mm- Subject: ANN: GNU libmatheval 1.1.0 release GNU libmatheval release 1.1.0 is out. Updates include: * Added support for number of mathematical functions (secants, cosecants, Heaviside step and Dirac delta functions). * Added method to get list of variables appearing in given expression. * Fixed issue with passing 32-bit, instead of 64-bit, pointers to Fortran code GNU libmatheval is a library (callable from C and Fortran) to parse and evaluate symbolic expressions input as text. It supports expressions in any number of variables of arbitrary names, decimal constants, basic unary and binary operators, and elementary mathematical functions. In addition to parsing and evaluation, libmatheval can also compute symbolic derivatives and output expressions to strings. === Subject: Origin of 360 degrees in a Circle? As an ex-RAF Navigator, the circle with its 360 degrees was fundamental to my daily life, in calculation of track, heading or drift. Bearing in mind the subdivisions are based on the number 60: 60 seconds () = 1 minute (Œ) 60 minutes (Œ) = 1 Degree Why then are there not 60 Degrees in a circle? Mechanical stopwatches measuring time have the hands rotate at the same 60:1 ratio around the circle. I agree we can¹t change it now but it just puzzles me as to WHY there are 360 degrees in a circle. Taking the decimal system as a comparison, EVERYTHING is based on the number 10 or its multiples. 1/10/100 etc.; now that is logical! But 60/60/360 is illogical. === Subject: Re: Origin of 360 degrees in a Circle? > As an ex-RAF Navigator, the circle with its 360 degrees was > fundamental to my daily life, in calculation of track, heading or > drift. > Bearing in mind the subdivisions are based on the number 60: > 60 seconds () = 1 minute (Œ) > 60 minutes (Œ) = 1 Degree > Why then are there not 60 Degrees in a circle? The 60s are from the Babylonians, who liked the number (aside from being large, it¹s a convenient base). The used a combination base 10 and base 60 system. Not surprisingly, their calendar had 360 days in a year; 360 being a multiple of 60 and close to the actual number of days in a year. (They had the occasional leap month to help things work out better). A degree, then, would be the amount the Earth circles the Sun in a day (or from an Earth-centric point of view, the amount the Sun moves relative to the Þxed background of stars). Jay === Subject: Re: Origin of 360 degrees in a Circle? Also they may have found it useful that 1, 2, 3, 4, 5, and 6 are all factors of 360, much as the ancient Egyptians used a rope with twelve equally spaced knots to form right angles in fabricating structures, etc. . WDA end > As an ex-RAF Navigator, the circle with its 360 degrees was > fundamental to my daily life, in calculation of track, heading or > drift. > Bearing in mind the subdivisions are based on the number 60: > 60 seconds () = 1 minute (Œ) > 60 minutes (Œ) = 1 Degree > Why then are there not 60 Degrees in a circle? > The 60s are from the Babylonians, who liked the number (aside from > being large, it¹s a convenient base). The used a combination base 10 > and base 60 system. > Not surprisingly, their calendar had 360 days in a year; 360 being a > multiple of 60 and close to the actual number of days in a year. > (They had the occasional leap month to help things work out better). > A degree, then, would be the amount the Earth circles the Sun in a > day (or from an Earth-centric point of view, the amount the Sun moves > relative to the Þxed background of stars). > Jay === Subject: Re: Origin of 360 degrees in a Circle? > .... the ancient Egyptians used a rope with twelve equally spaced > knots to form right angles in fabricating structures, etc.... There is no evidence whatever for that. It was purely a guess by the historian Moritz Cantor, which later writers carelessly copied as if it were a known fact. There¹s an interesting little discussion of it in B. L. van der Waerden, Science Awakening Vol. I, p.6. Ken Pledger. === Subject: Re: Origin of 360 degrees in a Circle? charset=iso-8859-1 > As an ex-RAF Navigator, the circle with its 360 degrees was > fundamental to my daily life, in calculation of track, heading or > drift. > Bearing in mind the subdivisions are based on the number 60: > 60 seconds () = 1 minute (Œ) > 60 minutes (Œ) = 1 Degree > Why then are there not 60 Degrees in a circle? > The 60s are from the Babylonians, who liked the number (aside from > being large, it¹s a convenient base). The used a combination base 10 > and base 60 system. > Not surprisingly, their calendar had 360 days in a year; 360 being a > multiple of 60 and close to the actual number of days in a year. > (They had the occasional leap month to help things work out better). > A degree, then, would be the amount the Earth circles the Sun in a > day (or from an Earth-centric point of view, the amount the Sun moves > relative to the Þxed background of stars). > Jay Add to what Jay has written the following observation: 2, 3, 4, 5, 6, 8, 9, 10 and 12 are all factors of 360: of the natural numbers not > 12 (and excluding unity, for obvious reasons), this omits 7 and 11 only. If you want divisibility by 7, look to money: in the UK, the guinea - 21 shillings. If you want divisibility by 11: either dream, or surprise me! John johnDOTmorrisonATtescoDOTnet -- Never wrestle a pig. You both get dirty and the pig likes it. - Paul Oldham --- Outgoing mail is certiÞed Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Origin of 360 degrees in a Circle? ... > If you want divisibility by 11: either dream, or surprise me! I seem to recall reading, in a discussion on the Duodecimal Society (a group advocating the use of base 12), a mention of different group advocating using a prime base (maybe 11). The suggestion in the However, in Martin Gardner¹s _Mathematical Magic Show_, Chapter 8, I found the following: Above 5, very few number systems have been based on primes. W.W. Rouse Ball, in _A Short Account of the History of Mathematics_ (fourth edition, 1908), cites only the 7-base system of the Bolas, a West African tribe, and the 11-base system of the early Maoris, although I cannot vouch for either assertion. Jay === Subject: Re: Origin of 360 degrees in a Circle? > Add to what Jay has written the following observation: 2, 3, 4, 5, 6, 8, > 9, 10 and 12 are all factors of 360: of the natural numbers not > 12 (and > excluding unity, for obvious reasons), this omits 7 and 11 only. If you want > divisibility by 7, look to money: in the UK, the guinea - 21 shillings. > If you want divisibility by 11: either dream, or surprise me! > John > johnDOTmorrisonATtescoDOTnet For divisibility by 7 and below, there is the now very often Œlistened to¹ sample rate, 44100 per second, the sample rate of Audio CDs. 44100 = 2*2*3*3*5*5*7*7. -- Jean-Pierre Moreau notuser@inexistant.nil (s/notuser/jpm-qc/; s/inexistant.nil/iquebec.com/) === Subject: Code needed for Mathematica re Pythagorean triples Does anyone have any code that will perform the following task? EqualAreaTriples[n] Returns all the Pythagorean triples (if any) that have area equal to n And similarly EqualPerimeterTriples[n] That will return all the Pythagorean triples (if any) that have perimeter equal to n Tony === Subject: Re: Code needed for Mathematica re Pythagorean triples >Does anyone have any code that will perform the following task? >EqualAreaTriples[n] >Returns all the Pythagorean triples (if any) that have area equal to n >And similarly >EqualPerimeterTriples[n] >That will return all the Pythagorean triples (if any) that have perimeter >equal to n Homework?... A.L. === Subject: Re: Code needed for Mathematica re Pythagorean triples >Does anyone have any code that will perform the following task? >EqualAreaTriples[n] >Returns all the Pythagorean triples (if any) that have area equal to n >And similarly >EqualPerimeterTriples[n] >That will return all the Pythagorean triples (if any) that have perimeter >equal to n > Homework?... > A.L. Hardly. I am a 49 year old mathematician. Tony === Subject: Re: [Find the Fallacy!!] -- ŒProof¹ of Monotheism.... > let x represent the number of idiots in the universe > it is obviously false that x=1 => x=0, > so, its negation, > ~(x=1 => x=0) > must be true. > now, if we consider the truth table: > p q p=>q ~p OR q > T T T T > T F F F > F T T T > F F T T > (for inclusive OR) > we see that > ~(x=1 => x=0) > must be equivalent to: > ~(~(x=1) OR (x=0)) > ~~(x=1) AND ~(x=0) > x=1 AND x!=0 > which implies that x=1 > therfore the number of idiots in the universe is one; > thus, there is one idiot and one idiot only. > However there are more than one idiots in the universe, so this > proofs that the proof is false. Fallacy found. I disagree! It should be obvious that what appear to be separate idiots are merely facets of the One True Idiot. -- http://hertzlinger.blogspot.com =?ISO-8859-1?Q?schaft?= === Subject: Re: Integral of Bessel function multiplied by exp > Hello everyone, > does anybody know how to compute analytically (at least, > approximately) > such integral if the upper limit of your integral is infty then you are lucky int_0^{infty} exp(-a*x) * J_0(b*sqrt(x)) * dx = frac{1}{a} * exp( - frac{b^2}{4*a}) (see eqn 24.94 in M.R.Spiegel, Handbook of Mathematics, SCHAUM¹s Outline Series, McGraw-Hill Book Company) with eqn 24.32 (you are near to the solution) I_n(x) = i^{-n} * J_n(i*x) you get int_0^y exp(-a*x_1) * J_0(i*sqrt(x_2)*sqrt(x_1)) * dx_1 = int_0^y exp(-a*x_1) * J_0(c*sqrt(x_1)) * dx_1 with c = i*sqrt(x_2) thus your integral gets int_0^{infty} exp(-a*x_1)*I_0(sqrt(c*x_1))dx_1 = frac{1}{a} * exp( - frac{c^2}{4*a} ) = frac{1}{a} * exp( - frac{i^2*sqrt(x_2)^2}{4*a} ) = frac{1}{a} * exp( - frac{-1*x_2}{4*a} ) = frac{1}{a} * exp( frac{x_2}{4*a} ) and so on for the next integrals > int_0^y exp(-a*x)*I_0(sqrt(b*x))dx, > where I_0 is the modiÞed 0-th order Bessel function of the Þrst > kind? > In fact, I need to compute > int_0^y ...int_0^y > exp(-a*(x_1+...+x_n))*I_0(sqrt(x_1*x_2))*...*I_0(sqrt(x_{n-1} *x_n))dx_n...dx_ 1, > where n is **very** big number. > P. Trifonov here are some additional books: F. Oberhettinger, Tables of Bessel Transforms, Springer Publications I.S. Gradshteyn, I.M. Ryzhik Tables of Integrals, Series and Products Academic Press I hope this helps & happy computing Herbert === Subject: Maple 9.5 I have recently received my Maple 9.5. Certainly there are some improvements relative to Maple 7. It seems it has much capabilities in complex computations. Although Maple solves lots of Riccati differential equations, still my methods are substantially different than the classical methods. I mainly solve one group as a class rather than handling in general (similar to my polynomials). Polynomial solving of Maple 9.5 is in very good shape having excellent packages such as SolveTools and DEtools. Overall this is a piece of software for the mathematics of the 21st century. Maple 9.5 solution of the following Riccati (in x(t), All others are parameters) is in terms of hypergeometric functions and it could not simplify the hypergeometric functions. DE := diff(x(t),t) = (960*m1*x(t)^2+48*m1*(20*m0+20*m1*t)*x(t)+48*m1*(4*m1*t* m0+20*n0+3*m1*t*h0))/(3600*n0-960*m1*t*h0+1920*m1*t*m0+1600* m1^2*t^2+225*h0^ 2- 900*m0*h0); My solution is all radical (will be posted later or in my lecture notes). Also Maple translates formulas with latex (expr), directly on the worksheet, which is very important tool for latex writings. I am beginning to write my lecture notes and hopefully after editing and writing with atex, I will provide my self-publishing materials. I have several lecture notes each 250 pages and I intend to present the Þrst one in 2005. Dr.Mehran Basti Happy holidays === Subject: Complex arithmetic I want to do arithmetic in rings such as Q]e^(Pi*I/n)], for some n, exactly. For example, I would like to do E(23) + E(45) * 4 * E(10) in GAP (where E(n) = e^(Pi*I*n)). However, I have many, many calculations to do so I want to do them extra quickly. the ring is known before the calculations are started(in fact, only 26 elements are needed to determine all the operations performed). So I want to do this extra quickly. How would I go about doing this? I¹ve thought about libPari, but it seems that pari is used for þoating point calculations. If anyone would know how to do it in libpari (or pari) --Jacques === Subject: Reverse Linear Regression I was wondering if there is an algorithm that can be used to create a set of n linear (X,Y) data points with a deÞned slope m, y intercept b, STD DEV m and STD DEV b. I could use it in making up lab questions i.e If a table of Voltage and Current data is given for an unknown resistor, plot the values Voltage vs Current and Þnd the value of resistor and the error in the resistance. Also determine how linear the data is. In some cases I have the more advanced students use their calculators to analyze the data or sometimes I have them use graph paper and eyeball it for the line of best Þt. I have done this by hand using my HP48GX calculator, but it would be nice to just type in the values for the slope, etc specify the number of points wanted and have it churn out the original data. I would like to do his in MAPLE if possible. I use it quite often, and my programming skills in other languages are either extremely rusty or nonexistent. It would also be nice to generate non-linear data for log or exponential functions, since these functions also Þt certain physical phenomena. Harold A. Climer Dept. Of Physics,Geology, and Astronomy U.T. Chattanooga 318 Grote Hall 615 McCallie Ave Chattanooga TN 37403 === Subject: Re: my work on polynomials > En ce qui concerne mon fran.8dais, je crois que les quelques dizaines de > messages que j¹ai d.8ej.88 envoy.8e .88 fr.sci.maths montrent qu¹il, n¹.8etant pas > parfait, est au moins acceptable. Tout a fait, et d¹ailleurs c¹etait juste une remarque de chauvinisme parfait pour les abords du 14 juillet. La forme d¹imperialisme qui consiste a utiiser l¹anglais, a dire que les etats unis sont le paradiss etc est malheureusement tres a l¹oeuvre en france. Pour la blague, on en est a acheter des cabernay californiens -- dont le seul but est d¹essayer d¹imiter des vins francais :-) --. Curieusement, les vins chiliens ou la biere chinoise ont moins d¹acheteurs :-))) Amities, Olivier === Subject: Re: my work on polynomials [ publication dans deux groupes, suivi vers fr.sci.maths seul ] [ crosspost with followup to fr.sci.maths only ] [ ich kann nicht auf Deutsch .9fbersetzen. Bitte entschuldigen Sie mir. ] >> By the way, learn some french also, good mathematics is not only >> written in german :-! Or don¹t post on a french forum. > Cette discussion a lieu dans deux newsgroups simultan.8ement: > aus.mathematics et fr.sci.maths. Ce n¹est pas ma faute se elle a .8et.8e > .8ecrite en englais d.8fs son d.8ebut. Oui, c¹.8etait normalement .88 Jon de positionner un suivi d.8fs son premier n¹irai pas jusqu¹.88 dire le devoir de le faire, m.90me si l¹id.8ee m¹a efþeur.8e. === Subject: indices Hi I am hopeless with maths, i seem to be struggling with the basics. Could someone please help me with the following, please. (2x-1) -x 3 =9 === Subject: Re: indices > Hi > I am hopeless with maths, i seem to be struggling with the basics. Could > someone please help me with the following, please. > (2x-1) -x > 3 =9 In ACII notation that should be 3^(2*x - 1) = 9^(-x) Since 9 = 3^2, 3^(2*x - 1) = 3^(-2*x) So 2*x = 1 = -2*x === Subject: Re: indices > Hi > I am hopeless with maths, i seem to be struggling with the basics. Could > someone please help me with the following, please. > (2x-1) -x > 3 =9 First , write (2x-1) - x 3 = 9 as 3^(2x-1) = 9^(-x). Replace 9 with 3^2 (so that there is one base instead of two) 3^(2x-1) = (3^2)^(-x). Simplify the right side 3^(2x-1) = 3^(-2x). [The following is the only noteworthy difference from the previous post.] The only way the last equation can be true is if the exponents are equal (because the two expressions are equal and have the same base), so 2x-1 = -2x. It follows that x = 1/4 I hope you Þnd this helpful. Kevin O¹Neill === Subject: Re: indices Hi (2x-1) -x 3 = 9 (2x-1) -2x 3 = 3 2x multiply equation with 3 (2x-1) 2x 3 * 3 = 1 (4x-1) 3 = 1 multiply equation with 3 4x 3 = 3 use logarithm (base = 3) to both sides of equation 4x = 1 x=1/4 sry for bad english :) bye Frank === Subject: two paths, one truth speaking of vectors, If you can Þnd two paths out of trouble, You have a better idea of where you¹ve really been. If you can Þnd two paths to reach your destination, You have a better idea of where you¹re heading. Sometimes what you¹re running from Is the exactly the same place you¹re heading, ... but we don¹t let them know. http://mypeoplepc.com/members/jon8338/polynomial/index.html === Subject: Re: two paths, one truth > speaking of vectors, > If you can Þnd two paths out of trouble, > You have a better idea of where you¹ve really been. > If you can Þnd two paths to reach your destination, > You have a better idea of where you¹re heading. > Sometimes what you¹re running from > Is the exactly the same place you¹re heading, > ... but we don¹t let them know. > http://mypeoplepc.com/members/jon8338/polynomial/index.html And when you take the same route more than once, you get to the same destination. (You¹re still making the same mistake you always have.) -- Christopher Heckman === Subject: Re: Solutions to Power Series > APPLICATION OF > http://mypeoplepc.com/members/jon8338/polynomial/id7.html > OK, Jon, now put your money where your mouth (and your webside) is: Solve t^5 + t^2 + 1 = 0 Iff you can do that, I¹ll have a closer look at your Œtheory¹. === Subject: Re: Solutions to Power Series > a correction is applied below: > If T*N+a[0] is the nth degree polynomial where T=(t,t^2,t^3,...,t^n), > N=(a[1],a[2],a[3],...,a[n]), if T is described from two orthogonal > systems, its linear requirement precipitates. T remains invariant > despite the coordinate system used to describe it. Consequently, my > revised web site showing the details, > > APPLICATION OF > http://mypeoplepc.com/members/jon8338/polynomial/id7.html > > Alright Jon, its time for some strong criticism. Take a look at this: t^3 - 2t^2 - 2t + 4 = 0 t = 2, sqrt(2), -sqrt(2) > (t-2)(t-3)(t-4)=0 > (t^2-5t+6)(t-4)=0 > t-4 > ------------- > t^3-5t^2+ 6t > -4t^2+20t-24 > ---------------- > t^3-9t^2+26t-24 > N = (1, -2, -2) > N=+/-(26,-9,1) > N*U[2]=0 at U[2]=(4,13,13) > i j k > U[3]=NxU[2]= 26 -9 1 =(-130,-334,374) > 4 13 13 > > N/|N|= > u[1]=( .944,-.327,.036)=(u[11],u[12],u[13]) > u[2]=( .212, .691,.691)=(u[21],u[22],u[23]) > u[3]=(-.249,-.640,.717)=(u[31],u[32],u[33]) > I have decided to try this polynomial with two sets of vectors one is marked with primes and the other is not. u[1] = (1/3, -2/3, -2/3) u[2] = (2/3, -1/3, 2/3) u[3] = (2/3, 2/3, -1/3) u¹[1] = (1/3, -2/3, -2/3) u¹[2] = (2/sqrt(5), 1/sqrt(5), 0) u¹[3] = (2/sqrt(5), 0, 1/sqrt(5)) > (-a[0]/|N|)u[1]=(.8717)u[1] > Q = +/-(-.823,.285,-.0314) > Q = -(4)/3 * (1/3, -2/3, -2/3) = (-4/9, 8/9, 8/9) Q¹ = -(4)/3 * (1/3, -2/3, -2/3) = (-4/9, 8/9, 8/9) > .212 -.249 1 0 0 > A = .691 -.640 I= 0 1 0 > .691 .717 0 0 1 > [ 2/3 2/3] A = [-1/3 2/3] [ 2/3 -1/3] [2/sqrt(5) 2/sqrt(5)] A¹ = [1/sqrt(5) 0 ] [0 1/sqrt(5)] > T = (AA^T -I)^-1 Q =(1.085,6.552,6.607)=(t,t^2,t^3) For the unprimed set of vectors, the expression T = (AA^T -I)^-1 Q does not exist since (AA^T -I) is not invertible in this case. This is an EXTREME problem. You can¹t even guarantee the equations generated to be solvable! For the primed set of vectors, T¹ = (4/9, -8/9, -8/9). Now Jon, since you did some rounding in your example, how would you round these to the correct values of 2, 1.41421356237309504880168872421, and -1.41421356237309504880168872421? > > t = 1.085 throw out > t = 2.560~3 > t = 1.876~2 > > > > n=8 > > T=[((AA^T-I)^-1)^T((AA^T-I)^-1)]^-1((AA^T-I)^-1)^T Q > correct above to, > T=[(AA^T-I)^T(AA^T-I)]^-1(AA^T-I)^T Q > > (Least Squares) > > LEAST SQUARES SOLUTION TO MATRIX > k a[k] SUMMARY > 0 -30 t^k |t^k| t ERROR > 1 9 2.384E-01 2.384E-01 2.384E-01 -2.731E+01 > 2 9 2.031E+01 2.031E+01 4.507E+00 1.590E+06 > 3 1 -1.902E+01 1.902E+01 2.669E+00 2.557E+04 > 4 3 -4.038E+00 4.038E+00 1.418E+00 2.165E+02 > 5 6 4.742E+00 4.742E+00 9.485E-01 1.830E+00 > 6 1 -3.753E+01 3.753E+01 1.830E+00 1.416E+03 > 7 1 3.320E+01 3.320E+01 1.649E+00 6.553E+02 > 8 9 3.010E+01 3.010E+01 1.530E+00 3.787E+02 > This has to be the worst, YOUR ERROR VALUES ARE UP TO HALF A MILLION TIMES LARGER THAN THE T VALUES!!!!!!!!!! When an engineer would see those kinds of error values, he would scream! > > Jon Giffen > jon8338@peoplepc.com === Subject: Re: Solutions to Power Series > a correction is applied below: > > > If T*N+a[0] is the nth degree polynomial where T=(t,t^2,t^3,...,t^n), > > N=(a[1],a[2],a[3],...,a[n]), if T is described from two orthogonal > > systems, its linear requirement precipitates. T remains invariant > > despite the coordinate system used to describe it. Consequently, my > > revised web site showing the details, > > > > APPLICATION OF > > http://mypeoplepc.com/members/jon8338/polynomial/id7.html > > > > > Alright Jon, its time for some strong criticism. I¹ve found it¹s not worth it telling him about this. I¹m posting replies to his, mainly to let other people know his methods don¹t work. -- Christopher Heckman === Subject: Re: Solutions to Power Series > If T*N+a[0] is the nth degree polynomial where T=(t,t^2,t^3,...,t^n), > N=(a[1],a[2],a[3],...,a[n]), if T is described from two orthogonal > systems, its linear requirement precipitates. T remains invariant > despite the coordinate system used to describe it. Consequently, my > revised web site showing the details, > > (t-2)(t-3)(t-4)=0 > (t^2-5t+6)(t-4)=0 > t-4 > ------------- > t^3-5t^2+ 6t > -4t^2+20t-24 > ---------------- > t^3-9t^2+26t-24 > > N=+/-(26,-9,1) REVOLUTIONARY !!! Why don¹t you tell C.F. Gauss or E. Galois; they would love to hear that you have easily solved a problem they have shown to be principally not solvable!!! === Subject: Re: Solutions to Power Series > > If T*N+a[0] is the nth degree polynomial where T=(t,t^2,t^3,...,t^n), > > N=(a[1],a[2],a[3],...,a[n]), if T is described from two orthogonal > > systems, its linear requirement precipitates. T remains invariant > > despite the coordinate system used to describe it. Consequently, my > > revised web site showing the details, > > > > (t-2)(t-3)(t-4)=0 > > (t^2-5t+6)(t-4)=0 > > t-4 > > ------------- > > t^3-5t^2+ 6t > > -4t^2+20t-24 > > ---------------- > > t^3-9t^2+26t-24 > > > > N=+/-(26,-9,1) > REVOLUTIONARY !!! > Why don¹t you tell C.F. Gauss or E. Galois; they would love to hear > that you have easily solved a problem they have shown to be > principally not solvable!!! Jon is evidently still using the same wrong procedure. He doesn¹t come up with 2, 3, or 4 as roots of the (cubic!) polynomial he gives. (He cheated by rounding.) -- Christopher Heckman === Subject: Re: Under what circumstances is a complete metric space Hilbertizable? Content-Length: 1652 Originator: rusin@vesuvius >>Given a complete metric space X (with no given linear space structure), >>under what conditions does there exist a Hilbert space structure on X which >>induces the given metric? To put it another way, under what circumstances is >>a complete metric space Hilbertizable? Further, how does one determine the >>Hilbert space structure on X which induces the given metric? >I¹ll assume this is a real Hilbert space (though I think complex ones can >be done as well). For a complex Hilbert space, Þrst deÞne the real Hilbert space structure with inner product (.,.) as in my previous posting. Take a real-linear isometry A of this real Hilbert space such that A^2 = -I: this will exist as long as the dimension is either even or inÞnite, since for any orthonormal basis {u_j}_{j in B} you can partition B into ordered pairs [j,j¹] and deÞne A u_j = u_j¹ and A u_j¹ = - u_j. Then you can deÞne complex scalar multiplication so that ix = A x, and the sesquilinear complex inner product so that = 1, = i, = -i, and = 0 otherwise (I¹m using the convention that makes the inner product conjugate-linear in the Þrst argument and linear in the second). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Under what circumstances is a complete metric space Hilbertizable? Content-Length: 935 Originator: rusin@vesuvius >Given any x <> 0 and t real, tx is the point (which must >be unique) such that d(0,tx) = |t| d(0,x) and d(tx,x) = |1-t| d(0,x). >Given any x and y, x+y is the point (which must be unique) such >that d(2x,x+y) = d(2y,x+y) = d(x,y). >So, up to the arbitrary choice of origin, the metric >determines the linear-space structure. Perhaps I should add that this uniqueness is true for a Hilbert space, or more generally for a strictly convex normed linear space, but not for a normed linear space in general. I don¹t know how to determine the linear-space structure from the metric in the case of a general normed linear space. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Under what circumstances is a complete metric space Hilbertizable? Content-Length: 1257 Originator: rusin@vesuvius | |>Given any x <> 0 and t real, tx is the point (which must |>be unique) such that d(0,tx) = |t| d(0,x) and d(tx,x) = |1-t| d(0,x). |>Given any x and y, x+y is the point (which must be unique) such |>that d(2x,x+y) = d(2y,x+y) = d(x,y). |>So, up to the arbitrary choice of origin, the metric |>determines the linear-space structure. | |Perhaps I should add that this uniqueness is true for a Hilbert |space, or more generally for a strictly convex normed linear space, |but not for a normed linear space in general. I don¹t know how to |determine the linear-space structure from the metric in the case |of a general normed linear space. how about this: for unequal x and y, deÞne the line through x and y to be the points z such that equidistance from x together with equidistance from y implies equidistance from z. then the true midpoint between x and y is the point on the line through x and y with x and y equidistant from it. i still didn¹t prove whether this actually works, though. -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Under what circumstances is a complete metric space Hilbertizable? Content-Length: 916 Originator: rusin@vesuvius >how about this: for unequal x and y, deÞne the line through x and y >to be the points z such that equidistance from x together with >equidistance from y implies equidistance from z. then the true >midpoint between x and y is the point on the line through x and y >with x and y equidistant from it. In this deÞnition, (x+y)/2 might not be on the line through x and y, and there might be no true midpoint. For example, consider R^3 with the norm ||x|| = sup(|x_i|: i=1..3). The two points p = [1,0,1] and z = [0,1/2,1] are equidistant from x = [0,0,0] and from y = [0,1,2], but are not equidistant from z. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Eigenvalue of symmetric matrix Content-Length: 1488 Originator: rusin@vesuvius > Sorry, I made a mistake on the question. > Correction to the problem: > Given is any real matrix A of size n by n. Let¹s denote the maximum > eigenvalue of (A+A¹) as c_max(A). > Is there any function f(A) which can be used as upperbound for > c_max(A), > i.e. c_max(A) leq f(A) > where f(A) is a function of eigenvalue(s) of A? > Hardy == Let A=||a(i,j)|| be a n x n real matrix , A¹=||a(j,i)||, B=A+A¹=||b(i,j)||, b(i,j)=a(i,j)+a(j,i) , M(A):= 2*tr(A)/n | b(2,2) b(2,3) | | b(i-1,i-1) b(i-1,i+1)| S(A):= | |+ SUM_{i=2 to i=n-1}| |+ | b(3,2) b(3,3) | | b(i+1,i-1) b(i+1,i+1)| | b(n-2,n-2) b(n-2,n-1) | + | | | b(n-1,n-2) b(n-1,n-1) | . Note that in your case b(i,j)=b(j,i). Further denote D(A)=4*(n-1)(tr(A))^2 -2n*S(A) . It may be shown that D(A)>= 0 . If we deÞne f(A) = M(A) + (1/n) sqrt((n-1)*D(A)) , c_max(A) =< f(A) where c_max(A):= maximum eigenvalue of B:=A+A¹ . Perhaps help,Alex === Subject: Re: subspace interpolation Epigone-thread: cungtoapimp Content-Length: 2930 Originator: rusin@vesuvius << Assume I have two vector sets {x} and {y} which contain vectors corresponding to realizations of multidimensional Gaussian processes. Assume now that we build a new vector set as follows: {z} where z = a x + b y is a linear combination of the original vector sets (a+b = 1). {x} and {y} can be both modelled by their mean vector and covariance matrix. Is there any result that allows you to compute the mean vector and covariance matrix of {z} only based on the knowledge of a, b, and mean and covariance of x and y? This can be seen as given a and b, and two subspaces deÞned by the eigenvectors of the respective covariance matrices, Þnd the subspace of the interpolation between the two subspaces. Alternativelly, it could be interpreted as, given two statistical gaussian models corresponding to two variables, Þnd the interpolated statistical model of the interpolation of the very same two variables. Here¹s one result that might be relevant: Given two d-dimensional subspaces P, Q of R^n, there is a well-deÞned continuous family of d=dimensional subspaces of R^n that interpolate between P and Q just as the shortest geodesic in the Grassmannian G(d,n) connects the points in G(d,n) deÞned by P and Q. This can be computed by a) Þrst Þnding the vectors v1,w1 in P, Q resp., that make the smallest angle with each other. b) Assuming we have found v_1,w_1,v_2,w_2,...,v_(k-1),w_(k-1) (k <= d), we continue by considering only the subspaces P_k, Q_k of P, Q resp., deÞned as the orthogonal complements of span(v_1,...,v_(k-1)) and span(w_1,...,w_(k-1)), resp. Now Þnd the vectors v_k and w_k in Pk, Qk, resp., having the closet possible angle. [If at any stage the vectors v_k,w_k with closest possible angle in P_k, Q_k are non-unique, then make an arbitrary choice among the pairs of angles sharing the smallest possible angle.] c) Ultimatly one gets orthonormal bases {v_1,...,v_d} and {w_1,...,w_d} of P and Q resp. d) Now, simultaneously in the 2-planes span(v_k,w_k) we rotate v_k at a constant speed s_k through an angle <= pi until it reaches w_k. WLOG we may assume that the s_k¹s are chosen so that the rotating v_k¹s (which we call {v_k(t)}) are all parametrized by [0,1]. e) It can be checked that for any t in [0,1], the vectors {v_k(t)} form an orthonormal basis for a k-dim subspace of R^n, and that the 1-parameter of d-planes in R^n given by P(t) = span{v_1(t),...,v_k(t)} for t in [0,1] is precisely the shortest geodesic between the points P, Q of the Grassmannian G(d,n) in its invariant Riemannian metric (unique up to a global positive factor, unless d = 2, n = 4, in which case we use the cartesian square of a round 2-sphere]. In terms of the original question, the interpolated subspace for a+b = 1, a,b, >= 0, is P(a). Dan Asimov === Subject: Riemann zeta at odd integers Content-Length: 445 Originator: rusin@vesuvius I think I may have discovered a new way of calculating the value of the Riemann zeta function at odd integers > 1; it¹s little too complicated to be represented in ASCII text, so it¹s available at http://www.helsinki.Þ/~peuha/rzeta.pdf (at the end, preceded by a sketch of a proof). Have someone found this before me, or is it a new discovery? -- Esa Peuha student of mathematics at the University of Helsinki http://www.helsinki.Þ/~peuha/ === Subject: The projection vectors in the Weiss projection from 2_21 to {3,4,3} Content-Length: 970 Originator: rusin@vesuvius Background H.S.M. Coxeter mentions an orthogonal projeciton of 2_21 to {3,4,3} discovered by Dr. Asia Weiss (see C.R.Math.Rep.Acad.Sci.Canada, VIII,6, December 1986). If one examines the Weiss projection of 2_21 to {3,4,3}, it will be seen that the projection vectors in this projeciton deÞne a {3}, i.e. an equilateral triangle. (By projection vectors I mean the vectors which one must subtract from the radius vectors whose endpoints are the vertices of 2_21 in Dr. Weiss¹ coordinatization of this polytope in order to obtain the radius vectors whose endpoints deÞne the vertices of a {3,4,3}.) Question Can anyone think of any other interesting cases in which a dim n -> m projection from one polytope to another (n < m) deÞnes a set of projection vectors which themselves determine a polytope belonging to a well-deÞned class of polytopes (e.g. regualr, Archimedean, etc.)? === Subject: New Surface Gallery Content-Length: 1520 Originator: rusin@vesuvius I believe many of you may already be familiar with the mathematical visualization software, 3D-XplorMath. I was for many years the developer, but a few years ago I was joined by an international group of mathematicians to advise and help me. (Hermann Karcher, Bonn; Patrick Iglesias, Marseille; Chuu-lian Terng, Boston; Martin Guest, Tokyo; Matthias Weber, Bloomington; Michael Murray, Australia; Xah Lee, Palo Alto). We refer to ourselves as the 3DXM Consortium. We have long had a small gallery of surfaces on the web to show what 3D-XplorMath could do, but it was fairly primitive. Now, Xah Lee , has taken over as the Consortium webmaster and has created a much expanded and more professionally designed Gallery that we have just put online. We invite you to visit it at: In addition to carefully designed full color images of over seventy important and interesting surfaces, the Gallery also contains stereo images (anaglyph and parallel view dual image), technical notes, animations, PDF Þles, and live rotation using a Java applet. Our aim is to make this the best gallery of mathematical surfaces on the Web, both for research and as a curriculum adjunct for those teaching or learning differential geometry. We welcome your suggestions for additions and improvements. Please address any messages either to Xah or to me (or both). Dick Palais === Subject: Re: Yet another number theory problem Epigone-thread: frizhempham Content-Length: 4492 Originator: rusin@vesuvius >>Given (r,s,t) a primitive Pythagorean triple with r^2 + s^2 = t^2 and >>r even. >Equivalently, suppose r = 2uv, s=u^2-v^2 for two coprime integers u,v. >>Let k > 1 be an integer. Is it possible to Þnd such a k such >>that(k-1)(k+1)^3 r^2 + (2k-1)^3 s^2 is a square? >So you want to Þnd integers k>1 and u,v (coprime) so that > F(k,u,v) = (k-1)(k+1)^3 ( 2uv) ^2 + (2k-1)^3 (u^2-v^2)^2 >is a square. Note that F is quartic in each variable separately, >so we can hold any two of them Þxed to constant values; the >question then asks whether a certain elliptic curve has any integer >points. >After a little experimentation (low values of k either lead to >quartics with no rational point because of p-adic restriction, or >lead to an elliptic curve of rank 0) we Þnd that when k=13 and v=1 >there is a _rational_ point of inÞnite order, namely with u=8750/14421. >Using homogeneity we thus get an _integer_ point with >k=13, u=8750, v=14421. In other words, the answer to the original >question is afÞrmative, with examples like > r,s,t = 252367500, -131402741, 284527741 > k = 13 >The expression equals 48651361375^2 . >dave I found this problem a useful means of restarting my brain after the summer break. The results I found might be of some interest. DeÞning x = u/v in Dave Rusin¹s formulation we want (2k-1)^3 x^4 + 2(2k^4-4k^3+12k^2-10k-1) x^2 + (2k-1)^3 to be a rational square. Clearly, if (2k-1) is itself a square this will happen with x=0, and this rational point implies the quartic will be birationally equivalent to an elliptic curve. (The example, with k=13, found by Dave is of this form) So, let (2k-1)=(2i-1)^2 so k = 2i^2-2i+1, then standard methods give the equivalent elliptic curve as g^2 = h ( h - A ) ( h - B ) with A = 64 i (i-1) (i^2-i+1)^3 B = 4 (2i^2-2i+1) (2i^2-2i-1)^3 and the transformation is x = u/v = 2 h (2i-1)^3 / g. Thus these curves have 3 Þnite torsion points none of which give a solution to the problem. Experiments suggest that there are no other torsion points but this would be difÞcult to prove. To Þnd a non-trivial solution we need the rank of the curve to be at least 1, and to know the coordinates of a point of inÞnite order. Using some of my own software and John Cremona¹s mwrank program we Þnd the following data i k Rank h-coord of point of inÞnite order 2 5 0 3 13 1 8605184/25 4 25 0 5 41 1 9670752 6 61 2 (1) 8052139130407533/9217344049 (2) 208683969849783936/1266007561 7 85 0 8 113 1 113288933664/169 9 145 1 35625440120129043520/5193075969 10 181 0 It is noteworthy, however, that these points do not always directly solve the original problem. (r,s,t) must be a primitive Pythagorean triple with r even, so s must be odd, so (u,v) must have opposite parities. The formulation above does not enforce this constraint so it is possible that the u/v = 2h(2i-1)^3/g transformation could lead to a fraction u/v with both u and v odd. This would give r and s both even and dividing by 2 would give r odd. This happens when i=5 (k=41), h=9670752, g=1102465728 and u/v=243/19. Adding this point to (0,0),(A,0) or (B,0) gives essentially the same result. To get opposite parities we must double the (h,g) point to (35277854976/361 , 5893895843297280/6859) leading to u=1197340, v=198531, and thus r=475418215080 and s=1394208517639. The question then arises whether values of k with (2k-1) non-square can lead to solutions. Experiments, again with my software and Cremona¹s ratpoint program, suggest that, for values of k<=200, the quartic is only everywhere soluble p-adically when (2k-1) is a square. This, however, is a good example, of the try, try, and try again philosophy as we soon Þnd examples of p-adically everywhere soluble quartics. These occur for k=241, 277, 445, 565, 577, 601, 661, 745, 865, 949 and 977. I found only one actual solution during moderate testing. k=865 gives a solution with u=420 v=13, leading to r=10920 and s=176231. I don¹t have the energy to investigate further. It would be nice to know how the problem arose. Allan MacLeod === Subject: Re: Elementary Þeld theory question Content-Length: 2552 Originator: rusin@vesuvius > Say we have two algebraic numbers x,y such that > [Q(x):Q] = m, [Q(y):Q] = n, (m,n) = 1. > Is it true that Q(x,y) = Q(x+y)? > I¹ve been stuck on this seemingly innocuous looking problem for a while. My crummy newsreader doesn¹t remember this far back so apologies for starting a new thread. for pointing me to MR0258803 (41 #3449) Isaacs, I. M. Degrees of sums in a separable Þeld extension. Proc. Amer. Math. Soc. 25 1970 638--641. which in fact essentially solves the problem with Q replaced by any Þeld. It¹s always true in characteristic zero, but there are some cases in characteristic p where one has to be careful (it¹s not always true in the char p case). I¹ll sketch the proof in the characteristic zero case (which is much simpler than the more delicate characteristic p arguments in the paper). Let E be the Galois closure of Q(x,y) and let G=Gal(E/Q). Let H be the subgroup of G corresponding to Q(x) and let K be the subgroup corresponding to Q(y). Then [G:H]=m and [G:K]=n so G:H intersect K]=mn and the conjugates of x+y are precisely x_i+y_j as x_i runs through the conjugates of x and y_j through the conjugates of y. As we had already established in the thread, we now have to rule out the possibility that x+y=x_i+y_j for some conjugates x_i not=x of x and y_j not=y of y. This equation implies that x-x_i=y_j-y=u is a non-zero element of E. Now here¹s the trick in Isaac¹s paper. Let V be the sub-Q-vector space of E generated by the conjugates of x, and let W be the sub-Q-vector space generated by the conjugates of y. Then u is in both V and W, and hence V, W, and V intersect W are all non-zero Q-vector spaces with a G-action. Now what can be we say about V intersect W? Well, G acts on the x_i via permutations and H is the stabiliser of x, so (as a representation of G) V is a subquotient of Ind_H^G(1). Similarly W is a subquotient of Ind_K^G(1), and one checks that (Ind_H^G(1),Ind_K^G(1))=1 e.g. by Mackey¹s decomposition theorem, because G=HK. Moreover the common irreducible representation giving rise to this 1 is easily seen to be the trivial representation. Hence G acts trivially on V intersect W! But this is a contradiction because it implies that x-x_i is a non-zero rational number t, so x is conjugate to x+t and hence to x+2t, x+3t,... . Lenstra for pointing me to the paper of Isaacs. Kevin Buzzard === Subject: Green¹s function branch point Originator: rusin@vesuvius Question: Is anything known about the order of the branch point for the Scrodinger equation Green¹s function when the energy spectrum is continuous? --------------------------------- Do you Yahoo!? Yahoo! Mail - Helps protect you from nasty viruses. [HTML attachment deleted; please post in þat text. --Mod.] === Subject: Parker Vectors, Permutation Groups and Generalized Conjugacy Search Content-Length: 1424 Originator: rusin@vesuvius In P.J.Cameron¹s, Permutation Groups, LMS 45 CUP (1999) p. 49 the following problems is posed: Problems: (1) What does the Parker Vector of a permutation group G tell us about G. (2) In particular,which groups or classes of groups are determined by P(G)? Generally what progress has been made on problems ? In R.A. Parker¹s paper The Computer Calculation of Modular Characters In Computational Group Theory ,ed M.D. Atkinson Academic Press (1984) p.272 Parker remarks that if A1 ,B1, generate a representation and A2 ,B2 generate another of the same dimensions then using the methods described in his paperthat a simultaneous solution to the equations X A1 X^(-1)= A2, X B1X^(-1)X can be computed should one such solution exist. This same equations arose in another context AAG 1999 I. Anshel,M. Anshel and D. Goldfeld, An Algebraic Method for Public-Key Cryptography Mathematical Research Letters 6,1-5,(1999) and interested readers are invited to contact me regarding AAG 1999. This researcher would like to thank Professor Alex Ryba for introducing Parker¹s ideas and methods in a recent conversation. _____________________________________________________________ ______ Professor Michael Anshel Department of Computer Sciences R8/206 The City College of New York New York,New York 10031 http://www-cs.engr.ccny.cuny.edu/~csmma/ csmma@cs.ccny.cuny.edu MikeAt1140@aol.com === Subject: Re: Problem with sets, functions, & cardinality Content-Length: 158 Originator: rusin@vesuvius about cardinality were incorrect, but that wasn¹t the problem. Mike Carroll Oro Valley, AZ === Subject: Paper published by Algebraic and Geometric Topology Originator: israel@math.ubc.ca (Robert Israel) The following paper has been published: Algebraic and Geometric Topology URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-26.abs.html Title: Links associated with generic immersions of graphs Author(s): Tomomi Kawamura Abstract: As an extension of the class of algebraic links, A¹Campo, Gibson, and Ishikawa constructed links associated to immersed arcs and trees in a two-dimensional disk. By extending their arguments, we construct links associated to immersed graphs in a disk, and show that such links are quasipositive. Secondary: 57M27 Keywords: Divide, graph divide, quasipositive link, slice Euler characteristic, four-dimensional clasp number Author(s) address(es): Department of Physics and Mathematics, Aoyama Gakuin University 5-10-1, Fuchinobe Sagamihara, Kanagawa 229-8558, Japan Email: tomomi@gem.aoyama.ac.jp === Subject: Re: quadratic forms invariants question Content-Length: 1450 Originator: rusin@vesuvius >> Let A be any symmetric integer matrix and consider the equivalence relation >> A --> P^t A P >> for invertible P. >> We have different equivalence relations depending on which P¹s are >> allowed. E.g.: >> 1) P might be any invertible matrix with rational entries. >> 2) P might be any invertible matrix with integer entries. >> 3) P might be any matrix with rational entries and det = +/- 1. >> I¹ve seen discussions of the complete set of invariants for equivalence >> relations #1 and #2. Equivalence relation #3 would seem to fall >> someplace in between these. Does anyone know what the invariants are in >> this case? >The invariants for 3) are precisely those for 2) except that the >determinants (equivalently, discriminants) must be the same, >not just the same up to squares. >Indeed, Þrst, if det(P) = 1, then det (A) will equal det (P^t A P). >Conversely, if B = P^t A P for some rational P and det(B) = det (A), >then det(P)^2 = 1. I think you mean 3) and 1), not 3) and 2). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: partial diff equation Epigone-thread: priplingspou Content-Length: 720 Originator: rusin@vesuvius >I would like to learn something about the differential equation >partial^2 g/(partial x_i partial x_j)= >partial f/(partial x_i) partial f/(partial x_j), >where f:U->R, Usubset R^n is given and smooth, and I am looking for g:U->R. >Please let me know if you know a good starting point in the literature. >Martin I am interested with Œyour Œdifferential¹ equation , to be sure of my understanding , tell me if f(x1,x2)=x1*x2 and g(x1,x2)=x1^2*x2^2 /4 - a very simple case indeed - Þt in. Do indexes i,j ...mean it must be satisÞed for g and f multivariate by any pair of x_k, x_p ? Friendly,Alain. === Subject: Re: Inverting elliptic integrals Originator: israel@math.ubc.ca (Robert Israel) >I need the inverse of the (incomplete) elliptic integral of the second >kind. It is well deÞned but I can¹t track down a solution or code >for it. >.... For a numerical representation of the inverse in terms of the angle phi, where E(phi,m)=int(theta=0..phi) sqrt(1-m*sin^2 theta) dtheta is the elliptic integral fo the second kind, one could expand E(phi,m) in a power series around phi=0, Emphi := phi -1/6*m*phi^3 +1/5*(1/6*m-1/8*m^2)*phi^5 +1/7*(-1/45*m+1/12*m^2-1/16*m^3)*phi^7 +1/9*(1/630*m-1/40*m^2+1/16*m^3-5/128*m^4)*phi^9 ... The expansion coefÞcients in front of the order phi^(n+1) (n=2,4,6,...) are [sum over k=2,4,..,n of U(k,n)*F(k,m)/k!]/(n+1)! where U(k,n) = (-1)^[(k+n)/2]/2^(k-n)*[sum l=0,1,...k of (-1)^l (l-k/2)^n*binom(k,l)] and F(k,m) = -m^(n/2)*[(k-1)!!]^2/(k-1) , with (k-1)!! = 1*3*5*7*...*(k-1) . Then invert this as outlined in chapt 3.6.25 of the book edited by M Abramowitz and I Stegun: phi := E(phi,m) +1/6*m*E(phi,m)^3 +1/120*m*(13*m-4)*E(phi,m)^5 +1/5040*m*(493*m^2-284*m+16)*E(phi,m)^7 +1/362880*m*(37369*m^3-31224*m^2+4944*m-64)*E(phi,m)^9 ... Another efÞcient ansatz is a higher order Newton method, since the derivatives of E(phi,m) with respect to phi are well known: d E(phi,m)/d phi = sqrt(1-m*sin^2 phi) This needs in addition a solid implementation of the original E(phi,m) itself. Richard J. Mathar, http://www.strw.leidenuniv.nl/~mathar === Subject: Re: Inverting elliptic integrals Originator: israel@math.ubc.ca (Robert Israel) >I need the inverse of the (incomplete) elliptic integral of the second >kind. It is well deÞned but I can¹t track down a solution or code >for it. >.... > For a numerical representation of the inverse in terms of the angle phi, > where E(phi,m)=int(theta=0..phi) sqrt(1-m*sin^2 theta) dtheta is the > elliptic integral fo the second kind, one could expand E(phi,m) in a > power series around phi=0, > Emphi := phi > -1/6*m*phi^3 > +1/5*(1/6*m-1/8*m^2)*phi^5 > +1/7*(-1/45*m+1/12*m^2-1/16*m^3)*phi^7 > +1/9*(1/630*m-1/40*m^2+1/16*m^3-5/128*m^4)*phi^9 ... > The expansion coefÞcients in front of the order phi^(n+1) (n=2,4,6,...) > are [sum over k=2,4,..,n of U(k,n)*F(k,m)/k!]/(n+1)! > where > U(k,n) = (-1)^[(k+n)/2]/2^(k-n)*[sum l=0,1,...k of (-1)^l > (l-k/2)^n*binom(k,l)] and > F(k,m) = -m^(n/2)*[(k-1)!!]^2/(k-1) , > with (k-1)!! = 1*3*5*7*...*(k-1) . > Then invert this as outlined in chapt 3.6.25 of the book edited by M > Abramowitz and I Stegun: > phi := E(phi,m) > +1/6*m*E(phi,m)^3 > +1/120*m*(13*m-4)*E(phi,m)^5 > +1/5040*m*(493*m^2-284*m+16)*E(phi,m)^7 > +1/362880*m*(37369*m^3-31224*m^2+4944*m-64)*E(phi,m)^9 ... Yes. That¹s the same series I posted last week. See my Out[2], in which z is the same as your E(phi,m). Of course, the coefÞcients of the polynomials in m are obtainable by reversion of series, as we both did. But does anyone know of a simpler way of calculating them? (I may submit the sequence of coefÞcients to the OEIS in the near future.) David Cantrell === Subject: Please name this symmetry or freedom Originator: israel@math.ubc.ca (Robert Israel) I have been studying differential geometry in an informal way. My grasp of math terminology has never been a strong point of mine. I would appreciate help in Þnding the name of the issue that follows, whether it is a symmetry, freedom, or something else. Here is the deÞnition of a covariant derivative: A_u;v = A_u,v - L^w_uv A_w where A_u;v is the covariant derivative, A_u,v is the derivative, L^w_uv is the Christoffel symbol of the second kind. Nothing has been speciÞed yet about either the potential or the metric tensor whose derivatives make up the Christoffel symbol. Let me present a concrete example of the issue that concerns me. Pretend that A^0;1 happens to equal 43. If one chooses a þat metric in Euclidean coordinates so the connection is zero, then all of the 43 would have to come from the derivative of the potential. If the derivative of the potential were zero, then all of the 43 would come from changes in the metric. There would be a continuous range of possible contributions from the derivative of the potential and the connection whose difference was 43. My question is what to call this continuous relationship between the derivative of the potential and the connection. It reminds me of gauge symmetry or the freedom to choose a guage. I don¹t know what to call this choice in differential geometry. doug quaternions.com === Subject: Re: some questions about the algebraic closure of Q_p Epigone-thread: twyswinply Originator: israel@math.ubc.ca (Robert Israel) >These questions will seem rather naive to ots of people, I bet. So I >apologize in advance. >Let K be the Þeld of algebraic p-adic numbers. (So it is a subÞeld >of Q_p, and it is the Henselization of Q with respect to the p-adic >valuation). >Let K_1 be the valued Þeld extension of K obtained by addition of the >roots of all X^(p^n)-X. So the residue Þeld of K_1 is the algebraic >closure of F_p. K_1 is the inertia Þeld of K. The valuation group of >K_1 is still Z with v(p)=1. >Let K_2 be the extension of K_1 obtained by addition of the roots of >all X^n-p. Let say we keep v(p)=1, and the valuation group of K_2 is Q. >The question is : is K_2 algebraically closed ? If not, why ? >Can someone give me an exemple, for a given p, of a polynomial which >does not split in K_2 ? >Other question : what happens if K=Q_p ? Your >rather naive< question made me thinking a lot - without a complete answer yet. Let¹s look at the simpler case of the p-adics Þrst. We can then collect the following facts: 1. Adjoining all roots of X^n-p, where n is prime to p, to the absolute inertia Þeld K_1 yields the absolute ramiÞcation Þeld V. Proof. It is known that tamely ramiÞed extensions of the inertia Þeld are generated by roots of polynomials X^n-c, n prime to p, c an arbitrary element of K_1. c=up^m, u a unit in the valuation ring of K_1, m an integer. By Hensel¹s Lemma u^(1/n) is an element of T. 2. K_1 contains the pth roots of unity. 3. K_2 contains the p^m-th roots of unity for all m. Let V_1 be the extension of V by all p^m-th roots of unity for all m. 4. K_2|V_1 is an abelian extension by Kummer theory. Moreover we know that 5. The Galois group of the algebraic closure Q_p^~ over V is a free pro-p-group of countably inÞnite rank. Until now I was not able to put the pieces together and obtain an answer to your question. The value group of V_1 is Q, because the value group of V is divisible by any n prime to p, and the p-divisibility is created by the cylotomic extension V_1|V. If V_1 is not already the algebraic closure of Q_p, then this Þeld has an interesting property: for every Þnite extension L|V_1 the fundamental equality of valuation theory does not hold. This is because the value group of V_1 is Q (thus e=1) and its residue Þeld is algebraically closed (thus f=1). Such Þelds are called defect Þelds. H === Subject: Re: Question re: the two 2_21¹s in 3_21 and series parallel posets Originator: israel@math.ubc.ca (Robert Israel) > ... > The notion of series-parallel DAGs or posets built up from smaller DAGs or > posets resonated for me at an odd harmonic because of Coxeter¹s observation > that 3_21 can be built (all but two vertices) from two 2_21¹s lying in > parallel in E8. So: > ************************************************************* **************** > Question: > Does any one know of any formal machinery that would relate the construction > of series-parallel posets to the construction of larger polytopes from > smaller parallel polytopes, or am I just making too much over two entirely > different different uses of the word parallel that can¹t be formally > related in any meaningful way ? > ************************************************************* *************** You are mixing two meanings of the word parallel and two meanings of the word lattice as well. Posets relate to algebraic lattices, which perhaps should be called something else to avoid confusion with n-dimensional periodic patterns, which are lattices in the geometric sense. In his book, Regular polytopes, Coxeter uses the subscript notation to represent the lengths of the three arms of the Dynkin diagram. The Þrst digit is for the arm with the marked vertex: the simplex vertex which corresponds to vertices of the polytope. Section 11.8 is about the 6, 7 and 8-dimensional polytopes, 2_21, 3_21 and 4_21. Distance Regular Graphs, by Brouwer, Cohen and Neumaier, discusses these polytopes as graphs in section 10.3. As a graph, 2_21 becomes the Schlaþi graph, 3_21, the Gosset graph, and 4_21, the root system graph of E8. The Gosset graph has two Schlaþi graphs as layers sandwiched between two individual vertices. Graphs with this two-layered structure are called Taylor graphs. The simplest Taylor graph is icosahedral, and the two layers are pentagons. In general, the two layers must be copies of a strongly regular graph in which each non-adjacent vertex shares half of the neighbors of any given vertex. Starting with the right kind of strongly regular graph, add a vertex which connects to all of its vertices. Then change the graph into a Taylor graph by doubling it, while putting crossed edges in place of the missing edges, thus making a double cover of a complete graph. Distribution matrices for 4_21, 3_21 and 2_21, and on the right, eigenvectors with eigenvalues above and multiplicities below: 56 28 8 -2 -4 1 0 56 0 0 0 1 1 2 7 28 42 56 1 27 27 1 0 56 1 1 1 -1 -3 126 0 12 32 12 0 126 1 0 -1 0 2 56 0 1 27 27 1 56 1 -1 1 1 -3 1 0 0 0 56 0 1 1 -2 7 -28 42 1 8 35 112 84 27 9 -1 -3 1 0 27 0 0 1 1 3 27 9 27 1 16 10 0 27 1 1 -1 -1 27 0 10 16 1 27 1 -1 -1 1 1 0 0 27 0 1 1 -3 27 -9 1 7 27 21 16 4 -2 1 0 16 0 1 1 4 40 16 1 10 5 16 1 1 -5 10 0 8 8 10 1 -2 4 1 6 20 === Subject: Properties of polynomials over Þnite Þelds Originator: israel@math.ubc.ca (Robert Israel) I¹m trying to understand some CS papers that make extensive use of multivariate polynomials over a Þnite Þeld F in their proofs. I¹m therefore looking for any text that deals with this subject. SpeciÞcally, they use the polynomial representation of some functions from the afÞne space F^k to the Þeld F and make various claims about interpolation of such k-variate polynomials based on their restriction to many afÞne subspaces of smaller dimension, e.g. F^{k-1}. So anything in that direction will be VERY helpful. Tim === Subject: MATLAB 7.0.1 Release 14 SP1, and eBooks MATLAB 7.0.1 Release 14 SP1, and eBooks MATLAB 7.0.1 Release 14 SP1 (c) MathWorks [3 CDs] MATLAB 7 Release 14 - MathWorks [2 CDs] CD NR 15 824, COMSOL_FEMLAB_V3.0A John.Wiley.Sons.Computational.Colour.Science.using.MATLAB.eBoo k Orchard.Publications.Signals.and.Systems.with.MATLAB.Applicati ons.eBook Orchard.Publications.Circuit.Analysis.II.with.MATLAB.Applicati ons.eBook 10/01/2001 WaveWarp Audio Toolbox v2.0.1 for MatLab please send e-mail, code_fu@pathÞnder.gr, astra35@freemail.gr, === Subject: Re: MuPad 3.0 Calculus of Variations - syntax > A bit more detail... > Here is a simple Calculus of Variations problem.... > Which curves can give an extremum of the functional > v(y(x)) = integral from 0 to 1 of ((y¹)^2+12xy)dx, y(0)=0, y(1)=1 ? > The answer is y=x^3 > Let¹s try to solve this.... > L := (diff(y(x),x))^2 + 12*x*y(x); > res := detools::euler(L, [x], y); > This gives the Euler equation 2.y([x,x])-12x (equation #1) > Now to check x^3 is the solution to equation #1 > y := x -> x^3; > op(res,1)+op(res,2); > This gives Error: Illegal operand [_power] during evaluation of y > 2*diff(y(x),x,x)-12*x; //gives 0 as expected. > This is a simple case, but I would like to be able to do this type of > veriÞcation of the Euler equation in more complicated cases without having > to type in the output from detools::euler. > Any help appreciated. It seems to work if you put y(x) := x^3 instead of with the ->. Is that the same? -- john === Subject: Re: MuPad 3.0 Calculus of Variations - syntax Hi John, > It seems to work if you put y(x) := x^3 instead of with the ->. > john Your suggestion suppresses the output of an error message, but it also doesn¹t evaluate res for y(x) := x^3 either. When you substitute y=x^3 into res it should calculate the 2nd derivative, which is y¹¹=6x, substitute this into res and return zero. All that happens is that it returns the expression for res unevaluated. I tried..... delete y; L := (diff(y(x),x))^2 + 12*x*y(x); res := detools::euler(L, [x], y); //res is Euler¹s eqn for L // All of these attempts below fail to evaluate res which should be zero. eval(subs(op(res,1)+op(res,2),y([])=x^3)); //1st try eval(subs(op(res,1)+op(res,2),y=x^3)); //2nd try eval(subs(op(res,1)+op(res,2),y(x)=x^3)); //3rd try y := x -> x^3; //4th try op(res,1)+op(res,2); In reality I am trying to set up equations along the lines... F := 1/sqrt((x^2+(y(x))^2))*sqrt(1+(diff(y(x),x))^2); res := Simplify(detools::euler(F, [x], y)); I take the numerator of res and re-type it as follows y(x)-x*(diff(y(x),x))^3+y(x)*(diff(y(x),x))^2+(x^2+(y(x))^2)* diff(y(x),x,x)- x*diff(y(x),x); and then substitute y := x -> coth(2*h)-sqrt((csch(2*h))^2-x^2); to see if it is a solution. I have a lot of these to do and it would be nice to just ask MuPad to evaluate res. Using MuPad in this way is still much better than doing the whole thing by hand. Brad === Subject: Re: MuPad 3.0 Calculus of Variations - syntax > Hi John, >>It seems to work if you put y(x) := x^3 instead of with the ->. >>john > Your suggestion suppresses the output of an error message, > but it also doesn¹t evaluate res for y(x) := x^3 either. > When you substitute y=x^3 into res it should calculate the 2nd > derivative, which is y¹¹=6x, substitute this into res and return zero. > All that happens is that it returns the expression for res unevaluated. > I tried..... > delete y; > L := (diff(y(x),x))^2 + 12*x*y(x); > res := detools::euler(L, [x], y); //res is Euler¹s eqn for L > // All of these attempts below fail to evaluate res which should be zero. > eval(subs(op(res,1)+op(res,2),y([])=x^3)); //1st try > eval(subs(op(res,1)+op(res,2),y=x^3)); //2nd try > eval(subs(op(res,1)+op(res,2),y(x)=x^3)); //3rd try > y := x -> x^3; //4th try > op(res,1)+op(res,2); > In reality I am trying to set up equations along the lines... > F := 1/sqrt((x^2+(y(x))^2))*sqrt(1+(diff(y(x),x))^2); > res := Simplify(detools::euler(F, [x], y)); > I take the numerator of res and re-type it as follows > y(x)-x*(diff(y(x),x))^3+y(x)*(diff(y(x),x))^2+(x^2+(y(x))^2)* diff(y(x),x,x)- > x*diff(y(x),x); > and then substitute y := x -> coth(2*h)-sqrt((csch(2*h))^2-x^2); > to see if it is a solution. > I have a lot of these to do and it would be nice to just ask MuPad to > evaluate res. > Using MuPad in this way is still much better than doing the whole thing by > hand. Brad, I¹m just treading water here, and I don¹t know if this applies or is practical, but... You seem to be saying that y([x,x]) is equivalent to diff(y,x,x), but muPad doesn¹t seem to accept the functional argument to the former. I tried using a substitution- delete x,y: L := (diff(y(x),x))^2 + 12*x*y(x): res := detools::euler(L, x, y) res:=subs(res, y([x,x]) = hold(diff(y,x,x))) y:=x^3 Then evaluating res gives 0, but if y is set to x^2 or x^4, it doesn¹t equal zero. I don¹t know if your results have y([x,x]) consistently enough for this to be useful. -- john === Subject: Re: MuPad 3.0 Calculus of Variations - syntax Hi John, > You seem to be saying that y([x,x]) is equivalent to diff(y,x,x), but > muPad doesn¹t seem to accept the functional argument to the former. I > tried using a substitution- > res:=subs(res, y([x,x]) = hold(diff(y,x,x))) > john This was the perfect answer to my problem!! I would not have thought of that in a million years. I notice in example 6 on subs there is an allusion to your suggestion, but I would not have picked that up. I had always looked upon subs as a means to modify input rather than output. The MuPad notebook for what I am doing now looks like..... delete x,y: //This is one of the functions of interest F := 1/sqrt((x^2+(y(x))^2))*sqrt(1+(diff(y(x),x))^2); //Calculate Euler-Lagrange equation - numerator only res := numer(Simplify(detools::euler(F, [x], y))); //Fix up 1st and 2nd derivatives res:=subs(res, y([x,x]) = hold(diff(y,x,x)), y([x]) = hold(diff(y,x))); //In twinkling of an eye, see if E-L is satisÞed by y := coth(2*h)-sqrt((csch(2*h))^2-x^2); Simplify(res); I just change the function of interest and run it again. You can see that it incorporates your suggestion. That has been incredibly helpful and again I express my gratitude. You can collect quite a few free beers if you ever get to Australia. Brad === Subject: Re: MuPAD 2.5 : Getting the values used to plot a function > I would like to write down into a Þle all the values used by the plot > algorithm of MuPAD when ploting a 2d function, with the syntax : > ---Begin > X_1 y_1 > x_2 y_2 > ... > X_n y_n > ---End > I can¹t Þgure out how to get these values. write the plot output in ASCII format into a Þle: plotfunc2d(sin(x), PlotDevice= [xxx.txt, Ascii]) This is not exactly the format you were looking for, but you just have to skip the Þrst 10 lines. -- *---* MuPAD -- The Open Computer Algebra System *---*| |*--|* Ralf Hillebrand (tonner@mupad.de) *---* === Subject: question on Mathematica package loading. I have posted this question 3 days ago (oct 27) to the Mathematica news group, and as of today oct 30, it still has not shown up, I assume it was not allowed on that group. It might be possible becuase I put the word Œbug?¹ in the subject. So I am removing the word Œbug¹ from the subject and trying this newsgroup. I am posting the question here in the hope someone can help. cheers, Bill ----- post ------- I am having hard time understanding why Mathematica is doing the following. To summarize, I create a simple package with one function test3[] in it. Load the package, and can call the function ok. I removed the Œusage¹ line, and try to call the function again, but it will not work AS expected. But this results in a new symbol added by the same name of the function, again as expected. But I then put the Œusage¹ line back into the package, clean the environment again, and reload the package, but I still can not access the package function again, it is still calling a fake symbol instead. ----- package Þle strange.m ---- BeginPackage[strange`] test3::usage=test3 <----- (1) Begin[`Private`] test3[x_]:=x^3; End[] EndPackage[]; ---- now test the package, we see we can call strange`test3[] <save) then did Remove[Global`*]; Remove[strange`*]; Names[strange`*] {} < I have posted this question 3 days ago (oct 27) to the Mathematica > news group, and as of today oct 30, it still has not shown up, I assume > it was not allowed on that group. > It might be possible becuase I put the word Œbug?¹ in the subject. > So I am removing the word Œbug¹ from the subject and trying this > newsgroup. > I am posting the question here in the hope someone can help. > cheers, > Bill > ----- post ------- > I am having hard time understanding why Mathematica is doing > the following. > To summarize, I create a simple package with one > function test3[] in it. Load the package, and can call > the function ok. I removed the Œusage¹ line, and try to call > the function again, but it will not work AS expected. But this > results in a new symbol added by the same name of the function, > again as expected. > But I then put the Œusage¹ line back into the package, > clean the environment again, and reload the package, > but I still can not access the package function again, it > is still calling a fake symbol instead. > ----- package Þle strange.m ---- > BeginPackage[strange`] > test3::usage=test3 <----- (1) > Begin[`Private`] > test3[x_]:=x^3; > End[] > EndPackage[]; > ---- now test the package, we see we can call strange`test3[] > < Names[strange`*] > {test3} > strange`test3[4] > 64 > ---------------------- > Now edit the Þle strange.m, and remove the line > marked by (1), so we now get > BeginPackage[strange`] > Begin[`Private`] > test3[x_]:=x^3; > End[] > EndPackage[]; > ---- Now remove all symbols, and reload, we see the > ---- name test3[] is not visible > Remove[Global`*]; Remove[strange`*]; Names[strange`*] > {} > Now load the packge again and try to call test3 > Get[strange.m]; Names[strange`*] > {} > You see, NO names in package. Now try to call strange`test3[], > this will cause a new symbol to be created instead as expected: > strange`test3[4] > test3(4) > Ok, now it seems to Þx this, all what I have to do is > to remove all the names in the strange context, and add the > usage line back to the package, and reload, and then > it should work. However, it still does not work. > Added line(1) back, SAVED the package to disk again (Þle->save) > then did > Remove[Global`*]; Remove[strange`*]; Names[strange`*] > {} > < Names[strange`*] > {test3} > -- so the name is back. OK. Now I would expect to be able > -- to call it again and it should work as before. BUT it > -- does not! > strange`test3[4] > test3(4) <----- This should come back as 64 > -- Why has not Mathematica call the correct function again?? > -- checked that there is NO global test3 function anywhere: > Names[Global`test3] > {} > -- So what is going on?? MathGroup was not updated for several days during that period. I don¹t see why your posting wouldn¹t have been accepted and using bug wouldn¹t have made any difference. I didn¹t try to follow everything that you did because it¹s a lot of work. But you always need a usage message for any function that is to be exported from a package. I don¹t understand why you would want to remove it and then see what happens. When working with packages I always work with a package in the .nb format, use initialization cells and use the Create Auto Save Package option when Þrst saving the notebook. It is easier to edit a notebook than a .m Þle. Then after changing the package I always 1) save it, 2) Quit the kernel and 3)reload the package. I never run into any difÞculties that way. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ === Subject: Re: Software for Clifford Algebra > Some years ago , I used a software > written by Prof. Parra (Spain), named Clifford. It worked > under Mathematica. I would like to know if > there exists an updated version of this program. > Yvon Siret A recent book Lectures on Clifford (Geometric) Algebras and Applications, editted by Rafal Ablamowicz and Garret Sobczyk, has an appendix that describes a number of software packages for for manipulating Clifford Algebras. Reference 42 gives: J. M. Parra and L. Rosello, Mathematica donloadable from URL: HTTP://www.birkhauser.com/detail.tpl?ISBN=0817639071 A different reference gives the identical URL for the 1995 version I just tried the URL and it is now redirected to Springeronline and gives the opportunity to buy Clifford Algebras with Numeric and Symbolic Computations I could not Þnd any downloadable software however. Richard Harke === Subject: Re: Software for Clifford Algebra > Some years ago , I used a software > written by Prof. Parra (Spain), named Clifford. It worked > under Mathematica. I would like to know if > there exists an updated version of this program. > Yvon Siret > A recent book Lectures on Clifford (Geometric) > Algebras and Applications, editted by Rafal Ablamowicz > and Garret Sobczyk, has an appendix that describes a number > of software packages for for manipulating Clifford Algebras. > Reference 42 gives: J. M. Parra and L. Rosello, Mathematica > donloadable from URL: > HTTP://www.birkhauser.com/detail.tpl?ISBN=0817639071 > A different reference gives the identical URL for the 1995 version > I just tried the URL and it is now redirected to > Springeronline and gives the opportunity to > buy Clifford Algebras with Numeric and Symbolic > Computations I could not Þnd any downloadable software > however. > Richard Harke There is a Mathematica package for Clifford Algebra done by Renan Cabrera. It is a subpackage for the Tensorial tensor calculus package on the Mathematica page at my web site below. The Clifford Algebra subpackage is still somewhat in the development stage. Renan is a student of William Baylis and is working pretty intensively on Clifford algebra. David Park djmp@earthlink.net http://home.earthlink.net/~djmp/ === Subject: Re: Software for Clifford Algebra are clifford algbra packages in Maple and Macsyma and Mupad and Axiom. Some of these systems may be available without buying anything. RJF === Subject: Differential Geometry with Maple 9 atlas - powerful Maple package for differential geometry calculations is now available for Maple 9 at http://www.graphtree.com/Maple/atlas/index.htm The package works with manifolds, mappings, embeddings, submersions, p-forms, tensor Þelds etc. The user may concentrate on geometrical problem not on the programming. Standard mathematical notations accepted in modern differential geometry are used in the package. All calculations are as coordinate free as possible. Some calculations with non-numerical dimension are available. Detailed help Þles, templates and examples are available. Any differential geometry problem can be solved by one and the same simple solving scheme. Besides that new differential geometry powerful tool (atlas 2D/3D Wizard) is available for Maple 9 at http://www.graphtree.com/Maple/atlasWizard/index.htm With this Wizard you can solve typical differential geometry problems in any 2D or 3D coordinate system: - plane curves - space curves - surface geometry -- Norb Bobrin, Graph Tree Ltd. Maple & Mathematica Packages, Programs & Math Arts http://www.graphtree.com/ === Subject: Re: looking for problems X-RFC2646: Format=Flowed; Response Here is my solution to the problem of how many members attended the Greenwich Village Knights and Knaves Club meeting. S P O I L E R S P A C E S P O I L E R S P A C E First, consider what we can deduce when we have two people A and B from the islands of Knights and Knaves and A says that B is a Knave. Now A must be either a Knight or a Knave. - If A is a Knight, then A is truthful and therefore B is a Knave. - If A is a Knave, then A is lying, so B is not a Knave; thus B must be a Knight. Note that in all cases, A and B are different types, i.e. if one of them is a Knight, the other is a Knave. So whenever A claims that B is a Knave (where A and B are from the islands of Knights and Knaves), then we can conclude that A and B are of different types. At the meeting, the reporter found that every attendee claimed that the member seated to his or her left was a Knave. This implies that every attendee is seated to the right of a member of a different type. In other words, the attendees were seated so that the Knights and Knaves were in strict alternation. Now the club president tells the reporter that there were 37 attendees. This is not possible, for you could not seat an odd number of club members around the table so that Knights and Knaves alternate. Therefore, the club president lied; she must be a Knave. The secretary says that the president is a Knave. This is a truthful statement, so the secretary must be a Knight. Ergo, the secretary¹s other statement, that there 40 members at the meeting, must be true. In short, the reporter can deduce that there were 40 members at the meeting (20 Knights and 20 Knaves). Moreover, she also knows that the club president is a Knave and that the club secretary is a Knight. === Subject: Re: volume help needed <4180e8af$0$33616$ed2619ec@ptn-nntp-reader02.plus.net> > >> http://www.djh.ukjournalists.co.uk/volume.htm > > Nothing shows up on my bowser. > I don¹t understand that - I¹ve just checked it and it seems ok > Read closely, nothing shows up on MY browser. Ah, got it, not your BOWSER then. ;) -- Odysseus === Subject: Re: volume help needed > >> http://www.djh.ukjournalists.co.uk/volume.htm > In glorious ASCII, the image is roughly > / > 16/ 16 > / > / > 5______/5 > 10 ^^ used the correct value > With the 16/5 corners and the 16/16 corner all right angles. They¹re not so marked, although they do look that way. We *were* warned that the diagram is rough, so I¹m not sure it¹s safe to assume anything not speciÞed. > The cross sectional area is 16*16 - (11*11)/2 = 195.5 > So the volume is 7038 cubic inches. > I have no idea what that is in other imperial measurements, as imperial¹s > a pile of crap. Yes, for me it¹s easiest to go _via_ metric: 7038 cu.in. * (0.254 dm/in.)^3 ~= 115.3 litres; 115.3 L / 4.546 L/Imp.gal. ~= 25.37 Imperial gallons. But FWIW converting to U.S. gallons is straightforward: 7038 cu.in. / 231 cu.in./USgal. ~= 30.47 U.S. gallons. -- Odysseus === Subject: Re: OOPS! A bit of a mistake > Mark Spahn a .8ecrit: > By calculus, I found that the area of circle B that does not > lie inside circle A works out to 1. But it looks like you have > a way to determine the answer without resort to calculus. > If so, please clearly state the solution. > Of course you don¹t need calculus. > As said, the searched area is half circle (A) minus area of > circle (B) *segment*, that is the part between chord PQ and > circle (B) > But that part is 1/4 of circle (B) minus triangle SPQ > wich is composed of two triangles with sides 1, > hence area(SPQ) = 1 (even no need of sqrt(2) !!!) > so Area = pi/2 - (2*pi/4 - 1) = 1 > Another Þne lune shape is also obtained in the following way > Draw a right triangle ABC, AB_|_AC, AB=1 AC=2. > Draw the two semicircles with diameters AB and AC, outside > the triangle. > Draw semicircle (ABC), with diameter BC. > This gives two lunes, whose total area is 1 > proof : > Area is > pi*AB^2 /8 + pi*AC^2 /8 the two semicircles > + AB*AC/2 triangle > - pi*BC^2/8 the big semicircle > and because of Pythagora BC^2 = AC^2 + AB^2, the > pi*xxx terms cancel and remains : > Total area of lunes is area of triangle = 1 > http://mathworld.wolfram.com/Lune.html > Search for Hippocrates of chios+lune for other similar shapes. > -- > philippe > (chephip at free dot fr) Philippe, When stated as clearly as you have, the derivation becomes obvious. book Journey Through Genius by William Denham, which is the best math book I have ever read. It has many clever, classic proofs of theorems that I remember from high school. In high school, they were simply stated without proof (e.g., that the area of a circle is pi*r^2). I assumed that the proof must have been very difÞcult, and it was not until I encountered Denham¹s book many years later that I found that the proof is straightforward enough that I could have understood it in high school. My reaction was, Why has this clever, understandable proof been withheld from me? What cruelty to act as if this information is beyond the capacity of high-school-age kids! -- Mark Spahn === Subject: Re: vertical spherical cap volume? > I thought it was possible and even easy to calculate, not FOR ME, but for persons trained to mathematics. > In practice, the total water volume in each cap il about 29 liters, less than one percent of the total volume (3000 liters); then, > my search was specially for mind satisfaction. However, I¹ll note down with as accuracy as possible the real shape of the cap when > it will be possible for me (in a few weeks). Then, I¹ll try to determine the known curve the more probable of the cap shape. > Yves Yves, I am curious. You have a tank whose shape is roughly _____ (_____). How did you determine its total volume? By Þlling the tank to the brim with a known volume of water? And how did you determine the volume of each cap? By tilting the tank upright and Þlling just one cap to its brim with a known volume of water? Here is a sub-problem: What is the most convenient way to determine the shape of each cap? Would it sufÞce merely to use a tape measure to measure the length between diametrically opposite points at the base of the cap, with the tape measure stretched across the middle of the cap? Or does some other meausement need to be taken also, such as the altitude of the cap above its base? (I don¹t know the answers to these questions.) And we haven¹t considered the thickness of the tanks walls and caps. -- Mark === Subject: Re: vertical spherical cap volume? Sorry, Mark, I have no time to answer you today, but only tomorrow. Have a good day. Yves > I thought it was possible and even easy to calculate, not FOR ME, but for > persons trained to mathematics. > In practice, the total water volume in each cap il about 29 liters, less > than one percent of the total volume (3000 liters); then, > my search was specially for mind satisfaction. However, I¹ll note down > with as accuracy as possible the real shape of the cap when > it will be possible for me (in a few weeks). Then, I¹ll try to determine > the known curve the more probable of the cap shape. > Yves > Yves, I am curious. You have a tank whose shape is roughly > _____ > (_____). How did you determine its total volume? By Þlling the > tank to the brim with a known volume of water? And how > did you determine the volume of each cap? By tilting the > tank upright and Þlling just one cap to its brim with a known > volume of water? > Here is a sub-problem: What is the most convenient way > to determine the shape of each cap? Would it sufÞce > merely to use a tape measure to measure the length > between diametrically opposite points at the base of > the cap, with the tape measure stretched across the > middle of the cap? Or does some other meausement > need to be taken also, such as the altitude of the cap > above its base? (I don¹t know the answers to these questions.) > And we haven¹t considered the thickness of the tanks > walls and caps. > -- Mark === Subject: orange problem > > > What do you calculate is the half-life of the environment? > That usually doesn¹t come up. What people want to know is whether > it¹s safe enough to build a baseball diamond or a factory on top > of a former oil reÞnery. The lake of gasoline þoating on the > water table caused by a hundred years of leaking storage tanks > isn¹t impacting the environment all that much because of > remediation and control efforts. They¹re working? How much of an ongoing effort is this? Who¹s paying for it? Is it a perpetual chore? What about the water? Is it not unusable, another aquifer ruined? Wouldn¹t it just be cheaper to give $10,000 to congress to be relieved of that chore? Thus save $10 million with a return of $1,000 for every dollar invested. It may look like a good return, but as buying politicians goes, there are better investments to make. My estimate for expected return on the purchase of politics is $1,000 to $10,000 for every dollar invested in a corruptocrate. > But the owners would like to re-use the land if possible, so there is a > lot of data being collected and analyzed. I¹m sorry to have to inform them, like a dry oil well, they used the land up. > I do the data collection, someone else does the risk assessment as I¹m > not a scientist either. Risk assessment is done by scientists? No, how can that be? Are they not done by economists and attorneys specializing in liability? > Will the > effects of the election cause any difference in your calculations? > How committed to cleaning up the environment do you think a > Republican oil man from Texas is? > Re-defeat Bush! Since 2000, I estimate the half life of the environment has been halved from the previous estimated of 2000. Ie, instead of running out of environment this century, we¹ll run out of environment the Þrst half of this century. === Subject: Re: orange problem >Since 2000, I estimate the half life of the environment has been halved >from the previous estimated of 2000. Ie, instead of running out of >environment this century, we¹ll run out of environment the Þrst half of >this century. You¹re not taking into account the half life of your estimate, which is four years. in 2008 you will have halved your present estimate to 2025. In 2008 it will become 2012, and you will suffocate before making another estimate. -- Patrick Hamlyn posting from Perth, Western Australia WindsurÞng capital of the Southern Hemisphere Moderator: polyforms group (polyforms-subscribe@egroups.com) === Subject: Improvements on electoral system? Every four years, there are numerous debates as to the strengths and weaknesses of the Electoral College system we use to elect our President. As you know, it is possible for someone to win the popular vote and still lose the election. drove the creation of a two-part Congress; a House to represent population and a Senate to represent individual states. It turns out that the Electoral College is very similar in one aspect, that being the number of votes. Each state gets 1 electoral vote for each member it has in the House and 2 electoral votes for its 2 Senators. But the analogy stops there. I¹ve played with various systems of vote distribution to see what, if any, would be an improvement to our current system. The State of Maine has what looks like a great system and my vote would be to use Maine¹s system for each state. What Maine does is give its 2 senate electoral-votes to the overall winner, and splits the remaining house electoral-votes as per the popular vote. Anyone think of a better system? === Subject: Re: orange problem > > > > What do you calculate is the half-life of the environment? > That usually doesn¹t come up. What people want to know is whether > it¹s safe enough to build a baseball diamond or a factory on top > of a former oil reÞnery. The lake of gasoline þoating on the > water table caused by a hundred years of leaking storage tanks > isn¹t impacting the environment all that much because of > remediation and control efforts. > They¹re working? Yes, the contamination is being conÞned to within the property boundaries and the nearby municipal wells are not affected. > How much of an ongoing effort is this? We get several million dollars a year for our part of the operation and we aren¹t the only ones working. > Who¹s paying for it? The reÞnery owners. In this case, although the reÞnery itself was closed and torn down, many of the storage tanks are still in place and the site still operates as a storage terminal. > Is it a perpetual chore? Yep. So far, at least 5 million gallons of oil have been recovered and once that slows to a trickle, further steps will be taken to deal with the stained soil such as vapor extraction, oil degrading bacteria, etc. > What about the water? Is it not unusable, another aquifer ruined? None of the proposed re-developement plans call for residential use. And the auqifer is not ruined. As I said earlier, the water outside the property line is not ruined. > Wouldn¹t it just be cheaper to give $10,000 to congress to be relieved of > that chore? Thus save $10 million with a return of $1,000 for every > dollar invested. It may look like a good return, but as buying > politicians goes, there are better investments to make. My estimate for > expected return on the purchase of politics is $1,000 to $10,000 for every > dollar invested in a corruptocrate. And as long as we¹re fantasizing, it would probably be cheaper to hire Superman to burn away the oil with his x-ray vision. > But the owners would like to re-use the land if possible, so there is a > lot of data being collected and analyzed. > I¹m sorry to have to inform them, like a dry oil well, they used the land > up. That may very well be the case. None of the proposed land re-use plans have been implemented yet, so it remains to be seen whether the land will ever be used again. > I do the data collection, someone else does the risk assessment as I¹m > not a scientist either. > Risk assessment is done by scientists? No, how can that be? Are they not > done by economists and attorneys specializing in liability? Financial risk is done by economists. Health risk is done by scientists. You should see the math involved in modeling ingestion pathways. > > Will the > > effects of the election cause any difference in your calculations? > How committed to cleaning up the environment do you think a > Republican oil man from Texas is? > Re-defeat Bush! > Since 2000, I estimate the half life of the environment has been halved > from the previous estimated of 2000. Ie, instead of running out of > environment this century, we¹ll run out of environment the Þrst half of > this century. Is there any science behind this estimate or is it as sensible as your political math? === Subject: Re: Improvements on electoral system? > Every four years, there are numerous debates as to the strengths and > weaknesses of the Electoral College system we use to elect our > President. As you know, it is possible for someone to win the popular > vote and still lose the election. > drove the creation of a two-part Congress; a House to represent > population and a Senate to represent individual states. > It turns out that the Electoral College is very similar in one aspect, > that being the number of votes. Each state gets 1 electoral vote for > each member it has in the House and 2 electoral votes for its 2 > Senators. But the analogy stops there. > I¹ve played with various systems of vote distribution to see what, if > any, would be an improvement to our current system. The State of Maine > has what looks like a great system and my vote would be to use Maine¹s > system for each state. > What Maine does is give its 2 senate electoral-votes to the overall > winner, and splits the remaining house electoral-votes as per the > popular vote. > Anyone think of a better system? There are many better systems. The best I can think of, offhand, would be to do away with the Electoral College completely, basing the election simply on the popular vote and using approval voting. (If you¹re not familiar with approval voting, do a Google search, or see .) David === Subject: Re: Improvements on electoral system? The only problem I see with an outright popular voting system is .. how would you manage recounts? Remember the Þasco in Florida? Imagine that on a national level. You¹d still need some kind of containing mechanism so that recounts could be limited to smaller areas. >>Every four years, there are numerous debates as to the strengths and >>weaknesses of the Electoral College system we use to elect our >>President. As you know, it is possible for someone to win the popular >>vote and still lose the election. >>drove the creation of a two-part Congress; a House to represent >>population and a Senate to represent individual states. >>It turns out that the Electoral College is very similar in one aspect, >>that being the number of votes. Each state gets 1 electoral vote for >>each member it has in the House and 2 electoral votes for its 2 >>Senators. But the analogy stops there. >>I¹ve played with various systems of vote distribution to see what, if >>any, would be an improvement to our current system. The State of Maine >>has what looks like a great system and my vote would be to use Maine¹s >>system for each state. >>What Maine does is give its 2 senate electoral-votes to the overall >>winner, and splits the remaining house electoral-votes as per the >>popular vote. >>Anyone think of a better system? > There are many better systems. The best I can think of, offhand, would be > to do away with the Electoral College completely, basing the election > simply on the popular vote and using approval voting. (If you¹re not > familiar with approval voting, do a Google search, or see > .) > David === Subject: RA Positions at UNR The Computer Vision Laboratory (CVL) at the University of Nevada, Reno (UNR) invites applications for research assistant positions starting in Spring2005/Fall 2005. Preference will be given to students who want to pursue a PhD degree in Computer Vision. Active research areas within CVL include object recognition, visual motion analysis, face detection and recognition, biometrics, tracking and pose estimation of human body/head/hand/eye-gaze, surveillance and activity recognition. CVL is currently funded by NSF, NASA, ONR, and Ford Motor Company. We are also collaborating with several government and industry laboratories. For more information, please visit http://www.cs.unr.edu/CVL Requirements: You must have a Þrst degree in either an Engineering subject, in Mathematics, in Physics, or in Computer Science. Good Mathematical background, programming skills in C or C++, and familiarity with Unix/Linux/Windows are necessary. Prior familiarity with Image Processing, Computer Vision, Pattern Recognition, and Machine Learning is desirable. Good communication and writing skills in English are essential. Interested students should send their CV by regular mail, e-mail, or fax to Dr. George Bebis (bebis@cs.unr.edu) or Dr. Mircea Nicolescu (mircea@cs.unr.edu) Dr. George Bebis Department of Computer Science & Engineering University of Nevada Reno, NV 89557, USA phone: (775) 784-6463 email: bebis@cs.unr.edu http://www.cs.unr.edu/~bebis Dr. Mircea Nicolescu Department of Computer Science & Engineering University of Nevada Reno, NV 89557, USA phone: (775) 784-4356 email: mircea@cs.unr .edu http://www.cs.unr.edu/~mircea === Subject: MATLAB 7.0.1 Release 14 SP1, and eBooks MATLAB 7.0.1 Release 14 SP1, and eBooks MATLAB 7.0.1 Release 14 SP1 (c) MathWorks [3 CDs] MATLAB 7 Release 14 - MathWorks [2 CDs] CD NR 15 824, COMSOL_FEMLAB_V3.0A John.Wiley.Sons.Computational.Colour.Science.using.MATLAB.eBoo k Orchard.Publications.Signals.and.Systems.with.MATLAB.Applicati ons.eBook Orchard.Publications.Circuit.Analysis.II.with.MATLAB.Applicati ons.eBook 10/01/2001 WaveWarp Audio Toolbox v2.0.1 for MatLab please send e-mail, code_fu@pathÞnder.gr, astra35@freemail.gr, === Subject: problem with using ATLAS(cblas_cgerc) aa = x * y^T I want to do the above multiplication. The code is listed below. But the result is not what I think. Could you help me? #include #include extern C{ #include } using namespace std; int main(int argc, char *argv[]) { complex *xx = new complex[10]; complex *yy = new complex[10]; complex (*aa)[10] = new complex[10][10]; for(int i = 0; i < 10; i ++) xx[i] = i; for(int i = 0; i < 10; i ++) yy[i] = i; complex alpha(1,0); cblas_cgerc(CblasRowMajor, 10, 10, &alpha, xx, 1, yy, 1, aa, 10); for(int i = 0; i < 10; i ++){ for(int j = 0; j < 10; j ++) std::cout << aa[i][j] << ; std::cout << std::endl; } } result: (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (19392,0) (34816,0) (63616,0) (119296,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (34816,0) (65536,0) (126976,0) (188416,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (63616,0) (126976,0) (192256,0) (257536,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (119296,0) (188416,0) (257536,0) (391168,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) (0,0) === Subject: Typo: g should be h in the system of equations === Subject: Re: hierarchical cluster analysis Did you look here? http://www.clustan.com/hierarchical_cluster_analysis.html http://rana.lbl.gov/EisenSoftware.htm http://www.cs.umd.edu/hcil/hce/ >My name is An. Please help me to solve somethings that I need! >I have a thesis subject about Hierarchical cluster analysis. Could >you show me about informations for this? EX: Book refference, or you >can teach me for this subject. >I try to search internet but I cann¹t collect any results. Please help >me ! I need more informations for my thesis. Please answer to me as >soon as possible! >An === Subject: Re: Þve-point Þnite difference > I am to assume f Œ(x0) = A*f(x0 - h) + B*f(x0 + h) + C*f(x0 + 2h) + > D*f(x0+3h) + O(h^4) > Find the coefÞcients AND explain why f(x0) is not necessary. > This is what I did Þrst: First I augmented this to include x0: > f Œ(x0) = A*f(x0 - h) + B*f(x0 + h) + C*f(x0 + 2h) + D*f(x0+3h) + E*f(x0) A possible answer: Consider that f:[a,b]-->R , x0 in (a,b), f¹(x) exists. A numerical differentiation formula is of form f¹(x0)= (1/h)*SUM_{k=1 to k=n}a_k*f(x0+h*b_k) + R[f] , n>= 2 , where R[f] is the remainder , h=/=0 , and b_1,...,b_n are such that a< x0+h*b_k < b , for k=1,2,...,n . Let us try to Þnd the coefÞcients and the ,,knots x0+h*b_k such that (*) R[h]= 0 for all polynomials of degree m , m being maximum possible. It¹s not difÞcult to see that m= n=number of knots. However, in this case we must have (1) P(b) := b_1*b_2*....*b_n=/=0 and at the same time (2) 1/b_1 +1/b_2 + ... + 1/b_n = 0 . Let us remark that (1) implies that x0+hb_k =/= x0. According to ,,optimality criterion (*) the set of all ,,optimal formulas are obtained as follows: i) Select the distinct numbers b_1,b_2,...,b_n such that (1)-(2) are veriÞed and moreover a< x0+h*b_k < b , k=1,2,...,n . ii) DeÞne a_k in the following manner : (3) a_k^{*} := (-1)^{n-1}P(b)/( w¹(b_k)*b_k^2 ) where w(x)=(x-b_1)(x-b_2)...(x-b_n). ii) Then f¹(x0)= (1/h)*SUM_{k=1 to k=n}a_k^{*} f(x0+h*b_k) +R[f] has the ,,algebraic degree of exactness m=n -(maximum possible). Moreover, when f is in C^{n+1}[a,b] then there exists a poinnt c in [a,b] such that R[f] = (-1)^nh^nP(b) f^{(n+1)}(c)/((n+1)!) that is R[f]= big O(h^n) on the space C^{n+1}[a,b]. For instance , Salzer¹s formula [see H.E.Salzer ,, Optimal points for numerical differentiation, Numerische Mathematik 2(1960) 214-227] f¹(x0)=approx= (4*f(x0+h)-3*f(x0)-f(x0 +2h))/(2*h) is not optimal/with respect to (*)/ (because one of b_1,b_2,b_3 is = 0). On the other hand f¹(x0)=approx= (32*f(x0+3h)-27*f(x0-2h)-5f(x0+6h))/(120*h) is optimal, and the remainder term satisÞes R[f]=big O(h^3) on the space C^{(4)}[a,b] . If you need some details,proofs, please inform me . Perhaps help, Alex === Subject: Re: Þve-point Þnite difference In message <5%igd.85082$kz3.66551@fed1read02>, tsmith >I am to assume f Œ(x0) = A*f(x0 - h) + B*f(x0 + h) + C*f(x0 + 2h) + >D*f(x0+3h) + O(h^4) >Find the coefÞcients AND explain why f(x0) is not necessary. S/+3/-2/ Should make the problem a bit more tractable. -- Martin Brown === Subject: Spline Þt through noisy data? I have values f(x), g(x) etc. that come out of a Monte Carlo algorithm. At every x I get 4 values, with known statistical errors (variance). The problem is that the useful information to be extracted comes from dividing these values (or taking difference of logs). info = f(x).g(x) / h(x).k(x) This gets awkward as f(x), g(x) etc. approach zero, because the relative errors will be large. One way to improve this is just to run the Monte Carlo integration longer, but errors will only decrease as 1 / sqrt(running time), which is not very nice. Now since I¹m running this for many values of x anyway, my idea is to improve the noise behavior by Þtting a curve for each function. It can be assumed that the functions are smooth with respect to x (i.e. many times differentiable etc.), and their derivatives are small. But their form is not known. It¹s not a linear Þt, or parabolic Þt or such. Therefore some sort of least-squares interpolation algorithm will be needed. Questions: - does this approach make sense? Will the errors be reduced? - if so, which algorithm should I take? Cubic spline? - is there quantitative info on the residual errors? It won¹t work if I need thousands of data points (between 20 and 100 should do) - is it feasible to program such an algorithm myself, or is there plain C source code available? I don¹t want to Þt a curve Œby hand¹ every time I run a Monte Carlo program. Arthur === Subject: Re: Spline Þt through noisy data? > Assuming you cannot calculate > info(x) = f(x).g(x) / h(x).k(x) > directly in your simulation, your approach seems plausible. But if you have > unbiased but noisy estimates for f, g, h, and k, you will not get an unbiased > estimate for info by simply using the above formula, since info is a NONLINEAR > function of your sub-functions. If f, g, h, and k are normal variables with > known means and covariance matrix, it should be possible to derive or simulate > an unbiased estimate for info. To simplify matters, maybe you can assume > the covariance matrix is diagonal. I think this is an important point. My textbook insists that I should Þrst calculate the means of f, g, h, k and then insert this into Œinfo(x)¹. This should be the best estimate for Œinfo¹. This opposed to calculating Œinfo¹ for every individual measurement f, g, h, k and then taking the average, which seems more intuitive to me. The error analysis is then done using jackknife or bootstrap analysis. As for the covariance matrix, I don¹t think I know about correlations between f, g, h, k. But as said, I have available all individual measurements of the functions. I¹ll have to think about it some more... maybe I should be worried about bias instead of noise. Arthur > There is spline code in C at http://www.nar-associates.com/nurbs/c_code.html I¹ll take a look. News==---- > http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups > ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: Spline Þt through noisy data? > Assuming you cannot calculate > info(x) = f(x).g(x) / h(x).k(x) > directly in your simulation, your approach seems plausible. But if you > have > unbiased but noisy estimates for f, g, h, and k, you will not get an > unbiased > estimate for info by simply using the above formula, since info is a > NONLINEAR > function of your sub-functions. If f, g, h, and k are normal variables > with > known means and covariance matrix, it should be possible to derive or > simulate > an unbiased estimate for info. To simplify matters, maybe you can assume > the covariance matrix is diagonal. > I think this is an important point. My textbook insists that I should Þrst > calculate the means of f, g, h, k and then insert this into Œinfo(x)¹. This > should be the best estimate for Œinfo¹. This opposed to calculating Œinfo¹ > for every individual measurement f, g, h, k and then taking the average, > which seems more intuitive to me. The error analysis is then done using > jackknife or bootstrap analysis. On second thought, I see that the textbook is right, as in the statistical limit you have averaged out the noise in f, g, h, k and the estimate for info(x) is unbiased. The Œnaive¹ way would introduce bias as a result of the noise. > As for the covariance matrix, I don¹t think I know about correlations > between f, g, h, k. But as said, I have available all individual > measurements of the functions. There is no correlation whatsoever with respect to x (simulation is restarted every time x is changed). Not within f(x), not between f(x), g(x) etc. So that¹ll be no problem. However I think that individual measurements of the 4 functions will give correlations among them. Not sure how to handle that, yet. - Arthur > I¹ll have to think about it some more... maybe I should be worried about > bias instead of noise. > Arthur > There is spline code in C at > http://www.nar-associates.com/nurbs/c_code.html > I¹ll take a look. > News==---- > http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 > Newsgroups > ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption > =--- === Subject: Re: Spline Þt through noisy data? > I have values f(x), g(x) etc. that come out of a Monte Carlo > algorithm. At every x I get 4 values, with known statistical > errors (variance). The problem is that the useful information to > be extracted comes from dividing these values (or taking > difference of logs). > info = f(x).g(x) / h(x).k(x) > This gets awkward as f(x), g(x) etc. approach zero, because the > relative errors will be large. One way to improve this is just to > run the Monte Carlo integration longer, but errors will only > decrease as 1 / sqrt(running time), which is not very nice. > Now since I¹m running this for many values of x anyway, my idea is > to improve the noise behavior by Þtting a curve for each > function. It can be assumed that the functions are smooth with > respect to x (i.e. many times differentiable etc.), and their > derivatives are small. But their form is not known. It¹s not a > linear Þt, or parabolic Þt or such. Therefore some sort of > least-squares interpolation algorithm will be needed. > Questions: > - does this approach make sense? Will the errors be reduced? > - if so, which algorithm should I take? Cubic spline? > - is there quantitative info on the residual errors? It won¹t work > if I need thousands of data points (between 20 and 100 should do) > - is it feasible to program such an algorithm myself, or is there > plain C source code available? I don¹t want to Þt a curve Œby > hand¹ every time I run a Monte Carlo program. > Arthur A least squares spline algorithm is given by Reinsch, Numerische Math. 10, 177 (1967), ib. 16, 451 (1971). -- ... To reply by email, make double u single in address ... === Subject: Re: Spline Þt through noisy data? posting-account=9V5Z1wwAAAALwTow5mg6lxY6M_VlBtOE i think this sounds like a great use for the acspline option of gnuplot. a regular cspline (cubic spline) will amplify any noise because it¹s forced to go through all the points. the acspline will weigh the points by their error and come close to the points, but it doesn¹t have to go through them. unfortunately the documentation on gnuplot is not good enough to describe exactly how acspline works or how it¹s implemented, but since it¹s open source i bet you could cut out what you wanted. maybe post this quesiton in the gnuplot newsgroup: comp.graphics.apps.gnuplot. someone may be able to help you out. === Subject: Got a website? Join the new Numerical Methods Webring. If you have a website on the subject of Numerical Methods, Analysis and Software and would like to increase visitors then one way of doing this is to join the Webring. It is completelNumerical Methods, Analysis and Software free. If you are interested then please follow the link http://www.science-books.net/webrings/nummeth.htm and follow the instructions. === Subject: **theory-edge** mailing list the **theory-edge** mailing list tracks technical advances & breakthroughs, particularly those relating to the interplay between - software - hardware - cyberspace - culture - theory sign up and/or browse messages here http://groups.yahoo.com/group/theory-edge/ - 6 years old! - 1K subscribers! - 10K messages! - fully searchable archives! - RSS feed! this link returns latest messages, most recent at bottom: http://groups.yahoo.com/group/theory-edge/messages/?expand=1 my theory-edge posts are written blog-style since its inception in 1998 by me, VZN, founder & moderator. we especially like to hear from sharp-minded researchers, academics, and industry professionals. we frequently scoop the mainstream media in spotting & discussing signiÞcant new technology trends. a large autosorted FAQ-like link collection built over years of web surÞng can be browsed here http://vznuri.orgspace.com/theory-edge/ - best links from all over the web! - collected titles, dates, and summaries! - 224 links, 24 categories! - live, autosorted pages via PHP/mysql code! - full of cutting edge categories & topics! listed below software algorithmics digital convergence, fractals, mosaic/netscape browser, java, extreme programming, digital physics simulation stock market analysis, social engineering, physics simulation, formula 1, transportation, betting, trafÞc engineering, molecular dynamics special fx movie based effects, synthespians, animation open source torvalds, linux, raymond, halloween docs bioinformatics biology + CS, protein folding problem, $100M IBM blue gene project hardware robotics mindstorms, neuromorphic computing, robot soccer, sony dreambot & qrio, bipeds, tilden, toys supercpus supercomputers, clusters, distributed computation, reconÞgurable computing qm computing quantum computing, photonics, spintronics, qm gates, molecules, theory, sites wartech black budget, ebombs, war robotics, military industrial porn complex nanotech nanotransistors, feynman, drexler, smalley, MEMS/microsystems cyberspace websearch infofreedom cyber independence, cyberjournals, open science, MIT open courseware, arXiv online preprint server, darknets, digital rights management, spam blogging denton, academic blogging, blogging vs newspapers, blogging as journalism, future of blogging cybercash economic warfare, gold, future of money, govt surveillance & control, wealth distribution, econophysics cybersecurity NSA, CIA, RSA, echelon, al-Qaida, cryptography, steganography, information warfare, cyberterrorism, homeland security, microsoft monoculture culture proÞles feynman, frankel, hillis, brockman¹s third culture & digerati, nelson, mandelbrot, conway, ramanujan, mathematicians, cranks, mead, nash, erdos gaming video games, game & movie convergence, game economies, nintendo, gaming AI, gaming math/physics simulation hacking hacker crackdown by sterling, hacker hall of fame, psychology of hacking, viruses egovt evoting, edemocracy, emergent democracy cypherpunk encryption, zimmerman, NSA, key escrow, denning, cypherpunks, sealand, DES crack singularity AI consequences, vinge, cyborg vision, joy¹s dystopian visions, kelly theory cs theory chaitin, chayes, theoretics CS, microsoft r&d, barabasi, comp.theory FAQ, satisÞability problem, phase transitions in CS, minesweeper math conquest P-time primality, $7M claymath awards, robbins conjecture automated thm proving, wiles-fermat thm, riemann conjecture, eternity puzzle math misc strogatz, godel, famous & challenge problems, math societies, math sites & magazines AI behavior analysis, loebner prize, cognitive science, evolutionary computing, webmind company, human vs computer chess, church-turing thesis, asimov laws graph theory barabasi & watts, small world graphs, hayes, physics of the web digital physics wolfram, fredkin, conway life game, gliders, rule 110, java life applets etc elsewhere other related sites, locations, egroups, mailing lists === Subject: eigenvalues X-RFC2646: Format=Flowed; Original hi, i¹m a bit rusty on this sort of stuff, but is there a theorem that allows you to tell just by looking at a square matrix, how many real and complex eigenvalues it will have? === Subject: Re: eigenvalues >hi, >i¹m a bit rusty on this sort of stuff, but is there a theorem that allows >you to tell just by looking at a square matrix, how many real and complex >eigenvalues it will have? The question about eigenvalues will have to be at least as hard as the corresponding question about roots of polynomials (every polynomial corresponds to a sparse matrix with roots of one becoming eigenvalues of the other). But even for polynomials the question is not easy; the best answer there is probably the use of Sturm sequences, which is probably not all that much easier than just Þnding the roots. But this analogy does suggest a question to which I do not know the answer: is there a technique for computing the number of real eigenvalues of a real matrix which, when applied to the companion matrix of a polynomial, reduces to the test using Sturm sequences? dave === Subject: Re: eigenvalues > hi, > i¹m a bit rusty on this sort of stuff, but is there a theorem that allows > you to tell just by looking at a square matrix, how many real and complex > eigenvalues it will have? The few I know are more a case of knowing the symmetry or form of the matrix. For example, if all components are real and the matrix is symmetric, then the eigenvalues are all real. If the matrix can be written as a product of another matrix and its transpose ( A=X¹X) then the eigenvalues are real and positive. You can generalize these to matrices with complex components. If the complex matrix is Hermitian (i.e equal to its transpose-complex conjugate), the eigenvalues are all real. There are more cases when the matrix is Normal or decomposable in certain ways. Check out books on linear algebra. -- Lou Pecora (my views are my own) === Subject: question about a norm on a trace space Good Morning, I have a domain of border Gamma divided on 2 parts dijoined : Gamma_D & Gamma_N. I note by H^{1/2}(Gamma) the space of the trace on Gamma, and i pose V = the space of u deÞned on Gamma_N whose extension on Gamma is such that its value on Gamm_D it¹s null. So i want to deÞne a norm on V . I will note norm_Gamma the norm of H^{1/2}(Gamma) deÞned as usual. Is this norm can be also a norm on V ? Zimar. === Subject: Thought problem X-NewsReader: GRn 3.2n February 9, 1999 [This problem/question arises from an FM radio receiving situation for those who care.] Given a (digitized) signal C which is a composite of signal A (weak), signal B (strong), and multiple occurrences of signal B¹ with various amplitudes and phases (echoes), is it mathematically possible to determine coefÞcients modelling the situation such that signal A can be extracted from the composite signal? It is known that this can be done with multiple devices, one for each echo/reþection path. The question is speculation as to whether a single device can do the job. === Subject: Solving linear equations, the parallel way I¹m not sure whether I post this in the right group but I guess I¹ll Þnd out. :-) Either way, here¹s my (somewhat short) story. I¹m working on a software application that connects to a server and then helps solving a set of linear equations which needs to be solved, this set is always a n*n matrix which can be solved. You could call this system some kind of distributed system, or a cluster. At the moment I¹ve designed it in a way so that each node helps calculating the inverse of the matrix A (thus A^-1) so the solution AX=B can be obtained by performing A^-1*B=X. However, as I mentioned, this project is still in design fase and I¹m afraid the current setup will use a lot of bandwith when we start talking about matrices with n>10^5 (containing 10^10 32bits elements results in a size of over 40 billion bytes). Let alone speaking about the possibility of calculating the determinant of any matrix with size (n-1)*(n-1). So my question to you is: What is the best (performance & bandwith) way to solve a set of linear equations on a distributed system? Any help on this would be greatly appreciated. :) ps: This application is being designed for a community having over 10.000 active members, so Þnding enough nodes won¹t be an issue in this case. Also there isn¹t any money involved in the development/distribution of this program; it will be distributed without charge and people can actually use the cluster for their own good. === Subject: Re: Solving linear equations, the parallel way > So my question to you is: What is the best (performance & bandwith) > way to solve a set of linear equations on a distributed system? Any > help on this would be greatly appreciated. :) Take a look at ScaLAPACK: http://www.netlib.org/scalapack/scalapack_home.html it requires either PVM or MPI to work if I recall correctly. You would have to really know what you are doing to get better performance.. and even then it would take a lot of time and effort. Also, if you are familiar with LAPACK then the calling structure is similar. Jason === Subject: Re: quadrature over a triangle > Given the coordinates of the 3 vertices (in 3D space) of a uniformly > charged linear triangular element how does one compute the potential > at a point not on the triangle. I¹m looking for an algorithm or C code > which holds even when the point is arbitrarily close to the surface ? > mark I think this is a linear problem, i.e. the potential at a point is the sum of the potentials of the inÞnitessimal surface elements that make up the triangular surface. Is this correct? If so, you just need a way of dividing the surface into a large number of small pieces. There are many ways to do this, and it shouldn¹t require much cleverness to Þnd a suitable method (unless, perhaps, you want to do the computation very fast.) Re-reading your question, I see that you want to be able to get arbitrarily close to the surface. I¹m not sure what the implications of this requirement are. Presumably to maintain accuracy you¹ll need to use a Þner subdivision of the surface for the region that is close to the speciÞed point. Maybe the linear size of a small piece should not exceed a speciÞed small fraction of the distance of the point from the surface. I¹d drop a normal from the point P to the surface, hitting the surface at the point C, then draw circles about C with radii chosen so that the distance between adjacent circles satisÞed the limit mentioned above. In other words, the width of the rings thus drawn would increase with distance from C. Then it would be a simple matter of adding the potential contribution from each ring (or partial ring). There may be a quasi-analytical method, based on Þrst determining the function f(r) that gives the fraction of the circle of radius r that falls within the triangle. If P(s,r)dr is the potential created by a complete ring of radius r, width dr, at a point s from the plane of the ring and on its axis, the result you want is Integral { P(s,r)f(r)dr } Gib === Subject: Re: Why doesn¹t MuPAD do...? > tell me why!!!! > function=-abs(x-2)/(x+1)^2-1/(x+1); > solve(function=0,x); > MuPAD doesn¹t give me the output....i¹m still waiting ... I don¹t have any idea why it keeps trying, but I think your expression is equivalent to f:=-abs(x-2)/(x+1) solve(f=1, x). When entered this way, after about 2.5 seconds, mupad gives an incomprehensible (to me) answer. The function goes to +/- inÞnity around x = -1, and there doesn¹t seem to be a solution for f(x) = 1. It approaches 1 asymptotically x becomes more negative. -- john === Subject: Re: Why doesn¹t MuPAD do...? >> tell me why!!!! >> function=-abs(x-2)/(x+1)^2-1/(x+1); >> solve(function=0,x); >> MuPAD doesn¹t give me the output....i¹m still waiting ... > I don¹t have any idea why it keeps trying, but I think your expression > is equivalent to > f:=-abs(x-2)/(x+1) > solve(f=1, x). > When entered this way, after about 2.5 seconds, mupad gives an > incomprehensible (to me) answer. both of you don¹t tell me which version you use, and I haven¹t tried out all of them. With MuPAD31, you get the empty set for both inputs. -- Stefan Wehmeier stefanw@math.uni-paderborn.de === Subject: Re: Why doesn¹t MuPAD do...? >tell me why!!!! >function=-abs(x-2)/(x+1)^2-1/(x+1); >solve(function=0,x); >MuPAD doesn¹t give me the output....i¹m still waiting ... >>I don¹t have any idea why it keeps trying, but I think your expression >>is equivalent to >>f:=-abs(x-2)/(x+1) >>solve(f=1, x). >>When entered this way, after about 2.5 seconds, mupad gives an >>incomprehensible (to me) answer. > both of you don¹t tell me which version you use, and I haven¹t tried out all > of them. With MuPAD31, you get the empty set for both inputs. I just installed 3.1, and it does give empty set for the o.p.¹s formula, but for the equivalent that I posted, it acts peculiarly. The Þrst attempt to solve produces another glob of incomprehensible solution. Then, repeating the attempt, it gives the null set. Opening another notebook, that sequence can be repeated. Looks as if there¹s still a bug. -- john === Subject: Re: Why doesn¹t MuPAD do...? >f:=-abs(x-2)/(x+1) >solve(f=1, x). >When entered this way, after about 2.5 seconds, mupad gives an >incomprehensible (to me) answer. >> both of you don¹t tell me which version you use, and I haven¹t tried out >> all of them. With MuPAD31, you get the empty set for both inputs. > I just installed 3.1, and it does give empty set for the o.p.¹s formula, > but for the equivalent that I posted, it acts peculiarly. The Þrst > attempt to solve produces another glob of incomprehensible solution. > Then, repeating the attempt, it gives the null set. Opening another > notebook, that sequence can be repeated. Looks as if there¹s still a bug. I just tried it out again. Due to problems with the internal sorting of variables, the (correct) result can not be simpliÞed to the empty set under some operating systems (Windows) while under others (Linux) everything works well. We will have to investigate this ... -- Stefan Wehmeier stefanw@math.uni-paderborn.de === Subject: Re: Why doesn¹t MuPAD do...? >tell me why!!!! >function=-abs(x-2)/(x+1)^2-1/(x+1); >solve(function=0,x); >MuPAD doesn¹t give me the output....i¹m still waiting ... >>I don¹t have any idea why it keeps trying, but I think your expression >>is equivalent to >>f:=-abs(x-2)/(x+1) >>solve(f=1, x). >>When entered this way, after about 2.5 seconds, mupad gives an >>incomprehensible (to me) answer. > both of you don¹t tell me which version you use, and I haven¹t tried out all > of them. With MuPAD31, you get the empty set for both inputs. 3.0. Looks like you must have Þxed it. -- john === Subject: MuPAD 3.1 Documentation in PDF format? The 3.1 version of the commercial symbolic math program MuPAD is now available (see www.mupad.com). Might anyone know when updated PDF documentation will also be available? Currently, downloadable documentation is available for versions 3.0 and 2.5 only (see http://www.mupad.com/index.php?menu=4&ID=66717). MuPad 3.1 does come with updated help Þles, in a proprietary format, but I¹d like to be able to produce printouts (nothing beats paper for storing information). Cordially, Richard Kanarek === Subject: Lateral Area of an Elliptical Cone by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iA4Eh3w06137; I would like to calculate the lateral area of right centered (not oblique) cone that has an elliptical base. Is there an analytical solution or must it be approximated? If so how? === Subject: Re: Lateral Area of an Elliptical Cone >I would like to calculate the lateral area of right centered (not >oblique) cone that has an elliptical base. Is there an analytical >solution or must it be approximated? If so how? Suppose your cone has the equation (z/h)^2 = (x/a)^2 + (y/b)^2 for 0 < z < h. Parametrize it as x = (az/h) cos(t), y = (bz/h) sin(t). In Maple 9.5: > with(VectorCalculus): A:= SurfaceInt(1, [x,y,z]=Surface(, z=0..h, t=0..2*Pi), inert); A:= simplify(A) assuming a>0,b>0,h>0; V:= simplify(value(A)) assuming a>0,b>0,h>0; / (1/2) (1/2) | / 2 2 | / 2 2 |h -b + a / | V := 2 h + b / EllipticE|-----------------| a | (1/2) | | / 2 2 | a h + b / / Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Symbolic eigenvectors > I understand that Mathematica calculates the eigenvectors of a matrix > containing symbolic elements using polynomial interpolation. Could someone > kindly point me to the algorithms used for this sort of thing? > Jerry. Actually for matrices of exact numbers or containing symbolic entities, Mathematica will do some form of pedestrian computation of the roots of the characteristic polynomial (that is, Det[matrix -lambda*identitymatrix]) in order to Þnd eigenvalues. For each of these a null space is computed to Þnd corresponding eigenvectors. When the matrix consists entirely of exact numbers e.g. rationals or Gaussians, Lagrange interpolation will be used to construct that characteristic polynomial To my way of thinking this is really an implementation detail (for efÞciency) rather than something intrinsic to the algorithm per se. Daniel Lichtblau Wolfram Research === Subject: GCD of multivariate polynomials by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iA5DjL902306; I wish to know if there is a way to compute the GCD of 2 multivariate polynomials. For eg, the gcd of abcd + abe+c^2d+ec+ecd+e^2 and abgh+abk+abl+cgh+ck+cl is ab+c+e. I constructed this by doing (ab+c+e)(cd+e) and (ab+c+e)(gh+k). Gagan === Subject: Re: GCD of multivariate polynomials > I wish to know if there is a way to compute the GCD of 2 multivariate > polynomials. > For eg, the gcd of abcd + abe+c^2d+ec+ecd+e^2 and > abgh+abk+abl+cgh+ck+cl is ab+c+e. > I constructed this by doing (ab+c+e)(cd+e) and (ab+c+e)(gh+k). (I think you meant abgh+abk+cgh+egh+ck+ek) Yes, it¹s possible. As with (plain old) integers, factoring is -not- the most efÞcient method (in general) for computing the gcd. What does work is a more generalized Euclidean gcd algorithm (divide the larger by the smaller, repeat with the remainder, stop when you get to 0, the remainder before you get 0 is the gcd). But larger and smaller are not so obvious here you have to order the variables (and so then also the terms). And there are other complications to deal with. For the details see any book on Groebner bases and Buchberger¹s algorithm (Cox, Little, O¹Shea: Ideals, Varieties, and Algorithms is nice). -- Mitch Harris (remove q to reply) === Subject: Re: GCD of multivariate polynomials >> I wish to know if there is a way to compute the GCD of 2 multivariate >> polynomials. 1. Yes, there is a way. Now for the questions you did NOT ask.. 2. Just about any computer algebra system has such a capability, which is necessary to simplify ratios of polynomials to simplest form. 3. There are public open-source programs with this capability. 4. It is a well-studied problem, and you can Þnd simple programs or far more elaborate programs that are efÞcient for large cases. 5. The names Reduced, Subresultant, Modular, EZGCD, Heuristic, Sparse, will be found if you Google for the subject. 6. If you are writing such a program, I recommend you NOT start with an exponential time procedure (Buchberger¹s algorithm) and hope that the special case of GCD will come out faster. You can write such a program more directly. Assuming you already know how to represent polynomials in several variables over the integers, and can add, multiply, and compute remainders, a decent GCD program, say subresultant, can Þt on half a page. A useful treatment can be found in Knuth¹s Art of Computer programming Volume 2. I think that the Œdetails¹ you Þnd in Cox etal, and books on Groebner bases will not address the computational task, but only the mathematical implications. If that is what you want, Þne. RJF >> For eg, the gcd of abcd + abe+c^2d+ec+ecd+e^2 and >> abgh+abk+abl+cgh+ck+cl is ab+c+e. >> I constructed this by doing (ab+c+e)(cd+e) and (ab+c+e)(gh+k). > (I think you meant abgh+abk+cgh+egh+ck+ek) > Yes, it¹s possible. As with (plain old) integers, factoring is -not- > the most efÞcient method (in general) for computing the gcd. > What does work is a more generalized Euclidean gcd algorithm (divide > the larger by the smaller, repeat with the remainder, stop when > you get to 0, the remainder before you get 0 is the gcd). > But larger and smaller are not so obvious here you have to order the > variables (and so then also the terms). And there are other > complications to deal with. For the details see any book > on Groebner bases and Buchberger¹s algorithm (Cox, Little, O¹Shea: > Ideals, Varieties, and Algorithms is nice). === Subject: Maple 8 | subs and matrices I think I asked something similar some time ago, but I¹ve misplaced the answer. I¹m using the linalg package in Maple, to do various calculations with one or more matrices. Suppose, for example, I have a matrix with symbolic structure [[a,b],[c,d]] What I want to be able to do is substitute in values for the variables a -> d, using the subs command. But, I don¹t seem to be able to Þgure out how one does this. If I try mat:=array([[a,b],[c,d]]); subs(a=0.5,b=0.2,c=1,d=0,mat); Nothing happens. But, if I use mat:=Matrix([[a,b],[c,d]]); subs(a=0.5,b=0.2,c=1,d=0,mat); Things work Þne. OK - there is a difference between array and Matrix. Fine - IF I explicitly deÞne a matrix using Matrix. But, what if I don¹t, or can¹t? For example, suppose I derive a Jacobian matrix, using (say) jac:=jacobian(a,[n[1],n[2]]); where a is the vector of functions, and n[1] and n[2] are the variables I want to differentiat the functions in a with respect to. Maple correctly gives me the Jacobian matrix, but I then want/need to be able to subs in the values of various parameters. Yes, I could subs them into the functions in a Þrst, then differentiate, but this is not always practical, or desirable (for some purposes). So, is there a way to subs things into matrices that are NOT explicitly created using the Matrix command? If not, then this is (IMO) a real limitation to Maple. === Subject: Re: Maple 8 | subs and matrices > I¹m using the linalg package in Maple, to do various calculations with > one or more matrices. Suppose, for example, I have a matrix with > symbolic structure > [[a,b],[c,d]] The linalg package is deprecated; you really should switch to the LinearAlgebra package. > What I want to be able to do is substitute in values for the variables > a -> d, using the subs command. But, I don¹t seem to be able to Þgure > out how one does this. > If I try > mat:=array([[a,b],[c,d]]); > subs(a=0.5,b=0.2,c=1,d=0,mat); Here are two ways to accomplish this map2(subs,{a=0.5,b=0.2,c=1,d=0},mat); subs(a=0.5,b=0.2,c=1,d=0, eval(mat)); > But, if I use > mat:=Matrix([[a,b],[c,d]]); > subs(a=0.5,b=0.2,c=1,d=0,mat); > Things work Þne. OK - there is a difference between array and Matrix. > Fine - IF I explicitly deÞne a matrix using Matrix. But, what if I > don¹t, or can¹t? For example, suppose I derive a Jacobian matrix, > using (say) > jac:=jacobian(a,[n[1],n[2]]); Try VectorCalculus:-Jacobian to work with Matrices. Okay, you¹re now wondering (or should be 8-), how am I supposed to Þgure that out? Well, I didn¹t know where it was, either. In fact, one of the reasons I continued to use linalg for a while was precisely this; I needed to compute jacobians and couldn¹t Þnd an equivalent in the LinearAlgebra package. The solution, at least in the Classic GUI, is to use the Full Text Search in the Help menu. Or ask in appropriate forum; comp.soft-sys.math.maple is better than sci.math.symbolic because it is speciÞc to maple. > where a is the vector of functions, and n[1] and n[2] are the > variables I want to differentiat the functions in a with respect to. > Maple correctly gives me the Jacobian matrix, but I then want/need to > be able to subs in the values of various parameters. Yes, I could subs > them into the functions in a Þrst, then differentiate, but this is > not always practical, or desirable (for some purposes). Joe Riel === Subject: Re: Maple 8 | subs and matrices >The linalg package is deprecated; you really should switch to the >LinearAlgebra package. OK - sounds like sage advice to start. >> What I want to be able to do is substitute in values for the variables >> a -> d, using the subs command. But, I don¹t seem to be able to Þgure >> out how one does this. >> If I try >> mat:=array([[a,b],[c,d]]); >> subs(a=0.5,b=0.2,c=1,d=0,mat); >Here are two ways to accomplish this >map2(subs,{a=0.5,b=0.2,c=1,d=0},mat); >subs(a=0.5,b=0.2,c=1,d=0, eval(mat)); >> But, if I use >> mat:=Matrix([[a,b],[c,d]]); >> subs(a=0.5,b=0.2,c=1,d=0,mat); >> Things work Þne. OK - there is a difference between array and Matrix. >> Fine - IF I explicitly deÞne a matrix using Matrix. But, what if I >> don¹t, or can¹t? For example, suppose I derive a Jacobian matrix, >> using (say) >> jac:=jacobian(a,[n[1],n[2]]); >Try VectorCalculus:-Jacobian to work with Matrices. Okay, you¹re now >wondering (or should be 8-), how am I supposed to Þgure that out? >Well, I didn¹t know where it was, either. In fact, one of the reasons >I continued to use linalg for a while was precisely this; I needed to >compute jacobians and couldn¹t Þnd an equivalent in the LinearAlgebra >package. The solution, at least in the Classic GUI, is to use the >Full Text Search in the Help menu. Or ask in appropriate forum; >comp.soft-sys.math.maple is better than sci.math.symbolic because >it is speciÞc to maple. the same time you posted your reply. And, as for the appropriate forum - good point. At one point, there wasn¹t a Maple-speciÞc list (or, if there was, my Unviersity news server didn¹t have it available). But, on the larger topic - I wonder if there is a Œrationale¹ reason for Maple not being consistent throughout - I don¹t see an obvious reason for subs to work for some things, but not others. === Subject: Re: Maple 8 | subs and matrices > And, as for the appropriate forum - good point. At one point, there > wasn¹t a Maple-speciÞc list (or, if there was, my Unviersity news > server didn¹t have it available). I have on-and-off over the years had particular trouble accessing comp.soft-sys.math.maple form my Uniersity¹s servers, and even when the group is accessible, sometimes postings, even my own posings, take a week or so to appear. I have had no such trouble with other groups. I wonder how that happens. During those periods of trouble, I used Google to access the group. There is always a several-hour delay with Google delay than that. Google is also useful when you want searchable archives or web-based access. > But, on the larger topic - I wonder if there is a Œrationale¹ reason > for Maple not being consistent throughout - I don¹t see an obvious > reason for subs to work for some things, but not others. The reason for this particular weirdness is explained on the page ?last_name_eval. However, please give MapleSoft credit where credit is due: This inconsistency was corrected years ago by the LinearAlgebra package. However, in order to support the running of user legacy code, the old package with its idiosyncracies must remain. One supports consistency by using the newer package. === Subject: Dynamic Rotation of 3D Graphics in Mathematica? X-RFC2646: Format=Flowed; Original Hello. Does anyone know of either a procedure, notebook, or some other technique that can allow one to dynamically rotate 3D Mathematica graphics using, say, a mouse? Something similar to the behavior of MathCAD or Maple. Or is their a feature within Mathematica besides ViewPoint that I am unaware of? I am currently running Mathematica 5.0.1. Darren === Subject: Re: Dynamic Rotation of 3D Graphics in Mathematica? > Hello. > Does anyone know of either a procedure, notebook, or some other technique > that can allow one to dynamically rotate 3D Mathematica graphics using, say, > a mouse? Something similar to the behavior of MathCAD or Maple. Or is > their a feature within Mathematica besides ViewPoint that I am unaware of? > I am currently running Mathematica 5.0.1. > Darren I love http://phong.informatik.uni-leipzig.de/~kuska/mview3d.html/ -- Peter Pein Berlin === Subject: Re: Dynamic Rotation of 3D Graphics in Mathematica? X-RFC2646: Format=Flowed; Response Darren >> Hello. >> Does anyone know of either a procedure, notebook, or some other technique >> that can allow one to dynamically rotate 3D Mathematica graphics using, >> say, a mouse? Something similar to the behavior of MathCAD or Maple. Or >> is their a feature within Mathematica besides ViewPoint that I am unaware >> of? I am currently running Mathematica 5.0.1. >> Darren > I love > http://phong.informatik.uni-leipzig.de/~kuska/mview3d.html/ > -- > Peter Pein > Berlin