate is about > *straight lines. The claim that Euclidean geometry cannot deal with > circles would sure get an amused reaction from the Greek geometers. > But you do *not* deal with circles by pretending that theyÕre straight > lines. > If I take a square and calculate its area assuming that it is a > triangle (base times height divided by two), IÕll be off by a factor of > two. The failing will be mine, *not* EuclidÕs. > You do have good posts, at times. This was *not* one of them. A straight line (geodesic path) on the surface of a sphere is a Great Circle. Elliptic (Bolyai-Lobechevsky, positive curvature) geometry then proceeds to spec. Euclid (except those parts independent of the Fifth Postulate) doesnÕt work on the surface of a sphere, or an ellipsoid in general. Hyperbolic (Riemann, negative curvature) surfaces similarly fail Euclid. A mile square encompasses than a square mile, a triangleÕs interior angles always sum to less than 180 degrees, and the ratio of circumference to diameter of all circles is greater than pi. In both cases, a straight line is defined as the path taken by a light ray in vacuum in that 2-D space. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Trying to unify axioms. > A axiom is a stated > unprovable assumption, It is indefensible for being an axiom - and > can be falsified by a single reproducible counterdemonstration. Falsifiability, indefensible - this has nothing to do with it. ItÕs simply a fact that in logic, every axiom of a theory T is provable in T. === Subject: Re: Trying to unify axioms. > A axiom is a stated > unprovable assumption, It is indefensible for being an axiom - and > can be falsified by a single reproducible counterdemonstration. > Falsifiability, indefensible - this has nothing to do with it. ItÕs > simply a fact that in logic, every axiom of a theory T is provable in T. Only one line can be drawn parallel to a given line through an exterior point. Everybody knows that from high school geometry. It is obviously true as EuclidÕs Fifth Postulate (here restated as PlayfairÕs Axiom). There is only one glitch: It is empirically wrong. There are no lines parallel to a given line on the EarthÕ surface. Pookie pookie. The are an infinite number of lines that can be drawn parallel to a given line through an exterior point on a hyperbolic surface. Euclid is incomplete for his fifth postulate. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Trying to unify axioms. .91[CapitalEth].91ñ.93»[EDouble Dot]±.93Ì.91[Micro] .93¡.93.b3.91Ë .91.b9.91¬.91ü.93á[EDo ubleDot].b9.91± > Pookie pookie. The are an infinite number of lines that can be drawn > parallel to a given line through an exterior point on a hyperbolic > surface. Euclid is incomplete for his fifth postulate. WhatÕs your fucking point already? Maybe that the Ancient Greeks should have done better than that? Lessee: They are sitting in the sun 2,500 years ago and they decide to find EVERYTHING is a game. Maybe they should have invented relativity instead. Wait, analyze and utilize the ergosphere of a couple of rotating black holes which they could not otherwise observe because they didnÕt have the appropriate apparatus and deduce how to have energy for free by putting those black holes in orbit around Earth by throwing beach pebbles in them. Maybe they should have dreamt of the Benzene molecule, so they could drive a Honda. Heck, why didnÕt they invent Hyperbolic Geometry, those fucking idiots? When was the last time you had your head OUT of your ass? > -- > Uncle Al -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === Subject: Re: Trying to unify axioms. > Only one line can be drawn parallel to a given line through an > exterior point. Everybody knows that from high school geometry. It > is obviously true as EuclidÕs Fifth Postulate (here restated as > PlayfairÕs Axiom). There is only one glitch: It is empirically > wrong. There are no lines parallel to a given line on the EarthÕ > surface. True or false is irrelevant. Every axiom of a theory T is provable in T. This is just a matter of terminology. === Subject: Re: Trying to unify axioms. > If an axiom was provable, what would you call the statements from which > the axiom is deduced? > In logic, an axiom is provable because it is deducible from itself. What a slick way to do away with critical thinking. Tommy Aquinas would be proud. IsnÕt this the basis of religious belief? === Subject: Re: Trying to unify axioms. > What a slick way to do away with critical thinking. You might as well say that 1+1=2 does away with critical thinking. === Subject: Re: Trying to unify axioms. >> What a slick way to do away with critical thinking. > You might as well say that 1+1=2 does away with critical thinking. Revisit my example. If we assume that this is a refrigerator, it follows that this is a refrigerator. That is, R->R. Or (R ^ R) v (-R ^ -R). Classic tautology, itÕs true whether R is true or false. So which is it, true or false? R canÕt resolve that on its own. So choose a postulate: 1a) R 1b) -R Once youÕve chosen one we can say until the cows come home that R->R or (-R)->(-R), but nothing can be decided until you pick. 1+1=2, on the other hand, is not just a restatment of a postulate. ThereÕs actually a chain of reasoning involved that comes to a conclusion thatÕs different from any of the postulates. -- clogging their in-boxes, but they fear a proposed national ÔDo Not SpamÕ registry will make it impossible to use e-mail as a marketing tool. === Subject: Re: Trying to unify axioms. > 1+1=2, on the other hand, is not just a restatment of a postulate. > ThereÕs actually a chain of reasoning involved that comes to a conclusion > thatÕs different from any of the postulates. Really? I thought 2 was in practice nothing more than a shorthand notation for 1+1, at least in traditional formal arithmetic. OG. === Subject: Re: Trying to unify axioms. >> 1+1=2, on the other hand, is not just a restatment of a postulate. >> ThereÕs actually a chain of reasoning involved that comes to a conclusion >> thatÕs different from any of the postulates. >Really? I thought 2 was in practice nothing more than a shorthand >notation for 1+1, at least in traditional formal arithmetic. > OG. Number theorists are wacky. IÕve seen the reasoning twice and never remember how it goes, but itÕs something about defining addition as the union of sets of empty sets. -- The preferred method of entering a building is to use a tank main gun round, direct fire artillery round, or TOW, Dragon, or Hellfire missile to clear the first room. -- THE RANGER HANDBOOK U.S. Army, 1992 === Subject: Re: Trying to unify axioms. Gregory L. Hansen says... >Revisit my example. If we assume that this is a refrigerator, it follows >that this is a refrigerator. >That is, R->R. >Or (R ^ R) v (-R ^ -R). >Classic tautology, itÕs true whether R is true or false. So which is it, >true or false? R canÕt resolve that on its own. If R is an axiom, then the axioms are sufficient to resolve the issue. Look, this is just a terminological issue. We can define a proof from a set of axioms T to be a sequence of statements S_1, S_2, ..., S_n such that for each j, S_j is either an axiom, or follows from earlier statements in the sequence by a valid rule of inference. We then say that S is deducible from T if there is a proof such that S is the last statement in the proof. By this definition, it is trivially true that every axiom of T is deducible from T. This isnÕt in any way contrary to the informal statement that an axiom is something that is assumed with proof. The confusion comes from the fact that there are conceptually two stages involved in formulating a theory: (1) Deciding what statements to include as axioms, and (2) Once we have the axioms, figuring out what statements are logically deducible from *them*. In stage 1, we pick our axioms mean when they say that an axiom is something assumed without proof---it is placed into the axiom set T without the need for any proof. But if a statement is in the axiom set, it is trivially deducible from *those* axioms. You could, I suppose, distinguish between the *proper* consequences of a theory and the *improper* consequences---a proper consequence is a statement that is not an axiom, but is logically deducible from the axioms. However, thatÕs really not a very useful distinction. The *same* theory can be described using different but equivalent axioms. Statements that are axioms in one description may be theorems in another description. For a trivial example, you could let x+0 = x be an axiom of arithmetic. Or you could prove it by induction from the axioms 0+x = x (y+1)+x = (y+x)+1 The mathematical attitude towards axioms is that they are simply a way to describe a theory. But nothing important about a theory depends on the way it is described. -- Daryl McCullough Ithaca, NY === Subject: Re: Trying to unify axioms. <0q4%b.127219$FO1.2443626@weber.videotron.net> Discussion, linux) > What a slick way to do away with critical thinking. >> You might as well say that 1+1=2 does away with critical thinking. > Revisit my example. If we assume that this is a refrigerator, it follows > that this is a refrigerator. > That is, R->R. > Or (R ^ R) v (-R ^ -R). > Classic tautology, itÕs true whether R is true or false. So which is it, > true or false? R canÕt resolve that on its own. Huh? Which is what? R -> R is clearly true, regardless of the truth value of R. WhatÕs the issue? > So choose a postulate: > 1a) R > 1b) -R > Once youÕve chosen one we can say until the cows come home that R->R or > (-R)->(-R), but nothing can be decided until you pick. Absolute nonsense. No matter whether R is true or false, *both* R -> R and ~R -> ~R are true. > 1+1=2, on the other hand, is not just a restatment of a postulate. > ThereÕs actually a chain of reasoning involved that comes to a > conclusion thatÕs different from any of the postulates. Oh? What are the postulates from which you derived 1+1=2? -- Jesse Hughes Besides, discoverers are too proud to kiss butt. Indiana Jones would never kiss some academicÕs ass to get published, and neither will I. --James Harris === Subject: Re: Trying to unify axioms. > What a slick way to do away with critical thinking. > You might as well say that 1+1=2 does away with critical thinking. It does if you proclaim it and fail to define the meaning of addition, the concept of number, etc., and then show how 1+1=2 follows from the definitions. Hey, by your way of definition, I could proclaim 1+1=4 as an axiom and get away with it. === Subject: Re: Trying to unify axioms. <0q4%b.127219$FO1.2443626@weber.videotron.net> Discussion, linux) > Hey, by your way of definition, I could proclaim 1+1=4 as > an axiom and get away with it. Of course you could. Why not? -- Jesse F. Hughes Radicals are interesting because they were considered ÔradicalÕ by modern mathematics depends on. --Another JSH history lesson === Subject: Re: Trying to unify axioms. > Hey, by your way of definition, I could proclaim 1+1=4 as > an axiom and get away with it. In a theory in which 1+1=4 is an axiom it is, trivially, provable that 1+1=4. === Subject: Re: Trying to unify axioms. > Hey, by your way of definition, I could proclaim 1+1=4 as > an axiom and get away with it. > In a theory in which 1+1=4 is an axiom it is, trivially, provable > that 1+1=4. And therefore it is an entertaining but useless endeavor. === Subject: Re: Trying to unify axioms. > And therefore it is an entertaining but useless > endeavor. Entertaining, useless, has nothing to do with it. You might as well comment on what is entertaining or useless in 1+1=2. ItÕs just a matter of accepted terminology in logic. === Subject: Re: Trying to unify axioms. >> If an axiom was provable, what would you call the statements from which >> the axiom is deduced? > In logic, an axiom is provable because it is deducible from itself. That doesnÕt even make sense. If we assume this is a refrigerator, it follows that this is a refrigerator. ThatÕs not a deduction, itÕs a restatement of the assumption. -- For every problem there is a solution which is simple, clean and wrong. -- Henry Louis Mencken === Subject: Re: Trying to unify axioms. >That doesnÕt even make sense. It makes perfect sense. ItÕs a simple observation about how provable is used in logic. === Subject: Re: Trying to unify axioms. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Trying to unify axioms. >> Nothing. >> >> You forever spewing fucking imbecile, an axiom by definition is >> irreducible and unprovable. >False. >First off, an axiom isnÕt necessarily irreducable, unless you want to >claim that reducing it makes the axiom not reducable and therefore not >an axiom. Sometimes we have an axiom that we find out can be reduced >into other axioms. Does that therefore disprove the axiom, or destroy >the usefulness of the axiom, or of taking it as such? No it doesnÕt. >Second, axioms are not unprovable. They CAN be proven. For an example >of this, consider the law of identity. Can you prove it? If not, then >how do we even know itÕs true? We do know itÕs true, and it IS an >axiom, so that just proves that axioms are not unprovable. >(...Starblade Riven Darksquall...) > If an axiom were reducible or proveable, it would be a conclusion, not an > axiom. The axioms would become the axioms used to reach that conclusion. So, then, the axioms in the system of the human mind are human perception and automatic thought processes, and the ability to learn? The fact is, the very idea that we need axioms is false. What we need are principles. Those can be proven but that does not mean they are therefore reducable to other facts. (...Starblade Riven Darksquall...) === Subject: Re: Trying to unify axioms. > Nothing. > > You forever spewing fucking imbecile, an axiom by definition is > irreducible and unprovable. >>False. >>First off, an axiom isnÕt necessarily irreducable, unless you want to >>claim that reducing it makes the axiom not reducable and therefore not >>an axiom. Sometimes we have an axiom that we find out can be reduced >>into other axioms. Does that therefore disprove the axiom, or destroy >>the usefulness of the axiom, or of taking it as such? No it doesnÕt. >>Second, axioms are not unprovable. They CAN be proven. For an example >>of this, consider the law of identity. Can you prove it? If not, then >>how do we even know itÕs true? We do know itÕs true, and it IS an >>axiom, so that just proves that axioms are not unprovable. >>(...Starblade Riven Darksquall...) >> If an axiom were reducible or proveable, it would be a conclusion, not an >> axiom. The axioms would become the axioms used to reach that conclusion. >So, then, the axioms in the system of the human mind are human >perception and automatic thought processes, and the ability to learn? >The fact is, the very idea that we need axioms is false. What we need >are principles. Those can be proven but that does not mean they are >therefore reducable to other facts. The neat thing about building a math with axioms, is that you can first build one, then go back and change just one axioms slightly. Go through the exercise of building the math again, and see the differences between the first and the second build. ItÕs fun to do. The neat thing about math is one doesnÕt have to include a reality check. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Trying to unify axioms. > Nothing. > > You forever spewing fucking imbecile, an axiom by definition is > irreducible and unprovable. >> >>False. >> >>First off, an axiom isnÕt necessarily irreducable, unless you want to >>claim that reducing it makes the axiom not reducable and therefore not >>an axiom. Sometimes we have an axiom that we find out can be reduced >>into other axioms. Does that therefore disprove the axiom, or destroy >>the usefulness of the axiom, or of taking it as such? No it doesnÕt. >> >>Second, axioms are not unprovable. They CAN be proven. For an example >>of this, consider the law of identity. Can you prove it? If not, then >>how do we even know itÕs true? We do know itÕs true, and it IS an >>axiom, so that just proves that axioms are not unprovable. >> >>(...Starblade Riven Darksquall...) >> >> >> If an axiom were reducible or proveable, it would be a conclusion, not > an >> axiom. The axioms would become the axioms used to reach that > conclusion. >So, then, the axioms in the system of the human mind are human >perception and automatic thought processes, and the ability to learn? >The fact is, the very idea that we need axioms is false. What we need >are principles. Those can be proven but that does not mean they are >therefore reducable to other facts. > The neat thing about building a math with axioms, is that you can > first build one, then go back and change just one axioms slightly. > Go through the exercise of building the math again, and see the > differences between the first and the second build. > ItÕs fun to do. The neat thing about math is one doesnÕt have > to include a reality check. > /BAH > Subtract a hundred and four for e-mail. Yup. Good point. (...Starblade Riven Darksquall...) === Subject: Re: Trying to unify axioms. Hi Gregory L. Hansen, You mentioned, If an axiom were reducible or provable, it would be a conclusion, not an axiom. . But if you kept reducing axioms ad infinitum, all the while keeping all axioms everywhere consistent ... then youÕd be a masochist, and your result would be convoluted. In short, youÕd be a string theorist. If an axiom works it works ... thatÕs good enough. No need to further reduce it. Physicalism is the only theory of everything, and it contains not one equation. Matter is the only reality. Time is perfectly spatial. The future is perfectly fixed but imperfectly known. === Subject: Re: Trying to unify axioms. > Hi Gregory L. Hansen, You mentioned, > If an axiom were reducible or provable, > it would be a conclusion, not an axiom. . > But if you kept reducing axioms ad infinitum, > all the while keeping all axioms everywhere consistent ... > then youÕd be a masochist, > and your result would be convoluted. > In short, youÕd be a string theorist. *Laughs* ThatÕs a good one. > If an axiom works it works ... thatÕs good enough. > No need to further reduce it. Good point. > Physicalism is the only theory of everything, > and it contains not one equation. > Matter is the only reality. > Time is perfectly spatial. > The future is perfectly fixed but imperfectly known. Actually thatÕs not true. Nature is fixxed, but nature is always changing. In that way we can say nature IS progress. The future is not set, but the means by which we must get to the future is essentially set. However, it also posesses degrees of freedom. (...Starblade Riven Darksquall...) === Subject: Re: Trying to unify axioms. >Hi Gregory L. Hansen, You mentioned, > If an axiom were reducible or provable, > it would be a conclusion, not an axiom. . >But if you kept reducing axioms ad infinitum, > all the while keeping all axioms everywhere consistent ... > then youÕd be a masochist, > and your result would be convoluted. ThatÕs true, but some stopping points are more useful than others. For instance, you might want to stop at something related to physically meaningful measurements, say the invariance of the speed of light, for example. If your postulates go beyond the measurable, you start to just make things up. > In short, youÕd be a string theorist. >If an axiom works it works ... thatÕs good enough. > No need to further reduce it. Well, system of axioms, really. >Physicalism is the only theory of everything, > and it contains not one equation. > Matter is the only reality. > Time is perfectly spatial. > The future is perfectly fixed but imperfectly known. -- Are those morons getting dumber or just louder? -- Mayor Quimby === Subject: Re: Trying to unify axioms. > Nothing. > > You forever spewing fucking imbecile, an axiom by definition is > irreducible and unprovable. > False. > First off, an axiom isnÕt necessarily irreducable, unless you want to > claim that reducing it makes the axiom not reducable and therefore not > an axiom. Sometimes we have an axiom that we find out can be reduced > into other axioms. Does that therefore disprove the axiom, or destroy > the usefulness of the axiom, or of taking it as such? No it doesnÕt. Uh, actually, axioms are irreducible. > Second, axioms are not unprovable. They CAN be proven. For an example > of this, consider the law of identity. Can you prove it? If not, then > how do we even know itÕs true? We do know itÕs true, and it IS an > axiom, so that just proves that axioms are not unprovable. The Law of Identity cannot be proven. It is taken to be an axiom of the first-order language with equality, and is only supported by our intuition. Try proving the axiom of commutativity. Oh wait, hereÕs a counter-example: Matrix multiplication isnÕt commutative! All axioms do is place constraints on what sorts of objects are being studied. Ôcid Ôooh === Subject: derivative in the sense of distributions Is it true that a function in L^2 is always differentiable in the sense of distributions? In other words, for any f in L^2, does it exists a function g in L^1_loc such that: - = for all v in C_0^infty I could find it in any book. Is there a good reference? Diogo === Subject: Re: derivative in the sense of distributions > Is it true that a function in L^2 is always differentiable in the sense of > distributions? In other words, for any f in L^2, does it exists a function > g in L^1_loc such that: > - = for all v in C_0^infty As David pointed out, every L^2 function (like every distribution) has a distibutional derivative, but the latter need not be in L^1 locally. For example, let f be the characteristic function of [0,1]. Then - = v(0) - v(1). So the distributional derivative of f is the singular measure delta_0 - delta_1, and thereÕs no way this is going to be given by a g in L^1_loc. === Subject: Re: derivative in the sense of distributions >Is it true that a function in L^2 is always differentiable in the sense of >distributions? Yes. >In other words, for any f in L^2, does it exists a function >g in L^1_loc such that: > - = for all v in C_0^infty No. Your in other words is wrong. Any element of L^2 has a derivative in the sense of distribiutions, but this derivative need _not_ be a function in L^1_loc. >I could find it in any book. Is there a good reference? _Any_ reference developing the theory of distributions from a mathematical point of view contains the following two facts: (i) any L^2 function defines a distribution (ii) every distribution has a derivative, in the sense of distributions. (On the other hand, youÕre not going to find a proof that the derivative of an L^2 function must be in L^1_loc in most books, because itÕs false.) >Diogo ************************ David C. Ullrich === Subject: e^(pi* i) I just read the e is transcendental thread and thought I would throw my hat in the ring. A few years ago while a freshman in college, I came up with this outrageous proof that took me a few days to finally prove incorrect. I leave it to you to prove false. I think itÕs pretty cool even if it is wrong :-) e^(pi*i) = -1 (e^(pi*i))^2 = (-1)^2 e^(2*pi*i) = 1 ln(e^(2*pi*i)) = ln(1) 2*pi*i = 0 This is where you say to yourself... WHOOPS! How can 2*pi*i be 0? Since 2, pi, and i are all constants and NONE are 0, we know from our multiplication laws that if a*b*c = 0 then a = 0 or b = 0 or c = 0 which is not the case here. So what happened? Enjoy! === Subject: Re: e^(pi* i) Anthony a .8ecrit dans le message de > I just read the e is transcendental thread and thought I would throw > my hat in the ring. A few years ago while a freshman in college, I > came up with this outrageous proof that took me a few days to > finally prove incorrect. I leave it to you to prove false. I think > itÕs pretty cool even if it is wrong :-) > e^(pi*i) = -1 In complex analysis, the principal logarithm of a complex number z is defined by: Log z=log |z| + i Arg z. Therefore Log[e^(2*pi*i)]= 0 + i Arg(2*pi)=0. Your mistake is that you take the principal log in the right side and another log in the left side! > ln(e^(2*pi*i)) = ln(1) > This is where you say to yourself... WHOOPS! > How can 2*pi*i be 0? Since 2, pi, and i are all constants and NONE > are 0, we know from our multiplication laws that if a*b*c = 0 then a = > 0 or b = 0 or c = 0 which is not the case here. > So what happened? > Enjoy! === Subject: Re: e^(pi* i) > I just read the e is transcendental thread and thought I would throw > my hat in the ring. A few years ago while a freshman in college, I > came up with this outrageous proof that took me a few days to > finally prove incorrect. I leave it to you to prove false. I think > itÕs pretty cool even if it is wrong :-) > e^(pi*i) = -1 This is where you say to yourself... WHOOPS! > How can 2*pi*i be 0? Since 2, pi, and i are all constants and NONE > are 0, we know from our multiplication laws that if a*b*c = 0 then a = > 0 or b = 0 or c = 0 which is not the case here. > So what happened? What happened is that you do not say what ln is nor what properties does it have. Jose Carlos Santos === Subject: easy topology problem... hello........ function f : X -> Y is continuous <=> for any B in Y , f^(-1){Fr(B)} in Fr{f^(-1)(B)} ------------------------------------------------ um......i canÕt prove.......... help...me, please. === Subject: Re: easy topology problem... > function f : X -> Y is continuous > <=> for any B in Y , f^(-1){Fr(B)} in Fr{f^(-1)(B)} > ------------------------------------------------ > um......i canÕt prove.......... Me neither; this is false. Take f:R --> R defined as f(x) = x^2 and take B=[0,1[. Then f^(-1)(Fr(B)) = f^(-1)({0,1}) = {-1,0,1} and Fr(f^(-1)([0,1[)) = Fr(]-1,1[) = {-1,1}. My guess is that what you meant was f^(-1){Fr(B)} contains Fr{f^(-1)(B)}. Am I right? Jose Carlos Santos === Subject: simple differentiable question. F(x,y,z) = (ax, by, cz) is differentiable <=> del F = (a,b,c) is exist -------------------------------- itÕs right?? I am anxious to this..... um.....let me advice...please...thank you. === Subject: Re: simple differentiable question. > F(x,y,z) = (ax, by, cz) is differentiable > <=> del F = (a,b,c) is exist > -------------------------------- > itÕs right?? > I am anxious to this..... > um.....let me advice...please...thank you. If F is a functions from R to R (reals) then F is differentiable iff delF is not zero... === Subject: Re: simple differentiable question. > F(x,y,z) = (ax, by, cz) is differentiable > <=> del F = (a,b,c) is exist del * F = (a,b,c) === Subject: Re: simple differentiable question. > F(x,y,z) = (ax, by, cz) is differentiable > <=> del F = (a,b,c) is exist > del * F = (a,b,c) in ASCII, it may be better to say div F and avoid confusion. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: simple differentiable question. >> F(x,y,z) = (ax, by, cz) is differentiable >> <=> del F = (a,b,c) is exist >> del * F = (a,b,c) >in ASCII, it may be better to say div F and avoid confusion. And even better to say div F = a + b + c ? --Lynn === Subject: Re: A slight generalization of WaringÕs problem > I_l = {n s.t. n=x_1^k+...+x_l^k does *not* have a solution}, >Now I wonder if anything is known about about the slightly more >general problem of finding how many k-th powers are required for the >representation of *most* numbers, i.e. of finding the least l for >which > (1/n)*card I_l intersects {1...n} -> 0 >as n->infty. I got no answers, any thoughts?!? Michele -- > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc === Subject: Re: A slight generalization of WaringÕs problem > I_l = {n s.t. n=x_1^k+...+x_l^k does *not* have a solution}, >Now I wonder if anything is known about about the slightly more >general problem of finding how many k-th powers are required for the >representation of *most* numbers, i.e. of finding the least l for >which > (1/n)*card I_l intersects {1...n} -> 0 >as n->infty. > I got no answers, any thoughts?!? > Michele Mathworld does make brief mention of this problem, with a couple of results the appropriate number of kth powers. === Subject: Re: the anticlassicalist }{ : mitchism for Daleks [...] > As I keep trying to point out, my focus is on education with these > series of posts. The theory of models and their logics is crucial > to all science. Oh, no it isnÕt! > Indeed, it rigorously formalises the notion of a theory. A philosopher of science may have a professional duty to give a toss about such a thing, but a hardworking biochemist or oceanographer by no means does. > Many models across disciplines have a Heyting structure, and these > are often unknown to practitioners in the various fields, so I use > that as my archetype for education reform. There are many cases > where the unfamiliarity with the logical structure of the models > used presents confusion to the students early on. Hello childrens! Today weÕre going to learn about counting! But first we will need a few set-theoretical preliminaries so as to avoid any confusion about the structure of the natural number system. Des has no Bourbaki today -- [T]he structural trend in linguistics which took root with the International Congresses of the twenties and early thirties [...] had close and effective connections with phenomenology in its Husserlian and Hegelian versions. -- Roman Jakobson === Subject: Re: the anticlassicalist }{ : mitchism for Daleks <103513r3gafocff@corp.supernews.com> <1035esdt30h4d38@corp.supernews.com> <1039dr5kpjloif1@corp.supernews.com> <6IvurhFpHeNAFwU6@baesystems.com> <103cn32aodr3m3a@corp.supernews.com> <103n8ds9mudn455@corp.supernews.com> In message <103n8ds9mudn455@corp.supernews.com>, galathaea >: OK. So if IÕm already unknowingly using this logic, what is the point >: youÕre trying to make? Believe me, it hasnÕt yet emerged from all the >: verbiage. >Well, from a practical point of view, understanding the logic of >orthomodular lattices that arises in propositions on a Hilbert space can >prevent a lot of early conceptual difficulties in the use of quantum >mechanics. Really? My experience is that the difficulties arise because quantum objects donÕt behave intuitively, i.e. like the objects of everyday experience. No amount of bafßegab about orthomodular lattices will change that. [snip another few hundred words...] >Extend to quantum field theories, strings, and such, and maybe it gets >clearer why I see the necessity to alter education of logic in physics >curricula. No. No clearer. >And this does not even touch upon the theory of causal sets and other >directions in physics where such an education would be beneficial. -- Richard Herring === Subject: simple question for cluster point. i saw a proposition increasing function have a cluster point(=limit point) at most --------------------- um........i have a doubt. i think that it must to exchage increasing function => increasing sequence or domain of increasing function is countable point. i think....... if f(x) = x for x in R, limit point is R. thus, proposition is not perfect. my thinking is right?? let me advice...please.......thank you === Subject: Solving a set of multivariate polynomial equations in Z/2Z I have a set of equations of the form sum of products of subsets of the variables equal to 0 or 1 in Z/2Z. Something like: x0.x1.x3 + x2.x3.x4 + x1 = 1 x1.x2.x3 + x3.x5 = 0 [etc] This is in Z/2Z, i.e. you can also see that as the variables being booleans, * = and, + = xor. I have 64 variables and 100-150 equations of maximal order 20 or so, and IÕm trying to solve the system. My questions are: - does this problem have a well known name? ItÕs very hard to find call it exactly. - are there known methods/algorithms/heuristics that work reasonably well, i.e. with which I can hope for an answer in max a week of cpu time? OG. === Subject: Re: Solving a set of multivariate polynomial equations in Z/2Z >I have a set of equations of the form sum of products of subsets of >the variables equal to 0 or 1 in Z/2Z. Something like: > x0.x1.x3 + x2.x3.x4 + x1 = 1 > x1.x2.x3 + x3.x5 = 0 > [etc] >This is in Z/2Z, i.e. you can also see that as the variables being >booleans, * = and, + = xor. I have 64 variables and 100-150 equations >of maximal order 20 or so, and IÕm trying to solve the system. >My questions are: >- does this problem have a well known name? ItÕs very hard to find > call it exactly. YouÕre looking for a SAT Solver. The problem you have is basically a boolean satisfiability problem. Your example: can you assign thruth values to P0, P1, ... such that (((P0 / P1 / P2) xor (P2 / P3 / P4)) xor P1) / not ((P1 / P2 / P3) xor (P3 / P5)) / etc. ? SAT Solvers (eg Chaff, SatZoo) like to have their input in conjunctive normal form and use some version of Davis-Putnam to cehck satisfiability (so you may have to translate to CNF). Maybe you should take a look at or . Peter -- Peter van Rossum, | Universal law of linearity: for all Dept. of Mathematics, New Mexico | f : R -> R and for all x, y in R: State University, Las Cruces, NM, USA. | f(x + y) = f(x) + f(y) === Subject: distributional derivative in L^2 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1PCZpG31596; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1PBECi25139 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1PBECt31601; Is it true that a function in L^2 is always differentiable in the sense of distributions? In other words, for any f in L^2, does it exists a function g in L^1_loc such that: - = for all v in C_0^infty I couldnÕt find it in any book. Is there a good reference? Diogo === Subject: Re: distributional derivative in L^2 > Is it true that a function in L^2 is always differentiable in the sense of > distributions? In other words, for any f in L^2, does it exists a function g in L^1_loc such that: > - = for all v in C_0^infty > I couldnÕt find it in any book. Is there a good reference? You (or someone else) has already posted that same question (under the name derivative in the sense of distributions) and David C. Ullrich has already provided an answer. Jose Carlos Santos === Subject: Re: IÕm guessing this is an easy problem - but I canÕt solve it! Finally found the solution myself... using (a^3 - b^3) = (a-b)(a^2 + ab + b^2) > Working through calculus... Stuck on a problem of factorisation: Please > help! > question is: > find lim u--> 1 f(u) > where f(u) = (u^4 - 1) / > (u^3 -1) > Obviously - I cannot just replace u with 1 as this will give me a division > by zero. I tried replacing (u^2 - 1) (u^2 +1) on top as this seemed most > logical - but I canÕt figure out what to do with the bottom now. > IÕm probably going to kick myself for not figuring this one out - but IÕve > spent hours on this and IÕm still lost. > Help! === Subject: determinant 1. Let A be a nxn real matrix such that A^t = -A, where A^t is the transpose of A. Show that det(A) >= 0. proof. Because A is real and det(xI - A) = det((-x)I - A), the set of roots of det(xI - A) is of the form {t_1, -t_1, ..., t_k, -t_k, c_1, conjugate(c_1), -c_1, -conjugate(c_1), ..., c_m, conjugate(c_m),-c_m, conjugate(c_m)}, where t_i are real and c_j are complex with nonzero imaginary part. Then I have no idea about this question. Please give me a hint. 2. Let A be a nxn real matrix. Let A^a = (b_ij) be the classical adjoint of A, that is, b_ij = (-1)^(i+j)det(A_ji), where A_ji is the (n-1)x(n-1) matrix obtained by deleting row j and column i from A. Prove that (A^a)^a = [(detA)^(n-2)]A. proof. If A is invertible, by the identity (A^a)A = (detA)I, the result is obvious. Otherwise, A is not invetible. In case A is of row-reduced echelon form, the result is easy to show. In general case of A, I have no idea about this question. Please give me a hint. -- === Subject: Re: determinant > 1. Let A be a nxn real matrix such that A^t = -A, where A^t is the > transpose of A. This is a skew-symmetric matrix. It has the property that all eigenvalues are purely imaginary. > Show that det(A) >= 0. > 2. Let A be a nxn real matrix. > Let A^a = (b_ij) be the classical adjoint of A, that is, > b_ij = (-1)^(i+j)det(A_ji), where A_ji is the (n-1)x(n-1) > matrix obtained by deleting row j and column i from A. > Prove that (A^a)^a = [(detA)^(n-2)]A. Remember that A A^a = A^a A = det(A)I and ask: what is det(A^a)? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: determinant > > 1. Let A be a nxn real matrix such that A^t = -A, where A^t is the > transpose of A. > This is a skew-symmetric matrix. It has the property that all eigenvalues > are purely imaginary. As an added hint, you might take note that iA is Hermitian, and Hermitian matrices necessarily have non-negative eigenvalues. What is the relationship between the eigenvalues of A and those of iA? Achava === Subject: Somewhat math-oriented web comic Since the artist wonÕt do plugs, a fan (i.e me) has to do it for him: http://home.comcast.net/~pshaughn/tandr.html I recommend to start with the archive and #1, youÕll either hate or love it immediately. -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de als man ankam wollte man werden, die geschichte schreiben, die doofen sollen sterben, der plan als man damals nach hamburg kam (Kettcar) === Subject: Re: Joint Probability Question >If I have two independant rv x and y, both drawn from different normal >distributions; how can I determine the probability that x will be less >than y? >IÕm leaving out the specifics of the distributions because this is >part of a homework question and I truely want to understand the >problem, not just get an answer. If your distributions are normal, then the mean of the difference is the difference of the means and the variance of the difference is the sum of the variances. Therefore, the probability of x being less than y, is a function of mean(Y)-mean(X) u = ------------------- sqrt(var(Y)+var(X)) It is pretty easy to guess which function. Rob Johnson take out the trash before replying === Subject: linear difference equations I want to find x(n) x(n+2) + 2*x(n+1) + c(n)*x(n) = 0 were c(n) is a given sequence and x(0) and x(1) are given. === Subject: Re: linear difference equations > I want to find x(n) > x(n+2) + 2*x(n+1) + c(n)*x(n) = 0 > were c(n) is a given sequence and x(0) and x(1) are given. First put n=0 to evaluate x(2). Then put n=1 to evaluate x(3). And so on. In this way you can find x(n) for any natural number n. === Subject: Re: linear difference equations You can also determine a ratio between terms and get a concise solution. Checkout Summation of Series, additive. The c(n) term may de-rail this, however if c(n+1) is related to c(n) perhaps not > I want to find x(n) > x(n+2) + 2*x(n+1) + c(n)*x(n) = 0 > were c(n) is a given sequence and x(0) and x(1) are given. > First put n=0 to evaluate x(2). Then put n=1 to evaluate x(3). > And so on. In this way you can find x(n) for any natural number n. === Subject: Re: e^(pi* i) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1PENJ408203; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1PEM5i08101 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1PEM5a05791; >I just read the e is transcendental thread and thought I would throw >my hat in the ring. A few years ago while a freshman in college, I >came up with this outrageous proof that took me a few days to >finally prove incorrect. I leave it to you to prove false. I think >itÕs pretty cool even if it is wrong :-) >e^(pi*i) = -1 This is where you say to yourself... WHOOPS! >How can 2*pi*i be 0? Since 2, pi, and i are all constants and NONE >are 0, we know from our multiplication laws that if a*b*c = 0 then a = >0 or b = 0 or c = 0 which is not the case here. >So what happened? >Enjoy! You are dealing with periodic functions. e^(2*pi*i) = e^0 = e^(4*pi*i) = etc. So you canÕt just equate the exponents. phil === Subject: Re: divergent series S_1^inf [1-(ln(n))/n]^n = inf how to prove this? I assume you mean oo --- ln(n) n > ( 1 - ----- ) [1] --- n n=1 -Yes, thatÕs exactly what I mean. When a < 1/2, log(1-a) > -a-a^2. Therefore, letting a = ln(n)/n, multiplying by n, and exponentiating, we get ln(n) n ( 1 - ----- ) > exp(-ln(n)-ln(n)^2/n) n 1 1 > - - [2] n 2 For all n >= 1, ln(n)/n < 1/2, therefore, we have [2] for all n >= 1. Since each term of [1] is greater than 1/(2n), compare with the harmonic series. your inequalities? === Subject: Re: divergent series >S_1^inf [1-(ln(n))/n]^n = inf >how to prove this? > I assume you mean > oo > --- ln(n) n > > ( 1 - ----- ) [1] > --- n > n=1 > When a < 1/2, log(1-a) > -a-a^2. Therefore, letting a = ln(n)/n, > multiplying by n, and exponentiating, we get > ln(n) n > ( 1 - ----- ) > exp(-ln(n)-ln(n)^2/n) > n > 1 1 > > - - [2] > n 2 > For all n >= 1, ln(n)/n < 1/2, therefore, we have [2] for all n >= 1. > Since each term of [1] is greater than 1/(2n), compare with the harmonic > series. Let me keep the thread going and pose: does the series sum_n 1/2^sqrt(n) converge or diverge? Curiously, I donÕt recall ever seeing this on a calculus exam. --Ron Bruck === Subject: Re: divergent series > Let me keep the thread going and pose: does the series > sum_n 1/2^sqrt(n) > converge or diverge? Curiously, I donÕt recall ever seeing this on a > calculus exam. > --Ron Bruck It converges. In fact, sum_n 1/a^(n^b) converges for any a>1, b>0, by comparison with sum_n 1/n^2, for example. This follows from the fact that n^b is eventually greater than 2 log n for any b>0. === Subject: Re: divergent series > Let me keep the thread going and pose: does the series > sum_n 1/2^sqrt(n) > converge or diverge? Curiously, I donÕt recall ever seeing this on a > calculus exam. > --Ron Bruck converges, by the integral test -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: divergent series > Let me keep the thread going and pose: does the series > > sum_n 1/2^sqrt(n) > > converge or diverge? Curiously, I donÕt recall ever seeing this on a > calculus exam. > > --Ron Bruck > converges, by the integral test Sounds good, but offhand I donÕt see how to integrate 1/2^sqrt(x). Is it really integrable by elementary functions? On the other hand, itÕs clear that 2^sqrt(n) is larger than, say, n^2, for all sufficiently large n. (Take logs, so need to know sqrt(n) grows faster than log(n).) Hence the series converges by the comparison test with sum of 1/n^2. More generally, if C > 1 and e > 0 are constants, then the sum of 1/C^(n^e) converges. Proof is the same. JoeS === Subject: Re: divergent series >> Let me keep the thread going and pose: does the series >> >> sum_n 1/2^sqrt(n) >> >> converge or diverge? Curiously, I donÕt recall ever seeing this on a >> calculus exam. >> >> --Ron Bruck >> converges, by the integral test >Sounds good, but offhand I donÕt see how to integrate 1/2^sqrt(x). Is >it really integrable by elementary functions? | sqrt(x) | 2 dx | | = | 2^y d(y^2) | | = | 2y 1/log(2) d(2^y) | | = 2/log(2) y 2^y - | 2/log(2) 2^y dy | = 2/log(2) y 2^y - 2/log(2)^2 2^y = 2/log(2)^2 2^y (log(2) y - 1) sqrt(x) = 2/log(2)^2 2 (log(2) sqrt(x) - 1) >On the other hand, itÕs clear that 2^sqrt(n) is larger than, say, n^2, >for all sufficiently large n. (Take logs, so need to know sqrt(n) >grows faster than log(n).) Hence the series converges by the >comparison test with sum of 1/n^2. >More generally, if C > 1 and e > 0 are constants, then the sum of >1/C^(n^e) converges. Proof is the same. Rob Johnson take out the trash before replying === Subject: Re: divergent series > > Let me keep the thread going and pose: does the series > > sum_n 1/2^sqrt(n) > > converge or diverge? Curiously, I donÕt recall ever seeing this on a > calculus exam. > > --Ron Bruck > > converges, by the integral test > Sounds good, but offhand I donÕt see how to integrate 1/2^sqrt(x). Is > it really integrable by elementary functions? > On the other hand, itÕs clear that 2^sqrt(n) is larger than, say, n^2, > for all sufficiently large n. (Take logs, so need to know sqrt(n) > grows faster than log(n).) Hence the series converges by the > comparison test with sum of 1/n^2. > More generally, if C > 1 and e > 0 are constants, then the sum of > 1/C^(n^e) converges. Proof is the same. Yes, itÕs integrable. Make the change of variable u = sqrt(x) in int 2^(-sqrt(x))dx to get int 2^(-u) 2u du and then integrate by parts. I confess this wasnÕt the way I solved the problem. That was just an order-of-magnitude calculation (show itÕs < 1/n^2 for n sufficiently large). And also the condensation test, that sum a_n is summable (given a_n decreasing) iff sum 2^n a_{2^n} is summable. --Ron Bruck === Subject: Re: Question about Incompleteness (Godel) Adjunct Assistant Professor at the University of Montana. [.snip.] >1. Is it a trivial corollary that the set of undecidable statements, >is not only non-empty, but infinite (or even uncountable) (In some >ÕinterestingÕ way - obviously ÔG & n=nÕ for all n is an infinite >undecidable set) Let A_0 be the original axiom set. Let G_0 be the Goedel statement for the theory with that axiom set. For each i>0, let G_i be the Goedel statement for the theory with axiom set A_i = A_{i-1} U {G_i-1}. Then G_i is expressible in the language of the original theory, makes a statement essentially distinct from G_0,...,G_{i-1}, and being formally undecidable from A_i, it is also formally undecidable from A_0. So the set {G_0,G_1,G_2,...} is a countable set of undecidable sentences for the original theory. Thus the set is infinite. We usually do not consider uncountable languages, because there is no point: any well-formed-formula can use only finitely many variables, so a countable set of variables will suffice. If the language is at most countable, then the entire set of WFFs is countable, so the set of undecidable statements must be at most countable. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Question about Incompleteness (Godel) ... > Huh? No, there are models in which the given Goedel statement is true, > and models in which it is false; one may adjoin it or its negation to > the axioms (provided certain conditions are met) and obtain a > consistent system. Yes, the new system has a ->new<- Goedel statement > which is formally undecidable in that system, but the old one is not > true but not provable. I looked at your response and didnÕt believe it for a minute, but during that minute I remembered the completeness theorem: if the Goedel statement were true in every model of T it would be deducible. Post in haste, repent at leisure ... === Subject: Re: Bayesian Class and Math/Stat Teaching... (Space shuttle o-rings) > ... If I recall, the point was that >just because there are many data points, one may still not have enough >information with which to proceed. All the data points they had indicated >that the o-ring was fine but the had no data points for how the o-ring might >behave in cold weather. > No. My fairly clear recollection is that they had enough data to be > positively suspicious that disaster might strike if they launched at a > temperature lower than any previous launch. This WASNÕT just a > general conservative view that one shouldnÕt push into unexplored > territory. [...] I believe you are correct. The original poster may be thinking of the more recent shuttle disaster. If I remember correctly, as part of the infamous Power Point presentation there is a graph showing some data about results of objects striking the wing, and there is a footnote on the graph that explicitly states, Estimated kinetic energy is greater than any yet measured (paraphrasing). When they did, later, carry out a simulation of the impact, it blew a hole in the wing. Robert Dodier === Subject: Re: Bayesian Class and Math/Stat Teaching Techniques > The way the business schools and law schools would teach that concept > (why does linear regression give the best straight line fit to model > data) would be to have a case where, say, someone using another method > got less than optimal results that ended in disaster and then show/teach > what the calculations might be with linear regression. That wouldnÕt teach *why* linear regression gives the best straight line. In law school, the case method of teaching is used for a couple of reasons. First, in many areas of law, cases are the *source* of the law. For example, consider copyright law. The statute is in many areas, such as fair use, very vague. It is simply not possible to decide from the statute in most situations whether or not something is fair use. You *must* read with the Supreme Court and other courts have said in actual cases, and from that you can develop a notion of what constitutes fair use. Second, because of the way the precedence system works, citations to previous cases are an important part of any legal argument. -- --Tim Smith === Subject: Re: Bayesian Class and Math/Stat Teaching Techniques > Sorry Russell. you are not a bozo. I was referring to JW. Glad to hear that. I thought that might be the case. > Maybe we should establish a business selling standard barrels of > herring for the national shoot-the-herring-in-the-barrel tournament? > Can we come up with a test to see whether or not the differences due > to Matjes vs Schmaltz herring are signficant w/r to herring > marknanship? . :-) I think to make it a viable business we have to have the contestants use a small enough caliber weapon that there is enough of the herring left to process for food. Russell > I am a geologist whose career has been based on > statistics. > IÕve worked in lots of fields myself. Right now my work is > in meteorology > and climatatology. Lots of applied statistics in all the > things IÕve done. > I enjoy theory, but it is difficult to find teachers or > books that relate > it well to practice, but they are out there and can be found > with some > effort. > I suffered through the theory classes and the > theorems. It did not help > that many of my classmates were physics majors who > were actually fairly > narrow probabalists. The background math / theory > has turned out to be > critical. Far too often the applied instructors > (in the business > college, in chemometrics, etc.) donÕt remember > (or care) that all > procedures have limited validity (robustness). A > prime advantage is to > understand where procedures make no sense because > the system of interest > does not cooperate. Hence the mean has limited > practical usefulness, as > does the standard deviation, regression > statistics, factors and the > like. Using the procedures without understanding > ( and investigating) > the underlying assumptions can be professionally > suicidal for an > applied statistician. One of my avocations is > evaluating the work of > bozos like you in lawsuits. > Why do think IÕm a bozo? > ItÕs like hitting herring in a barrel. > First, you have to get a barrel of herring. ;-) > Russell > A few weeks ago I posted a > message asking about books on > Bayesian > Statistics. I got some nice > Unfortunately I have since > dropped the class and am > wondering about whether > I should continue the degree > (Masters in Applied > Statistics) and would like > some thoughts from the > thoughtful people here. > I have a BS in Electrical > Engineering and an MBA, both > from the University > of Michigan. What I really > liked about the MBA program is > that it was almost > 100% applied. Probably 50-70%+ > of the classes was case > study classes, a > trend mostly propagated and > refined by Harvard Business > School where > supposedly 100% of the classes > are case study classes. > Apparently more and > more law programs have more > and more case study classes as > well. Case study > classes are really as applied > as you get because concepts > and theories are > learned in the context of real > world situations and > circumstances. I am also > the type of person who is a > very intuitive learner and has > a much easier > time learning when I see how > what I am learning relates to > challenges in > real life (eg: business, which > is what I do). > So given that the degree I am > pursuing is called Masters in > *APPLIED* > Statistics, I thought the the > courses would be heavily > applied and taught in > the context of solving real > world problems. No dice. Both > courses I took in > the first semester (part-time > evening program) had heavy > theory. The > Bayesian class was not even as > bad as the other one > (Mathematical > Statistics). There was > essentially no attempt on the > part of the professor > to relate the theory to real > world programs or to even give > real world > examples to illustrate the > concepts. It was formula, > theory, formula, > theory, theory, formula, etc. > I asked him about that and he > said thereÕs no > way around the theory. IÕm not > trying to get around the > theory but > theories and formulas mean > nothing to me without real > world context. IÕm not > stupid either -- I score in > the 99+ percentile on > quantitative SAT/GMAT/GRE > and 98+ percentile overall. > My thinking right now is that > my expectations were just off > and disciplines > like Math/Statistics are just > not as, ummm, progressive as > Business/Law when > it comes to teaching (please > -- no hate mail). Those > teaching Math/Stat may > also be too smart and are > not interested in mundane > day-to-day > business/industry problems > (hopefully that will stop the > hate mail!). So > whatÕs up with that? Why is a > degree called Masters in > Applied Statistics > so heavy in theory? IÕm not > interested in theory in the > absence of > application. I enrolled in > Masters in Applied > Statistics to learn how to > use statistical techniques to > solve real world problems, how > to use > statistical software to solve > real world problems, etc. and > not to learn > esoteric statistical theory in > the absence of application > that I will surely > forget an hour after the final > exam. > I am not trying to slam the > field. I am interested in some > opinions from > those in the field, especially > those teaching it, to help me > determine if I > should continue further. > I can relate to some extent. While in > theory I love theory, in > practice I often wonder how what IÕm > studying relates to what I want > to understand, whether the field is pure > The best math course I ever had was diff > eq using _Differential > Equations with Applications and > Historical Notes_ by George Simmons > as the text, in part because the > applications gave me something > concrete with which to relate. IÕm not > surprised that the Mathematical > Statistics course was heavily theory, > and in fact many of the stat > books IÕve looked at appeared tilted > that way. IÕve probably learned > more statistics from books on signal > processing than stat texts. > Someone might argue thatÕs why I know so > little about statistics. :-) > Perhaps it is just the particular > school/program youÕre in, but I donÕt > know. Maybe you should consider > economics and econometrics, which > can be heavy with stats and purport, at > least, to deal with real world > problems. Good luck. > Russell > -- -- Russell Martin R. L. Martin and Associates, Consultants in russell.martin@wdn.com Science and Technology http://www.rmartin.com All too often the study of data requires care. === Subject: Complete, but not Baire Hi all, Could someone give me an example of a complete uniform space which is not a Baire space? IÕve tried to construct such a space, but in vain. Furthermore, thereÕs no exemple of such a space in Steen & SeebachÕs Counterexamples in Topology. Of course, since IÕm not aware of the existence of a complete uniform space which is not a Baire space, perhaps the the question should be Is it true or false that... ?, but I would be very much surprised if it turned out the that every complete uniform space is a Baire space. Jose Carlos Santos === Subject: re:Complete, but not Baire Can you remind the definition of a Baire space and a uniform space? ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: stable manifold theorem Hi all, IÕm presenting the stable manifold theorem for non-linear odeÕs in an upcoming seminar (IÕm a grad student). When I first came across the stable manifold theorem, I asked myself: So itÕs a manifold, who cares? Why not just call it a set instead of a manifold? Well, given a manifold, we can setup a differential structure on it, and do calculus. What IÕd like to do in my seminar is give an example of a system where we exploit the fact that the stable manifold is indeed a manifold. Does anyone know of any nice examples? Matt === Subject: Re: stable manifold theorem >Hi all, >IÕm presenting the stable manifold theorem for non-linear odeÕs in an >upcoming seminar (IÕm a grad student). >When I first came across the stable manifold theorem, I asked myself: >So itÕs a manifold, who cares? Well, if you are a mathematician, you wouldnÕt say this. Since doing mathematics means to find structures. And a smooth manifold is a much nicer structure than just any old set. > Why not just call it a set instead of >a manifold? Well, given a manifold, we can setup a differential >structure on it, and do calculus. >What IÕd like to do in my seminar is give an example of a system where >we exploit the fact that the stable manifold is indeed a manifold. What do you mean by exploit? Since it is such a basic fact you use it all the time while doing dynamic systems theory. An interesting application is SmaleÕs example of a structurally stable diffeomorphism with infinitely many hyperbolic periodic points (the so-called horse shoe map). That is the case when the unstable and the stable manifold of a (saddle-type) fixed point have transversal intersections. Another very interesting system is the Henon equation (x,y) -> (a - x^2 + by , x). For a=1.4, b=0.3 (for example) this equation has two saddle-points. One saddlepoint has transversal homoclinic points. It is fascinating how this makes the unstable manifold to fold back on itself creating a fractal set. In this case it is conjectured, that the unstable manifold forms an attractor with chaotic dynamics. HTH Thomas >Does anyone know of any nice examples? >Matt === Subject: How to prove this probability Inequality X,Y,Z are random variables Try to prove E[EX-E(X|Z)]^2>=E[EX-E(X|Y,Z)]^2 === Subject: Re: a linear algebra thought by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1PH5Vr23773; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1PFLOi13429 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1PFLOC08857; >All, I have a question on linear algebra. Let V be a finite dimensianl >vector space, S be a noninvertable linear transformation on V. I >wonder if the Ker(S) is perpendicular to Range(S), or ker(S)+Range(S) >is direct sum and Ker(S)+Range(S)=V. First: in an abstract vector space one cannot talk about >being perpendicular< - this notion is only defined if the vector space comes equipped with a scalar product. Second: you are right that ker(S)+Range(S)=V, where the + denotes taking the direct sum (not the sum within V!). H === Subject: Metric Engineering, Dark Energy, Quintessence, Phantom Energy re: http://qedcorp.com/destiny/ www.stardrive.org/starship.shtml www.geocities.com/zcphysicsms/chap12.htm of his metric engineering section not understanding the w = -1 property of zero point energy and the (energy density)(1 + 3w) active source in EinsteinÕs GR from the /zpfguv term of the effective exotic vacuum stress-dark energy/matter tensor field. Hence he gets it wrong saying positive zero point energy density gravitates. In fact it anti-gravitates from the domination of the negative exotic vacuum quantum pressure. However, this error is easily fixed. Note w = 0 for ordinary matter at slow speeds. w = + 1/3 for EM radiation like the CDM w = - 1 for zero point vacuum ßuctuations w = -1/3 to > -1 for quintessent fields w < - 1 for phantom energy fields My vacuum coherence theory makes the strong Popper falsifiable prediction w = -1 on the nose, exactly. dark matter, i.e. dark matter detectors forever silent as a matter of principle except for false positive errors. Globs of w = -1 dark matter of positive pressure will be mistaken for w = 0 CDM when detected distantly from, e.g. gravity lensing. www.lerc.nasa.gov/WWW/bpp/TM-107289.htm www.ldolphin.org/hill.html === Subject: Book series Lecture Notes in Mathematics Are these books generally directed to people with advanced interests and degrees? So it seemed to me when I frequented my college library. David Ames === Subject: Re: Book series Lecture Notes in Mathematics > Are these books generally directed to people with advanced interests > and degrees? So it seemed to me when I frequented my college library. It depends on the editorial house (there are many Lecture Notes series!) but by and large, yes. They cover stuff that would be covered in an advanced course or seminar for which a textbook is not available. The Editorial Policy of the LMN series from Springer (the yellow books that I suspect is what you are referring to) states in part (taken from the overleaf front cover of the 1994 printing of MumfordÕs Red Book, LNM 1358): 1. Lecture Notes aim to report new developments - quickly, informally, and at a high level. The texts should be reasonably self-contained and rounded-off. Thus they may, and often will, present not only results of the author but also related work by other people. The back cover states: This series reports on new developments in mathematical research and teaching - quickly, informally, and at a high level. The type of material considered for publication includes 1. Research monographs 2. Lectures on a new field or presentations of a new angle in a classical field. 3. Seminars on topics of current research. Texts which are out of print but still in demand may also be considered if they fall within these categories. The London Mathematical Society Lecture Notes series does not seem to have an editorial policy printed on their books (at least the ones I have) but they seem to be of about the same scope, except that they also often publish conference proceedings. Arturo Magidin, sans .sig === Subject: Re: Book series Lecture Notes in Mathematics > Are these books generally directed to people with advanced interests > and degrees? Yes. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Book series Lecture Notes in Mathematics > Are these books generally directed to people with advanced interests > and degrees? Yes. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: simple differentiable question. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1PILlJ31484; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1PHbbi26708 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1PHbbI18140; >F(x,y,z) = (ax, by, cz) is differentiable ><=> del F = (a,b,c) is exist >-------------------------------- >itÕs right?? >I am anxious to this..... >um.....let me advice...please...thank you. Is that really what you meant to ask? Of course, (ax, by, cz) is differentiable and div F exists: F is linear. There is no <=> about it. IÕm wondering if you didnÕt mean to ask Is it true that a function F(x,y,z) of several variables is differentiable if and only if del F (i.e. grad F) exists. In that case, I would have to ask you to tell us precisely what definition of differentiable you are working with. I know some books in which the existence of grad F (in some neighborhood) IS the definition of differentiable. === Subject: Re: . The hardest of all hard facts . >Surely the anisotropy decreased over the 100 years from your first citation >to your last. The first citation considers the anisotropy of the speed of light in air and other gases, the last the anisotropy of the speed of light in vacuum. These are two different things. In theory the Lorentz contraction will ensure that the observed speed of light in vacuum is always equal to c, irrespectiver of whether or not the propagation of light occurs through an ether. This has been confirmed by experiment to a high degree of precision. In contrast the maths which Cahill worked out for gas filled Michelson-Morley type interferometers predicts that if the interferometer is in motion relative to quantum foam, the observed speed of light will vary depending on the direction of the light beam relative to the motion of the interferometer. This maths is explained in the first paper cited: R.T. Cahill http://arxiv.org/pdf/physics/0312082 This considered the Michelson-Morley experiment which in effect compared the speed of light in two orthogonal directions by using an air filled interferometer. A small anisotropy was observed. If my memory is correct, using Newtonion physics and ignoring the effect of the air, Michelson calculated that the anisotropy was equivalent to a speed through the ether of about 8 km/second. However, as he knew that the orbital velocity of the earth was 30 km / second relative to the sun, a speed of only 8 km/second relative to the ether did not make sense. Therefore he reported that the experiment had failed to detect motion relative to the ether. However when Cahill reanalysed the MichelsonÕs data so as to take into account the effects of the Lorentz contraction and the effect of the air (which required some new maths), he found the small anisotropy was equivalent to a speed of hundreds of km/second relative to space/quantum foam (whatever one likes to call it). He was able to obtain similar results when he reanalysed data from other M-M type experiments with gas filled interferometers. The last citation was: Modern Michelson-Morley experiments and gravitationally-induced anisotropy of c M. Consoli Istituto Nazionale di Fisica Nucleare, Sezione di Catania http://arxiv.org/pdf/gr-qc/0306105 Since this paper reports on experiments using vacuum interferomenters, no-one (not even Cahill) would have expected any anisotropy to have been observed. So I thought it was interesting that a very small anisotropy has now even been seen for a vacuum M-M experiment. Peter === Subject: Re: wedge product > But in more advanced differential geometry it is usually used for > exterior products/powers, especially of the cotangent spaces of > a manifold (leading to differential). As I donÕt have do CarmoÕs > book to hand I cannot tell what his convention is. A text at this > level could have either (or both) usages. He defines the wedge product like this: ``The vector product of u and v (in that order) is the unique vector u wedge v in R^3 characterised by (u wedge v)dot w = det (u,v,w) for all w in R^3.ÕÕ Then thereÕs a remark that says ``It is also very frequent to write u wedge v as u times v and refer to it as the cross product.ÕÕ Tobin Fricke === Subject: Simple counterexample needed (ODEs) IÕm looking for a function f(t, x, y) from R x R^2 to R^2 such that df/dt (at some time t0) is well defined for all (x, y), but not differentiable w.r.t. to (x, y). (Basically, IÕm after a case that fails to meet the conditions of the existence and uniqueness theorem for differential equations.) jill === Subject: Re: Simple counterexample needed (ODEs) > IÕm looking for a function f(t, x, y) from R x R^2 to R^2 such that > df/dt (at some time t0) is well defined for all (x, y), but not > differentiable w.r.t. to (x, y). The grammar is questionable here. Do you want f not differentiable w.r.t. to (x, y), or df/dt not differentiable w.r.t. to (x, y)? In either case you could consider f(t,x,y) = e^t*(g(x), g(y)), where g : R -> R is as bad as you want it to be. === Subject: Re: Simple counterexample needed (ODEs) > IÕm looking for a function f(t, x, y) from R x R^2 to R^2 such that > df/dt (at some time t0) is well defined for all (x, y), but not > differentiable w.r.t. to (x, y). > (Basically, IÕm after a case that fails to meet the conditions of > the existence and uniqueness theorem for differential equations.) > jill a classical example of an ODE with branching is yÕ = 3*y^(2/3) and the other variables can be dummy variables for an example you are looking for: f(t,x,y) = 3*x^(2/3) Here the partial df/dt is constant 0, the best of all differentiable functions. The branching: take a<0b This is a legitimate solution, besides the obvious constant zero solution, and the less obvious strictly increasing solution y=t^3. Hope it helps, ZVK(Slavek). === Subject: Re: How to prove this probability Inequality by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1PIt0802061; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1PIpTi01822 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1PIpT523699; >X,Y,Z are random variables >Try to prove >E[EX-E(X|Z)]^2>=E[EX-E(X|Y,Z)]^2 When is your homework due? === Subject: re: Metric Engineering, Dark Energy, Quintessence, Phantom Energy TYPO w = + 1/3 for EM radiation like the CDM should be w = + 1/3 for EM radiation like the CBR (Cosmic Background Radiation) === Subject: Re: Documents about multiplicative order in general and mersenne numbers > http://www.cerias.purdue.edu/homes/ssw/cun/ is the homepage for > the Cunningham project with is concerned with factoring numbers > of the form b^n + 1 and b^n - 1. > Material on multiplicative order can be found in pretty much > any book with a title like Introduction to Number Theory, > Elementary Number Theory, etc. Hi Gerry, thank you for this reply. IÕll look at the homepage of this project. Is it really true that everything which is known on the multiplicative order is written in introductory books to number theory? I already browsed some. Is this really all which is known about multiplicative order: Ord a (mod n) | Phi(n) Ord a (mod pq) = lcm(Ord a (mod p),Ord a (mod q)) And that for each p which is prime there exist Phi(p-1) primitive roots a where Ord a (mod p) = p-1 (if you take p and q as two distinct primes and n as an arbitrary integer greater than 1) Is this really all? Maybe I forgot some point, but this is what I found in two good introductory books about number theory. I can hardly believe that this is everything which was ever found out about multiplicative order. Could somebody shed a light, please? Juergen Bullinger === Subject: Re: Documents about multiplicative order in general and mersenne numbers > Is it really true that everything which is known on the multiplicative > order is written in introductory books to number theory? IÕm sorry. I didnÕt realize that you wanted to know everything that is known about multiplicative order. Since you didnÕt indicate that you knew anything about multiplicative order, I thought you wanted to know where to begin to learn about multiplicative order. Maybe it would work better if you let us know what it is that you actually want to know about multiplicative order. Then perhaps someone could tell you whether or not what you want to know is known, and, if it is, where to find it. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: re:Continuum hypotheses and Lebesgue measure Actually IÕm familiar with the Cantor-Benedixon theorem in the exact form I stated it. IÕm quite sure thatÕs the way Cantor saw the result. The generalization to Polish spaces was probably done later (perhaps by Benedixon). ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: re:Continuum hypotheses and Lebesgue measure By the Cantor-Benedixon theorem every uncountable closed set has a perfect subset (which thus must be of cardinality C). Hence every set of positive measure is of cardinality C. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: Continuum hypotheses and Lebesgue measure > By the Cantor-Benedixon theorem every uncountable closed set has a > perfect subset (which thus must be of cardinality C). Hence every set > of positive measure is of cardinality C. What the Cantor-Bendixon theorem says is that every Polish space (that is, a complete separable metric space) can be written as the union of a perfect set and a countable open set. But, of course, every compact subset of the reals is a Polish space and it follows from the theorem that it contains a perfect and compact subset with the same measure. Jose Carlos Santos === Subject: Re: Continuum hypotheses and Lebesgue measure >> Hi all, >> >> Consider this statement: >> >> If a Lebesgue-measurable subset M of the set R of real numbers has >> cardinal less than the cardinal of R then the Lebesgue measure of >> M is zero. >> >> This is trivial if we assume the continuum hypothsis. And what if we >> donÕt? ItÕs easy to see that if there were subsets of R whose cardinal >> was in between the cardinals of N (the naturals) and R, then, among >> those sets, there would be Lebesgue-measurable sets with Lebesgue >> measure zero. But would there be necessarily, among those sets, >> Lebesgue-measurable sets with Lebesgue measure greater than zero? >> >> (*) Every subset of R with cardinal < c is Lebesgue measurable >> [and therefore, as Ullrich noted, of measure zero] >> (*) is independent of ZFC. (*) is a consequence of MartinÕs Axiom. >> But there are also models of ZFC + notCH where (*) fails: >> were there are subsets of cardinal strictly between aleph_0 and c that >> have positive Lebesgue outer measure but (of course) zero inner >> measure. As I recall (It has been a long time...) the random reals >> type models do this. > Random reals? IÕve never heard of these. Could you provide some > information or sources of information? See Random and Cohen reals, by Kenneth Kunen; itÕs part of the Handbook of set-theoretic topology, edited by Kenneth Kunen and Jerry E. Vaughan. Jose Carlos Santos === Subject: Re: Continuum hypotheses and Lebesgue measure > Random reals? IÕve never heard of these. Could you provide some > information or sources of information? Do you know Boolean-valued models? For random reals you take Boolean- valued models, where the Boolean algebra is the measura algebra for [0,1]^A with product Lebesgue measure, for A a certain cardinal. I think A=aleph_2 does what was originally desired in this thread. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: re:How big can a manifold be? I read some of David GualdÕs paper and settled my conjectures about 1-manifolds. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: How big can a manifold be? > >> I reduced this claim to the following set-theoretic statement, but >>cannot prove it from the first glance: >>Consider a linearly ordered by inclusion family of sets, all of them >>having the cardinality of the continuum. Then their union has the >>cardinality of the continuum as well. > > > Because you canÕt prove it at first glance. You have to > take a second glance and assume the Axiom of Choice is true. > Then the proof is more trivial than the equivalent intersection > of sets of the power of the continuum. Since in order to > prove that it is true you have to *assume* that there is another set, A, > which also has the power of the continuum, > and for every set T in the ordered sets: > > T is a subset of A. > I donÕt follow: > The original question may be interpreted as follows: We know thatÕs it possible to interpret the question that way. Since the only thing Logicians are really useful for is interpreting the word cardinality any way they feel pre-disposed to interpret it as. And if you havenÕt noticed, most of them interpret cardinality to mean ordinality. Which is where the word function comes from, not from mathematics. Since the only thing mathemticians do full-time is interpret the word function to mean functional. Since Newton invented the word functional and itÕs simple. ItÕs so simple that even British and Latin Philosophers can understand it. And my question isnÕt at all ambiguous. Given that Lambda Calculi areÕt derived from (<,+,*,=,0) over R. They derived from ([,U,^,P,@,.EQ.) over Z. > Given an ordinal lambda, and a functional set f with domain > lambda, alpha < beta < lambda => f(alpha) subset f(beta) > and f(alpha) is of the cardinality of the continuum for each > alpha, the union of all f(alpha) is of the cardinality of the > continuum. > wlog, we can assume that if alpha < beta, then f(alpha) is not > equal to f(beta). But that still allows lambda to be as big as > 2^c. But that observation is irrelevent to Set Theory though, since itÕs only true because of philosophy, not science. Since itÕs a priori true that R is sufficiently large to allow anything that is not R to be as large as R. === Subject: Re: OT observation > Yeah, there seems to be some kind of internal order relationship implied in > a lot of the use of the negation of contraries (I think my use of the term > double-negative was likely improper, and accept the analyses of several > others on this point). > So when we say John is happy, it establishes a point happiness as being a > part of the state of John. When we say that John is unhappy, it establishes > a different point unhappiness as not being a part of the state of John. I asume you meant When we say that John is not unhappy because that makes a lot more sense for the argument with which you follow, but it also works this way. > Something like > happiness ------------------ unhappiness > and we seem to internally identify gradations between the states. The > difference between the statements then seems to be that the negation of > contraries does not assert the positive condition, it merely asserts that > the contrary position is not the case and that anywhere elsewhere on the > continuum is possible. Ok it sounds very reasonable BUT it has the major disadvantage of being worthless, because it works in this particular case but cannot be generalized. I can identify at least to levels at which this fails. The first is that for certain adjectives there can be no gradation. Take for example dead. The second is, that it also fails for any adjective depending on the context of the assertion. say: 1.- I am not unaware of the difficulties related with my argumentation. 2.- I am NOT unaware of my duties. For 1.- your hypothesis holds clearly. For 2.- it clearly doesnÕt, because not unaware in this case clearly means aware. > Not being certain of the affirmative position for > whatever reason is one exemplar of this usage. Natural languages are simply too slippery to cage with this kind of analysis. You keep trying to use this multivalue approach and oversee that though the problems that you attack display such structures, the approach itself doesnÕt benefit from them. Hilberts programm We can know and we will know was shattered by G.9adel. Much of science has since has been realising that it will never know everything. But there are some as yourself, who think its just a matter of sharpening the tools of science. I personally believe that we will always be limited in our ability to know by the limitations of our instruments. For mathematics at least, G.9adel showed that our instruments will always be blunt in the end. > This kind of negation can be modelled. One particularly nice thing I like > in Heyting models is that one already gets relationships such as > ~(~(a)) -> a > as an order relationship. === Subject: Re: OT observation > Yeah, there seems to be some kind of internal order relationship implied in > a lot of the use of the negation of contraries (I think my use of the term > double-negative was likely improper, and accept the analyses of several > others on this point). > So when we say John is happy, it establishes a point happiness as being a > part of the state of John. When we say that John is unhappy, it establishes > a different point unhappiness as not being a part of the state of John. > I asume you meant When we say that John is not unhappy because that > makes a lot more sense for the argument with which you follow, but it > also works this way. > Something like > happiness ------------------ unhappiness > and we seem to internally identify gradations between the states. The > difference between the statements then seems to be that the negation of > contraries does not assert the positive condition, it merely asserts that > the contrary position is not the case and that anywhere elsewhere on the > continuum is possible. > Ok it sounds very reasonable BUT it has the major disadvantage of > being worthless, because it works in this particular case but cannot > be generalized. > I can identify at least to levels at which this fails. The first is > that for certain adjectives there can be no gradation. Take for > example dead. The second is, that it also fails for any adjective > depending on the context of the assertion. > say: > 1.- I am not unaware of the difficulties related with my > argumentation. > 2.- I am NOT unaware of my duties. > For 1.- your hypothesis holds clearly. For 2.- it clearly doesnÕt, > because not unaware in this case clearly means aware. > Not being certain of the affirmative position for > whatever reason is one exemplar of this usage. > Natural languages are simply too slippery to cage with this kind of > analysis. You keep trying to use this multivalue approach and oversee > that though the problems that you attack display such structures, the > approach itself doesnÕt benefit from them. Hilberts programm We can > know and we will know was shattered by G.9adel. Much of science has > since has been realising that it will never know everything. But there > are some as yourself, who think its just a matter of sharpening the > tools of science. I personally believe that we will always be limited > in our ability to know by the limitations of our instruments. For > mathematics at least, G.9adel showed that our instruments will always be > blunt in the end. > This kind of negation can be modelled. One particularly nice thing I like > in Heyting models is that one already gets relationships such as > ~(~(a)) -> a > as an order relationship. Sorry for my carelessness. Yes not unhappy should be in place of unhappy. === Subject: Re: OT observation > > Sometime it seems that the exact mening of a statement is > not clear and that it is easier to rephrase the general meaning > in oneÕs own terms with a negative attached. > > ItÕs nice to see that John is happy? > Well, he certaintly isnÕt unhappy. > > (I donÕt know what you mean by happiness but > I agree that he seemed to be in better shape than > the last time we saw him.) > The opposite of a statement is the conTrapositive, not the negative. > Even Aristotle knew that. The contrapositive is logically equivalent to the original statement: so in what sense is it the opposite? > There are no unique antonyms in 1:1 > correspondence with adjectives. What is the antonym of tepid? Right. So therefore we must use the contrapositive instead: This cup of tea is tepid. The non-tepid does not include this cup of tea. Ah ... what was your point again? > Is your post related to mine? I said it was OT. DidnÕt it make > sense to you? OT^2: When the great coyote covered the ßowing spring waters and used to beget the net, wouldnÕt it have been nice if the nettles in their first naming frenzy could have chosen a name for off topic which was not also an abbreviation of on topic? === Subject: Re: OT observation Is your post related to mine? I said it was OT. DidnÕt it make >> sense to you? >OT^2: Are you sure this should be squared and not negated? > ..When the great coyote covered the ßowing spring waters and used >to beget the net, wouldnÕt it have been nice if the nettles in their >first naming frenzy could have chosen a name for off topic which was >not also an abbreviation of on topic? No, no, no. OT is for overtime. Re is on topic. /BAH Subtract a hundred and four for e-mail. === Subject: Re: OT observation > ~(~(a)) -> a This says it very clearly and gives me a chance to explain what I meant. The above is not true in all cases in the real world. And that is because the meaning of a can change and that change is what makes a kind of double negative useful. IÕm belaboring this poing because a guy who lead a Transactional Analysis group picked on a guy who did this and I thought the guy was right. This makes me think that since the leader of the group is a smart guy, that maybe this is a fine point. Anyway, maybe people have a different idea of what happy means. It might mean sexually satisfied. It might mean contented. It might mean a higher happiness as explained by Aristotle... So if A says that B is happy to C, then maybe C would say, well he isnÕt unhappy, which can be true of AÕs statement, which might be assumed because of oneÕs trust in AÕs statement. C can let A knows that he (C) doesnÕt know what A means by ÔhappyÕ, and can be agreeable in general without pretending to know what A means exactly. Sorry this is so convoluted and itÕs really not worth thinking about too much because if itÕs important you probably would have noticed it already, if you havenÕt. === Subject: Re: Public Notice: Copyright Violations > Let me make this simple for you. Complaining about copying usenet postings > is like complaining about copying toilet paper patterns. Actually, complaining about copying toilet paper patterns is a way to start some good lawsuits. For example, take Roger PenroseÕs lawsuit against the Kimberly-Clark corporation, for using a Penrose tiling on their Kleenex Quilted toilet paper. Pentaplex, a company that licenses PenroseÕs patterns, is also a plaintiff. A spokesman for Pentaplex says, ...when it comes to the population of Great Britain being invited by a multi-national to wipe their bottoms on what appears to be the work of a Knight of the Realm without his permission, then a last stand must be made. Googling for Penrose and toilet paper, brings up among many other interesting links. Is this a frivolous lawsuit? Well, letÕs just say that the mighty toilet paper empire was threatened enough to change the design of their Kleenex Quilted toilet paper. IÕve stepped into this discussion unawares, but I think my example is probably not what Richard Henry was thinking. I believe the purpose of his comment was to indicate that copying toilet paper patterns is perfectly fine. Thus, I couldnÕt resist pointint out this Penrose toilet paper incident. While I realize posting about all sorts of craziness is the norm for sci.math, I couldnÕt make this same assumption about all those other groups, so IÕve eliminated them from the header, even though my post can be said to be more math-related than the previous ones in the discussion ;-) === Subject: Re: Public Notice: Copyright Violations >Actually, complaining about copying toilet paper patterns is a way to >start some good lawsuits. For example, take Roger PenroseÕs lawsuit >against the Kimberly-Clark corporation, for using a Penrose tiling on >their Kleenex Quilted toilet paper. Pentaplex, a company that licenses >PenroseÕs patterns, is also a plaintiff. A spokesman for Pentaplex >says, ...when it comes to the population of Great Britain being >invited by a multi-national to wipe their bottoms on what appears to be >the work of a Knight of the Realm without his permission, then a last >stand must be made. Think how much more inappropriate it would have been for a Penrose tiling to have been used for the outer layer of a sanitary napkin, though. Lee Rudolph === Subject: Re: Public Notice: Copyright Violations <9OW_b.1170$h23.739@fed1read06> In message , Richard Henry >> I believe Google has been granted the legal right to archive Usenet >> Granted by whom? There are millions of copyright-owning authors on >usenet. >> Suppose you post through Google. Then your posts are subject to the >> following: >> By posting communications on or through the Service, you automatically >> grant Google a royalty-free, perpetual, irrevocable, non-exclusive license >> to use, reproduce, modify, publish, edit, translate, distribute, perform, >> and display the communication alone or as part of other works in any form, >> media, or technology whether now known or hereafter developed, and to >> sublicense such rights through multiple tiers of sublicensees. >> It seems likely that similar stipulations apply to all Usenet posts. If >> true, the millions of copyright-owning authors on Usenet have granted >> licenses to specific parties to use their work - basically as those >parties >> see fit. >It seems likely? I can offer only one data point: I for one have never signed any contract containing terms remotely like the above. -- Richard Herring === Subject: Re: Public Notice: Copyright Violations <9OW_b.1170$h23.739@fed1read06> <45D+nCZTNIPAFwmU@baesystems.comIn message , Richard Henry message > I believe Google has been granted the legal right to archive Usenet >> Granted by whom? There are millions of copyright-owning authors on >>usenet. > Suppose you post through Google. Then your posts are subject to the > following: > By posting communications on or through the Service, you automatically > grant Google a royalty-free, perpetual, irrevocable, non-exclusive license > to use, reproduce, modify, publish, edit, translate, distribute, perform, > and display the communication alone or as part of other works in any form, > media, or technology whether now known or hereafter developed, and to > sublicense such rights through multiple tiers of sublicensees. > It seems likely that similar stipulations apply to all Usenet posts. If > true, the millions of copyright-owning authors on Usenet have granted > licenses to specific parties to use their work - basically as those >>parties > see fit. >>It seems likely? >I can offer only one data point: I for one have never signed any >contract containing terms remotely like the above. Now read your Misoft licensing agreements. One no longer has to sign to cede intellectual property. Another magic incantation to make things go poof is EULA (I think IÕve spelled it correctly). /BAH Subtract a hundred and four for e-mail. === Subject: Re: Public Notice: Copyright Violations <9OW_b.1170$h23.739@fed1read06> <45D+nCZTNIPAFwmU@baesystems.com> <403c9483$0$3105$61fed72c@news.rcn.com> In message <403c9483$0$3105$61fed72c@news.rcn.com>, jmfbahciv@aol.com >>In message , Richard Henry >message >> I believe Google has been granted the legal right to archive Usenet >> Granted by whom? There are millions of copyright-owning authors on >usenet. >> Suppose you post through Google. Then your posts are subject to the >> following: >> By posting communications on or through the Service, you automatically >> grant Google a royalty-free, perpetual, irrevocable, non-exclusive >license >> to use, reproduce, modify, publish, edit, translate, distribute, >perform, >> and display the communication alone or as part of other works in any >form, >> media, or technology whether now known or hereafter developed, and to >> sublicense such rights through multiple tiers of sublicensees. >> It seems likely that similar stipulations apply to all Usenet posts. If >> true, the millions of copyright-owning authors on Usenet have granted >> licenses to specific parties to use their work - basically as those >parties >> see fit. >It seems likely? >>I can offer only one data point: I for one have never signed any >>contract containing terms remotely like the above. >Now read your Misoft licensing agreements. I donÕt think the validity of shrink-wrap licenses has been tested in this jurisdiction. > One no longer has >to sign to cede intellectual property. Another magic >incantation to make things go poof is EULA (I think IÕve >spelled it correctly). Indeed. -- Richard Herring === Subject: Re: Public Notice: Copyright Violations Discussion, linux) >> The fact that you have no idea what constitutes fair use says a lot >> here. >> And this fact is based on what exactly? >> You might start with the US Code, since it seems to apply in this case. >> http://www.copyright.gov/title17/92chap1.html#107 > Once again you appear to have reading difficulty. The alleged fact is > that JSH has no idea of what constitutes fair use. I doubt that the US > Code has commented on this allegation. It was in a footnote. The US Code gives an explicit (but partial) list of persons with no clue of fair use rights. James S. Harris appears in the footnote, three lines above Jesse F. Hughes. -- [R]eality has a fascinating ability to check us when we get a little too big for our britches... Make no mistake. There isnÕt a mathematician alive today that I canÕt now touch, and not a mathematical career on the planet that I canÕt now affect. --James Harris, render of worlds === Subject: Re: Public Notice: Copyright Violations Discussion, linux) > I believe Google has been granted the legal right to archive Usenet Really? Google has been granted this right by whom? When? If anything, IÕd guess that Google is relying on a presumed right to archive, but I donÕt know that there is any such right explicit in the law. I donÕt think Google has any more or less right to archive Usenet postings or web sites than anyone else. Some other archivist could do just what Google does with Usenet now (and as DejaNews did before Google). As far as I know, the Wayback Machine is on similarly shaky footing. (As is GoogleÕs caching and serving of web pages.) I donÕt think the legal status of these practices has been settled all that much. But, again, I donÕt know diddly here. If you have some good references, then IÕd be happy to see them. Popular journalism is preferable to technical legal documents for my addled brane. -- You lack the ability to reason, but instead get an idea in your head and hold on to it against all evidence. I donÕt find you credible, and reject your claims, as coming from a ßawed source. === Subject: Re: Public Notice: Copyright Violations > I believe Google has been granted the legal right to archive Usenet > Really? Google has been granted this right by whom? When? > If anything, IÕd guess that Google is relying on a presumed right to > archive, but I donÕt know that there is any such right explicit in the > law. I donÕt think Google has any more or less right to archive > Usenet postings or web sites than anyone else. Some other archivist > could do just what Google does with Usenet now (and as DejaNews did > before Google). > As far as I know, the Wayback Machine is on similarly shaky footing. > (As is GoogleÕs caching and serving of web pages.) > I donÕt think the legal status of these practices has been settled all > that much. But, again, I donÕt know diddly here. If you have some > good references, then IÕd be happy to see them. Popular journalism is > preferable to technical legal documents for my addled brane. Well I admit itÕs just a theory. ItÕs based on the idea that when you post to Usenet through Google, your posts are subject to the following: By posting communications on or through the Service, you automatically grant Google a royalty-free, perpetual, irrevocable, non-exclusive license to use, reproduce, modify, publish, edit, translate, distribute, perform, and display the communication alone or as part of other works in any form, media, or technology whether now known or hereafter developed, and to sublicense such rights through multiple tiers of sublicensees. IÕm thinking similar stipulations apply no matter which vehicle you choose to post to Usenet. Then IÕm thinking these vehicles have granted Google the right to archive. If you go to youÕll see ----- Google Groups - Terms and Conditions ... The contents within the Service are protected by copyright and other laws in both the United States and elsewhere. The Service includes both content owned or controlled by the Google as well as content owned or controlled by third parties and licensed to Google (collectively, the Materials). Google authorizes you to view and download a single copy of the Materials solely for your personal, non-commercial use. You may not sell or modify the Materials or reproduce, display, publicly perform, distribute, or otherwise use the Materials in any way for any public or commercial purpose without the written permission of Google. Special rules may apply to the use of certain software and other items provided via the Services, and are noted where appropriate. ---------- Google certainly makes it sound like they have a legal right to the archived posts. Perhaps itÕs all bluff. So my thought was Google has been granted licenses by lots of third parties (ISPs and other Usenet providers), who were granted licenses by the original posters in the manner given above. But I admit itÕs just a theory. === Subject: Re: Public Notice: Copyright Violations <871xojadfg.fsf@phiwumbda.org> Discussion, linux) > Well I admit itÕs just a theory. ItÕs based on the idea that when you post > to Usenet through Google, your posts are subject to the following: > By posting communications on or through the Service, you automatically > grant Google a royalty-free, perpetual, irrevocable, non-exclusive license > to use, reproduce, modify, publish, edit, translate, distribute, perform, > and display the communication alone or as part of other works in any form, > media, or technology whether now known or hereafter developed, and to > sublicense such rights through multiple tiers of sublicensees. I think that, without doubt, Google users have given Google the right you describe. > IÕm thinking similar stipulations apply no matter which vehicle you choose > to post to Usenet. Then IÕm thinking these vehicles have granted Google > the right to archive. If you go to > youÕll see > ----- > Google Groups - Terms and Conditions > ... > The contents within the Service are protected by copyright and other laws > in both the United States and elsewhere. The Service includes both content > owned or controlled by the Google as well as content owned or controlled by > third parties and licensed to Google (collectively, the Materials). > Google authorizes you to view and download a single copy of the Materials > solely for your personal, non-commercial use. You may not sell or modify > the Materials or reproduce, display, publicly perform, distribute, or > otherwise use the Materials in any way for any public or commercial purpose > without the written permission of Google. Special rules may apply to the > use of certain software and other items provided via the Services, and are > noted where appropriate. > ---------- > Google certainly makes it sound like they have a legal right to the > archived posts. Perhaps itÕs all bluff. So my thought was Google has been > granted licenses by lots of third parties (ISPs and other Usenet > providers), who were granted licenses by the original posters in the manner > given above. But I admit itÕs just a theory. I doubt that the various third parties have the ability to grant Google the right to archive and serve my posts. If Google has such a right, then it would surely have to come implicitly from the poster, who grants the right by posting. But if that argument holds water, then itÕs unclear why Google would implicitly have the right to archive and present Usenet posts and, say, Dik Decker or Rick Winter or whoever wouldnÕt have the same right (actually, Dik and Rick donÕt quote Usenet posts, I think). Well, as Barbie *should* have said, Copyright law is hard. I havenÕt any clue what rights one implicitly grants when posting to Usenet and what he doesnÕt. The situation would certainly be clearer if the X-no-archive header were replaced by an X-archive header (but then GoogleÕs archives would fit on a ßoppy). Dot sig serendipity strikes again. The quote below was selected randomly. -- Sale or rental of this disc is ILLEGAL. If you have rented or purchased this disc, please call the MPAA at 1-800-NO-COPYS. -- The MPAA begins a new anti-piracy program, found on a DVD purchased in China === Subject: Re: Public Notice: Copyright Violations Discussion, linux) >> IÕve noticed people with webpages extensively quoting from posts IÕve >> made and am now giving public notice that there is no right to copy >> extensively, that is, to go beyond fair use, from my writings without >> my express written permission. >> The fact that you have no idea what constitutes fair use says a lot >> here. > And this fact is based on what exactly? Well, who knows what page James is complaining about? I didnÕt see much quoting (none?) on RickÕs page, so it must be DikÕs page at . But, DikÕs page doesnÕt contain JamesÕs proof. It contains an altered proof of a different statement (a false statement). True, it does contain rather a lot of JamesÕs words, but it looks to me like a legitimate criticism of JamesÕs writing. Obviously, my opinion on these matters is worth squat, but just as obviously, IÕll give it anyway. DikÕs use of JamesÕs writing looks like fair use to me. It doesnÕt look excessive. In fact, it looks like a page or maybe a bit more, edited and with commentary. Maybe thereÕs some other page JSH has in mind, but he doesnÕt usually play his cards close to his chest. -- Jesse F. Hughes Ultimately, I can bring the entire mathematical establishment to its knees... Live in a fantasy world if you wish, but to me thatÕs just an expression of your intellectual inferiority. --James Harris === Subject: Re: Public Notice: Copyright Violations <67w_b.121$h23.35@fed1read06> Discussion, linux) >> How can I make this simple for you? He mentioned websites. Websites. >> Websites. Can you read? Websites are not Usenet. So whatever may be > common >> simple point escapes you escapes me. > Let me make this simple for you. Complaining about copying usenet postings > is like complaining about copying toilet paper patterns. Let me make this simple for you. If you donÕt regularly change your fuel filter, then your engine performance will decline. Well, it is as relevant to the discussion as your simple declaration. As Wade said, no one was complaining about copying Usenet postings. I am not sure what the fuel filter-changing habits are for the readers of sci.math, but it is likely that a few readers havenÕt given the matter any thought in some time. Therefore, while my advice was no more relevant than yours, it might have practical benefits for the readers. -- Jesse Hughes By definition m is a variable. By definition all then (sic) numbers represented by letters are variables--thatÕs algebra[,] Magidin. -- James Harris shows deep understanding of algebra === Subject: Re: Public Notice: Copyright Violations message >> How can I make this simple for you? He mentioned websites. Websites. >> Websites. Can you read? Websites are not Usenet. So whatever may be > common this >> simple point escapes you escapes me. > Let me make this simple for you. Complaining about copying usenet postings > is like complaining about copying toilet paper patterns. > Let me make this simple for you. If you donÕt regularly change your > fuel filter, then your engine performance will decline. > Well, it is as relevant to the discussion as your simple > declaration. As Wade said, no one was complaining about copying > Usenet postings. Sorry, but youÕre wrong. He was complaining about copying Usenet postings to a website. KeithK > I am not sure what the fuel filter-changing habits are for the readers > of sci.math, but it is likely that a few readers havenÕt given the > matter any thought in some time. Therefore, while my advice was no > more relevant than yours, it might have practical benefits for the > readers. > -- > Jesse Hughes > By definition m is a variable. By definition all then (sic) numbers > represented by letters are variables--thatÕs algebra[,] Magidin. > -- James Harris shows deep understanding of algebra === Subject: Re: Public Notice: Copyright Violations <67w_b.121$h23.35@fed1read06> <878yiradzk.fsf@phiwumbda.org> Discussion, linux) >> Well, it is as relevant to the discussion as your simple >> declaration. As Wade said, no one was complaining about copying >> Usenet postings. > Sorry, but youÕre wrong. He was complaining about copying Usenet postings > to a website. Yes, youÕre right. He was. Though it is still not at all obvious whether posting to Usenet gives an implicit right for others to present your posts on webpages -- especially webpages that are not part of Usenet propagation or archiving. So, I take it back. RichardÕs comment, Complaining about copying usenet postings is like complaining about copying toilet paper patterns, is not irrelevant. It is relevant, but depending on what the simile means, it is likely false. -- Jesse F. Hughes C is for Cookie. ThatÕs good enough for me. Cookie Monsters === Subject: solution Ok, I have this solution... AS someone say, it is weel know that the number of end-zeros of n! is f(n) = [n/5] + [n/5^2] + [n/5^3]... If n in base 5 is n = a0 + a1*5 + a2*5^2 + ... + ak*5^k = (a0 a1 a2 ... ak)_5 Then using the fact that [n/5^s] = as + a(s+1)*5 + ... =(as...ak)_5 and adding f(n) = (n - sum_i ai) / 5-1 < = n-1/4 like this, f(5) = 1, f(24) = (24-8)/4 = 4 , f(25) = (25 - 1)/4 = 6 ... usint the baund f(n) < = (n-1)/ 4 and solving n-1 / 4 = 153 we faund that 151 = f(617) < = 153 and we writte 617 in base 5 we have 617 = (4432)_5 which let us know that (4444)_5 = 624 have f(624) = (624-16)/4 = 152 end zeros and that 625 = (10000)_5 has f(625) = 624/4 = 156 zeros ... so there is no number integer n such that n! ends with 153 or 154 or 155 zeros... Bonito no? This formula and metoth let us know what are the numbers wich never would be the end-zero number of n! (like 5, 28, 29, 30,153, 154, 155...) Rogelio buedia@matem.unam.mx >... >> Anyhow counting 0Õs lets begin with n=5: >> 1*2*3*4*5 sequence will result with 1 0Õs : 5!= 120 >> 6*7*8*9*10 sequence will result with 1 0Õs >> 11*12*13*14*15 sequence will result with 1 0Õs too >> and so on but sequences >> 21*22*23*24*25 >> 46*47*48*49*50 >> 71*72*73*74*75 will result with 2 0Õs >> acc. to 4*25=100 6*50=300 and 75*8=600 >> and so on it will be till >> 91*92*93*94*95 sequence =95/5 + 3 = 21 0Õs totally >> 96*97*98*99*100 seguence will result with 2 0Õs >> 101*102*103*104*105 sequence will result with 1 0Õs >> and so on till >> 196*197*198*199*200 will result with 2 0Õs and totally 200/5 + 8 >> also 48 0Õs. >> now what with 153 0Õs ? it is like 625/5 + 25 = 150 0Õs >> then 630! turns to 151 0Õs >> 635! to 152 0Õs >> could 640! contain 153 0Õs or I am mistaken ? >> simply it looks for some extra 0Õs in some inside sequences: >> Not so easy to be correct in short time ! >No. 624!=152 zeros at the end, 625!=156 zeros, thatÕs because >625!=624!*625 = 624!*5^4, so you have 4 more 5Õs in the factors of >625! than in those of 624!. Combined with the 2Õs in the factors of >624! (you need equal amount of each because 10=2*5), thatÕs why you go >from 152 zeros to 156. === Subject: Re: 3 Squares Covering 1 Circle > I was asked (via e-mail) to explain this here in more detail. > If a sense, this is what I did when I found that JakeÕs > configuration yielded a saddle point. Three of the eigenlines > were hot -- 2 slightly, and 1 much more so. So I looked at > that hot line, and it suggested the twisted configuration that > produced the bona fide local maximum. > Hi Jim, > But there is still a large gap between the general explanation, > which you gave in a vivid manner, and the (I think) awkward > looking matrix itself, full of lengthy terms. > It would be a great pleasure for me, if you showed this beast, > which eventually gave you the best solution so far. Maybe I or > some of the interested lurkers feels so strong now as to try > by myself/herself/himself to play with the matrix and get an > improvement. Or else get a better feeling for why this should > be the best solution indeed. I canÕt really show you this beast. ItÕs too beastly. Even the formula for the area itself is complicated: itÕs a 15-gon whose vertices have messy formulas depending on 9 parameters + 3 circular segements. To then write down the 45 second derivatives of this formula . . . Ugh. What I can write down is the formula for the two-parameter family which yields the optimal configuration. The input parameters are r and phi (described in a previous post). The formula for the area of the resulting figure is then given as follows (in Mathematica notation): rc = Sqrt[2] r Cos[phi]; rs = Sqrt[2] r Sin[phi]; sm = 1 - (rc - rs)/2; sp = 1 - (rc + rs)/2; cm = Sqrt[1 - sm^2]; cp = Sqrt[1 - sp^2]; qm = Sqrt[1 + 2 sm cm]; qp = Sqrt[1 + 2 sp cp]; am = ArcSin[(sm - cm)/Sqrt[2]]; ap = ArcSin[(sp - cp)/Sqrt[2]]; xtra = 12 - 9 rc - 14 sm sp - r^2 (1 + 2 Sqrt[3]); area = (3/4) ( xtra + 2 (am + ap) + (1 - rc) (cm + cp) + (1 + rs) qm + (1 - rs) qp ); > I like these geometric puzzles very much and I want to say > to be a strong tool to achieve insight as well as results. You call the Hessian a strong tool. IÕd say the strong tool is Taylor series: f(x) = f(x0) + (grad f)(x0).x + (grad grad f)(x0):xx / 2 + ... Because (grad f)(x0) = 0 at the critical point (by definition of a critical point), the local behavior is just f(x) = f(x0) + (grad grad f)(x0):xx / 2 + higher-order terms, i.e., a constant plus a quadratic form. It would be interesting to see a multidimensional problem in which grad f, grad grad f, and grad grad grad f were all 0, so that one had to resort to analyzing a *quartic* form. Not as simple as good old eigenvalues and eigenvectors. Hmmm. > configuration so far, > Rainer Rosenthal > r.rosenthal@web.de === Subject: Re: ÔerfÕ function in C > It is possible to compress the code further at the expense of > clarity (warning: this is not for the faint of heart): > double Phi(double x) > {long double s=x,t=0,b=x,x2=x*x,i=1; > while(s!=t) s=(t=s)+(b*=x2/(i+=2)); > return .5+s*exp(-.5*x2-.91893853320467274178L); > } > This avoids the unnecessary computation of i+1 followed by the > unnecessary conversion of the result from int to long double, > eliminates the unnecessary variable pwr, and precomputes x*x. > Have you checked this algorithm in the tail? For, say, x = -50, I either > get inf/nan or an incorrect answer (0.5), depending on which compiler I > use. The compiler in my tests is gcc, specifically DJGPP or MingW. The > problem is overßow in s. For large arguments one might want to switch to asymptotics. I think, that is what Genz makes in his approx for cdf normal. === Subject: Re: ÔerfÕ function in C Yes, the paper cited below is the one I used to implement the inverse cumulative normal function. In that paper you recommended that such a function be used to generate random normally distributed numbers, instead of the Box-Muller method. 1) Would you still recommend this? 2) The approach suggested below seems to be aiming more for accuracy, rather than speed. On the other hand, the published algorithm is machine-dependent and therefore a bit of a headache to maintain. Is there another fast but portable algorithm with sufficient accuracy available? >>And what about the inverse cumulative normal distribution? >>I used to implement an algorithm you published, >>but do not like it because it is machine-dependent. >>What in your opinion is the best algorithm? > With the improvement on my method for > evaluating cPhi(x) as R(x)*phi(x), > when the Taylor series for R(x) is about zero, > (suggested by Bill Daly), the simple C function > double Phi(double x) > { long double t=0,b=1,s=x,pwr=x; > int i; > for(i=2;s!=t;i+=2) > { b/=(i+1); pwr=pwr*x*x; t=s; s+=pwr*b;} > return .5+s*exp(-.5*x*x-.91893853320467274178L); > should serve very well for solving the equation > Phi(X)=U > for positive X, given U>1/2, by NewtonÕs method: > Start with an initial estimate > x=sqrt(-1.6*ln(1-U^2)); > then repeat > x=x-(Phi(x)-U)/phi(x); > until you get no further changes in x. > The paper that you refer to must be where I want > to generate a normal X by means of solving > Phi(X)=U, given a random U, uniform in [0,1), > by getting an integer j, formed from certain > bits of the exponent part of the ßoating point > representation of U. Then use j to access tabled > values A[j] and B[j] to initalize the Taylor series > for Phi inverse, using the fractional part of U > as the argument in the series. It is designed for > speed in generating a normal variate directly > as Phi^(-1)(U), but only provides accuracy to 6/7 digits, > (As I recall, the Taylor series was easy and fast only for the first few > terms.) > It is described in > ``Normal (Gaussian) random variables in supercomputersÕÕ, (1991) > Journal of Supercomputing}, v 5, 49--55, > where the method is particularly suited for parallel computation. > George Marsaglia === Subject: This WeekÕs Finds in Mathematical Physics (Week 202) Also available at http://math.ucr.edu/home/baez/week202.html This WeekÕs Finds in Mathematical Physics - Week 202 John Baez This week IÕll deviate from my plan of discussing number theory, and instead say a bit about something else thatÕs been on my mind lately: structure types. But, youÕll see my fascination with Galois theory lurking beneath the surface. Andre Joyal invented these in 1981 - he called them especes de structure. Basically, a structure type is just any sort of structure we can put on finite sets: an ordering, a coloring, a partition, or whatever. In combinatorics people count such structures using generating functions. A generating function is a power series where the coefficient of x^n keeps track of how many structures of the given kind you can put on an n-element set. By playing around with these functions, you can often figure out the coefficients and get explicit formulas - or at least asymptotic formulas - that count the structures in question. The reason this works is that operations on generating functions come from operations on structure types. For example, in week190, I described how addition, multiplication and composition of generating functions correspond to different ways to get new structure types from old. JoyalÕs great contribution was to give structure types a rigorous definition, and use this to show that many calculations involving generating functions can be done directly with structure types. It turns out that just as generating functions form a *set* equipped with various operations, structure types form a *category* with a bunch of completely analogous operations. This means that instead of merely proving *equations* between generating functions, we can construct *isomorphisms* between their underlying structure types - which imply such equations, but are worth much more. ItÕs like the difference between knowing two things are equal and knowing a specific reason WHY theyÕre equal! Of course, this business of replacing equations by isomorphisms is called categorification. In this lingo, structure types are categorified power series, just as finite sets are categorified natural numbers. A while back, James Dolan and I noticed that since you can use power series to describe states of the quantum harmonic oscillator, you can think of structure types as states of a categorified version of this physical system! This gives new insights into the combinatorial underpinnings of quantum physics. For example, the discrete spectrum of the harmonic oscillator Hamiltonian can be traced back to the discreteness of finite sets! The commutation relations between annihilation and creation operators boil down to a very simple fact: thereÕs one more way to put a ball in a box and then take one out, than to take one out and then put one in. Even better, the whole theory of Feynman diagrams gets a simple combinatorial interpretation. But for this, one really needs to go beyond structure types and work with a generalization called stuff types. IÕve been thinking about this business for a while now, so last fall I decided to start giving a year-long course on categorification and quantization. The idea is to explain bunches of quantum theory, quantum field theory and combinatorics all from this new point of view. ItÕs fun! Derek Wise has been scanning in his notes, and a bunch of people have been putting their homework online. So, you can follow along if you want: 1) John Baez and Derek Wise, Categorification and Quantization. IÕd like to give you a little taste of this subject now. But, instead of explaining it in detail, IÕll just give some examples of how structure types yield some far-out generalizations of the concept of cardinality. This stuff is a continuation of some themes developed in week144, week185, week190, so IÕll start with a review. Suppose F is a structure type. Let F_n be the *set* of ways we can put this structure on a n-element set, and let |F_n| be the *number* of ways to do it. In combinatorics, people take all these numbers |F_n| and pack them into a single power series. ItÕs called the generating function of F, and itÕs defined like this: |F_n| |F|(x) = sum ----- x^n n! It may not converge, so in general itÕs just a formal power series - but for interesting structure types it often converges to an interesting function. WhatÕs good about generating functions is that simple operations on them correspond to simple operations on structure types. We can use this to count structures on finite sets. Let me remind you how it works for binary trees! ThereÕs a structure type T where a T-structure on a set is a way of making it into the leaves of a binary tree drawn in the plane. For example, hereÕs one T-structure on the set {a,b,c,d}: b d a c / / / / / / / / / / on an n-element set is n! times the number of binary trees with n leaves. Annoyingly, the latter number is traditionally called the (n-1)st Catalan number, C_{n-1}. So, we have: |T|(x) = sum C_{n-1} x^n where the sum starts with n = 1. ThereÕs a nice recursive definition of T: To put a T-structure on a set, either note that it has one element, in which case thereÕs just one T-structure on it, or chop it into two subsets and put a T-structure on each one. In other words, any binary tree is either a degenerate tree with just one leaf: X or a pair of binary trees stuck together at the root: ----- ----- | | | | | T | | T | | | | | ----- ----- / / / / We can write this symbolically as T = X + T^2 HereÕs why: X is a structure type called being the one-element set, + means exclusive or, and squaring a structure type means you chop your set in two parts and put that structure on each part. (I explained these rules more carefully in week190.) I should emphasize that the equals sign here is really an *isomorphism* between structure types - IÕm only using equals because the isomorphism key on my keyboard is stuck. But if we take the generating function of both sides we get an actual equation, and the notation is set up to make this really easy: |T| = x + |T|^2 In week144 I showed how you can solve this using the quadratic equation: |T| = (1 - sqrt(1 - 4x))/2. and then do a Taylor expansion to get |T| = x + x^2 + 2x^3 + 5x^4 + 14x^5 + 42x^6 + ... Lo and behold! The coefficient of x^n is the number of binary trees with n leaves! ThereÕs also another approach where we work directly with the structure types themselves, instead of taking generating functions. This is harder because we canÕt subtract structure types, or divide them by 2, or take square roots of them - at least, not without stretching the rules of this game. All we can do is use the isomorphism T = X + T^2 and the basic rules of category theory. ItÕs not as efficient, but itÕs illuminating. ItÕs also incredibly simple: we just keep sticking in X + T^2 wherever we see T on the right-hand side, over and over again. Like this: T = X + T^2 T = X + (X + T^2)^2 T = X + (X + (X + T^2)^2)^2 and so on. You might not think weÕre getting anywhere, but if you stop at the nth stage and expand out what weÕve got, youÕll get the first n terms of the Taylor expansion we had before! At least, you will if you count stages and terms correctly. I wonÕt actually do this, because itÕs better if you do it yourself. When you do, youÕll see it captures the recursive process of building a binary tree from lots of smaller binary trees. Each time you see a T and replace it with an X + T^2, youÕre really taking a little binary tree: ----- | | | T | | | ----- and replacing it with either a degenerate tree with just a single leaf: X or a pair of binary trees: ----- ----- | | | | | T | | T | | | | | ----- ----- / / / / So, each term in the final result actually corresponds to a specific tree! This is a good example of categorification: when we calculate the coefficient of x^n this way, weÕre not just getting the *number* of binary planar trees with n leaves - weÕre getting an actual explicit description of the *set* of such trees. Now, what happens if we take the generating function |T|(x) and evaluate it at x = 1? On the one hand, we get a divergent series: |T|(1) = 1 + 1 + 2 + 5 + 14 + 42 + ... This is the sum of all Catalan numbers - or in other words, the number of binary planar trees. On the other hand, we can use the formula |T| = (1 - sqrt(1 - 4x))/2 to get |T|(1) = (1 - sqrt(-3))/2 It may seem insane to conclude 1 + 1 + 2 + 5 + 14 + 42 + ... = (1 - sqrt(-3))/2 but Lawvere noticed that thereÕs a kind of strange sense to it. The trick is to work not with generating function |T| but with the structure type T itself. Since |T|(1) is equal to the *number* of planar binary trees, T(1) should be naturally isomorphic to the *set* of planar binary trees. And it is - itÕs obvious, once you think about what it really means. The number of binary planar trees is not very interesting, but the set of them is. In particular, if we take the isomorphism T = X + T^2 and set X = 1, we get an isomorphism T(1) = 1 + T(1)^2 which says a planar binary tree is either the tree with one leaf or a pair of planar binary trees. Starting from this, we can derive lots of other isomorphisms involving the set T(1), which turn out to be categorified versions of equations satisfied by the number |T|(1) = (1 - sqrt(-3))/2 For example, this number is a sixth root of unity. While thereÕs no one-to-one correspondence between 6-tuples of trees and the 1 element set, which would categorify the formula |T|(1)^6 = 1 there *is* a very nice one-to-correspondence between 7-tuples of trees and trees, which categorifies the formula |T|(1)^7 = |T|(1) Of course the set of binary trees is countably infinite, and so is the set of 7-tuples of binary trees, so they can be placed in one-to-one correspondence - but thatÕs boring. When I say very nice, I mean something more interesting: starting with the isomorphism T = x + T^2 we get a one-to-one correspondence T(1) = 1 + T(1)^2 which says that any binary planar tree is either degenerate or a pair of binary planar trees... and using this we can *construct* a one-to-one correspondence T(1)^7 = T(1) The construction is remarkably complicated. Even if you do it as efficiently as possible, I think it takes 18 steps, like this: T(1)^7 = T(1)^6 + T(1)^8 = T(1)^5 + T(1)^7 + T(1)^8 . . . = 1 + T(1) + T(1)^2 + T(1)^4 = 1 + T(1) + T(1)^3 = 1 + T(1)^2 = T(1) IÕll let you fill in the missing steps - itÕs actually quite fun if you like puzzles. If you get stuck, you can look up the answer in a couple of different places. While Lawvere was the first to figure this out, the first to write it up was Andreas Blass: 2) Andreas Blass, Seven trees in one, Jour. Pure Appl. Alg. 103 (1995), 1-21. Also available at http://www.math.lsa.umich.edu/~ablass/cat.html ThereÕs also a nice treatment based on more general results here: 3) Marcelo Fiore, Isomorphisms of generic recursive polynomial types, to appear in 31st Symposium on Principles of Programming Languages (POPL04). Also available at http://www.cl.cam.ac.uk/~mpf23/papers/Types/recisos.ps.gz In fact, Fiore and Leinster have a nice general theory that explains why the set T(1) acts so much like a sixth root of unity: 4) Marcelo Fiore and Tom Leinster, Objects of categories as complex numbers, available as math.CT/0212377. The idea is that whenever we have an object Z in a distributive category (a category with finite products and coproducts, the former distributing over the latter), and itÕs equipped with an isomorphism Z = P(Z) where P is a polynomial with natural number coefficients, we can associate to it a cardinality |Z|, namely any complex solution of the equation |Z| = P(|Z|) Which solution should we use? Well, for simplicity letÕs consider the case where P has degree at least 2 and the relevant Galois group acts transitively on the roots of this equation, so all roots are created equal. Then we can pick *any* solution as the cardinality |Z|. Any polynomial equation with natural number coefficients satisfied by one solution will be satisfied by all, so it wonÕt matter which one we choose. Now suppose the cardinality |Z| satisfies such an equation: Q(|Z|) = R(|Z|) where neither Q nor R is constant. Then the results of Fiore and Leinster say we can construct an isomorphism Q(Z) = R(Z) in our distributive category! In other words, a bunch of equations satisfied by the objectÕs cardinality automatically come from isomorphisms involving the object itself. This explains why the set T(1) of binary trees acts like it has cardinality |T|(1) = (1 - sqrt(-3))/2 or equally well, |T|(1) = (1 + sqrt(-3))/2 (Since the relevant Galois group interchanges these two numbers, we can use either one.) More generally, the set T(n) consisting of binary trees with n-colored leaves acts a lot like the number |T|(n). This has gotten me interested in trying to find a nice model of what I call the Golden Object: an object G in some distributive category thatÕs equipped with an isomorphism G^2 = G + 1 The Golden Object doesnÕt fit into Fiore and LeinsterÕs formalism, since this isomorphism is not of the form G = P(G) where P has natural number coefficients. But, it still seems that such an object deserves to have a cardinality equal to the golden ratio. James Propp came up with an interesting idea related to the Golden Object: consider what happens when we evaluate the generating function for binary trees at -1. On the one hand we get an alternating sum of Catalan numbers: |T|(-1) = -1 + 1 - 2 + 5 - 14 + 42 + ... On the other hand, we can use the formula |T| = (1 - sqrt(1 - 4x))/2 to get |T|(1) = (1 - sqrt(5))/2 which is -1 divided by the Golden Ratio. Of course, itÕs possible we should use the other sign of the square root, and get |T|(1) = (1 + sqrt(5))/2 which is just the Golden Ratio! Galois theory says these two roots are created equal. Either way, we get a bizarre and fascinating formula: - 1 + 1 - 2 + 5 - 14 + 42 + ... = (1 +- sqrt(5))/2 Can we fit this into some clear and rigorous framework, or is it just nuts? WeÕd like some generalization of cardinality for which the set of binary trees with -1-colored leaves has cardinality equal to the Golden Ratio. James Propp suggested one avenue. Following SchanuelÕs ideas on Euler characteristic as a generalization of cardinality, it makes sense to treat the real line as a space of cardinality -1. This will sound crazy unless you go back and read week147, so please do that! Anyway, using this idea it seems reasonable to consider the space of binary trees with leaves labelled by real numbers as a rigorous version of the set of binary trees with -1-colored leaves. So, we just need to figure out what generalization of Euler characteristic gives this space an Euler characteristic equal to the Golden Ratio. It would be great if we could make this space into a Golden Object in some distributive category, but that may be asking too much. Whew! ThereÕs obviously a lot of work left to be done here. HereÕs something easier: a riddle. WhatÕs this sequence? un, dos, tres, quatre, cinc, seis, set, vuit, nou, deu,... Now IÕd like to mention some important papers on n-categories. You may think IÕd lost interest in this topic, because IÕve been talking about other things. But itÕs not true! Most importantly, Tom Leinster has come out with a big book on n-categories and operads: 5) Tom Leinster, Higher Operads, Higher Categories, Cambridge U. Press, As youÕll note, he managed to talk the press into letting him keep his book freely available online! We should all do this. Nobody will ever make much cash writing esoteric scientific tomes - it takes so long, you could earn more per hour digging ditches. The only *financial* benefit of writing such a book is that people will read it, think youÕre smart, and want to hire you, promote you, or invite you to give talks in cool places. So, maximize your chances of having people read your books by keeping them free online! People will still buy the paper version if itÕs any good.... And indeed, LeinsterÕs book has many virtues besides being free. He gracefully leads the reader from the very basics of category theory straight to the current battle front of weak n-categories, emphasizing throughout how operads automatically take care of the otherwise mind-numbing thicket of coherence laws that inevitably infest the subject. He doesnÕt take well-established notions like monoidal category and bicategory for granted - instead, he dives in, takes their definitions apart, and compares alternatives to see what makes these concepts tick. ItÕs this sort of careful thinking that we desperately need if weÕre ever going to reach the dream of a clear and powerful theory of higher-dimensional algebra. He does a similar careful analysis of operads and multicategories before presenting a generalized theory of operads thatÕs powerful enough to support various different approaches to weak n-categories. And then he describes and compares some of these different approaches! In short: if you want to learn more about operads and n-categories, this is *the* book to read. Leinster doesnÕt say too much about what n-categories are good for, except for a nice clear introduction entitled Motivation for Topologists, where he sketches their relevance to homology theory, homotopy theory, and cobordism theory. But this is understandable, since a thorough treatment of their applications would vastly expand an already hefty 380-page book, and diffuse its focus. It would also steal sales from *my* forthcoming book on higher- dimensional algebra - which would be really bad, since I plan to retire on the fortune IÕll make from this. Secondly, Michael Batanin has worked out a beautiful extension of his ideas on n-categories which sheds new light on their applications to homotopy theory: 6) Michael A. Batanin, The Eckmann-Hilton argument, higher operads and E_n spaces, available as math.CT/0207281. Michael A. Batanin, the combinatorics of iterated loop spaces, available as math.CT/0301221. Getting a manageable combinatorial understanding of the space of loops in the spaces of loops in the space of loops... in some space has always been part of the dream of higher-dimensional algebra. These k-fold loop spaces or have been important in homotopy theory since the 1970s - see the end of week199 for a little bit about them. People know that k-fold loop spaces have k different products that commute up to homotopy in a certain way that can be summarized by saying they are algebras of the E_k operad, also called the little k-cubes operad. However, their wealth of structure is still a bit mind-boggling. James Dolan and I made some conjectures about their relation to k-tuply monoidal categories in our paper Categorification (see week121), and now Batanin is making this more precise using his approach to n-categories - which is one of the ones described in LeinsterÕs book. ThereÕs also been a lot of work applying higher-dimensional algebra to topological quantum field theory - thatÕs what got me interested in n-categories in the first place, but a lot has happened since then. For a highly readable introduction to the subject, with tons of great pictures, try: 7) Joachim Kock, Frobenius Algebras and 2D Topological Quantum Field This is mainly about 2d TQFTs, where the concept of Frobenius algebra reigns supreme, and everything is very easy to visualize. When we go up to 3-dimensional spacetime life gets harder, but also more interesting. This book isnÕt so easy, but itÕs packed with beautiful math and wonderfully drawn pictures: 8) Thomas Kerler and Volodymyr L. Lyubashenko, Non-Semisimple Topological Quantum Field Theories for 3-Manifolds with Corners, Lecture Notes in Mathematics 1765, Springer, Berlin, 2001. The idea is that if we can extend the definition of a quantum field theory to spacetimes that have not just boundaries but *corners*, we can try to build up the theory for arbitrary spacetimes from its behavior on simple building blocks - since itÕs easier to chop manifolds up into a few basic shapes if we let those shapes have corners. However, it takes higher-dimensional algebra to describe all the ways we can stick together manifolds with corners! Here Kerler and Lyubashenko make 3-dimensional manifolds going between 2-manifolds with boundary into a double category... and make a bunch of famous 3d TQFTs into double functors. Closely related is this paper by Kerler: 9) Thomas Kerler, Towards an algebraic characterization of 3-dimensional math.GT/0008204. It relates the category whose objects are 2-manifolds with a circle as boundary, and whose morphisms are 3-manifolds with corners going between these, to a braided monoidal category freely generated by a quasitriangular Hopf algebra object. (IÕm leaving out some fine print here, but probably putting in more than most people want!) It comes close to showing these categories are the same, but suggests that theyÕre not quite - so the perfect connection between topology and higher categories remains elusive in this important example. Answer to the riddle: these are the Catalan numbers - i.e., the natural numbers as written in Catalan. This riddle was taken from the second volume of StanleyÕs book on enumerative combinatorics (see week144). ------------------------------------------------------------- ---------- mathematics and physics, as well as some of my research papers, can be obtained at http://math.ucr.edu/home/baez/ For a table of contents of all the issues of This WeekÕs Finds, try http://math.ucr.edu/home/baez/twf.html A simple jumping-off point to the old issues is available at http://math.ucr.edu/home/baez/twfshort.html If you just want the latest issue, go to http://math.ucr.edu/home/baez/this.week.html === Subject: Re: Polynomial solutions to Pell eqn >The Pell equation and the obvious cubic generalisation >X^2 - DY^2 = ±1 and >X^3 + DY^3 +(D^2)Z^3 - 3DXYZ = 1 >have simple polynomial solutions >(n,1) for D = n^2 ±1 >(n^2, n, 1) for D = n^3±1 >In the square case, you can use continued >fractions to expand expression of the form, say, >SQRT(an^2 + bn + c), but in the cubic case >no such option is available. >Simple cases can be guessed >for example >D = n^3 ± 3 >X = n^6 ± 3n^3 ± 1 >Y = n^5 ±2n^2 >X = n^4 ± n >D = n^3 +2, n even >X = (9n^6)/4 + (9n^3)/2 +1 >Y = (9n^5)/4 + 3n^2 >Z = 3n(3n^3 +2)/4 >I was wondering if there is any >systematic approach to obtaining >solutions ? > Not real sure about a systematic approach, but looking at a few data points for > specific D may be useful. Consider the following: > D=10^3+5*10 > X=62591931727331611501 > D=100^3+5*100 > X=4910017588770392230083131681058781350001 > D=1000^3+5*1000 > X=489788270129887199471510415004171147924480921387800135000001 > D=10000^3+5*10000 > 00000001 > D=100000^3+5*100000 > X= 48977602682444006529246451274826892825824096005404164480666792 0000448092 > 0000138780000001350000000001 > D=1000000^3+5*1000000 > X= 48977602561224440064012924645120074826892800258240960000540416 4480006667 > 920000004480920000001387800000000135000000000001 > D=10000000^3+5*10000000 > X= 48977602560012244400640001292464512000074826892800002582409600 0000540416 > 44800000666792000000004480920000000013878000000000013500000000 000001 > D=100000000^3+5*100000000 > X= 48977602560000122444006400000129246451200000074826892800000025 8240960000 > 00005404164480000000666792000000000044809200000000001387800000 000000013500 > 00000000000001 > There is a pattern here that seems to indicate that a subset of {n^3+5*n} has a > polynomial solution to P3. IÕm not sure what the exact answer is, but it could > probably be worked out, given sufficient time and interest. There are lots of > other curiosities that can be found this way. Yes n^3 +3n and 3n^2 +3n +1 seem easy to approximate. n^ + Dn^3 where (n,D) >1 also seems to work. I did try expanding cubrt(n^6 +2n^3 +4) and cubrt(n^3 +2),say in steps cubrt(n^6 +2n^3 +4) = n^2 + y, etc. cubrt(n^3 +2) = n + z, etc and truncating the two expression so that they both have the same denominator. This sems to work if, say, y = 1/kn where k is an integer. This suggests continued fractions but simulataneous approximation of polys seems specially difficult. > Rich === Subject: Representing the ordering of the perception of colour? How can the order in the perception of colour be represented? Ie. in this ordering, red and orange are close to one another whle red and blue are far apart. Light red > medium red > dark red. The perception of colour is continuous. IÕm asking this because IÕm designing the colour layout in a CAD programme which needs to be determined in a logically consistent manner. So for a number of groups of objects, I need the colours to be as different as possible, but the objects within that group have to look visually related, by shades of colour as an example. Funky === Subject: Re: Representing the ordering of the perception of colour? > How can the order in the perception of colour be represented? > Ie. in this ordering, red and orange are close to one another whle red and > blue are far apart. Light red > medium red > dark red. The perception of > colour is continuous. > IÕm asking this because IÕm designing the colour layout in a CAD programme > which needs to be determined in a logically consistent manner. So for a > number of groups of objects, I need the colours to be as different as > possible, but the objects within that group have to look visually related, > by shades of colour as an example. The reason red-purple-blue seems like a reasonable sequence has to do with the response of the color receptors in the eye, not anything to do with the light wavelengths themselves. === Subject: Re: Representing the ordering of the perception of colour? > How can the order in the perception of colour be represented? > Ie. in this ordering, red and orange are close to one another whle red and > blue are far apart. Light red > medium red > dark red. The perception of > colour is continuous. Color spaces historically have been defined in various ways, e.g., a CIE chromaticity diagram and tristimulus coordinates, or Pantone chips. http:/www.pantone.com/ > IÕm asking this because IÕm designing the colour layout in a CAD programme > which needs to be determined in a logically consistent manner. So for a > number of groups of objects, I need the colours to be as different as > possible, but the objects within that group have to look visually related, > by shades of colour as an example. You might start by requiring that the B&W intensities be identical or not. Then worry about hue and saturation, etc. Edwin Land (of Polaroid) was obsessed with Mondrians and how adjacent colors changed perceived values. You could thus be mathematically on target and perceptually out to lunch. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Representing the ordering of the perception of colour? > How can the order in the perception of colour be represented? > Ie. in this ordering, red and orange are close to one another whle red and > blue are far apart. Light red > medium red > dark red. The perception of > colour is continuous.... Try googling for chromaticity coordinates. Ken Pledger. === Subject: Re: Full Beal Conjecture (revised) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1PLfLn19651; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1PLUfi18759 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1PLUfn04508; According to great interest at this question, I would like to discuss following classification: Once generally Beal Conjecture dissmiss x^a + y^b = z^c for x;y;z integers of gcd=1 and for a;b;c natural numbers >2 1.) For a;b;c as odd numbers weÕll have some fall of BealÕs conjecture, which might be proved once using similar method as aplicable for FLT conjecture for p=3 and bigger primes. ( this method uses also factorisation of numbers shaped as 3M+2 into factors as square numbers, what is applicated as 6n+1 shapes and turns into squares of all primes once beginning with 5 . I used to write down first lines of it at ca 17.02.04 ) 2.) For one, two or even three of exponents a;b;c as even numbers but bigger than 4 and so on containing some odd number bigger or equal to 3 it will be possible to turn such fall into (1)-st one: a=a1*a2; b=b1*b2; c=c1*c2 so (x^a2)^a1 + (y^b2)^b1 = (z^c2)^c1 where a1;b1;c1 odd numbers. 3.) For a;b;c as even numbers but discribed exclusively as 2^k or with the part a2;b2;c2 as 2^k where k=2 or bigger natural numbers so such fall is to collect into x^4 + y^4 = z^4 conjecture 4.) For only few combinations once one of a;b;c is equal to 4 but others odd or two of a;b;c are equal to 4 but the one resisting odd there is not possible to use the classification and method for odd exponents. 5.) Finally for a;b;c as even numbers but shaped in that way, that it is possible to rewrite conjectures: x^2 + y^4 = z^4 or x^4 + y^4 = z^2 You can see, that conjectures x^4 + y^4 = z^4 and x^2 + y^4 = z^4 or x^4 + y^4 = z^2 were already determinated firstly by P.Fermat. Once it is some hope for method for odd exponents so the resisting small gap can be closed too and so on it will be possible to express that any sum of powers bigger than 2 can not be substituted with the power bigger than 2. (Does Fermat expressed, that exponents of such powers are equal ?) Enjoy super games of powers Ro === Subject: dummy variable / integration help Alright, I clearly canÕt remember introductory calculus. Let p(x) = 1/sqrt(4 pi t) e^-(x^2/4t). Let T f (x) = INT p(x-y) f(y) dy (the integral, and all integrals henceforth, are assumed from -oo to oo). I have to show that T is a bounded operator from L_2 to L_2, so (let || || be the L^2 norm) ||T_t f||^2 = INT ( INT p(y) f(x-y) dy ) ^2 dx = INT (INT p(y)f(x-y) dy ) (INT p(z) f(x-z) dz) dz = INT INT INT p(y) p(z) f(x-y) f(x-z) dx dy dz = INT INT (INT f(x-y) f(x-z) dx ) p(y) p(z) dy dz. Now IÕd *like* to get ||f||^2 on the inside, but INT f(x-y) f(x-z) dx doesnÕt seem to be quite that, because of the (x-y) and (x-z), and any linear change of variable on x wouldnÕt solve the problem....where am I screwing up? === Subject: Re: dummy variable / integration help > Let p(x) = 1/sqrt(4 pi t) e^-(x^2/4t). > Let T f (x) = INT p(x-y) f(y) dy (the integral, and all integrals > henceforth, are assumed from -oo to oo). > I have to show that T is a bounded operator from L_2 to L_2, so (let || || > be the L^2 norm) > ||T_t f||^2 = INT ( INT p(y) f(x-y) dy ) ^2 dx Hint: Use JensenÕs Inequality on (INT p(y) f(x-y) dy)^2. === Subject: Re: dummy variable / integration help OK...I think also YoungÕs inequality / MinkowskiÕs integral inequality is directly on-point here, but...is there a way to do it directly? > Let p(x) = 1/sqrt(4 pi t) e^-(x^2/4t). > Let T f (x) = INT p(x-y) f(y) dy (the integral, and all integrals > henceforth, are assumed from -oo to oo). > I have to show that T is a bounded operator from L_2 to L_2, so (let || || > be the L^2 norm) > ||T_t f||^2 = INT ( INT p(y) f(x-y) dy ) ^2 dx > Hint: Use JensenÕs Inequality on (INT p(y) f(x-y) dy)^2. === Subject: Re: Request for comments on antiquated algebraic topology online-book > Well IÕve pretty much decided not to invest too much time with Lefschetz. > However, initially, I only need to learn a small subset of algebraic > topology (AT). Specifically, the same topics as found in the first chapter > of MunkresÕ ÔElements of ATÕ. The application is to distributed computing > where the evolution of finite asynchronous distributed protocols are modeled > as high dimensional simplicial complexes (if thats the correct terminology). > combinatorial. The seminal explanatory paper on the approach can be found > at: > http://www.cs.brown.edu/people/mph/HerlihyR96/sv.pdf > ThatÕs an interesting paper. I suppose the explanations are too brief > for you? If you read that paperÕs explanations on topology and combine > it with looking at things online like Wikipedia, I bet youÕll be > alright. See below also. > I have taken another look at Hatcher, and youÕre right - it doesnÕt seem > nearly as intimidating as it did before I bulked up on general topology a > bit. Hopefully I wonÕt need to shell out the cash for MunkresÕ book. > I think for you, it would be worth looking at the first chapter of > Munkres. Just check it out from your local university/college library. > There is no need to pay up all that money for something you only need a > piece of. Hi Chan-Ho. I started reading HatcherÕs book. I had been reading chapter 0 even though he notes it may be skipped intitially, and was having trouble with it. Chapter 1 was far more instructive and have been successfully making my way through it. l8r, Mike N. Christoff === Subject: Cardinality of pi(S^1) IÕm reading HatcherÕs book on Algebraic topology and he proves the following: The map v : Z -> pi(S^1) sending an integer n to the homotopy class of the loop w_n(s) = (cos 2*pi*n*s, sin 2*pi*n*s) based at (1,0) is an isomorphism. --- But isnÕt w_n hotomopic to w_m for all m,n in N, using a linear homotopy? ie: [w_n] = [w_m] meaning there is only a single homotopy class to map Z into. l8r, Mike N. Christoff === Subject: Re: Cardinality of pi(S^1) > IÕm reading HatcherÕs book on Algebraic topology and he proves the > following: > The map v : Z -> pi(S^1) sending an integer n to the homotopy class of the > loop w_n(s) = (cos 2*pi*n*s, sin 2*pi*n*s) based at (1,0) is an isomorphism. > --- > But isnÕt w_n hotomopic to w_m for all m,n in N, using a linear homotopy? > ie: [w_n] = [w_m] meaning there is only a single homotopy class to map Z > into. Actually, linear homotopies are only gauranteed in convex sets which S^1 is not. l8r, Mike N. Christoff === Subject: series summation Let Bin(k,m) be the binomial coeffient k choose m. The problem is to find the sum: Sum(k=m to infty) Bin(k,m)x^k where 0 WhatÕs the fastest-converging algorithm to compute ln(x)? And to avoid > answers like, Compute log_3(x) and multiply it by an appropriate > constant, the only acceptable operations in the algorithm are > addition, subtraction, multiplication, division, and, say sqrt. I > suppose we can allow rational constants and x, too :-) > Is there a quadratically converging method? I seem to recall there IS > one. > And how do you know, in general, whether thereÕs a quadratically > convergent method or not (not just for the ln problem, but for a > general function). > --Ron Bruck === Subject: Difficulty of calculus vs. discrete math I have taken college courses both in calculus and in discrete mathematics. What surprised me was the difficulties that other students were having with discrete mathematics. For some reason, they found calculus much easier. The discrete math was baby stuff: formal logic, divisibility, combinatorics, and the like. Why would one find calculus easy and discrete math difficult? === Subject: Re: Give me that old time ontology: (was: the anticlassicalist }{ i: linguistic negation) ... >There: IÕve used ontology in a sentence, and it wasnÕt so bad, was >it? ;-) >(Richard Herring vigorously spits and loudly calls for a pint). > Oh, IÕd call for a pint whatever you said ;-) > I think what I mostly donÕt like about the word is that it evokes > AnselmÕs argument for the existence of God. One false step and youÕll > believe your model really exists. Oh well ... I am ignorant of AnselmÕs argument, so I canÕt fall into _that_ trap, can I? ;-) ... > I think youÕre really talking about epistemology. Perhaps. But then I suppose IÕd have to own to the opinion that any meaning of ontology beyond one given to it by our episto-thingie was meaningless. (There ... IÕve said, if you strategically snip out some intervening words, that ontology ... was meaningless. _That_ should make you happy. ;-) === Subject: Re: Give me that old time ontology: (was: the anticlassicalist }{ i: linguistic negation) [snippa lotta selfadoring prose gooey astermakergrown man cry] Hey girls, I sure would like you to move this conversation to the womens loo or some such more appropiate place. I mean, get real! > I especially liked the punch line. Umberto Eco would be proud... Umberto Eco is never about self indulgent auto adoration, much as he enjoys his little games. If you were his characters he would give you some mean punishment, as he did to Belbo and Diotallevi for messinÕ around where they shouldnÕt been a messinÕ. You particularly Galathaea, remind me of that kraut neurotic bitch at the umbanda ceremony. Anyway a totally different question. Are you very attractive? I mean, is there a REAL reason why these creeps start drooling all around the place every time you say anything? IÕm still waiting for your explanation of how the proof of halting the problem relates to heyting particularly non boolean structures. I hope itÕs something more than the similitude between halting and heyting === Subject: Re: Give me that old time ontology: (was: the anticlassicalist }{ i: linguistic negation) > [snippa lotta selfadoring prose gooey astermakergrown man cry] > Hey girls, I sure would like you to move this conversation to the > womens loo or some such more appropiate place. I mean, get real! Yeah ... like IÕm going to be completely crushed being insulted by somebody supposedly named Guenther von Knakspott. <...> === Subject: Re: Give me that old time ontology: (was: the anticlassicalist }{ i: linguistic negation) > You particularly > Galathaea, remind me of that kraut neurotic bitch at the umbanda > ceremony. How darest thou, thou Teutonic blackguard, insult the fair damsel in such blatantly _ad puellam_ terms? My Gallic blood boils and gallantry makes me grow wings so that I may ßy to her rescue. > Anyway a totally different question. Are you very attractive? Er... on second thoughts... we might have here a pot-bellied old geezer with snot in his moustache, half his breakfast still on his shirt front (donÕt egg yolks make for a lovely colour?)... you take first go, mate. We Frogs have been at peace with you for more than half a century now, letÕs not waste that over a piece of stale tail. > I mean, > is there a REAL reason why these creeps start drooling all around the > place every time you say anything? It takes all kinds, all kinds. De gustibus non disputandum. If the editor of Maledicta was reading this, IÕd feel justified in posting here a few URLs of porn sites that would rid your stomach of its contents forthwith and without delay. Without his blessing, I shall refrain. In my view, satire and ridicule is the most effective way of dealing with his, her, or its, kind. And satyre too. Would you talk metaphysics with a broken 78rpm record? Or even... strawberry-pie recipes, or baby-bum wiping techniques? I donÕt. === Subject: Re: missing pages from SpanierÕs Algebraic Topology >I ordered a copy of SpanierÕs Algebraic Topology, and it arrived >today, minus page 13-14. (...) >Springer. Instead, I was wondering if someone with access to a >scanner would just scan these two pages (page 13 and 14 in the 1991 >printing--basically the first two pages of chapter 1) and email me the >scans at npenton@hotmail.com. (...) I have a scanner and the book, but itÕs a much older edition. If this does not make it useless for you, I can send you such a scan as a GIF image, but tell me first with what begins page 15 of your book, so that I can better guess what content you are missing. In my book the half-empty first page of chapter 1 is a short introduction, that continues on the next page before section 1; do you have this introduction ? Ulysse from CH === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? > Nathaniel Deeth > Age 11 > Math Genious If you are really a math genius, you would know that 11+5 = 16. === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? > I have to confess, this is homework, so I am not looking for a full > answer, just some hints where to look. > IÕm taking a math history class and the assignment is to write a paper > on whether Cantor was just misinformed and ultimately innocent despite > the harm he did mathematics... or whether he intentionally strove to > murder mathematics. > So far IÕve read into his biography and I am leaning toward the 2nd > case, since when he was young he was sexually abused by a math > teacher. That leads me to believe he had a big grudge against > mathematics in general. > If I adopt that route, one of my main arguments in this paper will be > that it would be impossible for anyone to be SO wrong about infinite > numbers purely be coincidence. I mean, heÕd have to be really tripping > on some hard stuff if weÕre to believe he REALLY believed the nonsense > he published about infinite numbers. > IÕm sure most of you have written a paper to this effect some time or > other in one of your mathematics history classes :-) > Nathaniel Deeth > Age 11 > Math Genious Welcome back, Nathaniel. YouÕve been an 11-year-old math genius for some years now, so I presume that you were born on Feb 29. I am too lazy (see previous posts) to figure out when you were born using this information. The truth is that Cantor intentionally strove to murder mathematics. He had been laughed at by mathematicians because, even as a child, he had a lousy singing voice in spite of the fact that his last name was Cantor. This rankled all the more as he was not Jewish. His family had converted some years before. He was also utterly incensed by the fools who insisted that .999999... was acually equal to 1.0, a pseudo-fact that any fool can see is false. He determined to go them one better. He started out trying to show that it was not possible to represent every number as a sum of at most n cubes (for any n), but he found the effort wearing, so he moved on to other pursuits. He published several proofs of GoldbachÕs conjecture and FermatÕs last theorem using only the mathematics of FermatÕs and GoldbachÕs day, but these efforts were met with scorn and derision. Finally he decided to get even. He invented a confusing and mystical doctrine of the infinite that has devolved into chaos and anarchy, just as had planned. He spent the remainder of his days addicted to zornication, and he finally died insane as a result. Achava === Subject: re:Cantor: ignorant, harmless fool or intentional liar? I conclude that the teacher is a son of a bitch. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? >I conclude that the teacher is a son of a bitch. And I conclude that the teacher does not even exist. --- === Subject: re:Cantor: ignorant, harmless fool or intentional liar? I have one question: was it a male or a female math teacher? ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? >> I have to confess, this is homework, so I am not looking for a full >> answer, just some hints where to look. >> IÕm taking a math history class and the assignment is to write a paper >> on whether Cantor was just misinformed and ultimately innocent despite >> the harm he did mathematics... or whether he intentionally strove to >> murder mathematics. >> So far IÕve read into his biography and I am leaning toward the 2nd >> case, since when he was young he was sexually abused by a math >> teacher. That leads me to believe he had a big grudge against >> mathematics in general. >> If I adopt that route, one of my main arguments in this paper will be >> that it would be impossible for anyone to be SO wrong about infinite >> numbers purely be coincidence. I mean, heÕd have to be really tripping >> on some hard stuff if weÕre to believe he REALLY believed the nonsense >> he published about infinite numbers. >> IÕm sure most of you have written a paper to this effect some time or >> other in one of your mathematics history classes :-) >> Nathaniel Deeth >> Age 11 >> Math Genious >Nathaniel, >This is supposed to be done on your own. You are not allowed to ask >anyone else for help, as this is cheating. >Your teacher, >Dr. Ben Zona And Craig Feinstein, or Dr. Ben Zona, proves that he is no genuine mathematician, as he did not even realize that Mike Deeth was lying when he said that it was homework. No class on the history on mathematics would set as homework the assignment which Mike Deeth claims to have been set, since no genuine history of mathematics Deeth has used here. In short, Deeth is a crank and a troll who wishes to espouse his own opinions and pretend that they are the mainstream. David McAnally At the moment, they (the Time Lords) are far from being all-powerful. ThatÕs why itÕs been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Re: Cantor: ignorant, harmless fool or intentional liar? Discussion, linux) > > I have to confess, this is homework, so I am not looking for a full > answer, just some hints where to look. > IÕm taking a math history class and the assignment is to write a paper > on whether Cantor was just misinformed and ultimately innocent despite > the harm he did mathematics... or whether he intentionally strove to > murder mathematics. > So far IÕve read into his biography and I am leaning toward the 2nd > case, since when he was young he was sexually abused by a math > teacher. That leads me to believe he had a big grudge against > mathematics in general. > If I adopt that route, one of my main arguments in this paper will be > that it would be impossible for anyone to be SO wrong about infinite > numbers purely be coincidence. I mean, heÕd have to be really tripping > on some hard stuff if weÕre to believe he REALLY believed the nonsense > he published about infinite numbers. > > IÕm sure most of you have written a paper to this effect some time or > other in one of your mathematics history classes :-) > > Nathaniel Deeth > Age 11 > Math Genious >>Nathaniel, >>This is supposed to be done on your own. You are not allowed to ask >>anyone else for help, as this is cheating. >>Your teacher, >>Dr. Ben Zona > And Craig Feinstein, or Dr. Ben Zona, proves that he is no genuine > mathematician, as he did not even realize that Mike Deeth was lying > when he said that it was homework. No class on the history on > mathematics would set as homework the assignment which Mike Deeth > claims to have been set, since no genuine history of mathematics > Deeth has used here. In short, Deeth is a crank and a troll who > wishes to espouse his own opinions and pretend that they are the > mainstream. Gosh, I thought that both Nathan the Great and the good Doctor were being humorous. So, Dr. Zona didnÕt know that Nathan was lying when he said it was homework? Does that mean Dr. Zona also forgot that heÕs *not* NathanÕs teacher? -- Jesse F. Hughes ItÕs easy folks. Just talk about my approach to your favorite mathematician. If they canÕt be interested in it, theyÕve demonstrated a lack of mathematical skill. -- James Harris === Subject: eigenvectors of the laplacian I am looking for information on the properties of the Laplacian on a bounded open set of R^n. In particular, if one consider a bounded open set O with a regular boundary, I would like to have a complete study of the properties of the eigenvectors {v1,v2,...} of the laplacian with Neumann boundary conditions. In fact, I am looking forward some approximation results. For example, given a C^1 regular function f on O, which is the convergence speed of the approximation sum{i vi} when N-->infinity ? This is in fact the study of the spectrum {} of a regular function f. At last (but not least), I would like to have some insight about discrete approximation results. For example, when I have a grid that approximates the domain O, and the laplacian is replaced by a finite difference scheme. Gabriel