mm-1299 === Subject: Re: Fermat's Last Theorem > y^2 = x(x - a^n)(x + b^n) (equ 4) > where > a^n + b^n = c^n (equ 5). > Frey does not derive equation 5 from equation 4; Frey and Wiles are > implying the existence of equation 5. No, they are *assuming* the existence of a solution to (5), in order to derive a contradiction. They then derive (4) (the Frey curve) on the assumption some particular positive integers a, b, c, and odd prime exponent n exist satisfying (5). > It's questionable that Wiles can assume the existence of equation 5 > without deriving it then basing his entire proof on an implied > equation. It's a proof by contradiction, and actually Ribet is the one assuming it. Fermat's equation is not dependent on an elliptic curve. Indeed not, except in the case n=3, where it *is* an elliptic curve. However, given a nontrivial (meaning in positive integers) solution, we obtain a corresponding elliptic curve, the Frey curve. This curve it it exists must not be modular, but that contradicts the Taniyama-Shimura theorem. Contradiction. Hence, proof by contradiction. > Using n=2, in equation 4, the following equation is formed, > y^2 = x(x - a^2)(x + b^2) (equ 6) > equation 6 is not Pythagoreans equations; therefore, equation 6 is > unrelated to Fermat's equation. Consequently, Wiles' proof of > Fermat's Last Theorem is invalid. We can form a curve for, for example, 3^2+4^2=5^2, which will be y^2 = x(x-9)(x+16). This is, indeed, an elliptic curve (it's nonsigular and hence has genus 1). There is nothing mysterious about this curve, and it is modular, with a modular parameterization x(q) = q^(-2)-3+32*q^2-30*q^4+290*q^6+162*q^8+1635*q^10+1890*q^12+8709*q^14 + ... y(q) = -q^(-3)+q^(-1)+32*q-93*q^3+932*q^5-188*q^7+7393*q^9+4103*q^11+48972*q^13+... === Subject: The record producer's problem Suppose you have a lot of recordings sitting in storage--for example, Dorati's complete Haydn symphonies; these all have a given fixed length. You want to bring out a complete edition, using as few CDs as possible, and assuming each CD can hold no more than 80 minutes. Subject to this condition, you want to make the variance in length between the recordings as small as possible. You can set this up in terms of a set of real numbers <= 1, {r1, r2, .. rn}, where you want a partion into subsets, such that the sum over each partion is <=1. You want to minimize the number of partitions, and subject to that, minimize some variance measure of the sum of partions. Anyone heard of this problem ever being considered? === Subject: Re: The record producer's problem >Suppose you have a lot of recordings sitting in storage--for example, >Dorati's complete Haydn symphonies; these all have a given fixed >length. You want to bring out a complete edition, using as few CDs as >possible, and assuming each CD can hold no more than 80 minutes. >Subject to this condition, you want to make the variance in length >between the recordings as small as possible. >You can set this up in terms of a set of real numbers <= 1, {r1, r2, >.. rn}, where you want a partion into subsets, such that the sum over >each partion is <=1. You want to minimize the number of partitions, >and subject to that, minimize some variance measure of the sum of >partions. >Anyone heard of this problem ever being considered? Yes, I asked a similar question a few years ago (fitting GIF thumbnails into a fixed-width image map) and was told it's the knapsack problem. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if you're afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: The record producer's problem > Suppose you have a lot of recordings sitting in storage--for example, > Dorati's complete Haydn symphonies; these all have a given fixed > length. You want to bring out a complete edition, using as few CDs as > possible, and assuming each CD can hold no more than 80 minutes. > Subject to this condition, you want to make the variance in length > between the recordings as small as possible. > You can set this up in terms of a set of real numbers <= 1, {r1, r2, > .. rn}, where you want a partion into subsets, such that the sum over > each partion is <=1. You want to minimize the number of partitions, > and subject to that, minimize some variance measure of the sum of > partions. > Anyone heard of this problem ever being considered? Without the variance condition, this is the bin=packing problem, isn't it? which is already NP-complete, isn't it? so you're taking a problem that's already impossible to solve, and trying to make it harder? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: The record producer's problem > Anyone heard of this problem ever being considered? > Without the variance condition, this is the bin=packing problem, > isn't it? which is already NP-complete, isn't it? so you're taking > a problem that's already impossible to solve, and trying to make it > harder? Bin packing it is, but the side condition is likely to give a unique optimal solution. I suppose taking a hard problem and making it harder is a dubious strategy. === Subject: Re: The record producer's problem > Suppose you have a lot of recordings sitting in storage--for example, > Dorati's complete Haydn symphonies; these all have a given fixed > length. You want to bring out a complete edition, using as few CDs as > possible, and assuming each CD can hold no more than 80 minutes. > Subject to this condition, you want to make the variance in length > between the recordings as small as possible. > You can set this up in terms of a set of real numbers <= 1, {r1, r2, > .. rn}, where you want a partion into subsets, such that the sum over > each partion is <=1. You want to minimize the number of partitions, > and subject to that, minimize some variance measure of the sum of > partions. > Anyone heard of this problem ever being considered? > Without the variance condition, this is the bin=packing problem, > isn't it? which is already NP-complete, isn't it? so you're taking > a problem that's already impossible to solve, and trying to make it > harder? NP-complete does not mean a problem is impossible to solve. === Subject: Re: The record producer's problem > Without the variance condition, this is the bin=packing problem, > isn't it? which is already NP-complete, isn't it? so you're taking > a problem that's already impossible to solve, and trying to make it > harder? > NP-complete does not mean a problem is impossible to solve. Yes, I'm aware of that, I was abusing the language in a way that is common but perhaps should be avoided when writing for people one doesn't actually know, so let me rephrase: If a problem is NP-complete then it will be a major mathematical breakthrough if you find an efficient algorithm for handling large instances of the problem. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: The record producer's problem >> Suppose you have a lot of recordings sitting in storage--for example, >> Dorati's complete Haydn symphonies; these all have a given fixed >> length. You want to bring out a complete edition, using as few CDs as >> possible, and assuming each CD can hold no more than 80 minutes. >> Subject to this condition, you want to make the variance in length >> between the recordings as small as possible. >> You can set this up in terms of a set of real numbers <= 1, {r1, r2, >> .. rn}, where you want a partion into subsets, such that the sum over >> each partion is <=1. You want to minimize the number of partitions, >> and subject to that, minimize some variance measure of the sum of >> partions. >> Anyone heard of this problem ever being considered? >Without the variance condition, this is the bin=packing problem, >isn't it? which is already NP-complete, isn't it? so you're taking >a problem that's already impossible to solve, and trying to make it >harder? With the variance condition and a fixed number of partitions, it's also NP-complete (since a special case would be where the numbers can all fit on n CD's, but only if all the CD's are full). See also the Makespan Scheduling and Set Partitioning problems. Although these problems are NP-complete, there are reasonably good heuristics and approximation methods. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Tough Fourier Transform Integral, need help! I am having difficulty solving the Fourier Transform of the following complex time signal; x(t) = exp(i*PI*exp(t))/sqrt(t), 1 < t < inf, i = sqrt(-1) This signal is a decaying complex chirp signal whose frequency increases with time. Any pointers welcome! Bob Adams === Subject: Tough Fourier Transform problem, part 2 I made an error in my previous post; please use this one instead! I am having difficulty solving the Fourier Transform of the following complex time signal; x(t) = exp(-t/2)*exp(i*PI*exp(t)), 1 < t < inf, i = sqrt(-1) This signal is a decaying complex chirp signal whose frequency increases with time. Any pointers welcome! Bob Adams === Subject: Re: Tough Fourier Transform problem, part 2 >I made an error in my previous post; please use this one instead! >I am having difficulty solving the Fourier Transform of the following >complex time signal; >x(t) = exp(-t/2)*exp(i*PI*exp(t)), 1 < t < inf, i = sqrt(-1) >This signal is a decaying complex chirp signal whose frequency >increases with time. Ah, this one can be done. Substitute t = ln(s): int_1^infty x(t) exp(-ikt) dt = int_e^infty s^(-1/2 - ik) exp(i pi s) ds which Maple does in terms of Whittaker M functions: *I^(-1/4+1/2*I*k)*Pi^(-1/4+1/2*I*k)*WhittakerM(5/4-1/2*I*k, -1/2*I*k+3/4, I*exp(1)*Pi)+I*Pi^(3/4+1/2*I*k)*exp(3/4+1/2*I*k-1/2*I*Pi*exp(1)) *I^(-1/4+1/2*I*k)*WhittakerM(1/4-1/2*I*k, -1/2*I*k+3/4, *Pi^(3/2+I*k)*(-1)^(1/4+1/2*I*k))*exp(-1/2-I*k)/((-4*Pi*k^2-8*I*Pi*k+3*Pi) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Tough Fourier Transform problem, part 2 >I made an error in my previous post; please use this one instead! >I am having difficulty solving the Fourier Transform of the following >complex time signal; >x(t) = exp(-t/2)*exp(i*PI*exp(t)), 1 < t < inf, i = sqrt(-1) >This signal is a decaying complex chirp signal whose frequency >increases with time. > Ah, this one can be done. Substitute t = ln(s): > int_1^infty x(t) exp(-ikt) dt > = int_e^infty s^(-1/2 - ik) exp(i pi s) ds > which Maple does in terms of Whittaker M functions: Does it really make any sense to talk about the Fourier transform on a semi-infinite interval, i.e. [1, +infinity[. I was under the impression that Fourer transforms live naturally on the entire real axis ]-infinity, +infinity[. On the other hand, a semi-infinite interval calls for the Laplace transform. Of course, this is just picking on words, the complexity of the solution will be the same. -Michael. === Subject: Re: Tough Fourier Transform problem, part 2 > Does it really make any sense to talk about the Fourier transform on a > semi-infinite interval, i.e. [1, +infinity[. I was under the impression that > Fourer transforms live naturally on the entire real axis ]-infinity, > +infinity[. On the other hand, a semi-infinite interval calls for the > Laplace transform. > Of course, this is just picking on words, the complexity of the solution > will be the same. > -Michael. Hello Michael, Actually from a signal point of view, a one sided transform gives us a wonderful relation between the real and imaginary portions of the Fourier transform. So from an analytic signal point of view, this actually makes sense, although we rather go from time = 0 instead of 1 up to infinity. Although the other can be had with a little modulation. Clay === Subject: Re: Tough Fourier Transform problem, part 2 >Does it really make any sense to talk about the Fourier transform on a >semi-infinite interval, i.e. [1, +infinity[. I was under the impression that >Fourer transforms live naturally on the entire real axis ]-infinity, >+infinity[. You're right, of course. I am assuming the original poster's function was 0 on (-infinity,1). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Tough Fourier Transform problem, part 2 If the spectrum of cisoids sum to zero on all points before your lower limit of integration then there is no difference. You are talking only of the Unilateral Laplace Transform. The Bilateral Laplace Transform covers +/- inf in the same way as the Fourier Transform. > Does it really make any sense to talk about the Fourier transform on a > semi-infinite interval, i.e. [1, +infinity[. I was under the impression that > Fourer transforms live naturally on the entire real axis ]-infinity, > +infinity[. On the other hand, a semi-infinite interval calls for the > Laplace transform. > Of course, this is just picking on words, the complexity of the solution > will be the same. === Subject: Re: Tough Fourier Transform problem, part 2 Try a Taylor expansion of exp(i*PI*exp(t)) and you get a much simpler transform to solve. Brad >I made an error in my previous post; please use this one instead! > I am having difficulty solving the Fourier Transform of the following > complex time signal; > x(t) = exp(-t/2)*exp(i*PI*exp(t)), 1 < t < inf, i = sqrt(-1) > This signal is a decaying complex chirp signal whose frequency > increases with time. > Any pointers welcome! > Bob Adams === Subject: Re: mappings: mathematical vocabulary issue/nonissue I think you all sort of missed my point. Several commonly used names: set, pair, list, element, point, function, you name it, are very short and make them suitable for naming things within computer programs. When I think of the concept in question, pair comes in closest, in being a full word (not a concatenation of abbreviations), which is also short. But it sais nothing about the for each... there exists a unique... property of functions. If length of words used to describe concepts are to be proportional to their frequency of use then mathematicians simply did not do a good job on coining a handy word for this concept which is used by so many computer programmers on a daily basis (such as for instance would have been 'maplet'). Mathematicians even adopted a notation for it (a |-> b) distinguishing X->Y. Why couldn't they have given it a name??? They gave X->Y the name function! Couldn't they have called it a|=>b a funclet and spread this word around textbooks like they did with all the other words out there??? Definition: Let f:X-Y be a function. A funclet is an ordered pair (a,b) such that ain X and bin Y and f(a) = b. I'd like to push this one... come one people, isn't it a cute word to adopt in the world of mathematics. Our sentences will all be so much concise and to the point!!! How hard is that??? :-D Neil === Subject: Re: mappings: mathematical vocabulary issue/nonissue > I think you all sort of missed my point. Several commonly used names: > set, pair, list, element, point, function, you name it, are very short > and make them suitable for naming things within computer programs. When > I think of the concept in question, pair comes in closest, in being > a full word (not a concatenation of abbreviations), which is also short. > But it sais nothing about the for each... there exists a unique... > property of functions. If length of words used to describe concepts > are to be proportional to their frequency of use then mathematicians > simply did not do a good job on coining a handy word for this concept > which is used by so many computer programmers on a daily basis > (such as for instance would have been 'maplet'). Mathematicians > even adopted a notation for it (a |-> b) distinguishing X->Y. > Why couldn't they have given it a name??? They gave X->Y the > name function! Couldn't they have called it a|=>b a funclet > and spread this word around textbooks like they did with all > the other words out there??? > Definition: > Let f:X-Y be a function. A funclet is an ordered pair (a,b) such > that ain X and bin Y and f(a) = b. > I'd like to push this one... come one people, isn't it a cute word > to adopt in the world of mathematics. Our sentences will all be so > much concise and to the point!!! > How hard is that??? :-D > Neil What exactly is wrong with element of f or ordered pair? If you want a single word, try double or pair or twins (the last courtesy of MathWorld). If you want to explicitely say that (a,b) is a member of f, then funclet does not help us any! You STILL have to make the context clear, that it's a funclet of THIS function rather than some other function just like you do with the other words. In this case, inventing a new word does not help us to do mathematics with fewer words, nor does it make anything more transparent. For these reasons, I doubt your word will take off, no matter how cute it is. --Robert === Subject: Frey-type curves for n=2 I mentioned this on another thread, and thought I'd expand a bit. If we take the 3-4-5 right triangle and the corresponding equation 3^2+4^2=5^2, we get a Frey-type elliptic curve (a Frey curve except for the fact that the exponent is 2) which is y^2 = x(x-3^2)(x+4^2). This has conductor 240, which means it can be parametrized by modular forms of level 240; I gave q-series for these in the previous thread. Note that this curve is *not* like an authentic Frey curve--not only is it modular (which we now know it must be) but it it is *not* semistable--the conductor 240 = 2^4 3 5, and hence it does not have multiplicative reduction at 2. However other curves of this sort can be semistable. In fact, lots is happening at prime 2. We have three rational points of order 2: 2[0,0] = 2[9,0] = 2[-16,0] = 0. We also have four rational points of order 4: 4[24, +-120] = 4[-6, +-30] = 0. This is group structure is typical of what we get for these curves. I suppose the bottom line about these curves is that they exist--Ribet's proof absolutely fails to prove there are no solutions of a^2+b^2=c^2. === Subject: Re: Special Pascal Line > .... > BTW I have had a lot of difficulties to reply here : > Your post had disapeared from the NG as seen by my news reader. > Still seen by another news reader I'm not accustomed to and from > which I'm replying now. > and a bug in my usual news reader (Netscape) ? > However I got the mail, this is why I searched for the NG post.... I've never before had that trouble with sci.math or any other news group except geometry.college and geometry.pre-college, where my posts frequently seem to disappear without trace. Does anyone else have such problems? Ken Pledger. === Subject: Re: Special Pascal Line > .... > BTW I have had a lot of difficulties to reply here : > Your post had disapeared from the NG as seen by my news reader. > Still seen by another news reader I'm not accustomed to and from > which I'm replying now. > and a bug in my usual news reader (Netscape) ? > However I got the mail, this is why I searched for the NG post.... > I've never before had that trouble with sci.math or any other news > group except geometry.college and geometry.pre-college, where my posts > frequently seem to disappear without trace. Does anyone else have such > problems? > Ken Pledger. Just once only on sci.math, usually never any loss of postings. === Subject: Re: Special Pascal Line Hello Ken, Ken Pledger a .8ecrit: >>.... >>BTW I have had a lot of difficulties to reply here : >>Your post had disapeared from the NG as seen by my news reader. >>Still seen by another news reader I'm not accustomed to and from >>which I'm replying now. >>and a bug in my usual news reader (Netscape) ? >>However I got the mail, this is why I searched for the NG post.... > I've never before had that trouble with sci.math or any other news > group except geometry.college and geometry.pre-college, where my posts > frequently seem to disappear without trace. Does anyone else have such > problems? Nope. It seems there is a problem with my Netscape 7.0. Or the use I make of it ;-( Some post can't be made visible through this program. post *is* visible, and some other are not... Strange ... I didn't try to unsubscribe then subscribe again, as the follow/ignore status of the threads would be lost. However I'll switch to an other newsreader ! (posted and mailed) -- philippe (chephip at free dot fr) === Subject: Re: Special Pascal Line Hello Ken, Ken Pledger a .8ecrit: >>.... >>BTW I have had a lot of difficulties to reply here : >>Your post had disapeared from the NG as seen by my news reader. >>Still seen by another news reader I'm not accustomed to and from >>which I'm replying now. >>and a bug in my usual news reader (Netscape) ? >>However I got the mail, this is why I searched for the NG post.... > I've never before had that trouble with sci.math or any other news > group except geometry.college and geometry.pre-college, where my posts > frequently seem to disappear without trace. Does anyone else have such > problems? Nope. It seems there is a problem with my Netscape 7.0. Or the use I make of it ;-( Some post can't be made visible through this program. post *is* visible, and some other are not... Strange ... I didn't try to unsubscribe then subscribe again, as the follow/ignore status of the threads would be lost. However I'll switch to an other newsreader ! (posted and mailed) -- philippe (chephip at free dot fr) === Subject: Re: how to find an ellipse' major axis and minor axis? > .... it's a good idea to start as > I suggested on sci.math in 1998 (slightly corrected). >> >> The general equation for a conic section is: >> >> ax^2 + bxy + cy^2 + dx + ey + f = 0 >> >> Let's say the coefficients a, b, c, d, e, and f are known and the >> equation >> forms an ellipse. >> >> major >> and minor axes, the ellipse's angle of orientation, and the >> coordinates of the ellipse's center?.... >> >> >> A lot of older geometry books give various formulae for this, but I >> think linear algebra gives a clearer and more modern view. >> >> First, I hope you don't mind a change to Salmon's notation with a >> factor 2 in some of the coefficients: >> >> ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0. >> >> This can be re-written using matrices: >> >> (a h g) (x) >> (x y 1) (h b f) (y) = 0. >> (g f c) (1) >> >> It's meant to be a 1 x 3 matrix, then a 3 x 3 symmetric matrix with >> a, >> b, c down the diagonal, then a 3 x 1. So the product is 1 x 1, i.e. >> essentially a scalar. >> If the 3 x 3 matrix is invertible, then the conic is proper. Otherwise >> it's degenerate (a line pair or a single point). >> >> To find a centre (X, Y), use the first two rows of the big matrix as >> coefficients in linear equations: >> >> aX + hY + g = 0 >> hX + bY + f = 0. >> >> Solve these equations for X and Y, if possible. >> A unique solution (X, Y) means that you have a central conic (ellipse, >> hyperbola or intersecting line pair). In this case, also calculate >> - k = gX + fY + c for later use. >> No solution means that you have a parabola. >> Infinitely many solutions (with one parameter) means that you have a >> parallel line pair, possibly a repeated line. >> >> Now concentrate on the 2 x 2 sub-matrix (a h) >> (h b). >> First, its determinant ab - h^2 says something about your conic. >> If ab - h^2 > 0, the conic is an ellipse (or a single point). >> If ab - h^2 = 0, the conic is a parabola (or a parallel line pair). >> If ab - h^2 < 0, the conic is a hyperbola (or an intersecting line >> pair). >> >> Next, find the eigenvalues and eigenvectors of the 2 x 2 matrix above. >> The eigenvectors point along the conic's axes, so telling you its >> orientation. >> To find how far from the centre the conic cuts an axis, calculate >> sqrt(k/(corresponding eigenvalue)). So the two values of this >> expression >> can be doubled to give you the lengths of the major and minor axes of an >> ellipse. For a hyperbola the eigenvalues have opposite signs, so the >> formula gives only one real distance along an axis. The other axis >> doesn't cut the hyperbola at all.... >> > .... > If you try the above method on > [1 1 -3; > 1 4 -6; > -3 -6 12] > f(x, y)=x^2+4*y^2+2*xy-6*x-12*y+12 > I've used your method: > 1*x+1*y-3=0 > 1*x+4*y-6=0 > to get x=2 and y=1, so(2, 1) is the center. Great! .... Shifting the origin to the centre at this stage will simplify the later algebra. So let x = 2 + x' and y = 1 + y'. You'll find that this transforms your x^2 + 2xy + 4(y^2) - 6x - 12y + 12 = 0 into (x')^2 + 2x'y' + 4*(y')^2 = 0. In general, it transforms a(x^2) + 2hxy + b(y^2) + 2gx + 2fy + c = 0 into a*(x')^2 + 2hx'y' + b*(y')^2 = k. > .... > then I get -k=-3*2-6*1+12=0... > .... If k = 0 then the conic is degenerate. You can check that your original matrix A is singular, so it can't give you anything better than a line-pair. In this case the two lines are complex, with only their point of intersection (2,1) having real coordinates. A simpler such example is x^2 + y^2 = 0 which is the imaginary line-pair (x + iy)(x - iy) = 0 whose only real point is (0,0). Completing the square in your example (x')^2 + 2x'y' + 4*(y')^2 = 0 gives (x' + y')^2 + 3*(y')^2 = 0 which gives real x' and y' only where x' + y' = 0 and y' = 0, at the single point (x',y') = (0,0), i.e. (x,y) = (2,1). You may like to study an example where the numbers come out simply: 5(x^2) - 4xy + 8(y^2) - 26x - 4y + 5 = 0 or, in matrix from, ( 5 -2 -13) (x) (x y 1) ( -2 8 -2) (y) = 0. (-13 -2 5) (1) You know how to use the first two rows to find the centre (3,1) and then the last row to find -k = -36. Shifting the origin to the centre by letting x = 3 + x' and y = 1 + y' then gives 5*(x')^2 - 4x'y' + 8*(y')^2 = 36, or, in matrix form, (x y) ( 5 -2) (x) (-2 8) (y) = 36. The 2x2 matrix ( 5 -2) (-2 8) has positive determinant 36 (nothing to do with the other 36), so the conic is an ellipse. The eigenvalues are 4 and 9. The corresponding eigenvectors are any non-zero scalar multiples of (2,1) and (-1,2). Hence (2,1) points along one of the axes, and the ellipse cuts it at a distance sqrt(36/4) = 3 from the new origin. Likewise (-1,2) points along the other axis, and the ellipse cuts it at a distance sqrt(36/9) = 2 from the new origin. That's enough for you to sketch the ellipse relative to either the new or old origin, and to work out any angles etc. that you want, without bothering to transform the equation by rotating axes (as most text-books do). Ken Pledger. === Subject: closed bounded subsets of C[0,1] by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iA81ms117164; A SHORT VERSION of my question is: what are examples of closed and bounded subsets of C[0,1] aside from f:[0,1] -> [a,b]? Specifically, can they be defined using some integral properties, e.g. a subset of functions such that int f(t) dt is positive and bounded and perhaps something else on int f(t)^2 or something? ---------------------------------------------- --------------------------------------------- A LONGER VERSION of the question is that I need existence of a solution of the following functional equation: f = Af To define operator A, consider positive definite, bounded, continuous, non-negative r(t,s) and define m(t) = int f(s)*r(s, t) ds. Then operator T is defined as follows Tf(t) = num(t)/den(t), where num(t): = int f(s)*m(s) ds + alpha^2 - m(t)*int f(s) ds den(t): = (int f(s)*m(s) ds + alpha^2)*r(t,t) - m(t)^2 alpha^2 is strictly positive, int := 'integral' and all integrals run from 0 to 1 === Subject: Re: closed bounded subsets of C[0,1] >A SHORT VERSION of my question is: what are examples of closed and bounded subsets of C[0,1] aside from f:[0,1] -> [a,b]? >Specifically, can they be defined using some integral properties, e.g. a subset of functions such that int f(t) dt is positive and bounded and perhaps something else on int f(t)^2 or something? >---------------------------------------------- >--------------------------------------------- >A LONGER VERSION of the question is that I need existence of a solution of the following functional equation: Are you really looking for closed and bounded subsets of C[0,1] or for _compact_ subsets? >f = Af >To define operator A, consider positive definite, bounded, continuous, non-negative r(t,s) and define >m(t) = int f(s)*r(s, t) ds. >Then operator T is defined as follows >Tf(t) = num(t)/den(t), where >num(t): = int f(s)*m(s) ds + alpha^2 - m(t)*int f(s) ds >den(t): = (int f(s)*m(s) ds + alpha^2)*r(t,t) - m(t)^2 >alpha^2 is strictly positive, int := 'integral' >and all integrals run from 0 to 1 ************************ David C. Ullrich === Subject: Re: closed bounded subsets of C[0,1] >A SHORT VERSION of my question is: what are examples of closed and >bounded subsets of C[0,1] aside from f:[0,1] -> [a,b]? Take any continuous functional g on C[0,1], any closed set K in R, and {f: g(f) in K} is closed. Intersections of closed sets are closed... >Specifically, can they be defined using some integral properties, e.g. a >subset of functions such that int f(t) dt is positive and bounded and >perhaps something else on int f(t)^2 or something? Yes, because these are continuous functionals. >A LONGER VERSION of the question is that I need existence of a solution >of the following functional equation: >f = Af >To define operator A, consider positive definite, bounded, continuous, >non-negative r(t,s) and define >m(t) = int f(s)*r(s, t) ds. >Then operator T is defined as follows Is this the same as A? >Tf(t) = num(t)/den(t), where >num(t): = int f(s)*m(s) ds + alpha^2 - m(t)*int f(s) ds >den(t): = (int f(s)*m(s) ds + alpha^2)*r(t,t) - m(t)^2 >alpha^2 is strictly positive, int := 'integral' >and all integrals run from 0 to 1 Yikes! How nonlinear! And what keeps the denominator away from 0? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: fp.h by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iA81ms417160; Anyone know where I can get the fp.h library for C++. I am trying to look at the following code to find roots of a cubic equation: http://www.worldserver.com/turk/opensource/FindCubicRoots.c.txt Jimmy === Subject: Re: Who first spoke of the Laws of Thermodynamics? > I've been trying to find out who first spoke of > the First and Second Laws of Thermodynamics? > I had assumed it was Clausius, > but have been unable to find any actual usage of these terms by him, > though I only have access to a partial translation of his 1850 paper. > Lord Kelvin in 1851 had Propositions 1 and 2, > which were statements of the two laws. > But who actually introduced the term Law? You may find it worth while to post this question to soc.history.science or to a physics news group. Ken Pledger. === Subject: Re: Two children paradox > Assume that we both know that Smith has two children, > but we don't know whether they are boys or girls. ... > If one is a boy and one is a girl, then I will flip a fair coin. >>That's cheating. Let's return to the idea that Smith has two >>children. Suppose you see Smith and one of his children, >>which, as it happens, is a boy. What is the probability that >>Smith's other child is a girl? Well, it's clearly 2/3. > On the contrary, it is clearly 1/2. > The child you have not seen could be a boy or a girl with equal > probability. Here you are trying to make a simple problem more > complicated that it is. That's so very wrong. Your coin flipping expedition made it more complicated than it had to be. If you know, a priori, that Mr. Smith has exactly 2 children, it's a conditional probability problem that may be analyzed as follows (from an earlier post): >>You can't just change this number to 1/2. It's been 2/3 >>for a long time now. Note: You have no information as to >>whether the child, you saw, is older or younger than the >>other child. So, you are stuck with BB, BG, GB. === Subject: Re: Two children paradox ^OX9W/.#XpUmm`>TD2zNE-t}emfPkFR.Z5`flY:3QYT$>dUwN^sm;MBV:F7aL9x*q!` ln!l}>Y6_45$%R|P7DSrBkEph@1-;P*s~F_28vO@e4p/'>}Pc?@rl8cz]d9RXOt, N. Silver > Assume that we both know that Smith has two children, > but we don't know whether they are boys or girls. ... > If one is a boy and one is a girl, then I will flip a fair coin. >>That's cheating. Let's return to the idea that Smith has two >>children. Suppose you see Smith and one of his children, >>which, as it happens, is a boy. What is the probability that >>Smith's other child is a girl? Well, it's clearly 2/3. > On the contrary, it is clearly 1/2. > The child you have not seen could be a boy or a girl with equal > probability. Here you are trying to make a simple problem more > complicated that it is. > That's so very wrong. Your coin flipping expedition made it more > complicated than it had to be. Wow, you really are making this more complicated than it has to be. Whether or not Smith's other child is a boy or girl is independent of the observed child's sex. The probability of the unobserved child's sex being female is 1/2. Let me put it this way: my sister was born before me. When I was conceived, my parents figured it was a 50-50 chance it would be a girl. You are saying that the probability they should have used is 2/3. > If you know, a priori, that Mr. Smith > has exactly 2 children, it's a conditional probability problem that > may be analyzed as follows (from an earlier post): >>You can't just change this number to 1/2. It's been 2/3 >>for a long time now. Note: You have no information as to >>whether the child, you saw, is older or younger than the >>other child. So, you are stuck with BB, BG, GB. Unfortunately, you've made an error in your analysis. We are not stuck with BB, BG, GB. Clearly BG and GB cannot both be possibilities. You've seen one child. That child cannot be both male or female (for the purpose of this mathematical problem, of course). It is irrelevant whether you know which is older. So the possible sample spaces, if the observed child is male, are {BB, BG}, or {BB, GB}. The probability of the unobserved child being female is 1/2. === Subject: Re: Two children paradox > Unfortunately, you've made an error in your analysis. > We are not stuck with BB, BG, GB. Clearly BG and > GB cannot both be possibilities. You've seen one child. > That child cannot be both male or female (for the purpose > of this mathematical problem, of course). It is irrelevant > whether you know which is older. We start with the premise that Mr. Smith has two children, which gives us these 4 equally- likely possibilities: {BB, BG, GB, GG}. Then we find out that at least one of Smith's children is a boy, which restricts the space to 3 equally-likely remaining possibilities: {BB, BG, GB}, which is an or statement, not an and statement. It's lucky that I'm not Marilyn Savant, because i that case you would have egg on your face. > So the possible sample spaces, if the observed child is male, > are {BB, BG}, or {BB, GB}. The probability of the > unobserved child being female is 1/2. === Subject: Re: Two children paradox > Unfortunately, you've made an error in your analysis. > We are not stuck with BB, BG, GB. Clearly BG and > GB cannot both be possibilities. You've seen one child. > That child cannot be both male or female (for the purpose > of this mathematical problem, of course). It is irrelevant > whether you know which is older. We start with the premise that Mr. Smith has two children, which gives us these 4 equally- likely possibilities: {BB, BG, GB, GG}. Then we find out that at least one of Smith's children is a boy, which restricts the space to 3 equally-likely remaining possibilities: {BB, BG, GB}, which is an or statement, not an and statement. It's lucky that I'm not Marilyn Savant, because in that case you would have egg on your face. > So the possible sample spaces, if the observed child is male, > are {BB, BG}, or {BB, GB}. The probability of the > unobserved child being female is 1/2. === Subject: Re: Two children paradox ^OX9W/.#XpUmm`>TD2zNE-t}emfPkFR.Z5`flY:3QYT$>dUwN^sm;MBV:F7aL9x*q!` ln!l}>Y6_45$%R|P7DSrBkEph@1-;P*s~F_28vO@e4p/'>}Pc?@rl8cz]d9RXOt Unfortunately, you've made an error in your analysis. > We are not stuck with BB, BG, GB. Clearly BG and > GB cannot both be possibilities. You've seen one child. > That child cannot be both male or female (for the purpose > of this mathematical problem, of course). It is irrelevant > whether you know which is older. > We start with the premise that Mr. Smith has > two children, which gives us these 4 equally- > likely possibilities: {BB, BG, GB, GG}. > Then we find out that at least one of Smith's children > is a boy, which restricts the space to 3 equally-likely > remaining possibilities: {BB, BG, GB}, which is an > or statement, not an and statement. > It's lucky that I'm not Marilyn Savant, because in that > case you would have egg on your face. No, that's not what we find out. You are missing the whole point of the paradox. We observe Mr. Smith with one of his children, a boy. We ask what the probability is of his other child being a girl. That is NOT the same as just asking what the probability is of Mr. Smith having a girl, based on knowing that he has at least one boy. You should read my previous post and Derek Holt's post before that very carefully. BTW, there is no need to email me, as you are wont to do, letting me know you've responded to the thread. === Subject: Re: Two children paradox >>Let's return to the idea that Smith has two children. >>Suppose you see Smith and one of his children, >> which, as it happens, is a boy. What is the probability that >> Smith's other child is a girl? Well, it's clearly 2/3. >> You can't just change this number to 1/2. It's been 2/3 >> for a long time now. Note: you have no information as to >> whether the child, you saw, is older or younger than the >> other child. So, you are stuck with BB, BG, GB. > Whether the child is elder or not is irrelevant. If B is elder, then we have BB or BG and P(G) = 1/2. > BB is no longer equally as probable as BG: actually BB is twice as > probable as BG, and twice as probable as GB. > Writing the older child first, and the first child we see as C', when we > know nothing we have these possible outcomes, each with probability 0.125: > {B'B}, {BB'}, {B'G}, {BG'}, {G'B}, {GB'}, {G'G}, {GG'}. > After we see the child, we have these possible outcomes, each with > probability 0.25: {B'B}, {BB'}, {B'G}, {GB'}. === Subject: Re: Two children paradox >If I tell you that I have two children and one is a boy, >you may infer that the probability that I have one boy and one girl >is 2/3. Assuming the gender of the other child is not just unknown, but 50% girl-boy probable! >And I can always tell you one of these two statements. Not if both are of ambiguous gender. Doug Goncz I love: Dona, Jeff, Kim, Kimmie, Mom, Neelix, Tasha, and Teri, alphabetically. I drive: A double-step Thunderbolt with 657% range. I fight terrorism by: Using less gasoline. === Subject: Re: Mathematical Rabbits In sci.math, Martin Penderis > Another rabbit problem: > A rabbit is in a box with two holes in the lid. After 1 second after > being put in, it pops its head out of one hole. After another 1/2 > second, it pops its head out of the other hole. After another 1/4 > second, it again pops its head out of the first hole. This pattern > continues. After what time will the rabbit be popping its head out of > both holes at the same time? > Theoretically, the answer is 2 seconds, but in practice it would be a > matter of splitting hares. > (Not original, but I cannot remember where the credit has to go) Ha ha very bunny... :-) Reminds me of Zeno's parajacks. Jack rabbits, that is. [rest crunched] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Mathematical Rabbits but the quantum rabbit can do just that. It can pop its head out of the two holes simultaneously. Quatum rabbits usually live in physics departments and are known not to eat grass. they obey the uncertainty principle: if you know how many you have you don't know where they are and if you know where they are, you don't know how many there are. the little worm > Another rabbit problem: > A rabbit is in a box with two holes in the lid. After 1 second after > being put in, it pops its head out of one hole. After another 1/2 > second, it pops its head out of the other hole. After another 1/4 > second, it again pops its head out of the first hole. This pattern > continues. After what time will the rabbit be popping its head out of > both holes at the same time? > Theoretically, the answer is 2 seconds, but in practice it would be a > matter of splitting hares. > (Not original, but I cannot remember where the credit has to go) > This week I've seen solutions to some problems that just seem more like > magic tricks than like science. As the professor works out the > solutions on the board, he seems like a magician pulling rabbits out of > a hat. Here are the mathematical rabbits I've seen this week: > Rabbit #1: (Analysis) Show that e is irrational. > Solution. He shows that for any positive integer p, > (1) 0 < p![e-(1+1/1!+1/2!+...+1/p!)] < 1. > Then if e=m/n for some integers m,n, there is a contradiction. > Ok, showing (1) (by a geometric series comparison on its tail) and the > contradiction (let p=n and the middle term in (1) becomes an integer) is > easy. But how the heck does he ever thought of to consider (1) to begin > with??? > Rabbit #2: (Complex) Show that there is no analytic bijective map from > 1<=|z|<=4 to 1<=|z|<=2. > Solution. Suppose f is such function. Now define (the rabbit) g(z) = > (f(z)^2)/z. Then on both |z|=1 and |z|=4, |g(z)|=1. Hence by maximum > principle, g is constant, say c. Thus, f = sqrt(cz). But sqrt is not > analytic around the origin if c != 0. (c cannot be 0 for f to be > injective.) > Rabbit #3: (Algebra) Show Aut Q_8 = S_4. That is, the automorphism > group of the quaternion group of order 8 is isomorphic to the symmetric > group of 4 elements. > Solution. This solution actually has two rabbits. > First rabbit: Let Q_8 = {+-1, +-i, +-j, +-k}. Now take cube and put > i,j,k on the faces so that i,j,k are on counterclockwise faces sharing a > vertex, and that (i, -i), (j, -j) and (k, -k) are on opposite faces for > each pair. Then this cube is the geometric representation of Q_8. > Thus, symmetry on the cube = symmetry on Q_8. Thus, Aut Q_8 = rotation > of the cube. Second rabbit: Now the rotation of the cube is just a > permutation of the 4 long diagonals of the cube. The professor shows > moreover that this association is actually an homomorphism. So, > rotation of the cube = S_4. Hence, Aut Q_8 = S_4. > Obviously, I took the loooong approach to list out the conjugacy classes > and all the possible automorphisms. Then I look for the conjugacy > classes of the automorphism group to show how it must be S_4. That took > hours and pages while the professor's solution took only 5 minutes and > just a quarter of the board. > Rabbit #4: (Combinatoric) Show that the # of combination of choosing k > items out of n items *with* replacement is the # of combination of > choosing k items out of n+k-1 items *without* replacement. > Solution. Label each of the n items by integers 1 to n. Suppose u've > chosen k items with replacement from n items. Now sort them in > ascending order according to their assigned integers. Add 1 to the > second item, 2 to the third, and so on. In this fashion, each set of k > non-unique items chosen out of n items corresponds to a set of k unique > items chosen out of n+k-1 items. Similarly, starting with choosing k > items without replacement from n+k-1 items, then ordering them in > ascending order, then subtracting 1 from the second item, 2 from the > third, and so on; thus each set of k unique items chosen out of n+k-1 > items corresponds to a set of k non-unique items chosen out of n items. > Hence, the # of combination of choosing k items out of n items *with* > replacement is the # of combination of choosing k items out of n+k-1 > items *without* replacement. > Lesson learned this week: If you know your subject well enough, *all* > proofs should be at most half a page long. Perhaps Fermat did know his > subject well, and did have a short proof for the FLT. > Kira === Subject: In probabilty, what is the difference between PDF f(x, y) and f(x|y)? Suppose I have jointly probability density functin f_XY(x, y), and contional pdf f_X|Y(x|y), now I want to look into the above two pdfs at y=2, what is the difference between f(x, Y=2) and f(x|Y=2)? Using gaussian variables as example? === Subject: Only Suzuki Groups? Suppose G is a nonabelian finite simple group for which (|G|,|Out(G)|)=1 (Inn(G) is a Hall subgroup of Aut(G)). Does it follow that G is a Suzuki group? ---- David To send me email, move the r from the beginning to the end of the part before the @ and insert alum. at the beginning of the part after the @. === Subject: Re: Only Suzuki Groups? >Suppose G is a nonabelian finite simple group for which >(|G|,|Out(G)|)=1 (Inn(G) is a Hall subgroup of Aut(G)). Does it follow >that G is a Suzuki group? No. As Jim Heckman pointed out, there are examples with |Out(G)|=1. But another example is G = PSL(2,32), with |Out(G)| = 5. In fact PSL(2,2^p) for any prime p >= 5 will work. Derek Holt. === Subject: Re: Only Suzuki Groups? > Suppose G is a nonabelian finite simple group for which > (|G|,|Out(G)|)=1 (Inn(G) is a Hall subgroup of Aut(G)). Does it follow > that G is a Suzuki group? Certainly not if you count the 14 sporadics with Out(G) = 1: M_{11}, J_1, M_{23}, M_{24}, Ru, Co_3, Co_2, Ly, Th, Fi_{23}, Co_1, J_4, B, M. Probably someone else will be able to answer for the groups of Lie type before I can tease it out of the tables in the _Atlas_. -- Jim Heckman === Subject: Re: do I understand co-variant vs. contra-variant? Ok, so what you are saying is if we have and if x |----> Ax = x', then what must y' be so that = ? And the answer would be y' = (A^{-1})^T y I agree that this is true and important in general, but it really doesn't address the nuts and bolts co-variant vs. contra-variant basis questions that I posted in late October and again on Nov 5. --Jeff === Subject: Re: do I understand co-variant vs. contra-variant? > > Ok, so what you are saying is if we have and if > x |----> Ax = x', then what must y' be so that = ? > And the answer would be y' = (A^{-1})^T y > I agree that this is true and important in general, but it really > doesn't address the nuts and bolts co-variant vs. contra-variant basis > questions that I posted in late October and again on Nov 5. > --Jeff What threw me, is that you quoted on Nov 5 tangent vectors (referring to the basis) transform contravariantly. I can't speak for everyone but I believe there is a general concensus in this NG that tangent vectors, denoted e>_u, transform COVARIANTLY and can rightly be called a covariant basis, or IOW's convariant unitary base vectors, the word unitary parlays because they aren't of magnitude 1 in general. An arbituary vector, I'll denote A> (I'm using the > to designate a vector quantity, much like texts use bold), is defined, A> = A^u e>_u , ( = A_u e>^u ) where A^u are *contravariant components* of A>. a bit colloquial, but I figure they mean components A^u, because a vector A> cannot have any covariant or contravariant notation until defined by a basis, I think, should be regarded as invariant. I don't have the text you are referring to, but you may direct your professor to my post, and if he has additional information, we would be pleased to hear from him. Ken S. Tucker === Subject: Re: SL_2(3) and other groups of order 24 > p = 2,3,5,7, mainly to look at PSL_2(p). > > I don't know what G = SL_2(3) is however. It has |G| = 24 = 8.3, > and the orders of the elements are > 4, 6, 3, 4, 6, 3, 1, 3, 3, 3, 4, 6, 3, 6, 4, 2, 6, 6, 6, 4, 3, 6, > 3, 4, or > (order, # elements) = (1,1), (2,1), (3,8), (4,6), (6,8) ==> 24 > elements. > > The small groups table for n = 24 gives (see Mathworld e.g.) > > C_24 > C_2 x C_12 > C_2 x C_2 x C_6 > S_4 > S_3 x C_4 > S_3 x C_2 x C_2 > D_4 x C_3 > Q x C_3 > A_4 x C_2 > T x C_2 > Five more nonabelian groups of order 24 > Reference: Burnside, pp. 157--161. > > I know that if m is odd, D_2m =~ C_2 x D_m. > > I'm not sure about D_12 though. But D_12 is not the group with the > above properties. > Is there another way to present this group, like the > above list, perhaps with semi-direct products? > SL(2,3) is the unique non-abelian semidirect product Q_8 x| C_3. D'oh. Make that: the unique non-*direct* semidirect such product. > It has a presentation a^c = b, b^c = ab>. > I am not used to the a^b notation. > Is a^b = bab^(-1), conjugation of a by b? I think it must be if > a and b generate Q. Yes. Or b^(-1)ab, depending on my mood. :-) > Yes, I just took a look at it. I have never seen this group before. > I still have to verify that it is in fact SL(2,3), but I will take > your word for it--you are seldom wrong in things like this > in my (limited) experience. Well, clearly SL(2,3) has a surjective homomorphism onto PSL(2,3) =~ A_4. There are only 2 such groups of order 24 -- the one under discussion and A_4 x C_2. And it's not too hard to show that SL(2,q) has only 1 involution for q odd. [...] -- Jim Heckman === Subject: Re: SL_2(3) and other groups of order 24 >> The small groups table for n = 24 gives (see Mathworld e.g.) >> >> C_24 >> C_2 x C_12 >> C_2 x C_2 x C_6 >> S_4 >> S_3 x C_4 >> S_3 x C_2 x C_2 >> D_4 x C_3 >> Q x C_3 >> A_4 x C_2 >> T x C_2 >> Five more nonabelian groups of order 24 >> Reference: Burnside, pp. 157--161. >> >> I know that if m is odd, D_2m =~ C_2 x D_m. >> >> I'm not sure about D_12 though. But D_12 is not the group with the >> above properties. >> Is there another way to present this group, like the >> above list, perhaps with semi-direct products? > > SL(2,3) is the unique non-abelian semidirect product Q_8 x| C_3. > D'oh. Make that: the unique non-*direct* semidirect such product. the unique non-*direct* semidirect ? What does this mean. I have never heard of it. Van === Subject: Re: SL_2(3) and other groups of order 24 <10orfstg9r7nt32@corp.supernews.com> <10otr2k8idcfr9d@corp.supernews.com> > Well, clearly SL(2,3) has a surjective homomorphism onto > PSL(2,3) =~ A_4. There are only 2 such groups of order 24 -- the > one under discussion and A_4 x C_2. And it's not too hard to > show that SL(2,q) has only 1 involution for q odd. My post above says there is only 1 element of order 2, and it is -1 (or p-1), the only element of Z*_p of order 2, and I see that 2 x 2 matrices with elements diag(k,k) with k in Z*_p usually have smaller order than other matrics, but again, this is far from a proof. Van === Subject: Re: Tensor Physics & WMD > [Saul-Paul Sirag] The issue of coordinate dependence of tensors is very > tricky. Perhaps the following statements (by Arthur Eddington, 1920) > will clarify the issue. Since that's obviously the same Eddington that proved that the fine structure constant equals 137, you should actually say that he tought geodesics were tricky. Since Tensors are so simple you can computerize them, since they have no coordinate dependence. Which is really Einstone used elevators in addition to trains in his gedankers, to show that geodesics are tensor-free algebras rather than calculus'. > 1. A tensor does not express explicitly the measure of an intrinsic > quality of the world, for some kind of mesh-system is essential to the > idea of measurement of a property, except in certain very special cases > where the property is expressed by a single number termed an invariant, > e.b. the interval, or the total curvature. Tensors are now and always been, the fairly simple calculus of transformations. Which is really why they were first used by Maxwell in EM theory to show that magnetic fields are coordinate invariant, and not by Einstone and astrologers. > 2.But to state that a tensor vanishes, or that it is equal to another > tensor in the same region, is a statement of intrinsic property, quite > independent of the mesh-system chosen. > Saul-Paul > Yes, a good start. Let's continue. > A tensor (or spinor or twistor) transformation is a multi-linear > homogeneous transformation on objects. > A spinor is a complex double-valued representation of some continuous > groups like SU(2) that covers O(1,3) of special relativity. > The objects are N-dimensional hyper-matrix arrays of elements of an > algebraic field. > Let X be the basic linear transformation. > For example, using summation convention on repeated pairs of identical > upper and lower indices, ^ denotes upper index > Au --> Au' = Xu'^uAu > is a simple linear transformation. > S ---> S' = S is a scalar invariant. > For example S = AuA^u > The simplest multi-linear transformation is > Fuv ---> Fu'v' = Xu'^uXv'^vFuv > X in general is a homomorphic image or representation of some GROUP > (maybe semi-group and maybe even up to a CATEGORY?) > Therefore we really have X(G) where G is some group. > All laws of physics must be covariant under the physical symmetry groups G. > This means that the laws of physics must be tensor equations in this > general sense. > What applies to the laws of physics DOES NOT APPLY to the fundamental > VACUUM SOLUTIONS of those laws! Or to any solution. > There is a lot of confusion on this issue. > In the SPECIAL CASE of Einstein's 1916 theory of gravity > G = GCT AKA DIFF(4), i.e. generally NONLINEAR LOCAL coordinate > transformations AT A FIXED POINT EVENT P. But in the special case of Einstein's 1919 theory of gravity, that is not true. Since by that time *he* had already disregarded the cosmological constant, in favor of a dynamic model of gravity. === Subject: Re: Simple group of order 60 > OK, Jyrki's excellent long post has inspired me to do the same > for the simple group of order 168, G =~ L(2,7) =~ L(3,2), but > not in as much detail. (I'm tired from staying up late last > night to watch election returns, and bummed out by the results.) > The number of Sylow 7-subgroups P_7's must be n_7 = 8, so 48 > elements of order 7 and |N(P_7)| = 21. > So G =~ a subgroup of A_8 > By the action of G by conjugation on the 8 Sylow 7-subgroups P_7, > right? Yes. Or equivalently, the action of G by multiplication on the 8 cosets of any one of the N(P_7)'s. > Also [G:N(P_7)] = 8 ==> |N(P_7)| = 21. I follow this part OK. > (But I have only studied A_4 and A_5, never A_n for n > 5, nor n = 8 > in particular.) > (then cyclic) there are too many elements. If Abelian then > there are 8 x 20 = 160 elements, ? Only if they intersect trivially, pairwise. Nothing to do with abelian. > which is too many since there > are at least 2 or order 3 and 7 in the Sylow 2-subgroup = 170 > 168, > so N(P_7) = is the non-Abelian group with |x| = 7, |y| = 3, > and yxy^(-1) = x^2. So they are the semidirect product of > C_3 and C_7 (which you called P_7). They're non-abelian because A_8 has no cyclic subgroup of order 21. Also because otherwise the normalizer of an intersection would have order >= 21*21/3 = 147. > By the argument above the > 8 groups N(P_7) must have a non-trivial intersection. > so N(P_7) is non-abelian and each intersects each other in a C_3 > I don't get how you know that they intersect in C_3. I see that > it is either C_3 or C_7, since 3 and 7 divide 21 and 168, > |N_i / N_j| = 3 or 7. I don't see how to eliminate the > 7, unless it is because they are the normalizers of different > sylow subgroups P_7. Yes, I think this argument works. Is this > right? So if we call the 8 normalizers N_i (i = 1,...,8), > we have > |N_i / N_j| = 3, i < j (say), since they normalize different > P_7's. Exactly. 2 N(P_7)'s can't share a P_7 or its normalizer would be larger than N(P_7). :-/ [...] > n_2 != 3 since > then G would a subgroup of S_3, and n_2 != 7 since then there'd > be a C_6, > I don't follow why this statement is true. > n_2 = 7 ==> |N(P_2)| = 3 x 8. Why would there be a C_6? Because every group of order 24 with a normal Sylow 2-subgroup has one: All groups of order 8, except the elementary abelian one E, have a characteristic involution, and E has 7 involutions so one of them must be normalized by a C_3 acting on E. -- Jim Heckman === Subject: Re: Simple group of order 60 > OK, Jyrki's excellent long post has inspired me to do the same > for the simple group of order 168, G =~ L(2,7) =~ L(3,2), but > not in as much detail. (I'm tired from staying up late last > night to watch election returns, and bummed out by the results.) > > The number of Sylow 7-subgroups P_7's must be n_7 = 8, so 48 > elements of order 7 and |N(P_7)| = 21. > So G =~ a subgroup of A_8 >> By the action of G by conjugation on the 8 Sylow 7-subgroups P_7, >> right? > Yes. Or equivalently, the action of G by multiplication on the 8 > cosets of any one of the N(P_7)'s. >> Also [G:N(P_7)] = 8 ==> |N(P_7)| = 21. I follow this part OK. >> (But I have only studied A_4 and A_5, never A_n for n > 5, nor n = 8 >> in particular.) >> (then cyclic) there are too many elements. If Abelian then >> there are 8 x 20 = 160 elements, > ? Only if they intersect trivially, pairwise. Nothing to do with > abelian. Quite true. I don't know what I was thinking. > They're non-abelian because A_8 has no cyclic subgroup of order > 21. Also because otherwise the normalizer of an intersection > would have order >= 21*21/3 = 147. I understand the statement re A_8, though this leads to questions about possible orders of elements of A_n and S_n. e.g., I know that products of disjoint cycles have order = lcm(orders of the cycles). Is that enough to determine the order of elements of A_n and S_n? Also, I need to get some notes and/or a text, but later for that. I don't understand the statement about the order of the intersection of the normalizer, but later for that too. >> By the argument above the >> 8 groups N(P_7) must have a non-trivial intersection. > so N(P_7) is non-abelian and each intersects each other in a C_3 >> I don't get how you know that they intersect in C_3. I see that >> it is either C_3 or C_7, since 3 and 7 divide 21 and 168, >> |N_i / N_j| = 3 or 7. I don't see how to eliminate the >> 7, unless it is because they are the normalizers of different >> sylow subgroups P_7. Yes, I think this argument works. Is this >> right? So if we call the 8 normalizers N_i (i = 1,...,8), >> we have >> |N_i / N_j| = 3, i < j (say), since they normalize different >> P_7's. > Exactly. 2 N(P_7)'s can't share a P_7 or its normalizer would be > larger than N(P_7). :-/ > [...] > n_2 != 3 since > then G would a subgroup of S_3, and n_2 != 7 since then there'd > be a C_6, >> I don't follow why this statement is true. >> n_2 = 7 ==> |N(P_2)| = 3 x 8. Why would there be a C_6? > Because every group of order 24 with a normal Sylow 2-subgroup > has one: All groups of order 8, except the elementary abelian > one E, have a characteristic involution, and E has 7 involutions > so one of them must be normalized by a C_3 acting on E. I am still not clear on this. I recently posted a question on the group SL(2,3) and it turned out that all the groups I looked at has an element of order 6, but I still don't see how to prove that all groups of order 24 have an element of order 6, in spite of what you say above. First, your statement about the characteristic involution of a group of order 8, I take to the non-identity element in the center, since the center has at least 2 elements (8 = 2^3 a p-group). Are you saying that this element in the center of the subgroup of order 8 also commutes with the element of order 3? This may be, but I still don't see how to prove it. (This is all pretty new to me, and I am self taught in algebra, is my excuse for being often so slow. I have also found my mind slowing down as I get older ;-) Van === Subject: Re: Simple group of order 60 >> OK, Jyrki's excellent long post has inspired me to do the same >> for the simple group of order 168, G =~ L(2,7) =~ L(3,2), but >> not in as much detail. (I'm tired from staying up late last >> night to watch election returns, and bummed out by the results.) >> >> The number of Sylow 7-subgroups P_7's must be n_7 = 8, so 48 >> elements of order 7 and |N(P_7)| = 21. > >> So G =~ a subgroup of A_8 > > By the action of G by conjugation on the 8 Sylow 7-subgroups P_7, > right? >> Yes. Or equivalently, the action of G by multiplication on the 8 >> cosets of any one of the N(P_7)'s. > Also [G:N(P_7)] = 8 ==> |N(P_7)| = 21. I follow this part OK. > (But I have only studied A_4 and A_5, never A_n for n > 5, nor n = 8 > in particular.) > > (then cyclic) there are too many elements. If Abelian then > there are 8 x 20 = 160 elements, >> ? Only if they intersect trivially, pairwise. Nothing to do with >> abelian. > Quite true. I don't know what I was thinking. >> They're non-abelian because A_8 has no cyclic subgroup of order >> 21. Also because otherwise the normalizer of an intersection >> would have order >= 21*21/3 = 147. > I understand the statement re A_8, though this leads to questions > about possible orders of elements of A_n and S_n. e.g., I know > that products of disjoint cycles have order = lcm(orders of the > cycles). Is that enough to determine the order of elements of A_n > and S_n? Also, I need to get some notes and/or a text, but later > for that. > I don't understand the statement about the order of the intersection > of the normalizer, but later for that too. > By the argument above the > 8 groups N(P_7) must have a non-trivial intersection. > >> so N(P_7) is non-abelian and each intersects each other in a C_3 > > I don't get how you know that they intersect in C_3. I see that > it is either C_3 or C_7, since 3 and 7 divide 21 and 168, > |N_i / N_j| = 3 or 7. I don't see how to eliminate the > 7, unless it is because they are the normalizers of different > sylow subgroups P_7. Yes, I think this argument works. Is this > right? So if we call the 8 normalizers N_i (i = 1,...,8), > we have > |N_i / N_j| = 3, i < j (say), since they normalize different > P_7's. >> Exactly. 2 N(P_7)'s can't share a P_7 or its normalizer would be >> larger than N(P_7). :-/ >> [...] >> n_2 != 3 since >> then G would a subgroup of S_3, and n_2 != 7 since then there'd >> be a C_6, > I don't follow why this statement is true. > n_2 = 7 ==> |N(P_2)| = 3 x 8. Why would there be a C_6? >> Because every group of order 24 with a normal Sylow 2-subgroup >> has one: All groups of order 8, except the elementary abelian >> one E, have a characteristic involution, and E has 7 involutions >> so one of them must be normalized by a C_3 acting on E. I thought about this some more after looking at your post on Q x| C_3, and thought of |S_4| = 24 but no element of order 6. So I now see that you are using the fact that N(P_2) = {x|x P_2 x^(-1) = P_2} is the normalizer of a group of order 8 to say that it always has an element of order 6. I still don't see why this is true though. Van > First, your statement about the characteristic involution of a group > of order 8, I take to the non-identity element in the center, since > the center has at least 2 elements (8 = 2^3 a p-group). > Are you saying that this element in the center of the subgroup of > order 8 also commutes with the element of order 3? > This may be, but I still don't see how to prove it. > (This is all pretty new to me, and I am self taught in algebra, > is my excuse for being often so slow. I have also found my > mind slowing down as I get older ;-) > Van === Subject: Re: Simple group of order 60 <10ott8f2ook637a@corp.supernews.com> >> >> n_2 != 3 since >> then G would a subgroup of S_3, and n_2 != 7 since then there'd >> be a C_6, >> > I don't follow why this statement is true. > n_2 = 7 ==> |N(P_2)| = 3 x 8. Why would there be a C_6? >> >> Because every group of order 24 with a normal Sylow 2-subgroup >> has one: All groups of order 8, except the elementary abelian >> one E, have a characteristic involution, and E has 7 involutions >> so one of them must be normalized by a C_3 acting on E. I thought I had proved the existence of N(P) having an element of order 6 when |N(P)| = 24, now I am not so sure. Does this look like the kind of reasoning that was being referred to? The def. of a characteristic subgroup, that f(P) = P for all autos f, in particular for inner autos which we consider here, and the fact that N(P) = {y|yPy^-1 = P} and |P| = 8 = 2^3, and that P is a normal subgroup of N(P) by def. of the normalizer, with |N(P)| = 24. Its easy to show the the center Z of P has an element z of order 2, and a subgroup of the center of G is characteristic, so that if |y| = 3, there is an element z in Z such that yzy^-1 = z, so that |yz| = 6. Van === Subject: Re: Simple group of order 60 > So I now see that you are using the fact that > N(P_2) = {x|x P_2 x^(-1) = P_2} is the normalizer of a group of > order 8 to say that it always has an element of order 6. > I still don't see why this is true though. > Van My step by step take: 1) So you have a group P of order 8 that is a normal subgroup of a group N of order 24. 2) The bigger group has an element x of order 3. 3) Conjugation by x gives an automorphism phi of P. 4) The order of phi in Aut(P) is either 1 or 3. 5) An element y of P is a fixed point of phi (i.e. an element with the property phi(y)=y), iff x and y commute. 6) If x and an element y of P commute, then the ord_N(xy) is 3*ord_N(y) 7) If y is a fixed point of phi, then so are all its powers. 8) If phi has a non-trivial fixed point it has a non-trivial fixed point of order 2. 9) If phi has a non-trivial fixed point, then N has elements of order 6. 10)If phi is of order 1, then all the elements of P are fixed points of phi. 11)If phi is of order 3, then phi has a non-trivial fixed point. 12)N has elements of order 6. Of the statements above, only #11 is not immediately obvious. That step depends on having a list of groups of order 8 (and their automorphisms) at hand. Students who have reached this level can usually reproduce the data on automorphisms in a couple hours at max (in the case they don't have it at their fingertips already). The cases of the quaternion group and the elementary abelian group are the most difficult ones, because they have quite a few automorphisms. If one is only interested about the fixed points (as is the case here), then e.g. the quaternion group is easy (how many elements of order 2 are there in Q_8?) Jyrki === Subject: Re: Simple group of order 60 <10ott8f2ook637a@corp.supernews.com> This had made it clear to me I need to buy another text. I got Hungerford's undergrad text, Intro to Abstract Algebra, and some notes online, and have learned the basics of algebra, but I was trained in physics, and my math eduation has big holes and shortcomings. (I can solve differential eqns, do numerical integration of PDEs, analyze data, but I am tired of all that). Any suggestions re a text on algebra and/or groups at the intro grad level? I don't get to a library any more. I recall looking at Lang's big Algebra book, and I could probably learn a lot from it, but its not really what I'm looking for. OTOH, his undergrad algebra book is too elementary, with what I have, and notes I have found. Van === Subject: fermat^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Fermat's Last Theorem Ben Ito 11-7-04 I will prove Fermat's Last Theorem. l. Introduction I will show that Fermat's n=4 is impossible, describe the elliptic curve derivation, and prove Fermat's n>2 theorem. 2. Fermat's Proof (n=4) Fermat is using the integer solutions of n=2 in a contradiction method to prove that n=4 does not form integer solutions. Fermat uses, A^4 + B^4 = C^2 (equ l). and A^2 = 2uv, B^2 = u^2 - v^2, and C = u^2 + v^2 (equ 2), (Shanks, p.141) to prove that n=4 does not form integer solutions. Fermat's contradiction method is correct; however, Fermat is not testing all integers. Fermat's contradiction method is only using the integers sequence described with equation 2. It is impossible to prove n=4 since it would require an infinite number of equations. 3. Elliptic Curve The derivation of the elliptic curve is described. The elliptic curve equation is derived using the integer solution equations of n=2 (Osserman, p.21), a = k(m^2 - n^2), b = k2mn, c = k(m^2 + n^2). (equ 3) The elliptic curve is only valid for n=2; therefore, an elliptic curve can not be used to prove FLT for n>2. 4. Fermat's Proof. A circle transformation (z = c) is used in Fermat's equation n=2, x^2 + y^2 = c^2. (equ 4) The integer solution equations of n=2, x = 2uv, y = u^2 - v^2, and z = u^2 + v^2 (equ 5a,b,c) are used in equation 4, (2uv)^2 + (u^2 -v^2)^2 = u^2 + v^2 = c.(equ 6) Consequently, equations 5c and 6 forms the transformation equation z=c which can only be derived when n=2; therefore, only n=2 forms integer solutions. 5. Conclusion Fermat's n=4 proof does not include all integers; therefore, Fermat's n=4 proof is incomplete.The proof of n=4 is impossible because it would require and infinte number of equations. Fermat's elliptic curves are derived using the integer solution equations of n=2; therefore, an elliptic curve can not be used to prove FLT when n>2. The n=2 equation is transformed into a circle equation then the integer solutions of n=2 are used to derive the transformation equation z=c which can only be formed when n=2. 6. References Robert Osserman. Fermat's Last Theorem (a supplement to the video). MSRI. 1994 Daniel Shanks. Solved and Unsolved Problems in Number Theory. Chelsea Pub. 1985. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Curves invariant under a rational map (Was: Re: Equivalent sequences) I'm changing the suuject line because this thread is diverging, but (IMHO) still interesting! Recall the prior thread was started by GE , who >Let's take the sequence u as : u(n+2)=(6-u(n+1))/u(n), and u(0), u(1) >having whatever given values. >If you plot on a plane the points whose coordinates are (u(n), u(n+1)) >you get a curve (closed or not), not a surface. >This suggests that there exist some kind of invariant like : > F(u(n), u(n+1)) = constant I found that this was correct for this particular recurrence relation, >There is a slight generalization to the recursion >u(n+2) = (a + b u(n+1))/u(n) >which has the invariant > (ab + (a+b^2)(x + y) + b (x^2 + y^2) + x y^2 + x^2 y)/(x y) But I wonder to what extent the generalizations can be pressed? I think I have a pretty good handle on the situation now. The description of what's possible and what's not is not all that complicated, in the end, but I'd like to develop this slowly, for easy reading. Let's start with a simple, familiar recurrence relation u(n+2) = u(n+1) + u(n) . When u(0)=0, u(1)=1 we get the familiar Fibonacci sequence. What happens if we follow GE's lead here? We get a sequence of points ( u(n), u(n+1) ) in the plane. Do they lie in an algebraic curve? Well, those who know this sort of thing probably know this one instantly but I didn't; I had to experiment a bit to discover that when x,y are consecutive Fibonacci numbers, we have x^2 + xy - y^2 = +- 1 so that all the points in our sequence lie in the (reducible) variety (x^2 + xy - y^2)^2 - 1 = 0 . With just a bit of algebra we find generally that substituting (x,y) = (y, x+y) simply changes the sign of x^2+xy-y^2 so that F(x,y) = (x^2 + xy - y^2)^2 is an invariant for this recurrence relation. Aha, now we're two-for-two; maybe there is a general result that the points in our set will lie in some simple algebraic curve!... In order to investigate further, I'd like to pull the trick of generalizing the question to simplify the work. So suppose that we more generally have any pair of rational functions, defining a map T: R^2 --> R^2 (well, it's defined on _most of_ R^2). Under what circumstances can we find a function F : R^2 --> R with the property that F o T = F ? Answer: always -- just take F = 0. Hm, I guess that wasn't the right question. What we'd like to have F do is to distinguish the orbits of T. That's not going to be possible, in general: the inverse images F^{-1} (p) will usually be nice curves, and in particular will usually be uncountable, whereas the orbits of T are countable, and so even if F^{-1} (p) contains whole orbits, it must contain uncountably many of them. Well, perhaps we want F^{-1}(p) to be limited to just the closure of one orbit, or something. Looks like we'll have to refine the question a bit as we go along. Let's start with an easy map T, such as T(x,y) = (x,-y). In this case the orbits contain at most two points -- they're not curves at all. The right way to define invariants for the orbits in this case is with PAIRS of functions F: we could use F1(x,y) = x and F2(x,y) = y^2, for example. The set of points with F1 = a and F2 = b consists precisely of the points in a single orbit. It turns out that what distinguishes this T is that it has finite order: T^2 = identity. There would be a similar situation with e.g. T(x,y) = (y,-x-y), or indeed any map T for which T^n = T^m for some n > m. We will have to watch out for maps T of finite order in the sequel. Next, let's consider the map T(x,y) = (x, y+1). In this case, two points are in the same T-orbit iff they have the same x-coordinate and the same _fractional part_ of their y-coordinate. Now, we can come pretty close to finding a pair of functions F = (F1, F2) which distinguishes orbits, as in the previous paragraph. For example, F = (x, sin(2 pi y)) nearly does what we want (the inverse image F^{-1} (p) will usually be the union of two orbits because sin(2 pi y) = sin(2 pi (1/2 - y)) ). On the other hand, if we limit our attention to _algebraic_ functions, we will be in trouble: there is no algebraic function to replace the sine function --- an algebraic function has finite inverse images. In this case, we would have to content ourselves with just F(x,y) = x --- this invariant has the property that F o T = F, although the inverse image of a single number is a vertical line, which includes uncountably many T-orbits. Well, we noted earlier that this is going to be a fact of life, so we will accept this invariant. Now how about the map T(x,y) = (x, y/2) ? This suffers from the same problems as we saw in the last paragraph: the invariant F(x,y) = x is constant on a single T-orbit, although it fails to distinguish between some distinct T-orbits. This example is a bit closer to the example in the OP's message: a single T-orbit is dense in at least one region, suggesting that the OP's curve really should be the vertical line after all, so that the use of the invariant F(x,y) = x is appropriate. Now let's consider a transformation which arises from a recurrence relation as in the OP's problem: T(x,y) = (y, -x/2 + 3y/2). We note that T is a linear map R^2 --> R^2, and it turns out to be easy to diagonalize; using a basis of eigenvectors we find that in the new coordinate system, T is precisely the map described in the previous paragraph. The geometry does not change at all (except for the location of the axes!), and we will still take those lines as the sought-for curves containing the orbits. The lines are distinguished by the same linear invariant F ; returning to the original (x,y)-coordinates, I make this invariant out to be F(x,y) = x - 2y . This trick of changing coordinates is useful. We can in particular use it whenever T is a diagonalizable linear map; then we may assume T(x,y) = (ax, by) for some a, b. That helps a lot. For example, what about the map T(x,y) = (2x, 4y) ? In this case there are no orbits which are lines. On the other hand, we note that the quantity F(x,y) = x^2 / y (defined almost everywhere) stays constant on orbits: F o T = F, so we use this as our invariant, and conclude that the orbits are in fact contained in the parabolas y = k x^2. More generally, if T(x,y) = ( c^m x, c^n y) for some real c and some integers m,n, then the quantity F(x,y) = x^n / y^m stays constant on orbits. Taking my cue from the previous examples, I will refer to F as the invariant, confident that there is really only one, unless c = +- 1 (which would mean T has finite order, giving finite orbits). Of course one could also use g o F for any function g of one variable; this would also be T-invariant. Naturally, if we want a rational function of x and y, then we must take g to be a rational function of F; in other words, the set of invariant functions we could use is the field R( F ) of all rational functions in F. Using the change-of-variables trick, we now see we have a solution for any diagonalizable linear T whose eigenvalues are powers of each other. But what about, say, T(x,y) = (2x, 3y) ? The best approximation to the previous result would be that F(x,y) = x^(log(3)/log(2)) / y . This is a well-defined invariant, and the curves y = k x^(1.58...) do indeed contain all the points in a single T-orbit. But this is of course not a rational function, and no rational function of F will be a rational function of x and y. I don't claim to have given a complete proof here, but I believe that with this simple example for T, we see that there is NO algebraic curve which contains a typical entire T-orbit. (Obviously the axes are algebraic curves containing whole orbits, but just as obviously they miss most of the plane between them!) It would be easy to masquerade this example as, say, a sequence arising in the OP's way: use the recurrence relation u(n+2) = -6 u(n+1) + 5 u(n), and you find that the curves containing an orbit are not algebraic curves. (You can even replace T by T^{-1} to get an example in which the orbit converges to a point -- the origin -- so that on a plot, the curve would definitely appear, but it's a transcendental curve.) While I am on the topic of linear T's, let me consider other types of linear maps. In some cases, T is diagonalizable, but not over the rationals; this is precisely what happens in the Fibonacci case. What enabled us to succeed in the Fibonacci case seems to be that the real curve x^(log(phi)/log(bar{phi})) / y actually IS algebraic: the two eigenvalues are (negative) inverses of each other, so the complicated exponent is just -1, making the invariant be 1/(xy) (or, just as well, xy itself). Here x and y refer to the coordinates in which the map T is diagonal. In a coordinate system where T(x,y) = (y,x+y), this translates to x^2+xy-y^2 (or a multiple thereof, depending on which eigenvectors are used to change to the other coordinate system). Well, I had to lie a little bit to get something so similar to the result from the top of this post: the two eigenvalues multiply to -1, not +1. But we can reconstruct the argument of the last two paragraphs, noting that if phi > 0 > bar{phi}, and if T(x,y) = ( phi x , bar{phi} y ), then F(x,y) = x^[ log( | bar{phi} | ) / log(phi) ] / y has the property that F o T = -F; this exponent really is -1 this time, so F(x,y) = 1/(xy) has the property that it changes sign when applying T, and so F^2 is a true invariant for this recurrence. But it seems that this trick works only if the recurrence is of the special forms u(n+2) = +-u(n+1) + k u(n) (so that the eigenvalues will be inverses of each other, up to sign), or u(n+2) = k u(n) (so that the eigenvalues will be negatives of each other). Here I'm restricting to rational coefficients and noting that if phi is a quadratic irrational and k is rational, then log(k)/log(phi) is irrational unless k=1 or phi = sqrtk^m) . I haven't looked too closely at the cases in which T is a linear map but not real-diagonalizable. You may have noticed that all the rational invariant functions F which I have found have been very simple. There's a good reason for this. We have considered only mappings T which are linear with integer coefficients. These mapping obvious carry integer points to integer points. It follows that if the algebraic curve F(x,y) = c has an integer point on it, then it has infinitely many of them. (We do need to check that no point is fixed under any iteration of the linear map T but that would imply that T has roots of unity among its eigenvalues, necessarily no more than sixth roots of unity in fact. We have been consistently setting aside cases in which T has finite order.) But that's a rare feature among algebraic curves: typically the number of integer points is _finite_. The only exception is for curves of genus zero, which is to say (rationally) parameterizable curves. For an irreducible nonsingular curve the genus will be zero only if the degree is 2. These other curves of the form y^m = k x^n which we have introduced have higher degree but nasty singularities at the origin which effectively make them of genus zero. The fact that these curves have genus zero makes them parameterizable with rational functions of one variable. This also leads to something I was alluding to in another post in a different branch of this thread: these curves can in a natural way be given a group structure, and using a real parameterization consistent with the group structure, there are natural maps from the curve to itself, given by addition by a fixed element of the group. That is, we may express the action of T on the curve it fixes, T : C --> C, as T(p) = p + q_T for some fixed point q_T in C. But that discrete transformation is just one stop along the way in a continuous transformation: there is for every real number t a transformation T_t : C --> C given by T_t ( p ) = p + ( t . q_T ) where + and . are the addition and scalar multiplication in this Lie group C. These other transformations T_t allow us to connect the dots between the iterates of T itself. (I am glossing over the non-invertible transformations and the orientation-reversing ones, which are obviously not just addition-by-something) To recap what we have done so far: for LINEAR maps T : C --> C we have found that the invariants are (a) pairs of (rational) functions, when T has finite order (b) single rational functions F(x,y) in some other cases (although F may be replaced with anything in the field R(F) ) (c) not algebraic at all in other cases. <*PUNCHLINE ALERT*> What I think is the key insight is in (b): what we want is not so much a particular rational function as a _field_ of rational functions, namely a subfield of R(x,y). The right question to ask is, what subfield is it? By Galois theory, we may view subfields as the fixed points under groups of automorphisms, and exactly the group we need is the group generated by T. Indeed this checks out in (a): when T has finite order, then the fixed field R(x,y) ^ will have finite index in the original field R(x,y), and in particular will still have transcendence degree 2. That's why the orbits must be described by _pairs_ of invariants. For example, when T(x,y) = (x,-y), the fixed field is R(x, y^2) . I admit I'm having a little trouble understanding case (c), that is, trouble transfering the previous examples convincingly to this field- theoretic setting. I'd be interested in hearing if someone has a nice algebraic proof that R itself is the only subfield of R(x,y) fixed under the action of the group generated by T(x,y) = (2x,3y) . Seems odd to have the fixed field drop down so low... Assuming that's correct, of course, that makes it believable that in a more complicated example like the one I concocted before: T(x,y) = (y, (3-x+2*y)/(11+x-y)) the orbits would not be contained in an _algebraic_ curve. We can at least impose some constraints on the type of algebraic curve which can arise. Starting in this case with any rational point (x,y), we observe that the iterates T^(n) (x,y) are also all rational. Excluding a comparatively small number of exceptional points (I don't know how to quantify that exactly), these orbits will all be infinite. Thus the curves we are looking for have infinitely many rational points. Again this is rare but not quite as limiting as in the case of integer points: what is true here is that an algebraic curve with infinitely many rational points has genus 0 or 1. When non-singular, that means the total degree must be at most 3, so we should expect either curves of low degree or containing singularities, just as with the genus 0 cases. (The remarks about fitting T into a continuous family of transformations still apply, since curves of genus 1 are also groups. I will spare you all the gory details.) As for the particular example which started this whole thread, T(x,y) = ( y, (6-x)/y ), I have to say I can't think of an easy way to put this into the current framework: I don't know an easy way to compute the fixed field even for a comparatively simple function like this. The invariant F(x,y) = (-y^2+x*y^2+x^2*y+7*y-6+7*x-x^2) / (x*y) which I found more or less by accident does indeed lead to (usually) non-singular curves F = c of genus 1. If someone else would like to chime in now, I would like to see how to compute for example the subfield of R(x,y) invariant under T(x,y) = ( y, (6-x)/y + 1 ) . Is it just R itself? Any takers? dave === Subject: Re: Curves invariant under a rational map (Was: Re: Equivalent sequences) > I'm changing the subject line because this thread is diverging, but > (IMHO) still interesting! >Let's take the sequence u as : u(n+2)=(6-u(n+1))/u(n), and u(0), u(1) >having whatever given values. >If you plot on a plane the points whose coordinates are (u(n), u(n+1)) >you get a curve (closed or not), not a surface. >This suggests that there exist some kind of invariant like : > F(u(n), u(n+1)) = constant > I found that this was correct for this particular recurrence relation, >There is a slight generalization to the recursion >u(n+2) = (a + b u(n+1))/u(n) >which has the invariant > (ab + (a+b^2)(x + y) + b (x^2 + y^2) + x y^2 + x^2 y)/(x y) > But I wonder to what extent the generalizations can be pressed? > I think I have a pretty good handle on the situation now. The > description of what's possible and what's not is not all that > complicated, in the end, but I'd like to develop this slowly, > for easy reading. > Let's start with a simple, familiar recurrence relation > u(n+2) = u(n+1) + u(n) . > When u(0)=0, u(1)=1 we get the familiar Fibonacci sequence. > What happens if we follow GE's lead here? We get a sequence of > points ( u(n), u(n+1) ) in the plane. Do they lie in an > algebraic curve? > Well, those who know this sort of thing probably know this one > instantly but I didn't; I had to experiment a bit to discover that > when x,y are consecutive Fibonacci numbers, we have > x^2 + xy - y^2 = +- 1 > so that all the points in our sequence lie in the (reducible) variety > (x^2 + xy - y^2)^2 - 1 = 0 . > With just a bit of algebra we find generally that substituting > (x,y) = (y, x+y) simply changes the sign of x^2+xy-y^2 so that > F(x,y) = (x^2 + xy - y^2)^2 is an invariant for this recurrence relation. > As for the particular example which started this whole thread, > T(x,y) = ( y, (6-x)/y ), > I have to say I can't think of an easy way to put this into the > current framework: I don't know an easy way to compute the fixed > field even for a comparatively simple function like this. The > invariant F(x,y) = (-y^2+x*y^2+x^2*y+7*y-6+7*x-x^2) / (x*y) > which I found more or less by accident does indeed lead to (usually) > non-singular curves F = c of genus 1. Very interesting, although no prior exposure to this topic... could not follow much even after reading several times, but your expository y,z(x,y)) is given, by what procedures are the solution and invariants obtained ? How is invariant x^2 + x y -y^2 pulled out of T(x,y)=(y,y+x)? and ( xy(x+y) + (x^2+y^2)+ 7 (x+y)- 6)/(x y) out of T(x,y)=(y,(6-x)/y) ? Is it just judicious trial and error? I tried to verify and guess the procedure for the Fibonacci case f(x,y)=x+y by assuming continuous functions for x and y in t along a curve ,because (OP,GE) was asking in terms of a 3D curve, not 3D surface, that is ordering functions x and y. > This suggests that there exist some kind of invariant like : > F(u(n), u(n+1)) = constant > If you plot on a plane the points whose coordinates are (u(n), u(n+1)) > you get a curve (closed or not), not a surface. So I took the x(t)= Fibonacci(t), y(t)=x(t+1)=Fibonacci(t+1). z(x,y)= F(x(t),y(t)) = x^2+ xy -y^2 should have a singular solution + - 1. This I verified by graphing it on x-axis for t. I got a perfect - cos(pi*t) instead of +- 1 as a continuous function for this Fibonacci invariant x^2 + x y -y^2 , +1 and -1 included as extreme singular solution values at integral t arguments. Now this can be obtained by (guessing still) eliminating t between F and delF/del t. As another example, (x-t)^2+y^2 =R^2 may be an invariant to a certain sequence with [as unit circles on x-axis having a constant] y=+- R envelope/singular solution. If this is really so, do we have to solve some PDEs ? If so, how then are the PDEs set up in terms of partial derivatives p,q,x,y,t etc.? If this is clear, T(x,y)= (not y,f(x,y)) cases can be attempted for further understanding in my case. === Subject: facts of limits, also of derivatives? Hello mathematicians Wondering about this for some time. If derivatives are basically taking the limit, then would it be right to say that the properties of limit should apply as properties of differentiation. Also facts about derivatives should have their roots in facts of limits. e.g. the power rule, how does it look in the subject of limit? === Subject: Re: facts of limits, also of derivatives? >Hello mathematicians >Wondering about this for some time. >If derivatives are basically taking the limit, then would it be right to >say that the properties of limit should apply as properties of >differentiation. >Also facts about derivatives should have their roots in facts of limits. >e.g. >the power rule, how does it look in the subject of limit? No one said ALL properties of limits had meaningin derivatives! No more than all properties of rational numbers are valid for the reals. Some properties of derivatives are direct results of properties of limits as shown in any decent calculus book. === Subject: Re: facts of limits, also of derivatives? I forgot to add: If the limit exists. === Subject: Re: facts of limits, also of derivatives? > Hello mathematicians > Wondering about this for some time. > If derivatives are basically taking the limit, then would it be right to > say that the properties of limit should apply as properties of > differentiation. > Also facts about derivatives should have their roots in facts of limits. > e.g. > the power rule, how does it look in the subject of limit? The definition of the derivative is: for a function f : R -> R df/dx(a) = lim_{h -> 0} [(f(a+h) - f(a))/h] e.g. df/dx(a^n) = lim [(a+h)^n - a^n)/h] expand via the binomial theorem, and you should be able to show that this is na^{n-1} === Subject: Re: Replicating A Result In Cohen And Harcort > [ Transparent lie - deleted. ] -- Mostly economics: r c 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: Mathematics for the Precocious? I'm wondering what you mathematicians think is the explanation for a rather puzzling phenomenon: the hugely disproportionate representation of mathematics majors among young people who have gained their undergraduate degrees at record-breaking young ages. It appears that more than half of the youngest ten students to earn an undergraduate degree in each of the past few decades (in the United States, and apparently also in Britain and in east Asia) earned degrees in mathematics. That appears to be equally true of female or male early college graduates. Yet mathematics is far from the most commonly pursued undergraduate major, so why not more English majors or history majors or psychology majors among the precocious? Is there something about mathematics that makes it especially well suited to the young college entrant? Or is there something about the young college entrant that makes him or her especially well suited to majoring in mathematics? Do you know of any examples of early college graduates in mathematics from previous decades? Should more young people follow that path, if able? -- Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345 Learn in Freedom (TM) http://learninfreedom.org/ remove .de to email === Subject: Re: Mathematics for the Precocious? > .... > It appears that more than half of the youngest ten students to earn an > undergraduate degree in each of the past few decades (in the United > States, and apparently also in Britain and in east Asia) earned > degrees in mathematics. That appears to be equally true of female or > male early college graduates. Yet mathematics is far from the most > commonly pursued undergraduate major, so why not more English majors > or history majors or psychology majors among the precocious? .... It's because mathematics is so easy. I've heard that adult education courses in the humanities (such as your examples English, history and psychology) often get a better response from older students, say, 30 and above, who have enough experience of life to appreciate the insights of those subjects. On the other hand, the sciences are often easier for younger people. Those are not my own ideas, but the experience of an adult education specialist. My own addition is that even the sciences still require grappling with difficulties thrown up by nature; whereas mathematics is a creation of human thought in which the progression of basic ideas seems natural and coherent, and even the difficult problems are clearly stated. For a bright youngster with aptitude for a variety of different subjects, it's not surprising that mathematics may be the easiest in which to make fast progress. Ken Pledger. === Subject: Re: Mathematics for the Precocious? >I'm wondering what you mathematicians think is the explanation for a >rather puzzling phenomenon: the hugely disproportionate representation >of mathematics majors among young people who have gained their >undergraduate degrees at record-breaking young ages. >It appears that more than half of the youngest ten students to earn an >undergraduate degree in each of the past few decades (in the United >States, and apparently also in Britain and in east Asia) earned >degrees in mathematics. That appears to be equally true of female or >male early college graduates. Yet mathematics is far from the most >commonly pursued undergraduate major, so why not more English majors >or history majors or psychology majors among the precocious? Is there >something about mathematics that makes it especially well suited to >the young college entrant? Or is there something about the young >college entrant that makes him or her especially well suited to >majoring in mathematics? >Do you know of any examples of early college graduates in mathematics >from previous decades? Should more young people follow that path, if >able? One reason is that one can progress much more rapidly in mathematics, as it does not require that much memorization and lots of clumsy writing. The school set up by Erdos in Hungary produced lots of young mathematics researchers, mostly in number theory, combinatorics, and set theory, for which the research problems are such as not to require much background to attack. It would be no real problem to teach those who can get good college degrees in any of the physical sciences more than is now required in mathematics for a degree with recommendation to go on by age 10-12, without spending any more time on it than is now spent on stupidly memorizing procedures to calculate. -- 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: Re: Mathematics for the Precocious? > I'm wondering what you mathematicians think is the explanation for a > rather puzzling phenomenon: the hugely disproportionate representation > of mathematics majors among young people who have gained their > undergraduate degrees at record-breaking young ages. > It appears that more than half of the youngest ten students to earn an > undergraduate degree in each of the past few decades (in the United > States, and apparently also in Britain and in east Asia) earned degrees > in mathematics. That appears to be equally true of female or male early > college graduates. Yet mathematics is far from the most commonly pursued > undergraduate major, so why not more English majors or history majors or > psychology majors among the precocious? Is there something about > mathematics that makes it especially well suited to the young college > entrant? Or is there something about the young college entrant that > makes him or her especially well suited to majoring in mathematics? Since most of the methods for measuring if a student is precocious are based on analytical and linguistic skill/talent, this doesn't seem surprising. I suspect you will see this in many of the hard sciences, as opposed to social sciences and humanities. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Mathematics for the Precocious? >Do you know of any examples of early college graduates in mathematics >from previous decades? One example I know is Charles Fefferman, who graduated from U. of Maryland in 1966 at age 17 and was a full professor at U. of Chicago at 22. An earlier example is Norbert Wiener, who graduated from Tufts in 1909 at age 14. > Should more young people follow that path, if >able? It worked out pretty well for those two, at least... Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Mathematics for the Precocious? > I'm wondering what you mathematicians think is the explanation for a > rather puzzling phenomenon: the hugely disproportionate representation of > mathematics majors among young people who have gained their undergraduate > degrees at record-breaking young ages. > It appears that more than half of the youngest ten students to earn an > undergraduate degree in each of the past few decades (in the United > States, and apparently also in Britain and in east Asia) earned degrees in > mathematics. That appears to be equally true of female or male early > college graduates. Yet mathematics is far from the most commonly pursued > undergraduate major, so why not more English majors or history majors or > psychology majors among the precocious? Is there something about > mathematics that makes it especially well suited to the young college > entrant? Or is there something about the young college entrant that makes > him or her especially well suited to majoring in mathematics? > Do you know of any examples of early college graduates in mathematics from > previous decades? Should more young people follow that path, if able? > -- > Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345 > Learn in Freedom (TM) http://learninfreedom.org/ > remove .de to email One can do well in mathematics without being sophisticated about the rest of our world (politics, geography, sociology, psychology, ...). Hence, the younger than usual student has a relative advantage in mathematics compared to more integrated disciplines such as medicine, history and law/polical science. -Bill97 === Subject: Re: Mathematics for the Precocious? > I'm wondering what you mathematicians think is the explanation for a > rather puzzling phenomenon: the hugely disproportionate representation of > mathematics majors among young people who have gained their undergraduate > degrees at record-breaking young ages. > It appears that more than half of the youngest ten students to earn an > undergraduate degree in each of the past few decades (in the United > States, and apparently also in Britain and in east Asia) earned degrees in > mathematics. That appears to be equally true of female or male early > college graduates. Yet mathematics is far from the most commonly pursued > undergraduate major, so why not more English majors or history majors or > psychology majors among the precocious? Is there something about > mathematics that makes it especially well suited to the young college > entrant? Or is there something about the young college entrant that makes > him or her especially well suited to majoring in mathematics? > Do you know of any examples of early college graduates in mathematics from > previous decades? Should more young people follow that path, if able? > -- > Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345 > Learn in Freedom (TM) http://learninfreedom.org/ > remove .de to email > One can do well in mathematics without being sophisticated about the rest of > our world (politics, geography, sociology, psychology, ...). Hence, the > younger than usual student has a relative advantage in mathematics compared > to more integrated disciplines such as medicine, history and law/polical > science. Similarly with music. It is surely for this reason that the fields in which one finds child prodigies - maths, music, chess - that these fields require little or no knowledge of the world, but are built upon foundations of pure thought. Or something. -- Larry Lard Replies to group please Remember, the letters in Archimedes Plutonium can be re-arranged to spell The Delicious Rump Man I Pound His Male Rectum Penis Could Mature Him Old-Time U.S. Urine Champ and many other delightful phrases > I am an idiot. === Subject: Re: periodic function on R > Hello > Suppose f:R->R is continuous, periodic and non-constant on R. Then, is > it possible that g(x) = f(x^2) is periodic on R? Here is another, almost elementary approach. Let f(x) be continuous and periodic with period T>0. Suppose that M=maxf(x), m=minf(x). Let's say that f(x) makes an inversion in the segment [a,b], if f({a,b})={m,M} and m T. Suppose now that g(x)=f(x^2) is periodic with period p>0. Then f(x^2) = f(x^2+2px+p^2), x in R. Making a change of the variable we get f(x) = f(x+2p.sqrt(x)+p^2), x> =0. Set for convenience h(x) = 2p.sqrt(x)+p^2, so f(x) = f(x+h(x)), x> =0. Consider now the map x -> x+h(x). It is clearly strictly monotone. Take some sufficiently large segment [0,N], then its image is the segment [h(0),N+h(N)]. Hence the above formula gives that the inversion indices of f(x) over these two segments should coincide: Inv(f(x), [0,N]) = Inv(f(x), [h(0),N+h(N)]). But this is impossible, as h(N) ->infinity so the difference between the lengths of these segments tends to infinity as well and the second index should increase by k.Inv(f(x),[0,T]), k .9a arbitrarily large. It is clear that this method may be extended to more general situations. Simeon === Subject: Re: Equivalent sequences >There is a slight generalization to the recursion >u(n+2) = (a + b u(n+1))/u(n) >which has the invariant > (ab + (a+b^2)(x + y) + b (x^2 + y^2) + x y^2 + x^2 y)/(x y) ... and still more generally, the recursion u(n+2) = (a + b u(n+1) + c u(n+1)^2)/(u(n)(c + d u(n+1) + e u(n+1)^2)) has the invariant (a + b(x + y) + c (x^2 + y^2) + d (x^2 y + x y^2) + e x^2 y^2)/(x y) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: the proof.... Proof that .99999999999... = 1 Let x = .9999999999... Then 10x = 9.99999999... subtract x from both sides of the equation... 9x = 9 x = 1 Thus, .9999999... = 1 *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: group theory fraud > I think its more likely that the orignal poster has no idea what he is > talking about. That may be so, but it is also irrelevant. The question one needs to ask is this: is there any possible interpretation of what the OP the following occurs to me: just as one can consider random Fourier series and random matrices, there is no reason why one can't consider random group laws on a set. One just needs to define a measure on the set of all group laws on the set or on a suitable subset of the set of all group laws. For a finite set, that is easy, in principle. For an infinite set, one could just use counting measure or one might be able to cook up some other ways. If X is a suitable topological space, maybe there is a nice topology on the set of group laws on X or on the set of group laws for which X is a topological group. Maybe it is too much to ask for an invariant measure on that set of group laws, but there might be something like stochastic measures on it. I'm not saying one can actually do it or that I have any special insight into how one might go about it, only that it is not obvious to me that something like that is not possible. Just for definiteness, what can one say about the set of all group laws on the Cantor set for which it becomes a topological group topologically isomorphic to the group of 2-adic integers? It seems that the set of all such group laws is a homogeneous space for the group G of homeomorphisms of the Cantor set onto itself, namely the space of cosets in G of the group of 2-adic units. Is there any kind of nice measure on the group G? If not, one can impose more conditions on the homeomorphisms to make it so. For example, consider all homeomorphisms which can be expressed by a convergent 2-adic power series a_1 z+ a_2 z^2 + ... Maybe if one uses the coefficients a_1,a_2,... of the power series (and maybe also the coefficients of the power series representing the inverse homeomorphism) as coordinates, one gets a nice compact topology from the cartesian product of countably many copies of the space of 2-adic integers. If so, then there is a Haar measure and one can speak of random group laws of this type. One can improvise indefinitely along these lines and run into lots of interesting possibilities. So what if the OP didn't know what he was talking about? -- Allan Adler * Disclaimer: I am a guest and *not* a member of the MIT CSAIL. My actions and * comments do not reflect in any way on MIT. Also, I am nowhere near Boston. === Subject: Re: group theory fraud > That may be so, but it is also irrelevant. The question one needs to > ask is this: is there any possible interpretation of what the OP > the following occurs to me: just as one can consider random Fourier > series and random matrices, there is no reason why one can't consider > random group laws on a set. One just needs to define a measure on the > set of all group laws on the set or on a suitable subset of the set > of all group laws. For a finite set, that is easy, in principle. > For an infinite set, one could just use counting measure or one > might be able to cook up some other ways. One possibility is a random quotient of the free group on n elements, for some value of random. We might take the set of all words, and assign a probability in such a way to each word so that with probability 1 we end up with a finite set of words, which we then take to be relations. Of course we pretty often are likely to end up with the uninteresting result that the group is trivial. === Subject: Re: convolving noise with noise will get Gaussian? what does that imply? > [...] > The CLT as I know it, applies to PDFs, not random variables. > > Do you mean that the convolution it speaks of is of the PDFs, not > the random variables themselves? > > That's exactly what I have been saying during this whole thread! > No, that's not exactly what you've been saying, and that is part > of my issue with you, Rune. You said, exactly, > The CLT as I know it, applies to PDFs, not random variables. > This statement does not mention convolution. C'm on, Randy. If you read the whole thread (including your own posts), you will find that the OP convolved the data in the first place. Secondly, when you add independent random variables, the distribution of the result is the convolution of the input variables' PDFs. in the post of November 4th (your first post in this thread). We agree in the basic properties of the CLT; why do you make such a fuzz about disagreeing with me now? > However, to > say that the CLT does not apply to random variables is pretty > much false in my book - it's all about random variables. > > OK, if we have to go nit-picking, > If I'm nit-picking, then so is Papoulis. His section on the CLT > begins like this: > Given n independent *RVs* x_i, we form their sum > x = x_1 + ... + x_n > This is an *RV* with mean ... and variance ... . ... Furthermore, if > the *RVs* x_i are of continuous type, ... the density f(x) of x > approaches a normal density ... . This important theorem ... . > [emphases mine]. He CLEARLY associates the CLT with RVs. I have never contested that. But if you want the CLT to work and produce Gaussian distributions, you need to work on the PDFs. > Now it is true that he also goes on to say The CLT can be expressed > as a property of convolutions ..., but it seems pretty clear that the > main interpretation and utility of the CLT is in association with RVs. > To divorce it from RVs and speak only of convolving positive > functions, while theoretically accurate, Make up your mind. Do you agree in tht what is convolved to produce results according to the CLT are PDFs, or do you not agree? > robs it of its real value: > explaining why randomness in nature is often Gaussian. No. The CLT is an ad hoc excuse for the analyst to stay with the nice and easily tractable Gaussian distributions instead of diving into the more tricky ones. The CLT does not make a non-Gaussian process Gaussian, it only provides some comfort in stating that one does not make a very big mistake if one chooses to work under the Gaussian hypothesis. > here's my 2c: A random process > generates random variables (or random data), RVs, that in some > way are characterized by a Probablility Density Function, PDF. > In that sense, the RV and the PDF are interconnected in that both > are associated with a random process. > Wow. Now that's rich, Rune. After two courses in Random Processes > and another two in basic probability theory, I've never heard anyone > condition the association of a RV and its PDF on an association > with a random process. I don't know where you've come up with that > idea, but it is completely unorthodox in my experience. Is a random process unorthodox to you? (OK, I should perhaps used the term stochastic process, but I didn't want to go pedantic on you...) Hey, Randy, this is a joke, right? > The CLT operator > Huh? Since when was anyone talking about a CLT operator? You've just > now introduced new language. The topic of discussion thus far has been > about a theorem, the Central Limit Theorem, NOT an operator! I'm not introducing new language. If you take a course on linear systems in maths, you'll find the term operator used all over the place. Particularly in the context of convolution integrals. If you express the CLT as a property of the expression y_CLT = y_1 (*) y_2 (*) ... (*) y_N where (*) means convolution and y_n are PDFs, the term CLT operator makes perfect sense. > takes multiple PDFs as input and produces one > PDF as output. When I look at the inner workings of the CLT, I see > PDFs, not RVs. I could have agreed with you if you said it's all > about random _processes_. You didn't. > No, I certainly did not, because the CLT (reverting to the terminology > that we've been using) at least as presented by Papoulis, is not about > a random process. It has NOTHING to do with random processes. Well, you may disagree with my approach to these matters and the exact way I interpret the problem and phrase my opionions. You should be very careful about how you state your objections, though. You might find yourself in a position you can not defend. > My point is that mentioning the CLT only makes sense when studying > PDFs. > I heartily disagree, for the reasons I've already explained above. Please, Randy, I know you don't mean this. Yes, the effects of adding several random variables is the reason why the CLT is interesting. Arguing *why* the CLT works, and *how*, requires the studying stochastic processes and the convolution of their PDFs. Not the random variables. For the simple reason that given a random vector, you don't know anything about its PDF. You can make up an opinion, based on a histogram, but you don't know. The concept of a PDF only makes sense in the context of a stochastic process. > The OP tried to link the CLT directly to the random variable. > As well he should. The only problem is, he apparently did so > improperly (i.e., via convolution of the RVs rather than the > sum of the RVs). The OP used random data (a single realization of a random variable) where a PDF should have been used. The exact nature of the PDF was never specified (not enven an estimate through a histogram), and no histogram of the resulting data were used. The important difference between a stochastic process generating random variables, and the random data as a realization of sucha random variable, was never grasped. The question was phrased in a way that disagreed just enough with standard terminology to cause confusion (denoting the random variable by the symbol f, which usually is reserved for PDFs). Apart from that, the OP did an excellent job in verifying the CLT. Rune === Subject: Analysis questions, please help... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iA8861u14673; Good day fellas, can someone help me out here? My questions are: oo 1. Let (a_n) be a real sequence that converges to a number L>0. n=1 Show that there exists a natural number N such that a_n > 0 whenever n >= N. 2. Suppose that f: R -> R is differentiable at c. Show that there are constants M in R and a > 0 such that |f(x) - f(c)| =< M |x - c| whenever |x - c|< a. [Thoughts] 1. Now the sequence given in the question converges to L >0, therefore, there exists a natural number N' such that 0 < |a_n - L| < epsilon whenever n >= N'. Now, 0 < |a_n - L|< epsilon <=> L - epsilon < a_n < L + epsilon How do I choose the N then such that L - epsilon > 0 so that I can derive a_n > 0 ? Or is there a another approach to this question? 2. How do I apply the Mean Value THeorem to this question, because this is the only approach I can think of? === Subject: Re: Analysis questions, please help... > Good day fellas, can someone help me out here? > My questions are: > oo > 1. Let (a_n) be a real sequence that converges to a number L>0. > n=1 > Show that there exists a natural number N such that a_n > 0 > whenever n >= N. > 2. Suppose that f: R -> R is differentiable at c. Show that there > are constants M in R and a > 0 such that |f(x) - f(c)| =< M |x - c| > whenever |x - c|< a. > [Thoughts] > 1. Now the sequence given in the question converges to L >0, > therefore, there exists a natural number N' such that 0 < |a_n - L| >< epsilon whenever n >= N'. You left out a very important part of the definition, namely that *for* *every* *epsilon>0* there exists a natural number N' such that.... As your statement stands, we don't know what epsilon is supposed to be. The whole point is that you get to choose any epsilon>0 that you happen to like. > Now, 0 < |a_n - L|< epsilon > <=> L - epsilon < a_n < L + epsilon > How do I choose the N then such that L - epsilon > 0 so that I can > derive a_n > 0 ? Or is there a another approach to this question? You don't choose an N. You choose an epsilon>0 so that L-epsilon > 0. > 2. How do I apply the Mean Value THeorem to this question, because > this is the only approach I can think of? The mean value theorem does not apply to this question. If you don't see why, you need to reread the hypotheses of the MVT and then look at what you are given. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Analysis questions, please help... > Good day fellas, can someone help me out here? > My questions are: > oo > 1. Let (a_n) be a real sequence that converges to a number L>0. > n=1 > Show that there exists a natural number N such that a_n > 0 > whenever n >= N. In other words, show that if the limit is positive, eventually the sequence is all positive. Go back to the definition of convergence. This should be almost trivial (in terms of Cauchy sequences, it's a one liner). > 2. Suppose that f: R -> R is differentiable at c. Show that there > are constants M in R and a > 0 such that |f(x) - f(c)| =< M |x - c| > whenever |x - c|< a. This is called Lipschitz continuous. It's stronger than continuous. Very important concept. > [Thoughts] > 1. Now the sequence given in the question converges to L >0, > therefore, there exists a natural number N' such that 0 < |a_n - L| > < epsilon whenever n >= N'. > Now, 0 < |a_n - L|< epsilon > <=> L - epsilon < a_n < L + epsilon Right. So long as epsilon How do I choose the N then such that L - epsilon > 0 The definition of convergence means that you choose the epsilon, and there exists an N for that epsilon. So choose epsilon < L, say epsilon = L/2 and you're done. You know N exists, you don't have to know what it is. > 2. How do I apply the Mean Value THeorem to this question, because > this is the only approach I can think of? Being differentiable at c means that (f(x)-f(c))/(x-c) converges to some value, call it g. See if you can go from convergence of the sequence of quotient values to the Lipschitz condition. - Randy === Subject: random vectors with a given covariance matrix. How can I generate n-dimensional random vectors with a given covariance matrix C (n x n) ? === Subject: Re: random vectors with a given covariance matrix. >How can I generate n-dimensional random vectors with a given >covariance matrix C (n x n) ? One way is to generate n independent (pseudo)random variables with variance 1, and transform: Y = C^(1/2) X. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Solution to x^pi - 1 = 0? I've been thinking about this problem for awhile now and I thought I had a good way to tackle it. I figured I'd start by factorizing it like you would for a polonomial, by dividing through by the first apparent root. This gives me something that looks like this, x^pi-1=(x-1)(Sum(N=1..Inf)(x^(pi-N)-x^N)) I know that Sum(N=1..Inf)(x^N) is just 1/(x-1) so I feel like I'm going in circles but I can't help thinking this is actually doing something. Does anyone have any thoughts? Dave === Subject: Re: Solution to x^pi - 1 = 0? >I've been thinking about this problem for awhile now and I thought I >had a good way to tackle it. I figured I'd start by factorizing it >like you would for a polonomial, I'd think a logarithmic solution would be easier. Obviously x<>0, so write x^pi = 1 pi*ln(x) = ln(1) pi*ln(x) = 0 ln(x) = 0 x = 1 Apologies if you're looking for something a lot more subtle, but in terms of real numbers the only solution is x = 1. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if you're afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: Solution to x^pi - 1 = 0? > I've been thinking about this problem for awhile now and I thought I > had a good way to tackle it. I figured I'd start by factorizing it > like you would for a polonomial, by dividing through by the first > apparent root. This gives me something that looks like this, > x^pi-1=(x-1)(Sum(N=1..Inf)(x^(pi-N)-x^N)) > I know that Sum(N=1..Inf)(x^N) is just 1/(x-1) so I feel like I'm > going in circles but I can't help thinking this is actually doing > something. Does anyone have any thoughts? > Dave You could try putting x = r.e^(i.a) then observing that x^pi = 1 = e^(2.i.n.pi) === Subject: How to translate Kontinuumssouce? Hermann Weyl (1885-1955), Hilbert's successor, coined the German expression Kontinuumssouce. He warned: We are less certain than ever about the ultimate foundations of mathematics. Who can translate the mentioned expression in English? Who can translate Weyl's pertaining rejection of large parts of set theory in hints to accepted newer publications? Is almagamization of infinity (into the electron mass by Feynman) a mathematical term? Eckard === Subject: Re: How to translate Kontinuumssouce? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Hermann Weyl (1885-1955), Hilbert's successor, coined the German > expression Kontinuumssouce. > He warned: We are less certain than ever about the ultimate foundations > of mathematics. > Who can translate the mentioned expression in English? First render the German spelling correctly, then we can think about translating it. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: How to translate Kontinuumssouce? >> expression Kontinuumssouce. > First render the German spelling correctly, then we can think about > translating it. Sorry for my typo. Sauce is originally a French word, perhaps related to the Latin word saucaptis denoting a spice. It has the same spelling in German and English, too. Russians write sous. Just Spaniards could mistake sauce like a willow. Kontinuum is also based on Latin origin. The first letter K instead of C is due to a reform of German spelling more than one hundred years ago. In all, it would not be very difficult to literally translate into something like sauce-like continuum. However, when Hermann Weyl coined this word in a bantering manner, he was respected enough as to express his opinion this way. I wonder if someone else also used such pronounced utterance revealing his displeasure of things like the zero set. There were very famous mathematicians who did not support at least some basics of Cantor's set theory: Gauss: protestiere gegen den Gebrauch einer unendlichen Gr.9a¤e als einer vollendeten, welche in der Mathematik niemals erlaubt ist. Leopold Kronecker blamed Cantor for his pernicious influence on young people. Henri Poincar.8e: Future Generations will consider set theory a disease. Emil.8e Borel did not accept the axiom of choice. Hausdorff, Banach, and also Tarski found pertaining paradoxes. Abraham Robinsohn: Any mention or purported mention, of infinite totalities is, literally, meaningless. Luitzen Brouwer disapproved set theory. Even Weierstrass, Cantor's teacher, clarified: Es ist unm.9aglich zwei unendlich gro¤e Zahlen a und b zu unterscheiden. === Subject: Re: How to translate Kontinuumssouce? > expression Kontinuumssouce. >> First render the German spelling correctly, then we can think about >> translating it. >Sorry for my typo. Sauce is originally a French word, perhaps related to >the Latin word saucaptis denoting a spice. It has the same spelling in >German and English, too. Russians write sous. Just Spaniards could >mistake sauce like a willow. >Kontinuum is also based on Latin origin. The first letter K instead of >C is due to a reform of German spelling more than one hundred years ago. >In all, it would not be very difficult to literally translate into >something like sauce-like continuum. However, when Hermann Weyl coined >this word in a bantering manner, he was respected enough as to express >his opinion this way. I wonder if someone else also used such pronounced > utterance revealing his displeasure of things like the zero set. >There were very famous mathematicians who did not support at least some >basics of Cantor's set theory: >Gauss: protestiere gegen den Gebrauch einer unendlichen Gre als einer >vollendeten, welche in der Mathematik niemals erlaubt ist. Did Gauss even know of Cantor? >Leopold Kronecker blamed Cantor for his pernicious influence on young >people. Kronecker is supposed to have stated that God created the integers; man created all else. However, it seems that there is more published in analysis than in the rest of mathematics, that rational numbers go back thousands of years, and that geometry and problems relating to the continuum more that 3000. How much of mathematics does not use more than integers? >Henri Poincar: Future Generations will consider set theory a disease. Alas, too many do. Even when set theory is not directly useful, it can give real insights, which are lost otherwise. >Emil Borel did not accept the axiom of choice. >Hausdorff, Banach, and also Tarski found pertaining paradoxes. It seems we are stuck with paradoxes, and have been since at least the time of Aristotle, who could not understand how a point could be an element of a line and divide that same line. We cannot get rid of the paradoxes, and we cannot even know if we have a consistent system. Anything large enough to get the integers, including addition, multiplication, and induction, already gets the paradoxes; sets are not needed for them. Borel and Lebesgue did not accept the axiom of choice. Hausdoff and Banach had no qualms about using it. Tarski was one of the major contributors to understanding it, as well as much else of set theory. None of these considered set theory as inappropriate of not of considerable use. >Abraham Robinsohn: Any mention or purported mention, of infinite >totalities is, literally, meaningless. ???? >Luitzen Brouwer disapproved set theory. I do not see that he disproved anything. He came up with alternate axioms, and intuitionistic logic and set theory have all of the Godel paradoxes. One cannot do ordinary probability in Brouwer's models. >Even Weierstrass, Cantor's teacher, clarified: Es ist unmglich zwei >unendlich groe Zahlen a und b zu unterscheiden. As Cantor was the first to show that this was not so, I am not surprised. Cantor opened a Pandora's box of problems, not all resolved yet. Remember that set algebra only goes back to the middle of the 19th century, although it would have been easily understood by Euclid's students, as would algebra, which they also did not know. -- This address is for information only. 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The candidate should be a team player with 0-3 years experience and experience in the likes of R, S, Matlab, C++, Kdb,... * Top Scientist, Engineers and More Wanted for Entry Positions: SciEssence, LLC, Telecommute. We are looking for Interns (Recent Graduates and Current Undergraduate and Graduate Students, and Post-Doctorals) to show off their scientific and engineering skills and knowledge, as well as their ability to work in a 21st Century Scientific & Engineering Firm while... === Subject: Re: Flaw in a Proof of Tychonoff's Theorem? >Here's my simple counterexample, in the product I x I. Let the family >F be: > { [0, 1/4] x [3/4, 1] u [3/4, 1] x [0, 1/4], > [0, 1/4] x [0, 1/4] u [3/4, 1] x [3/4, 1], > [0, 1] x [0, 1] } F does not have the F.I.P: the first two sets do not intersect. KP === Subject: Weird Bound I came across this bound in a paper without explanation: int_infty^infty 1/(1+|u^2-x^2|) du = o(1/(1+|x|)). where int is integration and infty is infinity. Can someone shed some Kira. === Subject: Re: Weird Bound >I came across this bound in a paper without explanation: > int_infty^infty 1/(1+|u^2-x^2|) du = o(1/(1+|x|)). >where int is integration and infty is infinity. Presumably it was actually int_{-infty}^infty 1/(1+|u^2-x^2|) du = o(1/(1+|x|)) or equivalently just int_0^infty 1/(1+|u^2-x^2|) du = o(1/(1+|x|)). >Can someone shed some >lights on how this bound is estimated please? The bound you state is not true: If x > 0 is large then the integrand is >= c > 0 for x - 1/x < u < x + 1/x, hence the integral is >= c/x. What's maybe true is that int_{-infty}^infty 1/(1+|u^2-x^2|) du = O(1/(1+|x|)), which, since the integral is finite (and bounded by _something_ for x in any bounded interval) is the same as int_{-infty}^infty 1/(1+|u^2-x^2|) du = O(x) for large positive x. A person can probably figure this out by noting what part of the demominator is important where. So for example int_{x-2/x}^{x+2/x} 1/(1+|u^2-x^2|) du <= int_{x-2/x}^{x+2/x} du = c/x, while (*) int_{x+2/x}^infty 1/(1+|u^2-x^2|) du <= int_{x+2/x}^infty 1/(u^2-x^2) du, which you can simply calculate. When I do that two ways I get something asymptotic to c log(x)/x both times. But it looks like (*) is only off by a constant factor, so I conclude that the integral is not O(1/x) either, what it really is is O(log(x)/x) for large x. The author lied. (Or you typed something wrong, or I lied here. But the o(1/x) is _certainly_ wrong...) >Kira. ************************ David C. Ullrich === Subject: Re: Weird Bound >>I came across this bound in a paper without explanation: >> int_infty^infty 1/(1+|u^2-x^2|) du = o(1/(1+|x|)). ... >a constant factor, so I conclude that the integral >is not O(1/x) either, what it really is is >O(log(x)/x) for large x. Explicitly, for x > 1, if p = sqrt(x^2-1) int_x^infty 1/(1+u^2-x^2) du = int_x^infty 1/(u^2-p^2) du = ln((p+x)/(x-p))/(2 p) = ln(2x)/x + O(ln(x)/x^3) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Bound on conformal map >If I have an analytic map of {z | |z| < 1} into itself, and f has a zero of >order 2 at z = 1/2, How can I obtain an upper bound on f(3/4)? What >principle(s) could I use? You could use the Schwarz Lemma to get a non-optimal upper bound, and you could use the _proof_ of the Schwarz Lemma to get the best possible bound. (There's Schwarz Lemmas and then there's Schwarz Lemmas. If the only one you know talks about functions that vanish at the origin then you need to apply it to the composition of f with a certain automorphism of the disk.) >Isaac ************************ David C. Ullrich === Subject: Re: Where can I find a list of non comutatives operations ? > I would need to established a reduced base of non commutative > operations. The operations should be grouped in 'rings' ( not > mathematical rings) in which no couple of operations shall comutate; > operations from different rings may commute, or even be identical. It seems to me that on the set of n-bit values, n>1, the operations OPj defined as x OPj y = (x xor j) + y mod 2^n are not commutative or associative, assuming 0 < j < 2^(n-1). I don't know your exact definition of operations that comutate, maybe picking distinct OPj defined as above would fit. Fran.8dois Grieu === Subject: Re: Where can I find a list of non comutatives operations ? -----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 |>I would need to established a reduced base of non commutative |>operations. The operations should be grouped in 'rings' ( not |>mathematical rings) in which no couple of operations shall comutate; |>operations from different rings may commute, or even be identical. | | | It seems to me that on the set of n-bit values, n>1, the | operations OPj defined as | x OPj y = (x xor j) + y mod 2^n | are not commutative or associative, assuming 0 < j < 2^(n-1). | I don't know your exact definition of operations that comutate, | maybe picking distinct OPj defined as above would fit. | | | Fran.8dois Grieu Tu as un nom bien francais mon ami :) about OPj : I did know that operation; the question is : is it possible to implement ( when coming about writing the C code ) OPj in various way so that one do not look like an other one ? if no => OPj is then one non comutative operations amongst the 10 000 I need. if yes => how different can the implementations be ? how many implementations can I write ? How much does it cost in time to reverse the operation ? ( to recover X given Y or the reverse ) The matter is not to find an operation which do not comutate with itself, but operations that do not comutate with other ones. What I need is a list of 1000 groups of at least 3 operations per group such as operations inside a group do not commutate. BUT operations inside a group SHALL NOT look familiar one to each other. ( I am sorry I forgot this point in my first message ) The 1000 groups shall not be redundant, ie, the elements of one group those 1000 groups of more than 3 operations each, I will deduce at least 10 000 sub groups of exactly 3 operations each, and all those sub groups will be elements of the 1000 mother groups. Does it sound more clear ? - -- DEMAINE Beno.94t-Pierre http:/www.demaine.info/ _o< apt-get remove ispell >o_/ There're 10 types of people: those who can count in binary and those who can't -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.2.5 (GNU/Linux) Comment: Using GnuPG with Thunderbird - http://enigmail.mozdev.org iD8DBQFBj2oRGWSTLbOSw8IRAhOxAKCyi13ZvNOfkTwbxXI6ywxrx+GrTwCfQlUM 0SoxoOrSMDp1jESx8k2+jm8= =Yvr1 -----END PGP SIGNATURE----- === Subject: Re: Different size infinities? >>Gauss, being a good physicist too, would have agreed with me anyway :-) >Sure enough, the definition of great thinker is as we suspected. >>I do not consider myself as a great thinker. That man doesn't understand when you make a joke. Not even when a :-) is included quite explicitly. I have said before that Mr. Ullrich is suffering from serious communication problems. > Hilarious. Of course you don't want to say out loud that you do > consider yourself a great thinker, because you realize how that > would sound. But it's evident that your _definition_ of great > thinker _is_ someone who agrees with me - that's the only > basis you have for your conclusion that those people you cite > from centuries ago were great thinkers, while there are no > great thinkers among living mathematicians (since your only > evidence for this is that you can't name one who agrees > with you). You are continuously distorting my arguments. In this case, you are mixing up cause and effect. How can someone who is dead and buried possibly agree with me? It's the reverse, of course. Sometimes I find that my thoughts stem from those of a great thinker. (And that certainly helps to find L.E.J. Brouwer a great thinker.) But it is not a necesary condition. I disagree with Aristotle on many issues; yet I find Aristotle a great thinker. >>I'm a humble programmer > Again, guffaw. You know that you're right about all this and > _every_ contemporary mathematician has it all wrong. But you > call yourself humble? As a programmer, I am humble. I am not So humble, though, because my which is titled The Humble Programmer and that piece of text is far from humble: http://www.cs.utexas.edu/users/EWD/ewd03xx/EWD340.PDF But I suppose so much depth (and pun) within common words must be too much of the good, for an illiterate person like you. [ .. rest deleted .. ] Han de Bruijn === Subject: re:Prime Quickie Not necessarily so. If you take n = 8 and the numbers 14, 15, ..., 21 then 7 which is between 4 and 8 devides two of these numbers. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Prime Quickie >by me. >Take a set of consecutive positive integers, say 3 4 5 6. >Call 5 an exclusive prime - it divides only one of the >numbers and none else. >?: All s.o.c.p.i. have at least one exclusive prime. >Attempted handwave solution (SPOILER)f Why the need for handwaving? Let the set have size n. If n is prime, we're done. Assume n is not prime (hence n => 4). Then your statement is true if there is a prime between n/2 and n. Exactly one number in your set will have that prime as a factor. -- I'm not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: re:psychology of geometric reasoning I think that to get a better intuition it's best to convince yourself using logic and observasions rather than calculations. I suspect it would also be closer to the way our mental intuition works. For example let O be the origin and A, B, C the adjacent vertices (1, 0, 0), (0, 1, 0), (0, 0, 1). By symmetry it is apparent that ABC is an equilateral triangle. Thus there are two rotations by 120 and 240 degrees that permute the vertices A, B, C. To see that these rotations leave O and thus the whole cube in place you only have to note that O is equidistant from A, B, C and thus must lie on the rotation axis which is perpendicular to the triangle ABC and goes through its center. I think when we use geometric intution, our brain analyzes data about symmetry and known geometric facts rather than make arithmetic calculations which I'm not sure it is capable of doing subconciously, although I'm not expert in the area. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: psychology of geometric reasoning > Rotating a cube on its diameter brings it into coincidence with itself > in three different ways. Things like this don't seem as obvious to me > as they apparently should. Consider the following plausibility > argument [Calculation cut.] > .... I'm interested in knowing anyone's introspections about the > rightness/wrongness of the way I convinced myself. I'm too lazy to check your calculation, but to make the result more obvious to yourself hold a real physical cube (say, a die) in your hands and turn it around. Fit it into the bottom corner of a room where two walls and the floor meet at right angles. Turn it about the appropriate diagonal and fit it there again. _See_ and _feel_ what the three-fold rotation is all about. Don't just try to imagine it, and worst of all don't try to handle it with two-dimensional pictures. Ken Pledger. === Subject: Re: Raging Storm Clouds on Mars > MARS ODYSSEY THEMIS IMAGES > All of the THEMIS images are archived here: > http://themis.la.asu.edu/latest.html === Subject: Re: Calculator wrong Ok , Lets say I have one apples and 3 pears to buy. The pears are two times more expensive as the apples. one apple does cost 1 euro So I pay 1 + 3 * 2 = 8 euro. Or 1*1+3*2 = 8 euro. ------------- A multiplicatin can be rewritten. 1+3*2 = 1+ (2+2+2) = 7 euro. Following the convention of RPN this proves RPN does not calculate correctly. At least this is my opinion. -------------- Rob. > I do not know about the school you went to, but at my school, reading > mathematical expressions from left to right was taught SIMULTANEOULY > with > the adoption of the classical order of operations. My grade school > teacher > explained it clearly why a convention is needed: to cut down the > parenthetical clutter. > Left-to-right, the expression tree <*, <+, 1, 2>, 3>, becomes 1+2*3 > With precedence, it becomes (1+2)*3 > One has more brackets than the other, the one you said has less clutter. > Care to explain that? You must genuinely believe it, as you mentioned it > twice. > As I'm sure you are well aware of this, any time there is an occurrence of > two INFIX binary operators (say, + and *) in the same expression, the > following question must be asked and resolved: Which operation should be > done first? To illustrate, consider the following lone expression under > infix interpretation of the operator symbols: (2+3*4) (where we will > agree, for the sake of this discussion, to drop the outermost pair of > parentheses). In the absence of any prior convention (no left-to-right > convention, no ranking of operators in one way or another) except infix > interpretation of binary operators, the following ambiguities in semantics > must be resolved: > (1) Read the expression 2+3*4 as (2+3)*4. > (2) Read the expression 2+3*4 as 2+(3*4). > (3) Read the expression in some other ways. > One may easily rule out (3) because we are under the assumption that + and > * are being read as infix operators. That leaves (1) and (2) to be > resolved. In the absence of any (explicit or implicit) convention, the > ambiguity can easily be resolved by inserting parentheses into the intended > places (recall: reading left-to-right *IS* already a convention by your > own standards). So, take your pick: (1) or (2). > The classical convention is to rank * higher than + so that one will > agree to drop the pair of parentheses when the intended meaning is > 2+(3*4). Now, of course, the opposite convention would achieve the goal > of eliminating one pair of parentheses equally well, and would work > perfectly fine if everybody agrees to use it, but a choice HAS BEEN made a > long time ago, and IMO, it is wasteful to force the masses (most of whom > couldn't care less about syntax trees) to confront this issue over and over, > and, even worse, it is highly counter-productive to encourage youngsters to > adopt the opposite convention. > That and the use of the word 'universal' in ALL CAPS several times. > Open your eyes and see that this notation is not universal, for pity's > sake. > Type 1 + 2 * 3 into Microsoft's calc.exe in standard mode. > There are 10^8 computers which demonstrate beyond any shadow of a doubt > that > the convention you claim is universal is anything but that. > Any human convention will only be in place if it is enforced or at least > encouraged by customs. If there are more and more rebels without a cause > like yourself to impress young children that adopting the opposite > convention is an intellectually superior and productive (spelled cool) > thing to do, the classical convention may become less and less universal > than any grade school mathematics teacher would have liked it to be. > Shedar === Subject: Re: Calculator wrong > Lets say I have one apples and 3 pears to buy. > The pears are two times more expensive as the apples. > one apple does cost 1 euro > So I pay 1 + 3 * 2 = 8 euro. Since when? > A multiplicatin can be rewritten. > 1+3*2 = 1+ (2+2+2) = 7 euro. > Following the convention of RPN this proves RPN does not calculate > correctly. Eh? -- 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: Calculator wrong > > Lets say I have one apples and 3 pears to buy. > > The pears are two times more expensive as the apples. > > one apple does cost 1 euro > > > > So I pay 1 + 3 * 2 = 8 euro. > Since when? Since HP invented RPN alogrithme I think. :-) > > A multiplicatin can be rewritten. > > > > 1+3*2 = 1+ (2+2+2) = 7 euro. > > Following the convention of RPN this proves RPN does not calculate > > correctly. > Eh? > -- > 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: Calculator wrong ----- Original Message ----- === Subject: Re: Calculator wrong > > I have here a casio FC100 financial calculator. > > > > If I type in the keys. > > 1+1/3= then the answer is 0.66666666 > > This is mathematical incorrect division should go first. > > Yup, on calculators that make that distinction. > > > The same with hpII10B (finanacial calculator). > > > > My casio fx4500p(for statistics) and my TI-36x solar does it correct. > > casio fx80 and fx81 also correct. > > They make that distinction indeed. > > But what convention is used depends on the calculator. *If* a calculator > has '(' and ')' keys it is likely that it will also know about the order > of evaluation. If it does not have those keys, it is likely that it does > *not* know about the order of evaluation. On calculators with those > keys you can distinguish (1 + 1)/3 from 1 + 1/3, on calculators without > them you can not (and the 1's involved can be expressions again). > That is funny I just bought a Casio fc100 (also financial calculator) > It uses RPN and doesn't have hooks !! > Meaning you can't calculate everything on it. > Consider this formula: > (1-(1+0.11)^(2/3))/(1+0.11-1) > The problem is the power of 2/3 > You first must calculate this and than the (1+0.11) > In this order you can't put the date in the calculator (without hooks) > The RPN sequence > 1 1 0.11 + 2 3 / ^ - 1 0.11 + 1 - / > works on my calculator. What's wrong with yours? answer should be -0.655006 What you type doesn't work on mine. Think you forgot some opperators. Rob === Subject: Re: Calculator wrong ... > But what convention is used depends on the calculator. *If* a > calculator has '(' and ')' keys it is likely that it will also > know about the order > of evaluation. If it does not have those > keys, it is likely that it does *not* know about the order of > evaluation. On calculators with those keys you can distinguish > (1 + 1)/3 from 1 + 1/3, on calculators without them you can not > (and the 1's involved can be expressions again). > > That is funny I just bought a Casio fc100 (also financial calculator) > It uses RPN and doesn't have hooks !! It does not have an enter button, so it does not do RPN. > Meaning you can't calculate everything on it. As I said, if you have a calculator that uses infix notation and does not have parenthesis, you can not calculate everything on it (and it will likely also not know about operator precedence. > Consider this formula: > > (1-(1+0.11)^(2/3))/(1+0.11-1) > The problem is the power of 2/3 > You first must calculate this and than the (1+0.11) > In this order you can't put the date in the calculator (without hooks) > > The RPN sequence > 1 1 0.11 + 2 3 / ^ - 1 0.11 + 1 - / > works on my calculator. What's wrong with yours? In proper RPN button notation, you should also show the use of the enter button (marked as E here): 1 E 1 E 0.11 + 2 E 3 E / ^ - 1 E 0.11 + 1 - / > What you type doesn't work on mine. > Think you forgot some opperators. Nope, your calculator is not RPN. -- 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: Calculator wrong >If I type in the keys. >1+1/3= then the answer is 0.66666666 >This is mathematical incorrect division should go first. The calculator doesn't know what you *intend*. Learn how it works. You enter 1+1/3 and you get (1+1)/3=2/3=0.66666... You enter 1/3+1 and you get 1.333, which is you want. The calculator isn't wrong, and neither are you. Tools do what they do. It's up to you to use them to acheive your goal. The calculator doesn't know what you *want* it to do, it just responds to buttons presses. Doug Goncz I love: Dona, Jeff, Kim, Kimmie, Mom, Neelix, Tasha, and Teri, alphabetically. I drive: A double-step Thunderbolt with 657% range. I fight terrorism by: Using less gasoline. === Subject: Re: Mathematical Rabbits by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iA8DIJ708664; Functional equation ............ f(x^4)=(f(x))^2 .... For x>1 TAKE f(x)=exp(sqrt(ln(x).v(ln(ln(x))) where v: ... Does not It also look rabbit-like ? What about mere maple or mu math yields! Alain. === Subject: Intuition about Lebesgue measure of the interval by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iA8DIJT08685; G'Day, A) The Lebesgue measure of the interval [0,1] is 1 by definition. B) We know that the measure of the rationals in [0,1] is zero. Thus, (A)+(B)=> C) the measure of the irrationals in [0,1] is 1. The above is a sort of negative proof that the measure of the irrationals in [0,1] is 1. Moreover, there is (at least for me) very little intuition gained from this approach. Does anybody have a method of directly showing that the measure of the irrationals in [0,1] is 1? Are there ways of building intuition here? My best attempt is to think about when an infinite set has non-zero measure. It seems that here must be such an overwhelmingly large number of points that they somehow add up to more than zero. ;{ Or perhaps, I am looking for more intuition than there is to be found. (A)+(B) => (C) Nothing more, nothingless. Phil. === Subject: re:Intuition about Lebesgue measure of the interval The intuition should be that the set of rational points is so insignificantly small that taking it off the full interval won't make a difference, which is of course equivalent to A + B => C *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Intuition about Lebesgue measure of the interval > A) The Lebesgue measure of the interval [0,1] is 1 by definition. Wrong. The Lebesgue measure of the interval [0,1] is 1 because one *proves* that the Lebesgue measure of each interval [a,b] is b - a. Also, please use the ENTER key once in a while. Jose Carlos Santos === Subject: Re: Intuition about Lebesgue measure of the interval >> A) The Lebesgue measure of the interval [0,1] is 1 by definition. > Wrong. The Lebesgue measure of the interval [0,1] is 1 because one > *proves* that the Lebesgue measure of each interval [a,b] is b - a. The proof that m([a,b]) = b - a depends on knowing what outer measure is, and also requires familiarity with compactness and the Heine-Borel theorem. If you don't know those things, then you probably won't find the measure-theoretic proof to be intuitive. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: logarithms I'm a high school math student and this problem has me stumped. Can someone maybe give me some hints? The problem is: log(x)/log(9) + log(8)/log(y) = 2 and log(9)/log(x) + log(y)/log(8) = 8/3 Solve for x and y. Please Help. === Subject: Re: logarithms >I'm a high school math student and this problem has me stumped. Can >someone maybe give me some hints? >The problem is: >log(x)/log(9) + log(8)/log(y) = 2 >and >log(9)/log(x) + log(y)/log(8) = 8/3 >Solve for x and y. Hint: Put u = log(x)/log(9); in that case what substitution do you make for log(9)/log(x)? And similarly for v = log(y)/log(8). Now you have simultaneous equations in u and v, with no logs. Solve them, then back-substitute to find x and y. (spoiler space) I get two solutions: (27,64) and (3,4). -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if you're afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: logarithms > I'm a high school math student and this problem has me stumped. Can > someone maybe give me some hints? > The problem is: > log(x)/log(9) + log(8)/log(y) = 2 > and > log(9)/log(x) + log(y)/log(8) = 8/3 > Solve for x and y. Put a = log(x)/log(9) and b = log(8)/log(y). Now, solve the system: a + b = 2 1/a + 1/b = 8/3. I hope that this helps. Jose Carlos Santos === Subject: arithmetic progressions containing squares? Don't know if this is a common/well-known question or not and, as for so much maths stuff, Google is too blunt a tool to dig out the facts. Anyway, I noticed that 3x + 2 doesn't produce a square for any integral x. It's pretty simple to prove, by showing that squares can only be congruent (mod 3) to 0 or 1, whereas 3x + 2 is always congruent to 2. What about the general case of ax + b - is there a simple relationship between a and b that tells you if this particular arithmetic progression contains squares or not, or is there nothing simpler than enumerating all the possible values mod a? I did a computer program to investigate this but no obvious pattern emerged. (Incidentally, I was also a bit surprised about how hard it was to write an efficient 'is this integer a square' function - the brute force check was painfully slow, and my feeble optimisation was to check if the integer had suitable values mod 3, 5, 7, 11 etc.) G.A. === Subject: Re: arithmetic progressions containing squares? >Don't know if this is a common/well-known question or not and, as for so >much maths stuff, Google is too blunt a tool to dig out the facts. >Anyway, I noticed that 3x + 2 doesn't produce a square for any integral x. >It's pretty simple to prove, by showing that squares can only be congruent >(mod 3) to 0 or 1, whereas 3x + 2 is always congruent to 2. >What about the general case of ax + b - is there a simple relationship >between a and b that tells you if this particular arithmetic progression >contains squares or not, or is there nothing simpler than enumerating all >the possible values mod a? I did a computer program to investigate this but >no obvious pattern emerged. (Incidentally, I was also a bit surprised about >how hard it was to write an efficient 'is this integer a square' function - >the brute force check was painfully slow, and my feeble optimisation was to >check if the integer had suitable values mod 3, 5, 7, 11 etc.) Check on quadratic residues. It is only necessary for something to be a quadratic residue mod 8 or mod odd primes, assuming that a and b have no common factors, and a little harder if they do. -- 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: Re: arithmetic progressions containing squares? > What about the general case of ax + b - is there a simple relationship > between a and b that tells you if this particular arithmetic progression > contains squares or not, or is there nothing simpler than enumerating all > the possible values mod a? Elementary number theory. In case a is prime, then just about half such arithmetic progressions contain squares. The question is equivalent to answering whether n^2 = b (mod a) has solutions. There is an efficient algorithm for deciding this based on the Jacobi symbol . See section 3.3 of Introduction to the Theory of Numbers by Niven, Zuckerman, and Montgomery, pages 142-147. Or consult another elementary number theory book and look up Jacobi symbol and quadratic reciprocity. === Subject: Re: arithmetic progressions containing squares? >> What about the general case of ax + b - is there a simple relationship >> between a and b that tells you if this particular arithmetic progression >> contains squares or not, or is there nothing simpler than enumerating all >> the possible values mod a? > Elementary number theory. In case a is prime, then just about > half such arithmetic progressions contain squares. The question > is equivalent to answering whether > n^2 = b (mod a) > has solutions. There is an efficient algorithm for deciding this based > on the Jacobi symbol . This is slightly misleading: as long as one knows the prime factors of a one can quickly decide whether this congruence is soluble. This isn't known to be the case otherwise. (When a is comopsite, the Jacobi symbol (b/a) may equal 1 when the congruence is insoluble). Nonetheless, quadratic reciprocity is where it's at ... -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: analysis...... hello.....doctor~ lim {(a_n) - 1} / {(a_n) + 1} = 0 n->00 find the lim a_n n->00 ---------------------------------------- um......i think........ for each e>0 there is an N for which |{(a_n) - 1} / {(a_n) + 1}| < e if n > N. so, |{(a_n) - 1}| < e*|{(a_n) + 1}| if n > N. so, -e*(a_n + 1) < a_n - 1 < e*(a_n + 1) if n > N. 1) -e*(a_n + 1) < a_n - 1 => a_n > (1-e)/(1+e) = 1 - {2*e/(1-e)} 2) a_n - 1 < e*(a_n + 1) => a_n < (1+e)/(1-e) = 1 + {2*e/(1-e)} so, 1 - {2*e/(1-e)} < a_n < 1 + {2*e/(1-e)} so, lim [1 - {2*e/(1-e)}] < lim a_n < lim [1 + {2*e/(1-e)}] n->00 and since e is any number, lim 2*e/(1-e) = 0 e->0 so, lim a_n = 1. right ?? um......i have a conscience about final step. so, i need your advice. thank you very much for your advice. === Subject: Re: analysis...... > lim {(a_n) - 1} / {(a_n) + 1} = 0 > n->00 Do notice, for all n, a_n + 1 /= 0 > find the lim a_n > n->00 > for each e>0 there is an N for which > |{(a_n) - 1} / {(a_n) + 1}| < e if n > N. > so, |{(a_n) - 1}| < e*|{(a_n) + 1}| if n > N. Then let e' = e/|a_n + 1| and you are done. ie, for each e', some N_e' with for all n > N_e' |a_n - 1| < e'|a_n + 1| = e and use N_e' for N. === Subject: Re: analysis...... Hello ... doctor (funny beans, as Americans would say) > hello.....doctor~ > lim {(a_n) - 1} / {(a_n) + 1} = 0 > n->00 > find the lim a_n > n->00 (a_n - 1)/(a_n + 1) = 1 - 2/(a_n + 1). Whence lim a_n = 1. -- Julien Santini === Subject: entropy Sorry for the so simple question. Let X be a discrete random variable that takes on one of the values x1, x2,..... , xn, with respective probabilities p1, p2, .... ,pn. The entropy of X is defined as: H(X) = -Sum{pi*log(pi)} in which log means log with base 2 and Sum means summation over all i. I wanna prove that H(X) is maximized when all of the pi are equal to 1/n. Using lagrange multipliers, I find that (1/n,...,1/n) is the only extremal point. But why this extremal point is a max? E. Farina. === Subject: Re: entropy > Sorry for the so simple question. > Let X be a discrete random variable that takes on one of the values x1, > x2,..... , xn, with respective probabilities p1, p2, .... ,pn. > The entropy of X is defined as: H(X) = -Sum{pi*log(pi)} in which log > means log with base 2 and Sum means summation over all i. > I wanna prove that H(X) is maximized when all of the pi are equal to > 1/n. Using lagrange multipliers, I find that (1/n,...,1/n) is the only > extremal point. But why this extremal point is a max? Try computing the value of H at some other point (where it's still easy to compute). === Subject: re:entropy Use the fact that x*log x is a convex function. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Skolem's Paradox and why is math the way it is? > |And i don't quite see whether ZF is strong enough to decide all > |'validity questions'. Are there IF-formulas whose IF-validity > |(interpreted as a sentence about sets) is independent from ZF? > No, the valid sentences of IF logic are not a recursively enumerable > set, so there isn't any formal system that suffices to prove all the > validities, let alone decide all of them (true and false). I thought Hintikka claimed that the sentances that had no winning strategy for either team could be listed by a Turing machine. And you'd expect that a Turing machine could list all the sentances (am I wrong about that?), so I thought the only hard part was separating the true sentances from the false sentances. J.E. === Subject: Re: context free grammar === Subject: context free grammar > context free grammar for generating all the strings > over {a,b} with the same number of a's and b's. > Here's the grammar: > S -> e | aSb | bSa | aSbbSa | bSaaSb (e is the empty string) Let D = words over { a,b } with same number of a's and b's. For the grammar G, I use the preferred S -> e | SS | aSb | bSa > An obvious induction shows that any string derived from S > using these productions has the same number of a's and b's. L(G) subset D. -- conversely If D not subset L(G), then some word in D, not in L(G) Let w be a word in D - L(G) with the smallest length. As w /= e, wlog let the first letter of w be a. Counting left to right, add 1 for each a and substract 1 for each b, until total is zero. Notice total is always positive until the last count and a zero will occure by len w or before and the earliest a zero could occure is at letter 2. If zero occures at n < len w, then the word u of the first n >= 2 letters of w and the word v of the remaining letters of w are in D. As len u, len v < len w, u,v in L(G). Thus S -> u, S -> v and finally S -> SS -> uv = w in L(G) But as this cannot be, the zero occures at len w. Now at len w - 1, as total is positive, total must be 1 and letter at len w is b. Thus some v in D with w = avb. Now as len v = len w - 2 < len w, v in L(G). Thus S -> v and as S -> aSb -> avb = w, w in L(G) But again this cannot be. Thus D - L(G) = nulset. D subset L(G) and finally D = L(G). QED > I would appreciate it if anyone could examine it > and let me know if I'm missing something obvious. > For the induction in the converse direction, The proof seems to be more complicated than needed. As it's style lacked in readiblity for cramming together stuff like T=aVa=aXbbYa and not putting longer equations on separate lines, it got scan consideration. ---- === Subject: Nash Equilibrium Hi can anyone recommend a beginner book about Nash Equilibrium for me? I am thinking about one question: for example in a match, that one side's strategy is somehow unpredictable is a best stragety I don't know any game theory, don't know how to prove it. === Subject: Re: Uniqueness of physical objects in the universe. > OK fellas, last time I was in here it was a veritable bloodbath of > back-and-forth mathematical snippets of highbrowed humor and cutting edge > sinicisms. I was forced to accept utter defeat at the hands of superior > intellects, but I have not forgotten my disgrace, and I am back with a > vengeance to regain my tattered dignity - to wit : > In our last episode I attempted to prove that No two objects in the > physical universe are identical. Apparently, this may be reducible to a > tautology. I am not sure that being a tautology matters much, because we are > talking about physical properties of real objects in the universe. If you > demonstrate a physical property of an object in the universe, then it dosent > matter how you did it - tautology or not. A physical property is a physical > property, and it matters not how you arrive at the proof of that property, > as long as the demonstration is valid. Tautologies are trivial within the > framework or abstract logical systems. If I say that It is a fast photon > because it is a fast photon - then at least you know that you have a > fast photon. Tautologies are not neccesarily trivial when you're talking > about real objects. > However, it seems that the original statement might be reducible to a > question of uniqueness. So, I have the following statement to work with : > ----------------------------------------------- > Every object in the physical universe is unique > ----------------------------------------------- > So, I would like to prove that statement. It is not possible to compare > every single physical object in the universe in a physics lab, so it must be > proved mathematically. > Definitions: > Physical object. > Any object in the physical universe which exists. This can be a person, > place or thing. A region of space/time is an object. A region of empty space > is an object. Locations are therefore objects. Events are objects, as per > relativity theory. If it exists in the physical universe then it is an > object. > Unique > A physical property of an object in the universe such that if an object > is unique, then there is no other object which is identical to that object. > There is a physical difference between objects which are distinct, and there > are no physical differences between objects which are identical. > Most uniqueness proofs require 2 things, first you demonstrate existence, > and then you demonstrate uniqueness. However, in this case, I cannot prove > that a physical object exists, it must be assumed (possibly via an axiom). > So, lets assume that objects really exist in the physical universe, and try > something like this- > ----------------------------------------------------- > Every object in the physical universe is unique > Proof > Suppose not > Let O1 and O2 be distinct objects in the physical universe which are > identical. > There are 2 possible cases, > 1) O1 and O2 are in separate locations > 2) O1 and O2 are in in the exact same location > Case 1) > If O1 and O2 are in two separate locations, then they are not identical, > and therefore they are both unique. > Case 2) > O1 and O2 are in the same location and they are also identical in every > possible physical respect. They cannot be distinct, because either O1 or O2 > is trivial and one of them does not really exist. > If you can have O1 and O2 in the same exact location, doing the same > exact thing, then let O3, O4, O(n) be identical to O1 and all in the same > exact location. You now have an infinite number of identical physical > objects in the exact same spot, which is obviously absurd. A contradiction. > QED > ----------------------------------------------------- > Is this science, or have I finally cracked ? > WillieK I agree with you. I am proposing a simple method to prove uniqueness. Below, I copied parts of your post and made some changes to prove correctness of your discovery. ----------------------------------------------- Every object in the physical universe is unique ----------------------------------------------- So, I would like to prove that statement. It is not possible to compare every single physical object in the universe in a physics lab, so it must be proved mathematically. Definitions: Physical object. Any object in the physical universe which exists. This can be a person, place, name, or thing. I give names to physical objects. If it exists in the physical universe then it is an object. Unique A physical property of an object in the universe such that if an object is unique, then there is no other object which is identical to that object. There is a physical difference between objects which are distinct, and there are no physical differences between objects which are identical. Most uniqueness proofs require 2 things, first you demonstrate existence, and then you demonstrate uniqueness. However, in this case, I cannot prove that a physical object exists, it must be assumed (possibly via an axiom). So, lets assume that objects really exist in the physical universe, and try something like this- ----------------------------------------------------- Every object in the physical universe is unique Proof Suppose not Let O1 and O2 be distinct objects in the physical universe which are identical. There are 2 possible cases, 1) O1 and O2 have different names 2) O1 and O2 have the exact same name Case 1) If O1 and O2 different names, then they are not identical, and therefore they are both unique. Case 2) O1 and O2 have the same name and they are also identical in every possible physical respect. They cannot be distinct, because I give unique names to every object and if O1 and O2 have the same name, then either O1 or O2 doesn't exist. If they both exist, then it violates my naming scheme. A Contradiction. Therefore, every object in physical universe is unique. === Subject: Re: Uniqueness of physical objects in the universe. > An Introduction to Complex Systems > Torsten Reil, Department of Zoology, University of Oxford > http://users.ox.ac.uk/~quee0818/complexity/complexity.html > Unfortunately, this is not what I'm after. > You cant talk about chaos as if it were a real physical phenomena But chaos and complexity science is about order...not chaos. Certainly you recognize the universe displays organization? > unless you > fill in the gaps which are clearly missing from current theory of physics. This is complete nonsense. The universe is understood through are relevant at all. > Where is the proof that you have points in spacetime ? > Where is the proof that you have lines in spacetime ? > Where is the proof that you have manifolds in spacetime ? > Where is the proof that you have solids in spacetime ? > All of the fundamentals are >absent<. There are new fundamental laws now. You are living in the dark ages. > And, unfortunately, the best anyone can do is say that they observed > something which looked like chaos, or it looked like a fractal, because > these things have very strict definitions which physics is unable to > satisfy. By insisting on proofs, you are limited to the simplest and least important aspects of reality. The emergent properties of a complex adaptive system, such as market forces or dark energy, are primary driving forces and far more important to understanding the real world. > Does chaos exist in nature ? I believe so. However, beliefs are one thing, > and science is another. It simply is not %100 valid to say that chaos exists > in nature unless it has been proved somehow, and I have never seen proof of > this. Chaotic behavior is merely motion that behaves according to the gas law. Of course that behavior exists, it is well accepted. The problem here is you don't know the meaning of the word. Chaos theory, now called complexity science, asserts that when static and chaotic behavior exist in an unstable equilibrium with each other, then the system begins to spontaneously organize. This is the mathematics of Darwinian evolution, but generalized to apply to non-living systems as well. You see, the material world organizes following the same rules as life. This is why life finds a way, because the evolution from matter to life is seamless and inevitable. As long as you cling to having two incompatible models for each you'll never see this simple truth. Or the true simplicity of the universe. Reducing to the smallest parts to understand reality is not the path, nature shows the way. As it is nature that defines reality....not...the other way around. Jonathan But nature is a stranger yet; The ones that cite her most Have never passed her haunted house, Nor simplified her ghost. To pity those that know her not Is helped by the regret That those who know her, know her less The nearer her they get. By E Dickinson s === Subject: Re: Uniqueness of physical objects in the universe. > I don't know if this will help, but it refers to physical geometry: > http://books.pdox.net/Math/Differential%20Synthetic%20Geometry/ > Interesting work, but he's got a classical approach. > I'm trying to make _general_ statements (as opposed to abstract statements) > about physical objects which can be justified rigorously. > If space is continuous, then it can be proven. That's all there is to it. Your understanding of this subtle point is amazing. (Hint - You can't *prove* anything in the sciences. You can only *describe*.) > Either from first principles of the universe, or by some combination of math > and experiment. First principals of the universe? What exactly are those? How do you propose to verify that the principles you choose to call the first principles are in fact true in every region of space-time? (Hint - even the Law of conservation of energy fails globally.) > I get very irritated when I hear about grainyness, and there is no real > proof that space/time is grainy. I need proof. Hope you feel the same & There's no proof that space-time is grainy or continuous. Nor will there be. What we do have are working grainy and continuous models of space-time with some predictive power. Try reading both volumes of Reality Rules by J. Casti. 'cid 'ooh === Subject: possible.... hello.....doctor~ if g(s-t)=f(s^2 - t^2, t^2 - s^2), f is differentiable. find the t*(dg/ds) + s*(dg/dt). -------------------------------------------- dg/ds = df/ds = (df/dx)(dx/ds)+(df/dy)(dy/ds) = fx(2s) + fy(-2s) (x = s^2 - t^2 , y = t^2 - s^2) dg/dt = similar~ is this right method ? and is g(s-t)=f(s^2 - t^2, t^2 - s^2) possible function ? g(s-t) => g(s,t) ...is this right ?? i need your advice. thank you very much for your advice. === Subject: Re: (Not quite) Cantor's diagonal proof > > > > Where did the theorem (as formally represented in MetaMath) come from > > - did MetaMath output it or did you enter it in by hand? > I entered it by hand. > > Where did the proof (as formally represented in MetaMath) come from - > > did MetaMath output it or did you enter it in by hand? > I entered it by hand So MetaMath doesn't create theorems, but only verifies proofs? Ok, verification is half of the problem. Let's look at your proof verification system. You give as an example on your home page Theorem 2p2e4 which has as its Expression |-(2+2)=4. The first line of its proof is df-2 with Expression |-2=(1+1). http://us.metamath.org/mpegif/df-2.html How does df-2 derive |-2=(1+1)? Each of the lines in df-2, including their links, only states an assertion concerning classes or wffs, as follows: c2 => class 2 c1 => class 1 caddc => class + co => class (1+1) wceq => wff 2=(1+1) In other words, since 2, 1, and + are classes, then 2=(1+1) is a wff. Then df-2 states as its Assertion that |-2=(1+1). 1. In df-2, how did we go from wff 2=(1+1) [last line of the proof] to |-2=(1+1) [Assertion]? The home page states that every step can be drilled down deeper and deeper until axioms will ultimately be found at the bottom . . . from 2+2=4 back to the axioms of set theory. Essentially everything that is possible to know in mathematics can be derived from a handful of axioms known as Zermelo-Fraenkel set theory. Walking back from |-(2+2)=4 I reach df-2 and the above 5 lines, not an axiom of ZF set theory. df-2 seems to be taking an arbitrary wff 2=(1+1) [the last line of its proof] and declaring it to be a theorem [with |-2=(1+1) as its Assertion.] Then can we do this (verify this proof) for any wff instead of 2=(1+1)? What is the general rule for which wffs MetaMath allows us to use here? 2. If we are saying that df-2 is a definition and that we also have definitions as well as axioms, then aren't you merely defining something close to the theorem? i.e., Theorem: |- (2+2)=4 Definition: |- 2=(1+1) It seems that you sneak definitions in instead of calling them axioms, and give as a definition an assertion that is very similar to the theorem. Then it only takes some substitution to transform the definition 2=(1+1) to the theorem (2+2)=4. How is this being derived from ZF set theory? ZF will tell us that certain sets exist, but what role does ZF play in the actual deduction of |-(2+2)=4? 3. If you add df-2 |-2=(1+1) simply as a definition, then you don't really have axioms for mathematics. You need a new definition (= axiom) for each such similar theorem. You can't just have the system derive theorems from a fixed set of axioms. You are manually setting up new axioms for each theorem. In fact, the arbitrariness of this substitution (wff => theorem) could be applied to false wffs as well. (Then every wff is provable and MetaMath is not sound or even meaningful.) We can conclude: c1 => class 1 c2 => class 2 wceq => wff (1=2) df-2 but use the above wff (1=2) instead of wff 2=(1+1) so => |-(1=2). Would this be verified by MetaMath? Why not? C-B === Subject: Re: (Not quite) Cantor's diagonal proof ... > http://us.metamath.org/mpegif/df-2.html > How does df-2 derive |-2=(1+1)? Each of the lines in df-2, including > their links, only states an assertion concerning classes or wffs, as > follows: > c2 => class 2 > c1 => class 1 > caddc => class + > co => class (1+1) > wceq => wff 2=(1+1) > In other words, since 2, 1, and + are classes, then 2=(1+1) is a wff. > Then df-2 states as its Assertion that |-2=(1+1). > 1. In df-2, how did we go from wff 2=(1+1) [last line of the proof] to > |-2=(1+1) [Assertion]? df-2 is a definition, not a theorem. It has a soundness justification (see below), but it does not have a proof. What you see under df-2 is a syntax breakdown to provide an understanding of how the wff corresponding to df-2 is constructed. It says Detailed syntax breakdown of Definition df-2, not Proof of Theorem df-2. It is provided to the reader merely for convenience, and is not a proof of df-2 in any sense. You can ignore it if you want. I added it because I thought that for more complex definitions it could be helpful to the reader. Older versions of those pages didn't even show it, and I was hesitant to introduce for the exact reason that people might mistake it for a proof. But I thought I had marked it clearly in the table's caption. Perhaps I will remove it in the future if it causes this kind of confusion. ... > 3. If you add df-2 |-2=(1+1) simply as a definition, then you don't > really have axioms for mathematics. You need a new definition (= > axiom) for each such similar theorem. You can't just have the system > derive theorems from a fixed set of axioms. You are manually setting > up new axioms for each theorem. The axioms do not include the symbol 2. We could equivalently work with (1+1), but that becomes tedious. So we introduce a definition, and when write down a theorem containing 2 we really mean that same theorem with 2 replaced by (1+1). It is nothing more than a shorthand to save us work when writing things down. The axioms also do not include a symbol for 1, which is in turn a shorthand for a more complex expression. By continuing in this fashion, we could ultimately produce an expession containing only the primitive variables and connectives of the set theory axioms. This final expression is the real theorem of set theory. All theorems shortened with definitions are not the real theorems of set theory but are convenient representations that allow humans to grasp them more easily. But each shortened expression represents, unambiguously, a unique real theorem of set theory. The justification for the definition 2=(1+1) is the following. (i) 2 is a new symbol not used before, and (ii) if we back-substitute what 2 represents into the definition itself, we end up with (1+1)=(1+1) which is a theorem of set theory, specifically an instance of cleqid. If a definition meets those criteria, it is said to be sound. More discussion on this is found at http://us.metamath.org/mpegif/mmset.html#definitions . Metamath currently does not enforce the soundness of definitions, which is a criticism that will disappear in its next generation Ghilbert being developed by Raph Levien http://ghilbert.org . However, all definitions in the Metamath database (set.mm) have been independently verified (by hand) as sound by two people other than myself. In addition, set.mm has been translated into Ghilbert and verified as correct (although strictly speaking the definitional soundness part of Ghilbert is still under development and as of this writing not yet complete). > In fact, the arbitrariness of this substitution (wff => theorem) could > be applied to false wffs as well. (Then every wff is provable and > MetaMath is not sound or even meaningful.) We can conclude: > c1 => class 1 > c2 => class 2 > wceq => wff (1=2) > df-2 but use the above wff (1=2) instead of wff 2=(1+1) so => |-(1=2). > Would this be verified by MetaMath? Why not? No, because what you have proved is that (1=2) is a wff, not a theorem. If you try to prove |- (1=2) from the existing axioms and definitions you will not be able to. To repeat, the syntax breakdown under df-2 is not a proof of it and does not justify its soundness. It should be ignored for that purpose. You cannot jump from wff 1=2 to |- 1=2 in an actual Metamath proof in the Metamath language; an error message will be issued if you try to do so. Here is an attempt to do that: 1eq2 $p |- 1 = 2 $= c1 c2 wceq $. Here is what the metamath program says: MM> verify proof 1eq2 1eq2 ?Error on line 64032 of file c:/n/mm/set.mm at statement 19004, label 1eq2, type $p: c1 c2 wceq $. ^^^^ The result of the proof (step 3) does not match the statement being proved. The result is wff 1 = 2 but the statement is |- 1 = 2. -- Norm nm at alum dot mit dot edu === Subject: Meaning of weight in general relativity This is much ado about trivia. What can you explain that is worth explaining? Nothing. You are clutching on to this one silly parody of an idea ignoring all the important stuff going on in modern relativity. Jack, if this is so trivial, then why did you put so much effort into refuting it? It's not trivial. Why? 1. We are friends. 2. Your general quest was worthwhile. 3. As it developed your rhetoric got more and more sounding like cliche crackpot Einstein-bashing or should I say MTW bashing? 4. Your confusions are common and worth explicating. Off the top of my head here is how I see them: I. You are confused about the relation of coordinate systems to reference frames. This also has to do with confusions about epistemological map vs ontological territory that leads to Magickal Thinking. First consider linear coordinate transformations, Galilean for simplicity. x -> x' = x - vt t -> t' = t This is a GLOBAL transformation. It can be physically approximated by two fleets of rocket ships out in free space far from gravitating sources. Each fleet moving uniformly relative to the other and internally relatively at rest to each other. All rocket engines off in that case! Similarly for a global constant acceleration transformation x - > x' - vt + (1/2)gt^2 t --> t' = t Note that this is nonlinear. Again one fleet fires its rockets. The other fleet keeps its rockets off. In this case we have a homogeneous inertial force in the S' global frame that is not asymptotically flat! This is not real gravity as Landau & Lifz point out in Ch 10. You have been disparaging of the asymptotic boundary condition BTW missing entirely its physical relevance to your problem. GCT in GR, u = 0,1,2,3 x^u -> x^u'(x^u) With Jacobian matrix X^u'u = x^u',u ,u is ordinary partial derivative - trivial flat zero connection field. Tensors transform locally and multilinearly with products of coefficients X at a fixed local physical event P that is not same as a formal manifold point p. This GCT can always be approximated by two fleets of rocket ships (exchanging EM signals understood) in arbitrary relative motion both intra-fleet and inter-fleet. Therefore, the relation between the coordinate system and the reference system is same in GR as it is in Galilean relativity except now there is complete local gauge freedom on how each rocket ship will fire its engines! In Galilean relativity, the fleets must be internally in lock-step like troops marching in formation. II. The {L-C} = GCT Tensor + Non-Tensor pseudo-problem This is like ruler and compass constructions leading to Galois finite group solvability and Godel undecidability. The problem here is, using ONLY the concepts within Einstein's original 1916 theory i.e. using only the {L-C} connection itself, with covariant partial derivatives ;u = ,u + {L-C} no second connection allowed, no torsion, no non-metricity etc. Can the above split be constructed entirely within the original 1916 GR considered as an axiomatic formal system? That's the FORMAL problem. All the additional machinery Alex brought in with a second affine connection different from the {L-C} connection is a completely different problem and is not relevant to the actual problem you first raised years ago without any notion in your mind of affine connections with torsion and non-metricity. Now it is obvious that you formulated this problem by confusing it with the following {L-C} transformation under GCT {L-C}^uvw -> {L-C}^u'v'w' = Xu'^uX^vv'X^ww'{L-C}^uvw + X^u'lY^lv'w' Y^lv'w' = x^l,v',w' Since the RHS is a tensor transformation + non-tensor piece This is obviously what gave you the idea but you garbled the categories mistaking a transformation for the object transformed. Finally, your trying to separate the vanishing g-force in a Local Inertial Frame (the LIF) into an objective tensor force minus an inertial force is a gross violation of the equivalence principle (WEP to be precise). Gravity is locally equivalent to an inertial force. All inertial forces disappear in an inertial frame. There is NO OBJECTIVE TENSOR gravity force! That is a complete confusion. The weight W^i, i = 1,2,3 we feel in a rest LNIF (Local Non-Inertial-Frame) is the electrical reaction force to the inertial force ~ gravity force, i.e. on the timelike non-geodesic: W^i = mc^2{^i00} = -F^i(EM reaction force) Note that only the mixed space-time components of the connection field contribute to weight. === Subject: Re: entropy by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iA8GfeP30260; >Sorry for the so simple question. >Let X be a discrete random variable that takes on one of the values x1, > x2,..... , xn, with respective probabilities p1, p2, .... ,pn. >The entropy of X is defined as: H(X) = -Sum{pi*log(pi)} in which log >means log with base 2 and Sum means summation over all i. >I wanna prove that H(X) is maximized when all of the pi are equal to >1/n. Using lagrange multipliers, I find that (1/n,...,1/n) is the only >extremal point. But why this extremal point is a max? >E. Farina. If p1 and p2 are unequal, what happens to the entropy when you replace both p1 and p2 by (p1+p2)/2 ? === Subject: Re: entropy >>Sorry for the so simple question. >>Let X be a discrete random variable that takes on one of the values x1, >> x2,..... , xn, with respective probabilities p1, p2, .... ,pn. >>The entropy of X is defined as: H(X) = -Sum{pi*log(pi)} in which log >>means log with base 2 and Sum means summation over all i. >>I wanna prove that H(X) is maximized when all of the pi are equal to >>1/n. Using lagrange multipliers, I find that (1/n,...,1/n) is the only >>extremal point. But why this extremal point is a max? >>E. Farina. > If p1 and p2 are unequal, what happens to the entropy when you > replace both p1 and p2 by (p1+p2)/2 ? ( of the type (0,0,1,0,...,0) ) are not found by the lagrange multipliers. Is this due to the non differentiability of H(X) in such points? === Subject: re:Entropy We can define a digit sequence without a pattern as one which cannot be generated by a finite Turing machine. Let's enumerate all the Turing machines that produce sequences of digits. Then consider the following sequence: The n-th digit is the n-th digit produced by the n-th (digit sequence producing) Turing machine plus one (modulo 10). If there was a Turing machine with number m that could produce this sequence its m-th entry would be (say) a and thus by the definition of the sequence it would have to be in fact a + 1. Contradiction. Thus the sequence has no pattern. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: re:Entropy > We can define a digit sequence without a pattern as one which cannot > be generated by a finite Turing machine. > Let's enumerate all the Turing machines that produce sequences of > digits. How? === Subject: Re: entropy >Let X be a discrete random variable that takes on one of the values x1, > x2,..... , xn, with respective probabilities p1, p2, .... ,pn. >The entropy of X is defined as: H(X) = -Sum{pi*log(pi)} in which log >means log with base 2 and Sum means summation over all i. >I wanna prove that H(X) is maximized when all of the pi are equal to >1/n. Using lagrange multipliers, I find that (1/n,...,1/n) is the only >extremal point. But why this extremal point is a max? >( of the type (0,0,1,0,...,0) ) are not found by the lagrange >multipliers. Is this due to the non differentiability of H(X) in such >points? These are extreme points of the domain, not critical points. I guess you could use the Karush-Kuhn-Tucker conditions (the version of Lagrange multipliers for inequalities as well as equalities). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Cauchy's Theorem Hello I have the gradient of a terrain (x and y slope of height-field) along a closed path and I want to make an estimate of the gradient at some point within the path. It occurs to me that I could use Cauchy's theorem to make this estimate but it's been a while since I looked at complex calculus and I'm not sure my understanding is up to scratch. If I assume that the surface is smooth (i.e. both gradients are continuous) then would I be correct in interpreting Cauchy's theorem as saying that the gradient at some point x',y' is given by the integral around the path of [( X(x,y) + iY(x,y))/((x-x') +i(y-y'))] (dx + idy) divided by 2pi.i where X(x,y) is the rate of change of height wrt x and Y(x,y) is same wrt y. Fred === Subject: Re: Cauchy's Theorem >I have the gradient of a terrain (x and y slope of height-field) along a >closed path and I want to make an estimate of the gradient at some point >within the path. It occurs to me that I could use Cauchy's theorem to make >this estimate but it's been a while since I looked at complex calculus and >I'm not sure my understanding is up to scratch. >If I assume that the surface is smooth (i.e. both gradients are continuous) >then would I be correct in interpreting Cauchy's theorem as saying that the >gradient at some point x',y' is given by the integral around the path of >[( X(x,y) + iY(x,y))/((x-x') +i(y-y'))] (dx + idy) divided by 2pi.i >where X(x,y) is the rate of change of height wrt x and Y(x,y) is same wrt y. Cauchy's theorem is for analytic functions of a complex variable. Does your terrain satisfy the Cauchy-Riemann equations? If not, Cauchy's theorem is of no use to you. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Cauchy's Theorem >I have the gradient of a terrain (x and y slope of height-field) along a >closed path and I want to make an estimate of the gradient at some point >within the path. It occurs to me that I could use Cauchy's theorem to make >this estimate but it's been a while since I looked at complex calculus and >I'm not sure my understanding is up to scratch. >If I assume that the surface is smooth (i.e. both gradients are continuous) >then would I be correct in interpreting Cauchy's theorem as saying that the >gradient at some point x',y' is given by the integral around the path of >[( X(x,y) + iY(x,y))/((x-x') +i(y-y'))] (dx + idy) divided by 2pi.i >where X(x,y) is the rate of change of height wrt x and Y(x,y) is same wrt y. > Cauchy's theorem is for analytic functions of a complex variable. > Does your terrain satisfy the Cauchy-Riemann equations? If not, > Cauchy's theorem is of no use to you. Actually, I'm not dealing with a physical terrain at all, just a 2D scalar field S(x,y) whose gradient (wrt x and y) is known along some closed loop - or put another way I have a vector field. I'm looking for a way of making the best estimate of the gradient of S(x,y) at some arbitrary point within the loop. Physical constraints mean that S(x,y) should be reasonably smooth - meaning it has a unique and well defined gradient at every point. I may have this wrong, but if I associate the components of the gradient of a smooth scalar field with the real and imaginary parts of a complex function C(x,y) then isn't C necessarily analytic? Fred === Subject: Re: Cauchy's Theorem >>I have the gradient of a terrain (x and y slope of height-field) along a >>closed path and I want to make an estimate of the gradient at some point >>within the path. It occurs to me that I could use Cauchy's theorem to >make >>this estimate but it's been a while since I looked at complex calculus >and >>I'm not sure my understanding is up to scratch. >> Cauchy's theorem is for analytic functions of a complex variable. >> Does your terrain satisfy the Cauchy-Riemann equations? If not, >> Cauchy's theorem is of no use to you. >Actually, I'm not dealing with a physical terrain at all, just a 2D scalar >field S(x,y) whose gradient (wrt x and y) is known along some closed loop - >or put another way I have a vector field. I'm looking for a way of making >the best estimate of the gradient of S(x,y) at some arbitrary point within >the loop. Physical constraints mean that S(x,y) should be reasonably >smooth - meaning it has a unique and well defined gradient at every point. >I may have this wrong, but if I associate the components of the gradient of >a smooth scalar field with the real and imaginary parts of a complex >function C(x,y) then isn't C necessarily analytic? No, it isn't. You need the Cauchy-Riemann equations: if C = u + i v then du/dx = dv/dy and du/dy = - dv/dx. Now if u = dS/dx and v = dS/dy you actually have du/dy = dv/dx, not - dv/dx, so you should take the complex conjugate of the gradient rather than the gradient itself: u = dS/dx and v = - dS/dy. But even then, you'll only have one of the Cauchy-Riemann equations and not the other one: du/dx = dv/dy says d^2 S/dx^2 + d^2 S/dy^2 = 0, i.e. S must be a harmonic function in order for this to work. For just an arbitrary smooth function S(x,y), you could easily have the gradient of S be 0 on your path but S not constant inside, e.g. think of a bump on an otherwise flat plane. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: EVIL FBI SADISTS should be KIDNAPPED and TORTURED for 3 YEARS Keith | Someone emailed and requested me to post it on Usenet newsgroups. | | Sadistic FBI agents should be tortured the way how FBI sadists and | perverts tortured this poor guy (non-muslim) for three years and | continue to torture him even after he left america tracking him with | implanted transponder chips and other methods. The FBI was too lenient. === Subject: Re: EVIL FBI SADISTS should be KIDNAPPED and TORTURED for 3 YEARS > Someone emailed and requested me to post it on Usenet newsgroups. > Would you please put OT: at the beginning of your subject line for > non-woodworking topics. This assists those of us who prefer to get our > conspiracy theories, politics and easy money in the correct forums. Why didn't you? This seems to be the troll that will not die. Stop talking about it! -- Peter T. Daniels grammatim@att.net === Subject: Re: EVIL FBI SADISTS should be KIDNAPPED and TORTURED for 3 YEARS > Would you please put OT: at the beginning of your subject line for > non-woodworking topics. This assists those of us who prefer to get our > conspiracy theories, politics and easy money in the correct forums. > Why didn't you? Ah, well, it is a good question isn't it? Seems hypocritical, right. Believe it or on, I omit it on purpose. Lots of people probably have that subject line in their kill file; if I go changing the subject line, it will now show up in quite a few readers, which does a disservice to the reader. > This seems to be the troll that will not die. Stop talking about it! Not a terribly good troll. In any case, I was partially being humorous (obviously missed the mark) and hey, you never know, he might even follow along. Sorry if I inconvenienced you; this would kinda suck 'cause it was what I was complaining about in the first place. Can't recall seeing this troll much before, but my eyes probably glazed over... PK === Subject: Re: EVIL FBI SADISTS should be KIDNAPPED and TORTURED for 3 YEARS > Someone emailed and requested me to post it on Usenet newsgroups. > Sadistic FBI agents should be tortured [snip 793 lines of trolled crap] Liar. Idiot. Boor. http://www.mazepath.com/uncleal/sunshine.jpg -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Normed Spaces by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iA8HHGg01025; I would appreciate your help with the following: i)A Haar measure for the unit disk exists, since every compact group has a Haar measure. Does any one know how this Haar measure is defined? ii)Let X be a normed space, N a closed subspace, (isn't every subspace closed, anyway?) let BallX be the unit ball centered at 0 in X. Let p be the standard quotient map: p:X-->X/N. Show that p maps BallX into Ball(X/N)=the unit ball in X/N. Use the norm ||p(x_0)||=d(x_0,N), which is well defined, since N is closed iii)Let X be Banach and Y any normed space. Let B(X,Y) be the set of bounded linear maps between X and Y. Let {T_n} be a sequence of linear maps in B(X,Y) such that Lim_n->00 T_n(x) exists for all x in X. , let this limit equal T_x , show that T is in B(X,Y). === Subject: Re: Normed Spaces > iii)Let X be Banach and Y any normed space. > Let B(X,Y) be the set of bounded linear maps between > X and Y. Let {T_n} be a sequence of linear maps in B(X,Y) > such that Lim_n->00 T_n(x) exists for all x in X. , let > this limit equal T_x , show that T is in B(X,Y). For this use the Banach - Steinhauss theorem. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Normed Spaces Originator: grubb@lola >i)A Haar measure for the unit disk exists, since every compact group >has a Haar measure. Does any one know how this Haar measure is defined? I assume you mean the unit circle, not thwe unit disk (since the unit disk is not a group). In that case dtheta does the job. >ii)Let X be a normed space, N a closed subspace, >(isn't every subspace closed, anyway?) let Definitely not! Let X be C([0,1]) and N={f:f is a polynomial}. Then N is a subspace which is not closed in X (in fact, it is dense). >BallX be the unit ball centered at 0 in X. Let p be >the standard quotient map: p:X-->X/N. >Show that p maps BallX into Ball(X/N)=the unit ball in >X/N. Use the norm ||p(x_0)||=d(x_0,N), which is well >defined, since N is closed If ||x||<=1, what can you say about ||p(x)||? >iii)Let X be Banach and Y any normed space. >Let B(X,Y) be the set of bounded linear maps between >X and Y. Let {T_n} be a sequence of linear maps in B(X,Y) >such that Lim_n->00 T_n(x) exists for all x in X. , let >this limit equal T_x , show that T is in B(X,Y). Uniform Boundedness Principle. --Dan Grubb === Subject: Re: Normed Spaces > ii)Let X be a normed space, N a closed subspace, > (isn't every subspace closed, anyway?) No, every finite dimensional (linear vector) subspace is closed; in the more general case, you can't tell (actually, any infinite dimensional normed vector space has a subspace which is not closed). let > BallX be the unit ball centered at 0 in X. Let p be > the standard quotient map: p:X-->X/N. > Show that p maps BallX into Ball(X/N)=the unit ball in > X/N. Use the norm ||p(x_0)||=d(x_0,N), which is well > defined, since N is closed If d(x_0,x_1) <= 1 then ||p(x_0)-p(x_1)|| = ||d(x_0,N)-d(x_1,N)|| <= d(x_0,x_1) <=1 > iii)Let X be Banach and Y any normed space. > Let B(X,Y) be the set of bounded linear maps between > X and Y. Let {T_n} be a sequence of linear maps in B(X,Y) > such that Lim_n->00 T_n(x) exists for all x in X. , let > this limit equal T_x , show that T is in B(X,Y). Trivial by the uniform boundedness principle (Banach Steinhaus theorem). === Subject: Re: Normed Spaces > i)A Haar measure for the unit disk exists, since every compact group > has a Haar measure. Does any one know how this Haar measure is defined? Consider the function f:[0,2pi] ---> S^1 t |-> exp(it) Given a subset A of S^1, the Haar measure of A is the Lebesgue measure of f^{-1}(A) divided by 2pi. Here, when I say *the* Haar measure I mean the only Haar measure defined in S^1 such that the measure of the whole S^1 is equal to 1. > ii)Let X be a normed space, N a closed subspace, > (isn't every subspace closed, anyway?) OF COURSE NOT! Take X = { (a_n)_n | a_n = 0 if n is large enough } with the norm ||(a_n)_n)|| = sum_n |a_n|. Now, define N as the set of those sequences such that a_1 + 2a_2 + 3a_3 + 4a_4 + ... = 0. Then N is a subspace, but it's not closed since the sequences (1,-1/2,0,0,0,...) (1,0,-1/3,0,0,...) (1,0,0,-1/4,0,...) ... all belong to N, but not their limit, which is (1,0,0,0,0,...). Jose Carlos Santos === Subject: Re: Uncle assAl: (SR) Lorentz t', x' = Intervals > : > > : > Let's cut to the quick. > : > When YOU have read the original paper, which is readily available > : > on the internet at > : > > : > http://www.fourmilab.ch/etexts/einstein/specrel/www/ > : > > : > then you can argue about what SR says or does not say. > : > I've no interest in your personal views of YOUR theory of SR, > : > or that of anyone else whose name is not Albert Einstein. > : > Until then, you ARE a crank. > : > > : > (rant snipped) > : > Androcles. > : I think I'll join Eric Gisse in remarking that Androcles fails to > : grasp the concept that Einstein is not special relativity and > special > : relativity is not Einstein. They are seperate entities. since ONLY > : Androcles treats webpages that A.E. did NOT put up, as the ONLY and > : audience who is now LONG dead, there ARE better ways to EXPLAIN the > : theory today, and there are better theories theories (like GR), and > : there are better WAYS to use SR, like Modern SR that are less prone > to > : abuse or error. > : You want to talk web pages? Well, I've read the ENTIRE webpage > : http://www.androc1es.pwp.blueyonder.co.uk/KoksDoppler.htm and I > think > : you should be interested in what I have to say about it: > : Koks, according to Androcles: Let us focus on what Stella and > Terence > : actually see with their own eyes. (Just to emphasize that we're > : talking about direct observation here, I'll put the verb see and > its > : brothers in the HTML strong font throughout this section.) > : [Androcles: I'll use capital where Koks uses bold.] > : No problems so far. > : Koks, according to Androcles: To make things interesting, we'll > equip > : them with unbelievably powerful telescopes, so each twin can WATCH > the > : other's clock throughout the trip. If each twin SAW the other clock > : run slow throughout the trip, then we would have a contradiction. > But > : this is not what they SEE. > : [Androcles: Oh yes it is what they SEE, and Koks' DOES have a > : contradiction.] > : Androcles lies. > Prove it. I say you have no idea what you are talking about. If you LOOK at the LIGHT bouncing off a clock that is moving AWAY from you, then you SEE the LIGHT being REDshifted, and you SEE the clock running slowler. EVEN Newtonian physics says this. If you LOOK at the LIGHT bouncing off a clock that is moving TOWARDS you, then you SEE the LIGHT being BLUEshift, and you SEE the clock running faster. EVEN Newtonian physics says this. So where does Newtonian physics and SR differ. In the dopplershift factor. US e the relativistic doppler for the SR prediction, use Newtonian doppler for the Newtonian physics prediction. Relativistically 0.8*c away from you means a redshift FACTOR of 1/3. 0.8*c towards you means a redshift FACTOR of 3. WHY does a bluesift of a redshift mean you SEE the clock run fast or slow? Assume monochrome light, the frequency of the light oscillated n times (at the surface of the clock being watched) while the clock was still before it's second hand moved from one position to another. When it is redshifted by a factor of D, the observer has to wait longer (1/D times as long) to see n oscillations of the light, so they have to wait longer (1/D times as long) to SEE the image of the clock move it's hand from the one position to another, to the image of the clock observered moves it's second-hand once every 1/D observer-seconds. Note how the time was independant of the color of the monochromatic light, so by the linearity and supperposition of E&M fields (true for Newton as well as A.E.), the clock's image as SEEN by the observer, runs slow (by a factor of 1/D). Androcles is so clueless that he likley has NO idea that what I'm describing BASIC and known to almost everyone other than himself that studies relativity. > : SR uses the relativisitic doppler, and says that > : Terrance SEES a redshifted signal for a long amount of terrance-time > : and a blue-shifted signal for a short amount of terrance-time. > Quantify it. Put some numbers in. The same numbers the idiot Koks > provided. Koks rounded. He said 14, I said 7+sqrt(14), he said c*0.99, I said c*sqrt(48/47), he said 1/14, I said 7-sqrt(48). I am NOT going to use Koks ROUNDED numbers. DO THE MATH. 0.99*0.99 is VERY close to (48/49). 7-sqrt(48) is VERY close to 1/14. 7+sqrt(48) is VERY close to 14. Use MY numbers because (for instanct turn-around), they are CORRECT. Koks was trying to avoid equations for the mathematically challenged. That's no you (I hope) Androcles, so use the EXACT numbers like I do. I put DETAILED numbers AND detailed calculations a half-dozen times for you Androcles. > Reference: >>http://www.physics.adelaide.edu.au/~dkoks/Faq/Relativity/SR/TwinParadox/tw in_intro.html >>Stella coasts along at (say) nearly 99 percent of light-speed. At 99percent, >>the time dilation factor is a bit over 7, so let's say the speed is just a >>shade under 99 percent and the time dilation factor is 7. >>Be prepared for me to prove you wrong. Wheras: Stella SEES a redshifted >>signal and a blueshifted signal for EQUAL amount of a stella-time (I'm going >>to assume instant turnaround). Androcles claims otherwise, I will give >>Androcles honesty and his math skills the benefit of the doubt, so I'll try to >>give an OBJECTIVE description of the twin-paradox situation followed by the >>actual: math (except I'll assume an instant turn-around). Terrance waits a >>[7-sqrt(48)] terrance-years (I'll call x=7-sqrt(48)) and then sends a light >>signal after Stella telling her to return, when Stella gets it she responds, >>I'll be right home and turns aroung and comes home at the same >>stella-terrance-speed stella-immediately. In this example Stella return >>exactlys 14 years after leaving, from which Terrance CALCULATES that she was >>going sqrt(48/49) times the speed of light each way. But what does Terrance >>SEE *during* the journey? He sees a signal from Stella that has it's >>frequency RED-shifted by a factor of x for the FIRST x terrance-years, and >>THEN he sends his come home signal, but then he waits 14-2*x MORE >>terrance-years before he gets the I'll be right home signal, so during that >>WHOLE time, (14-x terrance-years,) he sees a REDshifted signal, redshifted by >>a factor of x. For the LAST x terrance-years he sees a BLUE-shifted signal, >>that has it's frequency multiplied by 14-x instead of by x (i.e. f'=[14-x]*f, >>instead of f'=[x]*f). What does Stella SEE *during*the: journey? She sees a >>RED-shifted signal (f'=[x]*f) for ONE stella-year, then she gets the signal >>from Terrance telling her to turn around. She instantly breaks, sends a >>signal I'll be right home, and goes back at the same speed. >********** Then for the next stella-year: she sees a BLUE-shifted signal from >Terrance (f'=[14-x]*f).********** >That is where you screw up. Stella sends on the return trip (because Terence >receives) a whole lot more signals than one. Each signal she sends is one year >of her life. In the one year that Terence receives 14 signals, Stella has aged >14 years.Koks can't count to 14, and neither can you.Fault found, remainder >snipped.Androcles. Androcles. I provided all the CORRECT math, and all the calculations. Apparantly you ALSO want a JUSTIFICATION for EACH and EVERY step. I *will* do that for you, but FIRST you need to stop employing the Androcles-method of finding faults. I'll describe it for you. (A) Assume a Newtonian result and some the results of calculation of SR. (B) Be oblivious to the fact that they contradict each other. (C) See that a contradiction is about to appear and pick some RANDOM thing and assert that THAT thing is in error. (D) Deny that the error was caused by assuming a Newtonian result at the begining (E) Expect that someone, ANYone else finds your silly dance meaningful. You Newtonian assumption in this case is that Stella ages 14 years. But they ONLY way you can keep this assumption UP, IS to remainder snip, because if you analyzed your OWN predictions (like I went on to do) then EVEN YOU would find that it is CRAP. So let's use my math (with the sqrt(48) coming up a lot) not Kok's because you and I are smart enough and have good enough math skills to be exact, right? So let's look at what assumptions J.E. needs. (A) Stella is GONE for 14 terrance-years (B) If Terrance sends a LIGHT signal after Stella after waiting [7-sqrt(48)] terrance-years, that Stella recieves it at the same time as she turns around to come home. (C) The redshift factor x that Terrance SEES from Stella's signal for [7+sqrt(48)] terrance-years, is the same as the redshift factor that Stella SEES from Terrance's signal for the first part of her trip (before she sees herself turn around). (D) The blueshift factor y that Terrance SEES the blueshift factor that Stella SEES from Terrance's signal for the second part of her trip (after she SEES herself turn around, until after she SEES herself arrive back at home). (E) y=1/x (ALL the relativistic assumptions (equal redshifts, y=1/x, come from the fact that each CALCULATES the other as moving away at the same speed, v, and from the relativistic dopplershift formula). There is NO relativisttic assumption that Stella went less than the speed of light, THAT is taken care of by the assumption that the light signal Terrance sent CAUGHT UP with her to tell her to turn around, and that only happens when she is going less than the speed of light (as emitted from a source light Terrance). So PLEASE Androcles, don't later claim that Stella was going faster than c. PLEASE accept assumptions A,B,C,D E as consequences of the relatavistic dopplershift formula. (And PLEASE don't assume random other things that are ONLY true for Newtonian physics). If you can do that, they you can look for INTERNAL inconsistencies in SR, we ALREADY know that SR makes different PREDICTIONS than Newtonian physics, that WAS the point of SR. First I'll explain why these assumptions are reasonable. Terrance and Stella move at a relative speed of v=sqrt(48/49) times the speed of light. Why? Because if Stella moves at v, and Terrance waits 7-sqrt(48) terrance-years to send a signal, then it will take sqrt(48) terrance-years for the signal to go from Terrance to Stella. Calculation? After 7 years-terrance, Stella has traveled v*(7 terrance-years)=sqrt(48) terrance-light-years, but that's EXACLTLY how far that signal traveled in those 7 terrance-years, because he didn't SEND it for 7-sqrt(48) terrance-years, so it ONLY had sqrt(48) terrance-years to travel, so it traveled sqrt(48) terrance-light years. OK, then Stella turns around, and GETS BACK in seven more terrance-years, so then her velocity must be EQUAL in magnitude on the way back (opposite in direction) on her return journey. HOPEFULLY Androcles agress with me so far, since this paragrpah is ALL so TRUE for NEWTONIAN physics as well (For Newtonian physics, have Stella put on the breaks, THEN transmit the I'm coming home signal, which then CLEARY get to Terrance at 7 PLUS sqrt(48) years, and THEN have her turn back home at the same speed, for relativity, the speed of light is independant of the motion of the source, so it doesn't matter whether she breaks first, or transmit's home first. Note that at 7 terrance-years Terrance CALCULATES that Stella turns around, but he doesn't SEE her turn around unit 7 PLUS sqrt(48) terrance-years since she left. then she got a signal, pushed the turn around button, felt a big acceleration, then IMMEDIATELY saw Terrance's clock signals become blueshifted instead of redshifted. TERRANCE says that the relative velocity (a CALCULATED quantity) between him and Stella is equal in magnitude and opposite in direction for the two journeys. NEWTONIAN (and SR) physics says that therefore Stella COMPUTES that HER relative velocity is the same in magnitude and opposite in direction for the two parts. Now let's CONSIDER the Newtonian theory. Let's look at the signal Stella gives right BEFORE she puts on the breaks, WHEN she's stationary, and right AFTER she heads home. WHICH does Terrance SEE first? Androcles theory says that we actually SEE a BLURRY Stella, since we see stella's future BEFORE stella's past, so MULTIPLE clock NEWTONIAN physics TERRANCE doesn't HAVE a red-shifted before he SEES Stella turn around and blueshifted after he SEES her turn around. In Newtonian physics land, Terrance SEES just a redshifted Stella moving away, THEN he SEES Stella moving away AND towards him (both blue AND redshifted images of her signals), THEN he SEES her moving towards him, moving away, AND being stationary (that's at 7 PLUS sqrt(48) years) and since I'm not an expert on Newtonian redshifts for light (I'm mean the whole equations for E&M break down with Newtonian comes a terrance-time when he sees JUST a blue-shifted image of Stella moving ONLY towards him, but for all I know, with Newtonian physics, Terrance is STILL waiting to SEE Stella on her entire outbound journey even AFTER she arrives. I'd have a HARD job of figuring anything out in Newtonian theory since I don't the redshift, the laws of light, or much of anything. But we agree that it has a DIFFERENT prediction (blurry images as SEEN by observers), than SR. The SR theory says that light always moves at the speed of light, so we know with SR that the signal right BEFORE turn around appears BEFORE the signal AT turn around, which appears BEFORE the signal right AFTER turn around. So with SR, Terrance's VIEW goes from completely redshifted, to compeltely blueshifted, with no blurry images. And the change for Terrance happens at 7 PLUS sqrt(48) terrance-years. For Stella, the change (in what she SEES, either red or blue) happens when she gets the turn around signal and actually turns around. Androcles doubts whether Stella ages as much on to way out as on the way in. He seemed to doubt it because of a Newtonian assumption that she must age as much as Terrance did. So THIS time, I will FIRST compute how much she ages on the way back. We know that Terrance sends a signal with n=f*[7+sqrt(48)] terrance-years number of tick on his clock, and that Stella gets them ALL after getting her signal to turn around. Since Stella and Terrance had a relative velocity of sqrt(48/49)*c, and were moving towards each other, we can COMPUTE the relativistic doppler shift factor as 7+sqrt(48). So that means Stella saw n ticks in Y Stella-years and she say them at an OBSERVED freqeuncy f' where f'=f*[7+sqrt(48)]. So let's do the math Stella seeing every tick on Terrance's clock SINCE he sent the come home signal, and that she saw them all AFTER she turned around. n=f'*Y=f*[7+sqrt(48)]*Y (that's what she saw (frequency and time, since she turned around). But from Terrance we know that n=f*[7+sqrt(48)] terrance-years because that's what he sent (frequency and time, since he sent the come home signal). But Stella SEES the same number of ticks on TERRANCES clock as he SAW. So the dimensionless value n must be the same. So WHAT is the value of Y Androcles? (Yes I *can* do the math, I just wish that YOU would learn to do it consistently instead of RANDOM points invoking NEWTONIAN assumptions into a DIFFERENT theory.) I could carry on the calculations, but I've already done it time and time and time again. Androcles can't handle the truth, because he wants to make Newtonian assumptions and SNEAK them in to FOIL the theories of his NEMISI, the physicists, and he's going to get VERY frustrated that I refuse to allow him to do that. He'll also likely KEEP doing it, and KEEP not reading my posts in their entirety, even though it is clear that *I* read his entire posts. === Subject: Re: Uncle assAl: (SR) Lorentz t', x' = Intervals : > : > : > : > Let's cut to the quick. : > : > When YOU have read the original paper, which is readily available : > : > on the internet at : > : > : > : > http://www.fourmilab.ch/etexts/einstein/specrel/www/ : > : > : > : > then you can argue about what SR says or does not say. : > : > I've no interest in your personal views of YOUR theory of SR, : > : > or that of anyone else whose name is not Albert Einstein. : > : > Until then, you ARE a crank. : > : > : > : > (rant snipped) : > : > Androcles. : > : : > : I think I'll join Eric Gisse in remarking that Androcles fails to : > : grasp the concept that Einstein is not special relativity and : > special : > : relativity is not Einstein. They are seperate entities. since ONLY : > : Androcles treats webpages that A.E. did NOT put up, as the ONLY and particular : > : audience who is now LONG dead, there ARE better ways to EXPLAIN the : > : theory today, and there are better theories theories (like GR), and : > : there are better WAYS to use SR, like Modern SR that are less prone : > to : > : abuse or error. : > : : > : You want to talk web pages? Well, I've read the ENTIRE webpage : > : http://www.androc1es.pwp.blueyonder.co.uk/KoksDoppler.htm and I : > think : > : you should be interested in what I have to say about it: : > : : > : Koks, according to Androcles: Let us focus on what Stella and : > Terence : > : actually see with their own eyes. (Just to emphasize that we're : > : talking about direct observation here, I'll put the verb see and : > its : > : brothers in the HTML strong font throughout this section.) : > : : > : [Androcles: I'll use capital where Koks uses bold.] : > : : > : No problems so far. : > : : > : Koks, according to Androcles: To make things interesting, we'll : > equip : > : them with unbelievably powerful telescopes, so each twin can WATCH : > the : > : other's clock throughout the trip. If each twin SAW the other clock : > : run slow throughout the trip, then we would have a contradiction. : > But : > : this is not what they SEE. : > : : > : [Androcles: Oh yes it is what they SEE, and Koks' DOES have a : > : contradiction.] : > : : > : Androcles lies. : > Prove it. I say you have no idea what you are talking about. : : If you LOOK at the LIGHT bouncing off a clock that is moving AWAY from : you, then you SEE the LIGHT being REDshifted, and you SEE the clock : running slowler. EVEN Newtonian physics says this. If you LOOK at : the LIGHT bouncing off a clock that is moving TOWARDS you, then you : SEE the LIGHT being BLUEshift, and you SEE the clock running faster. : EVEN Newtonian physics says this. Of course, and it is symmetrical. : : So where does Newtonian physics and SR differ. In that stupid division by sqrt(1-v^2/c^2), of course. f' = f(1-cos(phi).v/c) --- straight Newton. f' = f(1-cos(phi).v/c) /sqrt (1-v^2/c^2) --- idiocy. In the dopplershift : factor. US e the relativistic doppler for the SR prediction, use : Newtonian doppler for the Newtonian physics prediction. : Relativistically 0.8*c away from you means a redshift FACTOR of 1/3. : 0.8*c towards you means a redshift FACTOR of 3. : : WHY does a bluesift of a redshift mean you SEE the clock run fast or : slow? Assume monochrome light, the frequency of the light oscillated : n times (at the surface of the clock being watched) while the clock : was still before it's second hand moved from one position to another. : When it is redshifted by a factor of D, the observer has to wait : longer (1/D times as long) to see n oscillations of the light, so they : have to wait longer (1/D times as long) to SEE the image of the clock : move it's hand from the one position to another, to the image of the : clock observered moves it's second-hand once every 1/D : observer-seconds. Note how the time was independant of the color of : the monochromatic light, so by the linearity and supperposition of E&M : fields (true for Newton as well as A.E.), the clock's image as SEEN by : the observer, runs slow (by a factor of 1/D). : : Androcles is so clueless that he likley has NO idea that what I'm : describing BASIC and known to almost everyone other than himself that : studies relativity. : : > : SR uses the relativisitic doppler, and says that : > : Terrance SEES a redshifted signal for a long amount of terrance-time : > : and a blue-shifted signal for a short amount of terrance-time. : > Quantify it. Put some numbers in. The same numbers the idiot Koks : > provided. : : Koks rounded. He said 14, I said 7+sqrt(14), he said c*0.99, I : said c*sqrt(48/47), he said 1/14, I said 7-sqrt(48). I am NOT : going to use Koks ROUNDED numbers. DO THE MATH. 0.99*0.99 is VERY : close to (48/49). 7-sqrt(48) is VERY close to 1/14. 7+sqrt(48) is : VERY close to 14. So if Stella takes one Terence-year to return she takes 14 Stella-years to return. That IS doing the math. LEARN TO COUNT. : : Use MY numbers because (for instanct turn-around), they are CORRECT. : Koks was trying to avoid equations for the mathematically challenged. : That's no you (I hope) Androcles, so use the EXACT numbers like I do. I know how to count to 14 for the 14 signals Terence receives that Stella sends one year apart that are received at 14 per Terence-year. 1/14 is the red shift, 14/1 is the blue shift. : : I put DETAILED numbers AND detailed calculations a half-dozen times : for you Androcles. And you, like Koks, can't count to 14. Year 0. S and T are together. T S Year 1.T send a birthday greeting to S. S is 0.99 light years from T, sends no birthday greeting, it's only been 26 days for S. T |-----S Year 2. T send a birthday greeting to S. 52 days for S, 1.98 light-years from T. Two greetings are chasing S. T |-----|-----S Year 3. T send a birthday greeting to S. 78 days for S, 2.97 light-years from T. Three greetings are chasing S. T |-----|-----|-----S Year 4. T send a birthday greeting to S. 104 days for S, 3.96 light-years from T. Four greetings are chasing S. T |-----|-----|-----|-----S Year 5. T send a birthday greeting to S. 130 days for S, 4.95 light-years from T. Five greetings are chasing S. T |-----|-----|-----|-----|-----S Year 6. T send a birthday greeting to S. 156 days for S, 5.94 light-years from T. Six greetings are chasing S. T |-----|-----|-----|-----|-----|-----S Year 7. T send a birthday greeting to S. 182 days for S, 6.93 light-years from T. Seven greetings are chasing S. T |-----|-----|-----|-----|-----|-----S TURNAROUND. Sanity check. Seven greetings on the way, six months, that's a frequency of 1/2 * 1/7 = 1/14. Here is where you and I are going to disagree. First inconsistency. Stella has travelled 6.93 light years in six months. That's a tad FTL. Pretending this distance somehow Lorentz contracts in Stella's inertial frame is pure nonsense. Stars don't move to suit human beings and then instantaneously leap back to their original positions. Stella will send a message I've arrived when she stops, and that will take 6.93 years to reach Terence. Now, we already know that T is going to see a frequency of 14, right? And he knows it will take Stella 7 years to return. That means he must see 14 * 7 = 98 signals from Stella. Apart from the I've arrived signal, Stella only sends one signal a year. She'll be dead before she can return, it will take 98 Stella years to get home. Only 7 Terence years, though. If I were Stella, I'd use the Newton method. Androcles. : : > Reference: : : >>http://www.physics.adelaide.edu.au/~dkoks/Faq/Relativity/SR/TwinParadox/tw i n_intro.html : >>Stella coasts along at (say) nearly 99 percent of light-speed. At : 99percent, : >>the time dilation factor is a bit over 7, so let's say the speed is : just a : >>shade under 99 percent and the time dilation factor is 7. : >> : >>Be prepared for me to prove you wrong. Wheras: Stella SEES a : redshifted : >>signal and a blueshifted signal for EQUAL amount of a stella-time : (I'm going : >>to assume instant turnaround). Androcles claims otherwise, I will : give : >>Androcles honesty and his math skills the benefit of the doubt, so : I'll try to : >>give an OBJECTIVE description of the twin-paradox situation : followed by the : >>actual: math (except I'll assume an instant turn-around). Terrance : waits a : >>[7-sqrt(48)] terrance-years (I'll call x=7-sqrt(48)) and then sends : a light : >>signal after Stella telling her to return, when Stella gets it she : responds, : >>I'll be right home and turns aroung and comes home at the same : >>stella-terrance-speed stella-immediately. In this example Stella : return : >>exactlys 14 years after leaving, from which Terrance CALCULATES that : she was : >>going sqrt(48/49) times the speed of light each way. But what does : Terrance : >>SEE *during* the journey? He sees a signal from Stella that has : it's : >>frequency RED-shifted by a factor of x for the FIRST x : terrance-years, and : >>THEN he sends his come home signal, but then he waits 14-2*x MORE : >>terrance-years before he gets the I'll be right home signal, so : during that : >>WHOLE time, (14-x terrance-years,) he sees a REDshifted signal, : redshifted by : >>a factor of x. For the LAST x terrance-years he sees a BLUE-shifted : signal, : >>that has it's frequency multiplied by 14-x instead of by x (i.e. : f'=[14-x]*f, : >>instead of f'=[x]*f). What does Stella SEE *during*the: journey? : She sees a : >>RED-shifted signal (f'=[x]*f) for ONE stella-year, then she gets the : signal : >>from Terrance telling her to turn around. She instantly breaks, : sends a : >>signal I'll be right home, and goes back at the same speed. : >********** Then for the next stella-year: she sees a BLUE-shifted : signal from >Terrance (f'=[14-x]*f).********** : >That is where you screw up. Stella sends on the return trip (because : Terence : >receives) a whole lot more signals than one. Each signal she sends is : one year : >of her life. In the one year that Terence receives 14 signals, Stella : has aged : >14 years.Koks can't count to 14, and neither can you.Fault found, : remainder : >snipped.Androcles. : : Androcles. I provided all the CORRECT math, and all the calculations. : Apparantly you ALSO want a JUSTIFICATION for EACH and EVERY step. I : *will* do that for you, but FIRST you need to stop employing the : Androcles-method of finding faults. I'll describe it for you. : (A) Assume a Newtonian result and some the results of calculation of : SR. (B) Be oblivious to the fact that they contradict each other. : (C) See that a contradiction is about to appear and pick some RANDOM : thing and assert that THAT thing is in error. (D) Deny that the error : was caused by assuming a Newtonian result at the begining (E) Expect : that someone, ANYone else finds your silly dance meaningful. : : You Newtonian assumption in this case is that Stella ages 14 years. : But they ONLY way you can keep this assumption UP, IS to remainder : snip, because if you analyzed your OWN predictions (like I went on to : do) then EVEN YOU would find that it is CRAP. : : So let's use my math (with the sqrt(48) coming up a lot) not Kok's : because you and I are smart enough and have good enough math skills to : be exact, right? : : So let's look at what assumptions J.E. needs. (A) Stella is GONE for : 14 terrance-years (B) If Terrance sends a LIGHT signal after Stella : after waiting [7-sqrt(48)] terrance-years, that Stella recieves it at : the same time as she turns around to come home. (C) The redshift : factor x that Terrance SEES from Stella's signal for [7+sqrt(48)] : terrance-years, is the same as the redshift factor that Stella SEES : from Terrance's signal for the first part of her trip (before she sees : herself turn around). (D) The blueshift factor y that Terrance SEES : the blueshift factor that Stella SEES from Terrance's signal for the : second part of her trip (after she SEES herself turn around, until : after she SEES herself arrive back at home). (E) y=1/x (ALL the : relativistic assumptions (equal redshifts, y=1/x, come from the fact : that each CALCULATES the other as moving away at the same speed, v, : and from the relativistic dopplershift formula). : : There is NO relativisttic assumption that Stella went less than the : speed of light, THAT is taken care of by the assumption that the light : signal Terrance sent CAUGHT UP with her to tell her to turn around, : and that only happens when she is going less than the speed of light : (as emitted from a source light Terrance). So PLEASE Androcles, don't : later claim that Stella was going faster than c. PLEASE accept : assumptions A,B,C,D E as consequences of the relatavistic dopplershift : formula. (And PLEASE don't assume random other things that are ONLY : true for Newtonian physics). If you can do that, they you can look : for INTERNAL inconsistencies in SR, we ALREADY know that SR makes : different PREDICTIONS than Newtonian physics, that WAS the point of : SR. : : First I'll explain why these assumptions are reasonable. Terrance and : Stella move at a relative speed of v=sqrt(48/49) times the speed of : light. Why? Because if Stella moves at v, and Terrance waits : 7-sqrt(48) terrance-years to send a signal, then it will take sqrt(48) : terrance-years for the signal to go from Terrance to Stella. : Calculation? After 7 years-terrance, Stella has traveled v*(7 : terrance-years)=sqrt(48) terrance-light-years, but that's EXACLTLY how : far that signal traveled in those 7 terrance-years, because he didn't : SEND it for 7-sqrt(48) terrance-years, so it ONLY had sqrt(48) : terrance-years to travel, so it traveled sqrt(48) terrance-light : years. OK, then Stella turns around, and GETS BACK in seven more : terrance-years, so then her velocity must be EQUAL in magnitude on the : way back (opposite in direction) on her return journey. HOPEFULLY : Androcles agress with me so far, since this paragrpah is ALL so TRUE : for NEWTONIAN physics as well (For Newtonian physics, have Stella put : on the breaks, THEN transmit the I'm coming home signal, which then : CLEARY get to Terrance at 7 PLUS sqrt(48) years, and THEN have her : turn back home at the same speed, for relativity, the speed of light : is independant of the motion of the source, so it doesn't matter : whether she breaks first, or transmit's home first. Note that at 7 : terrance-years Terrance CALCULATES that Stella turns around, but he : doesn't SEE her turn around unit 7 PLUS sqrt(48) terrance-years since : she left. : : then she got a signal, pushed the turn around button, felt a big : acceleration, then IMMEDIATELY saw Terrance's clock signals become : blueshifted instead of redshifted. TERRANCE says that the relative : velocity (a CALCULATED quantity) between him and Stella is equal in : magnitude and opposite in direction for the two journeys. NEWTONIAN : (and SR) physics says that therefore Stella COMPUTES that HER relative : velocity is the same in magnitude and opposite in direction for the : two parts. : : Now let's CONSIDER the Newtonian theory. Let's look at the signal : Stella gives right BEFORE she puts on the breaks, WHEN she's : stationary, and right AFTER she heads home. WHICH does Terrance SEE : first? Androcles theory says that we actually SEE a BLURRY Stella, : since we see stella's future BEFORE stella's past, so MULTIPLE clock : NEWTONIAN physics TERRANCE doesn't HAVE a red-shifted before he SEES : Stella turn around and blueshifted after he SEES her turn around. In : Newtonian physics land, Terrance SEES just a redshifted Stella moving : away, THEN he SEES Stella moving away AND towards him (both blue AND : redshifted images of her signals), THEN he SEES her moving towards : him, moving away, AND being stationary (that's at 7 PLUS sqrt(48) : years) and since I'm not an expert on Newtonian redshifts for light : (I'm mean the whole equations for E&M break down with Newtonian : comes a terrance-time when he sees JUST a blue-shifted image of Stella : moving ONLY towards him, but for all I know, with Newtonian physics, : Terrance is STILL waiting to SEE Stella on her entire outbound journey : even AFTER she arrives. I'd have a HARD job of figuring anything out : in Newtonian theory since I don't the redshift, the laws of light, or : much of anything. But we agree that it has a DIFFERENT prediction : (blurry images as SEEN by observers), than SR. : : The SR theory says that light always moves at the speed of light, so : we know with SR that the signal right BEFORE turn around appears : BEFORE the signal AT turn around, which appears BEFORE the signal : right AFTER turn around. So with SR, Terrance's VIEW goes from : completely redshifted, to compeltely blueshifted, with no blurry : images. And the change for Terrance happens at 7 PLUS sqrt(48) : terrance-years. For Stella, the change (in what she SEES, either red : or blue) happens when she gets the turn around signal and actually : turns around. : : Androcles doubts whether Stella ages as much on to way out as on the : way in. He seemed to doubt it because of a Newtonian assumption : that she must age as much as Terrance did. So THIS time, I will FIRST : compute how much she ages on the way back. We know that Terrance : sends a signal with n=f*[7+sqrt(48)] terrance-years number of tick on : his clock, and that Stella gets them ALL after getting her signal to : turn around. Since Stella and Terrance had a relative velocity of : sqrt(48/49)*c, and were moving towards each other, we can COMPUTE the : relativistic doppler shift factor as 7+sqrt(48). So that means Stella : saw n ticks in Y Stella-years and she say them at an OBSERVED : freqeuncy f' where f'=f*[7+sqrt(48)]. So let's do the math Stella : seeing every tick on Terrance's clock SINCE he sent the come home : signal, and that she saw them all AFTER she turned around. : n=f'*Y=f*[7+sqrt(48)]*Y (that's what she saw (frequency and time, : since she turned around). But from Terrance we know that : n=f*[7+sqrt(48)] terrance-years because that's what he sent (frequency : and time, since he sent the come home signal). But Stella SEES the : same number of ticks on TERRANCES clock as he SAW. So the : dimensionless value n must be the same. So WHAT is the value of Y : Androcles? (Yes I *can* do the math, I just wish that YOU would learn : to do it consistently instead of RANDOM points invoking NEWTONIAN : assumptions into a DIFFERENT theory.) : : I could carry on the calculations, but I've already done it time and : time and time again. Androcles can't handle the truth, because he : wants to make Newtonian assumptions and SNEAK them in to FOIL the : theories of his NEMISI, the physicists, and he's going to get VERY : frustrated that I refuse to allow him to do that. He'll also likely : KEEP doing it, and KEEP not reading my posts in their entirety, even : though it is clear that *I* read his entire posts. === Subject: Re: Uncle assAl: (SR) Lorentz t', x' = Intervals > Let's cut to the quick. > When YOU have read the original paper, which is readily available > on the internet at > http://www.fourmilab.ch/etexts/einstein/specrel/www/ > then you can argue about what SR says or does not say. > I've no interest in your personal views of YOUR theory of SR, > or that of anyone else whose name is not Albert Einstein. > Until then, you ARE a crank. > (rant snipped) > Androcles. > I think I'll join Eric Gisse in remarking that Androcles fails to > grasp the concept that Einstein is not special relativity and special > relativity is not Einstein. Before you waste more time on Androfart, perhaps it would be a good idea to find out what you are up against: http://users.pandora.be/vdmoortel/dirk/Physics/ImmortalFumbles.html This creature has only one purpose left in the last years of his miserable life: to annoy people as much as he possibly can. Enjoy - but do not think for a second that you will enlighten the Village Idiot :-) Dirk Vdm === Subject: Re: Uncle assAl: (SR) Lorentz t', x' = Intervals > Let's cut to the quick. > When YOU have read the original paper, which is readily available > on the internet at > > http://www.fourmilab.ch/etexts/einstein/specrel/www/ > > then you can argue about what SR says or does not say. > I've no interest in your personal views of YOUR theory of SR, > or that of anyone else whose name is not Albert Einstein. > Until then, you ARE a crank. > > (rant snipped) > Androcles. > I think I'll join Eric Gisse in remarking that Androcles fails to > grasp the concept that Einstein is not special relativity and special > relativity is not Einstein. > Before you waste more time on Androfart, perhaps it > would be a good idea to find out what you are up against: > http://users.pandora.be/vdmoortel/dirk/Physics/ImmortalFumbles.html > This creature has only one purpose left in the last years of > his miserable life: to annoy people as much as he possibly > can. > Enjoy - but do not think for a second that you will enlighten > the Village Idiot :-) > Dirk Vdm I understand that you are trying to be helpful, but as an educator I do believe that we are not teaching SR is the best way possible, and Androcles behavior confirms my theory. I think a cold hard focus on observables and modeling is the best route. Androcles does seem to object to modeling, so I'd like to see if a focus on observable quantities helps. Androcles went to the trouble to make a web page, and I don't think he did that just to annoy people, I think he has a serious intent, although I admit that when I read his posts, his spelling of Minkowksi does seem a bit whimsical, but spelling and the qualities of one's theories are not causally related. I wasn't fond of the axiomatic formulation of SR theory for pedagogical reasons, but it might work for someone like Androcles too. As a personal believer in the Gauge Theory of Gravity generalization of SR, I think a SR that focuses on observable quantities is a much better place to start. SR really is a clean and beautiful theory, and any mistakes Androcles makes are ones that a naive student COULD make on accident, so it's not like it's bad for me to study his responses. I am more than a bit frustrated that he has his own definitions for every word, like how I have to say separation instead of vector because the word vector has so much Newtonian baggage for Androcles that he just can't seem to shake off. But that could be true of any student, so again I don't see the harm. observable measurements is a good idea. Has someone already tried that with Androcles? I'm new here, so I wouldn't know. J.E. >Hi - I am looking for a adjacency-graph-to-triangle-corner >code for a class project in mesh generation. <...> > Isn't the set of all possible triangles only O(N^3)? > So checking them one by one is already faster. Since > your graph is sprase why not > for i = 1 to N do > for j > i and adjacent to i do > for k > j and adjacent to j do > if k adjacent to i > add triangle {i, j, k} > else if k > max adjacent to i > break; // early exit, might not be worth it. > This is O(N*Delta^2) where Delta = max degree. (Reposted with timings under code) That makes sense. Here is a Mathematica prototype implementation following your algorithm, before conversion to C: TriangleList[g_]:=Module[{T={},i,j,k,m,n,gi,gj}, For [i=1,i<=Length[g],i++, gi=g[[i]]; For [m=1,m<=Length[gi],m++, j=gi[[m]]; If [j<=i,Continue[]]; gj=g[[j]]; For [n=1,n<=Length[gj],n++, k=gj[[n]]; If [k<=j || !MemberQ[gi,k], Continue[]]; If [!MemberQ[T,{i,j,k}], AppendTo[T,{i,j,k}]]; ]]]; Return[T]]; Times for 32, 128, 512, 2048 and 8192 triangles: 0.03, 0.12, 0.45, 3.42, 42.62 sec. === Subject: Re: The real numbers, and general comments > But a model can always be written as a set of axioms - so the Godel > theorem also tells us that no _model_ can be complete, either. Your premise is wrong, and your conclusion is not even wrong. Part of the meaning of incompleteness is that *no* model can be written as a set of axioms. One more time: every sufficiently rich consistent system of axioms has multiple models, whose differing properties are undecidable from the axioms. You *can't* completely specify a unique model with axioms. Do you know what a model is yet? === Subject: Re: The real numbers, and general comments >Well, to me, this shows either that ZF is too limited or that these >questions really are not interesting. For, if they are, why no >develope a better foundation that can answer them? >>Most mathematicians prefer a minimal set of axioms, rather than a >>maximal set. Also, with the Godel incompleteness theorem, we know that >>there will always be questions we cannot answer, no matter how many >>axioms you provide. > But a model can always be written as a set of axioms - so the Godel > theorem also tells us that no _model_ can be complete, either. No. The axioms are not the model. A model is an example of something that satisfies the axioms. The axioms are not even *a* model. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: The real numbers, and general comments > Well, to me, this shows either that ZF is too limited or that these > questions really are not interesting. For, if they are, why no > develope a better foundation that can answer them? > > Most mathematicians prefer a minimal set of axioms, rather than a > maximal set. Also, with the Godel incompleteness theorem, we know that > there will always be questions we cannot answer, no matter how many > axioms you provide. > But a model can always be written as a set of axioms - so the Godel > theorem also tells us that no _model_ can be complete, either. > Andrew Usher Did you read my post where I pointed out the definition of the word model? === Subject: ? time-avg ODE soln Given an ODE du/dt = a*u+q(t) will the following two types of time-avg soln be always equivalent? (1) Solve u(t) explicitly and then let u_avg(t) = Integral( u(s), s = t..t+T )/T (2) Avg the source term first and then solve the ODE. That is, q_avg(t) = Integral( q(t), s = t..t+T )/T du_avg/dt = a*u_avg+q_avg(t) by Cheng Cosine Nov/08/2k4 UT === Subject: Re: ? time-avg ODE soln > Given an ODE du/dt = a*u+q(t) will the following two types of time-avg > soln be always equivalent? > (1) Solve u(t) explicitly and then let > u_avg(t) = Integral( u(s), s = t..t+T )/T > (2) Avg the source term first and then solve the ODE. > That is, q_avg(t) = Integral( q(t), s = t..t+T )/T > du_avg/dt = a*u_avg+q_avg(t) Yes, and not just for first-order ODE's but for any linear constant-coefficient ODE. If L is the averaging operator Lf(t) = 1/T int_t^{t+T} f(s) ds then L commutes with the derivative operator D, so if P(D) u = q (where P is some polynomial with constant coefficients) then P(D) Lu = L P(D) u = Lq. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: ? time-avg ODE soln > Given an ODE du/dt = a*u+q(t) will the following two types of time-avg > soln be always equivalent? > (1) Solve u(t) explicitly and then let > u_avg(t) = Integral( u(s), s = t..t+T )/T > (2) Avg the source term first and then solve the ODE. > That is, q_avg(t) = Integral( q(t), s = t..t+T )/T > du_avg/dt = a*u_avg+q_avg(t) Yes (with the proviso that you need to think how the initial conditions transform). Try substituting the definitions of u_avg and q_avg into the final equation. Mike Guy === Subject: Re: Analysis problem, need some help here... > So you are using the proof of contradiction, but how does the fact > that (a_n)_n has some convergent subsequence contradict anything, > so as to make the initial hypothesis false? I can be a lil slow at > times.. :) First of all, two things: 1) Please don't top-post. If you want to know what's that and why you shouldn't do it, read http://www.caliburn.nl/topposting.html or http://www.html-faq.com/etiquette/?toppost for instance. 2) While writing, please press the ENTER key once in a while. Now, concerning your specific question: let (b_n)_n be a convergent subsequence of (a_n)_n and let l = lim_n b_n. Then l belongs to [a,b] and lim_n f(b_n) = inf f = 0. But, since f is continuous at l, you have f(l) = f(lim_n b_n) = lim_n f(b_n) = 0, which is impossible. Jose Carlos Santos === Subject: This Week's Finds in Mathematical Physics (Week 208) Originator: baez@math-cl-n03.math.ucr.edu (John Baez) Also available at http://math.ucr.edu/home/baez/week208.html This Week's Finds in Mathematical Physics Week 208 Last weekend I went to a conference at the Perimeter Institute: 1) Workshop on Quantum Gravity in the Americas, http://www.perimeterinstitute.ca/activities/scientific/PI-WORK-2/ It was great to see the new building. I'd visited this institute before in its temporary location, which was a funky old hotel building complete with pool tables and a bar. The new building is very different: four stories of intensely modern architecture overlooking a lake, consisting mainly of an enormous atrium lined with walkways and glass-walled offices. There's also a big lecture theater, a couple of smaller seminar rooms, a library, a restaurant whose walls are all blackboards, a reflecting pool, and lots of little places to sit and talk, complete with espresso machines. In short, a theoretical physicist's idea of heaven! But perhaps the design of heaven shouldn't be left to theoretical physicists. Some aspects of the setup don't seem very comfortable. Like most modern architecture, the place is short on coziness - there's too much glass, metal and concrete for my taste. You also find yourself spending a lot of time climbing up and down uncomfortably narrow staircases. The last, at least, is no accident: they made the stairs skinny on purpose, so you have to say hello to anyone you meet going the other way. It'll be interesting to see how many collaborative papers come out of this. Abhay Ashtekar was supposed to give the first talk, but he got lost walking to the new building, so suddenly I had to give the first talk. Yikes! Jet-lagged and not fully awake, I sketched the problem of dynamics in quantum gravity: the problem of describing motion in a world where the geometry of spacetime is quantum-mechanical and interacts with matter. I gave a generally downbeat assessment of the progress so far in all known approaches: 2) John Baez, The problem of dynamics in quantum gravity, http://math.ucr.edu/home/baez/dynamics/ Even though the last few Weeks have been on quantum gravity conferences, I've been mainly working on n-categories lately, because I've been sort of fed up with quantum gravity. I did, however, sketch some avenues for progress - and later in this workshop I saw some work that really cheered me up! For example, I've always been fascinated by John Wheeler's old dream of matter without matter. In its original version, the idea was to an electric field going in one end and out the other, the ends will pair. But there were big problems with this idea: for example, More recently this idea was reincarnated in the spin network formalism by Lee Smolin, with spin network edges replacing wormholes: 3) Lee Smolin, Fermions and topology, available as gr-qc/9404010. A spin network is a gadget with vertices and edges, where the edges represent field lines - lines of the electric field or the analogous thing for other forces, including gravity. If a spin network edge goes between vertices that would otherwise be far apart, it acts a bit like a wormhole. It will be hidden from observers in the rest of they don't call them spin networks for nothing! A variant on this idea is to have spin networks with loose ends: edges that just fizzle out. This is more ad hoc, but easier to study in some ways. A decade ago, Kirill Krasnov and I showed how to describe 4) John Baez and Kirill Krasnov, Quantization of diffeomorphism- invariant theories with fermions, hep-th/9703112. However, the hard problem in quantum gravity is always dynamics. Does the dynamics of spin networks with loose ends actually mimic that this question in a toy model, 3-dimensional Lorentzian gravity: 5) Kirill Krasnov, Lambda<0 Quantum Gravity in 2+1 Dimensions I: Quantum States and Stringy S-Matrix, Class. Quant. Grav. 19 (2002) 3977-3998, also available as hep-th/0112164. Kirill Krasnov, Lambda<0 Quantum Gravity in 2+1 Dimensions II: 3999-4028, also available as hep-th/0202117. network ends - though you don't need to emphasize that viewpoint, since there are also other nice ways to think about what's going on, using hyperbolic geometry and complex analysis. It all fits together in a beautiful picture. In principle you can even calculate In this conference, Laurent Freidel explained how this idea works in 3-dimensional Riemannian gravity - a less physical but mathematically more tractable spin foam model. Some but not all of his work can be found here: 6) Laurent Freidel and David Louapre, Ponzano-Regge model revisited I: as hep-th/0401076. Laurent Freidel and David Louapre, Ponzano-Regge model revisited II: Equivalence with Chern-Simons, available as gr-qc/0410141. Freidel showed that if you take this theory and allow spin networks with automatically quantized. More surprisingly, so is their mass - and there's an upper bound on the mass! That's because when we quantize this theory, its gauge group automatically gets replaced by a quantum group. Physically, this means that spacetime becomes quantum-mechanical in such a way that it no longer makes sense to talk about times shorter to the rate at which its wavefunction oscillates, this puts an upper Mathematically, part of the point is that we can describe 3d Riemannian gravity as a gauge theory where the gauge group is the double cover of the 3d Euclidean group - the analogue of the Poincare group in this context. But when we quantize the theory, this gets replaced by a quantum group: the quantum double of SU(2). As with the 3d Euclidean group, unitary representations of this quantum group are classified by mass and spin... but now both mass and spin are discrete, and both are bounded above. Anyway, what's great about Freidel and Louapre's work is that it gives a simplified but mathematically rigorous testbed in which loose ends networks with hidden edges in this setup. So, we should be able to do calculations and see if a spin network with a hidden edge acts like pair. Unfortunately, all this work on gravity in 3d spacetime doesn't easily generalize to 4d spacetime. The reason is that gravitational waves are only possible when spacetime has dimension 4 or more... so 3d gravity all the fun comes from global topology, like wormholes. That's why 3d theories are easy to calculate with - we can use ideas from topological quantum field theory. The danger, though, is that these calculations are misleading it comes to real-world physics. Indeed, that's precisely the sort of thing I was worrying about in my talk. So, it really cheered me up when a young guy named Artem Starodubtsev spoke about a promising new spin foam model of quantum gravity in 4 dimensions! He's working on it now with Laurent Freidel. He has a couple of papers out that *hint* at the main ideas, but you'll have to wait to see what they're up to now: 7) Artem Starodubtsev, Topological excitations around the vacuum of quantum gravity I: The symmetries of the vacuum, available as hep-th/0306135. Artem Starodubtsev and Lee Smolin, General relativity with a topological phase: an action principle, available as hep-th/0311163. The basic idea is to treat 4d general relativity with positive cosmological constant as a perturbation of a topological quantum field theory. The topological theory has a single state, which corresponds to a quantum version of deSitter space: an exponentially expanding universe similar to the one we see today, but with no matter. To calculate in full-fledged gravity, we then use perturbation theory, getting answers as power series in a coupling constant. But the cool part is that unlike ordinary perturbative quantum gravity this perturbation theory is manifestly diffeomorphism invariant term by term. And each term is a sum over spin foams! Even better, the coupling constant in this theory is the cosmological constant in Planck units! That's an incredibly small dimensionless number: about 10^{-123}. Physicists like perturbation theory when the coupling constant is small, since then the first few terms tend to give reasonably accurate answers - even if the whole series diverges. For example, quantum electrodynamics gives high-precision answers because the fine structure constant is about 1/137, which is pretty small. But 10^{-123} is *really* small. I'd seen Starodubtsev talk about this in Marseille (see week206) but now he and Freidel have done calculations recovering Newton's law of gravity in an appropriate approximation from this theory. That may not seem like a big deal, but it's actually very cool to see Newton's law reemerge from a manifestly diffeomorphism-invariant theory of quantum gravity: no model had ever managed this feat before. For those of you hungering for technical details, I'll just say that the topological theory in question is BF theory with the symmetry group of deSitter spacetime, namely SO(4,1), as the gauge group. General relativity can be regarded as a perturbation of this BF theory by borrowing some ideas from the MacDowell-Mansouri formulation of general relativity. If you haven't heard of that, well, neither had I. It's a sort of old idea: 8) S. W. MacDowell and F. Mansouri, Unified geometric theory of gravity and supergravity, Phys. Rev. Lett. 38 (1977), 739-742. ... but here we aren't using anything anything about supergravity, just the fact that ordinary general relativity can be treated as a theory with gauge group SO(4,1) and a Lagrangian that breaks this symmetry down to the Lorentz group SO(3,1). The paper by Smolin and Starodubtsev listed above describes the details, but in the case of going from SO(5) down to SO(4). When we quantize BF theory in 4 dimensions we get a spin foam model called the Crane-Yetter model, where the spin foams are defined using the representation theory of a quantum group: a q-deformed version of the original gauge group. So, the real engine behind Freidel and Starodubtsev's calculations are spin foams involving the q-deformed version of SO(4,1), called SO_q(4,1). This is technically tricky because SO(4,1) is noncompact, and noncompact quantum groups are just beginning to be understood. So, there's probably still tons of mathematical work left to be done. But, the upshot is that Freidel and Starodubtsev calculate stuff as power series in the cosmological constant where each term is computed using SO_q(4,1) spin foams. It's sort of like a souped-up Feynman diagram expansion, but with spin foams replacing Feynman diagrams. Now that I've thrown around enough buzzwords to scare off the kids, I can tell you about Lee Smolin's talk, which was definitely X-rated: for adults only, people who can listen to speculations with just the right mixture of disbelief and open-mindedness. It was the last talk in the conference. And it was about the possible physical effects of spin networks with hidden edges! ahead and suggested that hidden edges can cause nonlocal effects in physics, like the force of gravity decaying more slowly than 1/r^2 - just as it does in MOND, the wacky but strangely accurate explanation of galactic rotation curves that uses a modification of Newtonian gravity instead of positing dark matter! (See week206 for more on MOND.) It's hard to make up sensible theories of forces that decay more slowly than 1/r^2, but nonlocal interactions would be one way to do it... and hidden spin network edges might cause those. There are a million things that could go wrong with this idea, but I like it, because it suggests a way quantum gravity might try to explain one of the big mysteries of physics - dark matter. And until we get our theories to make contact with experiment, it'll be very hard for us to tell if they're on the right track. Anyway, Smolin hasn't come out with a paper on this stuff yet, so we'll have to wait for more details. By the way: In what I've written this week, I've had to seriously downplay the cool math involved, to give (I hope) some inkling of the cool physics. Krasnov work on 2+1-dimensional Lorentzian gravity with positive cosmological constant uses the fact that the phase space of this theory is closely related to Teichmueller space - the space of complex structures mod diffeomorphisms that are connected to the identity. I talked about this space in week28, but I forgot to say that we can think of it as a space of flat SO(2,1) connections mod gauge transformations. Here SO(2,1) is just the Lorentz group in 3 dimensions. So, if we quantize 2+1 Lorentzian gravity with positive cosmological constant, we get a theory where states are described by SO_q(2,1) spin networks... but this is also a theory of quantum Teichmueller space. Again this is tricky because SO(2,1) is noncompact, of work started by Kashaev: 9) R. M. Kashaev, Quantization of Teichmueller spaces and the quantum dilogarithm, available as q-alg/9705021. 10) L. Chekhov and V. V. Fock, Quantum Teichmueller space, Theor. Math. Phys. 120 (1999) 1245-1259, also available as math.QA/9908165. You can get a sense of who's working on this stuff and what they're doing by looking at the references for this recent conference on 3d quantum gravity in Edinburgh, which unfortunately took place when I was in Hong Kong: 11) Workshop on physics and geometry of 3-dimensional quantum gravity, http://www.ma.hw.ac.uk/~bernd/references.html I should also add that people don't usually don't talk about the 3d Lorentz group SO(2,1) here; they talk about its double cover SL(2,R). Anyway, I'll quit here. The next conference on loops and spin foams will probably happen in Berlin at the Albert Einstein Institute in 2005, which happens to be the hundredth birthday of special relativity. I hope we can make a lot of progress before then and make Al proud. Quote of the week: The best way to have a good idea is to have a lot of ideas. - Linus Pauling (Not necessarily true, but worth keeping in mind.) ----------------------------------------------------------------------- mathematics and physics, as well as some of my research papers, can be obtained at http://math.ucr.edu/home/baez/ For a table of contents of all the issues of This Week's Finds, try http://math.ucr.edu/home/baez/twf.html A simple jumping-off point to the old issues is available at http://math.ucr.edu/home/baez/twfshort.html If you just want the latest issue, go to http://math.ucr.edu/home/baez/this.week.html === Subject: Natural Density Suppose we have two subsets A and B of the natural numbers, such that lim (N->00) #{n in A, n<=N}/N and lim (N->00) #{n in B, n<=N}/N exist. Does the same necessarily hold for the intersection of A and B? *-----------------------* www.GroupSrv.com *-----------------------* === Subject: re:Natural Density Obviously not, my bad... *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Natural Density > Suppose we have two subsets A and B of the natural numbers, such that > lim (N->00) #{n in A, n<=N}/N and > lim (N->00) #{n in B, n<=N}/N > exist. > Does the same necessarily hold for the intersection of A and B? No. Can you find an example, before one is provided for you? === Subject: Re: Functional analysis: space of bounded varation separable? > my course in functional analysis is so long ago. > So, here my question: > Is it true that the space of real valued functions on R of bounded > variation is separable? With what norm? Variation norm? Then the answer is no: the set of functions f_t, where f is zero up to t and 1 above t, is an uncountable set such that any two of them are far apart. === Subject: Re: Functional analysis: space of bounded varation separable? >> my course in functional analysis is so long ago. >> So, here my question: >> Is it true that the space of real valued functions on R of bounded >> variation is separable? >With what norm? Variation norm? Then the answer is no: >the set of functions f_t, where f is zero up to t and 1 above t, >is an uncountable set such that any two of them are far apart. There is no problem giving topologies in which it is separable. -- 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: Re: Cantor's proof that #(Evens) = #(Naturals) is inconsistent > David Ferguson, PAY ATTENTION to the following two paragraphs since you > have not understood the points that are made in those paragraphs. > > As I have commented before, you seem to have the idea that if two sets > A and B are constructed in different manners, then they are distinct sets, > First of all E and E* do not have the same elements. The set E* has > twice as msnay elements as does set E As a reminder, E = {m in N : m is even} and E* = {2n : n in N}. Theorem: E* is a subset of E. Proof: Suppose that m is an element of E*, then m = 2n for some natural number, n. As the product of two natural numbers (recall that 2 is a natural number), m is a natural number. As m is divisible by 2, then m is even. So m is an even natural number, and so an element of E. Since m is an arbitrary element of E*, then every element of E* is an element of E, and hence E* is a subset of E. Q.E.D. Since you do not believe that E* is a subset of E, then you must believe that there is a mistake in the above proof. So tell me where the mistake is, and exactly what the mistake is (be explicit about the nature of the error). This means that you CANNOT use your E* has twice as many elements as E argument. Your response must deal only ELSE. > Set E can be associated with the set of points on the number line: > 0, 2, 4, 6 ... > Set E* can be associated with the set of points on the number line: > 0, 1, 2, 3, 4, 5, 6 ... > There is no doubt whatsoever that there are two elements of set E* for > every element of E. No, there isn't. And if you had bothered paying attention to what I elements as E. E* also has 1034966 times as many elements as E. E has twice as many elements as E*. E has 1,000,000 times as many elements as E*. E and E* have exactly the same number of elements (hardly surprising since they are equal sets). The rules for finite sets do not all work for infinite sets. Any proof you have seen of what you are trying to use will rely on the finiteness of the sets involved. > I am impressed by your apparent conviction. You, however, are in the > position one would be in if one were defending the earth centered > solar sytem versus the sun centered solar system. Your crytal spheres > rotating in crystal spheres is outdated and does not describe reality. You appear to be looking in the mirror, since you are describing yourself. > The fact is and allways shall be that a set which has two elements for > each element of another set has twice as many elements as the smaller > set. I agree with this statement, except for your completely unjustified use of the word smaller. Replace the ridiculous use of the word smaller by use of the word other, and your statement would be 100% correct. But your statement would NOT imply that the two sets do not have the same number of elements, which is the additional unjustified conclusion which you are trying to force on us. > The best thing I can do here is print out your comments and decide if > they are worth a reply here or if you should read my forthcomming book > Sets and Numbers A book that will evidently be beset by logical errors throughout from its first page to its last. David ----- === Subject: Reflexive spaces In a book I read: Let A and B be two normed vector spaces that are isometrically isomorphic. Then A is reflexive => B is reflexive. In my opinion, a continuous isomorphism between A and B should suffice for the result to hold. Am I right ? Thks, -- Julien Santini === Subject: Re: Reflexive spaces >In a book I read: Let A and B be two normed vector spaces that are >isometrically isomorphic. Then A is reflexive => B is reflexive. >In my opinion, a continuous isomorphism between A and B should suffice for >the result to hold. Am I right ? Yes, of course. Reflexivity is invariant under isomorphism, not just isometry. I don't know why the book would include the word isometrically. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Homeomorphic non-isomorphic normed vector spaces My question is in the topic: does there exist any two normed vector spaces that are homeomorphic but not isomorphic ? -the example should be infinite dimensional since R^n is not homeomorphic to R^m whenever m<>n; the hypothesis of a norm is essential since the question is trivial for a topological vector space-. -- Julien Santini === Subject: Re: Homeomorphic non-isomorphic normed vector spaces >My question is in the topic: does there exist any two normed vector spaces >that are homeomorphic but not isomorphic ? Yes. Amazingly enough, all separable infinite-dimensional Banach spaces are homeomorphic. See M.I. Kadec, Function Anal. App. 1 (1967) 53-62. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Homeomorphic non-isomorphic normed vector spaces Originator: grubb@lola >My question is in the topic: does there exist any two normed vector spaces >that are homeomorphic but not isomorphic ? -the example should be infinite >dimensional since R^n is not homeomorphic to R^m whenever m<>n; the >hypothesis of a norm is essential since the question is trivial for a >topological vector space-. You're going too quickly. There are norms on R^2 that give the same topology but are not isometric. For example, the usual Euclidean norm and the sup norm. --Dan Grubb === Subject: Re: Homeomorphic non-isomorphic normed vector spaces You're going too quickly. There are norms on R^2 that give the same topology but are not isometric. Isometric, okay, but R^2 is always isomorphic to R^2. (My initial question was: does there exist any two normed vector spaces that are homeomorphic but not isomorphic ?) - Julien Santini === Subject: Re: The Nature of Mathematics - a plea for help >What troubles the physicists is the appearance of magic. I'm just >magically pulling definitions out of the air. Are these steps--to >demonstrate that we may arrive at a conclusion with fewer >assumptions than previously thought possible--mathematically >and logically admissible? > Bilge replied: >>I can do it even fewer steps by fiat. Watch: >>gamma = 1/sqrt(1-(v/c)^2) >>Presto. I've derived the result using your personal method of >>induction, with zero assumptions and a 1 line proof. > Bilge, > You have identified yourself as a physicist. You call > 1/sqrt(1-(v/c)^2) a result. You believe that function is pregnant > with physical meaning. You're certain that the function labeled > gamma contains a statement about physics. > Unless you are posting a retraction, you are missing my point. > I am asking a mathematical question. Do I have a legitimate > derivation? Are my steps mathematically and logically > admissible? The very step that Bilge outlined clearly for you, is inadmissible. A premise isn't a conclusion. It's called petitio principii. petitio principii (pe-t.94shÇ.90-oÇ pr.94n-s.94pÇ.90-.90Ç, -.90-.93Ç) noun Logic. The fallacy of assuming in the premise of an argument that which one wishes to prove in the conclusion; a begging of the question. [Medieval Latin pet.93tio principi.93 : Latin pet.93tio, request + Latin principi.93, genitive of principium, beginning.] Excerpted from The American Heritage¬ Dictionary of the English Language, Third Edition © 1996 by Houghton Mifflin Company. Electronic version licensed from INSO Corporation; further reproduction and distribution in accordance with the Copyright Law of the United States. All rights reserved. Richard Perry === Subject: Re: The Nature of Mathematics - a plea for help > I have a trivial question concerning mathematics and logic and I > would like to receive a resounding answer that is both unanimous > and clear from practicing mathematicians and logicians. The cross- > posting is for the benefit of the physicists who have read my paper > http://www.everythingimportant.org/relativity/special.pdf and are > stumbling over its elementary logic. Here's the logic that the > physicists find troubling. Suppose you begin a derivation by assigning > a clear and undisputed physical meaning to real parameters x, x', T, > claim it's legitimate to pick a real number 1/c and then define a new > quantity v by the equation u=v/sqrt(1 -v^2/c^2). The point is to then > write everything in the ensuing derivation in terms of v and not u. > I also believe it's legitimate to define, without any justification > whatsoever, the function gamma(v) = 1/sqrt(1 -v^2/c^2). It appears that you are making a definition. That's perfectly ok. Next you have to defend any interpretations you may put on that definition. I suspect they have an objections someplace else. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: The Nature of Mathematics - a plea for help Suppose you begin a derivation by assigning > a clear and undisputed physical meaning to real parameters x, x', T, > claim it's legitimate to pick a real number 1/c and then define a new > quantity v by the equation u=v/sqrt(1 - v^2/c^2). The point is to then > write everything in the ensuing derivation in terms of v and not u. > I also believe it's legitimate to define, without any justification > whatsoever, the function gamma(v) = 1/sqrt(1 - v^2/c^2). There is no requirement to give a physical meaning to any of the above, though of course that would be easy to do. You've defined u as a function of v, for real v with |v| < c. You can certainly do that if you choose, and you can also invert the function so that v = u/sqrt(1 + u^2/c^2), where u can be any real number. u and v are related by a plane curve of degree four and genus 0, which makes things nice and simple. You are also free to define the function gamma, and you'd not be the first one to do so. You might also consider the transformation r = c arctanh(v/c), rapidity, which has nice additivity properties. None of this is physics as yet; I'll leave criticism of your paper to those who have read it. However, the math involved in special relativity is not unduly difficult. Minkowski made the key observation that t^2 - (x^2 + y^2 + z^2)/c^2 is a quadratic form giving the geometric structure of what is now called Minkowski space, and it all flows from that. This happened back in 1907. We're coming up on the 100th aniversary of special relativity, with Einstein's Zur Elektrodynamik bewegter K.9arper from his magical year of 1905 and closing, more or less, with Minkowski's address to the Assembly of German Natural Scientists and Physicians on Sept 21, 1908. === Subject: Re: The Nature of Mathematics - a plea for help > I have a trivial question concerning mathematics and logic and I > would like to receive a resounding answer that is both unanimous > and clear from practicing mathematicians and logicians. The idiot squeaks. > The cross- > posting is for the benefit of the physicists Bull. [snip crap] > What troubles the physicists is the appearance of magic. [snip more crap] http://www.apa.org/journals/psp/psp7761121.html http://insti.physics.sunysb.edu/~siegel/quack.html -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: The Nature of Mathematics - a plea for help > Your ignorance is showing, Eugene. Nothing to do with religion of > any sort. There has NEVER been a prediction of Special Relativity > that was contradicted by an observation. NEVER! And millions of specific experiments have been done and hundreds of million of implicit experiments have been done. Every time a high supposed to behave, that is yet another corroberation of STR. Every electronic device that works as it should supports QED which is, in part, based on STR. Physics is empirical right down to the basement level. The final and true judgement on a physical theory is this: does it predict correctly. That is it! There is nothing else! Bob Kolker === Subject: FLT Fermat's Last Theorem Ben Ito 11-8-04 I will prove Fermat's Last Theorem. l. Introduction I will show that Fermat's n=4 proof is invalid and prove Fermat's n>2 theorem. 2. Fermat's Proof (n=4) Fermat is using the integer solutions of n=2 to prove that n=4 does not form integer solutions. Fermat uses, A^4 + B^4 = C^2 (equ l). and A^2 = 2uv, B^2 = u^2 - v^2, and C = u^2 + v^2 (equ 2), (Shanks, p.141). Fermat's n=4 proof forms integer solutions. Example, using u=2 and v=l, A^2 = 4, B^2=3 and C= 5 (equ 3). Inserting the results of equation 3 in equation l 4^2 + 3^2 = 5^2. (equ 3) Fermat's n=4 proof is invalid. 3. Elliptic Curve The derivation of the elliptic curve is described. The elliptic curve equation is derived using the integer solution equations of n=2 (Osserman, p.21), a = k(m^2 - n^2), b = k2mn, c = k(m^2 + n^2). (equ 4) The elliptic curve is only valid for n=2; therefore, an elliptic curve can not be used to prove FLT for n>2. 4. Fermat's Proof. A circle transformation (z = c) is used in Fermat's equation n=2, x^2 + y^2 = c^2. (equ 5) The integer solution equations of n=2, x = 2uv, y = u^2 - v^2, and z = u^2 + v^2 (equ 6a,b,c) are used in equation 5, (2uv)^2 + (u^2 -v^2)^2 = u^2 + v^2 = c.(equ 7) Consequently, equations 6c and 7 forms the transformation equation z=c which can only be derived when n=2; therefore, only n=2 forms integer solutions. 5. Conclusion Fermat's n=4 proof forms integer solutions; therefore, Fermat's n=4 proof is invalid. Fermat's elliptic curves are derived using the integer solution equations of n=2; therefore, an elliptic curve can not be used to prove FLT when n>2. The n=2 equation is transformed into a circle equation then the integer solutions of n=2 are used to derive the transformation equation z=c which can only be formed when n=2. 6. References Robert Osserman. Fermat's Last Theorem (a supplement to the video). MSRI. 1994 Daniel Shanks. Solved and Unsolved Problems in Number Theory. Chelsea Pub. 1985. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: FLT I will NOT prove Fermat's Last Theorem www.microscitech.com Richard Miller > Fermat's Last Theorem > Ben Ito > 11-8-04 > I will prove Fermat's Last Theorem. > l. Introduction > I will show that Fermat's n=4 proof is invalid and prove Fermat's > n>2 theorem. > 2. Fermat's Proof (n=4) > Fermat is using the integer solutions of n=2 to prove that n=4 does > not form integer solutions. Fermat uses, > A^4 + B^4 = C^2 (equ l). > and > A^2 = 2uv, B^2 = u^2 - v^2, and C = u^2 + v^2 (equ 2), > (Shanks, p.141). Fermat's n=4 proof forms integer solutions. Example, > using u=2 and v=l, > A^2 = 4, B^2=3 and C= 5 (equ 3). > Inserting the results of equation 3 in equation l > 4^2 + 3^2 = 5^2. (equ 3) > Fermat's n=4 proof is invalid. > 3. Elliptic Curve > The derivation of the elliptic curve is described. The elliptic curve > equation is derived using the integer solution equations of n=2 > (Osserman, p.21), > a = k(m^2 - n^2), b = k2mn, c = k(m^2 + n^2). (equ 4) > The elliptic curve is only valid for n=2; therefore, an elliptic curve > can not be used to prove FLT for n>2. > 4. Fermat's Proof. > A circle transformation (z = c) is used in Fermat's equation n=2, > x^2 + y^2 = c^2. (equ 5) > The integer solution equations of n=2, > x = 2uv, y = u^2 - v^2, and z = u^2 + v^2 (equ 6a,b,c) > are used in equation 5, > (2uv)^2 + (u^2 -v^2)^2 = u^2 + v^2 = c.(equ 7) > Consequently, equations 6c and 7 forms the transformation equation z=c > which can only be derived when n=2; therefore, only n=2 forms integer > solutions. > 5. Conclusion > Fermat's n=4 proof forms integer solutions; therefore, Fermat's n=4 > proof is invalid. > Fermat's elliptic curves are derived using the integer solution > equations of n=2; therefore, an elliptic curve can not be used to > prove FLT when n>2. > The n=2 equation is transformed into a circle equation then the > integer solutions of n=2 are used to derive the transformation > equation z=c which can only be formed when n=2. > 6. References > Robert Osserman. Fermat's Last Theorem (a supplement to the video). > MSRI. 1994 > Daniel Shanks. Solved and Unsolved Problems in Number Theory. Chelsea > Pub. 1985. > *-----------------------* > www.GroupSrv.com > *-----------------------* === Subject: question about a contour integral I was reading a paper on approximation theory and Diophantine equations in which the following integral plays a key role: I_n= 1/(2*pi*i) * INT_C{ (1+xz)^{n+1/3}/(z^2-1)^{n+1} dz}, where n is a positive integer and C is a simple closed counter-clockwise curve enclosing both the points 1 and -1. This integral is well-defined if C does not enclose the point z = -1/x . Unfortunately, contour integration is not one of my strong points, so I humbly ask: What happens if C does enclose the point z = -1/x? Why is the integral not well-defined in that case? === Subject: Re: question about a contour integral > I was reading a paper on approximation theory and Diophantine equations in >which the following integral plays a key role: >I_n= 1/(2*pi*i) * INT_C{ (1+xz)^{n+1/3}/(z^2-1)^{n+1} dz}, >where n is a positive integer and C is a simple closed counter-clockwise >curve enclosing both the points 1 and -1. >This integral is well-defined if C does not enclose the point z = -1/x . >Unfortunately, contour integration is not one of my strong points, so >I humbly ask: >What happens if C does enclose the point z = -1/x? >Why is the integral not well-defined in that case? When you write (1+xz)^(n+1/3), you're taking a fractional power of a complex number. That's a multivalued function. If C doesn't enclose -1/x, you can choose a branch of the function that is analytic in a neighbourhood of C and the region it encloses, and then you can use residues to evaluate the integral. If C does enclose -1/x, whatever branch of the function you use there will be a branch cut on which the function is discontinuous. The result is an integral which 1) can't be evaluated using the residue theorem 2) may depend on the location of the branch cut and where it intersects the path C. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Countable Choice >Can anyone give a brief account on what can be done with DC which >cannot be done with AC_N or give a reference (preferably on the net) >on the subject? I suggest you read the book by Paul Howard and Jean E. Rubin, _Consequences of the Axiom of Choice_. -- 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: Re: Countable Choice >> Anyway if you can please address my ordered sets question: >> Does the countable AC garantee that every partially ordered set >> without maximal elements contains an increasing sequence. >Why not, as long as set isn't empty. >Thus some x0 in S. As x0 not maximal, some x1 with x0 < x1, >etc., making countably many choices for an increasing sequence. This is not the countable axiom of choice, but the principle of dependent choices, considerably stronger. With the countable axiom of choice, it is necessary to have the countable list of sets form which one has to choose first. -- 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: Re: Countable Choice >By a k-tuple he means a subset with k elements. >The union of countably many finite sets is countable which can be >shown by an inductive construction. Not always true. The union of a well-ordered family of explicitly well-ordered sets is well-ordered. -- 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: Re: Countable Choice >>Assuming the countable axiom of choice, how to prove that every >>infinite set has a countable subset? >> Let X be the given set, and let Y_k be the set of all >> ordered k-tuples of elements of X. Let u_k be the chosen >> element from Y_k; then there is a natural equivalence >> between the integers and the union of the u_k. >Would you elaborate? >Each u_k is an ordered k-tuple in Y_k = X^k. >What is for example, u_k / u_(k+1) ? >What are the elements of the union of two different size tuples? >Now from the union of all u_k, how is there a countable subset of X? >What for example, if some x in X with for all k in N, u_k = { x }^k? Here are two proofs. One is easy, but obscures the power of what can be done. Define x_k, w_k recursively as follows. w_0 is the empty set. For k >= 1, x_k is the first element of u_k not in w_{k-1}; w_k = w_{k-1} union {x_k}: Then the x's form an explicit infinite sequence, as u_k has k elements, and w_{k-1} has only k-1. Another method is to form H = U u_k, and then order this set as follows: If x, y in H, let i be the first k with x in u_k, and j be the first k with x in u_k. Then x <= y if i < j or (i = j and x <=_i y), where <=_i is the ordering of u_i. It is easy to show that H so ordered is order-isomorphic to the set of positive integers. -- 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: re:Countable Choice > You should really either start using software that posts > replies correctly (or start using it correctly, making replies > to the post you're replying to instead of always replying to > the top-level post) or include some quoted text like everyone > else, or both. There's really no way to tell what the heck > you're talking about here without reading the entire thread > and then guessing which point you're talking about. I am using the groupsrv.com site and it has many problems (many threads come up messed up), particularily with the quote function. If you know a way around it please share. Otherwise all I can offer is my apology for confusing you and a promise to do my best to make myself clear in the future. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Countable Choice >> You should really either start using software that posts >> replies correctly (or start using it correctly, making replies >> to the post you're replying to instead of always replying to >> the top-level post) or include some quoted text like everyone >> else, or both. There's really no way to tell what the heck >> you're talking about here without reading the entire thread >> and then guessing which point you're talking about. >I am using the groupsrv.com site and it has many problems (many >threads come up messed up), particularily with the quote function. If >you know a way around it please share. Well, one way or another you managed to quote the text you were replying to in _this_ post, making things much more intelligible. There are a lot of free alternatives to this groupsrv.com thing, you know. A lot of people post through Google Groups, which is far from ideal but seems to work better than what you're using. There's also free news readers out there (for example Agent on Windows) and even free news servers that you could point a news reader to (for example gives 250,000 hits.) >Otherwise all I can offer is my apology for confusing you and a >promise to do my best to make myself clear in the future. >*-----------------------* > www.GroupSrv.com >*-----------------------* ************************ David C. Ullrich === Subject: Re: Countable Choice >I understand your point, David, although of course choosing just >k-tuples is insufficient, they have to be k-tuples with distinct >members. You should really either start using software that posts replies correctly (or start using it correctly, making replies to the post you're replying to instead of always replying to the top-level post) or include some quoted text like everyone else, or both. There's really no way to tell what the heck you're talking about here without reading the entire thread and then guessing which point you're talking about. >Probably it would be best to speak about injections from [1, ..., k] >to the given set. >Anyway if you can please address my ordered sets question: >Does the countable AC garantee that every partially ordered set >without maximal elements contains an increasing sequence. Well I don't know, sorry. >*-----------------------* > www.GroupSrv.com >*-----------------------* ************************ David C. Ullrich === Subject: question about a contour integral Oops: Slight correction to my last post: The last statement should read: This integral is well-defined for |x| < 1 if we take C so that it does not enclose the point z = -1/x. Sorry! Ray Steiner === Subject: Linear system of equations and inequations I have a mixed system of N linear equations and linear inequations (a system in which some are equations, some inequations, N in total) with N unknowns. The inequations have the additional constraint that Xi >= 0. Is there an iterative method for solving this system? If so, where can I find it (book / web)? Cristian === Subject: Re: Linear system of equations and inequations > I have a mixed system of N linear equations and linear inequations (a system > in which some are equations, some inequations, N in total) with N unknowns. > The inequations have the additional constraint that Xi >= 0. > Is there an iterative method for solving this system? If so, where can I > find it (book / web)? > Cristian This sounds a lot like a Linear Programming problem. Except you don't have an objective funtion that needs to be minimized. I wonder if you can think up a fake obj function and apply the simplex method. T === Subject: Re: Interesting Question >>No. Take S_t as the circle with center t/2 and radius t. >And how is the union of those for 0 <= t < 1 different from the open disk >with center 1/2 and radius 1? Pedantic detail, but ... The union includes the origin; the disk does not. Mike Guy === Subject: Re: Interesting Question >Pedantic detail, but ... >The union includes the origin; the disk does not. Oops! Wrong again - I was looking at the wrong picture. Mike Guy === Subject: Re: New countable infiniity logic Let me start by saying that I appreciate your critical analyses of my post. Most of your criticisms (which you would or do call my mistakes) seem to arise from the concept of the number of digits associated with the list. I was trying to draw conclusions (using this concept) that you seem to know to be wrong. Since I am not a mathematician, I accept that there is a high probability you are correct, yet I am trying to understand where my logic relating to the number of digits associated with the list is wrong. > Nowhere in the text that you quote does he use the word > infinitesimal. You are correct. I was trying in one word to convey the meaning I ... 0.999... and its difference from 1 is less than *every* positive epsilon, all at once. If such use of the word infintesimal initially reflected poorly on Mike Oliver to a nondiscerning reader, unlike yourself, I extend my apologies to Mike and thank you for pointing out what was my error. >It seems to me you do not accept that the infinite set of naturals >(or the set of naturals of infinte extent) must produce a list >infinitely long in extent for this function. > What gives you that idea? Here, what I stated was clearly incorrect. With regard to the issue of digits, let me explain why I believe it is pertinent to this post and then you can explain why I am being inconsistent or mathematically incorrect. Hopefully, I will be able to understand. I will again introduce the function that is described by: Convert the natural number directly to its natural decimalic expansion by placing a decimal point in front of the number; unless it ends with a zero, in which case all the consecutive ending zero digits are moved from the end of the number to become placholders behind (to the right of) the decimal point and in front of the remaining part of the number. 1 becomes 0.1 3 becomes 0.3 9 becomes 0.9 10 becomes 0.01 11 becomes 0.11 12 becomes 0.12 19 becomes 0.19 30 becomes 0.03 31 becomes 0.31 89 becomes 0.89 90 becomes 0.09 91 becomes 0.91 99 becomes 0.99 100 becomes 0.001 101 becomes 0.101 109 becomes 0.109 110 becomes 0.011 111 becomes 0.111 123 becomes 0.123 300 becomes 0.003 314 becomes 0.314 899 becomes 0.899 900 becomes 0.009 901 becomes 0.901 999 becomes 0.999 : becomes : n, becomes 0.X1X2X3...Xn : becomes : where n grows without bound, n is infinte in extent, the number of naturals approaches infinity, or there are an infinite number of positive integers (naturals). These all seem to me to convey the same concept. For a specified finite integer, the output or range of this function produces a finite decimalic expansion (terminating rational). No repeating rationals or irrational numbers seem to be present on this list when the set itself is finite. Does this mean that none are present when the set is infinite and why? The positive integers are in a one to one correspondence with this set that is known to contain *at least* all of the numbers represented by the terminating rationals. This list from the function (which I believe is also a set) is defined and can now be examined for its properties. One property of this list or set is associated with the number of digits d. For each power of ten of the naturals, an extra digit d is added to the terminating rationals. Thus when n = 1-9, there is 1 digit, when n = 10-99, there are 2 digits, when n = 100-999, there are 3 digits, when n = 1000-9999, there are 4 digits, : : when n grows without bound, the number of digits d grow without bound, when n is infinite in extent, the number of digits d is infinite in extent, when n approaches infinity, the number of digits d approaches infinity, when there are an infinite number of positive integers n => omega, there must be an infinite number of digits d or the number of digits must approach infinity. The decimalic expansion representation for *any arbitrary* number whose number of digits grows without bound, are infinte in extent, or approaches infinity, (or for which there are an infinite number of digits) is given by X1X2X3 to Xn...æ where each Xn represents the value of the nth digit (approaching infinity) and the *to* may allow concise identification of some particular number. For example the particular number 0.333..., this corresponds to the sequence from the following numbers pulled from the list. 3 produces 0.3 33 produces 0.33 333 produces 0.333 : : n approaches infinity using only threes for this particular n, produces 0.333... where the number of threes (total number of digits) approaches infinity. Similar logic for the decimalic expansion of Pi/100 or any other irrational number. How does representation 0.333... from the function (infinite set), where the total number of 3s in the decimalic expansion approaches infinity, differ from 0.333... = 1/3? Are not the two representations the same? If not, how do they differ? If so, do they represent different numbers? Please explain what I am missing. Did I make a mistake in representation of an arbitrary decimalic expansion whose digits approach infinity? > SPECIFIC FUNCTION *YOU* DEFINED DOES NOT GENERATE DECIMAL FRACTIONS > WITH AN INFINITE NUMBER OF NONZERO DIGITS. If the function creates an infinite set, is that set precluded from having an infinite number of elements or a number of elements that approaches infinity, some of which (an infinite number) have an infinite number of digits or at least whose number of digits approaches infinity? If so, please explain how the number digits is precluded from being infinite or approaching infinity for such an infinite set? >Note: because the only way the list for this specified function >(where a new digit is added for each succesive power of ten >numbers inserted into the function) can be infinite in extent is >if the number of digits themselves are infinite in extent. > As I said, you're confusing the number of entries in a list with what > is in one entry. Each entry on the list that *YOU* defined has only a > finite number of nonzero digits. The fact that there are an infinite > number of entries doesn't change that. However, when considering the set as a whole (in its entirety) corresponding to the range, based on the one to one correspondence with the natrurals or the induction method to an infinite set for the naturals, I believe one must conclude that it is also an infinite set. Is not a property of an infinite set that the number of elements (members) is infinite or at least approaches infinity. If the number of elements is infinite, approaches infinity, grows without bound, then the number of digits associated with the elements in the power of ten (associated with n growing without bound or n approaching infinity) is also infinite or approaches infinity. If the number of digits is not infinite, then the number of digits terminates and is of a finite number, turning this infinite set of outputs into a finite set. Am I misunderstanding the concept of an infinite set? I probably made an error somewhere. Hopefully, not many again!! Don Whitehurst === Subject: Re: New countable infiniity logic :> As I said, you're confusing the number of entries in a list with what :> is in one entry. Each entry on the list that *YOU* defined has only a :> finite number of nonzero digits. The fact that there are an infinite :> number of entries doesn't change that. : However, when considering the set as a whole (in its entirety) : corresponding to the range, based on the one to one correspondence : with the natrurals or the induction method to an infinite set for the : naturals, I believe one must conclude that it is also an infinite set. : Is not a property of an infinite set that the number of elements : (members) is infinite or at least approaches infinity. Loosely speaking, the number of elements in an infinite set is infinity. The number of elements in a set does not approach anything because the number of elements in a set is fixed. : If the number of elements is infinite, approaches infinity, grows : without bound, then the number of digits associated with the elements : in the power of ten (associated with n growing without bound or n : approaching infinity) is also infinite or approaches infinity. If the : number of digits is not infinite, then the number of digits terminates : and is of a finite number, turning this infinite set of outputs into a : finite set. : Am I misunderstanding the concept of an infinite set? You are confusing the set with its elements. Each natural number is finite. Each natural number can be represented by a finite number of digits. There are an infinite number of natural numbers, but each natural number is finite. For each natural number x there exists a number d such that x can be represented in decimal with d digits. However there does not exist a d such that for each natural number x, x can be represented in decimal with d digits. The fact that no such d exists does not change the fact that all natural numbers are finite. : I probably made an error somewhere. Hopefully, not many again!! All natural numbers are finite. There are an infinite number of finite natural numbers. That is what you need to understand. There is no relation between the size of a set and the size of the elements of the set. We can have a finite set of finite elements, a infinite set of finite elements, a finite set of infinite elements, or an infinite set of infinite elements. Stephen === Subject: where is error? I want to calculate I(u)=int(exp(-u*x^2/(1-x^2)),x=-1..1) I have found with Maple I(u)=sqrt(pi)*KummerU(1/2,0,u) I would like find again this result with differential equation of KummerU(a,b,z) function. KummerU(a,b,z) function is solution of equation: z*w''(z)+(b-z)*w'(z)-a*w(z)=0 So, I write I(u), I'(u), I''(u) and I introduce these functions in equation u*I''(u)-u*I'(u)-I(u)/2=0 (1) I(u)=int(exp(-u*x^2/(1-x^2)),x=-1..1) I'(u)=int(exp(...)*(-x^2)/(1-x^2),x=-1..1) I''(u)=int(exp(...)*(x^4)/(1-x^2)^2,x=-1..1) I obtain: int(exp(...)*[u*x^4/(1-x^2)^2+u*x^2/(1-x^2)-1/2],x=-1..1) =>int(exp(...)*[u*x^4/(1-x^2)^2+u*x^2/(1-x^2)-1/2],x=-1..1) =>int(exp(...)*[u*x^4+u*x^2*(1-x^2)-(1-x^2)^2/2]/(1-x^2)^2,x=-1..1) =>int(exp(...)*[u*x^4+u*x^2-u*x^4-1/2+x^2-x^4/2]/(1-x^2)^2,x=-1..1) =>int(exp(...)*[u*x^2-1/2+x^2-x^4/2]/(1-x^2)^2,x=-1..1) But, the expression [u*x^2-1/2+x^2-x^4/2]/(1-x^2)^2 would be equal to 0. isn't it? Where is error? === Subject: Re: where is error? >I want to calculate I(u)=int(exp(-u*x^2/(1-x^2)),x=-1..1) >I have found with Maple I(u)=sqrt(pi)*KummerU(1/2,0,u) (for u > 0) >I would like find again this result with differential equation of >KummerU(a,b,z) function. >KummerU(a,b,z) function is solution of equation: >z*w''(z)+(b-z)*w'(z)-a*w(z)=0 >So, I write I(u), I'(u), I''(u) and I introduce these functions in >equation u*I''(u)-u*I'(u)-I(u)/2=0 (1) >I(u)=int(exp(-u*x^2/(1-x^2)),x=-1..1) >I'(u)=int(exp(...)*(-x^2)/(1-x^2),x=-1..1) >I''(u)=int(exp(...)*(x^4)/(1-x^2)^2,x=-1..1) >I obtain: >int(exp(...)*[u*x^4/(1-x^2)^2+u*x^2/(1-x^2)-1/2],x=-1..1) >=>int(exp(...)*[u*x^4/(1-x^2)^2+u*x^2/(1-x^2)-1/2],x=-1..1) >=>int(exp(...)*[u*x^4+u*x^2*(1-x^2)-(1-x^2)^2/2]/(1-x^2)^2,x=-1..1) >=>int(exp(...)*[u*x^4+u*x^2-u*x^4-1/2+x^2-x^4/2]/(1-x^2)^2,x=-1..1) >=>int(exp(...)*[u*x^2-1/2+x^2-x^4/2]/(1-x^2)^2,x=-1..1) So far so good. >But, the expression [u*x^2-1/2+x^2-x^4/2]/(1-x^2)^2 would be equal to 0. No, it isn't. Hint: integrate by parts. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Fourier Transform of exp(-2*x^2/1-x^2), -1 Je would like to calculate Fourier Transform of f(x): > if -1 I have found that I(u)=int(exp(-u*x^2/(1-x^2)),x=-1..1) is equal to: > I(u)=sqrt(pi)*KummerU(1/2,0,u) > with KummerU(a,b,z)solution of z*w''(z)+(b-z)*w'(z)-a*w(z)=0 In fact, this function can be simplified: I(u)=u*exp(u/2)(K1(u/2)-K0(u/2)) with K0 and K1 Bessel functions. > I wanted to use this result to calculate Fourier transform but I don't > know how to do. > I don't know if it can be used. > any idea ? === Subject: Evolving the Weaire-Phelan Structure using Surface Evolver Hello! I am using the file phelanc.fe (that comes with Ken Brakke's surface evolver) to evolve the Weaire-Phelan Structure. The structure should start as Voronoi cells on centers 0 0 0 1 1 1 0.5 0 1 and so on. To my surprise all the vertices in phelanc.fe seem to be slightly shifted. For example one of the vertices in phelanc.fe should have the coordinates: 1 0 1.5 But what you find is: 0.999205 0.000988 1.500509 Can anyone explain this? Robert Bitsche === Subject: Lexicons In the mathworld.com entry under Real Number, there is a statement: Almost all real numbers are lexicons, meaning that they do not obey probability laws such as the law of large numbers (Gruber 1991; Calude and Could somebody please explain this? Barnaby === Subject: Re: hanging cable with discrete loads > I have a need to compute the height of loads (specifically, traffic > signals and signs) hung from a suspension cable. Googling found a lot > of catenary stuff, which I was already aware of, and some about > uniform loading, but (AFAICT) nothing helpful. > > I'd appreciate any clues. > > TIA, > George > See section 1.3 of M. Irvine, Cable Structures, Dover, 1981; > subsection Response to Many Point Loads Also from way back in the Dark Ages I seem to recall that Hildebrand (F.B.) in his 'Methods of Applied Math' covered this problem in some detail. Cal