mm-3219 === The second-brightest object in the sky next to the Sun is the Earth's moon. Since time immemorial, humans and other hominids have watched the Moon wax and wane over the skies of Earth. The lunar orbit takes about 27 1/3 solar days to complete one sidereal revolution. That's more than two days faster than the lunar synodic month of about 29 1/2 days, and reveals to the observer that the phases of the Moon are a function of the Earth and Moon together orbiting the Sun. The Moon appears to be almost exactly the same size as the Sun as viewed from Earth, since the Moon is in fact 400 times smaller and 400 times closer than the Sun on average. This is easy to see during a total solar eclipse, during which time the Sun's corona becomes distinguishable from the dark background, and the Moon is seen to be closer to Earth than the Sun is. The relative distances of the Sun and Moon to Earth don't appear to change all that much and both Sun & Moon orbit Earth in a fairly linear fashion, albeit the Moon's orbit with the Earth is more complex. As a result, we can be sure that both Sun & Moon orbit the Earth sidereally. We already know from observation that the planets orbit the Sun, and we've also deduced the length of every planet's sidereal year and average synodic period. We've learned to count these repeating intervals by their observable averages in days, months and years, by their repeating synodic aspects relative to the Sun, and by their repeating sidereal aspects to fixed stars on the Earth's caelestial firmament. There- by we see that the Moon's phases take consistently lon- ger than the Moon's sidereal month by about 2 1/6 days. We have learned that the difference between the Moon's sidereal and synodic month is directly attributable to the length of a sidereal year of the Earth-Moon system orbiting the Sun verses the length of a sidereal month. We view the Moon orbiting Earth on a long-term average of some 13 1/3 sidereal lunar orbits per sidereal year, i.e. as seen against the fixed background of stars. Re- member, that the distant stars care not that Earth and the other planets orbit the Sun or more accurately the solar system barycenter along with the Sun, since even nearby stars a few dozen lightyears away are yet 1000s of times more distant than Earth is to the Sun. To wit, one lightyear is the linear equivalent of about 63,240 astronomical units where 1 AU equals ~92,956,230 miles, and just one lightyear equals ~5,878,482,160,000 miles. Even Spica (alpha Virginis at 29 Virgo -2, magnitude 1) at 262 lightyears is some 1 1/2 quadrillion miles from Earth...16.5 million times Earth's distance to the Sun. Hence Spica's trigonometric parallax is barely one one- hundredth of an arcsecond or nearly 5000 times smaller than one arcminute ergo undiscernable to the naked eye by a factor of five-thousand to one. If Spica were but one-twentieth of a single lightyear from Earth then we might barely be able to discern one arcminute parallax. At 2160 miles in diameter, and ~239,000 miles distance from Earth, the Moon's diurnal parallax can be as high as one arcdegree from its geocentric position. For the Moon, annual parallax has no effect, since the Moon is forever orbiting Earth epicyclically some 13 1/3 times per sidereal Earth-year. Thus the Earth-Moon system is averaging 360 divided by 13 1/3 makes about 27 degrees per sidereal month on average that the Earth-Moon bary- center has advanced in its heliocentric sidereal orbit. This means that the Sun advances on average 27 degrees per sidereal month through the caelestial zodiac along the plane of the ecliptic, and Moon orbits Earth about 387 sidereal degrees per synodic month totalling 12.37 synodic months per sidereal Earth-year. Eclipse cycles are tied to where the lunar orbit crosses the ecliptic, also called the nodical month, and is about 27.21 days. The anomalistic month averages 27.55 days which is the approximate interval between lunar perigees or apogees, and which determines an annular or total solar eclipse. There are numerous eclipse cycles running concurrently due to the complexity of the lunar orbit. Most popular of these are Saros cycles of 18 tropical years plus 10 or 11 days. Next comes the Moon's nutation cycle every 18.61 years, and is the average time that it takes for the head of the dragon, or north lunar node to regress full circle through the caelestial zodiac. Next is the so-called Metonic cycle, although Meton was far from the first one to have discovered this very predictable 19-year cycle. The ancient calendars revealed intimate knowledge of this cycle by its affect on intercalation, i.e. when an extra lunar month was computed in advance to make every second or third year have 13 months, not just 12 months as usual. This intercalation cycle runs every 19 lunisolar calendar years, following a pattern so perfect that it is off +two hours every 19 tropical years like clockwork. Two hours each 19-years added up to one extra day every 220 years or about ten days per zodiacal age ergo 10 extra days in 2135 tropical years. This enabled precise calculation of a calendar decades in advance by the tribal mathematician, the astrologer. This allowed ritual observance of seasonal holidays or holy-days in season and at the right phase of the Moon, all imperative considerations for calendar calculation, for planting, harvesting and all religious observances. We've seen how Earth's tilted axis results in tropical years, such that 1508 Mayan Haabs is commensurate with 1507 tropical years, meaning 1508 contiguous intervals of 365 mean solar days apiece total 550,420 solar days per every 1507 mean tropical years. This intercalation interval of the ancient Mayan calendar makes 365.24220 average solar days per average tropical year, which is an hairbreadth from modern averages i.e. less than one second in time difference per year between ancient and modern averages for a tropical year. 1 second per year. ^ ^^^^^^ ^^^ ^^^^ We've seen how the heliacal risings of stars & planets attest to planetary motion in our solar system visible to the unaided human eye. We've learned how to use the human hand to estimate subtended angles between planet and star readily to within one degree of longitude and latitude, and we know from experience that the best of naked-eye astronomers can discern as little as one arc- minute between objects. We've seen how a man can count by days, months and years to estimate repeating cycles of the planets, chiefly sidereal years relative to the Sun and distant stars, and by synodic aspects to Earth relative to the Sun. We've learned from experience how one sidereal year on Earth is barely 20 minutes longer than a tropical year, and that this tiny disparity can add up over time to the tune of one constellation--one zodiacal age--every 2,135 tropical years plus 208 days. There are longer eclipse cycles, like the Inex some 29 years minus 20 days. After this is the exeligmos cycle of 54 Years plus 34 days, basically three Saros cycles. There's an eclipse cycle each 58 Years less 40-42 days. Similar types of eclipses return every 65 Years plus 0 to 3 days. Then there's the triple Inex cycle every 87 years less 61 days. Beyond this is solar totality each approximately 410 years. Above this, 18 Inex cycles is 521 Years plus or minus 1 or 2 days and is very useful for making very long-term eclipse predictions. There's one longer luni-solar cycle which is called the grand century of the Moon, and repeats like clockwork every 800 years. Lunar eclipses occulting the bearded star Regulus at 5 Leo recur in predictable, 800-year cycles. The last series ended in 1943, thus 2510 marks a point 567 years into this 800-year period, which is when the next eclipsed-Moon occulting Regulus series begins and lasts 233 more years, with a 19-year plus a 65-year se- paration between eclipse intervals which overlaps with metonic cycles of 19 years (235 solunar synods) and by similar eclipse-type cycles that repeat every 65 years. So 2510 AD marks the point 567 years + 233 years since the beginning of the last series in 1710. Backtracking in time shows that the last complete series ended 1143 AD, preceded by 343 AD, 458 BC, etc. === Subject: Re: 2D geometry question >I am looking for a way to determine container relationship of a point >vs rectangle in 2 dimension. x, y coordinates of rectangle and of >right news group i am posting message to. >- ap If what you want is determine if a point is inside/outside a rectangle or on an edge, then see the comp.graphics.algorithms newsgroup and their faq, subject 2.03: http://www.faqs.org/faqs/graphics/algorithms-faq/ I think the programming code there, is still faster than calculating the signed distance between the point and the 2 perpendicular lines of the rectangle. === Subject: Re: Can an analytic function on a unit disk have an infinite number of zeros? >Can an analytic function on an open unit disk have an infinite number of >zeros in the open unit disk? >Zdenek sin(z) has infinitely many zeros but has an unbounded domain. Try sin(f(z)) where f is defined on the unit disk and where its range includes much of the real line. -- Wanted: Experts at choosing the best of 100+ applicants for a position. Register as a California voter by September 22, and vote on October 7. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: Re: Can an analytic function on a unit disk have an infinite number of zeros? I am stupid :-))) It is getting late here:-))) Z. > Can an analytic function on an open unit disk have an infinite number of > zeros in the open unit disk? > Zdenek === Subject: Re: Asymptotic Formula for Trinomial Coefficients Re: Asymptotic Formula for Trinomial Coefficients in sci.math The formula is exact in n, for binomial and polynomial expansions where r, the number of terms, is the order of the polynomial and n the power. This answer comes from Finite Math with Business Applications, by Kemeny, Schleifer, Snell, and Thompson, Prentice-Hall, Englewood Cliffs, NJ, 1962. I am sure it hasn't changed since then. I can't go through the whole thing but it will be easy for you to understand. It has to do with getting comfortable with the notions of counting, permutations, and partitions of sets. >arbitrary coefficents of such expressions. That's what we'll have in a few minutes. ------------------------ We turn now to the problem of expanding the trinomial (x + y + z) ^ 3 = (x + y + z)*(x + y + z)*(x + y + z). This time we choose either an e or y or z from each of the three factors. Our choise is now represented by a three-cell partition of the set of numbers {1,2,3}. The first cell has teh numbers corresponding to factors from which we choose an x, the seond cell teh numbers corresponding to factors from which we choose a y, and the third thosee from which we choose a z. For example, the partition [{1,3},e,{2}] corresponds to a choice of x from teh first and third factors, no y's, and z from teh second factor. The term obtained is xzx = x^2z. The coefficeient of the term x^2z in the expansion is thus the number of three cell partitions with two elements in teh first cell, none in teh second, and one in the third. There are ( 3 ) ( 2,0,1 ) = 3 of such partitions. (That double paren represents a large paren around one number over three numbers without a dividing line, and is explained in the text along with the formula.) In general the coefficient of the term of the form x^ay^bz^c in the expansion of (x+y+z)^3 will be ( 3 ) ( a,b,c ) = 3! / (a! * b! * c!) ----------------------------- That's what you asked for, isnt' it, the coefficient of a particular term? Now it is HILARIOUS that they used THREE terms to the THIRD power! Pardon me shouting, but that makes the result very hard to follow. They give a result: (for you r = 3 so this one can be used, and their n is your n.) ---------------------- Multinomial Theorem: The expansion of (x1 + x2 + ... + xr) ^ n is found by adding all terms of the form ( n ) ( ) * x1^n1 * x2^n2 * ... * xr&nr ( n1, n2, ... nr) where (and this is the key) n1 + n2 + ... nr = n so this includes those nasty terms having just an x or an x^2y. It includes any term, and the formula you want is the one in the big parens on the left. So the term for the coefficient x^a * y^b * z^c where a + b + c = n is: n! / ( a! * b! * c! ) I typed all this in because I am studying the case for n = 3, using this book. Typing it in helps. For me, n will eventually take on any value, as does your n. It wouldn't be very obvious for r = 3 and n = 2, the fourth coefficient would be 2, as it is in your example, but that's it there. See: 3! = 6 but the term x is NOT just the term (0,1,0) in your special case. There is one other hidden term: (0,1,1). See it now? And since both violate the rule on which the answer is based, they an invalid coefficient of zero if misapplied in the obvious way. Remember a+b+c = n and n =2 in your case. On page 99, in Chap 3, Sec. 4 the key is disclosed. Instead of what you'd think, which is that 0!=0, we must say For reasons which will be clear later, we define 0! = 1. Thus we can say there are n! permutations of n distinct objects. So, while there is obviously one permutation of one object, there is also one permutation of zero objects, and that's what throws everyone off in statistical and quantum mechanics: you're counting things that aren't even there! When you do get used to that, it all comes out. And my bookmark is still on the _first page_ of the chapter in which all the above appears: Chapter 3: Partitions and Counting. It helps understand certain aspects of the quantum mechanics of indistinguishable (indistinguishible?) self-reproducing machine tools is a population of functionally identical, and if you don't look to closely, indistinguishible individuals So far it seems you have to start with two, which is where a every machine shop I have ever worked in has gotten wrong. You have to have two of everything! Not just two cheap bench grinders, but two identical expensive lathes and two of each tooling accesory, and two expensive mills, and two of each measuring instruments. And one operator, for safety's sake. Not like Dean Kamen's shop at all. It's Goncz's Postulate: You Need Two Of Everything! I should add here that by starting with two $29 collet indexers, two $29 four jaw chucks, two $39 cross vises, and two $59 drill presses, I was able to sell one self-reproducing machine tool for $300 and keep the other. I sold it to a guy who lives deep in the mountains. I don't have my reciept book with me. So now I am trying to get to self-reproducing machine tool #3. So n=3 for me. Quantum mechanics is a little farther along than statistical mechanics, which is nicely explained in Leonard B. Leob's classic Kinetic Theory of Gases, which I had the pleasure of borrowing from VMI for a few weeks via interlibrary loan, and almost got to buy ( the NVCC copy, which isn't available via ILL) at the NVCC-Alexandria library book sale. Missed it by about three days. I don't like the glitzy, politically correct modern textbooks. One reason I study at ODU via video streaming is so I can get up and go to my bookcase without disturbing other students, or having to walk back to a dorm room and miss part of a lecture. I have Alistair M. Ray's Quantum Mechanics, and am on systems, which appears near the end. Yours, Doug Goncz, Replikon Research, Seven Corners, VA The hormones work at different speeds: In a fight-or-flight scenario, glucocorticoids are the ones drawing up blueprints for new aircraft carriers; epinephrine is the one handing out guns. === Subject: Zeno's paradoxes (was Re: reciting N and other sets) [...] > Few people in this learned community have not heard of the famous > paradox of Achilles and the tortoise. I feel there is no need to > recite this so-called paradox here, for I have great confidence that > you are already a master of the problem. > [snipped much less interesting stuff] Actually I would say that many people do not understand Zeno's paradoxes. As an example, I offer the hordes of calculus teachers who resolve the Achilles paradox by explaining that infinite series can converge. This makes little sense when one realizes that the Achilles paradox relies on the fact that a particular series converges. === Subject: Re: Zeno's paradoxes (was Re: reciting N and other sets) > Actually I would say that many people do not understand Zeno's paradoxes. > As an example, I offer the hordes of calculus teachers who resolve the > Achilles paradox by explaining that infinite series can converge. This > makes little sense when one realizes that the Achilles paradox relies on > the fact that a particular series converges. Pray tell then, what is the paradox? === Subject: Re: Zeno's paradoxes (was Re: reciting N and other sets) the paradox seems to be a demonstration; whether it preceded or followed any notion of convergence of infinite series is of less interest. the parttial sums always increase, but they never get there. on the other hand, the hare *eventually* will catch the tortoise. > Actually I would say that many people do not understand Zeno's paradoxes. > As an example, I offer the hordes of calculus teachers who resolve the > Achilles paradox by explaining that infinite series can converge. This > makes little sense when one realizes that the Achilles paradox relies on > the fact that a particular series converges. > Pray tell then, what is the paradox? --A church-school McCrusade (Blair's ideals?): Harry-the-Mad-Potter want's US to kill Iraqis?... http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX === Subject: Re: Zeno's paradoxes (was Re: reciting N and other sets) > the paradox seems to be a demonstration; > whether it preceded or followed any notion > of convergence of infinite series is of less interest. > the parttial sums always increase, but > they never get there. on the other hand, > the hare *eventually* will catch the tortoise. I see, the paradox is that Achilles was a hare! === Subject: Re: Zeno's paradoxes (was Re: reciting N and other sets) > Actually I would say that many people do not understand Zeno's paradoxes. > As an example, I offer the hordes of calculus teachers who resolve the > Achilles paradox by explaining that infinite series can converge. This > makes little sense when one realizes that the Achilles paradox relies on > the fact that a particular series converges. I think you're being a little unfair here. If someone stated the Achilles paradox and asked what is wrong with such reasoning, is it not sufficient to point out one step in the argument that is false? In this case, the false assumption seems to be that an infinite series of positive numbers cannot converge *because* it is infinite. You might find an explicit evaluation of the particular series to be a more satisfying resolution, but I think that's a quibble over the meaning of resolution, not evidence of any greater or lesser understanding. === Subject: Name this operation? Can anyone tell me if the following operation has a special name in math, or possibly other fields? A(x), B(x) and C(x) are discrete time series. T(a, b) is some function taking and returning numbers. C(x) = T(A(x), B(x) ) I found something similar in fuzzy sets called 'fuzzy conjunction' but that had limitations on what T could be. With functions instead of time series, I would call it function composition. Is there such a thing as time series composition? Mike === Subject: Re: Why not anonymous publication? I can just see it at conferences. The lecturer would stand behind a screen so no one could identify him and write computer file so no one could identify his/her voice. But then we couldn't hold conferences anyway; we would have no idea whom to invite. And professorial hiring would be random since there would be no way of identifying who was doing what. Actually, I get a high when I prove something difficult or make a fascinating new discovering, but I also get a nice, if muted, frisson when, as happened recently, a paper crossed my desk titled something like, A new proof of Barr's theorem. So the motivations are many, but certainly everlasting fame is part of it. Do I look at the author's name before I decide to read a paper? You better believe it. There are people who have earned the right to have me read their papers and people who haven't and the latter will likely go into a slush pile unless the intro is extraordinary interesting. That is just a fact of life. (As a corollary, I always tell beginners that the intro is the most important part of any paper.) === Subject: Re: [Primes|Assymptotics] What is the order of Sum_{p <= n, p prime} Sum_{i=1..n} [n/p^i] p > Hi to everyone, > I have the following sum: > S(n) = Sum_{p <= n, p prime} Sum_{i=1..n} [n/p^i] p > where [a] is the integer part of a (that is the biggest integer r such > that r <= a < r+1). > I want to know if S(n) can be expressed as beautiful expression and > what is it, and what is the assymptotics of S(n) > Xan. Hi Xan, Your sum is known at the On-Line Encyclopedia of Integer Sequences : http://www.research.att.com/projects/OEIS?Anum=A025281 Note that Sum_{i=1..n} [n/p^i] is the power of the prime factor p in the factorization of n! so that Sum_{p <= n, p prime} Sum_{i=1..n} [n/p^i] ln(p) should simply be ln(n!) To speed up numerical evaluations note too that [n/p^i]=0 for i> ln(n)/ln(p) Hoping it helped a little, Raymond === Subject: Re: [Primes|Assymptotics] What is the order of Sum_{p <= n, p prime} Sum_{i=1..n} [n/p^i] p > > Hi to everyone, > > I have the following sum: > > S(n) = Sum_{p <= n, p prime} Sum_{i=1..n} [n/p^i] p > > where [a] is the integer part of a (that is the biggest integer r such > that r <= a < r+1). > > I want to know if S(n) can be expressed as beautiful expression and > what is it, and what is the assymptotics of S(n) > > Xan. > Hi Xan, > Your sum is known at the On-Line Encyclopedia of Integer Sequences : > http://www.research.att.com/projects/OEIS?Anum=A025281 > Note that Sum_{i=1..n} [n/p^i] is the power of the prime factor p in > the factorization of n! so that Sum_{p <= n, p prime} Sum_{i=1..n} > [n/p^i] ln(p) should simply be ln(n!) > To speed up numerical evaluations note too that [n/p^i]=0 for i ln(n)/ln(p) > Hoping it helped a little, > Raymond [...] so that Sum_{p <= n, p prime} Sum_{i=1..n} [n/p^i] ln(p) should simply be ln(n!) Why do you put ln(p) instead of p? Is this a typo error? Can you explain me why S(n) is ln(n!). In the number sequence link that you gave me, I find that the assymptotics is a(n)~(Pi^2/12)*n^2/ln(n). Is this a contradiction with your reasoning? Xan. === Subject: Re: [Primes|Assymptotics] What is the order of Sum_{p <= n, p prime} Sum_{i=1..n} [n/p^i] p > [...] so that Sum_{p <= n, p prime} Sum_{i=1..n} [n/p^i] ln(p) > should simply be ln(n!) > Why do you put ln(p) instead of p? Is this a typo error? ln(n!)= Sum_{p <= n, p prime} Sum_{i=1..n} [n/p^i] ln(p) This is different of S(n). Let's suppose that n!= p1^e1*p2^e2*...pk^ek (the pi are prime numbers between 2 and n with exponent ei) then : ln(n!)= sum_i ei*ln(pi) and S(n)= sum_i ei*pi (pi <= n) (S(n) is the nolog(n!) function of my previous link) I gave this indication to help you to understand the nolog function of the link and allow you to write S(n)= sum_i ei*pi > Can you explain me why S(n) is ln(n!). In the number sequence link > that you gave me, I find that the assymptotics is > a(n)~(Pi^2/12)*n^2/ln(n). Is this a contradiction with your reasoning? No this should be right (I have no simpler expression for S(n)) Hoping this will be clearer, Raymond > Xan. === Subject: How can I tell if a point is enclosed in a triangle? I have a triangle with verts (x1,y1),(x2,y2) and(x3,y3). The center of the triangle is (x0,y0). Is there an equation which tells me if the point (x4,y4) lies inside the triangle? === Subject: Re: How can I tell if a point is enclosed in a triangle? > I have a triangle with verts (x1,y1),(x2,y2) and(x3,y3). The center of the > triangle is (x0,y0). Is there an equation which tells me if the point > (x4,y4) lies inside the triangle? The algorithm used in finite element work is: [ 1 1 1] [L1] [ 1] let [T] = [x1 x2 x3], compute [L2] = [T]^(-1) [x4] [y1 y2 y3] [L3] [y4] If each Li (i=1,2,3) is between 0 and 1, the point is inside. (Actualy only 2 tests are required because L1+L2+L3=1.) The Li are called barycentric coordinates. Note that x0 and y0 are not used. The algorithm extends trivially to tetrahedra in 3D, and to simplices in n dimensions. === Subject: Re: How can I tell if a point is enclosed in a triangle? > I have a triangle with verts (x1,y1),(x2,y2) and(x3,y3). The > center of the triangle is (x0,y0). Is there an equation which > tells me if the point (x4,y4) lies inside the triangle? I've got a method that works in two dimensions, and it seems like it ought to work in n dimensions, but I don't see how to extend it. Let A = (x1,y1), B = (x2,y2), C = (x3,y3), D = (x4,y4). If D = s A + t B + u C, with s + t + u = 1, and all s, t, u >= 0, then D is in triangle(ABC), strictly > 0 for D in the interior of triangle(ABC). (This part should be true for n dimensions.) So, all you have to do is solve for s, t and u, and check them against zero. u = 1 - s - t D = s A + t B + (1 - s - t) C s (A - C) + t (B - C) = D - C [ x1-x3 x2-x3 ][ s ] = [ x4-x3 ] [ y1-y3 y2-y3 ][ t ] [ y4-y3 ] [ s ] = [ x1-x3 x2-x3 ]^-1[ x4-x3 ] [ t ] [ y1-y3 y2-y3 ] [ y4-y3 ] Let u1 = x1-x3, v1 = y1-y3, ... , v4 = y4-y3. [ s ] = [ u1 u2 ]^-1[ u4 ] [ t ] [ v1 v2 ] [ v4 ] [ s ] = [ v2 -u2 ] [ u4 ] [ t ] [ -v1 u1 ]/(u1 v2 - u2 v1)[ v4 ] (Sorry about the notation.) If (u1 v2 - u2 v1) = 0, two edges of triangle(ABC) are parallel. Depends on your situation whether this is an error, a fail, or a special circumstance. s = (+ u4 v2 - u2 v4)/(u1 v2 - u2 v1) t = (- u4 v1 + u1 v4)/(u1 v2 - u2 v1) If s >= 0, t >= 0, and u = 1 - s - t >= 0, then D is in triangle(ABC). Jim Burns === Subject: Re: How can I tell if a point is enclosed in a triangle? > I have a triangle with verts (x1,y1),(x2,y2) and(x3,y3). Is there an equation which tells me if the point (x4,y4) lies inside the triangle? Connect the inside point 4 to the vertices. When the non-zero triangle areas A total up to area of the given triangle, i.e., A(4,1,2) +A(4,2,3)+ A(4,3,1) - A(1,2,3) = 0 , point 4 lies inside the triangle bounded by straight lines between 1,2 and 3. > The center of the triangle is (x0,y0). is not needed here. === Subject: Re: How can I tell if a point is enclosed in a triangle? HP Pennypacker> I have a triangle with verts (x1,y1),(x2,y2) and(x3,y3). The center of the > triangle is (x0,y0). Is there an equation which tells me if the point > (x4,y4) lies inside the triangle? If we identify the vertices with complex numbers P1,P2,P3, and the other point with P4, the condition is that these three real numbers: Im((P2-P4)/(P1-P4)) Im((P3-P4)/(P2-P4)) Im((P1-P4)/(P3-P4)) all have the same sign. It extends easily enough from a triangle to any _convex_ polygon. For an arbitrary polygon, we can adapt a formula from complex analysis, which gives the winding number of a curve with respect to a point. The integral is replaced by a finite sum. A triangle has no well-defined center unless it is equilateral. The barycenter, incenter, orthocenter, and various others are in general distinct. === Subject: Re: How can I tell if a point is enclosed in a triangle? > HP Pennypacker> I have a triangle with verts (x1,y1),(x2,y2) and(x3,y3). > The center of the > triangle is (x0,y0). Is there an equation which tells me if the point > (x4,y4) lies inside the triangle? > If we identify the vertices with complex numbers P1,P2,P3, and the other > point with P4, the condition is that these three real numbers: > Im((P2-P4)/(P1-P4)) > Im((P3-P4)/(P2-P4)) > Im((P1-P4)/(P3-P4)) > all have the same sign. It extends easily enough from a triangle to any > _convex_ polygon. For an arbitrary polygon, we can adapt a formula > from complex analysis, which gives the winding number of a curve with > respect to a point. The integral is replaced by a finite sum. > A triangle has no well-defined center unless it > is equilateral. The barycenter, incenter, orthocenter, > and various others are in general distinct. This is probably a stupid question, but couldn't one define the center of a triangle as being the intersection of the three lines that extend from each corner of the triangle and end at the center of the edge opposite to that corner? If that doesn't work (ie: there is no single point of intersection), then apply the same idea on the triangle created by the three intersecting lines. Repeat until a single point is found or use the limit of this recursive process. The area of the center triangle should approach zero and hence define a single point in the limit. I hope that makes sense. Excuse me - I'm a little tipsy. :) l8r, Mike N. Christoff === Subject: Re: How can I tell if a point is enclosed in a triangle? >> HP Pennypacker> I have a triangle with verts (x1,y1),(x2,y2) and(x3,y3). >> The center of the >> triangle is (x0,y0). Is there an equation which tells me if the point >> (x4,y4) lies inside the triangle? >> If we identify the vertices with complex numbers P1,P2,P3, and the other >> point with P4, the condition is that these three real numbers: >> Im((P2-P4)/(P1-P4)) >> Im((P3-P4)/(P2-P4)) >> Im((P1-P4)/(P3-P4)) >> all have the same sign. It extends easily enough from a triangle to any >> _convex_ polygon. For an arbitrary polygon, we can adapt a formula >> from complex analysis, which gives the winding number of a curve with >> respect to a point. The integral is replaced by a finite sum. >> A triangle has no well-defined center unless it >> is equilateral. The barycenter, incenter, orthocenter, >> and various others are in general distinct. >This is probably a stupid question, but couldn't one define the center of a >triangle as being the intersection of the three lines that extend from each >corner of the triangle and end at the center of the edge opposite to that >corner? One _could_ define that to be the center. But one doesn't, because there are lots of other points with equally valid claims to being the center... >If that doesn't work (ie: there is no single point of intersection), then >apply the same idea on the triangle created by the three intersecting lines. >Repeat until a single point is found or use the limit of this recursive >process. The area of the center triangle should approach zero and hence >define a single point in the limit. I hope that makes sense. >Excuse me - I'm a little tipsy. :) >l8r, Mike N. Christoff ************************ David C. Ullrich === Subject: Re: How can I tell if a point is enclosed in a triangle? Michael N. Christoff > Larry Hammick > A triangle has no well-defined center unless it > is equilateral. The barycenter, incenter, orthocenter, > and various others are in general distinct. > This is probably a stupid question, but couldn't one define the center of a > triangle as being the intersection of the three lines that extend from each > corner of the triangle and end at the center of the edge opposite to that > corner? They do meet; that one's called the centroid, or barycenter. (I think centroid is the preferred name.) LH === Subject: Re: How can I tell if a point is enclosed in a triangle? > HP Pennypacker> I have a triangle with verts (x1,y1),(x2,y2) and(x3,y3). > The center of the > triangle is (x0,y0). Is there an equation which tells me if the point > (x4,y4) lies inside the triangle? > If we identify the vertices with complex numbers P1,P2,P3, and the other > point with P4, the condition is that these three real numbers: > Im((P2-P4)/(P1-P4)) > Im((P3-P4)/(P2-P4)) > Im((P1-P4)/(P3-P4)) > all have the same sign. It extends easily enough from a triangle to any > _convex_ polygon. For an arbitrary polygon, we can adapt a formula > from complex analysis, which gives the winding number of a curve with > respect to a point. The integral is replaced by a finite sum. > A triangle has no well-defined center unless it > is equilateral. The barycenter, incenter, orthocenter, > and various others are in general distinct. > This is probably a stupid question, but couldn't one define the center of a > triangle as being the intersection of the three lines that extend from each > corner of the triangle and end at the center of the edge opposite to that > corner? > If that doesn't work (ie: there is no single point of intersection), then > apply the same idea on the triangle created by the three intersecting lines. > Repeat until a single point is found or use the limit of this recursive > process. The area of the center triangle should approach zero and hence > define a single point in the limit. I hope that makes sense. > Excuse me - I'm a little tipsy. :) not to worry, sure but its A center, what about the average point? Herc === Subject: Re: How can I tell if a point is enclosed in a triangle? > I have a triangle with verts (x1,y1),(x2,y2) and(x3,y3). The center of the > triangle is (x0,y0). Is there an equation which tells me if the point > (x4,y4) lies inside the triangle? Given triangle ABC and a point D, let N be a vector perpendicular to the vector AB (if AB = (a,b), we can take N = (-b,a)). If the dot products and have different signs, stop; D is not inside the triangle. If they have the same sign, continue to the vector BC and repeat the process. D is inside the triangle iff the three pairs of dot products computed as above all have the same sign. === Subject: Re: How can I tell if a point is enclosed in a triangle? >>I have a triangle with verts (x1,y1),(x2,y2) and(x3,y3). The center of the >>triangle is (x0,y0). Is there an equation which tells me if the point >>(x4,y4) lies inside the triangle? > Given triangle ABC and a point D, let N be a vector perpendicular to the > vector AB (if AB = (a,b), we can take N = (-b,a)). If the dot products > and have different signs, stop; D is not inside the triangle. > If they have the same sign, continue to the vector BC and repeat the > process. D is inside the triangle iff the three pairs of dot products > computed as above all have the same sign. that should work fine. if you are interested in speed maybe aragorn. === Subject: Re: Sides of a right triangle given only area and hypotenuse? Is this subject of the newsgroup not working or what? I still see none of my postmessages about this hypotenuse area problem. In fact I see no posts of any of its responses except the original post, or the last one I tried to quote and comment for. G C === Subject: Re: Sides of a right triangle given only area and hypotenuse? > None of my messages seem to appear on the newsgroup about my responses to > this. > > > The problem can be solved. I made a sign error in my previous try. > hypotenuse h and area A are known; the sides x and y are unknown. > > (1/2)*x*y=A > h*h=x*x+y*y > from which obtain two expressions for y: > > y=(h^2 - x^2)^(1/2) > y=2*A/x > > combine by equating these both equal to y: > > 2*A/x = (h^ - x^2)^(1/2) > then simplify: > > x^4 - (h^2)(x^2) + 4*A^2 = 0 > > This can have one or two solution, since both x and y, and h and A must be > greater than zero. > > You can continue that degree 4 equation into: > > x^2 = [h^2 +|- sqrt(h^4 - 16*A^2)]/2 > > One may make the best sense of this when you have starting values for A and > h; > you could even try graphing the two equations for y, which become a half > circle > from -h to h and something that resembles a hyperbola in the first > quadrant > which may or may not touch the half circle at one or two points. > > G C > === >Subject: Sides of a right triangle given only area and hypotenuse? >Message-id: Calculate the sides of a right triangle given only the area and hypotenuse >- >is this possible? > > For an Area, A, hypoteneuse, h, and sides x and y, one must have > that 0 < 4*A <= h^2 in order to have real x and y, and then, > assuming 0 < x <= y, one gets > x = [sqrt(h^2 + 4*A) - sqrt(h^2 - 4*A)]/2 and > y = [sqrt(h^2 + 4*A) + sqrt(h^2 - 4*A)]/2 > This form of solution also shows clearly that when h^2 = 4*A, the > triangle is isosceles with x = y = h/sqrt(2) > One may verify that these are solutions to the equation cited obove, > x^4 - (h^2)(x^2) + 4*A^2 = 0, by direct substitution, and they are, > in fact, merely alternate forms of the positive solutions to the > equation x^2 = [h^2 +|- sqrt(h^4 - 16*A^2)]/2 shown above. AS the form of solutions , assuming x <= y, x = min{x,y} = [sqrt(h^2 + 4*A) - sqrt(h^2 - 4*A)]/2 y = max{x,y} = [sqrt(h^2 + 4*A) + sqrt(h^2 - 4*A)] are not in the usual form expected from solving something like x^4 - (h^2)(x^2) + 4*A^2 = 0, one might wonder how to find them in that form. One way to get those solutions is to linearize the original system of equations for the problem, { 2*A = x*y, x^2 + y^2 = h^2 }. This can be done by assuming 0 < v <= u and setting max{x,y} = sqrt(u) + sqrt(v) and min{x,y} = sqrt(u) - sqrt(v). Then x*y = u - v and x^2 + y^2 = 2*u + 2*v so one only needs to solve the system { u - v = 2*A , 2*u + 2*v = c^2 } for u and v and then back substitute to get x and y. Such neat linearizations are not always possible. === Subject: Re: Sides of a right triangle given only area and hypotenuse? Carmine (Bottom posters be darned!) > It's actually a programming project and I am so far past high school that my > algebra is long forgotten. Trig, I am still pretty good at; algebra - no > way. :) > posted in an earlier newsgroup message. > This one seems to produce the correct solution for one side of the triangle > with the other side easily found using the obvious method: > a=SQRT(2)*(c^2-SQRT(c^4-16*p^2))*SQRT(SQRT(c^4-16*p^2)+c^2)/(8*p) > where 'c' is the hypotenuse and 'p' is the area. > Now, as much as I hate to ask, is there a solution where the hypotenuse and > the *perimiter* are known? >Calculate the sides of a right triangle given only the area and > hypotenuse - >is this possible? >It is certainly possible. It's even easy, but knowing the path from here >>to there could be tricky if you're not up on your (high-school) algebra. >>Convince me it isn't homework, and I'll post the simple solution. >>Dale > Question: > Now, as much as I hate to ask, is there a solution > where the hypotenuse and the *perimiter* are known? > Answer: > Yes. > Dale > Spoiler follows (many lines down for the do-it-yourself crowd): >>SPOILER ALERT<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< >SPOILER ALERT<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< >>SPOILER ALERT<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< > Let a,b,c be the lengths of the three sides of the > right triangle, with c being the hypotenuse: > a^2 + b^2 = c^2. > We suppose that the following are known: > c, p = a+b+c > Note that p^2 can be expanded: > p^2 = (a + b + c) = a^2 + b^2 + c^2 + 2(ab + bc + ac) > But we know c^2 = a^2 + b^2, so this is > p^2 = 2 c^2 + 2(ab + bc + ac) > = 2( c^2 + ab + bc + ac) > Next, bring the terms containing c as a factor all together > and factor out c: > p^2 = 2( (a + b + c) c + ab) > and notice that p = a + b + c appears, as if by magic. > Take advantage of that to get > p^2 = 2 (p c + ab) > and find > 2 ab = p^2 - 2pc > Thus, we can find the area: A = ab/2 = (p^2 - 2pc)/4. > Now, we know the hypotenuse and the area. Here's the > solution I hid from yesterday's post: > Using (again) a^2 + b^2 = c^2, note that the square > (a + b)^2 = a^2 + 2 ab + b^2 > is secretly equal to > = c^2 + 2 ab > Thus (A = area = ab/2): > 4A = (a + b)^2 - c^2 > Turning this around (and using the fact that a,b,c are all > non-negative), we find: > a + b = sqrt(c^2 + 4A) > ab = 2A > leading to the quadratic equation > S^2 - sqrt(c^2 + 4A) S + 2A = 0. > The solutions are > S = (sqrt(c^2 + 4A) + sqrt(c^2 - 4A))/2 > and > S = (sqrt(c^2 + 4A) - sqrt(c^2 - 4A))/2 > In terms of the original problem, recall that A is > equal to > A = p (p - 2c)/4 > to get the following formulas for the sides a and b > S = (sqrt(c^2 + p(p-2c)) + sqrt(c^2 - p(p-2c)))/2 > and > S = (sqrt(c^2 + p(p-2c)) - sqrt(c^2 - p(p-2c)))/2 === Subject: Re: Help with exercise in Dunford & Schwartz Linear Operators >> For example, given r, choose f infinitely differentiable with compact >> support such that f^(r) <> 0. Then f cannot be in the range of >> irI - T. Suppose to the contrary that g is an element of D and >> g' - ir g = f. If we knew that g was, say, in L^1 and actually >> absolutely continuous then we could conclude that >> (g' - ir g)^(r) = 0, giving a contradiction. We don't know >> that g is that nice but this argument convinces me that >> g' - ir g cannot equal f regardless... >> >> Ah. g' - ir g = f shows that >> >> (exp(-irt) g(t))' = exp(-irt) f(t). >> >> But the integral of (exp(-irt) g(t))' from -infinity to infinity >> must be zero, since exp(-irt) g(t) is in D, while the integral >> of exp(-irt) f(t) is nonzero. >So then I just need to remember that g(t) must go to zero at plus and >minus infinity to conclude that the first integral you mention above vanishes, >right? > Yup. >Since you explained that one to me several years ago, > I did? Huh. >I guess I'm > So here's a quiz question: Where does the argument use the > fact that r is real? (The notation f^(r) assumes that, but > that's not essential - for any complex r one can find a suitable > f so that the integral of exp(-irt) f(t) is non-zero.) If r is not real then exp(-irt) is not bounded and exp(-irt)g(t) may not be in L^p and therefore not in D. === Subject: Re: Help with exercise in Dunford & Schwartz Linear Operators > For example, given r, choose f infinitely differentiable with compact > support such that f^(r) <> 0. Then f cannot be in the range of > irI - T. Suppose to the contrary that g is an element of D and > g' - ir g = f. If we knew that g was, say, in L^1 and actually > absolutely continuous then we could conclude that > (g' - ir g)^(r) = 0, giving a contradiction. We don't know > that g is that nice As long as we're beating this to death (maybe I should say as long as _I'm_ beating it to death) I should point out that this was a little dense - actually it's very easy to see that g has compact support, hence it _is_ in L^1 and _is_ absolutely continuous: Say f is supported in [a,b]. Then there exists c such that g(t) = c exp(irt) for t > b; since g is in L^p c must be zero. Similarly for t < a, so g is supported in [a,b]. >but this argument convinces me that > g' - ir g cannot equal f regardless... > > Ah. g' - ir g = f shows that > > (exp(-irt) g(t))' = exp(-irt) f(t). > > But the integral of (exp(-irt) g(t))' from -infinity to infinity > must be zero, since exp(-irt) g(t) is in D, while the integral > of exp(-irt) f(t) is nonzero. >>So then I just need to remember that g(t) must go to zero at plus and >>minus infinity to conclude that the first integral you mention above vanishes, >>right? >> Yup. >>Since you explained that one to me several years ago, >> I did? Huh. >>I guess I'm >> So here's a quiz question: Where does the argument use the >> fact that r is real? (The _notation_ f^(r) assumes that, but >> that's not essential - for any complex r one can find a suitable >> f so that the integral of exp(-irt) f(t) is non-zero.) >If r is not real then exp(-irt) is not bounded and exp(-irt)g(t) may not be in L^p and >therefore not in D. ************************ David C. Ullrich === Subject: Re: Open Problem 7 Points - any Progress? see http://www.isthe.com/chongo/tech/math/n-cluster/index.html You might like to check this reference out. I did the programming work and found the examples, and had Landon check my work. The existance of 7 points might be checked, if we had an efficient way to find Heron triangles sharing a common side. You simply pick 5 such triangles and then check conditions on the 3rd point not on the common side, so that none are on a circle or line. Conditions get a bit more sticky as we climb up the dimensions. We did a lot of searching, Landon had access to Amdahl supercomputers, so the search space was quite extensive, I believe the extents went past 27,000. Randall === Subject: Re: Open Problem 7 Points - any Progress? > see http://www.isthe.com/chongo/tech/math/n-cluster/index.html > You might like to check this reference out. the people in the related threads I mentioned. All my questions were answered: Q1 is answered by your definition of prime clusters: prime n-cluster: An n-cluster where the greatest common divisor of the mutual distances = 1 Q2 had a serious typo. I wanted to ask for a 6-cluster. And again I want to say thank you for (0,0) (546,272) (132,720) (960,720) (546,-1120) (1155,-540) and the list of others you provide. Q3 has the same answer Rainer Rosenthal r.rosenthal@web.de === Subject: Re: Open Problem 7 Points - any Progress? > prime n-cluster: An n-cluster where the greatest > common divisor of the mutual distances = 1 Oh well, on second thought this was not really what I was asking for with my question Q1. A prime n-cluster is just one, which is not constructed out of a smaller n-cluster by simply magnifying by an integer factor. And for n > 4 there is no n-cluster such that all mutual distances x, y have gcd(x,y)=1, since there are at least two with common factor 3. Repeating my question Question #1 But I wonder whether general position is meant in a broader way than (1)? So that *any* special positions are ruled out. I can now give a better explanation for it. Let me define The special points of a circle are those on the periphery and the one in the center And let me define An n-cluster is called euclidean whenever there are 3 points on a straight line or 4 special points of a circle. For the moment I am quite happy with these definitions, since they express my feeling about general positions without childishly trying to redefine something already defined in a precise manner. Any point set satisfying the non-euclidean condition is automatically an n-cluster, while there are n-clusters which are euclidean. This might be of some interest when looking at this fascinating topic as a tool for investigating euclidean geometry. To say it plainly: wouldn't it be nice, if one of the well know lovely theorems of plane geometry could tell, why there are no n-clusters with n=7? Rainer Rosenthal r.rosenthal@web.de === Subject: Plotting a graph Suppose I have recorded some distances (they're actually pointer distances between two objects in memory). Since these distances could be very large, I have separated them into buckets which are at intervals of Log base 2. So I then have buckets such as: 0, 2, 4, 16, ... to 2147483648. Distance data such as: 34, 16, 15, 100 etc will be in buckets 32, 16, 16 and 64 respectively. I then keep a count for each bucket to know the number of distances that belong to each one of them (actual distances are not required). I want to be able to plot this information on a graph. On the y-axis I need to place these distances and on the x-axis I want to keep the number of times these distances are sampled (i.e. the total number of pointer mutations.. this can go to millions!). Now someone adviced me that I normalize the y-axis (i.e. the distances that should be recored as percentages) in terms of the largest heap size encountered (i.e. the denominator in the normalization should be the same for all points on the x axis). What exactly does this mean? What should I do? Should I divide the bucket sizes by the largest heap size, and then depending upon the number of distances that were measured for that particular bucket I take a percentage of the overall mutations that there were? Could someone guide me as to how I should do this and use the given Tiff === Subject: Re: Map two triangles by transforming bounding rectangle david macneil ... > _____A_____________ > | x | > | x x | > | x x | > | x x | > | x x |C > |x x | > B x------------------- > What I now need to do is to work out the new locations for the corners > of the original bounding rectangle such that I can map the points A, B > & C onto any given new locations to create a new triangle of known > coordinates. I might not completely understand the context here, but I would just look at the two triangles first. Suppose first A and its image A' are just (0,0). We get two linear equations to describe B' and C' in terms of B and C, say in the form of a matrix M. Next, for arbitrary A, let T be the translation which takes A to the origin, and U the translation which takes the origin to A'. The desired mapping or matrix will be UMT. The bounding rectangle of A'B'C' is now of course determined by inequalities in terms of the coefficients of UMT. Sorry for the loose and hasty writing :) LH === Subject: Re: Mathematica: LCM for variable length list? Are you using Mathematica 5? In M5, you can type LCM[a_1,a_2,...]. > [this question originally posted to comp.soft-sys.math.mathematica] > What is the syntax to have Mathematica calculate the LCM for a list? > For example: if myList = {2, 3, 4, 5, 6}, I want to get 60 back. > The list length varies at runtime so I can't list the elements > explicitly in the code like LCM[ arg1, arg2, arg3, ... ]. > I tried passing the list like this: LCM[myList], but it returns the > original list instead of a single number (perhaps the LCM is operating > on each element of the list individually). > Any help greatly appreciated. > Mark > -- > Mark Lookabaugh > mlookabaugh (at) cox.net > USS Brewton FF-1086 Home Page > http://www.ussbrewton.com === Subject: Re: Mathematica: LCM for variable length list? >In various functional programming languages you'd want to use >apply here. So I looked it up, and sure enough that's what you >want in Mathematica: >Apply[LCM, yourList] Mark -- Mark Lookabaugh mlookabaugh (at) cox.net USS Brewton FF-1086 Home Page http://www.ussbrewton.com === Subject: Re: One Giant Leap > One Giant Leap > A bue soldier OMYGODTHESHRUBBERYISREVOLTING! -- If you have had problems with Illinois Student Assistance Commission (ISAC), please contact shredder at bellsouth dot net. There may be a class-action lawsuit in the works. === Subject: Re: Surprising people, my prime counting Excerpts from A crank infests sci.math It is a Mathematical Amateur And he posteth one of three. 'Why do you start so many threads Concerning F L T? It's already proved, by Andrew Wiles, Just a few years back. I trust it's true, but twixt me and you, The proper skills I lack.' 'With nothing more than High School Math, There is a Proof.' quoth he. 'Hold off, you crackpot loon, That cannot possibly be!' The newsgroup is spellbound by the trolling of the Amateur. The audience opens every post, They cannot resist the lure. The nonsense flows and flows and flows . . . The Amateur makes a discovery. The primes are here. The primes are there. The primes are all around. 'My Functional Defination counts All that can be found.' A Mathematician offers support and constructive criticism. At length replied Odlyzko Who agrees the counting's right. 'But correctness is not enough, The value is new insight.' The Amateur inhospitably calleth the Mathematician a liar. 'God save me' cries the Amateur, That they'll lie for social reasons, Even great Odlyzko.' . . . Knowledge, knowledge everywhere, The Internet's filled with links. Knowledge, knowledge everywhere, Alas, he never thinks. The Amateur becomes deranged. The core is rotten: O Christ! That ever should this be! They're teaching faulty math At the university. . . . Oops, I hear the phone ringing. Gotta go. -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Surprising people, my prime counting > Excerpts from *applauds* Jack Rudd === Subject: Re: Surprising people, my prime counting Here's my prime number counting method. ;-) http://www.primenumbertingbear.com/ > Excerpts from > *applauds* > Jack Rudd === Subject: Re: Surprising people, my prime counting > And yes I'm publicly calling Professor Odlyzko a liar. > Where did he lied? > #1 The main problem I had with your prime counting function is that I > did not see how it helped gain any significant insight.... That's probably the easiest whopper to focus on, and it's so obvious that it's almost sad. Here the difference between my discovery and *any* other found in human history is that it leads to a partial differential equation. Now it turns out that mathematicians have been trying to work out exactly how the prime counting distribution connects to continuous functions for well over a hundred years. They've come up with connections with the best and most important work by a guy called Chebyshev (spellings vary), but Riemann gets a lot of attention, and still they work with approximations that tantalize. The Riemann Hypothesis is now considered to be one of the major unsolved math problems, with a million dollars attached to solving it, but what was Riemann doing? He was looking for a connection between the prime distribution--the count of prime numbers like how 2, 3, 5, and 7 are the 4 primes up to 10--and continuous functions like x/ln(x), but he didn't find exactly what he sought. His now famous hypothesis was basically an afterthought, which he didn't consder all that important. The gist of what mathematicians have accomplished is that they can place the count of prime numbers in a box. So now here I come with something NEVER SEEN BEFORE IN MATH HISTORY, which is a prime counting function which leads to a partial differential equation, and Odlyzko gives his lie. But you see, I *may* have discovered the reason. That is, maybe it is THE reason that mathematicians looked for over such a long time. Why would they lie about it then? Well, oddly enough, it may not be the reason they wanted. James Harris === Subject: Re: Surprising people, my prime counting > > And yes I'm publicly calling Professor Odlyzko a liar. > > Where did he lied? > > #1 The main problem I had with your prime counting function is that I > did not see how it helped gain any significant insight.... > that it's almost sad. Professor Odlyzko stated an *opinion*: ... I did not see how it helped gain any significant insight ... An opinion cannot possibly be a lie. The opinion may be wrong, but it is subjective and cannot be disproved. The person thinks what he thinks. It is by no definition a lie. What's sad is that you malign Dr. Odlyzko by calling his personal opinion a whopper. > Here the difference between my discovery and *any* other found in > human history is that it leads to a partial differential equation. The implications here are provably false, for three reasons: 1. You derive a difference equation. That leads to your differential equation. You are absolutely, definitely not the first in human history to derive a differential equation from a difference equation. 2. You have errors in the derivation of your differential equation. I have mentioned this before and you corrected one of the errors. A very important error however still remains. Hint: chain rule. 3. In general, the solution to a differential equation derived from a difference equation does not provide a solution to the original difference equation. In general they are not strongly related. Examples have been given to you which demonstrate this. You have not the faintest wisp of a proof that the solution of your differential equation is related to the prime counting function or Riemann's function. In the utter absence of proof you continue to pretend you have made a great discovery. Odlyzko is exactly right: you have not provided any significant insight. That's MY opinion. > Now it turns out that mathematicians have been trying to work out > exactly how the prime counting distribution connects to continuous > functions for well over a hundred years. > They've come up with connections with the best and most important work > by a guy called Chebyshev (spellings vary), but Riemann gets a lot of > attention, and still they work with approximations that tantalize. The prime number theorem has been proved, and there is even a so-called elementary proof by Erdos and Selberg. For many purposes that is good enough. Littlewood proved a truly remarkable theorem which says essentially that the answer given by the prime number theorem is very good for large enough n. > The Riemann Hypothesis is now considered to be one of the major > unsolved math problems, with a million dollars attached to solving it, > but what was Riemann doing? > He was looking for a connection between the prime distribution--the > count of prime numbers like how 2, 3, 5, and 7 are the 4 primes up to > 10--and continuous functions like x/ln(x), but he didn't find exactly > what he sought. > His now famous hypothesis was basically an afterthought, which he > didn't consder all that important. > The gist of what mathematicians have accomplished is that they can > place the count of prime numbers in a box. > So now here I come with something NEVER SEEN BEFORE IN MATH HISTORY, > which is a prime counting function which leads to a partial > differential equation, and Odlyzko gives his lie. Odlyzko was exactly right: first, because he was stating his OPINION, not an absolute fact; second, because you provided no hint of proof that the solution to your D.E. approximates the prime counting function; third, because you don't have the slightest idea how to solve your D.E.. And anyway, as I noted above, you made errors in deriving it. > But you see, I *may* have discovered the reason. That is, maybe it is > THE reason that mathematicians looked for over such a long time. > Why would they lie about it then? No lying is involved. You have gone from a recursive difference equation to a differential equation. Let me temporarily ignore the fact that your derivation of the latter is incorrect. There is no general theorem connecting the solution to such a D.E. back the the original difference equation. This has been proved by counterexamples, including one from your most loyal follower, David Ullrich. Not only is there no general theorem, you do not have a shred of proof for your specific D.E. In short, what you have is an erroneous germ of an idea which probably is not going to work, even if you find a solution to your D.E. For this you want credit ??? Andrzej > Well, oddly enough, it may not be the reason they wanted. > James Harris === Subject: Re: Surprising people, my prime counting > Here the difference between my discovery and *any* other found in > human history is that it leads to a partial differential equation. What partial differential equation? Please post it and put your evidence where your mouth is. Otherwise, admit that you do not know what you are talking about. (Tacit admission by failure to respond will be considered conclusive!) -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Surprising people, my prime counting >> And yes I'm publicly calling Professor Odlyzko a liar. >> Where did he lied? >> #1 The main problem I had with your prime counting function is that I >> did not see how it helped gain any significant insight.... >that it's almost sad. He said he did not _see_ how your function led to any significant insight. To know that this is a lie you'd have to be able to read his mind. But anyway: >Here the difference between my discovery and *any* other found in >human history is that it leads to a partial differential equation. You have _never_ given _any_ indication of why the solution (if there is one) to that pde should have anything whatever to do with counting primes. (Also it's _not_ a partial differential equation, although that's a much less important point.) >Now it turns out that mathematicians have been trying to work out >exactly how the prime counting distribution connects to continuous >functions for well over a hundred years. >They've come up with connections with the best and most important work >by a guy called Chebyshev (spellings vary), but Riemann gets a lot of >attention, and still they work with approximations that tantalize. >The Riemann Hypothesis is now considered to be one of the major >unsolved math problems, with a million dollars attached to solving it, >but what was Riemann doing? >He was looking for a connection between the prime distribution--the >count of prime numbers like how 2, 3, 5, and 7 are the 4 primes up to >10--and continuous functions like x/ln(x), but he didn't find exactly >what he sought. >His now famous hypothesis was basically an afterthought, which he >didn't consder all that important. >The gist of what mathematicians have accomplished is that they can >place the count of prime numbers in a box. >So now here I come with something NEVER SEEN BEFORE IN MATH HISTORY, >which is a prime counting function which leads to a partial >differential equation, and Odlyzko gives his lie. >But you see, I *may* have discovered the reason. That is, maybe it is >THE reason that mathematicians looked for over such a long time. >Why would they lie about it then? >Well, oddly enough, it may not be the reason they wanted. Or maybe the things you say are _maybe_ true are not so. >James Harris ************************ David C. Ullrich === Subject: Re: Surprising people, my prime counting > And yes I'm publicly calling Professor Odlyzko a liar. > Where did he lied? > #1 The main problem I had with your prime counting function is that I > did not see how it helped gain any significant insight.... > that it's almost sad. > Here the difference between my discovery and *any* other found in > human history is that it leads to a partial differential equation. I explained that many times to you - your equation of is not a PDE, because it does not fit the definition of PDE - http://mathworld.wolfram.com/PartialDifferentialEquation.html Is is so hard to understand? If you use sqrt(y) or x/y as function arguments - it is not a PDE. There are lots of examples of PDEs of that page - anyone can see the truth. === Subject: Re: Surprising people, my prime counting >> And yes I'm publicly calling Professor Odlyzko a liar. >> Where did he lied? >> #1 The main problem I had with your prime counting function is that I >> did not see how it helped gain any significant insight.... > That's probably the easiest whopper to focus on, and it's so obvious > that it's almost sad. An important subtlety: If I actually believe in leprechauns and I say I think leprechauns exist I am not lying, even though leprechauns don't actually exist. It is merely my opinion, not a 'truth.' For it to be a lie, I would have to actually deep down not believe in leprechauns while voicing the claim that I actually do. Likewise, Professor Odlyzko has merely related his opinion of your discovery. For him to be lying, he would have to secretly actually know that there is a significant insight. In order to tell that it was a significant insight, he would almost certainly have to know what that insight was. However, you yourself have admitted that you don't know what this significant insight is, just that it may lead to one. Unless Professor Odlyzko knows more about your own discovery than you, he doesn't know either. Hence he does not know what your significant insight is, his statement is an accurate reflection of his opinion, and he is therefore not lying. These kinds of things are important when you want to throw around accusations like that. (followup adjusted) === Subject: Re: Surprising people, my prime counting > However, I'm somewhat pessimistic as I remind, people *trust* > mathematicians. > And for good reason, jerk. The mathematicians present their proofs > publically so can be read and cricized by anyone. > Bob Kolker Yeah, like you actually check and understand. It's become a religion. When someone like me challenges the status quo people repeat the dogma. So the dogma is that mathematicians publish their work, as if that proves it must be correct, but then, only other mathematicians usually check. Who polices them? Now I found a way to count prime numbers. I'm not asking for a ticker tape parade, or to be crowned king of the realm. I'm asking for mathematicians to do what most people would probably expect, which is dutifully record the result. Instead they try to act like I didn't find *anything* at all worth recording!!! So someone like this Robert Kolker posts angrily and calls me a name, and I say to other readers that something has gone terribly wrong in the math world!!! Mathematicians are behaving in a way that shows they *know* the power they have to control what you believe is true. Now what if your kid found a way to count prime numbers that others had missed? Wouldn't you want mathematicians to do their jobs? But what if they didn't. What could you do? Think about it. They have the power, and so far it might seem easier to them to use it. But the public gave them the power. James Harris === Subject: Re: Surprising people, my prime counting >> However, I'm somewhat pessimistic as I remind, people *trust* >> mathematicians. >> And for good reason, jerk. The mathematicians present their proofs >> publically so can be read and cricized by anyone. >> Bob Kolker >Yeah, like you actually check and understand. Uh, yes we do. The fact that _you_ cannot understand what you read in math journals (or would be unable to if you read something in a math journal) does not imply that all the rest of us are incapable of this. >It's become a religion. > When someone like me challenges the status quo people repeat the >dogma. >So the dogma is that mathematicians publish their work, as if that >proves it must be correct, but then, only other mathematicians usually >check. >Who polices them? >Now I found a way to count prime numbers. I'm not asking for a ticker >tape parade, or to be crowned king of the realm. >I'm asking for mathematicians to do what most people would probably >expect, which is dutifully record the result. >Instead they try to act like I didn't find *anything* at all worth >recording!!! >So someone like this Robert Kolker posts angrily and calls me a >name, and I say to other readers that something has gone terribly >wrong in the math world!!! >Mathematicians are behaving in a way that shows they *know* the power >they have to control what you believe is true. >Now what if your kid found a way to count prime numbers that others >had missed? Wouldn't you want mathematicians to do their jobs? >But what if they didn't. What could you do? Think about it. They >have the power, and so far it might seem easier to them to use it. >But the public gave them the power. >James Harris ************************ David C. Ullrich === Subject: Re: Primes and the Collatz conjecture! > Another interesting find is --- > Starting with powers of 2 where > n = > (((2^4)*10)-1)/3 = 53 > (((2^8)*10)-1)/3 = 853 > (((2^10)*10)-1)/3 = 3413 > (((2^16)*10)-1)/3 = 218453 > (((2^20)*10)-1)/3 = 3495253 > (((2^22)*10)-1)/3 = 13981013 where n= a prime starting numbers in a > Collatz path. 2^4 * 10 is the same as 2^5 * 5. Why the distinction? Because 5 is the root of the branch. When you look at a branch in binary, the root is 101 and every number higher up is formed by simply appending a 0: 10100000000000 1010000000000 101000000000 10100000000 1010000000 101000000 10100000 1010000 101000 10100 1010 101 The root pattern 101 is preserved across all members of the branch. Since 5 == 2 (mod 3) and under multiplication by 2, the successor of 2 (mod 3) is 1 (mod 3) and the successor of 1 (mod 3) is 2 (mod 3) the numbers in the branch alternate between 2 (mod 3) and 1 (mod 3). Sub-branches only attach at 1 (mod 3), so every alternate member of the 5 branch has a sub-branch attached, starting with the second. 10100000000000_ 1010000000000 101000000000_ 10100000000 1010000000_ 101000000 10100000_ 1010000 101000_ 10100 1010_ 101 Each successive sub-branch is related to the previous one by 4n+1, and the successor rules for 4n+1 are - the binary pattern appends 01 - n (mod 3) is succeeded by n+1 (mod 3) Once we know that the first sub-branch is 3 (0 (mod 3)), we know all the rest 10100000000000_110101010101 2 (mod 3) 1010000000000 101000000000_1101010101 1 (mod 3) 10100000000 1010000000_11010101 0 (mod 3) 101000000 10100000_110101 2 (mod 3) 1010000 101000_1101 1 (mod 3) 10100 1010_11 0 (mod 3) 101 > and continuing on -- avoiding 2^6, 2^12, 2^18, 2^24, 2^30,... these > will not produce a prime because n == 0 mod 3. > These prime starting # (seed) produce the following similar sequences Each new sub-branch is going to pass through all the numbers of the previous sub-branch. That's why the sequences are similar. > --- > E.g. > 53, 160, 80, 40, 20, 10 , 5, 16, 8, 4, 2, 1 > 853, 2560, 1280, 640, 320, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 > 13981013, 41943040, 20971520, 10485760, 5242880, 2621440, 1310720, > 655360, 327680, 163840, 81920, 40960, 20480, 10240, 5120, 2560, 1280, > 640, 320, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 > I am checking to see if there are numerous primes for larger powers of > 2. > If there is then these primes could be called Collatz seed primes. Where I thought this was going was some insight into the relationship between prime numbers and binary patterns. But when you list the COMPLETE tree structure, there is no apparent pattern. It looks like you just got lucky in that a bunch of the lower sub-branches happened to be prime. That pattern evaporates as you move up the branch, even allowing for the fact that all 0 (mod 3) are not prime. n (n - 1)/3 prime? 005629499534213120 002814749767106560 00000938249922368853 == 0 (mod 3) 001407374883553280 000703687441776640 00000234562480592213 No 1009 * 232470248357 000351843720888320 000175921860444160 00000058640620148053 No 733 * 80000846041 000087960930222080 000043980465111040 00000014660155037013 == 0 (mod 3) 000021990232555520 000010995116277760 00000003665038759253 Yes 000005497558138880 000002748779069440 00000000916259689813 No 13 * 70481514601 000001374389534720 000000687194767360 00000000229064922453 == 0 (mod 3) 000000343597383680 000000171798691840 00000000057266230613 No proven composite 000000085899345920 000000042949672960 00000000014316557653 No 41 * 349184333 000000021474836480 000000010737418240 00000000003579139413 == 0 (mod 3) 000000005368709120 000000002684354560 00000000000894784853 No proven composite 000000001342177280 000000000671088640 00000000000223696213 No 13 * 17207401 000000000335544320 000000000167772160 00000000000055924053 == 0 (mod 3) 000000000083886080 000000000041943040 00000000000013981013 Yes 000000000020971520 000000000010485760 00000000000003495253 Yes 000000000005242880 000000000002621440 00000000000000873813 == 0 (mod 3) 000000000001310720 000000000000655360 00000000000000218453 Yes 000000000000327680 000000000000163840 00000000000000054613 No 13 * 4201 000000000000081920 000000000000040960 00000000000000013653 == 0 (mod 3) 000000000000020480 000000000000010240 00000000000000003413 Yes 000000000000005120 000000000000002560 00000000000000000853 Yes 000000000000001280 000000000000000640 00000000000000000213 == 0 (mod 3) 000000000000000320 000000000000000160 00000000000000000053 Yes 000000000000000080 000000000000000040 00000000000000000013 Yes 000000000000000020 000000000000000010 00000000000000000003 Yes, == 0 (mod 3) 000000000000000005 > The next one is at (((2^40)*10)-1)/3 = 3665038759253 is also prime. But what is the significance of that? Can you predict which of the next 100 sub-branches are prime? I thought you had something interesting until I noticed you only mentioned the successes and glossed over the failures. To properly understand something, you should show ALL the data. The patterns formed by the failures can be just as significant as those formed by the successes. > Dan -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Primes and the Collatz conjecture! > Another interesting find is --- > Starting with powers of 2 where > n = > (((2^4)*10)-1)/3 = 53 > (((2^8)*10)-1)/3 = 853 > (((2^10)*10)-1)/3 = 3413 > (((2^16)*10)-1)/3 = 218453 > (((2^20)*10)-1)/3 = 3495253 > (((2^22)*10)-1)/3 = 13981013 where n= a prime starting numbers in a > Collatz path. > 2^4 * 10 is the same as 2^5 * 5. Why the distinction? Because 5 is the > root of the branch. When you look at a branch in binary, the root is > 101 and every number higher up is formed by simply appending a 0: > 10100000000000 > 1010000000000 > 101000000000 > 10100000000 > 1010000000 > 101000000 > 10100000 > 1010000 > 101000 > 10100 > 1010 > 101 > The root pattern 101 is preserved across all members of the branch. > Since 5 == 2 (mod 3) and under multiplication by 2, the > successor of 2 (mod 3) is 1 (mod 3) and the > successor of 1 (mod 3) is 2 (mod 3) > the numbers in the branch alternate between 2 (mod 3) and 1 (mod 3). > Sub-branches only attach at 1 (mod 3), so every alternate member of > the 5 branch has a sub-branch attached, starting with the second. > 10100000000000_ > 1010000000000 > 101000000000_ > 10100000000 > 1010000000_ > 101000000 > 10100000_ > 1010000 > 101000_ > 10100 > 1010_ > 101 > Each successive sub-branch is related to the previous one by 4n+1, and > the successor rules for 4n+1 are > - the binary pattern appends 01 > - n (mod 3) is succeeded by n+1 (mod 3) > Once we know that the first sub-branch is 3 (0 (mod 3)), we know all the > rest > 10100000000000_110101010101 2 (mod 3) > 1010000000000 > 101000000000_1101010101 1 (mod 3) > 10100000000 > 1010000000_11010101 0 (mod 3) > 101000000 > 10100000_110101 2 (mod 3) > 1010000 > 101000_1101 1 (mod 3) > 10100 > 1010_11 0 (mod 3) > 101 > and continuing on -- avoiding 2^6, 2^12, 2^18, 2^24, 2^30,... these > will not produce a prime because n == 0 mod 3. > These prime starting # (seed) produce the following similar sequences > Each new sub-branch is going to pass through all the numbers of the > previous sub-branch. That's why the sequences are similar. > --- > E.g. > > 53, 160, 80, 40, 20, 10 , 5, 16, 8, 4, 2, 1 > > 853, 2560, 1280, 640, 320, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 > > 13981013, 41943040, 20971520, 10485760, 5242880, 2621440, 1310720, > 655360, 327680, 163840, 81920, 40960, 20480, 10240, 5120, 2560, 1280, > 640, 320, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 > > I am checking to see if there are numerous primes for larger powers of > 2. > If there is then these primes could be called Collatz seed primes. > Where I thought this was going was some insight into the relationship > between prime numbers and binary patterns. But when you list the > COMPLETE tree structure, there is no apparent pattern. It looks like > you just got lucky in that a bunch of the lower sub-branches happened to > be prime. That pattern evaporates as you move up the branch, even > allowing for the fact that all 0 (mod 3) are not prime. > n (n - 1)/3 prime? > 005629499534213120 > 002814749767106560 00000938249922368853 == 0 (mod 3) > 001407374883553280 > 000703687441776640 00000234562480592213 No 1009 * 232470248357 > 000351843720888320 > 000175921860444160 00000058640620148053 No 733 * 80000846041 > 000087960930222080 > 000043980465111040 00000014660155037013 == 0 (mod 3) > 000021990232555520 > 000010995116277760 00000003665038759253 Yes > 000005497558138880 > 000002748779069440 00000000916259689813 No 13 * 70481514601 > 000001374389534720 > 000000687194767360 00000000229064922453 == 0 (mod 3) > 000000343597383680 > 000000171798691840 00000000057266230613 No proven composite > 000000085899345920 > 000000042949672960 00000000014316557653 No 41 * 349184333 > 000000021474836480 > 000000010737418240 00000000003579139413 == 0 (mod 3) > 000000005368709120 > 000000002684354560 00000000000894784853 No proven composite > 000000001342177280 > 000000000671088640 00000000000223696213 No 13 * 17207401 > 000000000335544320 > 000000000167772160 00000000000055924053 == 0 (mod 3) > 000000000083886080 > 000000000041943040 00000000000013981013 Yes > 000000000020971520 > 000000000010485760 00000000000003495253 Yes > 000000000005242880 > 000000000002621440 00000000000000873813 == 0 (mod 3) > 000000000001310720 > 000000000000655360 00000000000000218453 Yes > 000000000000327680 > 000000000000163840 00000000000000054613 No 13 * 4201 > 000000000000081920 > 000000000000040960 00000000000000013653 == 0 (mod 3) > 000000000000020480 > 000000000000010240 00000000000000003413 Yes > 000000000000005120 > 000000000000002560 00000000000000000853 Yes > 000000000000001280 > 000000000000000640 00000000000000000213 == 0 (mod 3) > 000000000000000320 > 000000000000000160 00000000000000000053 Yes > 000000000000000080 > 000000000000000040 00000000000000000013 Yes > 000000000000000020 > 000000000000000010 00000000000000000003 Yes, == 0 (mod 3) > 000000000000000005 > > The next one is at (((2^40)*10)-1)/3 = 3665038759253 is also prime. > But what is the significance of that? Can you predict which of the > next 100 sub-branches are prime? I thought you had something > interesting until I noticed you only mentioned the successes and > glossed over the failures. To properly understand something, you > should show ALL the data. The patterns formed by the failures can > be just as significant as those formed by the successes. > > Dan on any math involved here or any other math-related subject I am sure. My only objective here was basically to show these special primes that reside in this special sequence as the starting (seed) integer for this one path in the Collatz tree. Where each node off this path is only one and that is where all of these special odd primes and 0 (mod) 3's and other odd composites with prime factors reside thus creating only 3 odd integers, the (seed), prime 5, and 1 in their entire path starting with (seed) 3. I am just picking the primes related by one node to this one path. Also there is no 0 (mod) 5 in any of these odd composites (seeds). As in the Mersenne primes, not all-prime exponents of 2 create another prime. The same is true here, but with a twist, not all-even exponents (n) where (2^n*10-1)/3 will create another prime. You also have to consider the *10 factor of the Collatz (seed) primes when comparing densities of the two. Sorry if I misled anyone here into thinking there was some kind of pattern, but I never had that intention. I am doing a comparison table of the Mersenne primes and the Collatz (seed) primes. Yes, there is many more even exponents creating the Collatz (seed) prime then the Mersenne prime exponents, so that will explain the density factor between the two. A question also remains, is they're a Mersenne prime or Mersenne primes member(s) of this path in the Collatz tree? Dan electron-dot-cloud are galaxies === Subject: Re: muon catalyzed fusion: stickyness + muon radii = tokamak Coulomb Re: Coulomb barrier becomes Fusion Barrier Principle; compounding of Maxwell Equations Such that in tokamak fusion we have the Gauss law which is the Coulomb barrier. So we have one sphere of the Coulomb barrier and we have the enclosed cylinder of the Tokamak which can be generalized as Faraday's Law. So the maximum volume, surface area is 2/3 which translates into a maximum breakeven of 2/3. But in Muon catalyzed fusion we have the Coulomb barrier even though it is ameliorated by the muonic atom radius. So we have one sphere as the Coulomb barrier. But we have a second sphere of Stickyness because stickyness is also the Gauss law of Coulomb force. So we can equate the tokamak-coulomb with the two coulombs of muon fusion. Where is Faraday's or Ampere's Law in muon fusion? It is in the fact of creating muons in the first place. One can think of a muon as a electron that is power-packed. You take an ordinary electron and commit Faraday's or Ampere's Law upon that ordinary electron and turn it into a muon. Muon is sort of like a capacitor that stores extra energy coming from Faraday's law. So in a sense, muon catalyzed fusion replaces the tokamak machinery and imbues the energy from Faraday's or Ampere's laws into changing ordinary electrons into muons. Archimedes Plutonium, a_plutonium@hotmail.com whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: A non algebraic system of equations Boy, you guys sure make this math stuff seem easy. :) turned it into an impossible one. Impossible to me anyway. I thought the answer would have involved some horrible equations involving arcsins and arccos and imaginary numbers divided by transcendentals. -Paul Klemstine === Subject: Re: Face facts, many Americans *like* idea of killing > Ha. Yeah, they did sort of tinker with the details a bit. Guess > that's how it has to be when you only have 128 pages or so to > work with. > Ed Howdershelt - Abintra Press > Science Fiction and Semi-Fiction > A little off topic from this particular thread, but when did > Semi-Fiction officially become a recognized genre? Is it > restricted to science specifically or can the mix include a > combination of any or all genre's? Just curious. >> Genre? Recognized? By definition, perhaps. >> Don't know if a 'Semi-Fiction' genre exists. >> Don't really care, either. :) >> I use the term 'Semi-Fiction' to describe those of my works in >> which only names of people and certain places have been changed to >> avoid lawsuits. See titles Kim, Mindy, Anne, and Field >> Decision. >> Ed Howdershelt - Abintra Press >> Science Fiction and Semi-Fiction >> http://abintrapress.tripod.com >> think of it? It seems that changing the names of people and places >> isn't always enough. I was going to jokingly mention a couple of >> semi-celebs in my book but now I'm not sure if it's worth the risk. >> http://www.publishlawyer.com/carousel4.htm >> 2poor > The segment about privacy issues would pertain to my stuff. > I've cleared Kim, Mindy, and Anne with the women involved and accepted > many of their suggested revisions, and none of them have called their > lawyers. > Field Decision's key female character died in 2001, but she had no > objections to what I sent her for review. > Guess I'll wait and see if anyone sues. > If nothing else, there may be a book in it. > Ed Howdershelt - Abintra Press > Science Fiction and Semi-Fiction > http://abintrapress.tripod.com Yeah, I can see how it could get a bit tricky, especially when a lawyer starts out with: Well, maybe. I think I'm safe, though. Talk show hosts joke about public figures all the time. 2poor === Subject: Re: This Week's Finds in Mathematical Physics (Week 197) Originator: baez@math-cl-n01.math.ucr.edu (John Baez) >> tmf(n) is the space of supersymmetric conformal field theories >> of central charge -n. >Negative central charge means that the representations of the >Virasoro and super-Virasoro algebras are not unitary. Is this not >an issue here? Aaron Bergman's wonderful reference to a supersymmetric conformal field theory of central charge 24 with the Monster group as symmetries, when I had been implicitly wishing for one with central charge -24, makes this even more plausible. Part of the problem is that there are two sign conventions concerning the weights of modular forms: a modular form of weight k is also said to have degree -k. There's also a nasty factor of two floating around: some people say the weight is k/2 when others say it's k. This stuff got me all defining the central charge of a conformal field theory to be *minus* the weight of its partition function for the torus. But maybe it's just the weight, or twice the weight, or something... If someone could take me out of this agony, I'd be grateful. Let's see: in week126 I convinced myself that a 1-component massless free scalar field in 2 dimensions is a conformal field theory whose partition function for the torus is the famous Dedekind eta function: eta(q) = q^{1/24} sum_n (1 - q^n) and staring at some other stuff on the web, it looks like the the 24th power of this function is a modular form of weight 12, according to my favorite normalization, where the Eisenstein series E_n has weight n. So, eta(q) is sort of like a modular form of weight 1/2. That doesn't quite make sense, but it's okay for straightening out these normalizations. What's the central charge of this field theory? Well, browsing the web some more - a pathetic way to find this information, but I'm far from my favorite books - it sounds like people usually declare it to be 1. So, it sounds like a conformal field theory with central charge c has a partition function for the torus which is a modular form of weight w = c/2. So, I maybe should have said: tmf(n) is the space of supersymmetric conformal field theories of central charge 2n. Then everything seems to fit better. ................................................................ === Subject: The Magic of 24 A few months ago, John Baez had a great series of TWF where he discussed the origin of the 26 in bosonic string theory, eventually tying in the cancellation of the conformally anomaly (the sum of the central charges of the gauge-fixing bc CFT and the matter CFT) into monstrous moonshine. I'm not sure I got all of that, but having worked my way up to chapter 6 in Polchinki's book, I'm curious about the magic of the number 24. This seems to me, at first glance to be a disparate issue. It seems like it should tie into the 26, but I can't see off the top of my head how. Regardless, the famed Dedekind eta function appears in the calculation of the partition function on the torus. More specifically, we get (from Pol. eq 7.2.9) Z(t) = (4pi^2alpha't_2)^{-1/2} |eta(t)|^{-2} Now, because eta is a modular form (for something like Gamma_0(4)) of weight (1/2,0), and t_2 is a non-holomorphic form of weight (-1,-1), this has a total weight of zero thus being invariant under the modular group as one needs in string theory. Now, the eta function is quite an interesting object. The expression that is arrived at in Polchinski's book is the infinite product q^{1/24} Product_{n=1}^{Infinity} (1 - q^n) Now, what surprised me upon seeing this is how the weird but necessary q^(1/24) appears. Now, I have to admit that the 24 here seems rather magical even without the string theory figuring in. Now, when one deals with nice simple modular forms, it turns out the the space of forms is 1-dimensional, spanned by the Eisenstein series, up to the 12th dimension where there is the discriminant (2Pi)^{12} Delta = ----------- (E_4(z)^3 - E_6(z)^2) 12^3 (using the notations and normalizations of Zagier's book). This formula is overflowing with 12s, which appears to me because 12 is the least common multiple of 4 and 6, the first two nontrivial forms. That's dreadfully bland, though. (Twelve also appears, of course, in the zeta-function regularization of Sum n. Any connection?) Anyways, Zagier introduces the Dedekind eta function seemingly out of nowhere and shows that its 24th power is the discriminant using its transformation properties and the previous derivation of the dimension of the space of weight 12 forms. I find this horribly unelightening. Any suggestions? I don't remember finding Gunning's book any more helpful. Back to the physics. The partition function on the torus is derived in Polchinski 7.2: Consider the path intergral with no vertex operators, <1>_{T^2(t)} defeq Z(t). We can think of the torus with modulus t [really a tau] as formed by taking a field theory on a circle, evolving for Euclidean time 2pi t_2, translating in sigma^1 by 2pi t_1, and then identifying the ends. In operator language this gives a trace, (7.2.5) Z(t) = Tr[ exp(2pi i t_1 P - 2pi t_2 H] = (qqbar)^{-d/24} Tr(q^{L_0} qbar^{Lbar_0}) The L's here are the Virasoro generators and Polchinski has used the expressions P = L_0 + Lbar_0 and H = L_0 + Lbar_0 + (1/24)(c+cbar). This is where the magic 24 comes from, which makes everything nice and modularly invariant. So, now, we hit conformal field theory. My knowledge of that subject is unfortunately fairly sparse, limited to chapter 2 of Polchinski and a few _very_ brief forays into Di Francesco et al. I think I can use the stuff in Polchinski to formulate my question. The above relations for P and H ultimately related to the non-tensorial transformation of the stress-energy tensor in a conformal field theory, Polchinski's equation 2.4.26: (@_z z')^2 T'(z') = T(z) - (c/12) {z',z} where the funny bracket thing an ugly thing called the Schwarzian derivative which is left markedly unmotivated in Polchinski's book. Anyways, that 12 there comes from the TT OPE and the Ward identity and ultimately comes from the fact that 12 = 2*3!. An extra factor of two comes from {e^z, z} and, poof, 24. Eek. What does this all mean? It feels like there should be something deep going on here. In the process of writing this post, I noticed that if I squint _really_ hard, the Schwarzian derivative looks a bit like the reciprocal of the j-function. Or maybe not. It's all very weird, so, after hopefully not insulting anyone's intelligence with all this exposition (which I also hope is all correct) : What's so magic about 24? Aaron -- Aaron Bergman === Subject: Re: TOBILL.com has 1 bid http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=2748838713&category=11153 > TOBILL.com > If bill.com is waiting for $10,000,000 this will take its place meantime. > TOBILL.com very short generic financial com domain ideal for money service, > banking, debt collection, online bills payment . Buyer needs to issue a transfer > (costs under $10 from GoDaddy and adds one year rego), registered to Aug > Please bid now ROFL, you really are house at domain squatting. === Subject: Re: TOBILL.com has 1 bid [---spam---] Send those complaints to: 1cust64.tnt2.townsville.au.da.uu.net 210.84.108.64 UUNET Technologies, Inc. (UU-DOM) 3060 Williams Drive Ste 601 Fairfax, VA 22031 US === Subject: Re: TOBILL.com has 1 bid Hattan sent a radio signal across the vast expanse of deep space -._.--._.--._.--._.--._.--._. >TOBILL.com >If bill.com is waiting for $10,000,000 this will take its place meantime. >TOBILL.com very short generic financial com domain ideal for money service, >banking, debt collection, online bills payment . Buyer needs to issue a transfer >Please bid now > Spam reported to |-|erc's ISP. I recommend others do the same. We'll see happens with his ISP tonight on the Truman Show. Be sure to tune in ;-) -- apatriot #1, atheist #1417, Chief EAC prophet - Evil Atheist Conspiracy http://members.optusnet.com.au/~pk1956/ Shhh. Be very quiet, I'm hunting automorons. Heh heh. Properly read, the Bible is the most potent force for atheism ever conceived. - Isaac Asimov Fingerprint for PGP Keys at key server or go to http://members.optusnet.com.au/~pk1956/ RSA - 71 BA 7C 45 B5 4A 5F EA 72 DB EC 7F 7F A8 70 99 DSS - 9217 21A9 9C3F EB0B E302 AD0E 69C5 0F06 402E 0943 === Subject: Re: TOBILL.com has 1 bid http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=2748838713&category=11153 > TOBILL.com > If bill.com is waiting for $10,000,000 this will take its place meantime. > TOBILL.com very short generic financial com domain ideal for money service, > banking, debt collection, online bills payment . Buyer needs to issue a transfer > (costs under $10 from GoDaddy and adds one year rego), registered to Aug > Please bid now > Herc Dropped on your head as a baby were you ? Boc === Subject: Re: TOBILL.com has 1 bid why ? it's a valid Auction, you can't report someone for being dumb >TOBILL.com >If bill.com is waiting for $10,000,000 this will take its place meantime. >TOBILL.com very short generic financial com domain ideal for money service, >banking, debt collection, online bills payment . Buyer needs to issue a transfer >(costs under $10 from GoDaddy and adds one year rego), registered to Aug >Please bid now Spam reported to |-|erc's ISP. I recommend others do the same. --- John Hattan Grand High UberPope - First Church of Shatnerology john@thecodezone.com http://www.shatnerology.com === Subject: Re: TOBILL.com has 1 bid >why ? it's a valid Auction, you can't report someone for being dumb Correct, but I can report him for spamming items in unrelated newsgroups. Most TOS departments like to enforce newsgroup charters. --- John Hattan Grand High UberPope - First Church of Shatnerology john@thecodezone.com http://www.shatnerology.com === Subject: Re: TOBILL.com has 1 bid At some point in the past, John Hattan slavered, and posted this: >> why ? it's a valid Auction, you can't report someone for being dumb > Correct, but I can report him for spamming items in unrelated > newsgroups. Most TOS departments like to enforce newsgroup charters. What's actually surprising is that there is a bid on this item lol. -- Doug Semler http://home.wideopenwest.com/~doug_semler a.a. #705, BAAWA. EAC Guardian of the Horn of the IPU (pbuhh). I hate spam, standard email address munging applied. 42 DNRC o- === Subject: Re: TOBILL.com has 1 bid > At some point in the past, John Hattan slavered, > and posted this: >> why ? it's a valid Auction, you can't report someone for being dumb > Correct, but I can report him for spamming items in unrelated > newsgroups. Most TOS departments like to enforce newsgroup charters. > What's actually surprising is that there is a bid on this item lol. > -- > Doug Semler ever hear of schrill bidding ? Boc === Subject: Re: TOBILL.com has 1 bid @nnrp1.ozemail.com.au: > ever hear of schrill bidding ? Yeah, it's kinda high-pitched. Mostly done by women, and effeminate gay men. G === Subject: Re: TOBILL.com has 1 bid At some point in the past, Boc slavered, and posted this: >> At some point in the past, John Hattan slavered, and posted this: >> why ? it's a valid Auction, you can't report someone for being dumb > Correct, but I can report him for spamming items in unrelated > newsgroups. Most TOS departments like to enforce newsgroup charters. >> What's actually surprising is that there is a bid on this item lol. > ever hear of schrill bidding ? I think the term is shill. :) -- Doug Semler http://home.wideopenwest.com/~doug_semler a.a. #705, BAAWA. EAC Guardian of the Horn of the IPU (pbuhh). I hate spam, standard email address munging applied. 42 DNRC o- === Subject: Re: TOBILL.com has 1 bid > At some point in the past, John Hattan slavered, > and posted this: >> why ? it's a valid Auction, you can't report someone for being dumb > Correct, but I can report him for spamming items in unrelated > newsgroups. Most TOS departments like to enforce newsgroup charters. > What's actually surprising is that there is a bid on this item lol. I'm one of a million people who transfer $100s each month through postbillpay.com.au I'd be just as content if not more so to use tobill.com Its one of the best debt collectors names, I searched debt collectors but unfortunately its not an esatablished web enterprise yet, most of the results that show up are credit agencies. What would you value gy.net? ouch.com? 100.org? mule.com? Try several thousand dollars each. Financial names can be very valuable, its good enough to have a legitimate web presence, I'm only skimming potential buyers on newsgroups I know but like I said my power gets cut in 4 days, since its beyond anyones diginity to stand against the all tortuing US government that claims it owns me and you're all brainwashed something chronic by them to all turn on me. 100,000 people in townsville for the last year have all tuned in to the myriad of pointers that media all follows one man, how every story leads in and interacts, I've pointed out numerous explicit examples, you're all blind absolutely blind. But my ISP, but my electricity, totally ignore 1000's of post directly from the truman asking for assistance for years, if my alias was Bob it would have been solved already, you're all over war or something over me trying to see a ING LADY Herc === Subject: Re: TOBILL.com has 1 bid > At some point in the past, John Hattan slavered, > and posted this: > why ? it's a valid Auction, you can't report someone for being dumb Correct, but I can report him for spamming items in unrelated > newsgroups. Most TOS departments like to enforce newsgroup charters. What's actually surprising is that there is a bid on this item lol. > I'm one of a million people who transfer $100s each month through postbillpay.com.au > I'd be just as content if not more so to use tobill.com > Its one of the best debt collectors names, I searched debt collectors but unfortunately > its not an esatablished web enterprise yet, most of the results that show up are > credit agencies. What would you value gy.net? ouch.com? 100.org? mule.com? > Try several thousand dollars each. Financial names can be very valuable, its > good enough to have a legitimate web presence, I'm only skimming potential > buyers on newsgroups I know but like I said my power gets cut in 4 days, since > its beyond anyones diginity to stand against the all tortuing US government that > claims it owns me and you're all brainwashed something chronic by them to all > turn on me. 100,000 people in townsville for the last year have all tuned in to the > myriad of pointers that media all follows one man, how every story leads in and > interacts, I've pointed out numerous explicit examples, you're all blind absolutely blind. > But my ISP, but my electricity, totally ignore 1000's of post directly from the truman > asking for assistance for years, if my alias was Bob it would have been solved > already, you're all over war or something over me trying to see a ING LADY > Herc aaah that's right - this is the clown that thinks he is in his own Truman show in Townsville - and everyone in the town knows of him. Seen anyone in white uniforms and carrying butterfly nets at your door ? Boc === Subject: Re: TOBILL.com has 1 bid > At some point in the past, John Hattan slavered, > and posted this: > why ? it's a valid Auction, you can't report someone for being dumb Correct, but I can report him for spamming items in unrelated > newsgroups. Most TOS departments like to enforce newsgroup charters. What's actually surprising is that there is a bid on this item lol. I'm one of a million people who transfer $100s each month through > postbillpay.com.au > I'd be just as content if not more so to use tobill.com > Its one of the best debt collectors names, I searched debt collectors but > unfortunately > its not an esatablished web enterprise yet, most of the results that show > up are > credit agencies. What would you value gy.net? ouch.com? 100.org? > mule.com? > Try several thousand dollars each. Financial names can be very valuable, > its > good enough to have a legitimate web presence, I'm only skimming potential > buyers on newsgroups I know but like I said my power gets cut in 4 days, > since > its beyond anyones diginity to stand against the all tortuing US > government that > claims it owns me and you're all brainwashed something chronic by them to > all > turn on me. 100,000 people in townsville for the last year have all tuned > in to the > myriad of pointers that media all follows one man, how every story leads > in and > interacts, I've pointed out numerous explicit examples, you're all blind > absolutely blind. > But my ISP, but my electricity, totally ignore 1000's of post directly > from the truman > asking for assistance for years, if my alias was Bob it would have been > solved > already, you're all over war or something over me trying to see a ING > LADY > Herc > aaah that's right - this is the clown that thinks he is in his own Truman > show in Townsville - and everyone in the town knows of him. > Seen anyone in white uniforms and carrying butterfly nets at your door ? > Boc aus.tv know it! I'm from Townsville and YOU ARE the Truman! http://tinyurl.com/iky5 I was in Townsville over the weekend, and I heard him. Very spooky! http://tinyurl.com/iky8 >phone someone in Townsville, half of you must know someone there, >every day I go out people say THERES THE TRUMAN I'm in Townsville. We're sick of you. http://tinyurl.com/iky9 http://tinyurl.com/iky4 You rule Truman! http://tinyurl.com/fufb 1998 : The Truman Show starring Jim Carrey http://tinyurl.com/fpg4 2002 : The Majestic (The Truman Show 2!!) Jim Carrey costars Laurie Holden Now remember Jarod the Pretender is the series version of The Truman Show movie, and Jarod always gets back people who go against him. I might be under 2 years of continuous hideous torture from the Center right now, but they don't leave marks, they just want full control, but all I need is cash flow to get away from their mind torture. I video conference with the tv and I'm every hollywood stars idol for 3 years all doing their artistic best to portray my emotion, then when it comes down to it they all put it on and noone wants me to decide when enough torture is enough. All the wealthy rely on the government to remain entrenched, all the common folk have been blitzed by their free world front, there's no people parachuting down to me with you're on camera, when Jim Carrey left the dome city the millions of onlookers follow their tv trance, don't associate with the truman or we can't give you tv. let me spell it out 1 you're all stupid 2 you can't examine black and white evidence 3 you all think every person is the same 4 you're all cruel 5 its 2 minutes effort from a minority of people on the internet and my TORTURE ends!!!!!! 6 for $5,000 I could be away from continuous interogation and with my new family within a year 7 its been PUBLIC 2 years already 8 the truman will be 35 before he meets jen again, the only autogyrating hip man on earth forced into celibacy by the government from age 26 to 32 and counting. 9 these are 8 of 100 simple and equally absurdly straight forward facts that compell me into disbelief NONE OF YOU can examine what US government is doing to me Herc go on complain some more someone posted in your newsgroup, i didn't spell interrogation right so it doesn't count, the the government introduced corporal punishment continously every second for 2 years then it must be good, someone didn't say atheism is the bees knees quick phone the ISP quick you ing girls what a bunch of ing jerks Spam reported to |-|erc's ISP. I recommend others do the same. Biffo, you'll cost me ANOTHER 6 months of greater torture than you can handle for 2 minutes, a satelite picks up the residual trace of your thought to your voice box, and warps it back , you have to listen to everything, i cant SEEE i have to continuously hear every second, 5 women tell me off every morning in the toilet, WTF for BUSH you ing loony 100,000 witnesses you ing stupid internet morons === Subject: Re: TOBILL.com has 1 bid Restrict: binaries to pr0n only |-|erc loose on on a computer... and the unfortunate result was: :::: At some point in the past, John Hattan :::: slavered, :::: and posted this: ::::: :::::: why ? it's a valid Auction, you can't report someone for being :::::: dumb ::::: ::::: Correct, but I can report him for spamming items in unrelated ::::: newsgroups. Most TOS departments like to enforce newsgroup ::::: charters. ::::: :::: :::: :::: What's actually surprising is that there is a bid on this item lol. :::: ::: ::: I'm one of a million people who transfer $100s each month through :: postbillpay.com.au ::: I'd be just as content if not more so to use tobill.com ::: ::: Its one of the best debt collectors names, I searched debt ::: collectors but :: unfortunately ::: its not an esatablished web enterprise yet, most of the results ::: that show :: up are ::: credit agencies. What would you value gy.net? ouch.com? 100.org? :: mule.com? ::: Try several thousand dollars each. Financial names can be very ::: valuable, :: its ::: good enough to have a legitimate web presence, I'm only skimming ::: potential ::: buyers on newsgroups I know but like I said my power gets cut in 4 ::: days, :: since ::: its beyond anyones diginity to stand against the all tortuing US :: government that ::: claims it owns me and you're all brainwashed something chronic by ::: them to :: all ::: turn on me. 100,000 people in townsville for the last year have ::: all tuned :: in to the ::: myriad of pointers that media all follows one man, how every story ::: leads :: in and ::: interacts, I've pointed out numerous explicit examples, you're all ::: blind :: absolutely blind. ::: ::: But my ISP, but my electricity, totally ignore 1000's of post ::: directly :: from the truman ::: asking for assistance for years, if my alias was Bob it would have ::: been :: solved ::: already, you're all over war or something over me trying to see a ::: ING :: LADY ::: ::: Herc :: :: :: :: aaah that's right - this is the clown that thinks he is in his own :: Truman :: show in Townsville - and everyone in the town knows of him. :: :: Seen anyone in white uniforms and carrying butterfly nets at your :: door ? :: :: Boc :: :: : : aus.tv know it! : : I'm from Townsville and YOU ARE the Truman! : http://tinyurl.com/iky5 : : : I was in Townsville over the weekend, and I heard him. : Very spooky! : http://tinyurl.com/iky8 : : :: phone someone in Townsville, half of you must know someone there, :: every day I go out people say THERES THE TRUMAN : I'm in Townsville. We're sick of you. : http://tinyurl.com/iky9 : : : http://tinyurl.com/iky4 : You rule Truman! : : : : http://tinyurl.com/fufb 1998 : The Truman Show starring Jim Carrey : looks like Laurie Holden * http://tinyurl.com/fpg4 2002 : The : Majestic (The Truman Show : 2!!) Jim Carrey costars Laurie Holden : : : : Now remember Jarod the Pretender is the series version of The Truman : Show movie, : and Jarod always gets back people who go against him. : : I might be under 2 years of continuous hideous torture from the : Center right now, : but they don't leave marks, they just want full control, but all I : need is cash flow to : get away from their mind torture. I video conference with the tv and : I'm every : hollywood stars idol for 3 years all doing their artistic best to : portray my emotion, : then when it comes down to it they all put it on and noone wants me : to decide when : enough torture is enough. All the wealthy rely on the government to : remain entrenched, : all the common folk have been blitzed by their free world front, : there's no people parachuting down to me with you're on camera, : when Jim Carrey left the dome city : the millions of onlookers follow their tv trance, don't associate : with the truman or we can't give you tv. : : let me spell it out : : 1 you're all stupid : 2 you can't examine black and white evidence : 3 you all think every person is the same : 4 you're all cruel : 5 its 2 minutes effort from a minority of people on the internet and : my TORTURE ends!!!!!! 6 for $5,000 I could be away from continuous : interogation and with my new family within a year : 7 its been PUBLIC 2 years already : 8 the truman will be 35 before he meets jen again, the only : autogyrating hip man on earth forced into celibacy by the : government from age 26 to 32 and counting. 9 these are 8 of 100 : simple and equally absurdly straight forward facts that compell me : into disbelief NONE OF YOU can examine what US government is doing to : me : : Herc : : : go on complain some more someone posted in your newsgroup, i didn't : spell interrogation right so it doesn't count, the the government : introduced corporal punishment : continously every second for 2 years then it must be good, someone : didn't : say atheism is the bees knees quick phone the ISP quick you ing : girls : : what a bunch of ing jerks : : Spam reported to |-|erc's ISP. I recommend others do the same. : : : Biffo, you'll cost me ANOTHER 6 months of greater torture than you : can : handle for 2 minutes, a satelite picks up the residual trace of your : thought : to your voice box, and warps it back , you have to listen to : everything, : i cant SEEE i have to continuously hear every second, 5 women tell : me : off every morning in the toilet, WTF for BUSH you ing loony : : 100,000 witnesses you ing stupid internet morons Hey, I believe you! === Subject: Re: Circulant matrices over GF(2) > What is the number of n X n invertible circulant matrices over GF(2) ? > Did you know the following? > ID Number: A003473 (Formerly M0875) > URL: http://www.research.att.com/projects/OEIS?Anum=A003473 > Sequence: 1,2,3,8,15,24,49,128,189,480,1023,1536,4095,6272,10125, > 32768,65025,96768,262143,491520,583443,2095104,4190209, > 6291456,15728625,33546240 > Name: Generalized Euler PHI function. > Comments: a(n) is the number of n X n circulant invertible matrices over > GF(2) . - > References J. T. B. Beard Jr. and K. I. West, Factorization tables for x^n-1 > over > GF(q), Math. Comp., 28 (1974), 1167-1168. > See also: Sequence in context: A007919 A069752 A004731 this_sequence > A056802 > A026698 A011946 > Adjacent sequences: A003470 A003471 A003472 this_sequence A003474 > A003475 A003476 > Keywords: nonn,new > Offset: 1 > Author(s): njas No, what is the relation between Euler phi and circulant matrices ? === Subject: Re: Circulant matrices over GF(2) > What is the number of n X n invertible circulant matrices over GF(2) ? > Did you know the following? > ID Number: A003473 (Formerly M0875) > URL: http://www.research.att.com/projects/OEIS?Anum=A003473 > Sequence: 1,2,3,8,15,24,49,128,189,480,1023,1536,4095,6272,10125, > 32768,65025,96768,262143,491520,583443,2095104,4190209, > 6291456,15728625,33546240 > Name: Generalized Euler PHI function. > Comments: a(n) is the number of n X n circulant invertible matrices over > GF(2) . - > References J. T. B. Beard Jr. and K. I. West, Factorization tables for x^n-1 > over > GF(q), Math. Comp., 28 (1974), 1167-1168. > See also: Sequence in context: A007919 A069752 A004731 this_sequence > A056802 > A026698 A011946 > Adjacent sequences: A003470 A003471 A003472 this_sequence A003474 > A003475 A003476 > Keywords: nonn,new > Offset: 1 > Author(s): njas > No, what is the relation between Euler phi and circulant matrices ? I'm not sure about the definition of generalized Euler phi function, but in this case the number of n x n non-singular circulants over GF(2) is equal to the number of degree n polynomials in GF(2)[x] that are relatively prime to x^n + 1. So I would guess that the generalized euler phi function of the polynomial f(x) gives the number of polynomials of degree at most equal to the degree of f(x) which are relative prime to f(x). === Subject: Re: How to transform spherical/polar coordinates to Cartesian for n-dimensions? Please don't top post. I've restored the correct order below. in message : > in message <2rh1b.31872$0u4.12349@news1.central.cox.net>: > I am looking for the method to transform spherical coordinates of > arbitrarily high dimensions (n) to Cartesian coordinates (of the same > number > of dimensions of course). I have found how to go from polar (2d) to > Cartesian, and spherical (3d) to Cartesian, but I have not yet found > the > n-dimension solution. It is my intent to implement this in Java for a > program I am writing. Any help or links would be greatly appreciated. > The usual scheme, modulo renumbering of indices, is: > x_1 = r sin(a_1) > x_2 = r cos(a_1) sin(a_2) > x_3 = r cos(a_1) cos(a_2) sin(a_3) > ... > x_n = r cos(a_1) cos(a_2) cos(a_3) ... cos(a_{n-1}) > Of course, r >= 0, -pi/2 <= a_i <= pi/2 except -pi < a_{n-1} <= pi > Was that last line supposed to be: > x_n = r cos(a_1) cos(a_2) cos(a_3) ... cos(a_{n-1}) > or > x_n = r cos(a_1) cos(a_2) cos(a_3) ... sin(a_{n-1}) > or > x_n = r cos(a_1) cos(a_2) cos(a_3) ... cos(a_{n-1}) sin(a_n) The first. More specifically: x_n = r cos(a_1) cos(a_2) ... cos(a_{n-2}) cos(a_{n-1}) > ? It looked like in the x_1 through x_3 that the last factor was always > going to be sin of a_n... Yes. That's true for all x_i except the last one. For each x_i after the first, you change the sine at the end of the previous x_{i-1} to a cosine, and for all x_i except the last one you add the sine of a new angle a_i. To answer your other question, x_1 = x, x_2 = y, x_3 = z, ... However you want to label the Cartesian coordinates doesn't really matter. For more than 3 or 4 dimensions, it's usually more convenient to use numbered indices instead of different letters. -- Jim Heckman === Subject: Re: How to transform spherical/polar coordinates to Cartesian for n-dimensions? Forgive this short top post :) I understand it now - thank you very much! Now I will post a question about n-dimensional simplexes... Craig > Please don't top post. I've restored the correct order below. > in message : in message <2rh1b.31872$0u4.12349@news1.central.cox.net>: I am looking for the method to transform spherical coordinates of > arbitrarily high dimensions (n) to Cartesian coordinates (of the same > number > of dimensions of course). I have found how to go from polar (2d) to > Cartesian, and spherical (3d) to Cartesian, but I have not yet found > the > n-dimension solution. It is my intent to implement this in Java for a > program I am writing. Any help or links would be greatly appreciated. The usual scheme, modulo renumbering of indices, is: x_1 = r sin(a_1) > x_2 = r cos(a_1) sin(a_2) > x_3 = r cos(a_1) cos(a_2) sin(a_3) > ... > x_n = r cos(a_1) cos(a_2) cos(a_3) ... cos(a_{n-1}) Of course, r >= 0, -pi/2 <= a_i <= pi/2 except -pi < a_{n-1} <= pi > Was that last line supposed to be: > x_n = r cos(a_1) cos(a_2) cos(a_3) ... cos(a_{n-1}) > or > x_n = r cos(a_1) cos(a_2) cos(a_3) ... sin(a_{n-1}) > or > x_n = r cos(a_1) cos(a_2) cos(a_3) ... cos(a_{n-1}) sin(a_n) > The first. More specifically: > x_n = r cos(a_1) cos(a_2) ... cos(a_{n-2}) cos(a_{n-1}) > ? It looked like in the x_1 through x_3 that the last factor was always > going to be sin of a_n... > Yes. That's true for all x_i except the last one. For each x_i > after the first, you change the sine at the end of the previous > x_{i-1} to a cosine, and for all x_i except the last one you add > the sine of a new angle a_i. > To answer your other question, x_1 = x, x_2 = y, x_3 = z, ... > However you want to label the Cartesian coordinates doesn't > really matter. For more than 3 or 4 dimensions, it's usually > more convenient to use numbered indices instead of different > letters. > -- > Jim Heckman === Subject: Re: How to transform spherical/polar coordinates to Cartesian for n-dimensions? Also, that just computes the x value? What about the other n-1 dimensional values? Craig > in message <2rh1b.31872$0u4.12349@news1.central.cox.net>: > I am looking for the method to transform spherical coordinates of > arbitrarily high dimensions (n) to Cartesian coordinates (of the same > number > of dimensions of course). I have found how to go from polar (2d) to > Cartesian, and spherical (3d) to Cartesian, but I have not yet found the > n-dimension solution. It is my intent to implement this in Java for a > program I am writing. Any help or links would be greatly appreciated. > The usual scheme, modulo renumbering of indices, is: > x_1 = r sin(a_1) > x_2 = r cos(a_1) sin(a_2) > x_3 = r cos(a_1) cos(a_2) sin(a_3) > ... > x_n = r cos(a_1) cos(a_2) cos(a_3) ... cos(a_{n-1}) > Of course, r >= 0, -pi/2 <= a_i <= pi/2 except -pi < a_{n-1} <= pi > -- > Jim Heckman === Subject: Cartesian coordinates for verticies of a n-dimensional simplex? I am want to construct simplexes for n-dimensions. A Simplex is the simplest shape that can define a dimension. For example, a 2d simplex is an equilateral triangle. A 3d simplex is a 3-sided pyramid with a base, etc. I am interested only in a regular, unit simplexes, i.e., all vertices are equidistant from all other vertices, and each vertex is exactly 1 unit away from the origin that the simplex is centered on. I have found a broken link to David Anderson's paper on the n-dimensional simplex, but that doesn't Craig Garrett === Subject: quaternions-- what's the point? it seems to me that a quaternion is nothing more than a simple ordered quadruplet. And that multiplication/addition is simply extended in such a way that it is defined over such ordered quadruplets. It's as if I were to write a geometry book where I called points binarinions and write them as a#{(b}) instead of (a,b).. the point being that it would be arbitrary and accomplish nothing and, in fact, greatly obfuscate something for no reason. This, it seems, is the entire nature of quaternions. === Subject: Re: quaternions-- what's the point? > it seems to me that a quaternion is nothing more than a simple > ordered quadruplet. And that multiplication/addition is simply > extended in such a way that it is defined over such ordered > quadruplets. It's as if I were to write a geometry book where I > called points binarinions and write them as a#{(b}) instead of > (a,b).. the point being that it would be arbitrary and accomplish > nothing and, in fact, greatly obfuscate something for no reason. > This, it seems, is the entire nature of quaternions. So Hamilton spent years pondering a trivial problem, and subsequently came up with a entirely trivial result? The point is rotations. H is a generalization of C. While the unit ball in C rotates R^2, the unit ball in H rotates R^3. This comparison naturally extends to the corresponding rotatios groups. While the unit ball in C is isomorphic to SO(2), the unit ball in H is isomorphic to SU(2), which is a two fold cover of SO(3). When considered as algebras, C and H are instances of the Clifford algebra, which rotates quadratic spaces of all dimensions. M.T. === Subject: Re: quaternions-- what's the point? >it seems to me that a quaternion is nothing more than a simple >ordered quadruplet. And that multiplication/addition is simply >extended in such a way that it is defined over such ordered >quadruplets So far, so good. But then you go way off into left field: >It's as if I were to write a geometry book where I >called points binarinions and write them as a#{(b}) instead of >(a,b).. No, because there you are inventing nonstandard nomenclature for an existing concept. WRH invented a new concept, and needed new nomenclature to describe it. The fact that in the 21st Century we have more useful tools than Quaternions does not take away any of the importance the concept had when Hamilton invented them. >the point being that it would be arbitrary and accomplish nothing >and, in fact, greatly obfuscate something for no reason. Your nomenclature, yes; Quaternions, no. >This, it seems, is the entire nature of quaternions. No, just the nature of your understanding of them. At the time they were a useful tool, better than what was available, and even today they are useful in special cases. Instead of prejudging their utility and rationale from ignorance, you would do better to do some reading. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: quaternions-- what's the point? > it seems to me that a quaternion is nothing more than a simple > ordered quadruplet. And that multiplication/addition is simply > extended in such a way that it is defined over such ordered > quadruplets. It's as if I were to write a geometry book where I > called points binarinions and write them as a#{(b}) instead of > (a,b).. the point being that it would be arbitrary and accomplish > nothing and, in fact, greatly obfuscate something for no reason. > This, it seems, is the entire nature of quaternions. If you try it with triples or quintuples, you won't be able to define multiplication to obtain good properties. === Subject: Re: quaternions-- what's the point? > it seems to me that a quaternion is nothing more than a simple > ordered quadruplet. And that multiplication/addition is simply > extended in such a way that it is defined over such ordered > quadruplets. ... arbitrary and accomplish > nothing and, in fact, greatly obfuscate something for no reason. > This, it seems, is the entire nature of quaternions. > If you try it with triples or quintuples, you won't be able to define > multiplication to obtain good properties. Um, yes you can. Define (a,b,c,d)(e,f,g,h)=(x0,x1,x2,x3) where x0 = ae-bf-cg-dh x1 = af+be+ch-dg x2 = ag-bh+ce+df x3 = ah+bg-cf+de Boom. Suddenly quaternions are a completely pointless invention. === Subject: Re: quaternions-- what's the point? as noted, for mechanics, they can be used to account for pitch/yaw/roll with position. they can also be formulated as pairs of complex numbers (and, as I recall, the octonions as pairs of quaternions, where the story seems to end .-) I didn't follow your algebraic example: what is the point? > If you try it with triples or quintuples, you won't be able to define > multiplication to obtain good properties. > Um, yes you can. > Define (a,b,c,d)(e,f,g,h)=(x0,x1,x2,x3) where > x0 = ae-bf-cg-dh > x1 = af+be+ch-dg > x2 = ag-bh+ce+df > x3 = ah+bg-cf+de --A church-school McCrusade (Blair's ideals?): Harry-the-Mad-Potter want's US to kill Iraqis?... http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX === Subject: Re: quaternions-- what's the point? >as noted, for mechanics, they can be used >to account for pitch/yaw/roll with position. > they can also be formulated >as pairs of complex numbers (and, as I recall, >the octonions as pairs of quaternions, >where the story seems to end .-) Not so , there are also sedenions and there I think it ends Patrick === Subject: Re: quaternions-- what's the point? >as noted, for mechanics, they can be used >to account for pitch/yaw/roll with position. > they can also be formulated >as pairs of complex numbers (and, as I recall, >the octonions as pairs of quaternions, >where the story seems to end .-) > Not so , there are also sedenions and there I think it ends > Patrick Where it ends depends on what structure you want and what you are willing to give up. If you are not willing to give up an order such that one of x > 0, x = 0, x < 0 holds and x > 0, y > 0 imply x + y > 0 and x*y > 0 then it ends at R. If you are not willing to give up commutativity then it ends at C. If you are not willing to give up associativity then it ends at H. If you want a division algebra it ends at O. -- G.C. === Subject: Re: quaternions-- what's the point? > it seems to me that a quaternion is nothing more than a simple > ordered quadruplet. And that multiplication/addition is simply > extended in such a way that it is defined over such ordered > quadruplets. ... arbitrary and accomplish > nothing and, in fact, greatly obfuscate something for no reason. > This, it seems, is the entire nature of quaternions. > If you try it with triples or quintuples, you won't be able to define > multiplication to obtain good properties. > Um, yes you can. It depends what good properties are. The only associative, quadratic real algebras without divisors of zero are R, C and H. The only Banach division algebras are R, C and H. > Define (a,b,c,d)(e,f,g,h)=(x0,x1,x2,x3) where > x0 = ae-bf-cg-dh > x1 = af+be+ch-dg > x2 = ag-bh+ce+df > x3 = ah+bg-cf+de > Boom. Suddenly quaternions are a completely pointless invention. If you wish to prove the four squares theorem, quaternions come in handy. -- G.C. === Subject: Re: quaternions-- what's the point? > > it seems to me that a quaternion is nothing more than a simple > ordered quadruplet. And that multiplication/addition is simply > extended in such a way that it is defined over such ordered > quadruplets. ... arbitrary and accomplish > nothing and, in fact, greatly obfuscate something for no reason. > This, it seems, is the entire nature of quaternions. > > If you try it with triples or quintuples, you won't be able to define > multiplication to obtain good properties. > Um, yes you can. > Define (a,b,c,d)(e,f,g,h)=(x0,x1,x2,x3) where > x0 = ae-bf-cg-dh > x1 = af+be+ch-dg > x2 = ag-bh+ce+df > x3 = ah+bg-cf+de > Boom. Suddenly quaternions are a completely pointless invention. Quadruples work...yes, indeed. That is what Hamilton discovered after years of effort. But, as stated before, triples or quintuples don't work. === Subject: Re: quaternions-- what's the point? >it seems to me that a quaternion is nothing more than a simple >ordered quadruplet. And that multiplication/addition is simply >extended in such a way that it is defined over such ordered >quadruplets. It's as if I were to write a geometry book where I >called points binarinions and write them as a#{(b}) instead of >(a,b).. the point being that it would be arbitrary and accomplish >nothing and, in fact, greatly obfuscate something for no reason. >This, it seems, is the entire nature of quaternions. Quaternions provide an example of a skew-field structure on R^4. They have applications in number theory and mechanics. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Help with complex exponentiation >Hi all! >> Please help me with this (not homework): >> n^(p+qi) = a + bi >> a = f(n,p,q)= ? >> b = g(n,p,q)= ? >'exp','log','sin' and 'cos'? Without exp, yes. All the others you need. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Transform of 1/(1+X), where X~N(mu,sigma^2)? >I want to transform X->1/(1+X), where X~N(mu,sigma^2). Unfortunately, >1/(1+X) is not differentiable at the origin (the mean of the >tranformed RV) or I would use a Taylor expansion. Do you mean you want to determine the distribution of the transformed random variable? If so, there is no problem in applying the standard transformation formula; see any book on basic probability. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Transform of 1/(1+X), where X~N(mu,sigma^2)? > I want to transform X->1/(1+X), where X~N(mu,sigma^2). Unfortunately, > 1/(1+X) is not differentiable at the origin (the mean of the > tranformed RV) or I would use a Taylor expansion. > Any ideas? let f(x) = 1/(1+x) f'(x) = -1/(1+x)^2 f'(0) = -1 1/(1+x) = 1 - x + x^2 - x^3 + ... (x^2 < 1) === Subject: Re: a channel from God >even christnet just calls me a pagan for daring to say god is a man now.< Yeah, when we all know he's a white, middle class, elderly man with a beard. jc --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: a channel from God In sci.math, |-|erc : > I was just working on my space invaders game and blindly shooting > and looked up to notice a message on the screen. Your application has performed an illegal operation and will be shut down? > A simple message but if you check my site www.adamskingdom.com > on numerology it is very straightforward. Oh. Numerology. Taking it seriously automatically qualifies you as a Grade 3 Moron. (There are of course some recreational puzzles which depend on various aspects of numbers, our representation syste, or both, such as SEND + MORE = MONEY, but those aren't quite in the same genre.) To aspire to Grade 4 status you'll have to add two numbers incorrectly. In order to make this a fair test you are allowed to pick the two numbers. Were I to pick them and you add them incorrectly you'd only aspire to Grade 1 1/2, as I can pick some very hard numbers to add (for example, add e^(i * pi) and the square of the reciprocal of the number of military superpowers in the world who still have some money left :-) ). If you really want to be considered more intelligent, do something intelligent, like explain why God [*] would in fact use numerology when far more straightforward methods are available to do God's bidding. For example, if God really wanted someone to kill [insert victim type here], God would materialize that individual near a large crowd [insert victim type here] with a [insert wide-area nasty weapon lethal to everyone in the immediate vicinity here], and even that would be too complicated, as God could merely wave his magic [insert disgusting body part, fluid, or attribute here] and they'd all be dead or vanished from Earthly existance, no barfy muss, no fuss. A miracle! But wait, you say. Wouldn't people notice? Of course not. God would simply wave his other [insert disgusting body part, fluid, or attribute here] and everyone would immediately forget about the whole thing. If God wished peace on Earth he'd do a variant of the second. I'll admit the notion is nice but not borne out by events, recent or otherwise. It's a pity for I prefer peace, but it's clear that war is sometimes the only option, deadly as it is. Perhaps that will change with the advent of new technology such as brain-freeze implants. The idea here is a simple one: if one commits thoughtcrime (doubleplusungood) a small unit inserted in the back of one's head will simply stop the brain from thinking (the idea might be along the lines of an epileptic seizure although not nearly as debilitating) sufficiently long for the authorities to gain entrance to his location and subdue him using slightly less technologically advanced methods (the unit would also double as a locator beacon). In a pinch the authorities could send an area effect freeze command, and sort out the hostages from the hostage-takers, at their leisure. There are of course many problems with this particular implementation of such a system -- one of them being whether those with frozen brains will ever be unfrozen again. (There is a certain chemical that destroys the motor control system, leaving the user unable to do anything, even breathe. Certain drugs -- Ecstasy? -- can cause this effect over long enough usage. Even pot appears to have some effect on memory. Presumably a packet of this stuff could be broken on command. Not a pleasant, erm, thought.) I predict this idea will take quite awhile to catch on amongst the general populace (I doubt even John Ashcroft or J. Edgar Hoover would seriously entertain or have seriously entertained this idea for long), and I somehow doubt one would need a crystal ball, magic doodlebug, omnidirectional dowsing stick, mystic yoga meditation, or a revelation from above to conclude such. :-) [rest snipped] [*] I'm assuming here a vaguely Christian variant performing miracles, with omnipotence or near-omnipotence. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: The sum = product sequence in the complex plane [reply address is baloglouAToswego.edu] Consider the sequence An defined by A1*A2*...*An = A1+A2+...+An and, consequently, A(n+1) = A1*A2*...*An/(A1*A2*...*An-1). When A1 is real, it is possible to show that the infinite sum/product converges to 0 for A1 < 1 (with An --> 0) and diverges to +oo for A1 > 1 (with An --> 1). For complex An the convergence region seems to be less tractable, and I have already noticed some oscillation at A1 = 1/2 + i/2; there could be an 'oscillation curve' enclosing the convergence region, which may not be easy to determine analytically: could someone do so graphically? George Baloglou SUNY Oswego === Subject: Beyond Math READY TO HAVE YOUR DELUSIONS CHALLENGED? Humans are notoriously impressionable. Dress up just about any idea with a good dose of emotion and most of us willingly swallow it hook, line and sinker. In fact, a large portion of the ideas, assumptions and beliefs floating around in our heads right now are very likely false. History, science and logic all dictate the ideas we strongly believe and cherish today will fade into nothing more than yesterday's delusions for a more enlightened future. WARNING: LogicalReality contains ideas that may seem extreme, unusual and even a bit unsettling. When you stop and think about it, what other response could anyone be expected to have to the in-your-face shock of being exposed to new and different concepts even if those concepts are more rational. Most of us have developed a certain level of comfort with what we have learned to believe and do not take well to having our personal delusions challenged. Practically everything on LogicalReality.com has been purposely designed to help you reconsider the rationality of what you have learned to believe. At LogicalReality our goal is to see beyond the current level of social irrationalism and outright crazy ideas so we can get a more realistic look at today's issues in the cool light of unbiased reason. Are you ready for a more rational view? Fasten your seatbelt and prepare your mind for a reality shock! http://LogicalReality.com http://LogicalReality.com/Advanced.php http://RightSexWrongSex.com === Subject: Two identities 105^3=181^2+104^3 1456^3=2521^2+1455^3 http://www.primepuzzles.net/conjectures/conj_031.htm Tapio === Subject: Re: Two identities > 105^3=181^2+104^3 > 1456^3=2521^2+1455^3 So what? 8^3 = 7^3 + 13^2 and many many more. 1/3 + 1/2 + 1/3 = something I'm suppose to guess about > http://www.primepuzzles.net/conjectures/conj_031.htm === Subject: Re: Two identities > 105^3=181^2+104^3 > 1456^3=2521^2+1455^3 > So what? 8^3 = 7^3 + 13^2 and many many more. This was mentioned already 2^9=7^3 + 13^2. How many more? > 1/3 + 1/2 + 1/3 = something I'm suppose to guess about > http://www.primepuzzles.net/conjectures/conj_031.htm === Subject: Re: Two identities Tapio Hurme escribi.97 en el > 105^3=181^2+104^3 > 1456^3=2521^2+1455^3 >> So what? 8^3 = 7^3 + 13^2 and many many more. > This was mentioned already 2^9=7^3 + 13^2. > How many more? Infinity more solutions of the form (a+1)^3 = a^3 + k^2 because from that k^2 = 3a^2 + 3a + 1 That is a second degree equation in two variables, that possibly has infinite many integer solutions. The next are 20723^3 = 20722^3 + 35113^2 282360^3 = 282359^3 + 489061^2 But observe that 1/3 + 1/2 + 1/3 = 7/6 > 1. Then, it no refute the Fermat-Catalan conjecture. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Two identities Hugo Pfoertner escribi.97 en el >> Tapio Hurme escribi.97 en el > 105^3=181^2+104^3 > 1456^3=2521^2+1455^3 > So what? 8^3 = 7^3 + 13^2 and many many more. > This was mentioned already 2^9=7^3 + 13^2. > How many more? >> Infinity more solutions of the form >> (a+1)^3 = a^3 + k^2 >> because from that >> k^2 = 3a^2 + 3a + 1 >> That is a second degree equation in two variables, that possibly has >> infinite many integer solutions. The next are >> 20723^3 = 20722^3 + 35113^2 > Don't believe this one, should be > 20273^3 = 20272^3 + 35113^2 Certainly ... I copied it manually from DERIVE and misstyped the number. Better 'copy & paste', althought only be five digits ... By the other hand, the restriction 1/p + 1/q + 1/r <= 1 in the conjecture is precisse? It's says, an equation as x^2 + y^3 = z^4 or x^2 + y^3 = z^5 can has infinitely many integer solutions? -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Two identities [...] > By the other hand, the restriction 1/p + 1/q + 1/r <= 1 in the conjecture is > precisse? It's says, an equation as > x^2 + y^3 = z^4 or x^2 + y^3 = z^5 For the first equation infinitely many with y=0 and also 28^2 + 8^3 = 6^4, 27^2 + 18^3 = 9^4 63^2 + 36^3 = 15^4 ...? Second equation: 104^2 + 28^3 = 8^5 ...? Hugo > can has infinitely many integer solutions? > -- > Ignacio Larrosa Ca.96estro > A Coru.96a (Espa.96a) > ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Two identities > Tapio Hurme escribi.97 en el > 105^3=181^2+104^3 > 1456^3=2521^2+1455^3 > So what? 8^3 = 7^3 + 13^2 and many many more. > This was mentioned already 2^9=7^3 + 13^2. > How many more? > Infinity more solutions of the form > (a+1)^3 = a^3 + k^2 > because from that > k^2 = 3a^2 + 3a + 1 Correct and the solutions can be solved for example by using http://www.alpertron.com.ar/QUAD.HTM > That is a second degree equation in two variables, that possibly has > infinite many integer solutions. The next are > 20723^3 = 20722^3 + 35113^2 > 282360^3 = 282359^3 + 489061^2 > But observe that 1/3 + 1/2 + 1/3 = 7/6 > 1. Then, it no refute the > Fermat-Catalan conjecture. This is also true. Anyway, it's worth to remember that there are infinite solutions to the problem: The difference of the consecutive cubes equals to square. :-) Tapio > -- > Ignacio Larrosa Ca.96estro > A Coru.96a (Espa.96a) > ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Two identities > 105^3=181^2+104^3 > 1456^3=2521^2+1455^3 So what? 8^3 = 7^3 + 13^2 and many many more. > This was mentioned already 2^9=7^3 + 13^2. > How many more? > 1/3 + 1/2 + 1/3 = something I'm suppose to guess about > http://www.primepuzzles.net/conjectures/conj_031.htm Why not cite the few lines from the primepuzzles link (which is difficult to display on some browser with ... my Netscape 4.73 even crashes) The Fermat-Catalan Conjecture x^p + y^q = z^r has only a finite number of solutions if x, y & z are coprimes and 1/p+1/q+1/r<=1. As a matter of fact nowadays only 10 solutions are known: 1+2^3 = 3^2 (Catalan) 2^5+7^2 = 3^4 7^3+13^2 = 2^9 2^7+17^3 = 71^2 3^5+11^4 = 122^2 17^7+76271^3 = 21063928^2 1414^3+2213459^2 = 65^7 9262^3+15312283^2 = 113^7 43^8+96222^3 = 30042907^2 33^8+1549034^2 = 15613^3 Questions: 1) Can you find an eleventh solution to x^p + y^q = z^r or demonstrate that there are no more solutions? Hugo Pfoertner === Subject: Re: Weakly dense in linear order > Let S be a linear order. Claim: > S is order isomorphic to a subset of reals iff > some countable C subset S with for all x,y in S, > if x < y, then some z in C with x <= z <= y. > ie., C is weakly dense in S. An easy exercise for the weekend: Is there always a topology on S such that C weakly dense in S <==> C dense in the topology. Nemo === Subject: Re: Weakly dense in linear order > Let S be a linear order. Claim: > S is order isomorphic to a subset of reals iff > some countable C subset S with for all x,y in S, > if x < y, then some z in C with x <= z <= y. > ie., C is weakly dense in S. > An easy exercise for the weekend: > Is there always a topology on S such that > C weakly dense in S <==> C dense in the topology. I fail to see the significance of this ambigously stated problem. If S linear order and C subset S, then there's a topology T such that C is weakly dense in S iff C is topological dense in (S,T) provided S is a multi-point space, in which case T may be taken as: indiscrete topology when C is weakly dense in S discrete topolgy when C not weakly dense in S. If S is single point, then C = nulset is counterexample. -- If S linear order, then there's a topology T such that for all C subset S C is weakly dense in S iff C is topological dense in (S,T) ? No, S = {0} and C = nulset counterexample. To build another counter example Let S = {0,1,2}. {0,2}, {0,1}, {1,2} and {1} weakly dense in S {0}, {2} not weakly dense in S If {0} not dense, some open nonnul U with 0 not in U If {2} not dense, some open nonnul V with 2 not in V If {1} dense, for all open nonnul W, 1 in W. Thus 1 in U,V and {1} = U/V is open. Hence {0,2} is not dense. ---- === Subject: Re: Triples search >If any two edge lengths of an ND block have a common prime OR NONPRIME factor >greater than 2, the entire Lissajous serialization is sparse. The way this happens is that the line of serialized pixels crosses itself at right angles and at 45 degrees to the borders, forming squares. Inside each square, the pixels are missed, On the border of each square, those pixels have been serialized. These lines either reflect into and out of two corners, as in the 3x5 case, or form a loop as in the 4x6 case. Each edge of f pixels is shared between two squares, and each intersection is shared between four square. But there is an easier way to analyze these patterns. If you take an image, rotate it up through space around its right vertical axis, and flop down the reversed image, you have a 2X by Y image. If you do this along the horizonta axis, you have a 2X by 2Y image. If the image is, say, 137 X by 91 Y, and you tile-repeat the double-flopped image on x 91 times, and y 137 times, then the diagonal on this square pixels is of length 2XY but has a duplicate beginning at coordinate (XY, XY), so it is really of length xy and serializes the picture there. I'm not sure but I think this is by something like the pigeonhole principle. At that duplicate, it reflects. Now if you start with a picture of X and Y coprime, 2X by 2Y pixels, and quadruple it as above by flopping it both ways, then tile-repeat that image Y times on the x axis and X times on the y axis, you have a square with a diagonal of 4XY. That is the right number of pixels, but the trace is spare with density 1/2. However, if you start with an offset of one pixel on either axis, you miss the point of duplication and you have a closed loop serializing a 2X by 2Y picture, X and Y coprime. If you start with a 3X by 3Y picture, and most of the above is the same, you find that the first trace, starting in the corner, has 6XY pixels, has a duplicate, but needs 9XY, so it is sparse with density 1/3. If you offset that trace 1 OR 2 pixels on either axis, then it has length 6XY, no duplicate, and is sparse with denisty 2/3. By combining these traces you can achieve full serialization of any size picture or block, but you include steps of more than one pixel (that is, a distance of greater than sqrt 2 between two serialized pixels). In the complete perfect Lissajous serialization of a suitable candidate multidimensional data block, the steps between pixels are either: sqrt 1, sqrt 2, sqrt 3, ..... sqrt n and this is one of two classes of self-avoiding random walk. In this class, the pixels don't repeat but the lines between pixels cross. In the usual self-avoiding random walk on a lattice grid, the lines don't cross. So you can say the serialization is self-avoiding but its trace, that is the line in continuous space, is not. In general, on a 2D picture with common factor F between X and Y, the trace beginning in the corner has density 1/F and is open (reflects) and the F - 1 traces adjacent have density 2/F and are closed (are loops). I haven't worked out the denisty of a sparse 3D or ND block. It is probably the product of the densities of the various representative faces. In that case, if each face has density 1, then the block has denisty 1. Isn't that delightful? Remove Pee Dot Mil Yours, Doug Goncz, Replikon Research, Seven Corners, VA The hormones work at different speeds: In a fight-or-flight scenario, glucocorticoids are the ones drawing up blueprints for new aircraft carriers; epinephrine is the one handing out guns. === Subject: UFO Warp Drive (corrections) More corrections Corrections As a toy model take a generic static spherically symmetric metric without a black hole horizon since UFOs are obviously not black holes. The static spherical symmetry assumptions are also no good except as a very crude first approximation. Alcubierre's free float geodesic warp drive metric is neither static nor spherically symmetric. The warp drive is roughly similar to a self-accelerating (from the external observer's POV) dipole with dark energy and dark matter exotic vacuum boundary layers analogous to positive and negative charge in the electric case. Of course, unlike exotic vacua dipoles, electric dipoles do not self-accelerate. The dark energy is what Bondi, Terletski and Forward called negative matter (Journal of Propulsion, Vol 6, No 1 pp 28 - 37 Feb 1990, Robert Forward) The differential element of curved spacetime is then of the generic form ds^2 = - e^2phi(r)(cdt)^2 + [1 - b(r)/r]^-1dr^2 + r^2d(spherical polar angle line element) phi(r) is the red shift function, b(r) is the shape function of the UFO, r is the usual Schwarzschild curvature coordinate not the isotropic radial coordinate. Suppose we have an exotic vacuum /zpf boundary layer of thickness d with zero point energy density field (c^4/8piG)/zpf with /zpf a constant in the thin boundary layer. I did that too fast. This choice is wrong. The square root is wrong. Basically c^2/zpf acts like Grho in Newtonian gravity and in GR. So we need to solve the problem of a slab of area A thickness d. The effective G is very strong of course. Hence phi ~ (/zpfAd/r)f(d/r, angles) where d ~ 10^-4 cm is my guess and f is essentially a multipole expansion. The angular dependence destroys the spherical symmetry of course. For a signal originating inside the exotic vacuum boundary layer, the effective r is the spatial coherence length L of the signal. Assume a flat metric outside the exotic vacuum boundary layer For r >> d this happens automatically and smoothly because of the exponential dependence of the redshift function on d/r Therefore, the shift in proper time dtau (period of oscillation of the signal wave) comparing signals from the exotic vacuum region /zpf =/= 0 to the normal vacuum region /zpf = 0 where your tape recorder is located is (within the crude spherical symmetry approximation) dtau (exotic vacuum)/dtau(normal vacuum) = e^2 (/zpfAd/L) where d/r << 1 in the normal vacuum where the detector is located to receive the signal from the exotic vacuum boundary layer at the UFO. For a geometrodynamic universal red shift of all wave signals of any physical propagating energy field from metrically engineered space-time warping dtau (exotic vacuum)/dtau(normal vacuum) < 1 corresponding to the dark matter /zpf < 0 at the bow of the UFO On the other hand, at the stern of the UFO where we have dark energy /zpf > 0 we get a universal blue shift. Remember, the dark energy universally repels, the dark matter universally attracts. Put the two together in the right way and you get a net unidirectional timelike geodesic motion. Objects in the UFO are in weightless free float, though to the outside observer the object appears to accelerate and to traverse distances with effectively superluminal average speed. The former was shown by Bondi and Terletskii, apparently independently, some 40 years ago. The latter was shown by Alcubierre about 10 years ago. Alcubierre's warp drive metric (Class Quantum Gravity 11 (1994) is, for vacuum propeller self-motion along the x axis where the UFO position is xs(t) ds^2 = - (A^2 - BiB^i)(cdt)^2 + 2Bidx^idt + gijdx^idx^j i,j = x,y,z Note the non-static cross term 2Bidx^idt A = 1 is the free float timelike geodesic condition for the lapse function Bx = -vs(t)f(rs(t)) By = Bz = 0 vs(t) = dxs/dt rs(t) = [(x - xs(t))^2 + y^2 + z^2]^1/2 f(rs(t)) = [tanhZ(rs + d) - tanhZ(rs - d)]/2tanh(Zd) d, Z arbitrary parameters, but d is essentially the thickness of the exotic vacuum boundary layer The metric then becomes ds^2 = - (cdt)^2 + (dx - vxf(rs)dt)^2 + dy^2 + dz^2 That must be made compatible as a solution of the exotic vacuum field equation Guv + /zpfguv = 0 The basic dipole distribution of exotic vacuum field /zpf is shown in a computer drawing of trace of extrinsic spatial curvature in Alcubierre's original paper where essentially /zpf ~ trace of the extrinsic spatial curvature to be continued === Subject: Re: UFO Warp Drive (corrections) I was trying to forget that; have you ever tried to navigate a wormhole with a vacuum-propeller x-axis self-motion beanie !?! are you working for Spielberg, again? > Remember, the dark energy universally repels, the dark matter > universally attracts. Put the two together in the right way and you get > a net unidirectional timelike geodesic motion. Objects in the UFO are in > weightless free float, though to the outside observer the object appears > to accelerate and to traverse distances with effectively superluminal > average speed. The former was shown by Bondi and Terletskii, apparently > independently, some 40 years ago. The latter was shown by Alcubierre > about 10 years ago. > Alcubierre's warp drive metric (Class Quantum Gravity 11 (1994) > is, for vacuum propeller self-motion along the x axis where the UFO > position is xs(t) > ds^2 = - (A^2 - BiB^i)(cdt)^2 + 2Bidx^idt + gijdx^idx^j > i,j = x,y,z > Note the non-static cross term 2Bidx^idt > A = 1 is the free float timelike geodesic condition for the lapse function > Bx = -vs(t)f(rs(t)) > By = Bz = 0 > vs(t) = dxs/dt > rs(t) = [(x - xs(t))^2 + y^2 + z^2]^1/2 > f(rs(t)) = [tanhZ(rs + d) - tanhZ(rs - d)]/2tanh(Zd) > d, Z arbitrary parameters, but d is essentially the thickness of the > exotic vacuum boundary layer > The metric then becomes > ds^2 = - (cdt)^2 + (dx - vxf(rs)dt)^2 + dy^2 + dz^2 > That must be made compatible as a solution of the exotic vacuum field > equation > Guv + /zpfguv = 0 > The basic dipole distribution of exotic vacuum field /zpf is shown in > a computer drawing of trace of extrinsic spatial curvature in > Alcubierre's original paper > where essentially /zpf ~ trace of the extrinsic spatial curvature > to be continued --A church-school McCrusade (Blair's ideals?): Harry-the-Mad-Potter want's US to kill Iraqis?... http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX === Subject: Re: UFO Warp Drive (corrections) just remember, the vacuum propeller only works if its made out of dark matter... like the stuff that nibbler on futurama produces. > I was trying to forget that; > have you ever tried to navigate a wormhole > with a vacuum-propeller x-axis self-motion beanie !?! > are you working for Spielberg, again? > Remember, the dark energy universally repels, the dark matter > universally attracts. Put the two together in the right way and you get > a net unidirectional timelike geodesic motion. Objects in the UFO are in > weightless free float, though to the outside observer the object appears > to accelerate and to traverse distances with effectively superluminal > average speed. The former was shown by Bondi and Terletskii, apparently > independently, some 40 years ago. The latter was shown by Alcubierre > about 10 years ago. > Alcubierre's warp drive metric (Class Quantum Gravity 11 (1994) > is, for vacuum propeller self-motion along the x axis where the UFO > position is xs(t) > ds^2 = - (A^2 - BiB^i)(cdt)^2 + 2Bidx^idt + gijdx^idx^j > i,j = x,y,z > Note the non-static cross term 2Bidx^idt > A = 1 is the free float timelike geodesic condition for the lapse function > Bx = -vs(t)f(rs(t)) > By = Bz = 0 > vs(t) = dxs/dt > rs(t) = [(x - xs(t))^2 + y^2 + z^2]^1/2 > f(rs(t)) = [tanhZ(rs + d) - tanhZ(rs - d)]/2tanh(Zd) > d, Z arbitrary parameters, but d is essentially the thickness of the > exotic vacuum boundary layer > The metric then becomes > ds^2 = - (cdt)^2 + (dx - vxf(rs)dt)^2 + dy^2 + dz^2 > That must be made compatible as a solution of the exotic vacuum field > equation > Guv + /zpfguv = 0 > The basic dipole distribution of exotic vacuum field /zpf is shown in > a computer drawing of trace of extrinsic spatial curvature in > Alcubierre's original paper > where essentially /zpf ~ trace of the extrinsic spatial curvature > to be continued > --A church-school McCrusade (Blair's ideals?): > Harry-the-Mad-Potter want's US to kill Iraqis?... > http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) > http://www.rwgrayprojects.com/synergetics/plates/plates.html > http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX === Subject: Re: UFO Warp Drive (corrections) Oh, Good Lord!!! In how many ways is that subject line ridiculous??? === Subject: Re: UFO Warp Drive (corrections) > Oh, Good Lord!!! > In how many ways is that subject line ridiculous??? yeah, everyone knows ufo's travel through the galaxy wide worm hole network not with a warp drive! === Subject: Re: UFO Warp Drive (corrections) David, : yeah, everyone knows ufo's travel through the galaxy wide worm hole network : not with a warp drive! It's the same thing!!! You can look up Warp Drive on GOOGLE!! -- Ken Iran would be dangerous if they have a nuclear weapon. - President Bushisms : : > Oh, Good Lord!!! : > In how many ways is that subject line ridiculous??? : yeah, everyone knows ufo's travel through the galaxy wide worm hole network : not with a warp drive! : : === Subject: Re: UFO Warp Drive (corrections) > David, > : yeah, everyone knows ufo's travel through the galaxy wide worm hole > network > : not with a warp drive! > It's the same thing!!! > You can look up Warp Drive on GOOGLE!! obviously wormholes and warp drives are different. Wormholes on Farscape have twisty and bright walls that you have to stay inside of, but when the Enterprise engages warp drive there are no walls, just a bright light and all the stars get stretched out. and then when the warp drive malfunctions they can end up in a wormhole, though theirs looks a bit different than Farscapes that could be because of distortion from the malfunctioning warp drive. geez, don't you know anything??? === Subject: Re: UFO Warp Drive (corrections) LOL Gib === Subject: Re: UFO Warp Drive (corrections) Silly me, I forgot Farscape and Enterprise! I thought Jack Sarfatti was reading Ultimate Alchemy!! -- Ken Iran would be dangerous if they have a nuclear weapon. - President Bushisms : : > David, : > : yeah, everyone knows ufo's travel through the galaxy wide worm hole : > network : > : not with a warp drive! : > It's the same thing!!! : > You can look up Warp Drive on GOOGLE!! : : : obviously wormholes and warp drives are different. Wormholes on Farscape : have twisty and bright walls that you have to stay inside of, but when the : Enterprise engages warp drive there are no walls, just a bright light and : all the stars get stretched out. and then when the warp drive malfunctions : they can end up in a wormhole, though theirs looks a bit different than : Farscapes that could be because of distortion from the malfunctioning warp : drive. geez, don't you know anything??? : : === Subject: Does Gaussian Random Walk Have Maximum in Interval? In general, does a one-dimensional Gaussian random walk over a specified interval have a maximum? Let me clarify the question. For an instance of a Gaussian random walk over the time interval [t0,t1], let x(t) be the position of the walk at any time t in [t0,t1]. Then (tm,x(tm)) is a maximum iff x(tm) >= x(t) for all t in [t0,t1]. In general, does such a point exist? It seems like a simple question, but my intuition is contradicting itself, and I haven't thought of a mathematical argument for either answer. (IINM, it's axiomatic that a least upper bound exists, but I don't know whether there is necessarily a point on the walk that has this value.) I think that part of the reason that I'm struggling with this is that I normally build my intuition for mathematical concepts by thinking of examples. However, I can't describe an example of a Gaussian random walk, because it contains an infinite amount of information, and I don't have quite that much time. Would anyone care to offer some advice on how to get my brain around this concept? === Subject: Re: Does Gaussian Random Walk Have Maximum in Interval? >For an instance of a Gaussian random walk over the time interval >[t0,t1], let x(t) be the position of the walk at any time t in [t0,t1]. >Then (tm,x(tm)) is a maximum iff x(tm) >= x(t) for all t in [t0,t1]. >In general, does such a point exist? If by Gaussian random walk you mean Brownian motion (or any other suitably well-defined stochastic process), sure. X(t) is axiomatically continuous, so M = max{X(t): t0 <= t <= t1} is a well-defined random varable. For standard Brownian motion, M is not bounded (i.e., P{M > c} > 0 for all c), and M - X(t0) has density x |-> 2/sqrt(2 pi (t1-t0)) exp(-x^2/[2(t1-t0)]) for x > 0. Let T = min{t: X(t) = M}. Off hand, I do not know the joint distribution of T and M. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Does Gaussian Random Walk Have Maximum in Interval? >>For an instance of a Gaussian random walk over the time interval >>[t0,t1], let x(t) be the position of the walk at any time t in [t0,t1]. >>Then (tm,x(tm)) is a maximum iff x(tm) >= x(t) for all t in [t0,t1]. >>In general, does such a point exist? >If by Gaussian random walk you mean Brownian motion (or any other >suitably well-defined stochastic process), sure. X(t) is axiomatically >continuous, so M = max{X(t): t0 <= t <= t1} is a well-defined random >varable. For standard Brownian motion, M is not bounded (i.e., P{M > >c} > 0 for all c), and M - X(t0) has density x |-> 2/sqrt(2 pi >(t1-t0)) exp(-x^2/[2(t1-t0)]) for x > 0. >Let T = min{t: X(t) = M}. Off hand, I do not know the joint >distribution of T and M. This is not at all difficult for Brownian motion, starting at 0 at t0. For any T0 < T1, E([(M-x(T0)(M-x(T1)]) = .5*(T1-T0), and we can compute P(x(t) < M, t0 < t < T0 | M, x(T0)) and P(x(t) < M, T1 < t < t1 | M, x(T1)). As the two events for which we are computing conditional probabilities are independent, and for T1 - T0 small the i-th probability is almost linear in M - X(Ti), this gives the result. This can also be done for the Brownian bridge, but for other bias functions, generally not. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: class of differential equations are there cases in which homogenous second-order equations of the form: f''[x]=(a0 + a1*x + a2*x^2 + .... + an*x^n + ...)*f[x] admit analytic solutions? I am particularly interested in the equation: f''[x]=(a0 + a2*x^2 + a4*x^4)*f[x] Here, an are, in general, complex constants. yuric === Subject: Re: class of differential equations |>are there cases in which homogenous second-order equations of the form: |>f''[x]=(a0 + a1*x + a2*x^2 + .... + an*x^n + ...)*f[x] |>admit analytic solutions? I am particularly interested in the equation: |>f''[x]=(a0 + a2*x^2 + a4*x^4)*f[x] This equation can be reduced to a Riccati DE by the substitution f'/f = g: f = exp(int(g(x) dx)+C) where g' + g^2 = a0 + a2 x^2 + a4 x^4 The case a0 = a2 = 0 can be solved using Bessel functions. In Maple 9: > dsolve((D@@2)(f)(x)=a4*x^4*f(x),f(x)); 1/2 3 1/2 (-a4) x f(x) = _C1 x BesselJ(1/6, -----------) 3 1/2 3 1/2 (-a4) x + _C2 x BesselY(1/6, -----------) 3 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: Looking for Spare Cycles We are hoping to get some spare cycles to complete our first Collatz Conjecture project. We are trying to improve on our grid program so that we can start running some better projects (Collatz is nice, but it will most likely run till the sun burns out). So, if you would like to throw some of your spare computer cycles our way, visit our website and download the client. It is a Windows only platform using .Net at this point, but we are working on a Linux Client. As for Teams, it would be cool if you all used a Team ID of SCI.MATH http://www.gridontap.com === Subject: Re: Looking for Spare Cycles === >Subject: Looking for Spare Cycles >Message-id: <7uL1b.1280$_V.570@news04.bloor.is.net.cable.rogers.comWe are hoping to get some spare cycles to complete our first Collatz >Conjecture project. We are trying to improve on our grid program so that we >can start running some better projects (Collatz is nice, but it will most >likely run till the sun burns out). >So, if you would like to throw some of your spare computer cycles our way, >visit our website and download the client. It is a Windows only platform >using .Net at this point, but we are working on a Linux Client. >As for Teams, it would be cool if you all used a Team ID of SCI.MATH >http://www.gridontap.com Here's a couple suggestions: - fix the ing website BEFORE you start soliciting clients pages that time out, downloads that abort after 10% complete will only drive - try placing a link to the download ON THE DOWNLOAD PAGE! Jesus H. Christ, is this site designed by imbeciles? Let me me know when you've got the site working and I'll reconsider joining. -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Looking for Spare Cycles - fix the ing website BEFORE you start soliciting clients > pages that time out, downloads that abort after 10% complete will only drive > potential users away. I can't spare any cycles IF I CAN'T DOWNLOAD THE We're running a T1, it *could* be your connection - or they are messing with the power around here again, we are in Toronto (black out area). > - try placing a link to the download ON THE DOWNLOAD PAGE! You should scroll to the bottom of the page. Hope to see you on the Sci.Math team. === Subject: Re: Looking for Spare Cycles === >Subject: Re: Looking for Spare Cycles >Message-id: > pages that time out, downloads that abort after 10% complete will only >drive >> potential users away. I can't spare any cycles IF I CAN'T DOWNLOAD THE >We're running a T1, it *could* be your connection - Yeah, it could be. But I'll point out that I have no compatibility problems with PROFESSIONALLY designed web sites. >or they are messing with >the power around here again, we are in Toronto (black out area). You only get one chance to make a first impression. For every nasty reply like mine you get, how many people never reply at all and just dismiss you as incompetent and never return? So don't take it the wrong way, I'm just trying to help. Sometimes you have to swear at people to shake them out of their complacency. >> - try placing a link to the download ON THE DOWNLOAD PAGE! >You should scroll to the bottom of the page. >Hope to see you on the Sci.Math team. I'm already on the team (that part worked). If only I could download the program... -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Looking for Spare Cycles >>You should scroll to the bottom of the page. Silly me, it was right there under my nose. It wasn't underlined. It was in a color indistinguishable from the body text. The only way to tell if it was a link is to see if the cursor changes to a finger when you hover over it. It still won't download, even on my Windows XP machine. -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Minimal ellipse knowing its center and enclosing a set of points? hi, I would need to solve this: Given a point O and a set of points S = { P1, P2, ..., Pn}, how do we determine the ellipse of smallest area centered at O and enclosing all the points in S? Ideas? -- dave === Subject: Re: Minimal ellipse knowing its center and enclosing a set of points? > I would need to solve this: > Given a point O and a set of points S = { P1, P2, ..., Pn}, how do we > determine the ellipse of smallest area centered at O and enclosing all > the points in S? > Ideas? Knowing this to be an important and well-studied problem, I did a Google search for you using enclosing ellipse points minimal One reference found was Look at the section The Bounding Ellipsoid and the corresponding references. David === Subject: Re: Minimal ellipse knowing its center and enclosing a set of points? >Given a point O and a set of points S = { P1, P2, ..., Pn}, how do we >determine the ellipse of smallest area centered at O and enclosing all >the points in S? >Ideas? One idea: First translate the points so that O is the origin. The equation of an ellipse centered at the origin is A x^2 + B x y + C y^2 = D where A and 4A C - B^2 are both positive. Sorry, I do not know the formula for the area f(A,B,C,D) of the ellipse from these coefficients, but I am sure you can find it (or someone else here will tell us). Once you have it, you can set up the quadratic program, min f(A,B,C,D) s.t. A x_i^2 + B x_i y_i + C y_i^2 <= D for i = 1, 2, ..., n A >= 0 4A C - B^2 >= 0 (You can throw in C >= 0, D >= 0 if that helps your q.p. algortihm.) Another rough idea, though I am not sure it will work: Can you just determine the ellipse whereon lie the four points at greatest distance from O? You have to make adjustments so that no three of these four points are colinear. Actually, now that I think about it, this will not work in this rough form, but it might be a start towards an algorithm. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Minimal ellipse knowing its center and enclosing a set of points? > Given a point O and a set of points S = { P1, P2, ..., Pn}, how do we > determine the ellipse of smallest area centered at O and enclosing all > the points in S? > Ideas? > One idea: First translate the points so that O is the origin. The > equation of an ellipse centered at the origin is > A x^2 + B x y + C y^2 = D Actually, you can set D to 1. [...] -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Plotting a graph Sorry for posting this twice but I didn't get a reponse and was Suppose I have recorded some distances (they're actually pointer distances between two objects in memory). Since these distances could be very large, I have separated them into buckets which are at intervals of Log base 2. So I then have buckets such as: 0, 2, 4, 16, ... to 2147483648. Distance data such as: 34, 16, 15, 100 etc will be in buckets 32, 16, 16 and 64 respectively. I then keep a count for each bucket to know the number of distances that belong to each one of them (actual distances are not required). I want to be able to plot this information on a graph. On the y-axis I need to place these distances and on the x-axis I want to keep the number of times these distances are sampled (i.e. the total number of pointer mutations.. this can go to millions!). Now someone adviced me that I normalize the y-axis (i.e. the distances that should be recored as percentages) in terms of the largest heap size encountered (i.e. the denominator in the normalization should be the same for all points on the x axis). What exactly does this mean? What should I do? Should I divide the bucket sizes by the largest heap size, and then depending upon the number of distances that were measured for that particular bucket I take a percentage of the overall mutations that there were? Could someone guide me as to how I should do this and use the given Tiff === Subject: Re: Polar Equation of a circle > r^2 - 2 r h cos(theta) + h^2 = R^2 And further on, using T^2= h^2-R^2 one obtains r^2 - 2 r h cos(theta) + T^2 = 0 . T^2 is the power of the circle or square of tangent and product of radius vectors or secants from origin and we suddenly find focus shifting from R to T in polar co-ordinates although they are related. Also, it is the constant term in A x^2+ 2 C x y + B y^2 + 2 f x + 2 g y + T^2 =0. In cartesian representation T^2 is just a constant without physical appreciation that its square root is the geometric mean of extreme radii in polar representation. === Subject: Longest chain in GL(n,2) What is the longest chain of subgroups in the group GL(n,2) ? === Subject: Re: Longest chain in GL(n,2) >What is the longest chain of subgroups in the group GL(n,2) ? Simpleminded notational question from a non-practitioner: Does GL(n,2) mean the group of n by n matrices over Z/2Z with determinant 1? Lee Rudolph === Subject: Re: Longest chain in GL(n,2) >What is the longest chain of subgroups in the group GL(n,2) ? > Simpleminded notational question from a non-practitioner: > Does GL(n,2) mean the group of n by n matrices over Z/2Z with > determinant 1? > Lee Rudolph Yes , Dan === Subject: Re: What cardinality may a ring have? > Given a non-empty set S of cardinality kappa (finite or infinite) can > one always define +, * and 0 so that is a ring? I realsise that this has already been answered, but it is also a concequence of the L.9awenheim Skolem function (upward). This is the countable language version: If T is a first order theory in a countable language with a model of size kappa, then for any cardinality C greater than or equal to kappa, T has a model. Well, it is trivial to exhibit a ring of cardinality n for finite n, and to write the ring axioms in a first order way. Now Q is a countable ring so there exists a ring of any cardinality greater than or equal to |Q|. === Subject: Re: What cardinality may a ring have? > In my old-fashioned world, all rings had at least two members, an > additive identity and a distinct multipicative identity. Wow, you must be *old*. In my 50-year-old copy of Van der Waerden's _Modern Algebra_ that requirement had already been abolished: The integers form a ring C with identity; the even numbers a ring without identity. There are also rings with one or more right identities but without a left identity, or vice versa. === Subject: Re: What cardinality may a ring have? >> In my old-fashioned world, all rings had at least two members, an >> additive identity and a distinct multipicative identity. >Wow, you must be *old*. In my 50-year-old copy of Van der Waerden's >_Modern Algebra_ that requirement had already been abolished: The >integers form a ring C with identity; the even numbers a ring without >identity. There are also rings with one or more right identities but >without a left identity, or vice versa. When this came up in another thread, I checked the only two algebra texts I have (Van der Waerden and Herstein), and couldn't find anything about a multiplicative identity. A more interesting question to me is whether {0} should be considered a ring. I can't decide off the top of my head which would cause more difficulty... in one case one would have to put 'non-trivial' in for some statements, in the other case, it would sometimes be necessary to say 'ring or {0}'. Perhaps some insist on a multiplicative identity so they don't have to deal with {0}. If I were concerned about that, I would have (R,+) is an Abelian group with identity 0. (R/{0}',*) is a semigroup and distributivity as the definition. This would guarantee that there are at least two elements. Larry (this space unintentially left blank ..... make obvious deletion for email === Subject: UFO Warp Drive Metric Engineering You can see me lecturing on this in Paramount Pictures Star Trek IV Special Collector's Edition DVD Star Trek IV: The Voyage Home Disk 2 Time Travel: The Art of the Possible http://www.genreonline.net/Star_Trek_IV_SCE_DVD.html Picture of the non-spherically symmetric non-static dipole exotic vacuum zero point energy density /zpf field for the case of Alcubierre's toy model UFO vacuum propeller metric is at http://qedcorp.com/APS/warpdrive1.jpg Think of the vertical axis as the strength of the (c^4/8piG*)/zpf(x) zero point energy density warp drive local unified dark energy/matter scalar field distribution of exotic vacuum stuff. The UFO moves to the right in the picture along a free float (weightless) self-generated timelike geodesic no matter how accelerated the UFO trajectory looks to USAF tracking radar. The negative /zpf < 0 dark matter exotic vacuum region (boundary layer) of positive quantum pressure is at the bow where space is literally locally contracting causing the anomalous red shift opposing the normal motional Doppler blue shift of an approaching object for an observer in front of the UFO. In contrast, the positive /zpf > 0 dark energy exotic vacuum boundary layer of negative quantum pressure is at the stern where space is locally literally expanding behind the UFO (just like the universe is accelerating on the larger scale > 10 megaparsecs) causing the anomalous blue shift opposing the normal motional Doppler red shift of a receding object. Therefore, we see that the reverse Doppler effect is a good signature of an advanced military technology based upon practical metric engineering of the spacetime continuum. No wonder, Herman Bondi and Yakov Terletski worked on this problem 50 years ago. They did not get very far of course because we had to wait several decades for the discovery first of dark matter and then dark energy in 1999 from type 1a supernovae standard candles. As Above So Below. Advanced super technology Dyson-Thorne metric engineering civilizations know how to bottle dark energy/matter on a small scale contrary to More corrections Corrections As a toy model take a generic static spherically symmetric metric without a black hole horizon since UFOs are obviously not black holes. The static spherical symmetry assumptions are also no good except as a very crude first approximation. Alcubierre's free float geodesic warp drive metric is neither static nor spherically symmetric. The warp drive is roughly similar to a self-accelerating (from the external observer's POV) dipole with dark energy and dark matter exotic vacuum boundary layers analogous to positive and negative charge in the electric case. Of course, unlike exotic vacua dipoles, electric dipoles do not self-accelerate. The dark energy is what Bondi, Terletski and Forward called negative matter (Journal of Propulsion, Vol 6, No 1 pp 28 - 37 Feb 1990, Robert Forward) The differential element of curved spacetime is then of the generic form ds^2 = - e^2phi(r)(cdt)^2 + [1 - b(r)/r]^-1dr^2 + r^2d(spherical polar angle line element) phi(r) is the red shift function, b(r) is the shape function of the UFO, r is the usual Schwarzschild curvature coordinate not the isotropic radial coordinate. Suppose we have an exotic vacuum /zpf boundary layer of thickness d with zero point energy density field (c^4/8piG)/zpf with /zpf a constant in the thin boundary layer. I did that too fast. This choice is wrong. The square root is wrong. Basically c^2/zpf acts like Grho in Newtonian gravity and in GR. So we need to solve the problem of a slab of area A thickness d. The effective G is very strong of course. Hence phi ~ (/zpfAd/r)f(d/r, angles) where d ~ 10^-4 cm is my guess and f is essentially a multipole expansion. The angular dependence destroys the spherical symmetry of course. For a signal originating inside the exotic vacuum boundary layer, the effective r is the spatial coherence length L of the signal. Assume a flat metric outside the exotic vacuum boundary layer For r >> d this happens automatically and smoothly because of the exponential dependence of the redshift function on d/r Therefore, the shift in proper time dtau (period of oscillation of the signal wave) comparing signals from the exotic vacuum region /zpf =/= 0 to the normal vacuum region /zpf = 0 where your tape recorder is located is (within the crude spherical symmetry approximation) dtau (exotic vacuum)/dtau(normal vacuum) = e^2 (/zpfAd/L) where d/r << 1 in the normal vacuum where the detector is located to receive the signal from the exotic vacuum boundary layer at the UFO. For a geometrodynamic universal red shift of all wave signals of any physical propagating energy field from metrically engineered space-time warping dtau (exotic vacuum)/dtau(normal vacuum) < 1 corresponding to the dark matter /zpf < 0 at the bow of the UFO On the other hand, at the stern of the UFO where we have dark energy /zpf > 0 we get a universal blue shift. Remember, the dark energy universally repels, the dark matter universally attracts. Put the two together in the right way and you get a net unidirectional timelike geodesic motion. Objects in the UFO are in weightless free float, though to the outside observer the object appears to accelerate and to traverse distances with effectively superluminal average speed. The former was shown by Bondi and Terletskii, apparently independently, some 40 years ago. The latter was shown by Alcubierre about 10 years ago. Alcubierre's warp drive metric (Class Quantum Gravity 11 (1994) is, for vacuum propeller self-motion along the x axis where the UFO position is xs(t) ds^2 = - (A^2 - BiB^i)(cdt)^2 + 2Bidx^idt + gijdx^idx^j i,j = x,y,z Note the non-static cross term 2Bidx^idt A = 1 is the free float timelike geodesic condition for the lapse function Bx = -vs(t)f(rs(t)) By = Bz = 0 vs(t) = dxs/dt rs(t) = [(x - xs(t))^2 + y^2 + z^2]^1/2 f(rs(t)) = [tanhZ(rs + d) - tanhZ(rs - d)]/2tanh(Zd) d, Z arbitrary parameters, but d is essentially the thickness of the exotic vacuum boundary layer The metric then becomes ds^2 = - (cdt)^2 + (dx - vxf(rs)dt)^2 + dy^2 + dz^2 That must be made compatible as a solution of the exotic vacuum field equation Guv + /zpfguv = 0 The basic dipole distribution of exotic vacuum field /zpf is shown in a computer drawing of trace of extrinsic spatial curvature in Alcubierre's original paper where essentially /zpf ~ trace of the extrinsic spatial curvature to be continued === Subject: Re: Factorial/Exponential Identity, Infinity I examine the products of pairs of a set of integers {1, ..., n}. For example, for a set {1, 2, 3}, the pairs are {1, 2}, {1, 3}, and {2, 3}. The number of pairs is (n^2-n)/2. I put the products in a list. Each list element is a number, prime or composite, multiplied by the sum of a series of numbers. For example, the sum of these two list elements: 2*(1+2 +4) 3*(1+2 +4) is the sum of the products of the pairs of {1, 2, 3, 4}. There seems to be a rational way to enumerate the list elements as I products and sort them to compare to how an algorithm would enumerate them on a list. Here are the working lists for various values of n: 2: 2*(1) 3: 2*(1) 3*(1+2) 4: 2*(1+2 +4) 3*(1+2 +4) 5: 2*(1+2 +4) 3*(1+2 +4) 5*(1+2+3+4) 6: 2*(1+2+3+4 +6) 3*(1+2 +4 +6 +8) 5*(1+2+3+4 +6) Notice that the multipliers increment by one to an even number and then sometimes increment by two. 7: 2*(1+2+3+4 +6) 3*(1+2 +4 +6 +8) 5*(1+2+3+4 +6) 7*(1+2+3+4+5+6) 8: 2*(1+2+3+4 +6) 3*(1+2 +4 +6 +8) 5*(1+2+3+4 +6 +8) 7*(1+2+3+4+5+6 +8) 8*(1+2+3+4 +6) The number 8 is not prime. It is the cube of a prime, and product of a prime and a composite. 9: 2*(1+2+3+4 +6) 3*(1+2 +4 +6 +8) 5*(1+2+3+4 +6 +8) 7*(1+2+3+4+5+6 +8) 8*(1+2+3+4 +6) 9*(1+2+3+4+5+6+7+8) The number 9 is not prime. It is the square of a prime. 10: 2*(1+2+3+4+5+6) 3*(1+2 +4 +6 +8 +10) 5*(1+2+3+4 +6 +8 +10) 7*(1+2+3+4+5+6 +8 +10) 8*(1+2+3+4 +6) 9*(1+2+3+4+5+6+7+8 +10) 20*(1+2+3+4) The number 20 is greater than n. It's used instead of 10, the product of two primes or a semiprime, to have the multipliers follow the sequence starting with 1. Notice that the multipliers, (1+2+...), increment by one and then increment by two. 11: 2*(1+2+3+4+5+6) 3*(1+2 +4 +6 +8 +10) 5*(1+2+3+4 +6 +8 +10) 7*(1+2+3+4+5+6 +8 +10) 8*(1+2+3+4 +6) 9*(1+2+3+4+5+6+7+8 +10) 20*(1+2+3+4) 11*(1+2+3+4+5+6+7+8+9+10) Eleven is a prime, its list entry is each multiple of 11 less than 11. 12: 2*(1+2+3+4+5+6) 3*(1+2 +4 +6 +8 +10 +12) 5*(1+2+3+4 +6 +8 +10 +12) 7*(1+2+3+4+5+6 +8 +10 +12) 8*(1+2+3+4 +6) 9*(1+2+3+4+5+6+7+8 +10 +12) 20*(1+2+3+4 +6) 11*(1+2+3+4+5+6+7+8+9+10 +12) 12*(1+2 +4 +6 +8) Twelve has as factors 1, 2, 3, 4, 6, and 12. 13: 2*(1+2+3+4+5+6) 3*(1+2 +4 +6 +8 +10 +12) 5*(1+2+3+4 +6 +8 +10 +12) 7*(1+2+3+4+5+6 +8 +10 +12) 8*(1+2+3+4 +6) 9*(1+2+3+4+5+6+7+8 +10 +12) 20*(1+2+3+4 +6) 11*(1+2+3+4+5+6+7+8+9+10 +12) 12*(1+2 +4 +6 +8) 13*(1+2+3+4+5+6+7+8+9+10+11+12) Thirteen is a prime number, its list entry has multipliers of the sum from 1 to n-1. Calculating the list entry for thirteen is evaluating 13*(12+1) 12/2 = 13^2 *6. Compare that to the list entry for three. Most of the incrementations in the multipliers of three are increments by two. An expression of the list entry for three is the sum of the multipliers times three, and as well a partial sum of the mutipliers tmes three plus the remainder sum of the multipliers times three. This is about how (1+2+4+6+8+10+12) is 1+ 2*(1+2+3+4+5+6) = 1+2*(6+1)*6/2 = 43. That's different than the list entry for 7, most of its multiplier increments are by one. (1+2+3+4+5+6+7+8+9+10+11+12) - (7+9+11), or (12+1)*12/2 - 3*9 = 51. To calculate the list entry for 3 takes four multiplications/divisions and two additions/subtractions, for list entry 7 there are three multiplications and two additions, besides knowing the sequence of multiples and when they increment. 14: 2*(1+2+3+4+5+6) 3*(1+2 +4 +6 +8 +10 +12 +14) 5*(1+2+3+4 +6 +8 +10 +12 +14) 7*(1+2+3+4+5+6 +8 +10 +12 +14) 8*(1+2+3+4 +6) 9*(1+2+3+4+5+6+7+8 +10 +12 +14) 20*(1+2+3+4 +6) 11*(1+2+3+4+5+6+7+8+9+10 +12 +14) 12*(1+2 +4 +6 +8) 13*(1+2+3+4+5+6+7+8+9+10+11+12 +14) 14*(1+2 +4 +6 +8 +10 +12) Fourteen is semiprime, its factors besides itself and one are primes. As I get into higher numbers it gets a bit more complicated. The prime is the simple case, but as I add composites I want to preserve aspects of the multiplier progression. For example, with fifteen, it is in some ways like a prime in its multiplier series changes, yet it is a semiprime, it is also odd. Sixteen, a 4'th power of a prime and a square of a square of a prime and a composite of a prime and a cube of a prime, might lead to increments by four. This is about recurrence relations, or recurrence equations or difference equations. Calculating the sum of the elements of a set {1, ..., n} given n is a matter of evaluating (n+1)n/2, the product of all elements of the set is n!. This strategy explores the sum of the products of the pairs (triples/trios, quadruples, etc.) of a set {1,..., n} or equivalently {0, ..., n}. The outer product of a pair of vectors is a matrix with the entries of the matrix being the products of each i,j of the two vectors. That's somewhat different from the above where the diagonal and the upper lower elements are zero, eg, for (1, 2, 3, 4): 1 2 3 4 1 1 2 3 4 2 2 4 6 8 3 3 6 9 12 4 4 8 12 16 1 2 3 4 1 2 2 3 3 6 4 4 8 12 The sum of the elements of the outer product of {1,2,3,4} and itself is 1*(1+2+3+4) 2*(1+2+3+4) 3*(1+2+3+4) 4*(1+2+3+4) That is sum n squared, or (n+1)n/2 *(n+1)n/2, or 10*10=100. Each row and column sums to a multiple of ten, the sum of 1, 2, 3, and 4. The sum of the products of the pairs of {1, 2, 3, 4} is (2+3+4)+(6+8)+12 = 9+14+12 = 35. It would be possible to calculate the values of the diagonal of the outer product and then subtract that from the sum of the outer product, and then divide that result by two. The diagonal has the value sum for i from 1 to 4 of x^2, that being 1+4+9+16, 30. 100, calculated from (sum n) squared, minus 30, calculated from sum (n squared), equals 70, dividing that by two would give the result of 35. Now then, there is the question of deriving the sum of the squares, heh, sum of squares, in a way besides adding them together. Products of pairs for (1, ..., 48) Number of pair products: (48^2 - 48)/2 = 1128 Sum of pair products: 672476 Generated products: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 144, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 132, 136, 140, 144, 148, 152, 156, 160, 164, 168, 172, 176, 180, 184, 188, 192, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200, 205, 210, 215, 220, 225, 230, 235, 240, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120, 126, 132, 138, 144, 150, 156, 162, 168, 174, 180, 186, 192, 198, 204, 210, 216, 222, 228, 234, 240, 246, 252, 258, 264, 270, 276, 282, 288, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196, 203, 210, 217, 224, 231, 238, 245, 252, 259, 266, 273, 280, 287, 294, 301, 308, 315, 322, 329, 336, 72, 80, 88, 96, 104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224, 232, 240, 248, 256, 264, 272, 280, 288, 296, 304, 312, 320, 328, 336, 344, 352, 360, 368, 376, 384, 90, 99, 108, 117, 126, 135, 144, 153, 162, 171, 180, 189, 198, 207, 216, 225, 234, 243, 252, 261, 270, 279, 288, 297, 306, 315, 324, 333, 342, 351, 360, 369, 378, 387, 396, 405, 414, 423, 432, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250, 260, 270, 280, 290, 300, 310, 320, 330, 340, 350, 360, 370, 380, 390, 400, 410, 420, 430, 440, 450, 460, 470, 480, 132, 143, 154, 165, 176, 187, 198, 209, 220, 231, 242, 253, 264, 275, 286, 297, 308, 319, 330, 341, 352, 363, 374, 385, 396, 407, 418, 429, 440, 451, 462, 473, 484, 495, 506, 517, 528, 156, 168, 180, 192, 204, 216, 228, 240, 252, 264, 276, 288, 300, 312, 324, 336, 348, 360, 372, 384, 396, 408, 420, 432, 444, 456, 468, 480, 492, 504, 516, 528, 540, 552, 564, 576, 182, 195, 208, 221, 234, 247, 260, 273, 286, 299, 312, 325, 338, 351, 364, 377, 390, 403, 416, 429, 442, 455, 468, 481, 494, 507, 520, 533, 546, 559, 572, 585, 598, 611, 624, 210, 224, 238, 252, 266, 280, 294, 308, 322, 336, 350, 364, 378, 392, 406, 420, 434, 448, 462, 476, 490, 504, 518, 532, 546, 560, 574, 588, 602, 616, 630, 644, 658, 672, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375, 390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, 272, 288, 304, 320, 336, 352, 368, 384, 400, 416, 432, 448, 464, 480, 496, 512, 528, 544, 560, 576, 592, 608, 624, 640, 656, 672, 688, 704, 720, 736, 752, 768, 306, 323, 340, 357, 374, 391, 408, 425, 442, 459, 476, 493, 510, 527, 544, 561, 578, 595, 612, 629, 646, 663, 680, 697, 714, 731, 748, 765, 782, 799, 816, 342, 360, 378, 396, 414, 432, 450, 468, 486, 504, 522, 540, 558, 576, 594, 612, 630, 648, 666, 684, 702, 720, 738, 756, 774, 792, 810, 828, 846, 864, 380, 399, 418, 437, 456, 475, 494, 513, 532, 551, 570, 589, 608, 627, 646, 665, 684, 703, 722, 741, 760, 779, 798, 817, 836, 855, 874, 893, 912, 420, 440, 460, 480, 500, 520, 540, 560, 580, 600, 620, 640, 660, 680, 700, 720, 740, 760, 780, 800, 820, 840, 860, 880, 900, 920, 940, 960, 462, 483, 504, 525, 546, 567, 588, 609, 630, 651, 672, 693, 714, 735, 756, 777, 798, 819, 840, 861, 882, 903, 924, 945, 966, 987, 1008, 506, 528, 550, 572, 594, 616, 638, 660, 682, 704, 726, 748, 770, 792, 814, 836, 858, 880, 902, 924, 946, 968, 990, 1012, 1034, 1056, 552, 575, 598, 621, 644, 667, 690, 713, 736, 759, 782, 805, 828, 851, 874, 897, 920, 943, 966, 989, 1012, 1035, 1058, 1081, 1104, 600, 624, 648, 672, 696, 720, 744, 768, 792, 816, 840, 864, 888, 912, 936, 960, 984, 1008, 1032, 1056, 1080, 1104, 1128, 1152, 650, 675, 700, 725, 750, 775, 800, 825, 850, 875, 900, 925, 950, 975, 1000, 1025, 1050, 1075, 1100, 1125, 1150, 1175, 1200, 702, 728, 754, 780, 806, 832, 858, 884, 910, 936, 962, 988, 1014, 1040, 1066, 1092, 1118, 1144, 1170, 1196, 1222, 1248, 756, 783, 810, 837, 864, 891, 918, 945, 972, 999, 1026, 1053, 1080, 1107, 1134, 1161, 1188, 1215, 1242, 1269, 1296, 812, 840, 868, 896, 924, 952, 980, 1008, 1036, 1064, 1092, 1120, 1148, 1176, 1204, 1232, 1260, 1288, 1316, 1344, 870, 899, 928, 957, 986, 1015, 1044, 1073, 1102, 1131, 1160, 1189, 1218, 1247, 1276, 1305, 1334, 1363, 1392, 930, 960, 990, 1020, 1050, 1080, 1110, 1140, 1170, 1200, 1230, 1260, 1290, 1320, 1350, 1380, 1410, 1440, 992, 1023, 1054, 1085, 1116, 1147, 1178, 1209, 1240, 1271, 1302, 1333, 1364, 1395, 1426, 1457, 1488, 1056, 1088, 1120, 1152, 1184, 1216, 1248, 1280, 1312, 1344, 1376, 1408, 1440, 1472, 1504, 1536, 1122, 1155, 1188, 1221, 1254, 1287, 1320, 1353, 1386, 1419, 1452, 1485, 1518, 1551, 1584, 1190, 1224, 1258, 1292, 1326, 1360, 1394, 1428, 1462, 1496, 1530, 1564, 1598, 1632, 1260, 1295, 1330, 1365, 1400, 1435, 1470, 1505, 1540, 1575, 1610, 1645, 1680, 1332, 1368, 1404, 1440, 1476, 1512, 1548, 1584, 1620, 1656, 1692, 1728, 1406, 1443, 1480, 1517, 1554, 1591, 1628, 1665, 1702, 1739, 1776, 1482, 1520, 1558, 1596, 1634, 1672, 1710, 1748, 1786, 1824, 1560, 1599, 1638, 1677, 1716, 1755, 1794, 1833, 1872, 1640, 1680, 1720, 1760, 1800, 1840, 1880, 1920, 1722, 1763, 1804, 1845, 1886, 1927, 1968, 1806, 1848, 1890, 1932, 1974, 2016, 1892, 1935, 1978, 2021, 2064, 1980, 2024, 2068, 2112, 2070, 2115, 2160, 2162, 2208, 2256, Sorted products: 2, 3, 4, 5, 6, 6, 7, 8, 8, 9, 10, 10, 11, 12, 12, 12, 13, 14, 14, 15, 15, 16, 16, 17, 18, 18, 18, 19, 20, 20, 20, 21, 21, 22, 22, 23, 24, 24, 24, 24, 25, 26, 26, 27, 27, 28, 28, 28, 29, 30, 30, 30, 30, 31, 32, 32, 32, 33, 33, 34, 34, 35, 35, 36, 36, 36, 36, 37, 38, 38, 39, 39, 40, 40, 40, 40, 41, 42, 42, 42, 42, 43, 44, 44, 44, 45, 45, 45, 46, 46, 47, 48, 48, 48, 48, 48, 50, 50, 51, 52, 52, 54, 54, 54, 55, 56, 56, 56, 57, 58, 60, 60, 60, 60, 60, 62, 63, 63, 64, 64, 65, 66, 66, 66, 68, 68, 69, 70, 70, 70, 72, 72, 72, 72, 72, 74, 75, 75, 76, 76, 77, 78, 78, 78, 80, 80, 80, 80, 81, 82, 84, 84, 84, 84, 84, 85, 86, 87, 88, 88, 88, 90, 90, 90, 90, 90, 91, 92, 92, 93, 94, 95, 96, 96, 96, 96, 96, 98, 99, 99, 100, 100, 102, 102, 104, 104, 105, 105, 105, 108, 108, 108, 108, 110, 110, 111, 112, 112, 112, 114, 114, 115, 116, 117, 117, 119, 120, 120, 120, 120, 120, 120, 123, 124, 125, 126, 126, 126, 126, 128, 128, 129, 130, 130, 132, 132, 132, 132, 133, 135, 135, 135, 136, 136, 138, 138, 140, 140, 140, 140, 141, 143, 144, 144, 144, 144, 144, 145, 147, 148, 150, 150, 150, 152, 152, 153, 154, 154, 155, 156, 156, 156, 160, 160, 160, 160, 161, 162, 162, 164, 165, 165, 168, 168, 168, 168, 168, 170, 170, 171, 172, 174, 175, 175, 176, 176, 176, 180, 180, 180, 180, 180, 180, 182, 182, 184, 184, 185, 186, 187, 188, 189, 189, 190, 190, 192, 192, 192, 192, 195, 195, 196, 198, 198, 198, 200, 200, 200, 203, 204, 204, 205, 207, 208, 208, 209, 210, 210, 210, 210, 210, 215, 216, 216, 216, 216, 217, 220, 220, 220, 221, 222, 224, 224, 224, 225, 225, 228, 228, 230, 230, 231, 231, 232, 234, 234, 234, 235, 238, 238, 240, 240, 240, 240, 240, 240, 242, 243, 245, 246, 247, 248, 250, 252, 252, 252, 252, 252, 253, 255, 256, 258, 259, 260, 260, 261, 264, 264, 264, 264, 266, 266, 270, 270, 270, 270, 272, 272, 273, 273, 275, 276, 276, 279, 280, 280, 280, 280, 282, 285, 286, 286, 287, 288, 288, 288, 288, 288, 290, 294, 294, 296, 297, 297, 299, 300, 300, 300, 301, 304, 304, 306, 306, 308, 308, 308, 310, 312, 312, 312, 315, 315, 315, 319, 320, 320, 320, 322, 322, 323, 324, 324, 325, 328, 329, 330, 330, 330, 333, 336, 336, 336, 336, 336, 338, 340, 340, 341, 342, 342, 344, 345, 348, 350, 350, 351, 351, 352, 352, 352, 357, 360, 360, 360, 360, 360, 360, 363, 364, 364, 368, 368, 369, 370, 372, 374, 374, 375, 376, 377, 378, 378, 378, 380, 380, 384, 384, 384, 385, 387, 390, 390, 390, 391, 392, 396, 396, 396, 396, 399, 400, 400, 403, 405, 405, 406, 407, 408, 408, 410, 414, 414, 416, 416, 418, 418, 420, 420, 420, 420, 420, 423, 425, 429, 429, 430, 432, 432, 432, 432, 434, 435, 437, 440, 440, 440, 442, 442, 444, 448, 448, 450, 450, 450, 451, 455, 456, 456, 459, 460, 460, 462, 462, 462, 464, 465, 468, 468, 468, 470, 473, 475, 476, 476, 480, 480, 480, 480, 480, 481, 483, 484, 486, 490, 492, 493, 494, 494, 495, 495, 496, 500, 504, 504, 504, 504, 506, 506, 507, 510, 510, 512, 513, 516, 517, 518, 520, 520, 522, 525, 525, 527, 528, 528, 528, 528, 532, 532, 533, 540, 540, 540, 540, 544, 544, 546, 546, 546, 550, 551, 552, 552, 555, 558, 559, 560, 560, 560, 561, 564, 567, 570, 570, 572, 572, 574, 575, 576, 576, 576, 578, 580, 585, 585, 588, 588, 589, 592, 594, 594, 595, 598, 598, 600, 600, 600, 602, 608, 608, 609, 611, 612, 612, 615, 616, 616, 620, 621, 624, 624, 624, 627, 629, 630, 630, 630, 630, 638, 640, 640, 644, 644, 645, 646, 646, 648, 648, 650, 651, 656, 658, 660, 660, 660, 663, 665, 666, 667, 672, 672, 672, 672, 675, 675, 680, 680, 682, 684, 684, 688, 690, 690, 693, 696, 697, 700, 700, 702, 702, 703, 704, 704, 705, 713, 714, 714, 720, 720, 720, 720, 720, 722, 725, 726, 728, 731, 735, 736, 736, 738, 740, 741, 744, 748, 748, 750, 752, 754, 756, 756, 756, 759, 760, 760, 765, 768, 768, 770, 774, 775, 777, 779, 780, 780, 782, 782, 783, 792, 792, 792, 798, 798, 799, 800, 800, 805, 806, 810, 810, 812, 814, 816, 816, 817, 819, 820, 825, 828, 828, 832, 836, 836, 837, 840, 840, 840, 840, 846, 850, 851, 855, 858, 858, 860, 861, 864, 864, 864, 868, 870, 874, 874, 875, 880, 880, 882, 884, 888, 891, 893, 896, 897, 899, 900, 900, 902, 903, 910, 912, 912, 918, 920, 920, 924, 924, 924, 925, 928, 930, 936, 936, 940, 943, 945, 945, 946, 950, 952, 957, 960, 960, 960, 962, 966, 966, 968, 972, 975, 980, 984, 986, 987, 988, 989, 990, 990, 992, 999, 1000, 1008, 1008, 1008, 1012, 1012, 1014, 1015, 1020, 1023, 1025, 1026, 1032, 1034, 1035, 1036, 1040, 1044, 1050, 1050, 1053, 1054, 1056, 1056, 1056, 1058, 1064, 1066, 1073, 1075, 1080, 1080, 1080, 1081, 1085, 1088, 1092, 1092, 1100, 1102, 1104, 1104, 1107, 1110, 1116, 1118, 1120, 1120, 1122, 1125, 1128, 1131, 1134, 1140, 1144, 1147, 1148, 1150, 1152, 1152, 1155, 1160, 1161, 1170, 1170, 1175, 1176, 1178, 1184, 1188, 1188, 1189, 1190, 1196, 1200, 1200, 1204, 1209, 1215, 1216, 1218, 1221, 1222, 1224, 1230, 1232, 1240, 1242, 1247, 1248, 1248, 1254, 1258, 1260, 1260, 1260, 1269, 1271, 1276, 1280, 1287, 1288, 1290, 1292, 1295, 1296, 1302, 1305, 1312, 1316, 1320, 1320, 1326, 1330, 1332, 1333, 1334, 1344, 1344, 1350, 1353, 1360, 1363, 1364, 1365, 1368, 1376, 1380, 1386, 1392, 1394, 1395, 1400, 1404, 1406, 1408, 1410, 1419, 1426, 1428, 1435, 1440, 1440, 1440, 1443, 1452, 1457, 1462, 1470, 1472, 1476, 1480, 1482, 1485, 1488, 1496, 1504, 1505, 1512, 1517, 1518, 1520, 1530, 1536, 1540, 1548, 1551, 1554, 1558, 1560, 1564, 1575, 1584, 1584, 1591, 1596, 1598, 1599, 1610, 1620, 1628, 1632, 1634, 1638, 1640, 1645, 1656, 1665, 1672, 1677, 1680, 1680, 1692, 1702, 1710, 1716, 1720, 1722, 1728, 1739, 1748, 1755, 1760, 1763, 1776, 1786, 1794, 1800, 1804, 1806, 1824, 1833, 1840, 1845, 1848, 1872, 1880, 1886, 1890, 1892, 1920, 1927, 1932, 1935, 1968, 1974, 1978, 1980, 2016, 2021, 2024, 2064, 2068, 2070, 2112, 2115, 2160, 2162, 2208, 2256 There's a simpler form for the product of the pairs, it's ((sum n)^2 - sum (n^2))/2, but there might be use of an algorithm for mechanistic computation of that value. I consider then the next function, that to generate the product of each triple. Consider a three-dimensional cubic array, and the values under the diagonal. Ross === Subject: Re: My (Un)Originality (was Re: Reminder: Wages, Employment Not Determined By Supply, Demand) ***>I suspect accounting identities hold in dynamic systems too.<*** They do not. The identities are coherent rules and definitions such as Debits = Credits and Assets = Liabilities + Capital. Outputs equal inputs is mere assertion imposed on real world phenomena. ***>But I made no claims about dynamic systems in my reference to Leontief above. Even if Leontief's work could not be applied in any dynamic model, his work might still be useful for examining some aspect of the real world.<*** What would that be? Since Leontief's method imposes a demonstrably false assumption from the get-go, it is completely useless for examining any aspect of the real world. Worse than useless because it leads to false conclusions. > ***>If you want to see what economists have done with > my math when looking at the real world, Leontief's > work is interesting.<*** > Really? > Yes. > This is from a standard text: > > ***>We now use a key principle in building the > [Leontiefian] economic model: No sector > 'manufactures' money. For each sector, the total > value of its inputs (from other industries and > primary inputs sectors) exactly matches the total > value of its outputs (to other industries and final > demand). Note that industries may make profit. But > the profit is paid to someone! It goes to small > business proprietors, shareholders, etc.<*** > > To paraphrase: Inputs equal outputs; outputs equal > inputs. > > An impossibility for a dynamic system. > I suspect accounting identities hold in dynamic systems too. > But I made no claims about dynamic systems in my reference to > Leontief above. Even if Leontief's work could not be applied > in any dynamic model, his work might still be useful for > examining some aspect of the real world. > So if Leontief is an example of your math, then > your math is all wrong too in respect to its > applicability to the real world. > Nicholas Georgescu-Roegen has some interesting things to > say, which I do not fully understand, critiquing and > extending Leontief input-output analysis. But I fear that > to understand Georgescu-Roegen's writings on entropy and > economics would take more intellectual seriousness than > Mr. Ryan manifests above. > Anyways, there's plenty of literature relating my > unoriginal ideas to economic dynamics. For example, > Barkley Rosser, Jr. says reswitching can be viewed > as a cusp catastrophe, whatever that means. And then > there's always Richard Goodwin's paper about swinging > along the von Neumann turnpike. > Consider Appendix B in > function.) I think the bifurcations of equilibria shown point to > the possibility of complex (non-linear) dynamics. > -- > Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html > To solve Linear Programs: .../LPSolver.html > r c A game: .../Keynes.html > v s a Whether strength of body or of mind, or wisdom, or > i m p virtue, are found in proportion to the power or wealth > e a e of a man is a question fit perhaps to be discussed by > n e . slaves in the hearing of their masters, but highly > @ r c m unbecoming to reasonable and free men in search of > d o the truth. -- Rousseau === Subject: Our differences are limited to smooth functions. I was accused of deliberately attempting to indiscriminately mix continuous and discrete operations into a mish-mash of conflicting concepts, but I proved this allegation to be wrong in my post titled, Prove the unprovable or accept the absurd. Same way, I was also accused of proclaiming that certain calculations are impossible. I never tried to prove that certain calculations are impossible. I only tried to prove what Einstein had explicitly stated: As far as propositions of mathematics refer to reality they are not certain, and as far as they are certain they do not refer to reality. In algebra you say that X is a variable. Your variable in symbolic logic is such that at no time (never) and nowhere it is a variable. But it is not a constant! It means you call the unknown (but the knowable) constant as variable. Let us take the quadratic equation aX^2 + bX + c=0. Here X can have one or two real values or one real and two imaginary values. Is X a variable here? That is, if F(X)=0, then is X a variable? Definitely not. If Y=F(X), then you say X is the independent variable, meaning it follows no rule by itself, you assign its value at your will and pleasure. You assign different values to X, and each time you calculate the value of Y and get a definite value of Y/X. That is, you have a series of actual values of Y/X independent of each other. Therefore X, Y and Y/X are never and nowhere variables. This is the reason why in geometry lines are porous in the form of disconnected points. (Here connection is similar to connection between cause and effect.) In the quadratic equation aX^2 + bX + c=0, if values of the constants a, b and c are known then X satisfies a certain condition. The condition that the value of X satisfies is that it makes the size (or number of numbers, or, number of primary objects) on the LHS equal to that in the RHS; here it is 0. This is tautology. In simultaneous equations you first assign value to X so that you are left with F(Y)=0. This is again tautology. All these prove that you use the term variable to indicate finite number of values of initially unknown (but calculable) values of a set of symbols. OUR DISAGREEMENT IS LIMITED TO SMOOTH FUNCTIONS. In all cases where you assign value to the independent variable and get the value of dependent variable (including in geometry) the ratio of dependent and independent variables (or Y/X) follows no rule – Y/X is neither continuous nor it has any order. You cannot develop any expression, which shows that Y/X or slope m has continuous order. If m has continuous order the order can only be expressed in the form of dm=F(m). This is the only way you can express all internal changes and all smooth functions. If X is a continuous variable then it means X is undergoing internal change and all the variables which are dependent on X must be WITHIN X itself and not without it, and the change must be expressed as dX=F(X). dX=F(X) is a smooth/continuous function. A smooth function must have MOMENTUM of it own. In L=kT, L has constant momentum because motion of T cannot be stopped. In dX=F(X) momentum goes on increasing or decreasing by self-reference. Numbers do not have momentum of their own that is the reason it is not possible to express a continuous/smooth function in the form of Y=F(X). I also said that a smooth function that occupies a limited region in space or time is incommunicable. The reason is all smooth function must be of the form dX=F(X) and this expression is beginning less like time/field and contains no information! The expression of a smooth function that occupies a limited region is space-time is Omniscience. === Subject: Re: Our differences are limited to smooth functions. > I was accused of deliberately attempting to indiscriminately mix > continuous and discrete operations into a mish-mash of conflicting > concepts, but I proved this allegation to be wrong in my post titled, > Prove the unprovable or accept the absurd. No, you didn't prove anything of the sort. You only proved that you don't understand limiting processes -- and that you are more than willing to accept the absurd. > Same way, I was also > accused of proclaiming that certain calculations are impossible. I > never tried to prove that certain calculations are impossible. Oh yes, you did! > I only tried to prove what Einstein had explicitly stated: As far as > propositions of mathematics refer to reality they are not certain, > and as far as they are certain they do not refer to reality. You did not succeed in proving that. > In algebra you say that X is a variable. Your variable in symbolic > logic is such that at no time (never) and nowhere it is a variable. > But it is not a constant! > It means you call the unknown (but the knowable) constant as variable. > Let us take the quadratic equation aX^2 + bX + c=0. Here X can have > one or two real values or one real and two imaginary values. > Is X a variable here? That is, if F(X)=0, then is X a variable? > Definitely not. X is a solution to the equation F(X)=0. > If Y=F(X), then you say X is the independent variable, meaning it > follows no rule by itself, you assign its value at your will and > pleasure. You assign different values to X, and each time you > calculate the value of Y and get a definite value of Y/X. That is, > you have a series of actual values of Y/X independent of each other. I will assign values to X and calculate values of Y if I want to build a table with discrete entries. On the other hand, I might examine the relationships between X and Y to determine their properties as continuous variables -- one independent and the other dependent. > Therefore X, Y and Y/X are never and nowhere variables. ??? What does that mean ??? > This is the > reason why in geometry lines are porous in the form of disconnected > points. Are they? I doubt if Euclid would agree. > (Here connection is similar to connection between > cause and effect.) > In the quadratic equation aX^2 + bX + c=0, if values of the constants > a, b and c are known then X satisfies a certain condition. > The condition that the value of X satisfies is that it makes > the size (or number of numbers, or, number of primary objects) on the > LHS equal to that in the RHS; here it is 0. This is tautology. > In simultaneous equations you first assign value to X so that you are > left with F(Y)=0. This is again tautology. All these prove that > you use the term variable to indicate finite number of values of > initially unknown (but calculable) values of a set of symbols. It proves nothing of the sort. > OUR DISAGREEMENT IS LIMITED TO SMOOTH FUNCTIONS. No it isn't. One major disagreement concerns your promise to depart from this newsgroup permanently. You said you would. You haven't. I disagree with your decision to persist in haunting this newsgroup. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: proper place to put an acknowledgement in a math paper I was wondering where the proper place is to put an acknowledgement in I see this section used exclusively for grant and funding information and a separate section{Acknowledgement} containing the acknowledgement. Which one is preferable? === Subject: Re: proper place to put an acknowledgement in a math paper Visiting Assistant Professor at the University of Montana. >I was wondering where the proper place is to put an acknowledgement in >I see this section used exclusively for grant and funding information >and a separate section{Acknowledgement} containing the >acknowledgement. Which one is preferable? Some journals prefer one to the other. There is also sometimes a sections. In my own personal and rather limited experience, the support. This work was done while the author was visiting at... I thank the University of Salsipuedes for their hospitality during my stay, This work was partially supported by grant XYZ-123 from the Ministry of Disinformation, that sort of thing; while the Acknowledgement section is used to thank specific people for help in the actual body of the paper or for developing some ideas in the paper. But, again, you should check the journal's Guide to Authors. Some journals require that both types of acknowledgements be given in a separate section at the end with a specific title; others ask that any It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: proper place to put an acknowledgement in a math paper >I was wondering where the proper place is to put an acknowledgement in >I see this section used exclusively for grant and funding information >and a separate section{Acknowledgement} containing the >acknowledgement. Which one is preferable? Different publishers have (very) different rules. Until and unless the paper is submitted for publication, put the acknowledgments where you please. When and if you submit it, if you can determine the preference of the journal to which it's being submitted, then you can move it if you want to. I'm quite sure no editor (except, perhaps, that fiend-in-human-form J-P S---e) would refuse to read a paper with misplaced acknowledgments, on that ground alone. Lee Rudolph === Subject: Re: Paraboloidal Coordinates > I've been looking at some coordinate systems on Mathworld, and for some of > them we get x, y, z in terms like > > x^2 = (a^2-lambda)(a^2-mu)(a^2-nu)/(b^2-a^2) > > etc. > > What's with the squared quantities on the left-hand side? We're only > supposed to work in an octant of three-dimensional space? Can't > distinguish x=10 from x=-10? > This is true for many coordinate systems (bipolar for instance). Like > all things in math (and life) choice of a suitable coordinate system > requires some judgement. These systems are only useful for symmetrical > graphs, so f(10) = f(-10) and the distinction just doesn't matter. If > the symmetry is not present, just choose a different coordinate > system. > Darren If I wanted to find a metric, Christoffel symbols, divergence, and other such things, is there any problem with a naive application of the standard methods in, e.g., Arfken or a general relativity text? I assume I should first take x=(+/-)sqrt(stuff) and then I'll have a bunch of +/- symbols floating around in the results. === Subject: Re: one-way function (VMPC) > This inverse is not necessarily unique. > When n = 8, here are six values of P yielding the same Q: > P (0)(4)(176)(253) (03)(176524) (04)(176532) > x 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 > (...) In fact. The problem of inverting the function can be simplified to dinding ANY such P, which generates a given VMPC(P). It is unbelevable to be so hard while the function is so simple. In fact it's the addition operation only, which produces the function's resistance to inverting... > This Q is a composition of four applications of P, plus additions > of 1 and of 2 modulo n. The resulting Q will always be an odd permutation > if n is even, and an even permutation if n is odd, whether > the input P is odd or even. The original definition, with only three > references to P, does not have this problem. Hm, interesting observation. However it's not going to make the inversion easier? -- Bartosz Zoltak http://www.vmpcfunction.com qpbzoltak@vmpcfunction.com without qp === Subject: Re: Is there someone to help me PLEASE... > 1-The first qustion is to find the largest set > of natural numbers that if we choose two members a and b then ab+1 > is square of a natural number.(is there any limit for the number of > members of the set?) > I conjecture four (counting only the positive numbers). > Martin Gardner, in one of his puzzle books, mentions the following > set > {1, 3, 8, 120} > as having the interesting property that any two members of the set, > multiplied together, are always one less than a perfect square. > He then asks for a fifth integer which can be added to the set > without destroying its unique property. > Solution left as exercise for the reader. ;-) That was exactly the problem solved (in the negative) by Baker and Davenport in 1969, i.e. they showed there was no fifth integer that could be added with the augmented set preserving the property. The proof wasn't elementary, and as I recall it used the theory of log-linear forms, which had been developed by Prof Alan Baker et al for studying transcendental numbers. --------------------------------------------------------------------------- John R Ramsden (jr@adslate.com) --------------------------------------------------------------------------- A stockbroker is someone who invests your money until it's all gone. Woody Allen === Subject: Re: Is there someone to help me PLEASE... 1-The first qustion is to find the largest set > of natural numbers that if we choose two members a and b then ab+1 > is square of a natural number.(is there any limit for the number of > members of the set?) > I conjecture four (counting only the positive numbers). > Martin Gardner, in one of his puzzle books, mentions the following > set > {1, 3, 8, 120} > as having the interesting property that any two members of the set, > multiplied together, are always one less than a perfect square. > He then asks for a fifth integer which can be added to the set > without destroying its unique property. > Solution left as exercise for the reader. ;-) > That was exactly the problem solved (in the negative) by Baker and > Davenport in 1969, i.e. they showed there was no fifth integer that > could be added with the augmented set preserving the property. No; while I'm not familiar with that result, I am almost certain that Baker and Davenport must have shown that there was no fifth *positive* integer that could be added, etc. Gardner's trivial answer: The fifth integer is zero. -Arthur === Subject: Time scale calculus in number theory [I copied this to sci.math and sci.math.research a couple of months ago, and it never showed up in either. Strange!] See http://www.newscientist.com/inprint/ The erratic stop-start of the real world has given mathematicians a headache, says Vanessa Spedding. Now they have a way to deal with anything from mosquitoes to market traders p.28 I was wondering if this nascent field might be applicable in the theory of numbers, as another means, maybe even more revealing and powerful than existing methods, to link the discrete and erratic behaviour of prime numbers for example with continuous functions such as the zeta function. Sounds like good Phd thesis material if true (and maybe a megabuck payout if successful ;-) --------------------------------------------------------------------------- John R Ramsden (jr@adslate.com) --------------------------------------------------------------------------- A stockbroker is someone who invests your money until it's all gone. Woody Allen === Subject: Re: Time scale calculus in number theory >[I copied this to sci.math and sci.math.research a couple of >months ago, and it never showed up in either. Strange!] >See http://www.newscientist.com/inprint/ > The erratic stop-start of the real world has given > mathematicians a headache, says Vanessa Spedding. > Now they have a way to deal with anything from > mosquitoes to market traders p.28 >I was wondering if this nascent field might be applicable in the >theory of numbers, as another means, maybe even more revealing >and powerful than existing methods, to link the discrete and >erratic behaviour of prime numbers for example with continuous >functions such as the zeta function. http://br.groups.yahoo.com/group/ciencialist/message/26411 contains a subscription, unlike New Scientist. The bad news--its in Portugese! Chasing other references, http://at.yorku.ca/cgi-bin/amca/cajw-01 Fourth International Conference on Dynamic Systems and Applications do a search on: Time Scales and Applications It will be interesting to see what this can do that Digital Signal Processing (cf Real Time applications, by Embree) does not. John Bailey http://home.rochester.rr.com/jbxroads/mailto.html === Subject: (For James)How to derive a prime-counting function in a way requiring no ingenuity Let H(x) = 1 iff x = 0, else H(x) = 0 for x != 0. Now define a function which takes as input two numbers and a true or false statement. Let's call it f and define it thus: f(a,b,P) = a if P true, b if P false Lemma 1 (just some trivial computational results): 1a. f(a,b,P) = f(b,a, not P) 1b. f(a,b,P) = f(a-b,0,P) + b 1c. f(a,b,P) = f(1,b/a,P) * a assuming a != 0 1d. Let x be real, p and q be natural. 1da. f(1,0,x=0) = H(x) 1db. f(1,0,p|q) = H(q/p - floor(q/p)) The proofs are extremely easy, just consider the 2 separate cases where the given P is false and then true. I leave them to the reader. Lemma 2: Suppose P is a statement about natural numbers which is either true or false depending on the number. Alternatively let P:N->{0,1}. Then the number of integers m between 1 and natural M inclusive such that P(m) is true (or P(m)=1) is equal to sum ( i = 1,2,3,...,M ) f(1,0,P(i)) Proof: Let P be fixed and suppose that the statement is true for all M <= k. Let n be the number of m between 1 and k inclusive s.t. P(n) true. Let N be the number of m between 1 and k+1 inclusive s.t. P(n) true. Clearly, if P(k+1) is true then N = n+1, else N = n. But sum ( i = 1,2,3,...,k+1 ) f(1,0,P(i)) is equal to sum( i = 1,2,3,...,k ) f(1,0,P(i)) + f(1,0,P(k+1)) which, by the inductive assumption, is equal to n + f(1,0,P(k+1)). If P(k+1) is true then by definition of f, this becomes n+1, else again by definition of f this becomes n. This shows that sum( i = 1,2,3,...,k+1 ) = N and we're done here. For M = 1 the lemma is obviously true. Thus by induction we're finished. Suppose P(n) is TRUE iff n is prime, else P(n) FALSE. Clearly, the number of naturals between 1 and n inclusive which divide n equals 2 iff n is prime. Therefor the statements n is prime and the number of naturals between 1 and n inclusive which divide n equals 2 are equivalent. By Lemma 2, the number of naturals between 1 and n inclusive which divide n is given by sum ( i = 1,2,...,n ) f(1,0,i|n), which by Lemma 1 (part 1db) becomes: sum ( i = 1,2,...,n ) H(n/i - floor(n/i)) For simplicity, let phi(n) = the number of naturals between 1 and n inclusive which divide n. Thus n is prime iff phi(n) = 2, and phi(n) = sum( i=1,2,...,n ) H(n/i - floor(n/i)) By Lemma 2, the number of primes between 1 and x inclusive is: sum ( i = 1,2,...,x ) f(1,0,P(i)) = sum ( i = 1,2,...,x ) f(1,0,phi(i) = 2) = sum ( i = 1,2,...,x ) f(1,0,phi(i) - 2 = 0) = sum ( i = 1,2,...,x ) H(phi(i)-2) < Lemma 1 part 1da Expanding phi, we see that the number of primes between 1 and x inclusive is: sum( i = 1,2,...,x ) H( -2 + sum ( j = 1,2,...,i ) H(i/j - floor(i/j)) ) [We could simplify this into sum( i = 1,2,...,x ) H( sum( j = 2,3,4,...,i-1 ) H(i/j - floor(i/j)) ) The details are left to the reader] I stumbled upon this when I was 14 years old. Thought at the time it was something incredibly neat and spiffy. I have since come to realize that it is useless. Computationally, it is probably about the most inefficient prime counter concievable. And it offers no insight whatsoever, sheds no analytical light at all on primes. Exercises for the reader: 1. Using the lemmas above, find a formula for the nth prime. 2. Using the lemmas above, find a formula for the nth twin prime, assuming that n twin primes exist. 3. Using the lemmas above, find a formula for the number of twin primes between 1 and n inclusive. 4. Assume that 0^0 = 0 and find a formula for H that involves nothing but basic arithmetic (including exponents) and absolute values. 5. Repeat number 4 but this time assume 0^0 = 1. 6. Plug the results of 4 and 5 into the results of 1,2, and 3. *Harder problems 7. Find a general version of the formula for H which works no matter what 0^0 is defined as, so long as it is defined as a real or complex number at all. Plug this into 1,2, and 3. *VERY hard problems 8. Find an example of a textbook where the above formulas are stated. This exercise is believed to be impossible. Does this mean that the author is now a world-famous mathematician (the same would then apply to James Harris)? Discuss. === Subject: Re: 27 straight lines on a cubic surface |My impression is that |the all-27-lines-real locus in the space of real cubic surfaces, |when compared to the various other non-empty loci defined by how |many of the lines are real, is `small' in some non-precise sense Surely it's usual for some of them to be complex lines. |(similar to the sense in which, I think, it's observed that among |real affine cubic curves, one of has an oval, has no oval |predominates--I don't remember which, and I won't even swear it's |true, but I think it's been asserted in my hearing). If it's of the form y^2=x^3+ax+b, then it gets the oval when the cubic has three real roots. I'd tend to say that in some vague sense cubics tend to have just one real root. For it to have three, the x-axis has to pass through its graph in the part where it changes direction temporarily (and there has to be such a part). Keith Ramsay === Subject: Re: 27 straight lines on a cubic surface |> I'm sure that you such surfaces do exist. If memory serves me right, |> there is a (clay ?) sculpture of one such example in the mathematical |> art collection at MSRI made by Mrs Hirzebruch. | |There used to be one in the bowels of DPMMS in Cambridge. And on the first floor of the math building at MIT, along with a model of a ruled surface and various other little models. Keith Ramsay === Subject: Re: 27 straight lines on a cubic surface |I was reading in a history text that Cayley and Salmon proved that any |cubic surface has precisely 27 distinct straight lines lying |completely in it. There can be degeneracies in which some of the lines coincide, similar to the way a polynomial of degree n has n roots, but some might coincide. Keith Ramsay === Subject: need help in algebra !!! 1.solve equiation in set of natural numbers 1/x + 1/y + 1/z = 1 2. Find natural numbers divisable by 30, so that they have exactly 30 dividers. 3.Do m and n exist, writen with same digits (as 1234 and 4132) such that m - n = 1995 4. f(n-1) | f(n^n - 1) f je Euler's function === Subject: Re: need help in algebra !!! I'm assuming this is homework but hopefully the problem's been turned in by now. As they say, use it or lose it. :-) In sci.math, Goran : > 1.solve equiation in set of natural numbers 1/x + 1/y + 1/z = 1 That's an interesting one. Of course, that can be rewritten xy + yz + xz = xyz or x^2 + y^2 + z^2 + 2xy + 2yz + 2xz = x^2 + y^2 + z^2 + 2xyz (x + y + z)^2 = x^2 + y^2 + z^2 + 2xyz I'm not sure either of those help much. Or one can simply enumerate the solutions. If x=2, for example, y=4 and z=4 is one, but so is y=3 and z=6. (y=2 and z=infinity is not a solution although it's close; the problem is infinity is not a natural number.) If x = 3, then y=3 and z=3 is the only other solution unless 2 is included somewhere, which we've already enumerated. If x,y,z > 3 there are no solutions, for hopefully obvious reasons. > 2. Find natural numbers divisable by 30, so that they have > exactly 30 dividers. Another mildly interesting puzzler. If you mean proper dividers (divisors?), then this means that there has to exist unique prime values p_i and integer exponents e_i > 0 such that product(all i) (p_i^e_j) = N for some N, which is a multiple of 30, and product(all i) (e_i + 1) = 31 (mostly because this product is the number of combinations of prime powers in the product). Since 31 is prime there is only one i, which is clearly impossible as that requires 30 to be a power of a prime, which it's not. If you include the improper divider of the number itself, then product(all i)(e_i+1) = 30. 30 = 2 * 3 * 5, so there is exactly one solution: p1 * p2^2 * p3^4 for some value of p1, p2, and p3. Since the requirement is for the number to be a multiple of 30, p1, p2, and p3 must be selected from the primes 2, 3, and 5 (putting in an additional prime would muck up things). This means there are 6 such numbers; the computation of these I'll leave to the reader apart from the lowest one: 5 * 3^2 * 2^4 = 720. > 3.Do m and n exist, writen with same digits > (as 1234 and 4132) such that m - n = 1995 No, as 1995 is not divisible by 9. The reason is that the difference is the sum of terms a * (10^i - 10^j) for non-zero integers a and i >= 0, j >= 0. Since 10^i - 10^j is always a multiple of 9 (the proof is left to the reader) the problem as stated is impossible. Had you stated 1998, there would be a number of solutions; the simplest one might be arguably 3001 - 1003; another one might be 3421 - 1423. > 4. f(n-1) | f(n^n - 1) > f je Euler's function This problem specification is too ill-formed for me to process further. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: need help in algebra !!! > 1.solve equiation in set of natural numbers 1/x + 1/y + 1/z = 1 {2,3,6} > 2. Find natural numbers divisable by 30, so that they have > exactly 30 dividers. > 3.Do m and n exist, writen with same digits > (as 1234 and 4132) such that m - n = 1995 m - n == 0 (mod 9), but 1995 == 6 (mod 9). > 4. f(n-1) | f(n^n - 1) > f je Euler's function === Subject: Re: James Harris [professional mathematicians as a class] may feel that electic lights, and maybe electricity itself is a personal affront -- mea culpa; are you joining ELF? learn to do a geometrical proof. that is not irreconcilable with any notion of scientific method or an attack upon any fad of mathematical physics. constructive geometry a la the Pythagoreans is the first desktop science, requiring only that flat surface, paper and a pair of compasses. (I still don't know exactly what that is supposed to mean, but it's a specific kind of compass that is not as simple to use.) > Well, for the bulk of my postings that isn't necessarily a bad > assessment. > ************1996**************** > * First 49 days of 2002: 411 posts, or an avg of 8.4 per day > Wow. I've been quite dedicated. > Apparently to these people you see *mathematics* has nothing to do > with thinking, so advanced problem solving techniques aren't > necessary, and in fact are forbidden. --A church-school McCrusade (Blair's ideals?): Harry-the-Mad-Potter want's US to kill Iraqis?... http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX === Subject: Re: James Harris James Harris skrev i melding > Here's a summary of the JSH lamebrain's FLT postings to sci.math: > Well, for the bulk of my postings that isn't necessarily a bad > assessment. > I keep mentioning that my degree is in PHYSICS in the hopes that some > of you might show a recognition of the scientific method. > Other times, figuring some of you might hate science, I hope that at > least you appreciate the benefits, like lights, so I mention Thomas > Edison. > ************1996**************** > *Feb. 27th: first post, and it contains a mild hint that he already solved > FLT. > *Feb to April: strongly suggests he proved FLT. > *April: realizes he's a bonehead and blowhard, apologizes for being an > idiot, and unsubscribes from sci.math > ------------------------------------------------------------------ > ...obviously my attempt at a proof isn't even close. > ...obviously, I was suffering from a bit of delusion.... > ...the mild release making outrageous claims has given me > is finally surmounted by a sense of shame; so I'll quit. > ....My apologies to anyone concerned or in any way interested. > ------------------------------------------------------------------- > *June: rested for a month and a half, he's back > *August: decides Pierre done beat him this time: My last post was an > apology for annoying you with my attempts at proving FLT with simple algebra > which is obviously an impossible task, and such attempts have been a source > of irritation for mathematicians for hundreds of years. I was forcing > myself off because I was scared that I was insane and hopelessly obsessed > with a Quixotic approach, and was just making a complete fool of myself.... > Of course, posters on sci.math weren't so nice as to *suggest* but > would not only repeatedly call me insane, they'd talk about ways that > I should disappear, or call me inhuman! > Now why the math community would find posts about mathematics--albeit > often flawed--evidence of a *lack* of humanity is something on which > I've wondered. > *September: posts 4 different proofs > *October: on two consecutive days, posts a Correct Proof, a Corrected > Proof, and a Completed Proof, each one including an apology for the > previous post being incorrect. His errors included ...a brain fart. > ....I reworked everything, including taking out some embarrassing errors or > unimportant statements that had stayed in most of the previous posts. > *November: one post > *December: no posts > ************end 1996****************** > * 1997: 65 posts (approx 1.25 per week) > * 1998: 517 posts (approx. 1.4 per day) > One post includes: > I shouldn't be suprised that every time I think I'm finished with FLT there > seems to be a little bit to be added. > * 1999: 1,030 posts (approx. 2.8 per day) > * 2000: 393 posts (approx. 1 per day) > * 2001: 1,140 posts (approx. 3 per day) > * First 49 days of 2002: 411 posts, or an avg of 8.4 per day > Wow. I've been quite dedicated. > Still I see posters confidently, and proudly making posts that > apparently are bent on attacking me for following one of the greatest > ideas in human history: the scientific method. > Apparently to these people you see *mathematics* has nothing to do > with thinking, so advanced problem solving techniques aren't > necessary, and in fact are forbidden. > But you see words like brain-storming, creativity, problem > solving, and perseverance DO actually point to activities that > society finds appealing. > But then again, it's been well-known that mathematicians pride > themselves on NOT being practical. They may feel that electic lights, > and maybe electricity itself is a personal affront. > James Harris Mr. Harris I have degrees in physics (geophysics) and mathematics. When you talk about the scientific method, I suddenly realize that you are trying to use this on mathematics. It can be done only when we are talking about conjuctures, not theoremes. The theoremes are prooved. It's no way you can change that! All your work can be controlled by the theoremes. If, as has been shoved, your work does not comply with these, you are wrong! You can't just redefine the way mathematics is dealt with. In mathemathics we proove things, in natural sciences we can only unverify things (nothing can be prooved or verified). Karl-Olav Nyberg === Subject: Re: James Harris My American wife (NJ) has just told me how bad my spelling and grammar was in this post. My apoligy for that! === Subject: Re: James Harris > My American wife (NJ) has just told me how bad my spelling and grammar was > in this post. My apoligy for that! That is unfortunate, oh well you can always move to Australia :) Herc === Subject: Re: James Harris >My American wife (NJ) has just told me how bad my spelling and grammar was >in this post. My apoligy for that! ApolOgy accepted. :-) English spelling is very hard. === Subject: Re: James Harris Watch that spelling! What are mathemathicans or mathema- tisions ? === Subject: Measure question Say S is an abstract space, G is a subset of S and a Borel field, and u(X) is a measure on G. For each n, Q_n is in G, and the union of all Q_n=S. Also for each n, u(Q_n) is finite. How does one prove that for each A in G: limsup u(Q_n intersection A)= u(A) All help appreciated. === Subject: Probability-bayes formula I'm a bit confused with this problem. You have trouble detecting 0's and 1's. The prob. that you detect 0 correctly is .9 The prob that you detect 1 correctly is .85 I need to calculate the prob that you are correct given that you did see a 1. Now I know that (1)P(A | B) P(B) + P(A | B* ) P( B* ) = 1, (2)P ( AB ) + P( AB* ) = P( A ) and that (3)P (A | B) + P( A* | B) = 1 If I assume that the prob. that you detect 0 correctly is .9 means P( you detect 0 and you are correct) = .9and that the prob that you detect 1 correctly is .85 means P( you detect 1 and you are correct) = P( you do not detect 0 and you are correct) =.85 then by (2) we get P( you are correct ) = 1.75 which is absurd. If I assume that the prob. that you detect 0 correctly is .9 means P( you detect 0 | you are correct) = .9 and that the prob that you detect 1 correctly is .85 means ( you detect 1 | you are correct) = .85 then by (3) we get .9 + .85 = 1 which is also absurd. So I guess that the prob. that you detect 0 correctly is .9 means P( you are correct | you detect 0) = .9 and that the prob that you detect 1 correctly is .85 means P( you are correct | you detect a 1) = .85 I'm not sure why to conclude these last two equations or if in fact they're correct. If they're correct, then the answer to the problem is simply .85?? === Subject: Re: Probability-bayes formula > I'm a bit confused with this problem. > You have trouble detecting 0's and 1's. > The prob. that you detect 0 correctly is .9 > The prob that you detect 1 correctly is .85 > I need to calculate the prob that you are correct given that you > did see a 1. Here's how I would define the events of interest. This *assumes* that there are no exceptions like bit unrecognized possible. D1 = detected 1 D0 = detected 0 = ~D1 S1 = sent 1 S0 = sent 0 = ~S1 That would give me p(D0/S0) = 0.90 p(D1/S1) = 0.85 and I would want to find p(S1/D1) = ? You seem to be slicing things up differently: C = bit detected correctly ~C = bit not detected correctly I'm not sure how to put that into my terms. I can split C up into separate cases, C1 = detected 1 correctly = detected 1 given 1 was sent C0 = detected 0 correctly = detected 0 given 0 was sent but then I'm not sure how to combine them. This makes sense to me, somewhat: C = C1&S1 or C0&S0 ~C = ~C1&S1 or ~C0&S0 p(C) = p(C1/S1) p(S1) + p(C0/S0) p(S0) p(~C) = p(~C1/S1) p(S1) + p(~C0/S0) p(S0) Notice that, the way I'm using them, C1 and C0 act the same as D1 and D0. > Now I know that > (1)P(A | B) P(B) + P(A | B* ) P( B* ) = P(A), [as amended] > (2)P ( AB ) + P( AB* ) = P( A ) and that > (3)P (A | B) + P( A* | B) = 1 > If I assume that the prob. that you detect 0 correctly is .9 means > P( you detect 0 and you are correct) = .9 and that > the prob that you detect 1 correctly is .85 means > P( you detect 1 and you are correct) = > P( you do not detect 0 and you are correct) =.85 then by (2) > we get P( you are correct ) = 1.75 which is absurd. You don't want p(E&C) or p(~E&C) because you already _know_ 1 or 0 was detected _correctly_. Therefore C should go in the given slot, if it goes anywhere. > If I assume that the prob. that you detect 0 correctly is .9 means > P( you detect 0 | you are correct) = .9 and that > the prob that you detect 1 correctly is .85 means > ( you detect 1 | you are correct) = .85 then by (3) we get > .9 + .85 = 1 which is also absurd. I agree that there is an interpretation possible that looks like that, except for the numbers not checking out. In my terms, it would look like P( detect 0 | correct) = p(D0&S0)/p(D0&S0 or D1&S1) P( detect 1 | correct) = p(D1&S1)/p(D0&S0 or D1&S1) Apart from the numbers not working, though, I don't thing much of it because that's not something that is very useful for you to know. It's the fraction of successfully sent bits that are zero or one, and that will change depending on what's sent. Also, that means you have no information about the error rate of the bits being sent. > So I guess that the prob. that you detect 0 correctly is .9 means > P( you are correct | you detect 0) = .9 and that > the prob that you detect 1 correctly is .85 means > P( you are correct | you detect a 1) = .85 > I'm not sure why to conclude these last two equations or if > in fact they're correct. > If they're correct, then the answer to the problem is simply .85?? No, you're right to be skeptical of this. You can't have a probability that you detect 0 with you detect 0 in the given slot, not usefully, anyway. The interpretation that seems most natural to me is that some measurement has been made of some communication device. Zeros are sent in one end, and the fraction of zeros seen at the other is recorded -> 90%. Similarly for ones -> 85%. After the measurements are made, the device is actually used. A one is received; how sure are we that a one was sent? That's why I divided things up that way. I don't think we really care about being correct as such. We just want to know what was sent. There is another issue here, too. If you go with my interpretation, you'll need some sort of value for p(S0) and p(S1). Even if you do it your way, you'll at least need p(C) to get some definite answer. There may be more information available that you haven't paid attention to. However, in case there isn't, I suggest you make a reasonable assumption about, say, p(S1), make clear it's an assumption, and justify it as best you can. Alternatively, I suppose you could leave it unknown, and report p(S1/D1) as a function of p(S1). Jim Burns === Subject: Re: Probability-bayes formula Eqn (1) should read P(A | B) P(B) + P(A | B* ) P( B* ) = P( A ). > I'm a bit confused with this problem. > You have trouble detecting 0's and 1's. > The prob. that you detect 0 correctly is .9 > The prob that you detect 1 correctly is .85 > I need to calculate the prob that you are correct given that you did see a > 1. > Now I know that (1)P(A | B) P(B) + P(A | B* ) P( B* ) = 1, > (2)P ( AB ) + P( AB* ) = P( A ) and that > (3)P (A | B) + P( A* | B) = 1 > If I assume that the prob. that you detect 0 correctly is .9 means P( you > detect 0 and you are correct) = .9and that > the prob that you detect 1 correctly is .85 means P( you detect 1 and you > are correct) = P( you do not detect 0 and you are correct) =.85 then by (2) > we get > P( you are correct ) = 1.75 which is absurd. > If I assume that the prob. that you detect 0 correctly is .9 means P( you > detect 0 | you are correct) = .9 and that > the prob that you detect 1 correctly is .85 means ( you detect 1 | you are > correct) = .85 then by (3) we get .9 + .85 = 1 which is also absurd. > So I guess that the prob. that you detect 0 correctly is .9 means P( you are > correct | you detect 0) = .9 and that > the prob that you detect 1 correctly is .85 means P( you are correct | you > detect a 1) = .85 > I'm not sure why to conclude these last two equations or if in fact they're > correct. > If they're correct, then the answer to the problem is simply .85?? === Subject: Re: 3x-1 > I was messing around with the 3x+1 (Collatz Conjecture) and flipped the sign > so that the algorithm was 3x-1. The results are similar, but not the same. > I found for all integers less than 1,500,000 the set always resulted in a > pattern, I chose the following integers to end the sequence 1, 7, and 17 (a > Special number, and two primes). > Is 3x-1 the same type of problem as 3x+1? > Mike Curry Hi Mike. I like to messa'round with these 3x(+/-) Y myself. Yes indeed there are systems that work with a minus. One can try 1x+y or 1x-y 3x, 5x and so on. I think 7x-y has systems that converge I think it has been seen that some systems of Ax+y escape and some settle down. there is a difference between x+y and x-y that is true.. Remember zero is even. Does that answer anything? Ernst === Subject: Re: Dark Matter Puzzle A pile of loon doodoo. Jack, for crying out loud you're approaching seventy (or so). Don't you think it is time to put the reefer down and try and join up with reality? === Subject: Re: Dark Matter Puzzle > A pile of loon doodoo. > Jack, for crying out loud you're approaching seventy (or so). Don't > you think it is time to put the reefer down and try and join up with > reality? Now _there's_ a pot calling a kettle black... <*!PLONK!*> === Subject: Re: Dark Matter Puzzle > > A pile of loon doodoo. > > Jack, for crying out loud you're approaching seventy (or so). Don't > you think it is time to put the reefer down and try and join up with > reality? > Now _there's_ a pot calling a kettle black... > <*!PLONK!*> Gordon, poor lad. You think you've got a handle on how the universe works but all you've got a hold of is your own ass. I suggest you pull your head out of it before you suffocate. CC. === Subject: Re: Thoughts on the Collatz conjecture > I am still convinced that the conjecture is true, because after > hastily throwing together a computer program that I believe makes the > case. > > My argument here is, as each level of the Collatz tree grows new > branches are born derived from a smaller integer start number and so > does the density bands (see below) and their symmetry. This causes > a symmetrical squeeze play so to speak for other possible counter > examples start numbers and trees. > I did a computer program where integer start numbers are entered in > order where start number n = 1,2,3,4,5,6..n with there associated > sequences. Each start # turns on a corresponding numbered pixel and > all the path members turn on their respected pixels. These pixels stay > on. > Naturally some of these pixels are on when they get hit again for an > on because of the branching tree affect and also returning back to > 4,2,1 and terminating on 1. What happens as each new starting (n) the > (5) density bands out ahead of the all white band become more apparent > after about 32 y rows of all on (white) pixels. This would be an > integer start number the size of n =32*640 = 20480. > I set this thing up for reading across --- > 640 pixels where x ( n) (start #)= 1 to 640 and y=1 > Then where x = 1 to 640 and n=641 to 1280 and y=2 etc. > Y is set for a max. of 350 pixels. It bypasses any on pixel that is > not viewable on screen where y >350. > > It creates an interesting effect with 6 distinct bands with each of > the 5 bands having distinct symmetrical density patterns out ahead of > (n) the seed that is the all white band behind the seed. > This density band effect could be do to certain delta factors out > ahead of the seed or something to do with 2^n? > The short basic program is listed below with plenty of documentation > so someone can translate to Java, c++ or any other language. > > ? = Docs > 4 ? A Collatz conjecture pixel evaluation > 5 ' This program turns on the appropriate pixel for each starting > integer and all its sequence members. > Pixels once on, stay on. > 10 CLS > 12 Screen 9: ? Set graphics screen mode to 350 X 640 pixels > 15 DEFDBL A: ?Double precision for any variable beginning with A > 20 A=1:A3=A:A4=1:Y=1:A5=640:A6=640: PSET(A,Y): ? A is starting > integer (seed) and turns on pixel x(A) = 1 and Y = 1: This line never > used again. > 30 A1$=STR$(A): ? Line 30-54 checks to see if integer is odd or even. > 40 A2=LEN(A1$) > 50 J$=MID$(A1$,A2,1): IF J$= 1 THEN GOTO 200 > 51 IF J$= 3 THEN GOTO 200 > 52 IF J$= 5 THEN GOTO 200 > 53 IF J$= 7 THEN GOTO 200 > 54 IF J$= 9 THEN GOTO 200 > 60 A3=A3/2:A=A3:GOSUB 500: IF A =< 1 THEN 320 ELSE 30:' This line > handles even integers and goes to subroutine that evaluates the > correct x and y pixel to turn on. > 200 A3=(A*3) +1:A=A3:GOSUB 500:GOTO 60: ' Handles odd integers of seed > and its sequence. Ect. > 320 A4=A4+1:A=A4:A3=A4:GOSUB 500:GOTO 30: 'Retrieves the next seed and > repeats the whole process creating a new sequence from that seed. > 500 IF A>A6 THEN A6=A6+A5:Y=Y+1:ELSE 530: ' Sets Up A for right row > (Y) > 510 If Y>350 THEN Y=1:GOTO 540: ' If integer value in any sequence is > (350*640) then this line bypasses the pixel command (PSET) because > pixel will not be in a viewable area of the screen. > 515 IF A>A6 THEN 500: ' Go back to line 500 and add another 640 to > variable A6 > 520 IF A=< A6 AND Y>1 THEN > Y=Y-1:A7=Y*A5:A8=A-A7:Y=Y+1:PSET(A8,Y):Y=1:GOTO 540: ' Sets up x(A8) > value when y>1 and thus the correct x,y coordinates for any applicable > integer with a value > 640. > 530 If Y=1 THEN PSET(A,Y): GOTO 540:' A branch from 500 where Else > means Y=1 > 540 A6=A5:Y=1:RETURN: ' Resets variables and returns for next > integer. > 600 END > Please excuse the hastily thrown together code. Should have done a > renumber also! > You have to think of each row of 640 pixels as rows cut off at that > point and then stacked on each other where you can easily view how > these density patterns out ahead of the seed number are formed. This > would probably go unnoticed if the line stayed continuous as in the > number line. > This creates 5 distinct and fascinating density band patterns that > grow in width as the first solid white band or seed band grows in > width. > Please note, when first starting out the bottom (last) density pattern > starts to show a checkerboard pattern on an angle. > > If nothing more, its interesting! > As always, any evaluations or comments are welcome. > Dan Replying to the last of this thread ( at my point of view at this time ). I have new data. Is anyone interested in forming a group? It would be great to crack this problem. We should lay down the pride and work together. I think we can crack this problem. Ernst === Subject: Dice and probability Hello... I'm new to this group, and I'm not a math wiz, so please forgive me if this is an inappropriate question. If I have 10 dice, each a different color, and I roll all of them together 10 times, what is the chance that the 10 die would come up the numbers that they did, in the order that they did? Also, is there an equation that I can plug in a variable number of dice and variable numbers of throws? TIA, D. === Subject: Re: Dice and probability > Hello... > I'm new to this group, and I'm not a math wiz, so please > forgive me if this is an inappropriate question. > If I have 10 dice, each a different color, and I roll all > of them together 10 times, what is the chance that the 10 > die would come up the numbers that they did, in the order > that they did? > Also, is there an equation that I can plug in a variable > number of dice and variable numbers of throws? > TIA, > D. Do you really mean to roll the 10 dice 10 times?? If you roll the 10 dice once then the answer is 1/(2^10). === Subject: Re: Dice and probability === >Subject: Re: Dice and probability >Message-id: Hello... >> I'm new to this group, and I'm not a math wiz, so please >> forgive me if this is an inappropriate question. >> If I have 10 dice, each a different color, and I roll all >> of them together 10 times, what is the chance that the 10 >> die would come up the numbers that they did, in the order >> that they did? >> Also, is there an equation that I can plug in a variable >> number of dice and variable numbers of throws? >> TIA, >> D. >Do you really mean to roll the 10 dice 10 times?? If you roll the 10 dice >once then the answer is 1/(2^10). Care to make a small wager on that? -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Dice and probability === >Subject: Re: Dice and probability >Message-id: Hello... >> I'm new to this group, and I'm not a math wiz, so please >> forgive me if this is an inappropriate question. >> If I have 10 dice, each a different color, and I roll all >> of them together 10 times, what is the chance that the 10 >> die would come up the numbers that they did, in the order >> that they did? >> Also, is there an equation that I can plug in a variable >> number of dice and variable numbers of throws? >> TIA, >> D. >Do you really mean to roll the 10 dice 10 times?? If you roll the 10 dice >once then the answer is 1/(2^10). > Care to make a small wager on that? No I don't. How about 1/6^10. > -- > Mensanator > 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Dice and probability === >Subject: Dice and probability >Message-id: <4LS1b.450$oh3.88745@news.uswest.netHello... >I'm new to this group, and I'm not a math wiz, so please >forgive me if this is an inappropriate question. >If I have 10 dice, each a different color, and I roll all >of them together 10 times, what is the chance that the 10 >die would come up the numbers that they did, in the order >that they did? Probablity is the total number of possible successes divided by the total number of possible outcomes. For a particular number to appear on a single die, say 6, there is only one outcome that is 6 out of six faces on the die. So the probability for one die is 1/6 For a pair of numbers to appear on a pair of distinct (different colors) dice, it is 1/6 * 1/6 For n distinct dice it is (1/6)^n That's the probability for _a_ particular set of dice. For m disticnt rolls of n dice, the probability is ((1/6)^n)^m >Also, is there an equation that I can plug in a variable >number of dice and variable numbers of throws? >TIA, -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Empirical Proof of NP=P >I was wondering if current state of the art SAT solvers >provide an empirical proof of NP=P. > [...] >There are existing state of the art solvers are much better than mine. >Has anyone found a solvable problem that can't be solved >in polynomial time by these programs? > The fastest current solvers use backtracking plus > clause learning, and on unsatisfiable inputs can > be simulated by resolution. There are many families > of instances on which they provably require > exponential time. These include random formulas, > pigeon-hole formulas, and a slew of others. > There is no such thing as a satisfiable instance > that cannot be solved by them in polytime -- any > finite set of instances can be solved in time n^k > for some fixed k. > Here's a more reasonable question you might ask instead: > Is there any known, infinite, set of satisfiable CNF > formulas for which; > - there is no known polytime algorithm to construct > satisfying assignments, and > - no current solver appears to solve them in polytime > with high probability. (Here, you must pick k s.t. > n-variable instances are solved typically in > less then n^k time, and 2n-variable instances are > solved consistently in much less then (2n)^k, for > whatever values of n my choice of solver works.) > Constructing such a set would be considered a useful > contribution -- it's not easy. Failure to construct > such sets is not very strong evidence that P=NP. I think we can show that it is impossible to construct such a set. I agree that this is not enough to prove P=NP. Let me define some terms. Define a N variable truth table to be a Boolean expression. A CNF instance represents a Boolean expression. One expression can be represented by a lot of different instances. There are at least 2^N instances for a given expression. Another way to think of this is to look at how many ways a CNF instance can be renumbered. Any solvable 3-SAT problem can be renumbered so that its solution is setting all variables false. I can define a very simple SAT solver. Call it the Stupid Solver (SS). SS looks to see if setting all variables to false is the solution. If so, SS returns true, else SS returns false. In theory, SS can solve any set of expressions because every expression has at least one instance where SS will return true. Take the factoring problems suggested by one poster. The polytime formula that encodes the factoring problem could, in theory, always produce CNF instances that are solvable by Stupid Solver. Russell - 2 many 2 count === Subject: Re: Empirical Proof of NP=P >> The worst case was 7,523,876 clauses examined. >> This is approximately 100^3.4. > [and] >The average number of clauses examined was about 63,500. >This is around 50^2.8. > So for 50 vars you take n^2.8, but > for 100 vars you take n^3.4. > What about 200 vars? > If you're letting the exponenent increase with n, > you'll have a hard time convincing anyone you're > talking about polynomial time. I don't think this line of thinking is helpful. In fact I think it's misleading the OP. For problems in NPC like SAT there is currently no known algorithm which solves all (every single and not one omitted) instances in polynomial time. The fact that you can produce an algorithm that operates efficiently on a handful (10, 20, 100, a finite amount) of instances says absolutely nothing about a solution to the general problem. Consider that in the case of factoring which is no harder than SAT it's easy to devise an algorithm which efficiently solves infinitely many instances (2n), and yet doesn't have impact on the general problem. As a related question, is it the case that there is a p-time algorithm which solves any finite subset of instances of any problem in NPC? I would think yes. My reasoning is that a finite set of problem instances could be answered by a (possibly large) look up table. > DM > -- > Dave === Subject: Re: Empirical Proof of NP=P 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > | > I was wondering if current state of the art SAT solvers > | > provide an empirical proof of NP=P. > | You can not test this empirically by looking only at problems of a > | certain fixed size, large though it may be (and 100 variables is > | well within the range these algorithms can solve) -- you need to > | plot average runtime against problem size for different problem > | sizes and fit a curve to it. > Even that's wouldn't be sufficient. An algorithm that runs in O(n^5 + > 2^{n/10000}) steps would exhibit polynomial behavior for any realistic > value of n. On the other hand, an algorithm that ran in O(n^{100}) > time would be effectively indistinguishable from exponential. (Add > zeros as necessary.) > The most you can hope for is an empirical proof that it doesn't matter > whether or not P=NP. Notice I didn't say empirical proof, I don't think those two words go together very well. But a plot such as I suggested (and for which Will Naylor has provided data) would be a lot more convincing than the evidence we started this thread with in Easterly's post. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: Empirical Proof of NP=P Distribution: inet Originator: daw@mozart.cs.berkeley.edu (David Wagner) >Is there any known, infinite, set of satisfiable CNF >formulas for which; > - there is no known polytime algorithm to construct > satisfying assignments, and > - no current solver appears to solve them in polytime > with high probability. (Here, you must pick k s.t. > n-variable instances are solved typically in > less then n^k time, and 2n-variable instances are > solved consistently in much less then (2n)^k, for > whatever values of n my choice of solver works.) >Constructing such a set would be considered a useful >contribution -- it's not easy. On the contrary: I think it is quite easy. The factoring problem can be expressed with poly-size CNF formulas, and there is no known polytime algorithm to solve the problem. If you found an efficient problem that solves this problem, you could win tens of thousands of dollars (see RSA's Factoring Challenge). === Subject: Re: Empirical Proof of NP=P Distribution: inet >>Is there any known, infinite, set of satisfiable CNF >>formulas for which; >> - there is no known polytime algorithm to construct >> satisfying assignments, and >> - no current solver appears to solve them in polytime >> with high probability. (Here, you must pick k s.t. >> n-variable instances are solved typically in >> less then n^k time, and 2n-variable instances are >> solved consistently in much less then (2n)^k, for >> whatever values of n my choice of solver works.) >>Constructing such a set would be considered a useful >>contribution -- it's not easy. > On the contrary: I think it is quite easy. The factoring problem can be > expressed with poly-size CNF formulas, and there is no known polytime > algorithm to solve the problem. If you found an efficient problem > that solves this problem, you could win tens of thousands of dollars > (see RSA's Factoring Challenge). Hey, no fair reavealing my secret application! :-) Seriously, that's exactly what I had in mind in an earlier posting when I said that a solution that worked on 2000-4000 variables would be interesting. However, I was counting wrong -- I was thinking that factoring a 1000-bit number would need 2000 bits of input (the number and then the non-deterministic inputs for the factors), but actually the number itself is hard-coded into the formula and isn't represented by variables. So I'll half my earlier statement and say that if anyone can handle 1000-2000 variable formulas then we can do something very interesting! -- Steve Tate - srt[At]cs.unt.edu | A computer lets you make more mistakes faster Dept. of Computer Sciences | than any invention in human history with the University of North Texas | possible exceptions of handguns and tequila. Denton, TX 76201 | -- Mitch Ratliffe, April 1992 === Subject: Bob's Positive Integer Pages http://my.tbaytel.net/forslund/index.html Enjoy - Bob === Subject: Re: Bob's Positive Integer Pages === >Subject: Bob's Positive Integer Pages >Message-id: Enjoy - Bob I enjoyed it greatly. You're as funniy as James Harris and your work is equally a load of rubbish. The reference to Nico Benshop really made my day. Good luck promoting your new number system. -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Bob's Positive Integer Pages It is not my intent to promote the alternate number system as a replacement for our existing system. However, i DO feel it is important that people are aware that a number system can exist without the need for the digit zero. Bob === >Subject: Bob's Positive Integer Pages >Message-id: Enjoy - Bob > I enjoyed it greatly. You're as funniy as James Harris and your work is equally > a load of rubbish. The reference to Nico Benshop really made my day. Good luck > promoting your new number system. > -- > Mensanator > 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Bob's Positive Integer Pages Glad you enjoyed it! FYI, ANS was also published in the American Mathematical Monthly by another author after i published it in the Southwest Journal of Pure and Applied Mathematics. Both are peer reviewed journals. === >Subject: Bob's Positive Integer Pages >Message-id: Enjoy - Bob > I enjoyed it greatly. You're as funniy as James Harris and your work is equally > a load of rubbish. The reference to Nico Benshop really made my day. Good luck > promoting your new number system. > -- > Mensanator > 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Bob's Positive Integer Pages === >>Subject: Re: Bob's Positive Integer Pages >>Message-id: Glad you enjoyed it! >>FYI, ANS was also published in the American Mathematical Monthly by >>another author after i published it in the Southwest Journal of Pure and >>Applied Mathematics. Both are peer reviewed journals. > Were they aware of your claim that zero is not a number? > Who were the peers, guys like Nico Benschop? > Are you sure you didn't mean beer reviewed? Don't forget that even Benschop had a bogus paper published in a peer reviewed journal: Benschop, N. F. Powersums representing residues mod $psp k$, from Fermat to Waring. Comput. Math. Appl. 39 (2000), no. 7-8, 253--261 -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Bob's Positive Integer Pages > http://my.tbaytel.net/forslund/index.html > Enjoy - Bob An Alternate Number System (ANS) - A number system that has no need for the digit zero. Fair enough, the ANS has no need for the digit zero; but _we_ do to symbolize the number nought. -- G.C. === Subject: Re: Bob's Positive Integer Pages Zero is Not a number. It is the absence of number. IMHO. Just as infinity is not a number but is rather the idea of unlimited. I do not wish to discuss this any further. There are two schools of thought that will argue endlessly just as they argue that 0.99999=1 > http://my.tbaytel.net/forslund/index.html > Enjoy - Bob > An Alternate Number System (ANS) - A number system that has no need for > the digit zero. Fair enough, the ANS has no need for the digit zero; > but _we_ do to symbolize the number nought. > -- > G.C. === Subject: Re: Bob's Positive Integer Pages > Zero is Not a number. It is the absence of number. IMHO. > Just as infinity is not a number but is rather the idea of unlimited. > I do not wish to discuss this any further. > There are two schools of thought that will argue endlessly just as they > argue that 0.99999=1 Also, I don't know of any school that argues that 0.99999 = 1. Actually, 0.99999 = 99999/100000 -- at least in the rational or real number systems. http://my.tbaytel.net/forslund/index.html Enjoy - Bob > An Alternate Number System (ANS) - A number system that has no need for > the digit zero. Fair enough, the ANS has no need for the digit zero; > but _we_ do to symbolize the number nought. > -- > G.C. === Subject: Re: Bob's Positive Integer Pages > Zero is Not a number. It is the absence of number. IMHO. You contradict yourself. I quote from your pages: When I use negative operators such as subtraction and division, I choose to restrict them such that the results are integers greater than zero. Hmmm. If zero is Not a number, then what does it mean to say an integer is greater than zero? To compare objects with 'less than' requires that the objects be members of the same set -- in this case, the integers. > Just as infinity is not a number but is rather the idea of unlimited. > I do not wish to discuss this any further. > There are two schools of thought that will argue endlessly just as they > argue that 0.99999=1 http://my.tbaytel.net/forslund/index.html Enjoy - Bob > An Alternate Number System (ANS) - A number system that has no need for > the digit zero. Fair enough, the ANS has no need for the digit zero; > but _we_ do to symbolize the number nought. > -- > G.C. === Subject: Re: Mean Reverting Process It's been a couple of days and there has been no response. Can someone please help me on this problem? === Subject: Re: Mean Reverting Process > It's been a couple of days and there has been no response. Can someone > please help me on this problem? I'm not sure whether I can help, but I have some thoughts on this. In the context of mathematical finance mean-reverting stochastic differential equations appear when modeling interest rate dynamics. For example the Vasicek model is mean-reverting. It is characterized by the following SDE: dr(t)=(b-ar(t))dt+sdW(t) Here r is the interest rate you're modeling, in this case the short rate. b and a are parameters to be estimated from the market, W is a standard Brownian motion under the martingale measure (which is not unique) and s is the volatility. This process is now reverting to the mean level b/a. (Note that you can write the SDE in integral form to be able to see what kind of process it is. The SDE is solvable and the solution has a Gaussian distribution.) So sure, it depends on the parameters to which level the process is reverting. However, where this is explained in a more general context I don't know. By the way, you speak of *the* mean reverting process. Which process is that? === Subject: probability and/or logic puzzle Yes, this came out of a textbook . . . but I passed that class about 25 years ago, so I'm not cheating on my homework. Start with a normal deck of playing cards. Discard all but the aces and kings. Now you have eight cards left. Your friend draws two cards and hides them from you. He truthfully tells you that at least one of his cards is an ace. What is the probability that he holds two aces? He replaces the cards and you shuffle them well. Your friend again draws two cards, and truthfully tells you that one of them is the ace of spades. What is the probability that he now holds two aces? (Hint: this isn't the same answer as above!) Okay . . . I don't get it. Why isn't it the same answer both times? Also, in case it matters, this isn't a math book. It's a (non-mathematical) logic book. Ted Shoemaker shoemakerted@yahoo.com === Subject: Re: probability and/or logic puzzle I'm going to try to answer this one...this may be wrong, it is late August after all... > Start with a normal deck of playing cards. > Discard all but the aces and kings. Now you > have eight cards left. Your friend draws > two cards and hides them from you. He truthfully > tells you that at least one of his cards is an ace. > What is the probability that he holds two aces? > Either he has two aces or he has an ace and a king. Since there's 4 aces in all, and it doesn't matter which ace comes first, there's 6 ways he can have 2 aces (4*3/2) Since there's also 4 kings, there's 16 ways he can have an ace and a king. (4 * 4) So the chance of him having two aces(instead of an ace and a king) is 6/22 = 3/11 > He replaces the cards and you shuffle them well. > Your friend again draws two cards, and truthfully > tells you that one of them is the ace of spades. > What is the probability that he now holds two aces? > (Hint: this isn't the same answer as above!) He has the ace of spades...there's 3 ways for him to have 2 aces (1*3) There's 4 ways for him to have an ace and a king. (1*4) So the chance of him having 2 aces is 3/7. >Okay . . . I don't get it. Why isn't it the same >answer both times? When you don't know exactly which specific object he has, the sample space for the number of ways he has an ace and a king becomes large ( = 4 ^ 2 instead of 4) faster than the sample space for the number of ways he has two aces (= 4 * 3 / 2 = 6 instead of 3) so the probability of 2 aces goes down. === Subject: Which came first: negative integers or 0? A history question: which came first: negative integers or nought (= zero)? -- G.C. === Subject: Re: Which came first: negative integers or 0? >A history question: which came first: negative integers or nought (= >zero)? 0. Introduced to European thinkers with the reconquest of Spain. Invented much earlier in India. Jon Miller === Subject: Re: Which came first: negative integers or 0? I believe that negative numbers actually came first. Zero and infinity were denied existence on religious and philosophical grounds. One implies the other. It goes back to the theory of creation. If the Greeks, and later the Catholic church, admitted the existence of zero, then that would throw a wrench into their theory of creation, as they refused to believe that the universe was created ex nihilo. Lurch > A history question: which came first: negative integers or nought (= > zero)? > -- > G.C. === Subject: Re: Which came first: negative integers or 0? > I believe that negative numbers actually came first. It seems like I've read that zero came first. Logically, it would almost have to. Consider what negative numbers are used for -- addition & subtraction, right? Now, first order of business: 1 + (-1) = ??? > Zero and infinity were > denied existence on religious and philosophical grounds. One implies the > other. It goes back to the theory of creation. If the Greeks, and later > the Catholic church, admitted the existence of zero, then that would throw a > wrench into their theory of creation, as they refused to believe that the > universe was created ex nihilo. > Lurch > A history question: which came first: negative integers or nought (= > zero)? > -- > G.C. === Subject: Re: Which came first: negative integers or 0? > I believe that negative numbers actually came first. Zero and infinity were > denied existence on religious and philosophical grounds. One implies the > other. It goes back to the theory of creation. If the Greeks, and later > the Catholic church, admitted the existence of zero, then that would throw a > wrench into their theory of creation, as they refused to believe that the > universe was created ex nihilo. That seems self-contradictory. To discuss the existence of zero they had to be acquainted with the concept of zero. Gib === Subject: Re: Which came first: negative integers or 0? denied existence on religious and philosophical grounds. One implies the > other. It goes back to the theory of creation. If the Greeks, and later > the Catholic church, admitted the existence of zero, then that would throw a > wrench into their theory of creation, as they refused to believe that the > universe was created ex nihilo. > That seems self-contradictory. To discuss the existence of zero they > had to be acquainted with the concept of zero. No contradiction. I can discuss the existence of an honest president and be acquainted with the concept of an honest president and still present evidence that the president isn't honest. Thus discussions about the existence of an honest president would conclude with a dire need for an independent invesitgation, such as team of UN inspectors, armed with cross examining and subponea powers. Another discussion to be had is does the president exist? Some think not, that the existence of the president is actually the existence of the current resident made out to look like the president. === Subject: Re: Which came first: negative integers or 0? Well, I am poorly paraphrasing. If you want the full story, read Zero: the birth of a dangerous idea Lurch > I believe that negative numbers actually came first. Zero and infinity were > denied existence on religious and philosophical grounds. One implies the > other. It goes back to the theory of creation. If the Greeks, and later > the Catholic church, admitted the existence of zero, then that would throw a > wrench into their theory of creation, as they refused to believe that the > universe was created ex nihilo. > That seems self-contradictory. To discuss the existence of zero they > had to be acquainted with the concept of zero. > Gib === Subject: Big gap Actual search results.. David C. Ullrich William Elliot Twentyman Bundgaard Jack Middleton Johnson jsavard@ecn.ab.ca Israel Mensanator Virgil << Prev 25 threads Next 25 threads >> This is an actual jump in Google.. Maybe it is a glitch Maybe it is the reality of the Neo Con's and patriot act. Yes do support patroit act.. No Money No rights! Interesting... Is this homeland security in action?? That information is to good for the people? Am I wrong here? I hope so. Ernst === Subject: Re: Big gap === >Subject: Big gap >Actual search results.. >David C. Ullrich >William Elliot >Twentyman >Bundgaard >Jack Middleton >Johnson >jsavard@ecn.ab.ca >Israel >Mensanator >Virgil ><< Prev 25 threads Next 25 threads >This is an actual jump in Google.. >Maybe it is a glitch Probably. >Maybe it is the reality of the Neo Con's and patriot act. Don't you think they would have tracked down and silenced traitors like James Harris if that were the case? > Yes do support patroit act.. No Money No rights! A glitch. > Interesting... > Is this homeland security in action?? I would hope they have bigger fish to fry than sci.math. > That information is to good for the people? > Am I wrong here? Yes. > I hope so. > Ernst -- Mensanator 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === Subject: Re: Serre's work <3f3e8483$0$266$626a54ce@news.free.fr> In , on at 11:57 PM, haberg.remove@matematik.su.se (Hans Aberg) said: >You know, the Nobel Prize is awarded for things like what Einstein >did (the photoelectric effect and other theoretical contributions to >physics) and the like. Its statutes require that one should pinpoint >some truly exceptional work. Nu, and you get a Fields medal for a pretty face? >In math, such an exceptional work might perhaps mean proving the FLT >or something like that. Proving FLT was, of course, spectacular, but what Serre did was more important. But then, you don't know or care what he did. >You are confusing success with making an exceptionally high >scientific achievement. Au contraire, you are the one who is confused. Proving FLT was a crowd pleasure, and certainly involved some high powered Mathematics, but it was not nearly as important as you seem to believe. In , on >There seems to be two pictures: Those of the Serre fan club, which >says that he was the one responsible for the Weil conjectures being >proved, could have proved the FLT if he only wanted, and so on. Why are you so fixated on FLT? Whate does it have to do with whether his work on Coherent Sheaves was important to Algebraic Geometry? >I am more inclined to believe the second picture, but you will have >to go to those interested in dealing with such historical matters to >sort that out. If you admit that you don't know enough to judge, then why are you judging, and why are you dismissing out of hand those that do know? >Einstein's collected works runs in the tens of thousands of pages, >scientific results, namely those on relativity, the photoelectric >effect and the Brownian motion. Statistical Mechanics? To say nothing of a few other odds and ends. >But if one merely points out that Serre never has done anything in >that category, one gets a lot of strong reactions like the one of >yours. No. If one makes the claim, without a shred of evidence, and insults those who disagrees, then one gets a lot of strong reactions from those who don't sufffer fools gladly. >That is pretty strange. No, what is strange is that you would make such strong statements from a position of ignorance. >It wasn't the personality, but the politics. If the politics follows >the ethical standards of science Like learning the history of his contributions before dismissing them? >I am nowadays interested in interdisciplinary science, and it makes >it hard for me to focus on this highly specialized scientific >circuit, An ethical person in that situation would refrain from making comments such as yours. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Affine Connection & Gauge Theory >Geometry on the space with affine connection is more general than >Riemanian geometry. I thought the Grassmann algebla on the former >space might be relevant with the gauge transformation. Namely the >structure equation on the Grassmann algebla corresponds to the gauge >field. No, because the gauge group is not limited to the linear transformations of the tangent space or of the Grassmann algebra on it. You need to deal with connections on an arbitrary vector bundle, not necessarily related to the tangent bundle. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Questions for James Harris > Well, you may *wish* to exclude 1/3 from the ring Z[1/2] but the > definition of the ring does NOT exclude it, and in fact, in the ring > 1/3 = 1/4 + 1/4^2 + 1/4^3 +.... > I'm curious to know how you think you might exclude it. There is no part of the definition of rings that says you have to include infinite sums. There is even no part of the definition of rings that says that the infinite sum above converges. Using some other kinds of convergence we can even get a field (the 2-adics) that contains Z[1/2] where 1/3 = -1 - 4 - 4^2 - 4^3 - ... -- 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: Questions for James Harris [snip to get to the heart of the problem] > Well, you may *wish* to exclude 1/3 from the ring Z[1/2] but the > definition of the ring does NOT exclude it, and in fact, in the ring > 1/3 = 1/4 + 1/4^2 + 1/4^3 +.... > I'm curious to know how you think you might exclude it. A ring is an additive group. That means that if you have two elements a1 and a2, then a1 + a2 is an element of the ring. Of course it is then easy to prove that if you add three elements of the ring, a1, a2, and a3, then a1 + a2 + a3 is also an element of the ring. This follows from the associative property of the ring. By induction you can prove that for any n, if a1, a2, ..., an are in the ring, then their sum a1 + a2 + ... + an is in the ring. The induction argument works for any FINITE n. A ring is closed under FINITE sums. There is no property of a ring that will guarantee that it is closed under INFINITE sums, whether they are convergent or not. A field is a ring. The rational numbers are a field. The rationals are NOT closed under infinite sums. If they were, you would conclude rationals = reals, which is not true. When you consider Z[1/2] as including infinite convergent sums, you are flaunting the definition of ring. Z[1/2] means: polynomials in x = 1/2. Polynomials are FINITE sums of powers of elements. Polynomials are not infinite series. Your insistence that Z[1/2] includes infinite series is the root of the disagreement here. By your definition, Z[1/2] includes all convergent *series* in powers of 1/2. Just to avoid confusion, let's call that not Z[1/2], but Z*[1/2]. Of course, as I believe you know, Z*[1/2] = reals. So it is not surprising that 1/3 is in Z*[1/2]. It is NOT true that Z[1/2] = reals. Moreover, Z[1/2] is not a field. Z*[1/2] is. It is also not surprising that 1/3 is NOT in Z[1/2]. The only numbers in Z[1/2] are integers and fractions which have a power of 2 in their denominators. The fact that Z*[1/2] = reals is not a new or interesting result. This ring contains even transcendental numbers like pi and e. It is a field. I cannot see how it helps you at all with respect to your object ring. All your claims to making a great discovery here that mathematicians want to ignore are bogus. What you have done is merely an abuse of notation. What you define as Z[1/2] is not correct according to the standard definition. What you have done is another way of defining the real numbers. It strongly suggests what a number of people have said here many times: you actually don't know the definition of ring, even after 8+ years of hearing it. > There is nothing specific in the ring definition that would exclude > that sum, False. Rings are closed under *finite* sums. In general an infinite series of elements in a ring is not in the ring. The ring of rational numbers is a perfect example. > and given that it's decidable, Another stupid abuse of terminology. What you call decidable is what mathematicians call convergent. There is another meaning already taken for the term decidable. You must have encountered the concepts of convergence and divergence of series in some math course you had as an undergraduate. Why do you need to invent new terminology, and then claim that mathematicians are narrow close-minded suppressors of the truth ? Why not use standard terminology? You will get to the nub of the disagreement much faster. *But maybe you don't want to*. Maybe you are trolling. You want to prolong the back-and-forth to maximize the attention you are getting. You know that these little tricks, inventing your own definitions when it is not necessary, piss mathematicians off. In that sense you are like a skunk cabbage, attracting flies because you smell bad. The flies are annoyed and disappointed when they find the plant does not really contain a dead skunk like they had hoped. Better to be a crank than a troll (if you have to choose). Better (i.e., more honest) to be a dead skunk than a skunk cabbage. The flies will be more appreciative and less angry. Better go back to your old substantive arguments, not this silly one based on an abuse of terminology. > that is, you can get a single > number answer, there is no rational reason to exclude it, pun > intended. > These and other clarifications (NOT obfuscations) would be > appreciated. > > ---- David > Well then, you can step forward once again to answer and NO > obfuscations would be appreciated. No, David is right. You are trolling here, flaunting definitions, abusing notation, and you are doing it KNOWINGLY. That amounts to obfuscation. If attention is all you want it is working. If you actually want to demonstrate new mathematical results you are of course wasting your time and everyone else's. Andrzej > James Harris === Subject: Re: Questions for James Harris clearly, Herr Doktor-Professor Harris is referring to his patented Object Ring. is the League of Ordinary Mathematicians going to sue him, for infringing upon the Reals (tm) ?? > 1/3 = 1/4 + 1/4^2 + 1/4^3 +.... > a1 + a2 + ... + an is in the ring. > The induction argument works for any FINITE n. > The rational numbers are a field. > The rationals are NOT closed under infinite sums. > If they were, you would conclude > rationals = reals, > Polynomials are not infinite series. > So it is not surprising that 1/3 is in Z*[1/2]. > It is NOT true that > Z[1/2] = reals. > Moreover, Z[1/2] is not a field. Z*[1/2] is. > It is also not surprising that 1/3 is NOT in Z[1/2]. > The only numbers in Z[1/2] are integers and fractions > which have a power of 2 in their denominators. > The fact that Z*[1/2] = reals is not a new or interesting > result. This ring contains even transcendental numbers like > pi and e. It is a field. I cannot see how it helps you at > all with respect to your object ring. --A church-school McCrusade (Blair's ideals?): Harry-the-Mad-Potter want's US to kill Iraqis?... http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX === Subject: Re: Questions for James Harris > ... you are like a skunk cabbage, attracting flies because you > smell bad. The flies are annoyed and disappointed when > they find the plant does not really contain a dead > skunk like they had hoped. > Andrzej Andrzej, -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Questions for James Harris >> I am trying to understand this 'Ring of Objects' of which you speak so >> much, but I haven't seen a definition. The closest thing that I have >> seen to a definition is The Object Ring is a commutative ring that >> includes all numbers such that 1 and -1 are the only members that are >> both a unit and an integer, and where no non-unit member is a factor >> of any two integers that are coprime. >> This raises more questions than it answers: >> 1. What kinds of things are members of your object ring? complex >> numbers? polynomials? functions? elements of some strange quotient >> ring? vectors of some kind? >My idea is to include any and all numbers that can be considered >objects versus numbers that are fractional like 1/2. >One thing that I find fascinating is that often people will respond to >me on that attacking my use of the word fractional, or if I say >fraction I get that every number is a fraction, where, they're >saying like 2/1 is a fraction, or 4/2 is a fraction. So define like 1/2 in a way that rules them out? What is a fraction? Is it a ratio of two integers? 2 is a ratio of two integers. Is it perhaps a ratio between two relatively prime integers? That leaves 5/3 and 2/15, but rules out 4/2. That would be what is usually meant at the elementary level by fraction, but you clearly want to include some irrationals while excluding others. Under what test? What is your definition? - Randy === Subject: Re: Questions for James Harris >> 4. What is the definition of integrality for members of this ring? For >> something like Z[1/2], we can use the usual notion of integrality >> (and, by the way, this ring is not a field since 3 is in it, 3 is not >> 0, and 1/3 isn't in it), and then things are all well and good. But >> what does it mean to say that an object is an integer? >Well, you may *wish* to exclude 1/3 from the ring Z[1/2] but the >definition of the ring does NOT exclude it, and in fact, in the ring > 1/3 = 1/4 + 1/4^2 + 1/4^3 +.... >I'm curious to know how you think you might exclude it. >There is nothing specific in the ring definition that would exclude >that sum, and given that it's decidable, that is, you can get a single >number answer, there is no rational reason to exclude it, pun >intended. I wonder what this would do to your object ring. I believe sqrt(2) is an object, so 1-sqrt(2) should be an object and we have (1-sqrt(2))/sqrt(2) = (1-sqrt(2)) + (1-sqrt(2))^2 + (1-sqrt(2))^3 + ..... (1-sqrt(2))/sqrt(2) = 1/sqrt(2) - 1 so 1/sqrt(2) should be an object and thus 1/2 should be an object. -- Wim Benthem === Subject: Re: Questions for James Harris Visiting Assistant Professor at the University of Montana. > 4. What is the definition of integrality for members of this ring? For > something like Z[1/2], we can use the usual notion of integrality > (and, by the way, this ring is not a field since 3 is in it, 3 is not > 0, and 1/3 isn't in it), and then things are all well and good. But > what does it mean to say that an object is an integer? >>Well, you may *wish* to exclude 1/3 from the ring Z[1/2] but the >>definition of the ring does NOT exclude it, and in fact, in the ring >> 1/3 = 1/4 + 1/4^2 + 1/4^3 +.... >>I'm curious to know how you think you might exclude it. >>There is nothing specific in the ring definition that would exclude >>that sum, and given that it's decidable, that is, you can get a single >>number answer, there is no rational reason to exclude it, pun >>intended. >I wonder what this would do to your object ring. >I believe sqrt(2) is an object, so 1-sqrt(2) should be an object >and we have >(1-sqrt(2))/sqrt(2) = (1-sqrt(2)) + (1-sqrt(2))^2 + (1-sqrt(2))^3 + ..... >(1-sqrt(2))/sqrt(2) = 1/sqrt(2) - 1 so 1/sqrt(2) should be an object >and thus 1/2 should be an object. Nice; much simpler than my proof that if it contains sqrt(n) for all integers, and is complete, then it must be all the complex numbers. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Questions for James Harris > I am trying to understand this 'Ring of Objects' of which you speak so > much, but I haven't seen a definition. The closest thing that I have > seen to a definition is The Object Ring is a commutative ring that > includes all numbers such that 1 and -1 are the only members that are > both a unit and an integer, and where no non-unit member is a factor > of any two integers that are coprime. > This raises more questions than it answers: > > 1. What kinds of things are members of your object ring? complex > numbers? polynomials? functions? elements of some strange quotient > ring? vectors of some kind? > > My idea is to include any and all numbers that can be considered > objects versus numbers that are fractional like 1/2. > > One thing that I find fascinating is that often people will respond to > me on that attacking my use of the word fractional, or if I say > fraction I get that every number is a fraction, where, they're > saying like 2/1 is a fraction, or 4/2 is a fraction. > Perhaps you should clarify yourself by saying something like > necessarily fractional or unavoidably fractional. Well, I've often heard mathematicians and posters on math newsgroups claim that 1/2 and 2 are both fractions, and they'll give something like 2/1, but I keep doubting that most people see it that way, while I know the technical basis of the claim. That is, yes, you can write 2/1 and say that 2 is a fraction, but there's an intuitive sense of a difference between 1/2 as a fraction and 2 as a fraction, which seems to be something that mathematicians quite deliberately attack as invalid, unless they're just repeatedly trying to annoy me by jumping on my use of fraction for 1/2, versus 2 as a whole number. > > To me, however, there IS a distinction between numbers like 1, 2 and 3 > and numbers like 1/2, 1/3 and pi. > And what is this distinction? How well can you define it? Logically, I can simply note that 1/2 is an instruction. Literally, it's one of two and there's not a practical use where it's not about a set containing two elements, like two subsets. If you split a pie, you have two pieces. Taking 1/2 of the pie, is taking one of two. So on my website with Object Mathematics I consider it to be an operator. See http://groups.msn.com/AmateurMath/objectmathematic.msnw So what's going on with Object Mathematics is that the mathematics is the way people think versus an emphasis on abstractions like forcing 1/2 to be a number, when no one uses it that way in real life. Now trying to push some rigor onto the idea, the primary distinction in object mathematics is that in the ring of objects, defined on the page, numbers can be coprime. Once you introduce numbers like 1/2 you lose that property. > 2. What do you mean by 'coprime' in this context? > > Once you have a clear idea of a number as an object, then it makes it > easy to define coprime as simply being when no two numbers share > anything but unit factors, which are factors of 1. > I know what a unit is. You shouldn't take offense as my posting is for the group which hopefully includes lots of readers from all over the world, otherwise I think it'd be a waste of time. That is, in posting, I'm not really concerned about you as an individual or your particular beliefs, but more with you as an abstraction representing a large group of people, some of whom I supposed would need to be told that a unit is a factor of 1. > For instance, 2 and 3 are coprime because they only share 1 or other > units as factors, while 2 and 4 are not because they share 2 as a > factor, and it's not a unit. > I know how all this works in Z. But some other rings aren't so nice. > How about the standard example of nonunique factorization, the ring > Z[(-5)^(1/2)]? There's no need to bring up non-uniqueness of factorization as that's a different topic. Here the issue is whether or not coprimeness in one ring, means coprimeness in a higher or all rings, and you've picked an example where it does not. > This is the ring of algebraic integers in Q((-5)^(1/2)). For instance, > 1+(-5)^(1/2) and 1-(-5)^(1/2) are coprime by your definition. Yet 2 > divides their product (their product is 6) and 2 divides neither of > them. Yet 2 cannot be factored further in this ring; it is > irreducible. You get the same thing with the ring of evens, where 2 and 6 are coprime, within the ring, because 3 is not in the ring. Mathematicians may have wished to find a way to get an *absolute* coprimeness versus what I've heard called a relative coprimeness, but a major point I've made is that that they failed. That is, the ring of algebraic integers has the problem that coprimeness is still relative, which I can prove with some basic algebra and an expression which can be seen as a polynomial with respect to one variable, but factored with respect to others. With the ring of objects, I just get to the nitty-gritty, and define the ring such that it is an absolute ring. The problem with algebraic integers is that the definition--focusing on monic polynomials--doesn't do the job. And in hindsight, there's really no reason to have believed that it did. > > 3. You seem to merely (not counting the other issues with what you are > saying, like the ones I raise in my other questions) assert the > existence of a ring with these properties without proving uniqueness > or uniqueness up to isomorphism of this ring, or proving something > like the existence of a maximal subring of C (the field of complex > numbers) with this property. > > That's because I'm focusing on a particular property of numbers, and > including those with that property. > This is not a property of individual numbers. This is a property of > _sets_ of numbers. Well, I think it's a property of the numbers, and I'll illustrate with the Bible story where King Solomon faces two women demanding a child. The way the story goes two women had claims on the child, and King Solomon couldn't figure out which one had the greater claim. So he talked of splitting the child and giving each one of them half. But you see, there's no such thing as 1/2 of a healthy child. The child's mother screamed out in horror and said to give the child to the other woman, who had been acting out of spite, and wanted to go along with the split. Solomon gave the child to the rightfully concerned mother. Now then, did the child have the property of wholeness or does the set of children? > Also, while we are on the topic of properties of different sets of > numbers, I would like to mention where you could run in to trouble: > A definition: If A is a ring, here I will denote by A* the group > formed by the units of A under multiplication. > Let P(x) be a polynomial with coefficients in Z which is irreducible > in Z[x]. Let alpha be a complex root of P(x). Then let O be the ring > of algebraic integers in Q(alpha). Then let T be the subgroup of O* > consisting of all elements of O* which have finite order. Then the > minimal number of elements of O* needed to generate O*/T is C-1, where > C is the number of irreducible factors that P(x) when factored in the > ring R[x], the ring of polynomials in x with real coefficients. > A consequence of this is that the only algebraic number fields (other > than the field Q of rational numbers) K such that the ring of integers > in K has a finite unit group are the imaginary quadratic fields. This > follows from the Dirichlet Unit Theorem as follows: I'll interject here that I see the use of the term field with rationals to be a social fossil. That's because mathematicians, needing to exclude non-rationals, rely on the definition of rational to exclude numbers that result from decidable operations. That is, you can get a *single* number from doing something like adding up the series 1/3 + 1/3^2 + 1/3^3+... so I say such operations are NOT excluded by the definition of a ring, while I've seen posters argue that you have to specifically allow convergent infinite series. Why? The answer to me is that mathematicians, in breaking coprimeness, break it across the board, but of course don't want irrational numbers in the field of rationals! Rather than discard the term and just use reals, they hold on to it. I say that you have reals and complex numbers. There is no field of rationals and the imaginary quadratic fields you mention must be the field of complex numbers. Basically, when you make 1/2 a number like 2, you break the rule of coprimeness across the board. The same thing happens if you use 1/3 or 6/52, or any such fraction. > Let the degree of P be D. Then since we are looking at algebraic > number fields other than Q, we have D>1. Other notation continues that > used in the statement of the Dirichlet Unit Theorem that I have > presented here. > Also note that since any irreducible polynomial in R[x] has degree<=2, > we have C>=D/2. Thus, when D>2, C>1 and O* has units of infinite > order. These must be different from 1 or -1 because 1 has order 1 and > -1 has order 2. > Therefore we are left with the D=2 case, which is the case where P is > quadratic. Now write P(x)=ax^2+bx+c. If b^2-4ac > 0, then we have 2 > real roots of P and therefore C>1 again. If b^2-4ac<0, then we have no > real roots so C=1. Since b^2-4ac<0, the field we get in this case will > be an imaginary quadratic field and the theorem is proven. > So you have to be very restrictive about membership to make sure that > only 1 and -1 are units in your ring. For instance, if your ring > contains 3^(1/3), then it must also contain b=(3^(1/3)^2-2=9^(1/3)-2. > But since your ring contains 3^(1/3), it also contains > c=4+3*3^(1/3)+2*9^(1/3). But b*c=1 and neither b nor c is +/-1, so you > have to exclude 3^(1/3). The definition I use follows. The Object Ring is a commutative ring that includes all numbers such that -1 and 1 are the only members that are both a unit and an integer, where no non-unit member is a factor of any two integers that are coprime. http://groups.msn.com/AmateurMath/objectmathematic.msnw What I did was cut to the chase, and get to the property I wanted. Now it might have seemed that the definition for algebraic integers was similarly all-inclusive, but I can prove that it is not. I say rather than try to find definitions like it, just get to the needed property. > Gauss began that process with gaussian integers which, of course, > are numbers a+bi where 'a' and 'b' are integers. > > Then Dedekind extended the work started by Gauss with algebraic > integers which, of course, are the roots of monic polynomials with > integer coefficients. > > Notice that the word integer keeps coming forward, even though many > of the numbers included look NOTHING like an integer. > So is your notion of integrality the usual notion of being an > algebraic integer? You still haven't told me what your notion of > integrality is. This is quite an obfuscation to me. I like that word integrality, and no, I don't associate it in and of itself with algebraic integers in the way you describe, as I know algebraic integers fail to be as inclusive as necessary. What I've done is focus on a set where ALL members are numbers that are integral, which includes gaussian integer, algebraic integers, and an infinity of other numbers as well. And I'll give the definition I use once more. The Object Ring is a commutative ring that includes all numbers such that -1 and 1 are the only members that are both a unit and an integer, where no non-unit member is a factor of any two integers that are coprime. http://groups.msn.com/AmateurMath/objectmathematic.msnw > > For instance (-1+sqrt(3)i)/2 is an algebraic integer. > Yes. I know that since I know it's a root of x^2 + x + 1. > So, are you excluding it from your ring? Yes or no? No. > Suppose we leave the land of radicals. Let a be the largest of the > three real roots of x^5-3125x-5. Is a included in your ring? Do you > consider it to be an integer? Yes or no? It's an algebraic integer, so it's included. The polynomial is irreducible over Q, so it's not an integer. If it's an algebraic integer, it's included, just like algebraic integers include gaussian integers, and I remind that I extended from Dedekind. > 4. What is the definition of integrality for members of this ring? For > something like Z[1/2], we can use the usual notion of integrality > (and, by the way, this ring is not a field since 3 is in it, 3 is not > 0, and 1/3 isn't in it), and then things are all well and good. But > what does it mean to say that an object is an integer? > > Well, you may *wish* to exclude 1/3 from the ring Z[1/2] but the > definition of the ring does NOT exclude it, and in fact, in the ring > > 1/3 = 1/4 + 1/4^2 + 1/4^3 +.... > > I'm curious to know how you think you might exclude it. > See below. > > There is nothing specific in the ring definition that would exclude > that sum, > Yes there is: the fact that that sum has infinitely many nonzero terms > and therefore requires a metric to be well-defined. Z[1/2] is defined > as the ring formed by _polynomials_ in 1/2 whose coefficients are in > Z. _Polynomials_, as opposed to _power series_, have necessarily only > finitely many nonzero terms. This ring is defined such a way that no > notion of analysis needs to appear in it or in work with it; infinite > series belong to analysis. Polynomials are the corollary to integers themselves. However, in putting in 1/2 as a number you don't have a polynomial corollary, and as an example of a corollary, consider 1/(x+1), and notice that if you try to resolve through division, you get an *infinite* number of terms. So you have a fundamental disconnect in your thinking, where you try to use properties of integers, when you've broken away from integers, with the connection being polynomials. Now in integers there are no infinite sums that are decidable. So it might seem logical to apply the same rule to polynomials, which works--as long as you don't include numbers like 1/2. I think to paraphrase a common phrase, the problem is that mathematicians try to have their cake and eat it too. > It converges in the metric completion of Q provided by R. However, if > you change your metric to the p-adic metric then (no matter which > prime p is chosen) that series diverges: the terms become indefinitely > large when p=2 and they all have p-adic absolute value 1 when p is > odd. However, there's nothing in the definition of a ring that needs all of that excess. Mathematicians have gilded the lily because they don't acknowledge the fundamental contradiction of trying to make 1/2 into a number like 2. So you put in 1/2, but then block using a rule from integers, to try and stop infinite series, but later add them back in, claiming that it's about convergence, and call the resultant set reals. Now if by use of a metric above you essentially look to 1 or an equivalent, which I'd call a *whole* object, then you're just trying to do two things at once. You either need to use only whole objects *consistently* or choose to use fractional objects, and that's why if you use even a single fraction like 1/2, you have the field of reals. I take it from what you said also that a metric greater than 1 will send you off into infinity. > and given that it's decidable, that is, you can get a single > number answer, > What in the world are you trying to say here?? Ok, let's talk about why infinity is not a number. Well, of course, given any number that someone thinks is the largest number you can just add 1, and so on. But say x equals infinity. Well 2x equals infinity, so 2x=x, 2=1, contradiction. So, infinite sums of non-zero *integers* are not decidable, as there is no largest number. The other case is something like 1 + -1 + 1 + -1 +... which is not decidable either as it can be 1, -1, or 0, depending on how many terms you check, and there is no answer out to infinity. That seals it with integers. But with rationals, you have infinite series that ARE decidable, like 1/2 = 1/3 + 1/3^2 + 1/3^3 + .... It turns out that it's simpler to just use decidability, as then you cover everything, with just ONE idea. It's efficient. > there is no rational reason to exclude it, pun > intended. > Yes there is a reason: it does not meet the criterion for membership > in the ring, as I described above. What happened is that mathematicians came back later and tried to fix things, but had to use unnecessarly complicated rules, as they sought to keep the field of rationals while they realized they had interesting numbers like transcendentals. Rather than use the simplest idea, they created a hodge-podge, including focusing on polynomials--analogs to integers--even when they were sticking in numbers like 1/2. Now then, can you give me your definition of a ring? Then can you specify what in that definition excludes infinite series? I consider your previous statements on the subject to be obfuscations. > These and other clarifications (NOT obfuscations) would be > appreciated. > > ---- David > > Well then, you can step forward once again to answer and NO > obfuscations would be appreciated. > Understood and accepted. Be sure to tell me when I am engaging in > obfuscation so I can clarify in a reply post, in case I have > considered something clear when it isn't so clear to you. Certainly. And I appreciate your doing so in this post. Hopefully I cleared up the area. James Harris === Subject: Re: Questions for James Harris > Now then, can you give me your definition of a ring? > Then can you specify what in that definition excludes infinite series? http://mathworld.wolfram.com/Ring.html As everybody can see, nothing in that definition *includes* infinite series. === Subject: Re: Questions for James Harris > 1. What kinds of things are members of your object ring? complex > numbers? polynomials? functions? elements of some strange quotient > ring? vectors of some kind? > My idea is to include any and all numbers that can be considered > objects versus numbers that are fractional like 1/2. So your idea is that Object Ring contain all numbers that can be considered objects. Isn't that kinda circular? How can I distingush an object number from a non-object number? You told us recently that sqrt(2) is an object. How do you know that - some secret algorithm? === Subject: Re: Are all mathematicians music lovers? convey a wrong image of religions due either to bias (which is >unacceptable but unfortunately very common) or lack of >knowledge/understanding of the reference text Or neither. An unbiased observer, noted that the preaching of the clergy differs from the writings might well conclude that the preachings of the clergy were more representative of the religion actually practiced than the writings were. Right now you are in the position of someone in 1492 saying that The Inquisition was not sanctioned by Christianity; perhaps it wasn't, but it had the blessing of the Pope. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Problem with derivate (RS+T) > When I enter the following > 2. SIN (X) > 1. X > [RS+T] > the result in Degree MODE -> (PI/180)*COS(X) > in Gradians -> (PI/200)*COS(X) > in RAD -> COS(X) This is totally correct! > Why there are 3 different solutions? Mathematically speaking, if you differeniate sin(X) while working in degrees, you must use the 'chain rule'. I don't have time to explain why at the moment, but I hope someone else can :-) cheers, Al === Subject: Re: Calculus is irrational? at 02:58 PM, mhelland@techmocracy.net (Mike Helland) said: >So God is a rational notion, eh? Of course it is. Whether it is a correct notion belongs in a different news group. >The bible-thumpers would love to hear this. I would no more judge religion by the behavior of the >But because I say observe No. You say rational, which is something quite different. >If this is true God must be somewhere thanking you. >How does my defintion of rationality say that I cannot reason about >unicorns? Are you now implying that you've seen unicorns? >I don't disagree with the second sentence. But how about >imagination bounded by logic and observation as rationality? Human beings reason about events that haven't happened. It's one of the things that makes us more successful than other animals. Such reasoning is often rational, at least by the definition that the rest of the world uses. It's also essential to our civilization. >Time is the boundry of what we observe, Are you under the imporession that you've said something either meaningful or rational? >really works. I'm sure that Kip and company will be impressed. >Now, if a >mathematician wishes, he can prove Lynds wrong by showing how >calculus can be used to determine what Lynds says does not exist. Why would a Mathematician bother; what you're talking about is Physics, not Mathematics. A Physicist would probably also not bother, for the same reason that my doctor doesn't waste her time arguing with the Christian Scientists. >The issue is that calculus isn't talking about the real world, where >Lynds is. No, the difference is that while neither is talking about the real world, Calculus has successfully been used to model the real world in theories that yielded testable and verified predictions. >I don't see why we just don't use the >word irrational to describe things that lie outside the real >world. For the same reason that we don't use the word submarine for them. >If we're clear on what is rational and what is not, You aren't. >Calculus starts with some irrationality No. >(a ball that bounces an infinite number of times) That wasn't Calculus; that was Mike Helland. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Calculus is irrational? > You're obviously talking about numerical infinity. > Which calculus and group theory heads don't know exist, > which is why mathematicans are obviously are always > given low-bugdet intergral jobs. And we save the logic > for people with brains. Miss a chance at the local glory hole? Don't worry, I'm sure he'll be back. 'cid 'ooh === Subject: Re: Calculus is irrational? > You're obviously talking about numerical infinity. > Which calculus and group theory heads don't know exist, > which is why mathematicans are obviously are always > given low-bugdet intergral jobs. And we save the logic > for people with brains. > > Miss a chance at the local glory hole? Don't worry, I'm sure he'll be back. Well, if you call calculus glory, yes. Since, the only thing in the universe, that has less to do with logic, than mathematicians, is calculus. === Subject: Re: Calculus is irrational? > Acid Pooh > Why resist the suggestion that one may be rational about non-physical > (and hence, inobservable) objects? Clearly, you understand that we > can reason about non-physical objects. > I resist because I think there is a subtle but important difference in > being rational and being reasonable. That's why I clarified my > defintion. > I'll make a leap here and > suggest that you further understand that mathematics uses logic to > form a (semi-) consistent whole. (That semi- might be provocative, > but it's irrelevant to the issue at hand. It is only included for > accuracy.) > Yes, here's the thing semi-consistent. The problem isn't that its > semi-consistent, the problem is when this system is used to discuss > something like Time. > Time is the boundry of what we observe, so I think it's fair to say > that Time is the boundy or our rationality. Along comes Peter Lynds > mathematician wishes, he can prove Lynds wrong by showing how calculus > can be used to determine what Lynds says does not exist. This is metaphysics, not mathematics. The fact that there exists a strong analogy between a function (say, Sin(x)) and the motion of a pendulum says nothing but that the function is a fairly good model of said motion. > The issue is that calculus isn't talking about the real world, where > Lynds is. You say to be accurate we need to call Calc > semi-consistent. Well of course a semi-consistent notion will > contradict a consistent notion. I don't see why we just don't use the > word irrational to describe things that lie outside the real world. > If we're clear on what is rational and what is not, it allows us to > sort through sticky subjects like Time much easier. The word Non-physical already exists to do that. Your intuition regarding consistency is wrong. You obviously don't know what I was refering to. In 1931, Kurt Goedel showed that arithmetic--and thus calculus--cannot prove its own consistency. And yet, stronger systems CAN prove the consistency of arithmetic. The word irrational already has other important uses. And if you open any textbook on metaphysics, you'll see that time continues to be a sticky problem, even though every distinction imaginable has been made in the field, including many you don't know exist. > But irrational has many > strong connotations--and what it denotes is not what you think it > does. > Depending on the degree of irrationality. Calculus starts with some > irrationality (a ball that bounces an infinite number of times) and > then follows a rational process to solve. So it's not completely > irrational. THere is quite a bit of rationality to it, but it's not > completely rational either. > Mike Helland Look Mike: here's a dictionary's definition of irrational. Your refusal to use the conventional definition seems to be a paradigmatic example. iraraationaal EE (EPE)EEPronunciation KeyE adj. 1. a. Not endowed with reason. b. Affected by loss of usual or normal mental clarity; incoherent, as from shock. c. Marked by a lack of accord with reason or sound judgment: an irrational dislike. 2. a. Being a syllable in Greek and Latin prosody whose length does not fit the metric pattern. b. Being a metric foot containing such a syllable. Thus, the word irrational does not denote what you think it does. You are inventing your own language and expect others to discuss your ideas on your terms. This is unrealistic. 'cid 'ooh === Subject: horseshoe map How to construct the periodic table for the horseshoe map for periods up to 6? -- === Subject: chaos f: R -> R. orbit{x, f(x), (f^2)(x), ...} Let {(f^i)(x)} be the rbit of x. {(f^i)(x)} is chaotic if (1) {(f^i)(x)} is not asymptotic periodic. (2) Lipunov exponent > 0. What does (1) mean? What is the definition of asymptotic periodic? ex. Let n be a positive integer and f(x) = nx (mod 1) on [0,1]. Which orbits of f are chaotic? Please give me a hint. -- === Subject: Can anyone put a formula to this problem? or point me in the right direction? Hi all, Apologises if this is the wrong area for this problem i am working on. But formulas isn't my forte and I hoping someone can help me out. I am convinced there should be a way to resolved this in a math lanuage. This is my first post ever so forgive me if this dosnt come accross well its a complex problem for me.......... Problem. A source gnerates a random number in a stream one after the other that is feed into a another source that receives the numbers one by one. Source Ouput----> 2534,35566,36567,3567,3445,7767----->input source For every 50 numbers generated from the source they belong to a specfic group for example group 'A' will contain a list of 50 numbers A = (7767,3445,3567,36567,35566,2534.....). Groups will contiune to be created each group holding 50 number and so on. The problem arises when an intermidiate source that takes a number before the numbers arrive at the input source and wants to find out what group the number belongs to. I like to call this a neighbourhood theroy or something like that because if the input source knows the neighbour numbers then it can figure out the group it belongs to. The problem gets worst if the input source dosnt receive its immediate neighbour numbers i.e. sometimes numbers don't make it therefore it has too look to its next neighbour number i.e. two neighbours away from the orginal number. In the above list for example if the intermidate source grabs number 3567 and asks ether the output source or input source what group it belows too the awnser should be group 'A' because the output source will know its neighbours of 36567 is its left neighbour and 3445 is its right neighbour.The output source will never loose its neighbour numbers. But if the intermidate sources were to ask the input source what group '3567' belows to too it will also know its neighbours of '36567' from the left side and '3445' from its right side but input source might be missing its first rightside neighbour of 3445 but it has its next one which is 7767. Therefor the input source should return group 'A'. It is possible that the input souce could be missing serveral of its neighbours therefore the formula would have to go to its next available number which could be N number from the orginal number that intermidate souce is asking for what group it belongs to. It is possible that the input souce could return more than one group response like group A & B & C because the input source was missing allot of its neighbours. Therefore the input put must return more than one group to ensure that the number was in one of the groups of A,B,C. The output source never looses its neighbours numbers so it would be better if the intermidate source where to ask the output source for the group identify but this group identity from the ouptut source may not be correct because some of the numbers may never make it too the input souce. If every number were to make it to the input souce then this would be fine. 90% of the time all numbers do make it to the input source but I need to guarentee a 100% group identity. So i could ask output source but I would have to verfiy this with the input souce before i return the group identity to ensure that group identity is correct. Sorry for long description.....confussing I know but the problem is doing my head in and I am trying to attempt a formula for it that will cover all possble situations...but i can't even get started..... :( Any help would be much appricated.... Jason. === Subject: Re: Nobel Prize in mathematics <1fzwzvi.1etjg77ltxziuN%medtib@alussinan.org> In , on at 08:30 PM, haberg.remove@matematik.su.se (Hans Aberg) said: >Here are the rules: Who are you to make the rules? >Create a Nobel Prize in mathematics, Only Nobel code do that? Would you settle for Fields? Abel? >with the same high standard statutes and committee review work, You mean put a doctor with knowledge of Geometric Optics in charge of evaluating candidates for the prize in Physics? The Nobel committees are far from being exemplars of excellence, and far too much politics is involved in their decisions. >importance relative rest of science, That doesn't sound much like the Nobel Prize to me. When the Nobel prize for Literature takes into account the importance of a novel relative to Medicine and the Novel prize for Physics takes into account importance relative to new concerti, then it will be time to discuss such an outlandish comparison. >I never declared anything; Google is your friend. You declared that Serre's work was not of Nobel caliber. >For a proper evaluation process, He's gone through several. Maybe the Fields and Abel committees know something you don't. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Publishing <3f40920e$0$16529$626a54ce@news.free.fr> <3f4102f7$0$1135$626a54ce@news.free.fr> <3f412a81$0$16543$626a54ce@news.free.fr> <3f422ca1$0$16184$626a54ce@news.free.fr> <3f428e65$0$267$626a54ce@news.free.fr> <3f436847$0$1159$626a54ce@news.free.fr> In , on at 07:57 PM, haberg.remove@matematik.su.se (Hans Aberg) said: >Serre is evidently considered the Immaculate, and those even hinting >at something else will get their motives questioned even before any >facts can be presented. Would it not be better to get the facts >before questioning them? The first fact that I would like to get is *who* considered Serre to be immaculate. If the answer is that nobody did or does, and you just made that up out of the whole clothe, than nothing else you may claim can be given any credence. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Publishing <3f40920e$0$16529$626a54ce@news.free.fr> <3f4102f7$0$1135$626a54ce@news.free.fr> <3f412a81$0$16543$626a54ce@news.free.fr> <3f422ca1$0$16184$626a54ce@news.free.fr> <3f428e65$0$267$626a54ce@news.free.fr> In , on >Poor excuse in the days of Internet broadband. Excellent justification, unless you are offering to pay for universal broadband. Some of us still have dial access. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: is there any Gram-Schimt techique for integer matrix? For some reason, I want to restrict the construction of an orthogonal matrix within the domain of integer matrix. Is there a Gram-Schimt kind of technique which can deal with integer matrix and get the result as near as orthogonal matrix as possible?(even better if the result can be orthonormal after a scaling factor for the whole matrix) I don't know if the following makes things easier... let's say there is originally an real matrix which is orthogonal/orthonormal. After some scaling and round off into integer matrix, it is no longer orthogonal. How to make some modification to restore its orthogonal as much as possible(within the integer domain) Please give me some advice on how to do this problem? If an earlier mathematician had worked on this problem, please point me to the resources or the similar places... maybe some mathematician had already worked it out... -Walala === Subject: Re: a matrix problem related to eigevalues >Are there theorems on the effect of row operations >(swapping rows, adding multiples of row i to row j) >and eigenvalues? Because it's immediately clear that >you can reduce R(n) to >[0 0 .... 0 1/n] >[0 0 ... 1/(n-1) 0] >[ ... ] >[0 1/2 ... 0 0] >[1 0 ... 0 0] >by subtracting each row from all the rows below it. The row operations correspond to matrix multiplications: A -> E A where E is an elementary matrix. Unfortunately there is no simple way to relate the nonzero eigenvalues of E A to those of A in general. 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: How do you discuss mathematics with a person who thinks mathematics is a part of physics? > ... He calls the mathematics that is done without any connection to the > physical world imaginary. > Like it or not, every major development in mathematics ( or in any > science for that matter} is either inspired/provoked by reference, > directly or indirectly, to a real physical situation. Like it or not, the only *important* development in mathematics is the one that is not inspired by physics, and the one that mathema-cretins have been avoiding for 5000 years. That is: How many dimensions does the universe have, and where the are they? And don't say: See Goedel. Since they are obviously *not* in either Classical Mechanics or Quantum head Mechanics. > This applies perhaps as well to already realized or systemematized > work. The form of equations in electomagnetic theory for example,( > that once made Boltzmann to remark if it was not God's work afterall ) > when physically/geometrically arranged on paper - are visualizations > of beautiful relationships and concepts with symmetry. If information > is ordered into a set, it is also a real world exercise. But, symmetry (either capitalized or uncapitalized) has always been nothing more than a mathematician's excuse for being both Chemically-Challenged and Entropically clueless. === Subject: Re: How do you discuss mathematics with a person who thinks mathematics is a part of physics? >Like it or not, every major development in mathematics ( or in any >science for that matter} is either inspired/provoked by reference, >directly or indirectly, to a real physical situation. Like it or not, that's not only wrong, but obviously wrong to anybody who knows any Mathematical History. Of course, just because you consider your work immune to practical application doesn't mean that it won't be applied, but that's ex post facto. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org