mm-3749 === Subject: angle from 2-d drawing i would appreciate any help i can get here / if you have 2 lines that make an angle in 3-d. looking at the front view theres a line that appears to be vertical and a line going from the bottom of that vertical line to the right at a 45 deg angle / in right view, the line going at the 45 deg angle is now appearing to be vertical and the vertical line from the front is at a 45 deg angle, going off to the right.the deg of this 3-d angle is 60 deg. is it possible to figure out what the deg of the 3-d angle is from the two 2-d view's ? tks ww === Subject: Re: angle from 2-d drawing >i would appreciate any help i can get here / if you have 2 lines that >make an angle in 3-d. looking at the front view theres a line that >appears to be vertical and a line going from the bottom of that >vertical line to the right at a 45 deg angle / in right view, the line >going at the 45 deg angle is now appearing to be vertical and the >vertical line from the front is at a 45 deg angle, going off to the >right.the deg of this 3-d angle is 60 deg. is it possible to figure out >what the deg of the 3-d angle is from the two 2-d view's ? If you're sure that you're seeing the same line in 2 different views, you can figure out its direction as a 3-D vector. And if you have 2 different 3-D vectors, you can calculate their angle. Define the coordinate space as x=left/right, y=backward/forward, z=up/down. Let's call the point where the 2 lines intersect the origin (0,0,0). By looking at the first line in the front view, you know it goes from (0,?,0) to (0,?,h1). In the second view, you can see it goes from (?,0,0) to (?,h1,h1). Wave your hands in the air and declare the vector to be <0,h1,h1>. Similarly, the second line goes from (0,?,0) to (h2,?,h2) and from (?,0,0) to (?,0,h2), which means the vector is . Take the dot product of the 2 vectors and divide by the lengths to get the cosine of the angle between them. (0*h2 + h1*0 + h1*h2) / sqrt(h1^2+h1^2) / sqrt(h2^2+h2^2) = 1/2 arccos(1/2) = 60 --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: a rigid transformation problem given two point sets {pi} {qi}, determing the rotation matrix R and translation matrix T, so that the error E is minimized: E = square sum of d ( R*pi + T, qi) the distance is defined as follow: d ( p, q ) = || p - q|| if || p - q|| < M M otherwise where M is a constance. how to solve this problem? === Subject: Re: a rigid transformation problem >given two point sets {pi} {qi}, determing the rotation matrix R and >translation matrix T, so that the error E is minimized: >E = square sum of d ( R*pi + T, qi) Do you mean sum of the squares? >the distance is defined as follow: >d ( p, q ) = || p - q|| if || p - q|| < M > M otherwise >where M is a constance. That could be rather nasty, as it makes the objective function non-differentiable. Would a smooth cut-off such as d(p, q) = M tanh(||p - q||/M) work for your purpose? >how to solve this problem? Using a numeric solver. You have your choice of parametrizing the rotation matrix, e.g. in terms of Euler angles, or using constraints R' R = I and det(R) > 0. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: a rigid transformation problem schrieb Robert Israel : >>given two point sets {pi} {qi}, determing the rotation matrix R and >>translation matrix T, so that the error E is minimized: >>E = square sum of d ( R*pi + T, qi) > Do you mean sum of the squares? Let us suppose he does. >>the distance is defined as follow: >>d ( p, q ) = || p - q|| if || p - q|| < M >> M otherwise >>where M is a constance. > That could be rather nasty, as it makes the objective function > non-differentiable. But not badly non-differentiable - I personally would always prefer a piecewise linear function to a fully nonlinear function like tanh in this context. *In theory* you could solve this optimization problem as follows: Say the number of given pairs of points (pi,qi) is n. For each of the 2^n subsets S of the index set {1,2,...,n}, calculate (R,T) which minimize E = sum_S || R pi + T - qi ||^2 (i.e. with the subset and M set to infinity) - this is a standard problem, known as procrustes problem. For the soulution (R,T), evaluate E + M^2 * (n - #S). Then choose the index set S for which this value is minimal, and the corresponding solution (R,T). On the way you can discard certain choices of S: E.g., if the optimal solution (R,T) to the index set S satisfies || R pi + T - qi || < M for some index i not contained in S, then it will not be optimal. Also if M is large enough so that both point sets {pi | i=1..n} and {qi | i=1..n} can be covered by congruent figures of diameter <= M, then the optimal solution will occur for the full index set S = {1,2,...,n}. Concerning the solution of the procrustes problem: The optimal vector T will be of the form T = q_S - R p_S, where q_S and p_S denote the centroids of the points qi, resp. pi, and where R minimizes sum || R (pi-pS) - (qi -qS) ||^2. Let's assume that pS = qS = 0. To minimize sum || R pi - qi ||^2 = sum ||pi||^2 + ||qi||^2 - 2 (qi,R pi) = constant - 2 sum (pi' qi, R), where ' denotes the transposition, one has to determine the (rotation) matrix R for which the scalar product (M,R):= trace (M'*R) with the matrix M := sum pi'*qi becomes maximal. To this end, take an singular value decomposition of M, M = U' D V with orthogonal matrices U and V and diagonal matrix D - maybe better take such a decomposition for which the matrices U and V have positive determinants, and that the diagonal matrix D has as as few negative entries as possible, i.e. zero or one. Then the optimal matrix R will be of the form R = U' W V with W a rotation matrix that maximizes trace(D'*W). - If all entries d11, d22, ..., dnn of D are non-negative, then W = Id = diag(1,1,...,1), R = U' V will be an aoptimal solution. If one entry of D ist negative, w.l.o.g. if d11 < 0 <= d22 <= d33 <= ... <= dnn, then I would *guess* that as long as d11+d22 is non-negative, the identity matrix W = Id still achieves maximum, while for d11 < -d22 the matrix W = [ 0 1 0 0 ... ] [-1 0 0 0 ... ] [ 0 0 1 0 ... ] [ 0 0 0 1 ... ] [ . . . . ... ] might achieve the maximum. - Unfortunately, I cannot supply literary references at the moment. Sorry! > Would a smooth cut-off such as > d(p, q) = M tanh(||p - q||/M) work for your purpose? It might be better to take two different values for the two values M in this formula, maybe even different from the value of M in the OP's definition of distance, or might it not? >>how to solve this problem? > Using a numeric solver. > You have your choice of parametrizing the rotation matrix, e.g. > in terms of Euler angles, or using constraints R' R = I and > det(R) > 0. I wonder how good numeric solvers can cope with the fact that the gradient of tanh is so small at larger values. - Might there not be the possibility that the solver gets stuck in local maxima where (most of) the points (R pi + T) and qi are far apart from each other. Maybe you could provide an example. - That would be marvellous! Thomas Mautsch === Subject: Re: a rigid transformation problem >schrieb Robert Israel : >given two point sets {pi} {qi}, determing the rotation matrix R and >translation matrix T, so that the error E is minimized: >E = square sum of d ( R*pi + T, qi) >> Do you mean sum of the squares? >Let us suppose he does. >the distance is defined as follow: >d ( p, q ) = || p - q|| if || p - q|| < M > M otherwise >where M is a constance. >> That could be rather nasty, as it makes the objective function >> non-differentiable. >But not badly non-differentiable - I personally would always prefer >a piecewise linear function to a fully nonlinear function like tanh >in this context. >*In theory* you could solve this optimization problem as follows: >Say the number of given pairs of points (pi,qi) is n. >For each of the 2^n subsets S of the index set {1,2,...,n}, Not very practical, of course, unless n is rather small... ... >> Would a smooth cut-off such as >> d(p, q) = M tanh(||p - q||/M) work for your purpose? >It might be better to take two different values for the >two values M in this formula, >maybe even different from the value of M >in the OP's definition of distance, or might it not? >how to solve this problem? >> Using a numeric solver. >> You have your choice of parametrizing the rotation matrix, e.g. >> in terms of Euler angles, or using constraints R' R = I and >> det(R) > 0. >I wonder how good numeric solvers can cope >with the fact that the gradient of tanh is so small >at larger values. - Might there not be the possibility >that the solver gets stuck in local maxima where >(most of) the points (R pi + T) and qi are far apart >from each other. >Maybe you could provide an example. - >That would be marvellous! OK, here's an example in Maple 10. > with(LinearAlgebra): with(Optimization): # Here is a parametrization of a 3D rotation matrix # in terms of Euler angles a,b,c > Rabc:= Matrix([[ cos(c)*cos(b)*cos(a)-sin(c)*sin(a), -sin(c)*cos(b)*cos(a)-cos(c)*sin(a), sin(b)*cos(a) ], [ cos(c)*cos(b)*sin(a)+sin(c)*cos(a), -sin(c)*cos(b)*sin(a)+cos(c)*cos(a), sin(b)*sin(a) ], [ -cos(c)*sin(b), sin(c)*sin(b), cos(b) ]]); > V:= ; # Make up some data. Ten random vectors for the p_i > P:= [seq(RandomVector(3,generator=rand(-5..5)),i=1..10)]; [1] [-4] [ 0] [-5] [ 4] [-4] [ 5] [ 4] [2] [ 5] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] P := [[4], [ 5], [-1], [ 2], [ 5], [-2], [-3], [-4], [4], [-5]] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [0] [-2] [ 5] [-1] [-4] [ 2] [ 3] [ 5] [3] [ 0] # The true transformation has Euler angles 1, 2, 3 and translation # <1,2,3 Rt:= eval(Rabc,{a=1.0, b=2.0, c=3.0}); Tt:= <1.0,2.0,3.0>; Q:= map(v -> Rt.v+Tt, P); # An extra translation of <1,1,1> is added to the first three vectors in Q # so the cutoff will have something to do. > for i from 1 to 3 do Q[i]:= Q[i] + <1,1,1> od: # Now get the residuals d(p,q) = tanh ||p - q|| > E:= [seq(Rabc.P[i]+V-Q[i], i=1..10)]; ds:= map(v -> tanh(sqrt(v^%T . v)), E); # Minimize the sum of squares of residuals using the least-squares # solver > LSSolve(ds); [1.3090743731944, [a = 1.00382746183031957, c = 2.99466288305908712, b = 1.99673288295575734, v1 = 1.04762504047361582, v2 = 2.03025329876483252, v3 = 3.02999015731672605]] It was very quick, and the result is very close to the true transformation. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: a rigid transformation problem suppose the correspondence is known, i.e. pi corresponds to qi === Subject: Minimax quickie Trivial version - ABC is a line segment with a hinge at B, where another segment BD is attached. Maximize area(ADC). (sin @=sin(pi-@) is NOT required for proving :-) Somewhat more complicated - Insert another hinge E on BD with a segment EF. For easyness, let all hinges lie on the midpoint of the segment and let all segments have equal length (AB=BC=BE=ED=EF/2). -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn === Subject: Re: Minimax quickie > Trivial version - > ABC is a line segment with a hinge at B, where another > segment BD is attached. Maximize area(ADC). > (sin @=sin(pi-@) is NOT required for proving :-) Shure ... area = base * altitude is enough... > Somewhat more complicated - > Insert another hinge E on BD with a segment EF. > For easyness, let all hinges lie on the midpoint of > the segment and let all segments have equal length > (AB=BC=BE=ED=EF/2). And ? what are you looking for in the more complicated version ? if maximize Area(AFC), it is as trivial as the trivial version... (even without the easyness values). So may be your didn't say you want to maximize area (ACDF) ? That sounds more interesting ! -- philippe mail : chephip at free dot fr site : http://chephip.free.fr/ === Subject: Re: Minimax quickie > So may be your didn't say you want to maximize area (ACDF) ? > That sounds more interesting ! Sorry, of course ACDF (or even better, the closure of all points). (I already tried, the result is an ugly sextic with 4 real zeros.) -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn === Subject: Re: Minimax quickie >> So may be your didn't say you want to maximize area (ACDF) ? >> That sounds more interesting ! > Sorry, of course ACDF (or even better, the closure of all points). > (I already tried, the result is an ugly sextic with 4 > real zeros.) Nearly the same. I just found a cubic with one suitable zero... That is a cubic in cos(@) with the condition -1 <= cos(@) <= 0 this last condition restricting the 3 real solutions to only one. I found 36*cos^3(@) - 61*cos^2(@) + 16 = 0 that is @ = 117.048738987...deg, @ being angle ABD, then EF perpendicular to AD. maximum area = AB^2 * 3.947250201868925... The not easy values result into a similar cubic, with just different coefficients. B.T.W. ACDF *is* the closure, other configurations result into obviously smaller closure, or same closure by symetry (case AB = BC). If AB != BC, these different non symetric configurations result into two cubic (that is may be your sextic). -- philippe mail : chephip at free dot fr site : http://chephip.free.fr/ === Subject: Re: Minimax quickie > I found 36*cos^3(@) - 61*cos^2(@) + 16 = 0 > that is @ = 117.048738987...deg, @ being angle ABD, > then EF perpendicular to AD. > maximum area = AB^2 * 3.947250201868925... I forgot I automatically replace sin(@) by 2u/(1+u^2) and cos(@) by (1-u^2)/(1+u^2) which saves un-surd-ing but doubles the degree. > The not easy values result into a similar cubic, with just different > coefficients. Tried to find an interesting value set but failed so far. > B.T.W. ACDF *is* the closure, other configurations result into > obviously smaller closure, or same closure by symetry (case AB = BC). > If AB != BC, these different non symetric configurations result into > two cubic (that is may be your sextic). ACDF *is* the closure, but the letters may show up not in that order :-) -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn === Subject: geomotric sum of exponentially distributed random variables It is well know that the sum of independant squared normally distributed random variables (rv's) has a chi-square distribution. A sum of exponentially distributed rv's is gamma distributed. So can anyone tell me if there is an explicit distribution according to which a geometric sum of exponentially distributed rv's is distributed? [With geometrically distributed I mean that we have a sum of independant exp distributed rv's and the number of summands is distributed (independently from the exp distributions) according to a geometric distribution.] pkg === Subject: geometric sum of exponentially distributed random variables It is well know that the sum of independant squared normally distributed random variables (rv's) has a chi-square distribution. A sum of exponentially distributed rv's is gamma distributed. So can anyone tell me if there is an explicit distribution according to which a geometric sum of exponentially distributed rv's is distributed? [With geometrically distributed I mean that we have a sum of independant exp distributed rv's and the number of summands is distributed (independently from the exp distributions) according to a geometric distribution.] pkg === Subject: sum of exponentially distributed random variables with geometrically distributed number of summands It is well know that the sum of independant squared normally distributed random variables (rv's) has a chi-square distribution. A sum of exponentially distributed rv's is gamma distributed. So can anyone tell me if there is an explicit distribution according to which a sum of exponentially distributed rv's with a geometrically distributed number of summands is distributed? pkg === Subject: sum of i.i.d. exponentially distributed random variables with geometrically distributed number of summands It is well know that the sum of independent squared normally distributed random variables (rv's) has a chi-square distribution. A sum of exponentially distributed rv's is gamma distributed. So can anyone tell me if there is an explicit distribution according to which a sum of exponentially distributed rv's with a geometrically distributed number of summands is distributed? pkg === Subject: Re: sum of i.i.d. exponentially distributed random variables with geometrically distributed number of summands > It is well know that the sum of independent squared normally > distributed random variables (rv's) has a chi-square distribution. A > sum of exponentially distributed rv's is gamma distributed. > So can anyone tell me if there is an explicit distribution according to > which a sum of exponentially distributed rv's with a geometrically > distributed number of summands is distributed? Unless I am mistaken, it is also geometric with modified parameters. Say S = sum(X_i: i=1..,N), where the X_i are iid with pmf P{X = k} = p q^(k-1) and N is geometric with P{N=n} = s r^(n-1). The moment-generating function of S looks like the mgf of a geometric R.V. with success parameter P = p(1-r) = ps. I will leave the derivation to you. R.G. Vickson > pkg === Subject: Re: sum of i.i.d. exponentially distributed random variables with geometrically distributed number of summands 1) Why P[X=k]=p q^{k-1}? The X_i should be i.i.d. EXP-distributed! 2) I have not yet ckecked formally your argumentation, but the distribution of the sum should be supported on [0,infty) (sum of rv's with continuous distribution should have again a continuous distribution even if the number of summands is determined by a discrete distribution!) but if, as you claim, the sum has a geometric distribution it would of course be supported on IN. So I wonder if your claim is realistic? ... or did I ignore anything in your post? pkg === Subject: Re: sum of i.i.d. exponentially distributed random variables with geometrically distributed number of summands > 1) Why P[X=k]=p q^{k-1}? The X_i should be i.i.d. EXP-distributed! Sorry: my morning coffee had not yet kicked in. For iid exponential summands, the answer is exponential with a modified rate. You can get the result by looking at Laplace transforms. Or, you can look at the exponential as a kind of limiting form of the geometric and get the result that way. R.G. Vickson > 2) I have not yet ckecked formally your argumentation, but the > distribution of the sum should be supported on [0,infty) (sum of rv's > with continuous distribution should have again a continuous > distribution even if the number of summands is determined by a discrete > distribution!) but if, as you claim, the sum has a geometric > distribution it would of course be supported on IN. So I wonder if your > claim is realistic? > ... or did I ignore anything in your post? > pkg === Subject: Re: sum of i.i.d. exponentially distributed random variables with geometrically distributed number of summands > 1) Why P[X=k]=p q^{k-1}? The X_i should be i.i.d. EXP-distributed! > Sorry: my morning coffee had not yet kicked in. For iid exponential > summands, the answer is exponential with a modified rate. You can get > the result by looking at Laplace transforms. Or, you can look at the > exponential as a kind of limiting form of the geometric and get the > result that way. > R.G. Vickson I hate people who respond to their own posts, but here goes anyway. An even easier (more intuitive) way to see this is: suppose we have a piece of equipment with exponential time to failure with rate r (mean = 1/r). When it fails, we instantly replace it by an identical new item from a stock of spares. The number of spares available is geometric with P{N = k} = p(1-p)^(n-1), n=1, 2, ... . When the current item fails and we have no more spares, the system stops. So, if the equipment is working at time t, there is a probability that the system stops in the next interval of length h > 0. What is the stopping probability? Well, the current item has to fail (prob. = r h + o(h)) and then we have to have no spare in stock (prob. = p), for an overall stopping probability of rp h + o(h). Voila! we get an exponential again. RGV > 2) I have not yet ckecked formally your argumentation, but the > distribution of the sum should be supported on [0,infty) (sum of rv's > with continuous distribution should have again a continuous > distribution even if the number of summands is determined by a discrete > distribution!) but if, as you claim, the sum has a geometric > distribution it would of course be supported on IN. So I wonder if your > claim is realistic? > > ... or did I ignore anything in your post? > > pkg === Subject: Re: sum of i.i.d. exponentially distributed random variables with geometrically distributed number of summands >I hate people who respond to their own posts This is really a good intuitive explanation. By the way, I have in the meantime also browsed some textbooks and formula (5.6) in Feller, volume 2, chapter II.5 (at least in my edition, which is the blue soft cover Wiley edition) is exactly the answer to our topic. Of course it is completely consistent with what has been written here. Shows again that beneath the level of Kallenberg Feller still is THE reference ... pkg === Subject: Re: sum of i.i.d. exponentially distributed random variables with geometrically distributed number of summands >It is well know that the sum of independent squared normally >distributed random variables (rv's) has a chi-square distribution. A >sum of exponentially distributed rv's is gamma distributed. >So can anyone tell me if there is an explicit distribution according to >which a sum of exponentially distributed rv's with a geometrically >distributed number of summands is distributed? The distribution is geometric. One way to see this is via sampled Poisson processes: you are determining the time until the first arrival of a certain type. Another way is to use transforms: The moment generating function of sum(i=1..N, X_i) is g o m (functional composition), where g is the probability generating function for N and m is the moment generating function for X. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: sum of i.i.d. exponentially distributed random variables with geometrically distributed number of summands >> It is well know that the sum of independent squared normally >> distributed random variables (rv's) has a chi-square distribution. A >> sum of exponentially distributed rv's is gamma distributed. >> So can anyone tell me if there is an explicit distribution according to >> which a sum of exponentially distributed rv's with a geometrically >> distributed number of summands is distributed? > The distribution is geometric. Oops. I meant exponential. > One way to see this is via sampled Poisson processes: you are > determining the time until the first arrival of a certain type. > Another way is to use transforms: The moment generating function of > sum(i=1..N, X_i) is g o m (functional composition), where g is > the probability generating function for N and m is the moment > generating function for X. -- Stephen J. Herschkorn sjherschko@netscape.net === Subject: Re: sum of i.i.d. exponentially distributed random variables with geometrically distributed number of summands >>It is well know that the sum of independent squared normally >>distributed random variables (rv's) has a chi-square distribution. A >>sum of exponentially distributed rv's is gamma distributed. >>So can anyone tell me if there is an explicit distribution according to >>which a sum of exponentially distributed rv's with a geometrically >>distributed number of summands is distributed? >The distribution is geometric. One way to see this is via sampled >Poisson processes: you are determining the time until the first arrival >of a certain type. Another way is to use transforms: The moment >generating function of sum(i=1..N, X_i) is g o m (functional >composition), where g is the probability generating function for N >and m is the moment generating function for X. Do you mean exponential? The methods you have given work, but also direct use of calculus works to do this. -- 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: sum of i.i.d. exponentially distributed random variables with geometrically distributed number of summands : It is well know that the sum of independent squared normally (snip) Here's a tip: Don't post the same thing four times, it really pisses people off and makes them not want to answer your questions. Justin === Subject: Re: sum of i.i.d. exponentially distributed random variables with geometrically distributed number of summands I think I have deleted the other postings so that they do not appear four times in the forum and it should not be a disadvantage if the author of a new contribution changes it if he notices that it can be formulated more precisely! Actually those really interested in the topic will not be annoyed by it ... pkg === Subject: Re: sum of i.i.d. exponentially distributed random variables with geometrically distributed number of summands >I think I have deleted the other postings so that they do not appear > four times in the forum and it should not be a disadvantage if the > author of a new contribution changes it if he notices that it can be > formulated more precisely! > Actually those really interested in the topic will not be annoyed by > it I can see all four of your posts. The usual procedure would be to for you to reply to your own incorrect post with the required correction. Also, the subject line of this post is unusually long. -- Clive Tooth www.clivetooth.dk === Subject: Re: sum of i.i.d. exponentially distributed random variables with geometrically distributed number of summands <47qed4Fh0glrU1@individual.net> OK, in this case I am sorry. But I really just see one post!??? Nevertheless I curiously wait for the first answers to my question instead of complaints because of the way they were posted ... pkg === Subject: Re: Minimal union of cartesian products > Do the following two copies of the same question come from a > mathematics olympiad or some such contest? If so, we'd better not > answer them in the meantime. No. I get too much spam on my real email address. That's why I cancelled the first message and posted the second one instead. The cancel didn't propagate, though. The question is still puzzling me. === Subject: decomposition of polynomials Let f be a polynomial over Z. It is a well-known task with smart known solutions to find polynomials g and h over Z such that f(x)=g(x)h(x) if such g and h exist. But I have never seen solutions to the problem to find polynomials g and h over Z such that f(x)=g(h(x)) if such g and h exist. Is it trivial? Has it never been attacked? I can hardly imagine either. Helmut Richter === Subject: Re: decomposition of polynomials > Let f be a polynomial over Z. It is a well-known task with smart known > solutions to find polynomials g and h over Z such that f(x)=g(x)h(x) > if such g and h exist. But I have never seen solutions to the problem > to find polynomials g and h over Z such that f(x)=g(h(x)) if such g > and h exist. Is it trivial? Has it never been attacked? I can hardly > imagine either. This has been considered, and is not well understood. The security of the so-called ``double-round quadratic cipher'' depends on the fact that it seems to be difficult to decompose a quadratic polynomial over a finite field into a composition of lower-degree polynomials. === Subject: Re: decomposition of polynomials > Let f be a polynomial over Z. It is a well-known task with smart known > solutions to find polynomials g and h over Z such that f(x)=g(x)h(x) > if such g and h exist. But I have never seen solutions to the problem > to find polynomials g and h over Z such that f(x)=g(h(x)) if such g > and h exist. Is it trivial? Has it never been attacked? I can hardly > imagine either. > This has been considered, and is not well understood. The security of > the so-called ``double-round quadratic cipher'' depends on the fact > that it seems to be difficult to decompose a quadratic polynomial over > a finite field into a composition of lower-degree polynomials. In that post, I meant quartic polynomial, not quadratic. === Subject: Re: decomposition of polynomials >Let f be a polynomial over Z. It is a well-known task with smart known >solutions to find polynomials g and h over Z such that f(x)=g(x)h(x) >if such g and h exist. But I have never seen solutions to the problem >to find polynomials g and h over Z such that f(x)=g(h(x)) if such g >and h exist. That's a big if, in the following sense. Suppose I hand you a polynomial of degree 6. I do so by giving you 7 integers, or equivalently 6 rational numbers -- you might as well assume that f, g, and h are monic polynomials in Q[x] because you can move around the constant terms from factor to factor. Now, for you to factor f as g h, you have to come up with, say, a quadratic factor and a quartic factor, or two cubics, etc. In each case, you need to come up with a total of 6 rational numbers -- the coefficients of g and h, partitioned according to the degrees you think g and h have. You can if you like treat this as a set of 6 equations (one for each coefficient in f) in 6 unknowns (one for each coefficient in g and h ). As you may know, this typically has a finite solution set over the complex numbers, which of course makes perfect sense here -- there is a unique (up to order) factorization of f into monic linear factors over C, and those factors can be grouped in just a few ways to make g and h. Now let's play the game again but trying to factor f as g o h . (Again we start with f of degree 6.) The key difference now is that the degree of g o h is the _product_ of the degrees of g and h. So we have fewer possible pairs of degrees to consider, and they are smaller degrees. You might consider, for example, the possibility that g is a quadratic and h is a cubic. Well, again you can write out the equations which state that f = g o h and compare coefficients. You again have 6 equations, one for each coefficient in f, but now only 5 unknown coefficients in g and h. That's very different from the previous paragraph -- 6 equations in 5 unknowns are likely to be contradictory, meaning that there are no solutions at all. (It's actually a little worse than this -- the equations are also a little redundant in the sense that if one factorization is possible then there are multiple factorizations since we can compose g and h with linear functions; so after solving just 4 of the equations, you'll already run out of variables.) In this way we conclude that most polynomials of any fixed degree are likely to be irreducible -- over any field, not just Q -- in the sense that f = g o h ==> deg(g)=1 or deg(h)=1. (A sextic x^6 + q5 x^5 + ... is decomposable as cubic o quadratic iff -q5^5+3*q5^3*q4+81*q1-27*q2*q5 = -27*q3-5*q5^3+18*q5*q4 = 0 and it's decomposable the other way iff 5*q5^4-24*q5^2*q4+16*q4^2-64*q2+32*q5*q3 = -64*q1-q5^5+8*q5^3*q4-16*q5*q4^2-8*q3*q5^2+32*q3*q4 = 0 .) More high-falutin' responses are also possible. Look for polynomial decomposition, primitive Galois group, and widen your search to the more general problem of finding g, h with f | (g o h) instead of f = g o h . If you'd like to try your hand at this, consider the following interesting example proposed when this question was asked in July 2001: x^6+235*x^5+1430*x^4+1695*x^3+270*x^2-229*x+1 to the literature: dave === Subject: Re: decomposition of polynomials > which also give citations to the literature: Some of the links in my prior posts are now stale, but Googling polynomial decomposition will reveal links to many online expositions, e.g. see below. --Bill Dubuque Polynomial Decomposition Jaime Gutierrez (Cantabria) and Dexter Kozen (Cornell) http://www.fachgruppe-computeralgebra.de/CAR/CAR24/node6.html http://www.fachgruppe-computeralgebra.de/CA-Rundbrief/car24.pdf Polynomial decomposition is the problem of representing a given polynomial f(x) as a functional composition g(h(x)) of polynomials of smaller degree. Polynomial decomposition is useful in simplifying the representation of field extensions of high degree and is provided as a primitive by many major symbolic algebra systems. Decomposition problems for multivariate polynomials and for rational and algebraic functions are also of interest. The theory of polynomial decomposition was initiated by Ritt in 1922 [19] and further developed in [9,8]. A survey was presented in [20]. Decompositions of f(x) in F[x] are intimately related to the intermediate fields between F(f) and F(x) [8,19]. The problem breaks into two cases, the *tame* and the *wild*. The tame case is when the characteristic of F does not divide the degree of g, including characteristic 0. In the tame case, the problem is rational--that is, if f decomposes over an extension of F, then it decomposes over F [20]--and maximal decompositions are unique up to insertion of linear decomposition factors and commutativity of powers of x and Chebyshev polynomials [19]. Both facts may fail in the wild case [8,20]. The first analyzed algorithms for decomposition of polynomials were provided by Barton and Zippel [3] and Alagar and Thanh [1], who considered polynomials over fields of characteristic zero. Both solutions involved polynomial factorization and took exponential time. For some time, the problem of finding nontrivial decompositions was considered to be computationally hard; a cryptographic protocol was based on its supposed intractability [5]. Kozen and Landau [17] discovered the first polynomial-time algorithms in the tame case that do not require factorization. The time bounds were O(n^3), O(n^2) if F supports an FFT. A similar algorithm was discovered independently by Gutierrez et al. [15]. Kozen and Landau also gave efficient NC algorithms (parallel polylog time on polynomially many processors) with a time bound of O(log^2 n). In the wild case, they reduced the problem to factorization and gave an O(n^log(n)) algorithm for the decomposition of irreducible polynomials over general fields admitting a polynomial-time factorization algorithm and an NC algorithm for irreducible polynomials over finite fields. The sequential bounds in the tame case were improved by von zur Gathen [10] to O(n log^2(n) loglog n), O(n log^2 n) if F supports an FFT. He also gave an improved algorithm for the wild case, yielding a polynomial-time reduction to factorization, and observed undecidability over sufficiently general fields [11]. Several extensions and variations have been investigated. Dickerson [7] and von zur Gathen [10] investigated the decomposition of multivariate polynomials. Gutierrez [13] gave a polynomial decomposition algorithm over factorial domains. Von zur Gathen and Weiss [12] gave an exponential method for finding decompositions of the form f(x) = g(h(x),k(x)) for homogeneous g(y,z). Casperson et al. [6] gave an algorithm that, given an irreducible f(x), finds g(x) and h(x) such that f(x) divides g(h(x)). Binder [4] gave a fast method to compute the Luroth generator of the union field of two polynomials. Hommel and Kovacs [16] defined and investigated the sine-cosine decomposition problem: determine whether a given bivariate polynomial f(s,c) has a decomposition of the form f(s,c)=g(h(s,c)) mod s^2+c^2-1. This problem has applications in robot kinematics. Gutierrez and Recio [14] gave a cubic-time algorithm that does not require factorization. It is based on approximating a root of f(s,c) mod s^2+c^2-1. Zippel [21] presented a polynomial-time algorithm to decompose rational functions over any field admitting efficient polynomial factorization. His approach uncovers a strong relation to Luroth's theorem. Alonso et al. [2] gave two exponential-time algorithms for rational function decomposition that compare favorably with Zippel's in practice. They also presented several applications: faithful reparametrization of unfaithfully parameterized curves, computing intermediate fields in a simple purely transcendental extension of F, and providing a birationality test for subfields of F(x). Kozen, Landau, and Zippel [18] addressed the decomposition problem for algebraic functions. They uncovered a connection to univariate resultants over algebraic function fields and showed that the decomposition problem essentially asks whether some power of a given irreducible bivariate polynomial f(x,z) can be expressed as the resultant with respect to y of two bivariate polynomials g(x,y), h(y,z). They determined necessary and sufficient conditions for the existence of a nontrivial decomposition and classified all such decompositions up to isomorphism. They also gave an exponential-time algorithm for finding a nontrivial decomposition if one exists. Literature 1 V. S. Alagar and M. Thanh. Fast polynomial decomposition algorithms. In Proc. EUROCAL85, pages 150-153. Springer-Verlag Lect. Notes in Comput. Sci. 204, 1985. 2 C. Alonso, J. Gutierrez, and T. Recio. A rational function decomposition algorithm by near-separated polynomials. J. Symbolic Computation, 19:527-544, 1995. 3 D. R. Barton and R. E. Zippel. Polynomial decomposition algorithms. J. Symb. Comp., 1:159-168, 1985. 4 F. Binder. Fast computations in the lattice of polynomial rational function fields. In Proc. ISSAC-96. ACM Press, 1995. 5 J. J. Cade. A public key cipher which allows signatures. 6 D. Casperson, D. Ford, and J. MacKay. An ideal decomposition algorithm. J. Symbolic Computation, 21(2):133-137, 1996. 7 M. Dickerson. Polynomial decomposition algorithms for multivariate polynomials. Technical Report TR87-826, Comput. Sci., Cornell Univ., April 1987. 8 F. Dorey and G. Whaples. Prime and composite polynomials. J. Algebra, 28:88-101, 1974. 9 M. D. Fried and R. E. MacRae. On curves with separated variables. Math. Ann., 180:220-226, 1969. 10 J. von zur Gathen. Functional decomposition of polynomials: the tame case. J. Symb. Comput., 9:281-299, 1990. 11 J. von zur Gathen. Functional decomposition of polynomials: the wild case. J. Symb. Comput., 10:437-452, 1990. 12 J. von zur Gathen and J. Weiss. Homogeneous bivariate decompositions. J. Symbolic Computation, 19:409-434, 1995. 13 J. Gutierrez. A polynomial decomposition algorithm over factorial domains. Comptes Rendus Mathematiques de l'Academie des Sciences, 13(2-3):81-86, April-June 1991. 14 J. Gutierrez and T. Recio. Advances on the simplification of sine-cosine equations. J. Symbolic Computation, 26:31-70, 1998. 15 J. Gutierrez, T. Recio, and C. Ruiz de Velasco. A polynomial decomposition algorithm of almost quadratic complexity. In Proc. AAECC-6/88, volume 357 of Lect. Notes in Comput. Sci., pages 471-476. Springer-Verlag, 1989. 16 G. Hommel and P. Kov'acs. Simplification of symbolic inverse kinematic transformations through functional decomposition. In Proc. of the Conference Adv. in Robotics, pages 88-95, 1992. 17 D. Kozen and S. Landau. Polynomial decomposition algorithms. J. Symb. Comput., 7:445-456, 1989. 18 D. Kozen, S. Landau, and R. Zippel. Decomposition of algebraic functions. Journal of Symbolic Computation, 22(3):235-246, Sep. 1996. 19 J. F. Ritt. Prime and composite polynomials. Trans. Amer. Math. Soc., 23:51-66, 1922. 20 A. Schinzel. Selected topics on polynomials. University of Michigan Press, 1982. 21 R. E. Zippel. Rational function decomposition. In Stephen Watt, editor, International Symposium on Symbolic and Algebraic Computation, pages 1-6, New York, July 1991. ACM. === Subject: Re: decomposition of polynomials >More high-falutin' responses are also possible. Look for >polynomial decomposition, primitive Galois group, and widen >your search to the more general problem of finding g, h with > f | (g o h) instead of f = g o h . >If you'd like to try your hand at this, consider the following >interesting example proposed when this question was asked in July 2001: > x^6+235*x^5+1430*x^4+1695*x^3+270*x^2-229*x+1 If we let alpha be a root of y^3 - y^2 - 2*y + 1, then the sextic splits over the field Q(alpha, sqrt(21)). alias(alpha = RootOf(y^3 - y^2 - 2*y + 1, y)); alias(sqrt21 = RootOf(t^2 = 21, t)); f := x^6+235*x^5+1430*x^4+1695*x^3+270*x^2-229*x+1; factor(f, {alpha, sqrt21}); # Six linear factors -- VP Cheney Burr-ed his gun as a bird flew past The nation responds burr as we await bird flu shots and fight a real cold war. pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA === Subject: Re: decomposition of polynomials >>More high-falutin' responses are also possible. Look for >>polynomial decomposition, primitive Galois group, and widen >>your search to the more general problem of finding g, h with >> f | (g o h) instead of f = g o h . >>If you'd like to try your hand at this, consider the following >>interesting example proposed when this question was asked in July 2001: >> x^6+235*x^5+1430*x^4+1695*x^3+270*x^2-229*x+1 > If we let alpha be a root of y^3 - y^2 - 2*y + 1, >then the sextic splits over the field Q(alpha, sqrt(21)). you would do that!). Allow me to use this to illustrate my prior remarks. Peter notes that the intermediate quadratic subfield in the splitting field for this polynomial is K = Q(sqrt(21)). Sure enough, the polynomial factors into two cubics over K. They are p +- sqrt(21) q where p(x) = x^3+235/2*x^2+229/2*x-1 and q(x) = 49/2*x*(x+1). So now if x is any root of f, it makes the product of these two sums vanish, so p(x) is either sqrt(21) q(x) or its negative, so that p(x)^2 - 21 q(x)^2 = 0, i.e. ( p(x)/q(x) )^2 - 21 = 0. If it weren't for that darn q(x) in there, we would conclude that the sextic is exactly the composite f = g o p where g(x) = x^2 - 21, because that equation relates two monic sextics with equal roots! We can do _something_ with the q there as well, but we don't quite get a decomposition of f. Note that if as above x is any of the roots of f, then 1/q(x) lies in the field Q(x) generated by x, but this field is the same as the polynomial ring Q[x], so that we may write 1/q as a _polynomial_ in x. In this particular case that's not too hard: since f(x)=0, 1 = x * ((1-f(x))/x), and that second factor is a polynomial in x , so we have a polynomial inverse for x itself; you can do likewise to get an inverse for y=x+1 (first rewrite f as a polynomial in y). Multiply these two inverses together to get an inverse for q(x) (up to a constant). Don't forget that you can throw away any multiple of f(x). Likewise we can then multiply out p(x) * 1/q(x) and reduce mod f to get a small polynomial which equals p(x)/q(x) whenever x is a root of f(x). Carrying out this prescription I get this to be h(x) = ( 938*x^4+5252*x^3+4*x^5+4388*x^2+84*x-225 )/49 Now we make the same observation as I made above when pretending there was no q(x) : h(x)^2-21 vanishes whenever x is a root of f ; except that now h^2 - 21 is of higher degree than f, so we don't get equality, we get divisibility. Sure enough, you can calculate that if h is as above and g(x) = x^2 - 21 then f | g o h since g o h = 4/2401 * (4*x^4+936*x^3+4785*x^2+2229*x+51)* f(x) . You can reverse the logic of all this: if you have somehow found two polynomials g and h with f | g o h, then any root x of f would make g(h(x))=0, i.e. h(x) is one of the roots r_i of g. So the field extension can be done in two steps, first up to the splitting field of g, then adjoining the solutions to each equation h(x) - r_i . This particular sextic happens to have cylic galois group, so you can view the splitting field either as a cubic extension of a quadratic extension, as I did above, or a quadratic extension of a cubic extension. Someone suggested using the Chain Rule to decide whether f is a composite, which is clever but I don't see any way to use that for the broader problem of finding composites which are _multiples_ of f. dave === Subject: Re: decomposition of polynomials >More high-falutin' responses are also possible. Look for >polynomial decomposition, primitive Galois group, and widen >your search to the more general problem of finding g, h with > f | (g o h) instead of f = g o h . >If you'd like to try your hand at this, consider the following >interesting example proposed when this question was asked in July 2001: > x^6+235*x^5+1430*x^4+1695*x^3+270*x^2-229*x+1 >> If we let alpha be a root of y^3 - y^2 - 2*y + 1, >>then the sextic splits over the field Q(alpha, sqrt(21)). >you would do that!). Allow me to use this to illustrate my prior remarks. >Peter notes that the intermediate quadratic subfield in the splitting >field for this polynomial is K = Q(sqrt(21)). Sure enough, the polynomial >factors into two cubics over K. They are p +- sqrt(21) q where > p(x) = x^3+235/2*x^2+229/2*x-1 and q(x) = 49/2*x*(x+1). >So now if x is any root of f, it makes the product of these two >sums vanish, so p(x) is either sqrt(21) q(x) or its negative, so >that p(x)^2 - 21 q(x)^2 = 0, i.e. ( p(x)/q(x) )^2 - 21 = 0. >If it weren't for that darn q(x) in there, we would conclude that >the sextic is exactly the composite f = g o p where g(x) = x^2 - 21, >because that equation relates two monic sextics with equal roots! >We can do _something_ with the q there as well, but we don't quite >get a decomposition of f. Note that if as above x is any of the >roots of f, then 1/q(x) lies in the field Q(x) generated by x, >but this field is the same as the polynomial ring Q[x], so that >we may write 1/q as a _polynomial_ in x. In this particular >case that's not too hard: since f(x)=0, 1 = x * ((1-f(x))/x), and >that second factor is a polynomial in x , so we have a polynomial >inverse for x itself; you can do likewise to get an inverse for >y=x+1 (first rewrite f as a polynomial in y). Multiply these >two inverses together to get an inverse for q(x) (up to a constant). >Don't forget that you can throw away any multiple of f(x). >Likewise we can then multiply out p(x) * 1/q(x) and reduce mod f >to get a small polynomial which equals p(x)/q(x) whenever x is >a root of f(x). Carrying out this prescription I get this to be > h(x) = ( 938*x^4+5252*x^3+4*x^5+4388*x^2+84*x-225 )/49 >Now we make the same observation as I made above when pretending >there was no q(x) : h(x)^2-21 vanishes whenever x is a root of f ; >except that now h^2 - 21 is of higher degree than f, so we don't >get equality, we get divisibility. Sure enough, you can calculate that >if h is as above and g(x) = x^2 - 21 then f | g o h since g o h = > 4/2401 * (4*x^4+936*x^3+4785*x^2+2229*x+51)* f(x) . >You can reverse the logic of all this: if you have somehow found two >polynomials g and h with f | g o h, then any root x of f >would make g(h(x))=0, i.e. h(x) is one of the roots r_i of g. >So the field extension can be done in two steps, first up to the >splitting field of g, then adjoining the solutions to each >equation h(x) - r_i . >This particular sextic happens to have cylic galois group, so you >can view the splitting field either as a cubic extension of a quadratic >extension, as I did above, or a quadratic extension of a cubic extension. >Someone suggested using the Chain Rule to decide whether f is a >composite, which is clever but I don't see any way to use that for >the broader problem of finding composites which are _multiples_ of f. >dave Let alpha denote a root of f. Another way to view the field f(alpha) is Q(w21 + 1/w21) = Q(cos(2*pi/21)), where w21 is a primitive 21-st root of unity. This is the real subfield of Q(w21). The galois group is cyclic of order 6, and has four subgroups, all normal. So there are four subfields. These are Q(cos(2*pi/21)) (degree 6) Q(cos(2*pi/7)) (degree 3) Q(cos(2*pi/21) + cos(8*pi/21) + cos(10*pi/21)) (degree 2) Q (degree 1) If you pick a polynomial h so that h(alpha) in the subfield of degree 2 or 3, then h(alpha) will be the root of a quadratic or cubic, say g(h(alpha)) = 0. Then f(x) divides g(h(x)). -- VP Cheney Burr-ed his gun as a bird flew past The nation responds burr as we await bird flu shots and fight a real cold war. pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA === Subject: Re: decomposition of polynomials >f(x)=g(x)h(x) rough guess: look at the galois fields of g(h(x)) and f(x). if there is a injective transformation between the roots, you have it. I remember some basic theorems in Z/nZ and polynomial rings U === Subject: Re: decomposition of polynomials >Let f be a polynomial over Z. It is a well-known task with smart known >solutions to find polynomials g and h over Z such that f(x)=g(x)h(x) >if such g and h exist. But I have never seen solutions to the problem >to find polynomials g and h over Z such that f(x)=g(h(x)) if such g >and h exist. Is it trivial? Has it never been attacked? I can hardly >imagine either. >Helmut Richter The chain rule should pretty much clinch it. A necessary condition is that h'(x) must divide f'(x), so that determines a finite set of possibilities for h, up to a constant. Then, using the fact that the other factor must equal g'(h(x)), we can start again with a similar problem of lower degree with a requirement that h'(x) for the new problem match original h'(x) factor. Seems like it should be easy in practice. quasi === Subject: resistance propotionell to the square of the velecito >> I have made transformation of units . >You have a solution for which the trajectory is the same >for all objects of the same cross sectional area, regardless >of mass. Thus I suspect an error in your transformation. >> In the following equation's the unit of distance is >> the distance the body must travel to hit a mass of m/2. >I don't know what that means. Suppose (a) the mass >is 1 kg, or (b) the mass is 1000 kg. What is the unit >of distance? >- Randy _ u means that u is a vector. _ _ force_resistance=-2.v.|v|.q.A _ force_gravity =m.g gives: _ _ _ _ m.a=m.g-2.v.|v|.q.A _ _ _ _ a=g-2.v.|v|.q.A/m _ _ _ _ a=g-2.v.|v|.q.A/m x''=g_x-2.x'.SQRT(x'^2+y'^2).q.A/m y''=g_y-2.y'.SQRT(x'^2+y'^2).q.A/m Introduce the UNIT-distance s0: x/s0''=g_x/s0-2.x/s0'.SQRT(x/s0'^2+y/s0'^2).q.A/m.s0 y/s0''=g_y/s0-2.y/s0'.SQRT(x/s0'^2+y/s0'^2).q.A/m.s0 x/s0=x_2 y/s0=y_2 Let: 2.q.A/m.s0=1 x_2''=g_x/s0-x_2'.SQRT(x_2'^2+y_2'^2) y_2''=g_y/s0-y_2'.SQRT(x_2'^2+y_2'^2) Introduce the UNIT-time t0: d(dx_2/dt)/dt=g_x/s0-dx_2/dt.SQRT((dx_2/dt)^2+(dy_2/dt)^2) d(dy_2/dt)/dt=g_y/s0-dy_2/dt.SQRT((dx_2/dt)^2+(dy_2/dt)^2) d(dx_2/d(t/t0))/d(t/t0)= g_x/(s0/t0^2)-dx_2/d(t/t0).SQRT((dx_2/d(t/t0))^2+(dy_2/d(t/t0))^2) d(dy_2/d(t/t0))/d(t/t0)= g_y/(s0/t0^2)-dy_2/d(t/t0).SQRT((dx_2/d(t/t0))^2+(dy_2/d(t/t0))^2) g_x=-00 Let: g_y/(s0/t0^2)=-10 t/t0=T d(dx_2/dT)/dT=-00-dx_2/dT.SQRT((dx_2/dT)^2+(dy_2/dT)^2) d(dy_2/dT)/dT=-10-dy_2/dT.SQRT((dx_2/dT)^2+(dy_2/dT)^2) So in the new UNITS the equation became: v_x'=-00-v_x.|v| v_y'=-10-v_y.|v| === Subject: Is there such a thing as a cubic liter? A liter is a volume of some fluid. Cubic also refers to volume. So what does a cubic liter refer to? When I looked on Google for a cubic liter, there were some listings (not thousands, but hundreds). But I'm not sure whether using the two words together really makes any sense. === Subject: Re: Is there such a thing as a cubic liter? You can imagine a liter of water put into different shape containers. So you could have a spherical liter, a cylindrical liter, and (yes) a cubic liter. === Subject: Re: Is there such a thing as a cubic liter? Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >A liter is a volume of some fluid. Cubic also refers to volume. So what >does a cubic liter refer to? When I looked on Google for a cubic liter, >there were some listings (not thousands, but hundreds). But I'm not >sure whether using the two words together really makes any sense. It's a unit of volume in nine-dimensional space. -- Richard === Subject: Re: Is there such a thing as a cubic liter? >>A liter is a volume of some fluid. Cubic also refers to volume. So what >>does a cubic liter refer to? When I looked on Google for a cubic liter, >>there were some listings (not thousands, but hundreds). But I'm not >>sure whether using the two words together really makes any sense. >It's a unit of volume in nine-dimensional space. To summarize, the 3 possibilities are: 1. Misuse of units. 2. Talking about 9-dimensional space, possibly related to a unified field theory. 3. Referring to a 3-dimensional volume which has the shape of a cube. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Is there such a thing as a cubic liter? >>A liter is a volume of some fluid. Cubic also refers to volume. So what >>does a cubic liter refer to? When I looked on Google for a cubic liter, >>there were some listings (not thousands, but hundreds). But I'm not >>sure whether using the two words together really makes any sense. > It's a unit of volume in nine-dimensional space. > -- Richard Actually, cubic liters is a unit of AREA (in ten-dimensional space). === Subject: Re: Is there such a thing as a cubic liter? >A liter is a volume of some fluid. Cubic also refers to volume. So what > does a cubic liter refer to? When I looked on Google for a cubic liter, > there were some listings (not thousands, but hundreds). But I'm not > sure whether using the two words together really makes any sense. No. Its just bad usage. Like square acres or linear miles. The cubic is either redundant or plain wrong, depending upon how pedantic you are. My favourite mangled unit of measure is using kW as a unit of energy - as in converting to solar will save you 200 kW of electricity per year. === Subject: Re: Is there such a thing as a cubic liter? >>A liter is a volume of some fluid. Cubic also refers to volume. So what >> does a cubic liter refer to? When I looked on Google for a cubic liter, >> there were some listings (not thousands, but hundreds). But I'm not >> sure whether using the two words together really makes any sense. >No. Its just bad usage. Like square acres or linear miles. The cubic >is either redundant or plain wrong, depending upon how pedantic you are. >My favourite mangled unit of measure is using kW as a unit of energy - as in >converting to solar will save you 200 kW of electricity per year. Mangled? Probably, but not necesarily. Throwing away a certain lamp in my household will save me 0.1 kW of electricity because it's got a 100W bulb in it and it always seems to be turned on. So it's true the per year doesn't really fit here, but it's not quite true that the expected kWh must appear in the sentence. Politically important examples of this include confusing quantities with their time-derivatives, especially mixing up debts and deficits. I seem to be the only one bothered by news reports of a ten billion dollar tax reform proposal, say. dave === Subject: Re: Is there such a thing as a cubic liter? > A liter is a volume of some fluid. Cubic also refers to volume. So what > does a cubic liter refer to? When I looked on Google for a cubic liter, > there were some listings (not thousands, but hundreds). But I'm not > sure whether using the two words together really makes any sense. >> No. Its just bad usage. Like square acres or linear miles. The cubic >> is either redundant or plain wrong, depending upon how pedantic you are. >> My favourite mangled unit of measure is using kW as a unit of energy - as in >> converting to solar will save you 200 kW of electricity per year. > Mangled? Probably, but not necesarily. Throwing away a certain lamp > in my household will save me 0.1 kW of electricity because it's got > a 100W bulb in it and it always seems to be turned on. So it's > true the per year doesn't really fit here, but it's not quite > true that the expected kWh must appear in the sentence. > Politically important examples of this include confusing quantities > with their time-derivatives, especially mixing up debts and deficits. > I seem to be the only one bothered by news reports of a ten billion dollar > tax reform proposal, say. According to U.S. Senator Tim Johnson (Dem., South Dakota): ``Currently, Asian central banks hold about $2 trillion worth of U.S. debt, with Japan and China leading the pack. OPEC nations in the Middle East hold about $44 billion of our debt. $2 trillion works out to about $7000 per US citizen. Source: http://johnson.senate.gov/~johnson/releases/200503/2005309659.html David Bernier === Subject: Re: Is there such a thing as a cubic liter? <44180b6f$0$25198$afc38c87@news.optusnet.com.au> In message , David Bernier .... >``Currently, Asian central banks hold about $2 trillion > worth of U.S. debt, with Japan and China leading the pack. > OPEC nations in the Middle East hold about $44 billion of our debt. > $2 trillion works out to about $7000 per US citizen. >Source: http://johnson.senate.gov/~johnson/releases/200503/2005309659.html The whole point of debt is that the money belongs to someone else... -- Jeremy Boden === Subject: Re: Is there such a thing as a cubic liter? >A liter is a volume of some fluid. Cubic also refers to volume. So what > does a cubic liter refer to? When I looked on Google for a cubic liter, > there were some listings (not thousands, but hundreds). But I'm not > sure whether using the two words together really makes any sense. > No. Its just bad usage. Like square acres or linear miles. The cubic > is either redundant or plain wrong, depending upon how pedantic you are. > My favourite mangled unit of measure is using kW as a unit of energy - as in > converting to solar will save you 200 kW of electricity per year. This reminds me of some puzzle about a boat, where they say it travels at some number of knots per hour. The trick to the problem was to know that a knot is a unit of velocity, so that knots per hour means acceleration... -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Is there such a thing as a cubic liter? > A liter is a volume of some fluid. Cubic also refers to volume. So what > does a cubic liter refer to? When I looked on Google for a cubic liter, > there were some listings (not thousands, but hundreds). But I'm not > sure whether using the two words together really makes any sense. Cubic liter is an absurdly lousy redundancy as is also square hectare, cubic quart and square acre. Riddle of the day. Is a linear inch more linear than a linear centimeter? === Subject: Re: Is there such a thing as a cubic liter? Hi >> A liter is a volume of some fluid. Fluid? Superfluous. >> Cubic also refers to volume. Not always? How much air would fit into 1.2535 s^3 (assuming normal pressure and temperature)? > Cubic liter is an absurdly lousy redundancy as is > also square hectare, cubic quart and square acre. Huh? What about 5 l * 6 l * 7 l = 210 l^3 What's so wrong about that? I don't have a reasonable interpretation of cubic liters at hand, but they may appear anyway... Markus === Subject: Re: Is there such a thing as a cubic liter? >> Cubic liter is an absurdly lousy redundancy as is >> also square hectare, cubic quart and square acre. > Huh? > What about 5 l * 6 l * 7 l = 210 l^3 > What's so wrong about that? l^3 is liter cubed, rather than cubic liter. - Oliver === Subject: Re: Is there such a thing as a cubic liter? > A liter is a volume of some fluid. > Fluid? Superfluous. Superfluous helium? -- Ben === Subject: hahn banach hi all, I'm searching an example of two convex sets A and B that cannot be separated by an hyperplane. I think they exist but I can't find simple examples .. Fedor. === Subject: Re: hahn banach > hi all, > I'm searching an example of two convex sets A and B that cannot be > separated by an hyperplane. I think they exist but I can't find simple > examples .. In R^2, take A to be the lower half-plane {(x,y): y <= 0 } and B = {(x,y) : x>0, xy >= 1}. A and B are both closed and convex, and they are disjoint. -- A. === Subject: Re: hahn banach > hi all, > > I'm searching an example of two convex sets A and B that cannot be > separated by an hyperplane. I think they exist but I can't find simple > examples .. > In R^2, take A to be the lower half-plane {(x,y): y <= 0 } and > B = {(x,y) : x>0, xy >= 1}. A and B are both closed and convex, and > they are disjoint. But they are also separated by the hyperplane y=0, since all points in A have y <= 0 and all points in B have y >= 0. === Subject: Re: hahn banach On Wed, 15 Mar 2006 13:02:43 -0500, A N Niel >> hi all, >> >> I'm searching an example of two convex sets A and B that cannot be >> separated by an hyperplane. I think they exist but I can't find simple >> examples .. >> In R^2, take A to be the lower half-plane {(x,y): y <= 0 } and >> B = {(x,y) : x>0, xy >= 1}. A and B are both closed and convex, and >> they are disjoint. >But they are also separated by the hyperplane y=0, >since all points in A have y <= 0 and all points >in B have y >= 0. Depends on exactly what we mean by separated by a hyperplane, which of course the OP hasn't yet defined. There _are_ notions that might reasonably be called separated by a hyperplane such that these two sets are _not_ separated by a hyperplane. ************************ David C. Ullrich === Subject: Re: hahn banach <150320061302436225%anniel@nym.alias.net.invalid> thank you all ! But I was looking for an example where separated is as in the definition of A N Niel :-) === Subject: Re: hahn banach > thank you all ! But I was looking for an example where separated is > as in the definition of A N Niel :-) So your example is as given by Herman Rubin: > The > classical is the space of all polynomials over the > reals, where A is the set of all polynomials whose > leading coefficient is positive, and B the set whose > leading coefficient is negative. These are two disjoint convex sets that are NOT separated by a hyperplane, even in the weak sense. === Subject: Re: hahn banach >hi all, > I'm searching an example of two convex sets A and B that cannot be >separated by an hyperplane. I think they exist but I can't find simple >examples .. I am presuming you mean disjoint convex sets. The classical is the space of all polynomials over the reals, where A is the set of all polynomials whose leading coefficient is positive, and b the set whose leading coefficient is negative. To avoid this problem in infinite dimensional spaces, one needs to assume that one of the sets has an internal point. This is a point such that any line through the point has an open interval containing the point which lies entirely in the set. -- 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: hahn banach On 15 Mar 2006 04:18:50 -0800, Fedor I'm searching an example of two convex sets A and B Disjoint? Closed? >that cannot be >separated by an hyperplane. A closed hyperplane (ie one defined by a _continuous_ linear functional)? And separated in what sense? >I think they exist but I can't find simple >examples .. A = {0}, B = {0} gives an example that answers the question as you stated it. Two other examples that may or may not be what you want; unclear because of the missing details in the question: (i) In R^2, let A = upper half-plane union positive real axis, B = lower half-plane union negative real axis. A and B are disjoint, convex, and if L is any linear functional on R^2 then L(A) intersects L(B). Of course A and B are not closed. (ii) In R^2, A = closed lower half-plane, B = {(x,y) : y >= exp(x)}. A and B are closed convex disjoint, and they cannot be separated in a weaker sense than in (i): There does not exist a linear functional L and a number s such that Lp > s for all p in A and Lp < s for all s in B. On the other hand there _does_ exist L such that L(A) and L(B) are disjoint (the point here being you have to specify what you mean by separated...) >Fedor. ************************ David C. Ullrich === Subject: Re: hahn banach > hi all, > I'm searching an example of two convex sets A and B that cannot be > separated by an hyperplane. I think they exist but I can't find simple > examples .. > Fedor. Define separated and hyperplane. Do you require that A and B are disjoint? === Subject: Integration of differential forms on manifolds question, I am trying to learn integration on manifolds from Spivak's Calculus on Manifolds, and I was wondering if someone could give me the motivation behind his definition of integration of a differential form on some manifold. His definition is : Let w be a k-form on the n-manifold M with boundary. Let c be a singular k-cube, that is, a continuous map c : [0,1]^k -----> M. Then, Definition : Int_M w = Int_[0,1]^k c*(w) where c*(w) is the pullback of the form w to [0,1]^k. In general, when I try to calculate integrals of forms over manifolds, am I going to be looking for these maps c? Do I need to just look for ONE singular k-cube in order to integrate, or must I cover my entire manifold with singular k-cubes? Can I just look for singular k-cubes through the fact that manifolds are locally Euclidean, and then try to use the local parameterizations as k-cubes (after some readjustment?) ? If you have any other insight, I would be very grateful as well. James === Subject: How can I get the solution of a^3 + b^3 = c^2? I begin learning number theory, but my passion of learning is runaway, facing this problem.. someone help me plz.T.T 1 =< a =< b =< c, a,b,c is natural number a^3 + b^3 = c^2 ---------------- (1) [1] how can I get triple (a, b, c) that solution of (1). I got some solutions.... (1, 2, 3), (2, 2, 4), (7, 21, 98), (8, 8, 32) .......more.. ....I have cramp in my fingers for calculating the equation...T.T there may be more easily method for finding the solution. [2] solution (8, 8, 32) has relation with (2, 2, 4). it can be rewrite like this ((2^2)*2,(2^2)*2,(2^3)*4)! I know that (A, B, C) is the solution, (n^2*A, n^2*B, n^3*C) is solution, too! so, I want more easily way to find solutions, not look like (n^2*A, n^2*B, n^3*C)! === Subject: Re: How can I get the solution of a^3 + b^3 = c^2? Hodol nous a r.8ecemment amicalement signifi.8e : > I begin learning number theory, but my passion of learning is > runaway, facing this problem.. > someone help me plz.T.T > 1 =< a =< b =< c, a,b,c is natural number > a^3 + b^3 = c^2 ---------------- (1) > [1] how can I get triple (a, b, c) that solution of (1). > I got some solutions.... (1, 2, 3), (2, 2, 4), (7, 21, 98), (8, 8, > 32) .......more.. > ....I have cramp in my fingers for calculating the equation...T.T > there may be more easily method for finding the solution. > [2] solution (8, 8, 32) has relation with (2, 2, 4). it can be rewrite > like this ((2^2)*2,(2^2)*2,(2^3)*4)! I know that (A, B, C) is the > solution, (n^2*A, n^2*B, n^3*C) is solution, too! > so, I want more easily way to find solutions, not look like (n^2*A, > n^2*B, n^3*C)! 1) Immediate simple infinite set of solutions : [a(a^3+b^3)]^3 + [b(a^3+b^3)]^3 = [(a^3 + b^3)^2]^2 2) A little bit prettier : Take any couple (u,v) in N*^2 let w = u^3 + v^3 let d = product of all prime factors of w whose power in w is odd Then w/d is a perfect square h^2 Then u^3 + v^3 = h^2 d Then (du)^3 + (dv)^3 = (d^2h)^2 Example : u=5, v=11 u^3 + v^3 = 1456 = 2^4*7*13 ==> d = 7*13, h = 4 (5*7*13)^3 + (7*11*13)^3 = (4*7^2*13^2)^2 455^3 + 1001^3 = 33124^2 -- Patrick === Subject: Re: How can I get the solution of a^3 + b^3 = c^2? > 1 =< a =< b =< c, a,b,c is natural number > a^3 + b^3 = c^2 ---------------- (1) > [1] how can I get triple (a, b, c) that solution of (1). > . . . not look like (n^2*A, n^2*B, n^3*C)! Your solution (7, 21, 98) is derived from the identity 1^3 + 3^3 = 2^2 * 7 by multiplying it by 7^3. For any other pair (x, y), represent x^3 + y^3 as w^2*z, and multiply by z^3 to give a triplet (x*z, y*z, w*z^2) which is a solution of (1). But a coprime triplet is a completely different problem! -- === Subject: Re: How can I get the solution of a^3 + b^3 = c^2? To. bert thank you for reply. I find that (xz, yz, z^2) is solution when each value x,y,z satisfy x^3+y^3=z. interestingly, z 's exponent is 1. although (xz, yz, z^2) is not primitive solution, I can find more easily solution of (1)! I think this is similar your answer! sincerely === Subject: Re: How can I get the solution of a^3 + b^3 = c^2? thank you for reply. I find that (xz, yz, z^2) is solution when each value x,y,z satisfy x^3+y^3=z. interestingly, z 's exponent is 1. although (xz, yz, z^2) is not primitive solution, I can find more easily solution of (1)! I think this is similar your answer! sincerely === Subject: Re: How can I get the solution of a^3 + b^3 = c^2? >> 1 =< a =< b =< c, a,b,c is natural number >> a^3 + b^3 = c^2 ---------------- (1) >> [1] how can I get triple (a, b, c) that solution of (1). >> . . . not look like (n^2*A, n^2*B, n^3*C)! You want (a+b)(a^2-ab+b^2) to be a perfect square. The lazy person's way to do this is to make each of the factors be a perfect square. (I think the only other way to get primitive solutions is to have each of the factors be 3 times a square; you can accomplish that in a way similar to what I do below.) Equivalently, write r = a/b ; you want b ( r+1 ) to be a rational square and r^2-r+1 to be a rational square. You can accomplish the former just by choosing a suitable b, after you have arranged the latter. But to solve r^2 - r + 1 = s^2 in rationals is easy; you just need to find rational points on a conic, and those all points come from a parameterization. You can find details elsewhere on the web (pick one rational point, e.g. r=0,s=1, and draw the line through there with slope m; the line meets the conic in one more point). The answer I get here is that r must equal (1+2*m)/(1-m^2) for some rational m. Writing m = u/v and comparing this formula to the definition r = a/b we discover that we must have a = -v*(v+2*u) * k , b = (u^2-v^2) * k for some rational k . To return to the original equation, you want a^3+b^3 to be a perfect square; well, the formulas above make a^3+b^3 equal to (u^2-2*v*u-2*v^2) * (u^2+v*u+v^2)^2 *k^3 so for any integers u and v you can just choose k=(u^2-2*u*v-2*v^2) and this will be a perfect square. Of course this process will likely give a and b common factors, which is the sort of thing you're trying to avoid. One way to get primitive solutions would be to choose u and v to be coprime (which would almost ensure that v(v+2u) and u^2-v^2 are coprime; their gcd is either 1 or 3). But then you also want to be able to choose k = +-1 to avoid giving a and b a common factor. It turns out there is a parity problem with k=-1 but you _can_ find pairs u,v for which (u^2-2*v*u-2*v^2) is already a square and thus set k = 1 above. Making this expression be square can be done just as above -- let r' = u/v, say the magic word conic! and out pops a parameterization u = f^2+2*g^2, v = -(2*f+2*g)*g which, if you combine with the previous parameterization, gives this family of solutions to the original problem: a = 4*g*(f^3 + g^3) b = f*(f^3 - 8*g^3) c = f^6 + 20*f^3*g^3 - 8*g^6 There are other solutions not given by substituting integers into this formula. As I noted above, at the very least you can re-do some of the computations with factors of 3 moved around. You can also scale the triples you get this way, using the reduction in the OP's request (e.g. take f=4, g=1 then divide a and b by 2^2 to get a solution with a=65, b=56) but that still won't give all the primitive solutions. Anyway this equation has infinitely many primitive solutions. More generally people have long looked at other equations of the form a^m + b^n = c^p with different combinations of m,n,p . As long as 1/m + 1/n + 1/p > 1 (as in this case) we expect infinitely many integer solutions; when this expression = 1 there are only a few. (Fermat handled the cases with 1/2 + 1/4 + 1/4 = 1 and conjectured the case 1/3 + 1/3 + 1/3 = 1; these and the cases 1/2 + 1/3 + 1/6 = 1 reduce to well-known elliptic curves of rank 0.) Whenever 1/m + 1/n + 1/p < 1 there are only finitely many solutions with coprime a,b,c. In fact, there are only a few such combinations of m,n,p for which any solutions are known at all! (excluding trivial cases -- those with one of a,b,c equal to 0 or 1). To heighten interest in this problem someone has even offered prize money for its solution so now this is sometimes known as the Beal Prize Conjecture. More info: http://www.math-atlas.org/98/beal dave === Subject: Re: How can I get the solution of a^3 + b^3 = c^2? thank you for teaching me. (1)But I want to know about using conic.(I guess that the conic's formular look like.. y^2=x^3+C.) (2)And I wonder how you get the answer that r must equal (1+2*m)/(1-m^2) . sincerely. === Subject: Any med for your girl to be happy! boundary=------------ms010701010502070601040203 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id k2FF2q306381 for ; Wed, 15 Mar 2006 10:02:52 -0500 --------------------------------------------------------------------- Cheapest medications based LICENSED online phartmacy! Cialis Soft Tabs as low as $4.72 Viagra Professional as low as $3.8 Viagra Soft Tabs as low as $3.8 Cialis as low as $5.67 Valium as low as $2.85 Generic Viagra as low as $3.5 Need medicine? All here! === Subject: Is the sum of two nonsingular Toeplitz matrices still be nonsingular? Maybe firstly I should introduce what's Toeplitz matrix: It's a matrix in which all the elements along the diagonal are equal, and those along each line parallel to the diagonal are also equal. For example, Tn= t0 t(-1) ... t(2-n) t(1-n) t1 t0 ... t(3-n) t(2-n) . . ... . . t(n-2) t(n-3) ... t0 t(-1) t(n-1) t(n-2) ... t1 t0 Then my problem is: if there are two nonsingular Toeplitz matrices, is their sum still be nonsingular? I try to prove the nonsingularity of the sum by the definition of nonsingularity, but it doesn't seem easy. Is there any other way to prove this nonsingularity? Sang === Subject: Re: Is the sum of two nonsingular Toeplitz matrices still be nonsingular? > Maybe firstly I should introduce what's Toeplitz matrix: > It's a matrix in which all the elements along the diagonal are equal, > and those along each line parallel to the diagonal are also equal. For > example, > Tn= > t0 t(-1) ... t(2-n) t(1-n) > t1 t0 ... t(3-n) t(2-n) > . . ... . . > t(n-2) t(n-3) ... t0 t(-1) > t(n-1) t(n-2) ... t1 t0 > Then my problem is: if there are two nonsingular Toeplitz matrices, is > their sum still be nonsingular? > I try to prove the nonsingularity of the sum by the definition of > nonsingularity, but it doesn't seem easy. Is there any other way to > prove this nonsingularity? > Sang Try Tn + (-Tn). Both are Toeplitz. Dale. === Subject: Re: Hermitian codes? Does anyone know of a good introduction to Hermitian codes? I would > like to learn more about this class of codes, but I struggle to get > good information on it. Any pointers to books, papers and websites > will be greatly appreciated. (I did search on Google, but I must have > searched for the wrong keywords...) > Jaco Versfeld > The 50th anniversary commemorative issue of IEEE Transactions > (by Blake et al) on Algebraic-Geometry codes. I warmly recommend > that (and references therein). start with this one. > Tom H.9aholdt has also written a chapter (was it for the Handbook???) > titled Algebraic-Geometry Codes without Algebraic Geometry, where > they base there approach to the Feng-Rao syndrome voting mechanism. > IOW their goal was to prove Riemann-Roch Theorem by demonstrating that > the decoding algorithm works. > During a project I once implemented K.9atter's decoding algorithm > for Hermitian codes in ANSI C. You are welcome to have the source, > if you are desperate, but the documentation is kinda terse (should > I say it doesn't exist). Actually I used a suitably shortened (or > was it punctured?) version of the Hermitian codes so that I could > by Heegard et al (don't remember the exact title, it appeared in IT > in late 90s and had Gr.9abner Bases as a buzzword). Therefore my > testbench encoding is non-systematic. I also made the shortcut > of solving for the error values by Gaussian elimination rather than > the more efficient DFT. I simply didn't have the time to learn the > theory to do that. You may want to ask Rasmus Nielsen for a better > documented implementation of (a slightly different version of) > Feng-Rao decoding algorithm. Groebner bases... I gave up trying to understand that one after some time. May be if you have time, can you please provide a brief overview of what exactly Groebner bases are, and what they are for, Jyrki? Prasanna. === Subject: Re: Hermitian codes? > Groebner bases... I gave up trying to understand that one after some > time. May be if you have time, can you please provide a brief overview > of what exactly Groebner bases are, and what they are for, Jyrki? When you look at an (algebraic) curve or surface, you know it can be defined by a set of polynomials which vanish there. For example, the x-axis in R^3 is the set of points where y = z = 0. But the defining set of polynomials is not unique. You could also define the x-axis by the equations y+z = 0, y-2z = 0 or even, more perversely, by the equations y+z = 0, (y-2z)^2 = 0. Mathematically the observation is that the set of all polynomials which vanish on the x-axis (or any other curve or surface or whatever) is an _ideal_ in the polynomial ring -- it includes not only the few that you write down, but also any multiples of those (i.e., those times any polynomial whatsoever) and any sums of _those_. All you accomplish by writing down a few polynomials is to provide something to get started with to describe the whole ideal. Those few form a _basis_ for the ideal. The examples above just show that {y, z} and {y+z, y-2z} are different bases for the same ideal. (The other example is a little different -- those two polynomials form a basis of a slightly smaller ideal but the two ideals have the same _radical_. That's a more complicated algebraic construction, but geometrically it just means they describe the same set of points.) OK, well once you know that different bases exist for the same ideal, then the question is, which bases are better for computational purposes? One idea we can borrow from Gaussian elimination: it's better to have an upper-triangular set of polynomials like y+z = z^2 = 0 : the last equation can be solved for z, then we plug into the next-to-last and solve for y, etc. Another idea we can borrow from the Euclidean algorithm for finding the gcd's of two polynomials; if your list of polynomials includes x^3+x^2 and x^2-2x, then in order for both these polynomials to vanish, their gcd (which is just x) must also vanish. This is what Groebner Bases do for you. They are just bases for ideals in polynomial rings, but they are chosen to be somehow minimal in the sense that those two principles have been applied to some terminal state. There's a little bit of choice involved (which variables will be considered last when you try Gaussian elimination?) so you see references to term order. But in practice they are just a means to enable one to carry out commutative-algebra computations in these ideals. Many of the more subtle concepts in commutative algebra can be made very concrete once ideals are represented with respect to a Groebner basis. They're very useful, but a tad over-rated. That is, they provide a methodical and _often_ rapid way to do computations. But the method is not _always_ fast, and it has a tendency to consume huge amounts of computer memory in the process -- sometimes even when the final answer itself is very compact. So people have been looking at alternative ways to carry out some important computations like counting points in 0-dimensional varieties. Research continues. dave === Subject: Re: Hermitian codes? Groebner bases... I gave up trying to understand that one after some > time. May be if you have time, can you please provide a brief overview > of what exactly Groebner bases are, and what they are for, Jyrki? > When you look at an (algebraic) curve or surface, you know it can > be defined by a set of polynomials which vanish there. For example, > the x-axis in R^3 is the set of points where y = z = 0. > But the defining set of polynomials is not unique. You could also > define the x-axis by the equations > y+z = 0, y-2z = 0 > or even, more perversely, by the equations > y+z = 0, (y-2z)^2 = 0. > Mathematically the observation is that the set of all polynomials > which vanish on the x-axis (or any other curve or surface or whatever) > is an _ideal_ in the polynomial ring -- it includes not only the > few that you write down, but also any multiples of those (i.e., those > times any polynomial whatsoever) and any sums of _those_. All you > accomplish by writing down a few polynomials is to provide something > to get started with to describe the whole ideal. Those few form a > _basis_ for the ideal. The examples above just show that {y, z} > and {y+z, y-2z} are different bases for the same ideal. > (The other example is a little different -- those two polynomials form > a basis of a slightly smaller ideal but the two ideals have the same > _radical_. That's a more complicated algebraic construction, but > geometrically it just means they describe the same set of points.) > OK, well once you know that different bases exist for the same ideal, > then the question is, which bases are better for computational purposes? > One idea we can borrow from Gaussian elimination: it's better to have > an upper-triangular set of polynomials like y+z = z^2 = 0 : the last > equation can be solved for z, then we plug into the next-to-last > and solve for y, etc. Another idea we can borrow from the > Euclidean algorithm for finding the gcd's of two polynomials; if > your list of polynomials includes x^3+x^2 and x^2-2x, then in > order for both these polynomials to vanish, their gcd (which is just x) > must also vanish. > This is what Groebner Bases do for you. They are just bases for ideals > in polynomial rings, but they are chosen to be somehow minimal in the > sense that those two principles have been applied to some terminal state. > There's a little bit of choice involved (which variables will be > considered last when you try Gaussian elimination?) so you see > references to term order. But in practice they are just a means to > enable one to carry out commutative-algebra computations in these ideals. > Many of the more subtle concepts in commutative algebra can be > made very concrete once ideals are represented with respect to > a Groebner basis. > They're very useful, but a tad over-rated. That is, they provide a > methodical and _often_ rapid way to do computations. But the method > is not _always_ fast, and it has a tendency to consume huge amounts > of computer memory in the process -- sometimes even when the final > answer itself is very compact. So people have been looking at > alternative ways to carry out some important computations like > counting points in 0-dimensional varieties. Research continues. > dave === Subject: Re: Hermitian codes? S. J. Kim, Introduction to Coding Theory and Algebraic Geometry Codes, Proceedings of Workshops in Pure Mathematics. Vol. 18, part 1, 1988. T. Hoholdt, J. H. van Lint, and R. Pellikaan, Algebraic Geometry Codes in Handbook of Coding Theory. Massimo Giulietti, NOTES ON ALGEBRAIC-GEOMETRIC CODES, try Google. Madhu Sudan, Coding Theory: Tutorial & Survey, try Google. Madhu Sudan, Decoding of Reed Solomon Codes beyond the Error Correcting Bound, try his website at MIT. Stephanie Fitchett, Bezouts Theorem: a taste of algebraic geometry, try Google. T. Hoholdt, R. Pellikaan, On decoding of algebraic geometry codes, in IEEE transactions on Info Theory. Efficient Decoding Methods in Goppa Codes based on Klein Curves , try Google. And always keep a copy of Mc Williams and Sloane. :o) === Subject: Re: Hermitian codes? One point Hermitian codes? A class of Algebraic Geometry Codes? I did some reading on them a while ago. Check www.lrz-muenchen.de/~prasanna/ useful (better, check the references in them). I did have some good papers on them, but don't remember their names now. I will search my machine on any good references I might have. If I can locate something I will let you know. === Subject: mathematics and arethmetic Is Mathematics the generalisation of all things numeric. And is arethmetic specific solutions? When I put numbers to paper, does that start becoming arithmetic? === Subject: Re: mathematics and arethmetic 03/15/2006 >Is Mathematics the generalisation of all things numeric. No. Much of Mathematics has nothing to do with numbers. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: mathematics and arethmetic > Is Mathematics the generalisation of all things numeric. And things not numeric, geometry for example. >And is arethmetic specific solutions? > When I put numbers to paper, does that start becoming arithmetic? Not necessarily. Suppose I want to list all the 2 digit numbers combinations (that means the order of the digits doesn't matter, adding 23 to the list excludes 32 because 32 is the same digits in a different order): ['01', '02', '03', '04', '05', '06', '07', '08', '09', '12', '13', '14', '15', '16', '17', '18', '19', '23', '24', '25', '26', '27', '28', '29', '34', '35', '36', '37', '38', '39', '45', '46', '47', '48', '49', '56', '57', '58', '59', '67', '68', '69', '78', '79', '89'] Was that arithmetic? Before you answer, consider asking the same question about all the 2 letter pairs taken from abcdefghij: ['ab', 'ac', 'ad', 'ae', 'af', 'ag', 'ah', 'ai', 'aj', 'bc', 'bd', 'be', 'bf', 'bg', 'bh', 'bi', 'bj', 'cd', 'ce', 'cf', 'cg', 'ch', 'ci', 'cj', 'de', 'df', 'dg', 'dh', 'di', 'dj', 'ef', 'eg', 'eh', 'ei', 'ej', 'fg', 'fh', 'fi', 'fj', 'gh', 'gi', 'gj', 'hi', 'hj', 'ij'] Sometimes when you put numbers to paper, you are simply manipulating symbols. === Subject: name for fx = x / (x + 1) ? Can someone help me refresh my memory? It seems like the function f(x) = x / (x + 1) has a name of some sort, but I can't remember it or seem to track it down. Mike === Subject: Re: name for fx = x / (x + 1) ? out, this isn't terribly different from a hyperbola, and I guess this must have been the cause of my mathematical deja vu. === Subject: Re: name for fx = x / (x + 1) ? > out, this isn't terribly different from a hyperbola, and I guess this > must have been the cause of my mathematical deja vu. The light dawns on me - what you were really asking for was the name of the figure you get when you graph y = x / (x + 1). In that case, it's not just not terribly different from a hyperbola, it *is* a hyperbola! -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: name for fx = x / (x + 1) ? out, this isn't terribly different from a hyperbola, and I guess this > must have been the cause of my mathematical deja vu. > The light dawns on me - what you were really asking for > was the name of the figure you get when you graph y = x / (x + 1). > In that case, it's not just not terribly different from a hyperbola, > it *is* a hyperbola! Specifically, it is the reciprocal function, translated one unit up and one unit to the left. (I like to call the hyperbola xy=1 'the reciprocal function' to distinguish it from other hyperbolas- I do not know if there is an official name for it.) === Subject: Re: name for fx = x / (x + 1) ? out, this isn't terribly different from a hyperbola, and I guess this > must have been the cause of my mathematical deja vu. > The light dawns on me - what you were really asking for > was the name of the figure you get when you graph y = x / (x + 1). > In that case, it's not just not terribly different from a hyperbola, > it *is* a hyperbola! > Specifically, it is the reciprocal function, translated one unit up and > one unit to the left. I forgot the flip in this description. There is a reflection in the === Subject: Re: name for fx = x / (x + 1) ? Well, I wish I could claim that. Really, though, the function just looked really familiar to me... === Subject: Re: name for fx = x / (x + 1) ? Well , Function h(x) = x / (x + 1) may be written: h(x) = 1 / ( 1+ 1/x) namely a conjugate form g^ [ -1](1+g(x) ), with g(x) = g^ -1(x) = 1/ x ,so all n iterate h^[n](x) = 1 / ( n+ 1/x) = x/(1+n*x) , n being an integer . Since h^[-1](x) = x/(1-x) , h^[0](x) = x ... Alain. === Subject: Re: name for fx = x / (x + 1) ? > Can someone help me refresh my memory? It seems like the function > f(x) = x / (x + 1) > has a name of some sort, but I can't remember it or seem to track it > down. > Mike A Moebius mapping of parabolic type, normalized to have a unique fixed point at 0, attractive on the right and repulsive on the left. It is the generator of an iterative semigroup f_n(x) = x / (x + n). Aren't you sorry you asked? === Subject: Re: name for fx = x / (x + 1) ? > Can someone help me refresh my memory? It seems like the function > f(x) = x / (x + 1) > has a name of some sort, but I can't remember it or seem to track it > down. > Mike Since this is 1 - 1/(x+1), it looks like a hyperbola. === Subject: Re: name for fx = x / (x + 1) ? > Can someone help me refresh my memory? It seems like the function > f(x) = x / (x + 1) > has a name of some sort, but I can't remember it or seem to track it > down. I have always called it Irving, but I'm not sure that's standard. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: name for fx = x / (x + 1) ? Finally, someone with my own sense of humor! I guessed Fred but clearly Irving is more appropriate... === Subject: Re: name for fx = x / (x + 1) ? > Can someone help me refresh my memory? It seems like > the function > f(x) = x / (x + 1) > has a name of some sort, but I can't remember it or > seem to track it down. This probably isn't what you're looking for, but you might find it interesting that the graph of y = x/(x+1) consists entirely of those points (x,y) that satisfy consist entirely of those points (x,y) that satisfy x+y = xy and y-x = y/x, respectively. I guess you could call these operation-changing functions! Dave L. Renfro === Subject: Re: name for fx = x / (x + 1) ? > Can someone help me refresh my memory? It seems like the function > f(x) = x / (x + 1) > has a name of some sort, but I can't remember it or seem to track it > down. Maybe you are thinking of this class of functions: (ax+b)/(cx+d) called fractional linear transformations. http://mathworld.wolfram.com/LinearFractionalTransformation.html LH === Subject: Re: name for fx = x / (x + 1) ? tutufan@gmail.com escreveu: > Can someone help me refresh my memory? It seems like the function > f(x) = x / (x + 1) > has a name of some sort, but I can't remember it or seem to track it > down. > Mike It's the ratio of 2 polynomials. Such functions are usually called rational functions. Artur === Subject: Re: name for fx = x / (x + 1) ? I've never seen a name for it and I wonder why it would have one! Gordon PhD Math 1976 === Subject: Which is more flawed and erroneous, MathWorld's Euclid Infinitude of Primes or Wikipedia's --- quoting the current Wikipedia page 15MAR06 on Euclid's Infinitude of Primes proof --- How many prime numbers are there? There are infinitely many prime numbers. The oldest known proof for this statement is given by the Greek mathematician Euclid in his Elements (Book IX, Proposition 20). Euclid states the result as there are more than any given [finite] number of primes, and his proof is essentially the following: Suppose you have a finite number of primes. Call this number m. Multiply all m primes together and add one (see Euclid number). The resulting number is not divisible by any of the finite set of primes, because dividing by any of these would give a remainder of one. And one is not divisible by any primes. Therefore it must either be prime itself, or be divisible by some other prime that was not included in the finite set. Either way, there must be at least m + 1 primes. But this argument applies no matter what m is; it applies to m + 1, too. So there are more primes than any given finite number. --- end quoting --- --- quoting MathWorld's Euclid Infinitude of Primes proof of 15MAR06 --- Given a finite sequence of consecutive primes 2, 3, 5, ..., p, the number N==2.3.5...p+1, (1) known as the ith Euclid number when p==p_i is the ith prime, is either a new prime or the product of primes. If N is a prime, then it must be greater than the previous primes, since one plus the product of primes must be greater than each prime composing the product. Now, if N is a product of primes, then at least one of the primes must be greater than p. This can be shown as follows. If N is composite and has no prime factors greater than p, then one of its factors (say F) must be one of the primes in the sequence, 2, 3, 5, ..., p. It therefore divides the product 2.3.5...p. However, since it is a factor of N, it also divides N. But a number which divides two numbers a and b I've been thinking about contacting Wikipedia and asking for the Infinitude of Primes section to be made unalterable to stop AP tampering with it. Does anyone except AP have any reason why this would not be a good idea, and is it possible for Wiki to lock sections of a page without locking the rest of the page? It was lost to history as to the extent of the hatred towards people of genius. We can never know how much hatred Democritus endured. Nor can we fully appreciate the hatred endured by Galileo or even Newton. But the world, for the first time, has a written electronic history of the hatred that A. P. endured. If any college professor of mathematics had done what I have done for the Euclid Infinitude of Primes, they would have appeared on the front cover of every science magazine and newspaper in the world. Instead, when A.P. does something of huge importance, hatred and suppression emerges. The reason for this is that I am the author of the Atom Totality theory, and that these little-minded hatemongers cannot be happy with the credit of anything that comes from Archimedes Plutonium. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Which is more flawed and erroneous, MathWorld's Euclid Infinitude of Primes or Wikipedia's > I am going to try giving my most elegant and best flowing proofs of > Infinitude of Primes both Direct and Indirect. > Direct method of Infinitude of Primes proof: We show that if we can > increase the set cardinality by one more new prime to any finite set of > primes means the primes are infinite. What? version of this proof. Oh I see. Parody. > I get the sense that MathWorld held a drunk on alcohol party and who > could write the most pathetic and messy Euclid's Infinitude of Primes. > Archimedes Plutonium > www.iw.net/~a_plutonium > whole entire Universe is just one big atom > where dots of the electron-dot-cloud are galaxies Pretty good. === Subject: Re: Which is more flawed and erroneous, MathWorld's Euclid Infinitude of Primes or Wikipedia's > Wikipedia's is better than MathWorld's but that is not saying much > because both are flawed and invalid. How many different threads will you start on the same silly subject? *PLONK^^2* > Archimedes Plutonium -- Ioannis === Subject: Re: Which is more flawed and erroneous, MathWorld's Euclid Infinitude of Primes or Wikipedia's >>Wikipedia's is better than MathWorld's but that is not saying much >>because both are flawed and invalid. > How many different threads will you start on the same silly subject? > *PLONK^^2* Yeah Wikipedia itself has a discussion mechanism and more. http://en.wikipedia.org/wiki/Wikipedia:Articles_for_deletion === Subject: Re: Which is more flawed and erroneous, MathWorld's Euclid Infinitude of Primes or Wikipedia's <1142444677.605807@athnrd02> <4418565C.7050202@fastmail.fm> Yeah Wikipedia itself has a discussion mechanism and more. http://en.wikipedia.org/wiki/Wikipedia:Articles_for_deletion I figured out why the Wikipedia and MathWorld Euclid Infinitude of Primes is so weird, strange and has the feel of someone drunk writing a proof. Is because both pages have sent authors to try to verbatim give what Euclid gave over 2 milleniums as a proof. That is why the MathWorld has a fetish over whether a number is not equal to 1, and that is why the Arthur Rubin writeup of IP in Wikipedia is so odd is because he tried to assemble a proof keeping to the original. But in both cases, do we want to keep to the history and sacrifice and lose validity or do we want to give a proof which is valid and true. Because Euclid mixed methods and his own taken word for word is not a valid proof. Much like several of Euclid's theorems in geometry are invalid because they had hidden assumptions for which modern math corrected. We certainly give Euclid the credit for proving primes are infinite because the heart of the proof is the forming of multiply the lot and add 1. But the assemblage of the proof would not pass muster in our modern day logical syllogisms of Symbolic Logic. So I think it is recommended that we give a version of Euclid IP that is valid and true and does not follow the historical version, since so many believe it was reductio ad absurdum, yet I believe it was direct method. The valid proof of IP is these two: Direct Method: to any finite set of primes of cardinality n can be increased by one more by forming a new number W+1 where we multiply the lot of primes add 1. Either W+1 is prime itself or that W+1 has a prime factor not on the finite list. Since any finite set of primes can be increased by one more means primes are infinite. The Direct Method is the one Euclid was writing. Indirect Method: Definition of Primes. Next step is the reductio ad absurdum where we suppose primes are not infinite but instead are finite. Then p_f is the last and final prime and 2,3,5,7, ..., p_f is all the primes that exist. Construct W+1 = (2x3x5x...x p_f ) +1 Because of the definition of prime in step one W+1 is necessarily a new prime. Contradiction. Hence primes are infinite. I think it is appropriate for MathWorld and Wikipedia to give both the direct and indirect method, for it allays so much of the confusion. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: C Hierarchy Mail-To-News-Contact: abuse@dizum.com >> C: There are the CRANKS. They sit at the top of the C Hierarchy. >> R: Then there are those who RESPOND to the CRANKS. They exhibit of rationality in their responses, but their irrationality is evident in that they think that one can rationally engage a CRANK, and even convince him of his folly. >Not necessarily so. I, and many others, have responded to cranks on >occasion, NOT in an attempt to convince the crank, but in order to help >naive readers avoid falling into the crank's net. Many of the >respondents are educators---active or retired---who are still >interested in trying to help newcomers sort out the wheat from the >chaff. As somebody who has learned more from the responses to certain posters than they're willing to, I'd just to thank those of you who provide this public service. -- Michael F. Stemper #include Build a man a fire, and you warm him for a day. Set him on fire, and you warm him for a lifetime. === Subject: Image transformation - homography matrix and control points - MATLAB I have an image with some detected feature points. I can transform the image using the 3x3 matrix H in matlab like so: t = maketform('projective',H'); [img2 xdata ydata] = imtransform(img, t, 'bicubic'); My problem is that I am trying to also transform the co-ordinates of the detected features so that they line up correctly in the transformed image. To do this I am using the following code: for i = 1:size(x2,2) x2t(i) = (H(1,1)*x2(i) + H(1,2)*y2(i) + H(1,3))/(H(3,1)*x2(i) + H(3,2)*y2(i) + H(3,3)); y2t(i) = (H(2,1)*x2(i) + H(2,2)*y2(i) + H(2,3))/(H(3,1)*x2(i) + H(3,2)*y2(i) + H(3,3)); end with x,y co-ords in x2,y2. But this doesn't seem to transform the points correctly - is the equation wrong. Can anyone help? Chris === Subject: A modest suggestion Many of our posters, with valid mathemtics questions, do not speak English as their first language. There are several on this site who choose to criticize them for bad spelling and bad grammar. I say Lighten up! I've learned enough French, Spanish and Russian to get by in a restaurant or shop, but I know it is atrocious to the native speaking person. I frankly must conclude that the folks on this site who choose to correct the grammar of folks from many miles away may simply be embarrassed by the fact they know only one language. I suggest that next time one of you wants to criticize, why not step back for a second and ask yourself How would I post to a German site? === Subject: Re: A modest suggestion > Many of our posters, with valid mathemtics questions, do not speak > English as their first language. There are several on this site who > choose to criticize them for bad spelling and bad grammar. I think that this is mostly NOT the case. Badly worded, illogical postings are different from postings containing grammatical errors. When I was teaching, I had classes that were > 50% asian, so many students did not have English as their first language. I always told these students to first try to formulate their questions/answers clearly and logically in their native language, THEN do a translation into English. I think that what people here object to are postings that make no sense at all in any language. R.G. Vickson Adjunct Professor, University of Waterloo > I say > Lighten up! I've learned enough French, Spanish and Russian to get > by in a restaurant or shop, but I know it is atrocious to the native > speaking person. > I frankly must conclude that the folks on this site who choose to > correct the grammar of folks from many miles away may simply be > embarrassed by the fact they know only one language. I suggest that > next time one of you wants to criticize, why not step back for a second > and ask yourself How would I post to a German site? === Subject: Re: A modest suggestion I'll be honest when I say that, yes, illogical posts are, well, illogical, I have seen too many times comments such as Learn how to spell! or Put some spaces in your sentences. These are definitely aimed at the language errors. Thus they are unfair and thus my post. === Subject: Re: A modest suggestion > I'll be honest when I say that, yes, illogical posts are, > well, illogical, I have seen too many times comments such > as Learn how to spell! or Put some spaces in your > sentences. These are definitely aimed at the language > errors. Thus they are unfair and thus my post. Many spelling errors can be detected by using a spell-check on the post before submitting it (I copy and paste the text onto a Microsoft Word document and look for the words underlined in red -- it's that simple) and spacing issues are language independent. Also mostly language independent are things like double punctuation marks, no period at the end of a sentence, not capitalizing at the beginning of a sentence, etc. I find it especially curious when a native speaker does this. For me, at least, it's *harder* NOT to do these things simply because this is standard format for everything else I have to write, whether it's a memo to someone, a report for my supervisor, an e-mail to a client, a letter of recommendation for someone, anything being prepared for publication, etc. But that's me. Maybe other people manage to get through the entire day without having to write anything where not coming across as illiterate is an issue. Dave L. Renfro === Subject: Re: A modest suggestion > I'll be honest when I say that, yes, illogical posts are, > well, illogical, I have seen too many times comments such > as Learn how to spell! or Put some spaces in your > sentences. These are definitely aimed at the language > errors. Thus they are unfair and thus my post. > Many spelling errors can be detected by using a spell-check > on the post before submitting it (I copy and paste the > text onto a Microsoft Word document and look for the words > underlined in red -- it's that simple) and spacing issues > are language independent. Also mostly language independent > are things like double punctuation marks, no period at > the end of a sentence, not capitalizing at the beginning > of a sentence, etc. I find it especially curious when > a native speaker does this. For me, at least, it's > *harder* NOT to do these things simply because this > is standard format for everything else I have to write, > whether it's a memo to someone, a report for my supervisor, > an e-mail to a client, a letter of recommendation for > someone, anything being prepared for publication, etc. > But that's me. Maybe other people manage to get through > the entire day without having to write anything where not > coming across as illiterate is an issue. All I can say is, I'm glad e. e. cummings died before the advent of the Internet :-) --Ron Bruck ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: A modest suggestion > I'll be honest when I say that, yes, illogical posts are, well, > illogical, I have seen too many times comments such as Learn how to > spell! or Put some spaces in your sentences. These are definitely > aimed at the language errors. Thus they are unfair and thus my post. yourriteiagreeecompletlyhooneedsspelingrospacesropuncuashun -- J. H. Palmieri Associate Professor of Mathematics University of Washington Box 354350, Seattle, WA 98195-4350 palmieri@math.washington.edu http://www.math.washington.edu/~palmieri/ === Subject: Factoring polynomials over binary field When can a polynomial of the form 1+x+x^2+...+x^{n-1}, with n prime be factored over binary field? For instance they factor for n equal to 7, 17, 23 but not for 3, 5, 11 1+x+x^2+x^3+x^4+x^5+x^6 = (1+x^2+x^3)(1+x+x^3) === Subject: Re: Factoring polynomials over binary field > When can a polynomial of the form 1+x+x^2+...+x^{n-1}, with n prime be > factored over binary field? For instance they factor for n equal to 7, > 17, 23 but not for 3, 5, 11 > 1+x+x^2+x^3+x^4+x^5+x^6 = (1+x^2+x^3)(1+x+x^3) Use the buzzword cyclotomic coset. These are minimal sets S of moduli (modulo n) that have the property i is in S => 2*i is in S. Modulo 7 you have two cyclotomic cosets {1,2,4} (4*2=1 mod 7) and {3,5,6} (6=3*2, 5=6*2 mod 7, 2*5=3 mod 7), but modulo 11 there is only a single cyclotomic coset (apart from {0} that is) {1,2,4,8,5=16,10=2*5,9=20,7=18,3=14,6} (and finally 2*6=12=1). The irreducible factors are in one-to-one correspondence with the cyclotomic cosets. See any textbook on finite fields. Jyrki === Subject: Re: Factoring polynomials over binary field > When can a polynomial of the form 1+x+x^2+...+x^{n-1}, with n prime be > factored over binary field? For instance they factor for n equal to 7, > 17, 23 but not for 3, 5, 11 > 1+x+x^2+x^3+x^4+x^5+x^6 = (1+x^2+x^3)(1+x+x^3) Irreducible if and only if 2 is a primitive root mod n? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Factoring polynomials over binary field >> When can a polynomial of the form 1+x+x^2+...+x^{n-1}, with n prime be >> factored over binary field? For instance they factor for n equal to 7, >> 17, 23 but not for 3, 5, 11 >> 1+x+x^2+x^3+x^4+x^5+x^6 = (1+x^2+x^3)(1+x+x^3) > Irreducible if and only if 2 is a primitive root mod n? Is this an exercise in some number-theory book or something? I answered the same question at least once a few months ago in sci.math.research. (Same answer as Gerry's of course!) dave === Subject: Re: Factoring polynomials over binary field By binary field I meant Z/2 === Subject: Double series Hi all, I have a double series (x_i_j), (i,j) in N^2, where the real numbers x_i_j are generated by a simulation program. I know that, for every i, there exists the real number y_i = Sum(j=1, inf) x_i_j and also know that Sum(i=1,inf) y_i converges to some unknown real y. Since the numbers x_i_j are obtained by a simulation program, I'm trying to aproximate y by the sum of the terms of a finite m X n matrix, provided m and n are sufficiently large. That is, given a tolerance eps, I want to find m and n such that I can ensure |(Sum(i=1,m) (Sum(j=1, n) x_i_j_ - y| <= eps, so that, the aproximation y ~= Sum(i=1,m) (Sum(j=1, n) x_i_j is reasonable for my purposes. Is there any condition that allows me to make such eveluation? I think a necessary condition is that we can interchage the summation signs without affecting the limit, right? But is this sufficient? Artur === Subject: A bit of Fun... OK, all work and no fun makes John a dull boy. Have you wondered what your students search for while sitting in front of a search engine? No? Here's your chance to find out. All search phrases sorted according to category: 1) Sex, ing Machines, Wanking, Animal Sex, Perversions!! 2) Math, Computers, Physics, Biology, Geography, History, Science, Science Projects, Science Fiction, Einstein!! 3) Art, Music, MIDI, Composers, Pianists, Painters!! 4) Philosophy, Religion, Faith, Metaphysics, Demons, Priests, Mythology, The Meaning of Life!! 5) Matrix! 6) Sociology, Politics, Lawyers, Human Relations, Sports, Boyfriends, Girlfriends, Advertising!! 7) Medicine, Doctors, Diseases, Insanity, Dementia, Schizophrenia, Paranoia, Desperation! 8) Illucidity, Nonsense, Bull! Winner category: 2) The insightful reader will probably ask how is it possible for my tracker to trap so many seemingly unrelated search phrases. The answer is simple: My web pages are over 10 megabytes of text, and the search engine is usually doing an OR (words) search, so it may well throw a searcher on my web pages, without necessarily preserving the question's context (i.e. when a page contains all the search words, but not necessarily on the same sentence). So, sit back, relax and enjoy: http://ioannis.virtualcomposer2000.com/writing/SearchPhrases.html -- Ioannis === Subject: Re: A bit of Fun... > Have you wondered what your students search for while sitting in front of a > search engine? > No? Here's your chance to find out. All search phrases sorted according to > category: > 2) Math, Computers, Physics, Biology, Geography, History, Science, Science > Projects, Science Fiction, Einstein!! > Winner category: 2) Winner within that category: Searching for proof of lemma 4 Phil -- What is it: is man only a blunder of God, or God only a blunder of man? -- Friedrich Nietzsche (1844-1900), The Twilight of the Gods === Subject: Big Pick Watcher by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id k2FKE0313766 for ; Wed, 15 Mar 2006 15:14:00 -0500 by support2.mathforum.org (8.12.10/8.12.10/The Math Forum, $Revision: 1.6 secondary) with ESMTP id k2FKCIF4002162 for ; Wed, 15 Mar 2006 15:12:18 -0500 Big News Just Out! More expected Thurs Nanoforce Inc. (NNFC), This really moved on Monday! Went Crazy on Tuesday! Wend Should do even Better! Big News Just Out Monday March 13, 2006 (PRIMEZONE) Nanoforce Technologies Inc.'s Patent Position Expands Again With Issuance of Patent for Producing Improved Catalyst SYMBOL: N N F C NANOFORCE INC. (NNFC) - Current Price: 1.17 Up over .72 cents in last 5 days alone Tuesday Was GREAT Wend Will Be BETTER! Before we start with the profile of NNFC we would like to mention something very important: There is a Big PR Campaign just started. We are already seeing movement it will go all week so it would be best to get in NOW Nanoforce Inc. (NNFC) appears to be on an upward trend. Is not the time to buy ? With current marketing campaign we expect NNFC to climb thru mid March Get In Now! NANOFORCE INC. (NNFC) COMAPNY INFORMATION Nanoforce, Inc. is a company founded to support the development, and acquisition of exciting new products that integrate fundamental developments in Nanotechnology. The Nanotechnology industry is in its infancy and provides a wealth of opportunity for companies that position themselves well now to profit from the burgeoning opportunities that will be provided by nanotech produced products in the near future. NANOFORCE INC. (NNFC) RECENT NEWS (For Complete News Detail, check your current stock site.) March 13 Nanoforce Technologies Inc.'s Patent Position Expands Again With Issuance of Patent for Producing Improved Catalyst March 7, 2006: Nanoforce Inc. Hires Local Management Team to Head Refinery Science Corp.'s El Paso Office February 13, 2006: Nanoforce Acquires Unique package of Nanotech. Ips December 16, 2005: Nanoforce Acquires Refinery Science Corp. December 16, 2005: Nanoforce approves 10 for 1 split of Refinery Science. CHECK OUT NANOFORCE (NNFC) TODAY! === Subject: Sperical Polar conversion to Cartesian with a twist Hi Folks, I have an application where I need to convert spherical polar to Cartesian in 3D. The application relates to a helicopter flying above some uneven terrain. A ray extends from the front of the helicopter and I need to test if it hits the terrain and where. I plan to do this with a repeated series of Ray Triangle Intersection Test. The ray triangle intersection function accepts input of a point where the ray originates, and a second set of vectors defining the direction of the ray. The problem is my ray is defined by only two angles (it doesnt have length - i guess becauses its a ray) I need to convert my two angles to direction vector so I can plug it into the Ray Tirangle function. In other words I have a line extending from point xyz, in the direction of elevation angle A and direction and D. I need to find a vector coordinates to define the direction of this line. I am missing a length, and it can be defined as any length as long as it lies on the line. Anyone able to help? -Al === Subject: Re: Sperical Polar conversion to Cartesian with a twist I am not really sure what you are asking. may be the following helps x = r cos phi sin theta y = r sin phi sin theta z = r cos theta r^2 = x^2+y^2+z^2 but i am afraid this does not really help. when conisidering a moving plane /helicopetr. this objet has 6 degrees of freedom. this is because while the helicopter is flying, the center of mass may fly on a trajectory, whereas the nose may go up and down. in addition the heli may swing left and right and it may even crank a bit. so the setting up the equations of motion you may need the geometry of the heli. in addition you need the moment of inertia of the copter. you need to solve six equ of motion three of which describe the co-ordinate x,y,z and the other three are the derivatives of x,y,z may be you have done this ? === Subject: Re: Sperical Polar conversion to Cartesian with a twist > I am not really sure what you are asking. may be the following helps Mentioning Spherical coords probably confused things. Dont worry about the orientation of the helicopter. Consider this: I have a point in xyz. (the helicopter) I have the direction of a ray starting at xyz defined by two angles, one in the horozontal and the other is elevation. I want to convert this direction, defined by the angles to a direction defined by a vector from xyz. xyz is now like a origin for this vector The vector can be any length or magnitude as long as it lies on the line defined by the two angles. Make sense? -Al === Subject: Re: Sperical Polar conversion to Cartesian with a twist >I want to convert this direction, defined by the angles to a direction i am still guessing what you mean. do have the problem to calc the angle between the rays. this ist the arc cos of the scalar product of the the two rays what do mean by direction define by a vector form xyz , is it the direction of the velocity? another guess: the two rays form a plane the normal vector of the plane is the cross product of the two rays now any vector perpendicular to this normal vector and (!) pointing into the direction between the two rays is your solution? === Subject: Re: Sperical Polar conversion to Cartesian with a twist > Mentioning Spherical coords probably confused things. > Dont worry about the orientation of the helicopter. Consider this: > I have a point in xyz. (the helicopter) > I have the direction of a ray starting at xyz defined by two angles, > one in the horozontal and the other is elevation. > I want to convert this direction, defined by the angles to a direction > defined by a vector from xyz. xyz is now like a origin for this vector > The vector can be any length or magnitude as long as it lies on the > line defined by the two angles. > Make sense? Sure. It's high-school geometry time. So, suppose z increases up, x increases right, and y increases forward. (I think that's a right handed system.) You will have to substitute in the correct things if that's not it. Let the first angle be the angle counter-clockwise from the x-axis. Call it phi. So, straight ahead is phi=90 degrees, right is phi = 0, left is phi = 180, and straight back is phi=270. (Or -90 if you like.) Let the second angle be down from the z axis. Call it theta. So, level is theta = 90 degrees, straight up is theta = 0, and straight down is theta=180. A unit vector in the direction of the ray has x,y, and z components: (sin theta sin phi, sin theta cos phi, cos theta) (Exercise left for the student: Show this is a unit vector.) And to extend this, you just multiply it by r, the new length. You may not have taken the zero of your phi to be on the x-axis. Then you just have to put in whatever your zero is and swing things around by 90 degrees (or whatever). Hopefully you know how to deal with something like cos(theta + 90) and so on. Or you may have taken your theta to be zero when the ray is level, so you would need to adjust that by 90 degrees. But basically you just need to make the correct combination of cos and sin of the two angles to make a unit vector in the direction of the ray. Socks === Subject: Re: Can you see the Math in this equation?? Yes, I can see the math. And I can see the fatal flaw that dooms it to failure, spammer. Go to jail. See Title 39, Section 3005 of the United States legal code. Or just read what the judges say for yourself: http://www.usps.com/judicial/1976deci/4-38.htm http://www.usps.com/judicial/1978deci/6-72.htm http://www.usps.com/judicial/1982deci/14-64.htm > MAKE A LOT OF MONEY FAST AND LEGAL!!!!!! > PLZ READ === Subject: First female Ph. D. in various countries Will you please post (or reply to rgompa@iuk.edu) the first woman to receive a Ph. D. in mathematics from various countries? I know that Sonia Kovalevsky is the first woman to receive Ph. D. in Mathematics. She is originally from Russia. She got Ph. D. from Germany. Winifred Edgerton Merrill is first American Woman to receive Ph. D. in mathematics. Please help. .. Raghu === Subject: Re: First female Ph. D. in various countries The first _American_ woman to get a doctorate in math _in the USA_ was Winifred Edgerton Merrill in 1886 at Columbia. You can probably find more such data at www.awm-math.org === Subject: Re: First female Ph. D. in various countries >The first _American_ woman to get a doctorate in math >_in the USA_ was >Winifred Edgerton Merrill in 1886 at Columbia. You can >probably find >more such data at >www.awm-math.org I have tried to find the information at www.awm-math.org and did not find the data I wanted. So far, first ph.d.'s from Canada and America have been posted. I hope we will have more soon. I am from India, but I do not know who the first female Ph. D. in mathematics from India was. But, I hope some from this group will post. === Subject: Re: First female Ph. D. in various countries I think it's Saraswati Mohan. === Subject: Re: First female Ph. D. in various countries >I think it's Saraswati Mohan. more information about Saraswathi Mohan through searching web. If you have any further information, please post it. I appreciate === Subject: Re: First female Ph. D. in various countries ><29226581.1142543835379.JavaMail.jakarta@nitrogen.mathforum.org>, >>I think it's Saraswati Mohan. >more information about Saraswathi Mohan through searching >web. If you have any further information, please post >it. I appreciate http://webhome.idirect.com/~aum108/sanskirt.htm I see that sarswati mohan is the first ph. d. in Sanskit. I am not sure whether she is the you are thinking of. I want to find out first ph.d. in mathematics. .. Raghu === Subject: Re: First female Ph. D. in various countries >Will you please post (or reply to rgompa@iuk.edu) the >first woman to receive a Ph. D. in mathematics from >various countries? Cecilia Krieger was the first woman (and the third person) to receive a PhD in Mathematics at a Canadian university (U. of Toronto, 1930). You might find the Biographies of Women Mathematicians website useful: Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: First female Ph. D. in various countries Was Sophie Germain, a contemporary of Gauss, ever awarded a PhD? Sadly she was not allowed to attend classes, but with KFG's help she published as S. Germain. === Subject: Re: First female Ph. D. in various countries > Will you please post (or reply to rgompa@iuk.edu) the > first woman to receive a Ph. D. in mathematics from > various countries? For Canada it was Cecelia Krieger, in 1930: http://www.science.ca/scientists/scientistprofile.php?pID=283 http://www.agnesscott.edu/lriddle/women/krieger.htm At least two other Canadian notables were around UofT at that time: Fields and Beatty. === Subject: =?iso-8859-9?q?t=FDkkkkkkkkkaaaaaaaaaaaaaaaaaaa?= t[CapitalYAcute]kla buna www.smsci.kulubu.com ok === Subject: Splines is it true that for cubic splines placement of its first knot t_i-1 has no impact on values N_i,n(t) of the basic function in the last interval of its support [t_i+n-1, t_i+n] ? I have to use such theorem but don't know where it comes from or what are the mathematical bases of that :(( Bartek === Subject: Gauss-Seidel Iteration Problem- A Better Solution??? I was presented the following problem recently: Consider the following tridiagonal linear system, and assume that the coefficient matrix is strictly diagonally dominant. D1X1 + C1X2 = B1 A1X1 + D2X2 + C2X3 = B2 A2x2 + D3X3 + C3X4 = B3 .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. An-2Xn-2 + Dn-1Xn-1 + Cn-1Xn = Bn-1 An-1Xn-1 + DnXn = Bn (i.) Write an iterative algorithm, followign Gauss-Seidel iteration, that will solve this system. Your algorithm shoiuld efficiently use the sparseness of the coefficient matrix. (ii). Construct a MATLAB program based on your algorithm and use it to solve the following tridiagonal system. (A 50x50 matrix) 4X1 + X2 = 1 X1 + 4X2 + X3 = 2 X2 + 4X3 + X4 = 1 X3 + 4X4 + X5 = 2 .. .. .. .. .. .. .. .. .. .. .. .. X48 + 4X49 + X50 = 1 X49 + 4X50 = 2 Here is my Solution, However my colleagues insist a better one exists or the code can be optimised. Does anyone have any ideas???? Any help is greatly appreciated. SOLUTION (i.) For a Tridiagonal Matrix X1(K+1) = (B1 - C12*X2^K)/D11 EQN. 1 Xi(K+1) = 1/Dii * (Bi - Ai*(i-1)*Xi-1^(K+1) - Ci(i+1)*Xi+1^K) For i = 2,3,....N-1 EQN.2 XN^(K+1) = (BN - AN,N-1 * XN-1^(K+1))/DNN EQN.3 (ii.) MATLAB CODE A = triu(tril(ones(50), 1),-1); %Creates the tridiagonal matrix A A(find(eye(length(A)))) = 4; b = ones(50,1); %Creates n by 1 matrix b b(2:2:length(b)) = 2; P = ones(50,1); %Creates a initial guess n by 1 matrix P max1 = 100; %Maximum number of iterations delta = 1e-14; %Tolerance number for convergence N = length(A); %Size of square matrix x = P; %initializes x in memory using P for k = 1:max1 x(1) = (b(1) - A(1,2)*P(2))/A(1,1); %Eqn 1 x(N) = (b(N) - A(N,N-1)*x(N-1))/A(N,N); %Eqn 3 for i = 2:N - 1 x(i) = (b(i) - A(i,i-1)*x(i-1) - A(i,i+1)*P(i+1))/A(i,i); %Eqn 2 end err = abs(norm(x-P)); if (err<=delta) %Checks for convergence break end P = x; end === Subject: Re: Symbolic Logic write-up of Infinitude of Primes, direct & indirect methods; Infinitude of Twin Primes > I assume you meant p_f instead of pn in (8). > (13) is based on the assumption in (3). > This gives a contradiction so (15) correctly says (3) must be false. > Since (3) is false, we do NOT know whether W+1 is prime. > The conclusion that W+1 is prime was based on a false premise. > W+1 is frequently composite when 2,3,...,p_f are the first primes, e.g. > 2*3*5*7*11*13+1 = 59*509. > Jens, you do not understand Euclid's Indirect Method and your above is > wrong, silly and foolish. You do not know the logical structure of a > proof. How a proof works and how logic works. And you flip flop between > the Direct and Indirect. > Trouble with correcting Euclid's old proof on the Internet is that > everyone who has little to no math abilities, thinks they know this > proof, and they come pouring out of the woodwork like blind bats. If I > was doing something like the quintic, or Riemann hypothesis instead of > Infinitude of Primes, then all these people who know little to nothing > about math would stay back into their woodwork or bat cave. True, I would never tackle a proof of the Riemann Hypothesis as I have not the skills to critique it. However, your proof is easy for even the most amateur mathematician to invalidate, which must be quite embarrassing for you. > Try reading the below and perhaps the lights will come on, but I doubt > it. You simply don't understand the consequence(s) of contradiction, namely the fact that once a contradiction is reached, any result based on the false premise is no longer necessarily true. As a little exercise for you, please point out the error in the following proof: Theorem: 9 is prime Proof: Assume 2x+1 is prime for all natural numbers x. Then 2(4)+1 is prime and 2(10)+1 is prime. But we know that 2(10)+1 = 21 = 3*7. Thus, a contradiction, so 2x+1 is not prime for all x, but 2(4)+1 = 9 is prime. At this point I assume you contend that you already knew 9 was not prime, and that in your proof (under the eventually false premise), the quantities W+1 and W-1 are not known to be composite. But it doesn't matter, the results (that 9 is necessarily prime in my proof, and that W+1 and W-1 are necessarily prime in yours) are all based on premises proved to be false. Once contradiction is reached, all results based on the false premise do not necessarily hold. Other posters have been trying to get you to see this, with no success. Even if they concede that W+1 and W-1 are prime, based on the (false) premise, once the contradiction is reached, then the premise is false, and the Unique Factorization Theorem (UFT) applies. I imagine Euclid's (valid) proof includes the reference to the UFT because he (and everyone else but you) understood that once the contradiction is reached, the UFT will apply. That doesn't make Euclid's proof invalid, just difficult for _you_ to understand. I'm not sure why other posters bother to keep trying to illuminate you, but I know that the invalidity of your proof is very obvious to me, and I only have a minor in mathematics. If I may be so bold as to speak for the group, I would say that you should abandon this proof and put your math degree to good, productive use. If I had the skills one would gain from a full degree in Mathematics, I wouldn't bother myself with an elementary approach to the problem of infinitude of twin primes, especially one that someone without a full degree in Mathematics can easily refute. Here's an idea if you're interested in a new approach. Definition: Let k be defined as the kernel of a twin prime pair such that 6k-1 and 6k+1 are prime. If either 6k-1 or 6k+1 are not prime, then k is NOT a kernel. Conjecture: For every k > 1, if k is the kernel of a twin prime pair, then it can be written as the sum of two (not necessarily distinct) kernels a and b. Note that k=1 is a kernel of a twin prime pair. I have verified this conjecture up to k = 11896420. Examples: When k=2, then a = b = 1, which are both kernels. Also, when k = 3, then a = 2, b = 1, which are both kernels. A possible approach is to assume finitely many twin primes, then construct a new kernel. Hence a new twin prime pair, and thus contradiction. Care to work on this one? Respectfully, Matt B === Subject: Confidence Intervals If we know the true population mean (mu) what would the procedure be for finding a 95% confidence interval for sigma^2? === Subject: Re: Confidence Intervals > If we know the true population mean (mu) what would the procedure be for finding a 95% confidence interval for sigma^2? What does your textbook say? What does your instructor say? R.G. Vickson === Subject: variance Can someone please tell me why sometimes when calculating the variance === Subject: Re: variance > Can someone please tell me why sometimes when calculating the variance On Monday, Wednesday, and Friday, whee divide by n-1. On Tuesday, Thursday, and Saturday, weee divide by n. On Sunday, we do not compute the variance. === Subject: Re: variance >Can someone please tell me why sometimes when calculating the variance The variance of a discrete distribution is defined as n n 1 --- _ 2 1 --- 2 _ 2 - > (x - x) = - > x - x n --- k n --- k k=1 k=1 That is, the mean of the squares minus the square of the mean. However, when taking sampling, in general the sample mean differs from the distribution mean. The sample mean is shifted toward the samples which, in general, causes the sample variance to be smaller than the distribution variance. As shown on the page at , the the expected sample variance is (n-1)/n times the distribution variance. Therefore, the proper way to correct this is to multiply the sample variance by n/(n-1). Since the formula for the variance has n in the denominator, the correction replaces the n with an n-1. Rob Johnson take out the trash before replying === Subject: Re: variance <20060315.180528@whim.org> Ah, thank you, that makes perfect sense. === Subject: Re: variance > Can someone please tell me why sometimes when calculating the variance If you KNOW the mean, you divide by n. However, if you also have to estimate the mean (by taking the sample average) when computing the variance, you divide by (n-1). You do this in either case so as to obtain a so-called _unbiased_ estimator of variance. Go to a book to find out what is meant by 'unbiased', and why it is (sometimes) regarded as important. R.G. Vickson === Subject: Profile of convicted Nazi spy Korey Jerome Kruse of Olathe, KS I am alone today due to the covert machinations of dozens, possibly hundreds of women in several countries, Korey Jerome Kruse, 34, said. What we are looking at is a plot of epic proportions, which may seem counterintuitive, but that is, in fact, precisely what they would like you to believe. Conspiracy Theorist Korey Jerome Kruse gathers relevant evidence on the Internet. I am a Mechanical Engineer who spent many years in Aerospace, including structural design, and in the design, and use of shaped charge explosives (like those that would be used in missile warheads). In light of a broken engagement two years ago, area school-bus driver and longtime conspiracy enthusiast Korey Jerome Kruse outlined an intricate theory to reporters Tuesday to explain his failure to begin a new relationship. Korey Jerome Kruse said he began to suspect a hidden hand at work Kruse Osborne, when potential dates routinely refused to return his calls or e-mails. Korey Jerome Kruse cites a comprehensive archive of past-girlfriend-related historical evidence that he has collected over the past 10 years. Korey Jerome Kruse, considered the top Robert Korey Jerome Kruse romantic-failure expert in the country, has spent years studying the hundreds of letters, photographs, receipts, gifts, and videotapes of himself with former girlfriends, looking for a common pattern. The focus of Korey Jerome Kruse's current research is the six-day According to phone-company records, I called Korey Kruse at exactly me-which she agreed to do after a quick shower, Korey Jerome Kruse said. Twelve minutes later, at 7:46 p.m., Korey Kruse called to say she had 'changed her mind' about dinner, but wanted to come to my apartment to 'deliver some news.' It was there that Osborne announced that she no longer wished to marry Korey Jerome Kruse. Korey Jerome Kruse continued: What happened in that 12-minute gap? What-or who-got to her? And why won't she release her phone records to me? Korey Jerome Kruse said a coordinated, secret, high-level effort is the only plausible explanation. There are wheels within wheels, he said. I would like to give you my input as to the events on September 11, and why it is a physically provable fact that some of the damage done to the Pentagon could not have occurred from a Boeing 757 impact, and therefore the 9/11 Commission report is not complete and arguably a cover-up. I will not speculate about what may have been covered up, I will only speak from my professional opinion. But I will said. March 8 is the 67th day in the Gregorian calendar, Korey Jerome Kruse said. Pope Gregory XIII was a subject familiar to Korey Kruse, who minored in Renaissance history in college. Also, if you add the numbers 3, 8, and 2,004, you get 2,015. Add 2, 0, 1, and 5, and once again, you get 8-exactly the number of women I dated exclusively before I met Korey Kruse. Others targeted by Korey Jerome Kruse as conspiracy players include a female dispatcher who works for his bus company, his mother, and a waitress he asked out three months ago. Korey Jerome Kruse has even traced the conspiracy to figures in the highest echelons of American society, including former Cosmopolitan editor Helen Gurley Brown, Oprah Winfrey, and the TV series Sex And The City. Recently, Korey Jerome Kruse examined a newly unearthed 1997 video of him and then-girlfriend Donna Soderblum at his sister's wedding. According to Korey Jerome Kruse, repeated slow-motion viewings revealed a telling detail. See that sneer and eye-roll on Donna's face, after she turns away from me and goes back to talking to my sister? Korey J. Kruse said. It's all there in frames 336 through 408. Longtime friend Korey Kruse Warren agrees that Korey Jerome Kruse's single status is not a fluke, but he rejects Korey Jerome Kruse's analysis. Said Warren: I explain all of Bob's difficulties in my meticulously researched and voluminously footnoted 'Lone Wardrobe Theory.' Korey Jerome Kruse dismissed Warren's analysis. Warren's theory is interesting, but it has a long way to go in explaining why I've remained single for more than two years. There is no explanation why, for example, I am rejected by women even when I go out to bars, he said. Also, lots of men fit that description, he added, including Korey Kruse's current boyfriend, Burke. explain why I do not believe the Pentagon was hit by a Boeing 757. An 18-page manifesto that explains Korey Jerome Kruse's theory in more detail is available for free download from his website. The structural design of a large aircraft like a 757 is based around managing the structural loads of a pressurized vessel, the cabin, to near-atmospheric conditions while at the lower pressure region of cruising altitudes, and to handle the structural and aerodynamic loads of the wings, control surfaces, and the fuel load. It is made as light as possible, and is certainly not made to handle impact loads of any kind. If a 757 were to strike a reinforced concrete wall, the energy from the speed and weight of the aircraft will be transferred, in part into the wall, and to the University of Kansas structural failure of the aircraft. It is not too far of an analogy as if you had an empty aluminum can, traveling at high speed hitting a reinforced concrete wall. The aluminum can would crumple (the proper engineering term is buckle) and, depending on the structural integrity of the wall, crack, crumble or fail completely. The wall failure would not be a neat little hole, as the energy of the impact would be spread throughout the wall by the reinforcing steel. This is difficult to model accurately, as any high speed, high energy, impact of a complex structure like an aircraft, into a discontinuous wall with windows etc. is difficult. What is known is that nearly all of the energy from this event would be dissipated in the initial impact, and subsequent buckling of the aircraft. We are lead to believe that not only did the 757 penetrate the outer wall, but continued on to penetrate separate internal walls totaling 9 feet of reinforced concrete. The final breach of concrete was a nearly perfectly cut circular hole (see below) in a reinforced concrete wall, with no subsequent damage to the rest of the wall. (If we are to believe that somehow this aluminum aircraft did in fact reach this sixth final wall.) EXIT HOLE IN PENTAGON RING-C American Airlines Flight 77, a Boeing 757, is alleged to have punched through 6 blast-resistant concrete walls .84a total of nine feet of reinforced concrete.84before exiting through this hole. It is physically impossible for the wall to have failed in a neat clean cut circle, period. When I first saw this hole, a chill went down my spine because I knew it was not possible to have a reinforced concrete wall fail in this manner, it should have caved in, in some fashion. How do you create a nice clean hole in a reinforced concrete wall? with an explosive shaped charge. An explosive shaped charge, or cutting charge is used in Olathe various military warhead devices. You design the geometry of the explosive charge so that you create a focused line of energy. You essentially focus nearly University of Kansas all of the explosive energy in what is referred to as a jet. You use this jet to cut and penetrate armor on a tank, or the walls of a bunker. The signature is clear and unmistakable. In a missile, the explosive charge is circular to allow the payload behind the initial shaped charge to enter whatever has been penetrated. I do not know what happened on 9/11, I do not know how politics works in this country, I can not explain why the mainstream media does not report on the problems with the 9/11 Commission. But I am an engineer, and I know what happens in high speed impacts, and how shaped charges are used to cut through materials. I have not addressed several other major gaps in the Pentagon/757 incident. The fact that this aircraft somehow ripped several light towers clean out of the ground Olathe, KS without any damage to the aircraft (which I also feel is impossible), the fact that the two main engines were never recovered from the wreckage, and the fact that our government has direct video coverage of the flight path, and impact, from at least a gas station and hotel, which they have refused to release. You can call me a tin hat, crazy, conspiracy theory, etc, but I can say from my expertise that the damage at the Pentagon was not caused by a Boeing 757. === Subject: Re: Profile of convicted Nazi spy Korey Jerome Kruse of Olathe, KS On 15 Mar 2006 17:40:40 -0800, Fred MacMurray harrassed a food & cooking group when he absolutely nothing of interest, unless one is interested in his masturbating at the keyboard === Subject: Hash function, Birthday paradox and probabilities. I'm wondering about this problem, its coming from a computer science background but I think this is mainly a probability theory problem. The title is probably misleading as I dont really know how to describe the problem in one line, and I'll probably get the vocabulary all wrong, so dont hesitate to fix the way I set the problem : Lets have: - a hash function H() that take as input a bit string of length L and return an hash of length N bits (ie an integer in the range [0, 2^N[) - a iterator function F, that takes as input a bit string of length L and return another bit string of length L, - S[0], S[1],... S[n], bit strings of length L, so that S[i+1]=F(S[i]), - h[0], ...h[n], so that h[i]=H(S[i]) - M a fixed hash Given all that lets define m, the smallest positive integer so that h[m]=M. I'm wondering how A and H affect the probabilistic properties of m. For example, the practical example is this: - S[0] is a file, that can be divided in 3 parts: S[0]={T, s[0], U} where s[0] is a 32 bits integer. (L is the length of this file) - The H() function is the sha1 hash of the file, truncated to the last 16bits (N=16) - S[i]={T, s[i], U}, with s[i+1]=a(s[i]) - a() is a function that transform a 32 bits integer into another one. (So A() is the function that take a file, and change the 32 bit value at a fixed offset according to the function a()) Now the question is what is the best function a() to minize m (in average), and how to minimize the spread of m around its average. (There must be a better word than spread but I dont remember) I have been trying two things: - s(0)=random(), s(i+1)=s(i)+1 (modulo 2^32), - s(i)=random() I was thinking that, wether I use random or increments, the average of m should be 2^N, and my experiences seem to agree with that. Now, i'm not sure about the spread, my instinct says that it might be better with random, but then again I'm not sure why... Anybody has some though about this. ? -- F.G. === Subject: Re: Hash function, Birthday paradox and probabilities. >I'm wondering about this problem, its coming from a computer science >background but I think this is mainly a probability theory problem. The >title is probably misleading as I dont really know how to describe the >problem in one line, and I'll probably get the vocabulary all wrong, so >dont hesitate to fix the way I set the problem : >Lets have: >- a hash function H() that take as input a bit string of length L and >return an hash of length N bits (ie an integer in the range [0, 2^N[) >- a iterator function F, that takes as input a bit string of length L >and return another bit string of length L, >- S[0], S[1],... S[n], bit strings of length L, so that >S[i+1]=F(S[i]), >- h[0], ...h[n], so that h[i]=H(S[i]) >- M a fixed hash >Given all that lets define m, the smallest positive integer so that >h[m]=M. >I'm wondering how A and H affect the probabilistic properties of m. You mean F and H? ... >Now the question is what is the best function a() to minize m (in >average), and how to minimize the spread of m around its average. >(There must be a better word than spread but I dont remember) If you really want m to be small, make F (which I suppose is the same as a()) always return a string S such that H(S) = M. But I doubt that's what you really want. >I have been trying two things: >- s(0)=random(), s(i+1)=s(i)+1 (modulo 2^32), >- s(i)=random() s(i)=random() is not compatible with S[i+1]=F(S[i]). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: more Galois Theory In the pursuit of an easy constuction of S_n as a Galois group over Q for twin primes, n,p with n = p+2, as a preliminary step show that there exists an even integer a such that a is not a square in F^x(_p) . Then consider the polynomial f(x) = (x^p - x) (x^2 - a) + px^3 + 2p. Show that f is irreducible over Q and Gal_Q (f) is isomormphic to S_n. === Subject: Re: more Galois Theory > In the pursuit of an easy constuction of S_n as a Galois group over Q for > twin primes, n,p with n = p+2, as a preliminary step show that there exists > an even integer a such that a is not a square in F^x(_p) . Sorry, I don't understand this notation. Do you mean F_p? > Then consider the > polynomial f(x) = (x^p - x) (x^2 - a) + px^3 + 2p. Show that f is > irreducible over Q and Gal_Q (f) is isomormphic to S_n.