³ ³ mm-2579 === Subject: Re: Why I am unwilling to use Mathematica if I can help it Okay. How is that an issue with Unicode? === Subject: Re: Why I am unwilling to use Mathematica if I can help it All of my requests for a simple WYSIWYG version of Unicode have been answered with pointers to such as Lyx which are not that. Where do I find the straightforward stuff? -- 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: Why I am unwilling to use Mathematica if I can help it There is a Unicode editor at http://www.yudit.org/ . I know nothing about it, other than it exists. === Subject: Re: Why I am unwilling to use Mathematica if I can help it There is a distinction between the ease of input and the quality of the output. TeX can produce the better math typesetting than its common competitors. BUT the ease of input is rather bad and an argument that for value received it is a bad bargain has much merit. I once heard comments from the floor about the deficiencies of LaTeX that had much weight as the criticism came from its author. At one time there was a rather slick math editor called MathWriter available for Macintosh which seems to faded away. Both of these suggest that developing good mathematical input is very hard. Perhaps something will come from the Fortress project at Sun. === Subject: Re: Why I am unwilling to use Mathematica if I can help it ... This is the new (? 30 years+) orthodoxy: full descriptive names. I suppose it's better than ls | rm. (I choose this specifically because it DOESN'T work, just in case someone tries it.) The point is that short names are quicker, and when you're immersed in a project, more efficient. But come back two months later... --Ron Bruck PS. When Wolfram was touring the country hyping the new Mathematica--I think it was at version 3?--I recall him making claims like, Mathematica's typesetting capabilities were superior to TeX's, and implying that Mma would replace TeX. Ha. Another aspect of Wolfram's mine is bigger mindset. Ah, if only they had made TeX the underlying typesetter! We might now have something like Display TeX instead of Display PostScript. Instead, we're stuck with the constantly recurring problem of a label in Greek, say, which looks fine in Mathematica, but comes out in Courier when translated to PDF. Sigh. Yes, we DO need our own display language. Meanwhile, instead of Unicode being the default [the bare minimum that we need], we're still stuck with 7-bit ascii. Maybe even 6-bit :-( === Subject: Re: Why I am unwilling to use Mathematica if I can help it As I understand it, the Unicode people seem unwilling to give us straight simple Unicode, but want to put in the frills. What we need is simple multifont WYSIWYG, and in most of the situations I face, fixed width as well. Alignment is difficult. -- 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: Why I am unwilling to use Mathematica if I can help it The problem is that the embed fonts (into the postscript if you are using the .ps to .pdf) options are so well hidden and so unmentioned that they might as well not exist. This may have been plausible when file space was tight (remember when a 10MB disk was big!) and files were rarely exchanged. The files always display correctly for the original producer as the display can find the fonts used but not embedded. But if you do not have the same font environment as the originator then you are out of luck. Some times I have asked the originator if it might be possible to have the file produced with the full embedding so that I would be able to read the mathematics. There are three types of responses - 1. no response 2. what is embedding? 3. loosely translated as #@!$% and stay lost! === Subject: Re: terrible mathematica bug ( //N ) but WHY?? Hi. By using just //N, you are asking that the calculation be done using machine precision. Note that the Cosh function returns a rather large number... N[Cosh[(43*Pi)/Sqrt[2]]] 1.5263010243767756*^41 You are loosing precision during the calculation in conjunction with the entire formula Notice that when you simplify the equation, you get 1 function Sech, when can be calculated to any precision. FullSimplify[Cosh[(43*Pi)/Sqrt[2]] + (1 - Cosh[43*Sqrt[2]*Pi])* Csch[43*Sqrt[2]*Pi]* Sinh[(43*Pi)/Sqrt[2]]] Sech[(43*Pi)/Sqrt[2]] You can get an answer to a much larger precision: N[Sech[(43*Pi)/Sqrt[2]], 40] One way to see this is to ask for more precision in your //N example... Here, we ask for 20, instead of machine precision... N[Cosh[(43*Pi)/Sqrt[2]] + (1 - Cosh[43*Sqrt[2]*Pi])* Csch[43*Sqrt[2]*Pi]* Sinh[(43*Pi)/Sqrt[2]], 20] You will see an error that the internal precision limit was reached during the calculation of Cosh function. Using your //N example, the program made the best rounding that it could. One workaround... $MaxExtraPrecision = [Infinity]; Now, you can do the above calculation to 20 digits. Another option when just using //N is to tell the program that your numbers carry more precision prior to using //N Cosh[(43`500.*Pi)...etc Here, we tell the equation that the number 43 has a precision of 500, which will be used when converted to machine precision when using just //N. You will get an error pointing to the other equation that now has this problem, so do the same there. etc. HTH -- Dana DeLouis === Subject: ACCPM Compilation Problems Sigh, I'm stumped so if anyone could help me it would be much appreciated. Ok I'll describe the problem as best I can. I am trying to compile the latest version of ACCPM with Microsoft Visual C++ 6.0 but I keep getting linking errors. Software is here for you to try for yourself http://www.hec.unige.ch/professeurs/VIAL_Jean-Philippe/pages_web/accpm/index .html When I try to compile any of the examples I am told that I am missing the library dfor.lib. I think this is because part of the libACCPM.lib that comes with ACCPM was written in fortran.. I just need to be able to compile it but I am fresh out of ideas... The only other piece of information I have is that I think the dfor.lib library was a part of something called Compaq Visual Fortran. I haven't been able to find that file anywhere though. Any help or direction would be appreciated. I have also read papers on the new version of ACCPM called Proximal-ACCPM but have been unable to find a downloadable version. Does anyone know where I can get it? Ryan === Subject: Problems with bounded linear least squares I've been using the Stark and Parker BVLS Fortran code to solve a bounded linear least squares problem. For the most part, I've been getting good solutions, but I'm running into situations where the solver bails out of the Kuhn-Tucker test after a single variable has become active. Needless to say, the resulting solution in such cases are far, far from optimal. Is anyone out there aware of strategies for getting around these dead solutions or possibly different algorithms for solving bounding linear least squares? Any help would be greatly appreciated. === Subject: Re: Problems with bounded linear least squares if you are sure the algorithm fails, try another one from the LS part of http://plato.asu.edu/guide.html such as Lawson&Hanson's code. Hans Mittelmann === Subject: Re: Problems with bounded linear least squares Try Gauss-Seidel iterations, minimizing w.r.t. each x_i in turn, and setting it to the boundary value if it is out of bounds. === Subject: Re: Problems with bounded linear least squares Do you mean peforming Gauss-Seidel iterations to warm-start the bvls algorithm? If so, do I want to keep the x_i values or just use the Gauss-Seidel iterations to determine the active set? I think the bvls code, during a warm start, sets any active variables values to (bu_i - bl_i)/2. Or are you talking about using Gauss-Seidel iterations to try and back-out of bad solutions? === Subject: Re: Problems with bounded linear least squares I had in mind doing the whole thing by Gauss-Seidel (mostly because that's the easiest way to do it, even though it's slow), but if you want to use GS to warm-start bvls then that's probably worth a try. === Subject: Re: Problems with bounded linear least squares I found a paper that explored using simple iterations (Gauss-Seidel and Projected Gradient) to speed-up active-set style algorithms. In the paper, they experimented with using simple iterations to determine an initial x (along with the active set) and for allowing the interchange of several active constraints in a single step. The results indicated that using GS iterations to determine an initial solution and active set had the greatest impact on the performance. Messing with multiple constraints worsened the convergence times and PG iterations did not do as well as GS iterations. I'll give an initial set of GS iterations a shot. In the paper they switched out of the GS iterations into the active set algorithm when the number of active variables does not change after four GS iterations. I would hope that in addition to reducing the number of bvls iterations, the warm-start will hopefully keep the algorithm from hitting a dead-end. By the way, the paper is called On Computational Aspects of Bounded Linear Least Squares Problems authored by Achiya Dax. I found it in the ACM Transactions on Mathematical Software === Subject: Re: Problems with bounded linear least squares After experimenting a bit more, I'm positive that BVLS is doing the correct thing. 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The APS provides a stimulating intellectual environment, and offers the successful candidate many... http://jobs.phds.org/jobs/tarodriguez/listing_2005_07_08 === Subject: Wallace Normal Random Generator was published in ACM-TOMS, 3/1996. More was done later, as late as Anyone have experience with Wallace's algorithms? (This is an efficient vectorized algorithm for generating normal variants. It does not require a uniform generator in the inner loop.) Best, DTB === Subject: DDASPK problem Hi all, I am trying to solve a system of DAEs with DDASPK, which arises from a control volume method problem. The algorithm works smoothly for small scale problems, but when I am trying to run it for a large number of equations (over 20000), which means that I have to increase the size of lrw and liw, which are the dimensions of the work arrays RWORK and IWORK respectively, I get compilation errors (fatal error LNK1248: image size (D2E76000) exceeds maximum allowable size (80000000)). Anyone knows how can I overcome this? Thanx in advance! Kostas P.S. I forgot to mention that I am using Intel Fortran Compiler 8. === Subject: Re: DDASPK problem i don't have that compiler but assume it is the problem you always have: stack and heap. normally all variables are on stack and stack size is limited much smaller than available memory size. if you place the work arrays in blank common the problem should vanish (you can take _a c t u a l_ subroutine parameters from common) hth peter === Subject: Re: DDASPK problem Hi Peter, First of all thanx a lot for your reply. I am not quite sure I understand what you mean, when you suggest to place the work arrays in a blank common and take the actual subroutine parameters from common. Could you be more specific please. Again, thanx for your time! Kostas === Subject: Re: DDASPK problem something like this C program main integer iwl,wl parameter ( iwl=100000, wl=10000000) C iw=integer work array , w=double work array integer iw(iwl) double precision w(wl) common w,iw C more instructions .......... call ddaspk(............,iwl,iw,wl,w) C this is quite different from omitting the line ''common w,iw'' C since now w,iw are on the ''heap'', not the dynamic stack hth peter === Subject: multiple lines copy in InputForm Can anyone tell me how to copy multiple lines in InputForm in Mathematica? For example, When I try to put 3 lines into same cell and at end of each expression I put //InputForm, it only returns the last item in the cell. Also, is there an automatic way to get ALL expressions entered as InputForm? instead of having to type //InputForm at end of each line??? In[59]:= ab + b // InputForm ab + b // InputForm c + d // InputForm Out[61]//InputForm= c + d === Subject: Re: A new kind of Calculus theory It is amazing that someone in this day and age still does not know how to write science and mathematics in English. English is now the universal language of science. Please write your work in English if you want more people to read it. === Subject: Looking for an algorithm: polynomial eigenvalue problem Suppose I have a system (Mathematica, in my case) that can solve basic eigenvalue problems for N x N matrices, do matrix inversions, powers, etc. What I am looking for is a short description of a basic algorithm that will make use of these basic facilities to solve the polynomial eigenvalue problem (PEP). The PEP problem is: A_0, A_1, ..., A_N, Find the eigenvalues lambda (scalars) and eigenvectors v (N-vectors) that satisfy (A_0 + lambda A_1 + lambda^2 A_2 + ... + lambda^p A_p) v = 0, where A_0 and A_p are not both singular. There are p times N solution pairs (lambda, v) If it matters, my problems will be of size, say N a few hundred or less and p = O(N). alan === Subject: Re: Looking for an algorithm: polynomial eigenvalue problem Have you tried a web search? Googling polynomial eigenproblems gives me 1510 hits, for example #3 is a nice presentation on linearization www.ma.man.ac.uk/~higham/talks/talk05_acme.pdf Some particular cases have been extensively studied because of important applications. E.g. N=2 appears in linearized flutter, also in linearized mechanical vibrations with nonproportional damping. Another name for that expression (by Peter Lancaster) is lambda matrix and he's got 2 books on the subject still in print, but do not deal with the computational side. === Subject: Re: Looking for an algorithm: polynomial eigenvalue problem Yes, I had done some web searches, but it wasn't until after my original post that I ran across a clear discussion of the linearization method -- the same paper by Mackey et al mentioned in your link. Also, I ran across one of Lancaster's books yesterday, and ordered it (very nice to find it as a Dover publication!) At this point, I have coded a straightforward version of a basic linearization in Mathematica and it seems to work fine on randomly generated problems of modest size. alan === Subject: Re: Looking for an algorithm: polynomial eigenvalue problem The very simple problem: {( 1 0 ; 0 1) + lambda*(0 1;0 0)}*v = 0 has only the trivial solution v=0 since the 2x2-Matrix (1 lambda ; 0 1 ) is regular for all lambda. ? Rainer === Subject: Re: Looking for an algorithm: polynomial eigenvalue problem you already got the hint how to transform the PEP into a p times p block matrix ordinary evp. (in matlab terms: polyeig) but this will subsequently involve solving a problem of dimension n*p times n*p which might be quite detrimental in your case (applying e.g. QR to the frobenius block companion matrix will produce a total fill in up to Hessenberg form). there is the alternative possibility directly to apply Newton's method to the system P(lambda)*x(lambda) = 0 x'(lambda)*x(lambda) = 1 . the books ''templates for the matrix eigenvalue problem'' by alan edelman and coworkers gives more hints in his direction. hth peter === Subject: Re: Looking for an algorithm: polynomial eigenvalue problem Sorry for a typo - I meant to say: A_0, A_1, ..., A_p, === Subject: Re: Looking for an algorithm: polynomial eigenvalue problem Since originally posting, I now see at least how to do the 'naive' linearization, and have coded that successfully in Mathematica. If this leads to problematically large systems in my application, I'll investigate further methods. Rainer, Perhaps I should have said that both A_0 and A_p should be non-singular (I am still learning about this problem). The method I coded certainly requires that the leading coefficient matrix A_p be non-singular, since I have to invert it. I expect you will find that there is a non-trivial solution to any simple p=1 and N=2 case (your example) as long as you ensure this. If you find an exception to that rule, I would certainly like to hear about it. alan === Subject: Re: Looking for an algorithm: polynomial eigenvalue problem Look at page 375 of http://www.library.cornell.edu/nr/bookcpdf/c9-5.pdf hth Roland === Subject: Re: Looking for an algorithm: polynomial eigenvalue problem Sorry, I read only the title and the answer does not match at all === Subject: where to find numerical examples hermite matrices? hello; I am learning about Hermite matrices and about positive/negative-definite properties of matrices. But I am having hard time seeing numerical examples of such matrices, or to generate such ones. For example, i'd like to find a function that generates a random say 3x3 postive definite matrix. Is it possible to generate a numerical example matrix of some order n with some specific eigenvalues ? say eigenvalues as specified in a list or a vector? Matlab has galley of many known matrices. it has functins to generate such matrices as Vandermonde, Hadamard, Hankel , Toeplitz,Hilbert , and many more, but can't see a function to generate say a matrix which is postive-definite for example? here is the web site http://www-ccs.ucsd.edu/matlab/techdoc/ref/gallery.html there is also the web site called the 'matrix reference manual', but again no numerical examples. only theory. There is a matlab toolbox called polynomial toolbox which from its docs seem to have a hermite function. but I do not have this toolbox. it is commerical. steve === Subject: Re: where to find numerical examples hermite matrices? take a random diagonal matrix D, with random numbers (of the distribution you want) in some interval, and some unitary matrix, for example V(i,j)=(sin(i*j*pi/(n+1)))*sqrt(2/(n+1)) and multiply A=V*D*V^t .^t transposition this has the only disadvantage that the eigenvectormatrix is fixed. but you might go further: fill a matrix B with random numbers. Then take C=(B-B^t)/2 C is skew symmetric. then set U=(I+C)*inv(I-C); this is real and orthonormal. i don't know what its distribution function is, given the generators distribution function, but it should also be pretty random. then A= U*D*U^t is real symmetric with the desired eigenvalues and eigenvectors known hth peter === Subject: Re: where to find numerical examples hermite matrices? What is the probability distribution? There are lots of ways to do this. Let D be a diagonal matrix with the characteristic values along the diagonal and 0 elsewhere. Then any such matrix must be P'DP, where P is orthogonal, assuming symmetry. -- 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: where to find numerical examples hermite matrices? It would not be difficult to write, but your specification is insufficient. Its like asking for a function which generates a random integer, or a random real number. I'll guess that this is not what you are asking for. Here is one simple way to generate a 3x3 non-negative definite matrix (but positive definite with a very high probability.) M = rand(3); M = M'*M; Yes. Pick some eigenvalues. Lets say this list of 10 arbitrary values: v = [2 3 5 7 11 13 17 19 23 29]; M = diag(v); The matrix M has the desired eigenvalues. This code is also quite fast, as you might imagine. HTH, John D'Errico === Subject: Re: where to find numerical examples hermite matrices? all zeros everywhere other than on the diagonal. but I guess it meets the specs :) Steve === Subject: Re: where to find numerical examples hermite matrices? Hey, Its every bit of what you asked for. ;-) You want more? They are never happy. Sigh. % choose some arbitrary eigenvalues V = [2 3 5 7 11 13 17 19 23 29]; % and then a random set of orthogonal vectors X = orth(rand(10)); % if we know both vectors and values, we can build % a matrix that has that decomposition M = X*diag(V)*X' M = 12.4699 1.9081 -4.9192 -3.1176 -2.0541 1.3638 -0.8812 -3.8060 -2.6574 4.6862 1.9081 18.4100 -1.0099 0.6535 -3.7012 -2.2513 1.0427 -4.6964 -5.0855 -1.0613 -4.9192 -1.0099 14.0282 2.2268 -3.6798 -3.8800 -2.9690 2.2955 -0.6786 -0.0692 -3.1176 0.6535 2.2268 12.5547 -0.3116 -0.5090 -3.1515 -3.1894 4.4199 -5.5880 -2.0541 -3.7012 -3.6798 -0.3116 12.8856 -0.5696 -0.5136 2.6078 -1.2635 -2.0395 1.3638 -2.2513 -3.8800 -0.5090 -0.5696 8.2645 -0.5984 1.9858 -0.0779 -1.1076 -0.8812 1.0427 -2.9690 -3.1515 -0.5136 -0.5984 10.8441 -2.7879 -1.4961 1.6686 -3.8060 -4.6964 2.2955 -3.1894 2.6078 1.9858 -2.7879 12.7666 -1.9656 -3.5172 -2.6574 -5.0855 -0.6786 4.4199 -1.2635 -0.0779 -1.4961 -1.9656 13.7109 -3.4597 4.6862 -1.0613 -0.0692 -5.5880 -2.0395 -1.1076 1.6686 -3.5172 -3.4597 13.0655 % now check to see if we succeeded. [vec,vals]=eig(M); % the eigenvalues are as specified, although not % sorted. eig does not ensure a sorted set. diag(vals) ans = 29.0000 23.0000 19.0000 17.0000 13.0000 11.0000 7.0000 2.0000 5.0000 3.0000 Is it random enough? John -- The best material model of a cat is another, or preferably the same, cat. A. Rosenblueth, Philosophy of Science, 1945 === Subject: EC-funded Research Visits: Application Deadline 31/08/2005 Are you a scientist of postgraduate level or above, working in EU or an Associated State? Do you require large computing power to improve your research? Would you like to visit a similar research group in Italy, Spain, UK, The Netherlands, France, or Germany? ++++++++++++++++++++++++++++++++++++++++++++++++++++++++ + + + EC-funded HPC-Europa Visitor programme is the answer + + + ++++++++++++++++++++++++++++++++++++++++++++++++++++++++ See what we can offer you: - Access to the best high performance computing systems in Europe - Technical support and consultancy - Scientific collaboration with a host researcher - Travel and living expenses - Logistical and administrative support See http://www.hpc-europa.org/ta.html for more information and online application form. The Programme runs until the end of 2007. ++++++++++++++++++++++++++++++++++++++++++++++++++++++ + Next deadline for applications + + is 31 August 2005 + ++++++++++++++++++++++++++++++++++++++++++++++++++++++ Appications for visits can be made to: CINECA (Italy), CEPBA (Spain), EPCC (UK), HLRS (Germany), IDRIS (France), SARA (The Netherlands) === Subject: Exponential regression y = exp(x) + b I look for any explanation on exponentail regression and program. === Subject: Re: Exponential regression y = exp(x) + b This is an example of nonlinear regression. My NLREG nonlinear regression program can handle this type of problem easily. The NLREG program would be: Variables x, y; Parameters b; Function y = exp(x) + b; Data; You may want to add another parameter as a coefficient to exp(x). You can download a free demonstration copy of NLREG from http://www.nlreg.com -- Phil Sherrod (phil.sherrod 'at' sandh.com) http://www.dtreg.com (decision tree modeling) http://www.nlreg.com (nonlinear regression) === Subject: Re: Exponential regression y = exp(x) + b As written it's a linear regression w/ the change of variable to z= exp(x)... That would seem to make more sense... === Subject: Re: Exponential regression y = exp(x) + b Prony's algorithm ?? === Subject: Re: Exponential regression y = exp(x) + b Generalised non-linear regression. For y = a + b exp(c x) there are plenty of codes - searching for ASCE Dissolved Oxygen procedure, to find some examples. In Excel, or most modern spreadsheets, set up a contiguous block to hold the three parameters a, b and c. Let's assume you have assigned them symbolic names A, B and C. Then, as an example, create in (say) Column A: x values Column B: = a + b * exp(c * [Column A]) Column C: y values Column D: Any weighting values - typical values are either 1 or 1/Y Column E: Weighted error = ([Column C] - [Column D]) * weight Column F: Squared value of Column E Column G: A single cell, summing F Then go to the 'Solver' add-in and specify that you wish to minimise the sum if squares value in G, changing the block of values occupied by A, B and C. === Subject: Re: Exponential regression y = exp(x) + b y = a*exp(x) + b === Subject: Re: Exponential regression y = exp(x) + b This is trivial. Maybe you want y = a* exp(c*x) + b A.L. === Subject: Radiation boundary condition in finite difference method Can anyone help me to locate a text (or maybe provide a short explanation) dealing with the discretization of a radiation heat transfer (T1^4 -Tinf^4) boundary condition. My goal is to be able to solve a steady state heat transfer problem using finite difference techniques. === Subject: Re: Radiation boundary condition in finite difference method I assume that you want to use an implicit method, hence you need to linearize this boundary condition so that you can solve your problem using a linear equation solver. A couple of choices: 1. Start with an initial guess of the surface temperature. Compute radiation heat flux. Solve problem with fixed heat flux at boundary. (Neumann boundary condition.) Get new surface temperature and iterate, etc. 2. Again start with an estimate of the surface temperature. Linearize radiation heat transfer equation about this temperature. You now have a mixed (Cauchy) boundary condition. Solve and iterate. I have personally had better luck with method 2. Method 1 seems to work also, but converges more slowly. If anyone has rigorously shown that these methods converge, I am unaware of it. The details of the implementation depend on how your grid is arranged (does the boundary coincide with the grid points, or is the boundary halfway between grid points?) and how you implement your boundary conditions (i.e., do you use the guard cell method?) Also, the iteration is required only if you are solving a steady-state problem. In a transient problem, you can get away with using the heat flux (method 1) or the linearization ( method 2) computed from the surface temperature at the previous time step. Olin Perry Norton === Subject: FEM - isoparametric FE because we use this FE and how to calculate the integrals through the isoparametric-FE transformated domain ??? --- === Subject: Re: FEM - isoparametric FE you can see any book concerning on the FEM written by mathematician , I suggest the book written by Riviart FINITE ELEMENT EMTHODS FOR ELLIPTIC SYSTEMS. === Subject: Optimize the condition number The problem is given two matrices A,B (both of them are ill-conditioned). Find a scaling factor c such that cond(A+cB) is minimal. (This is useful if we have 2 equivalent system of linear equations). Dong Ta. === Subject: Re: Optimize the condition number with the exception of quite special cases (for example A and B symmetric and commuting) this will result in a highly nonlinear optimization problem, whatever norm you might use, in many cases even nonsmooth. I can see no other way than brute force optimization, since also perturbation results will not be useful in this case. since it is a one parameter problem you might try a derivative free code, but since this involve minimizing and maximizing the norm you must use a method for global optimization. clearly an inversion code cannot be used here, so you have an additional trouble in minimizing norm((A+cB)x) over the unit sphere with respect to x. hth peter === Subject: Need help on matrix computations Could any body please provide some suggestions to the following question. Given that A is a real symmetric matrix of size nxn (postive definite, inverse exists), which is faster to compute: Eigen values of A or inverse of A. Could anybody also please give me the some figure on the number of flops required for computations: e.g. n^3, (4/3)n^3 etc where n is the number of rows of the matrix. The matrix A is small sized matrix: n is typically between 4-10. Amit. === Subject: Re: Need help on matrix computations Check Globu and Van Loan, Matrix Computation. === Subject: New mathematics/physical sciences positions at http://jobs.phds.org, July 18, 2005 New job listings at http://jobs.phds.org - Jobs for PhDs List your job at no cost! http://jobs.phds.org/jobs/post * Head of Market Risk Analytics: Phi Partners, London. Responsible for all quantitative needs (from valuation to market risk measurement to potential future credit exposure) of FMD, including model validation and testing. 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The successful candidate will collaborate on a DOE-funded project to... http://jobs.phds.org/jobs/tarodriguez/listing_2005_07_14 === Subject: bounded least squares problem(bvls)+ sum-to-one contraint I'm currently working on a convex linear-quadratic problem I have to solve th following problem: - min 2-norm Dx-Y, where D is a m by n known matrix, Y a known m-vector and x an unknown n-vector, I found the bvls routine which is working quiet well but doesn't satisfy the constraint sum(i) xi=1. I would like to modify the bvls routine in order to satisfy it...I need your help...Does somebody know how to do? Looking forward for your answer, ad' === Subject: Re: bounded least squares problem(bvls)+ sum-to-one contraint there are many comvex QP codes listed in the respective section of http://plato.asu.edu/guide.html Hans Mittelmann === Subject: the Hanging Cable problem Given the straight-line distance (dd) between two posts of equal height (h), find the length of a cable hanging from the two posts, and its minimum distance from the ground. As is well-known, the curve described by the cable is a catenary (y = a*cosh(x/a) for constant parameter 'a'). This curve is symmetric about the ordinate, with minimum height = a at the ordinate. Then, with d = dd/2 = distance from origin to post, the height h = a*cosh(d/a). There is a straightforward method of solution, utilizing Newton-Raphson -- but with a glitch. Writing: y/a = (y/x)*(x/a) = cosh(x/a) and substituting u = x/a: (y/x)*u = cosh(u), and in particular: (h/d)*u = cosh(u). However, the line y = (h/d)*u fails to intersect the curve y = cosh(u) when the slope is too small (< 1.5888). Is there a general solution to yield parameter 'a' to cover such a case? === Subject: Re: the Hanging Cable problem to find the length of the cable then there is a standard integral formula to find the length of any curve. you know what the curve 'looks' like as its a cosh function, the rest should be fairly straight forward. as for its distance from the ground then that would be given by the height of the post subtract the distance of the lowest point of the ggraph ie. its minima , again thats found easily using calculus === Subject: Re: the Hanging Cable problem : to find the length of the cable then there is a standard integral formula to : find the length of any curve. you know what the curve 'looks' like as its a : cosh function, the rest should be fairly straight forward. Wrong. The previous responder (Cantrell) was correct: given only post height and distance between posts, more info is required to find length. It is _not_ fairly straight forward since the length = a*sinh(dd/2a) and parameter 'a' is not known -- that is the whole problem. It turns out that 'a' depends upon the ratio of the horizontal tension at the lowest point and the weight per length. : as for its distance from the ground then that would be given by the height : of the post subtract the distance of the lowest point of the ggraph ie. its : minima , again thats found easily using calculus Wrong again. It is not even encessary to use calculus, since the cosh(x) function is known to achieve its minimum at x = 0, where the minimum height = a. But what is 'a'? === Subject: Re: the Hanging Cable problem obviously you'd need to know the values of constants, but my method is sound as far as i can see if you knew that the question couldnt be answered, then why pose it? === Subject: Re: the Hanging Cable problem Inadequate information. Given only dd and h, there are many possible different cable lengths (and correspondingly, many possible different minimum distances from the ground). For example, for fixed dd and h: If the cable were quite taut, then its length would be only slightly more than dd and its minimum distance from the ground would be only slightly less than h. OTOH, if the cable were slack enough, it could touch the ground at its vertex. But you might need to use the more general form y = a*cosh(x/a) + b. In other words, nothing guarantees that you must have b = 0. I must suppose you meant 1.50888 instead. Is such a case relevant? If so, please show us an example. BTW, you might find a thread from a few months ago to be helpful. See David Cantrell === Subject: Money TURN SIX DOLLARS INTO $6000 EASY AND HONESTLY!!! READING THIS MESSAGE WILL CHANGE YOUR LIFE FOREVER I found this on a bulletin board and decided to try it. A little while back, I was browsing through newsgroups, just like you are now, thousands of dollars within weeks with only an initial investment of $6.00! So I thought, Yeah right, this must be a scam, but like most of us, I was curious, so I kept reading. Anyway, it said that you send $1.00 to each of the 6 names and address stated in the (There are thousands) No catch, that was it. So after thinking it over, and talking to a few people first, I thought about trying it. I figuredwhat have I got to lose except 6 stamps and $6.00, right? Then I invested the measly $6.00. Well GUESS WHAT!?... within 7 days, I started getting money in the mail! I was shocked! I figured it would end soon, but the money just kept coming in. In my first week, I made about $25.00. By the end of the second week I had made a total of over $1,000.00! In the third week I had over $10,000.00 and it's still growing. This is now my fourth week and I have made a total of just over $42,000.00 and it's still coming in rapidly. It's certainly worth $6.00, and 6 stamps, I have spent more than that on the lottery!! Let me tell you how this works and most importantly, WHY it works... Also, make sure you print a copy of this I promise you that if you follow the directions exactly, that you will start making more money than you thought possible by doing something so easy! SuggestionRead this entire message carefully! (print it out or download it.) Follow the simple directions and watch the money come in! It's easy. It's legal. And, your investment is only $6.00 (Plus postage) IMPORTANTThis is not a rip-off; it is not indecent; it is not illegal; and it is 99% no risk - it really works! If all of the following instructions are adhered to, you will receive extraordinary dividends. PLEASE NOTE: Please follow these directions EXACTLY, and $50,000 or more can be yours in 20 to 60 days. This program remains successful because of the honesty and integrity of the participants. Please continue its success by carefully adhering to the instructions. You will now become part of the Mail Order business. In this business your product is not solid and tangible, it's a service. You are in the business of developing Mailing Lists. Many large corporations are happy to pay big bucks for quality lists. However, the money made from the mailing lists is secondary to the income which is made from people like you and me asking to be included in that list. Here are the 4 easy steps to success: STEP 1Get 6 separate pieces of paper and write the following on each piece of paper PLEASE PUT ME ON YOUR MAILING LIST. Now get 6 US $1.00 bills and place ONE inside EACH of the 6 pieces of paper so the bill will not be seen through the envelope (to prevent thievery). Next, place one paper in each of the 6 envelopes and seal them. You should now have 6 sealed envelopes, each with a piece of paper stating the above phrase, your name and address, and a $1.00 bill. What you are doing is creating a service. THIS IS ABSOLUTELY LEGAL! You are requesting a legitimate service and you are paying for it! Like most of us I was a little skeptical and a little worried about the legal aspects of it all. So I checked it out with the U.S. Post Office (1-800-725-2161) and they confirmed that it is indeed legal. Mail the 6 envelopes to the following addresses: #1) J. R. Sac 5018 Calhoun # 6 Houston, TX 77004 #2) Jeremy Patrick P O Box 570632 Houston, TX 77257 #3) Erfan Haroon 24335 Fairmount St. #4) Robbie Duncan, P.O Box 5363,Newcastle West, NSW 2302,Australia #5) Paola Alarcon Pasaje Los Libis 200. La Victoria. Lima13, Lima-Peru #6) Christian Huerta Barcelona 553. Pueblo Libre Lima21, Lima-Peru STEP 2Now take the #1 name off the list that you see above, move the other names up 1 place (#2 becomes #1, #3 becomes #2, etc...) and add YOUR name and address as number 6 on the list. least 200 newsgroups. (I think there are close to 24,000 groups) All you need is 200, but remember, the more you post, the more money you make! You won't get very much unless you post like crazy. :) This is perfectly legal! If you have any doubts, refer to Title 18 Sec. 1302 & 1341 of the Postal lottery laws. Keep a copy of these steps for yourself and, whenever you need money, you can use it again, and again. PLEASE REMEMBER that this program remains successful because of the honesty and integrity of the participants and by their carefully adhering to the directions. Look at it this way. If you are of integrity, the program will continue and the money that so many others have received will come your way. NOTEYou may want to retain every name and address sent to you, either on a computer or hard copy and keep the notes people send you. This VERIFIES that you are truly providing a service. (Also, it might be a good idea to wrap the $1 bill in dark paper to reduce the risk of mail theft.) So, as each post is downloaded and the directions carefully followed, six members will be reimbursed for their participation as a List Developer with one dollar each. Your name will move up the list geometrically so that when your name reaches the #1 position you will be receiving thousands of dollars in CASH!!! What an opportunity for only $6.00 ($1.00 for each of the first six people listed above) Send it now, add your own name to the list and you're in business! ---DIRECTIONS ----- FOR HOW TO POST TO NEWSGROUPS------------ Step 1) You do not need to re-type this entire letter to do your own posting. Simply put your cursor at the beginning of this letter and drag your cursor to the bottom of this document, and select 'copy' from the edit menu. This will copy the entire letter into the computer's memory. Step 2) Open a blank 'notepad' file and place your cursor at the top paste a copy of the letter into notepad so that you can add your name to the list. Step 3) Save your new notepad file as a .txt file. If you want to do your postings in different settings, you'll always have this file to go back to. Step 4) Use Netscape or Internet explorer and try searching for various newsgroups (on-line forums, message boards, chat sites, discussions.) message by highlighting the text of this letter and selecting paste from the edit menu. Fill in the Subject, this will be the header that everyone sees as they scroll through the list of postings in a particular group, click the post message button. You're done with your first one! Congratulations...THAT'S IT! All you have to do is jump to different newsgroups and post away, after you get the hang of it, it will take about 30 seconds for each newsgroup! **REMEMBER, THE MORE NEWSGROUPS YOU POST IN, THE MORE MONEY YOU WILL MAKE! BUT YOU HAVE TO POST A MINIMUM OF 200** That's it! You will begin receiving money from around the world within days! You may eventually want to rent a P.O.Box due to the large amount of mail you will receive. If you wish to stay anonymous, you can invent a name to use, as long as the postman will deliver it. **JUST MAKE SURE ALL THE ADDRESSES ARE CORRECT.** Now, each of the 5 persons who just sent me $1.00 make the MINIMUM 200 postings, each with my name at #5 and only 5 persons respond to each of the original 5, that is another $25.00 for me, now those 25 each make 200 MINIMUM posts with my name at #4 and only 5 replies each, I will bring in an additional $125.00! Now, those 125 persons turn around and post the MINIMUM 200 with my name at #3 and only receive 5 replies each, I will make an additional $625.00! OK, now here is the fun part, each of those 625 persons post a MINIMUM 200 letters with my name at #2 and they each only receive 5 replies, that just made me $3,125.00!!! Those 3,125 persons will all deliver this message to 200 newsgroups with my name at #1 and if still 5 persons per 200 newsgroups react I will receive $15,625,00! With an original list, you just take the latest posting in the newsgroups, and send out another $6.00 to names on the list, putting your name at number 6 again. And start posting again. The thing to remember isdo you realize that thousands of people all over the world are joining the now!! So, can you afford $6.00 and see if it really works?? I think so... People have said, what if the plan is played out and no one sends you the money? So what! What are the chances of that happening when there are tons of new honest users and new honest people who are joining the internet and newsgroups everyday and are willing to give it a try? Estimates are at 20,000 to 50,000 new users, every day, with thousands of those joining the actual internet. Remember, play FAIRLY and HONESTLY and this will really: work. === Subject: Simplifications of Ax=0 : what kinds are there? Given: A certain family of homogeneous linear systems A(n).x(n)=0 comprises M(n) equations in N(n) unknowns. These are produced by a linear equation solver program and, apart from being integers, there is no obvious pattern in the nonzero coefficients A(n). Observations: For n = 3, M(3) = 1, so there is only one equation; thus it's already as simple as possible. For n = 4, it's possible to combine visual inspection, manual trial-and-error, and some intuition to find that those equations are equivalent to a simpler system with the same number M(n) of equations but in which each equation involves exactly four of the variables x. To be precise: x_a - x_b + x_c - x_d = 0 for certain, symmetrically related, sets of indices {a,b,c,d}. Goal: also be simplified into such a form. (These are much too big to simplify with pen and paper.) So far: I've tried Matlab and GAP, and looked at LAPACK and some other packages, but their linear solvers all reduce A to a reduced row echelon form (RREF), which is evidently NOT what is needed, as it produces large numbers of nonzero coefficients in badly asymmetric sets of equations. The problem seems to be that Gaussian elimination addresses a different question than what's important here. I'm not trying to solve for the x(n), instead i'm trying to express their interrelationship using the simplest possible equations. Question: Is there a general algorithm for simplifying a homogeneous system of linear equations Ax=0 to an equivalent system Bx=0 that has a minimal set of nonzero coefficients? It should do so in a way that treats each equation and each unknown equally under law, so as to reveal symmetries, not obscure them. Furthermore, if A is an integer matrix, then so must B be. (No normalization of vectors, please.) solutions, Zoe Tropo. -- Choose life! === Subject: Many Digits Friendly Competition 2005 THE MANY DIGITS FRIENDLY COMPETITION 2005 4 October 2005 Radboud University, Nijmegen, The Netherlands Call For Participation In the Institute for Computing and Information Sciences at Radboud University Nijmegen we will organise a friendly competition/benchmark session for investigating the state of the art in the various implementations of exact real arithmetic. The event will be held on October 4th 2005, in conjunction with the Small TYPES workshop on Constructive analysis, types and exact real numbers The event will be akin to those held in CCA00 interactive software packages capable of handling arbitrary precision arithmetic on elementary functions are invited to participate. Moreover all libraries for handling arbitrary precision arithmetic in any programming language are welcome to participate The participants can choose to attend the event at Nijmegen, or they can choose to participate remotely via Internet. Friendly Competition The friendly competition consists of giving a solution to a set of problems that are posed by the organisers shortly (few hours) before the event. The problems cover the minimum abilities of a system capable of handling the elementary real functions and are expected problem that might be in this set is to calculate the first 10,000 digits of sin(sin(1)). A set of `practice problems' can be found on the website of the competition The solutions consist of the input needed to generate the answers to the problems plus the answers themselves. The competition criteria will be the accuracy (correctness) of the solution and the time needed for producing the solution. Apart from this, we will also compare the other aspects of the solutions (such as memory usage and readability and length of the input code) to give a better comparison. On the day of the competition the solutions will be executed by participants on one specific machine at the event site, under either Debian/GNU Linux or Windows XP. A more detailed description of the setup and format of the competition can be found on the website of the competition The outcome of this friendly competition is expected to give a comparison in performance of the systems that chose to participate for the selected problems. Remote Participation In case of remote participation, the participants will receive the set of problems by email. Then they will be given the opportunity to log into our local machine and perform their solution. Systems, Packages and Libraries Those who wish to participate in the Friendly Competition are expected to provide the software necessary for their solutions, so it can be run on the machine used for the competition. Schedule * August 26th deadline for registering for the competition * August 29th final list of registered participants on the web * September 23rd timings of the practice problems for the registered participants posted on the web * October 4th competition Registration If you are interested in participating in the event we will be happy to hear from you: 1.Team name, name and email of contact person(s) and the platform your system needs to run (GNU Linux or Windows XP). 2.Whether you plan to come to the meeting or whether you would like to participate remotely. We would prefer that participants (especially those whose system requires Windows) to send their intention to participate before August 26th, 2005. Local Organisers: Milad Niqui and Freek Wiedijk === Subject: Mathematical Movie!!! Tied Up & Strung Out:: Hollywood String Theory Movie!!! Looking For Extras!!! http://physicsmathforums.com/showthread.php?t=56 (pre-production photos) http://physicsmathforums.com/showthread.php?t=56 Tied Up & Strung Out: Hollywood String Theory Movie!!! Looking For Extras!!! FOR IMMEDIATE RELEASE: ALL TIED UP & STRUNG ALONG, a movie about String Theorists and their expansive theories which extend human ignorance, pomposity, and frailty into higher dimensions, is set to start filming this fall. Jessica Alba, John Cleese, Eugene Levie, Jackie Chan, and David Duchovney of which is still cheaper than String Theory itself, and will likely displace less physicists from the academy. As contemporary physics is about money, hype, mythology, and chicks, Ed Witten explained from his offices at the Princeton Institute for Advanced Study, The next logical step was Hollywood, although I thought Burt Reynolds should play me instead of Eugene Levy. Brian Greene, the famous String Theorist who will be played by David the truth is out there Duchovney, explained the plot: String theory's muddled, contorted theories that lack postulates, laws, and experimentally-verified equations have Einstein spinning so fast in his grave that it creates a black hole. In order to save the world, we String Theorists have to stop reformulating String Theory faster than the speed of light. We are called upon to stop violating the conservation of energy by mining higher dimensions to publish more BS than can accounted for with the Big Bang alone. Greene continues: At first my character is reluctant to stop theorizing, but when my love interest Jessica Alba is sucked into the black hole, I search my soul and find Paul Davies there, played by John Cleese. I ask him what he's doing in my soul, and he explains that the answer is containined in the mind of God, which only he is privy too, but for a small fee, some tax and tuition dollars, a couple grants here and there, and an all-expense-paid book tour with stops in Zurich and Honolulu, he can let me in on it. But as soon as he points out that we can make more money in Hollywood than printing more coffee-table books that recycle Einstein, Bohr, Dirac, Feynman, and Wheeler, I am converted. I agree to turn my back on String Theory's hoax and save Jessica Alba. But it's not that easy, as standing in Greene's way is Michio king of pop-theory-hipster-irony-the-theory-of-everything-or-anything-made- you-read-this Kaku, played by Jackie Chan. Kaku beats the crap out of Greene for alomst blowing the ironic pretense his salary, benefits, and all-expense paid trips depend on. I have better hair! Kaku argues as he delivers a flying back-kick, There can be ony ONE! I WILL be String Theory's front man!! But Greene fights back, Kaku! Kaku! (pronounced Ka-Kaw! Ka-Kaw! like Owen Wilson did in Bottle Rocket) It is theoretically impossible to build a coffee tables strong enough to support any more coffee-table physics books!!! And what the #&#%&$ does M stand for in M theory??? How does it all end? Does physics go bankrupt funding theories that have expanded our ignorance from four dimensions into ten, twenty, and thirty dimensions? Do tax payers revolt? Do young physicists overthrow the hand-waving, contortionist bullies and revive physics with a classical renaissance favoring logic, reason, and Truth over meaningless mathematical abstractions? Does Moving Dimensions Theory (MDT) prevail with its simple postulate? We'll all just have to wait! But in the meantime, how do you think it will play out? MDT's postulate: THE FOURTH DIMENSION IS EXPANDING AT A RATE OF C RELATIVE TO THE THREE SPATIAL DIMENSIONS IN QUANTIZED UNITS OF THE PLANCK LENGTH, GIVING RISE TO TIME AND ALL CLASSICAL, QUANTUM MECHANICAL, AND RELATIVISTIC PHENOMENA. http://physicsmathforums.com/showthread.php?t=56 (pre-production photos) === Subject: Least squares of 2 splines I have x-y data that often form a V shape (or inverted V) and so I would like to find the two best straight lines that fit them. I see it as 4 free parameters, the x and y position of the knot and the two slopes on either side. I would appreciate it greatly if you could direct me to an algorithm that would efficiently fit these or provide any helpful suggestions as to how to proceed. My languages of choice are R and VB. J === Subject: Re: Least squares of 2 splines Here is a MacGyver tentative solution that could inspire you (to build something more sophisticated ! ). It's written in mathematica and uses the Fit function for linear fitting. Using an initial guess of the knot, sorted data is seperated into two subsets for linear fitting. The intersection of the two lines gives a new knot and so on till precision is reached. In[1]:= myFit[xKnot_, yKnot_, data_List, precis_]:= Module[{x0,y0,x1,y1,d1,d2,fi1,fi2}, {x0,y0} = {xKnot,yKnot}; {x1,y1} = {x0,y0}+1; d1 = Select[data,#[[1]] < x0&]; fi1 = Fit[d1,{1,x},x]; fi2 = Fit[d2,{1,x},x]; xx = First@Solve[fi1 == fi2,x]; {x0,y0} = {x1,y1}; {x1,y1} = {x /. xx, fi1 /. xx}; ]; {x1,y1,fi1,fi2} ]; Toy example (data in shape of inverted V): In[2]:=data={{0,1.1},{1,1.5},{2,2.1},{3,3.2},{4,3.3}, {5,2.5},{6,1.4},{7,0.5},{8,0.1}}; In[3]:=myFit[3,4,data,0.001] Out[3]={3.69935, 3.49255, 0.94+0.69 x, 6.6-0.84 x} hth V.Astanoff === Subject: Re: Least squares of 2 splines Order the points in terms of x. Then fit two lines to each of the n-3 left-right partitionings of the n points -- (2,n-2),...,(n-2,2) -- and pick the partitioning that gives the smallest residual sum of squares. === Subject: Re: Least squares of 2 splines All great ideas. J === Subject: Re: Least squares of 2 splines write a function which evaluates the residual vector for those parameters, for example function res=vfit(para,xdat,ydat) % para = (xpt,ypt,slope1,slope2) describing a v-like % function graph with vertex (xpt,ypt) and slope1 for x<=xpt % res is the vertical residual vector for the data % points (xdat,ydat) n=length(xdat); if length(ydat) ~= n error('vfit: xdat and ydat have different length'); end if length(para)<4 error('vfit: para has not 4 components'); end res=zeros(n,1); for i=1:n if xdat(i) < para(1) res(i)=ydat(i)-(para(2)+para(3)*(xdat(i)-para(1))); else res(i)=ydat(i)-(para(2)+para(4)*(xdat(i)-para(1))); end end and feed this into a nonlinear least squares solver. given a reasonable initial guess for the parameters it should succeed in finding the optimal ones although the residual vector depends nonsmooth on the parameters (seen globally) . for nonlinear least squares codes see for example http://plato.la.asu.edu/topics/problems/nlolsq.html since this is a small dim problem , you should have no problem hth peter === Subject: Re: Least squares of 2 splines Let (u,v) be the coordinates of the knot and a,b the slopes, then your model can be formulated as m(x;a,b,u,v):= v + a*max(x-u,0) + b*max(u-x,0) Now the standard least squares function for your data points (x_i,y_i) is phi(a,b,u,v)= 0.5 * sum_{i=1}^n { (y_i-m(x_i;a,b,u,v))^2 } and you are looking for a minimum of this NONLINEAR functional. Unfortunately the functional is non-differentiable in u. Therefore you need an algorithm for non-differentiable optimization. You can use the free Scilab. It has a routine 'optim' which has a parameter for non-differentiable optimization. A compromise would be to approximate the 'max' function by a differentiable function, e.g. MaxD(0,x):= x*HD(x,delta) where HD is a differentiable approximation of the Heaviside function, e.g. 0 for x < -delta HD(x,delta) = 0.5*(sin((pi/2)*x/delta)+1) for -delta <= x < delta But beware, the corresponding 'mathematically differentiable' functional gets badly conditioned if delta is small. P.S. You can assume u to be known and compute directly the optimal values of a, b and v depending on u , since the functional is quadratic in a, b and v. Then there remains a 1D nonlinear, non-differentiable optimization problem (for u), which is much easier. -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany === Subject: Finding all local minima Hi I have a one-dimensional array of a large number of reals, and I wish to locate *all* the local minima in that array. Short of checking for the first and second numerical derivatives in a grid search, is there a way to do this efficiently ? === Subject: Re: Finding all local minima When state changes from state -1 to +1 you have left a local minimum. -- Julian V. Noble Professor Emeritus of Physics ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ For there was never yet philosopher that could endure the toothache patiently. -- Wm. Shakespeare, Much Ado about Nothing. Act v. Sc. 1. === Subject: Re: Finding all local minima The point i is a local minimum if and only if ((f(i)