mm-295 == Subject: mupad plot of plane I'm trying to use muPad to plot planes to help me visualize mycalculus homework.The plotfunc3d seems to expect an argument that expresses z in termsof x and y.So, for this equation of a plane, 2x - 2y + z = 12, I can rearrange itwith z as a function of x and y and plot it thus:plotfunc3d(12-2*x+2*y)But, x - y = 0 is a plane, parallel to z and at 45 deg. to x and yaxes. How could I get mupad to plot that? --john ==== Subject: multiple integrals and laplacehi-does anyone know of any sites where i can enter a double or tripleintergral and get the answer and the workings-and please the same forlaplace transforms.thankyou ==== Subject: Re: General Factorization Tool> I'm posting now to make several corrections.One more correction is needed. number such that 1<=a<=n, as a factor in any ring where you have the> single factorizationg_1...g_n = G.I'll call such a ring a complete ring.> It turns out that you need an additional condition, which is that with> j and k some counting numbers less than or equal to n, if> f_j(x)/f_k(x) is in the ring for some value of x, then it must be in> the ring for all possible values of x within that ring.> It's a consistency requirement between the factors.That doesn't work in general. I've stricken that section from thebasic Tool definition and now consider it an advanced topic.Basically with the polynomial-like expression for which the ffunctions are roots, there's a condition that the modular residue withrespect to K, or a k_j, where j is a counting number where 1<=j<=n ofthe coefficients, of that polynomial-like expression must be constantwith respect to x for a perfect factorization.The definition for a complete ring remains the same. ==== Subject: Mathematica 2.2 and EllipticIntegrate PackageForgive me for asking a question about such an old version!I recently purchased a used Mathematica v2.2, and it seems to run fineon Win XP. However, I cannot get the Calculus`EllipticIntegrate`package to work. It seems to load without incident, but I am stilunable to do the classic elliptic integrals like< Risch considers differential fields composed of functions. It is essentialy: > treating functions as formulas modulo simplification laws. So (x-1)/(x^2-1): > is equal to 1/(x+1).: It also means that functions with singularities are not trea properly,: e.g. atan(x)+atan(1/x), which I mentioned before. Since you can shift : this anywhere , eg atan(x-1)+ atan(1/(x-1)) and add up as many of : these as you would like, you will still have a constant. Do you really : want to claim that the integral is given by Risch?Indefinite integral? Yes. atan(x)+atan(1/x) really decribes many differentfunctions. Usual real variable conditions make us to choose branches so that the function has jump, but from the point of view of integration the function is just given by different rule for positive and negative x.If you look at your functions as (picewise) holomorphic then singularitiesare handled properly. Just gluing pices together is separate problem.: I think you misunderstand his results. You can encode diophantine : equations into real numbers which are crea by the composition of: functions like f(x):=x*sin(x) and g(x):=sin(x). No branches are: necessary.Yes, no need for branches. However Richardson needs absolute value, and that destroyes holomorphy, so his result do not work in a differential field.: Certainly it may give you a useless formula, but it might compute : forever, or crash your computer. ^^^^^^^Do you think that it is trapped in a loop. I would think that in most cases it will eventually terminate (and we just do not have enough cycles available).: > If you want precise answer you must limit what expression you accept: > as an answer.: This would make a computer algebra system very unpopular. Mathematica: does provide some options this way, but most users don't want to have: to guess the form of the answer before using Integrate[].I was too imprecise here. You = programmer. Clearly Mathematica can not give you as an answer a formula which is not expressible in Mathematica. And, IMHO it uses as possible integrals a small subsetof expressible formulas. : One can invent a new usefull class of expressions, but: > I do not expect current computer algebra systems to do so.: Actually, they do, if they express an answer in terms of what they: consider simpler integrals that may not (yet) have names.I consider that just as a single class of expressions (already invenand implemen). For me new class of expressions either should bebigger then known classes or have some properties that make it worthwhileto distinguish from other classes. For example, hypergeometric functionsform a distinguished class inside all functions expressible as recursive power series.: On the: > other hand knowing that integral is non-elementary can help finding: > a usefull answer.: I think it is more important practically to first determine that: the integral can't be done by very simple methods. In Joel Moses (1967?) : PhD project SIN, he uses something simpler to Risch, but not: as carefully done, as part of the 3rd phase. After you've determined: that the integral cannot be done by the Risch algorithm, I'm not: sure what you would do next. Maybe heuristic simplification, or: as I've argued, look for symmetries, look it up in integral tables.Table lookup is fast so AFAIK is done before Risch. If Risch run outof resources then simplification may help. If Risch proved that integralis non-elementary simplification will not help (assuming that simplifiedproblem is still equivalent to original). One can look for well knownnon-elementary integrals, but I doubt that a program may go really farwith that. On the other hand if integral is proved to be non-elementary then it is quite reasonable to declare it a special function.Of course, one can just go trough table of known algoriths (is it elliptic?, is it hypergeometric?, ...). But I think that most non-elementary integrals are really research problems. For practicalpurpose it is probably more usefull to find properties of such integrals then give formulas. And IMHO if one wants to evaluate an integral then numerical integration seem to be much moreefficient then symbolic one.: > What matters is if the program can handle: > your question. If problem is decidable you may still run out of resources: > (or the program may give you an answer even for undecidable problem). : Of course, a particular instance of a problem may have an answer even if: the general problem is not solvable. You can easily determine if SOME: Turing machines halt. Run them for 100 steps and see if they halt. If: they halt you've solved the halting problem for them.Exactly. And if your average machine stops after 100 steps then you have very efficient method. Now consider theorem proving. Suppose that youwant to know if a theorem have a completely formal proof that can be readin a normal humal life. That give you upper bound on proof length, so theproblem is decidable. In fact, naive implementation will happily run on apersonal computer, but nobody expect it to give an answer (since the computer will fail before the program finishes).: > In the other words: there is no warranty. You may hope for the best : > (and you frequently get it), but no warranty. : If you are in the business of writing a computer program, you get: extra credit if your program always terminates. If you KNOW that: your program might go on forever, then you must build a shell around: it that makes it quit after too much time or space. This is not: very nice, since you can never be sure if it would get the answer: if you only let it go for another 0.0001 second.A programmer can not work around the laws of nature. One can try to discover hard cases quickly and give up early. But with hard cases one will also reject some easy ones -- that is practicalside of complexity theory.: > : > Coming back to integration, I do not have an example at hand. But : > the integration algoritm in some cases have to handle polynomials: > of degree proportional to absolute values of coefficients of your function.: > For transcendental functions (logarithm included) those polynomials depend: > on multiple variables. So, the algorithm is at least doubly exponential.: There are other doubly-exponential algorithms, e.g. cylindrical : algebraic decomposition. They may or may not be useful.I spent some time (not much) looking for theoretical bounds. One (quite good) bound for cylindrical algebraic decomposition made me wonder why we are able to solve a set of linear equations: when I plugged numbers corresponding to set of 3 equations on R^3 I gotan astronomical number. I belive that cylindrical algebraic decompositionis usefull, just some some problems are really hard, an even for easyproblems it is hard to decide that the problem is easy without lookingat a specific instance.: > If your function contains irrational algbraic expressions then really: > heavy machinery kicks in. Practically, if complexity is exponential or : > bigger then pretty small inputs can be too big.: Too big for what? When Macsyma was written, it ran on a machine that : 1 Megahertz. There was a partial implementation (exp/log) case of the: Risch algorithm.Modern PC is few tousend times more capable than that. If you look atmax degree of dense polynomial in 10 variables that fits into the machineyou see that bigger machine does not help much. : > : > By the way, I was thinking of doing a simple implementation of Risch: > algorithm. However, I quickly realized that I will run of memory on : > quite simple examples unless I implement a lot of tricks to make intermediate: > expressions smaller. : I don't know what you were using for a programming foundation for this,: but if you were programming it in (say) Mathematica or Maple, it would: be simpler. I was using pari.: But a complete implementation has apparently not yet been: programmed in the 24 years since Risch's original papers, including in: Mathematica or Maple, so maybe there are other barriers than running : out of memory.Time complexity also. Risch gave a short summary of his full methodin 1969. Specialits say that the method is impractial. AFAIKS Rischpropses rather simple method where some steps are basically a bruteforce search (he comptes bound on order of a divisor and then wantsto check all powers). So my impression is that even today algebraicpart of original Risch algorithm is way beyond capacity of our computers. Bronstein version need less resources, but still quite a lot. There is one additional obstacle: after long calculations onemay get huge expression which is of little practical use. In principle full transcedental part should be doable. But to demonstratethe capability one needs sufficienly complex input function. Let merefer you to the function given by original poster. This function(IMHO not very complica) contains 6 transcendental parameters and3 transcendental monomials (2 logarithms and exponential). I do not knowwhy Mathematica can not handle this function (it looked like easy caseof Risch algorithm), but in general for similar functions one have to solve system of linear equations where coefficients are rationalfunctions of 9 variables. Even if original functions are of low degreeand pretty sparse elimination procedure could blow then up to enormous size(at least that is my feeling, you seem to have more experience with suchcalculations).: > My point was that: > typical computer algebra system does not show important information,: > even though the program have it. Claim that there are no elementary: > integral gives some usefull info.: The best way I know of to do this is to find the integral, as, for : example, demonstrating that it is: erf(x), or li(x) or si(x) or J_n(x): and that it doesn't reduce by special case to something else, and: then declare that it is not elementary.: If your integrand is in a canonical form expressed in an algebraic: field, then there are alternatives.: If you are writing a program for a computer algebra system, you have: to take what you are given. If you write an integration program that: says your expression does not appear to be a composition of elementary: functions of the sort that I know about, so I refuse to try to integrate: it unless you simplify it for me this would not be very popular.I will take a lession from Microsoft: just write Out of memory. I know that a user is likely to swich from program that plainly saysthat is incapable of solving his problem to one that is equally incapable, but does not admit that. Of course the user will typicallytake some substitute solution (integrate numerically or restate problem) and fighting with second program may help in this process. Still, I think that first program is better.: > : > : (ix) integral done, but in a form so peculiar as to be useless for any : > : further work, using (say) complex exponentials and algebraics when : > : elementary functions like cos and tan would do.: > : > Risch algorithm gives minimal differential field. : I think this is wrong. You have to give IT the minimal differential : field or else it might give the wrong answer. Some extra parts may come : out of the algorithm, but you have to start it out.By Liouville minimal differential field is field genera by our functionplus some logs. Risch wants to have a tower of monomial/algebraic extensions, so probably may introduce extra lower order terms. But it will not introduceextra algebraic functions. And it seem to give a smallest number of logarithms. AFAIR if one use results from 1979 (and/or similar resultof Rosenlicht from 1976) one can eliminate useless monomials from the tower, even if original formula contains such terms. So I think that reasonable implementation of Risch algorithm gives answers which typicallycan not be just simplified. I guess that Mathematica probably do not have full support for towers of number fields, so probably is not able to fullysimplify constants, but I would expect missed simplifcations to be quiterare.: >AFAIK it is possible : > to modify algorithm to give real answer if it exists. However, in : > general the user have to transform the answer to a usefull form --: > I think that the main problem is that algorithm gives expanded form: > which may be just too big.: > : The integration program --not the user-- should be responsible for : giving the result in a useful form. After all, the integration program : should contain transformations to restore sin/cos to exp(ix)... etc. ^^^^^^^^^^^^^^^^^^^^^^^^^^I thougt that you prefer real expressions. Should integration program recognize that a polynomial is an orthogonal polynomial from a known family? Is ((x^1000-1)/(x-1))^5 simpler thenexpanded form? Usefullness depends in intended use, and that is known tothe user, but hard to comunicate to a program. When one wants to checkto to formulas are equal a canonical form is preferred. In fact, formechanical comparison fully expanded form works quite well. Other properties are seen better from factored form. Sometimes transcendentalexpression makes analytical properties clearer. I wish programs bettercapabilities for simplification, but I do not expect miracles.: I think it is clear that our perspectives are quite different. I would : like to say what is it that a programmer writing an integration program : must do. Your perspective is two-fold that of a computer science : theoretician observing its complexity, or that of a mathematician asking : about integrability in closed form. As I have said previously, I have : not ever encountered any real user who cared about integrability in : closed form.Some mathematicians do care. They just want to put a formula in a paperand will work hard to find it. Personally, I do program. And I got some feeling what I need from a machineto run given program. Beeing a research mathematician I would happily programsomething to help solve research problems. But packages like Mathematicado not solve my problems, and for programming I want both low-level controland a clear picture what a package is doing for me. -- Waldek Hebischhebisch@math.uni.wroc.pl ==== Subject: I can calculate siNUS,cosiNUS but howcan I calculate aNUS ? ==== Subject: Re: Journals and my papers>Dear newsgroup:>The new list is as follows: Journals typically assign 1-3 individuals to review each paper.If a paper is accep subject to some changes, those individuals will check your revision too. Therefore, for each paper you submit,expect to be asked to review four other papers (or 4/n papersif your submission had n authors, with n > 1).The 3-author paper had two reviewers and is being accep-- I don't know the status of the others. I owe about 4*(1 + 0.33 + 0.25) = 6.3 reviewer reports of my own in payment for the time others spend looking at my papers.fifth paper awaiting my attention. One of the five is the seconddraft I have seen; the other four are first drafts by the author(s).Two of these five are for a conference or journal where one of my recent papers was submit, one for a conference or journal where I have published in the past, and two for conferencesor journals I have never published in, subscribed to, or attended. In one of the four submit cases, my report was one of three sentto the author(s), and all three of us made the same accept/rejectrecommendation, making the editor's work easier.I haven't heard back on the other three submit cases -- twoof the deadline dates have not arrived, so other reviewers may still be busy. If all ten of your papers have only one author, budget enough timeto submit about 40 reports on other individuals before you submit again. Peter Montgomery>1-Basti.M. Solving polynomials with differential equations. Submit>for publication to Journal of Algebra.>2-Basti.M. A class Polynomial. Submit for publication to Journal of>Algebra.>3-Basti.M. A class Polynomial II. Submit for publication to Central>European Journal of Mathematics.>4-Basti.M. A class Polynomial III. Submit for publication to>Journal of Algebra.>5-Basti.M.Polyomials and Riccati differential equations. Submit for>publication to the Asian Journal of Mathematics.>6-Basti.M. New Methods Of Solving Riccati Differential>Equations.Submit for publication to IMA Journal of Numerical>Analysis.>7-Basti.M. New Methods Of Solving Riccati Differential Equations II.>Submit for publication to IMA Journal of Numerical Analysis.>8-Basti.M. On Bessel Differential Equations. Submit for publication>Journal on Scientific computing.>10-Basti.M.On Associa Legendre. Submit for publication to the>Sincerely>Dr.Mehran Basti-- John Adams served two terms as Vice President and one as President, but lostreelection. Later his son became President despite losing the popular vote.That son lost his reelection attempt badly. Now history is repeating itself.pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California ==== Subject: Re: Journals and my papersDear Prof.Mihailovs:Thanks for your advice.I hope the time is now right. It has taken me two decades to reach in this stage.SincerelyM.Basti ==== Subject: Re: Journals and my papersDear Dr. Basti,In this situation, when you are the leader in a new field of applyingdifferential equations to solving classical algebra problems and there arenot so many people who understand that, it is natural that the editors andreferents have problems with accepting your papers. You might try anotherway of publishing them - give a talk at a conference that publishproceedings and your paper will be automatically published in the conferenceproceedings. Having several such publications, it would be much easier toBest wishes,Dr. Mihailovshttp://webpages.shepherd.edu/amihailo/ ==== Subject: Re: Journals and my papersDear newsgroup:Here are sites for the quadratic and cubic equations pos in thisnewsgroup, although the irreducible part is not here.The irreducible part was generalized for all odd n in the paper on Aclass polynomial II.Prof.Mihailovs can have his comments here and act as semi externalreferee for the Journal of Algebra.http://mathforum.org/epigone/sci.math.symbolic/ zinfloujen/9kmh49wrrzfv@forum.mathforum.comandhttp:// mathforum.org/epigone/sci.math.symbolic/sigulgrix/6b03hbss3808 @forum.mathforum.comEspecially the Central European Journal of mathematics should lessonto his remarks (if any).SincerelyDr.M.Basti ==== Subject: help neededI know the notions of pull back bundle, tensor product of two bundles over aI don't know how to construct the notion of tensor product of two differentbundles over two distinct manifolds, say M_1 and M_2 starting (i supposefrom the previous one notions).i ask you kindley to Help me please. ==== Subject: Re: Differentiate many variables> I have a system consisting of 6 atoms. They all interact with one another> (bonds and van der waals forces).> I have the equations for the interactions between atoms.> Each atom is in 3D space. I can alter the x,y,z for each atom.> I will have a function that depends on these 3*6 variables. Constructing> this function should be fairly easy.> (For ease of work, I presume that I should fix one of the atoms at the> origin).> I want to minimise this function (analytically).> So, How do I differentiate this 18 variable function?> I do have a degree in comp + math but it was a long time ago, and I don't> even know what this would be called. Multivariate differentiation? Partial> differentiation?> Anyone know how to do this (or have any links)> Also, I've got some tools that I'll be using once I got the method straight> in my head: Mathematica, mathcad and matlab.> Anyone got any idea how I would use one of the packages to help me?> (preferably mathematica, as I heard that the strongest symbolically)> Thanks for any advice.> ChocaI'm not sure what you have in mind by analytically. Often this meansgive an exact result (e.g. in terms of radicals) or give a resultin terms of parameters in the equations. (It is mainly with theadvent of hybrid symbolic-numeric methods that the former is not aproper subset of the latter, by the way.)Anyway, I'll point out a few possibilities that may include yourparticular needs.First, as you note, you can (and should) reduce the degrees offreedon. For this you can locate one point at the origin, another at{x1,0,0}, and a third at {x2,y2,0}. This brings you to 12 variables.Next, if you really just want a numeric result, then you can usenumeric optimization code built into whatever program you decide upon.In Mathematica your available routines would be FindMinimum orNMinimize, the latter being generally better (though slower) in caseswhere you do not have a good idea of starting values. You would supplyyour objective function, the variables, and starting values with thislast being optional if you use NMinimize.If you have parameters present, or wish to find an exact minimizer,then you really cannot use iterative methods such as those thatunderly the functions I mentioned above. Instead you might set all 12partial derivatives (with respect to x1, x2, y2, ...) to zero. Thisgives a system of 12 equations that will be satisfied at all criticalpoints, which include local minima. You then might try to solve thissystem using an exact solver (in Mathematica this would be Solve).But... unless the equations are polynomial it is unlikely that anexact solver will know how to handle them. And if they are polynomialbut nonlinear, it is quite possible that the exact method will exhausteither memory or your patience. It is hard to know in advance when asystem of equations will do this, so you will have to try it to findout for certain.If you go the exact route, and there are no parameters, then it is nowjust a matter of testing the objective function values at each pointin the solution set to see which is a minimizer. Note that forgeneric objective functions, (I think) the reduction to 12 variablesshould suffice to give you a finite solution set, which is anothergood reason beside reduction of complexity to get rid of 6 variables.If instead you have one or more parameters present, then you will needto do further numeric work in order to see which parametrized solutionspaces correspond to minimizers. Depending on the form of solution(e.g. in terms of Mathematica Root[] functions) this may even jumpfrom one solution to another as critical point functions cross pathsin the parameter space. This can be either an ugly detail in need ofspecial handling, or a beautiful aspect of dynamics, depending on yourpoint of view.Daniel LichtblauWolfram Research ==== Subject: Re: two differential equations to solve === > .... this is the differential equation(s)> (3) (+/-) y^2 - y'^2 - 1 = 0> I am interes in what functions satisfy (3). Thanks! It yields to a standard elementary trick. Since x is missing, express it in the formdy/dx = y' = an expression involving y only.Then dx/dy = 1/(dy/dx) = another expression in y.So x = the integral of that expression with respect to y. Ken Pledger. ==== Subject: Re: two differential equations to solveHow about if you differentiate each equation wrt x, andsolve for the second derivative? The resulting equationin each case should be simple enough to solve byinspection. If you have arbitrary constants of integration,they are rela - substitute the general solution of the2nd order result in the original equation, and solvefor one constant in terms of the other.This works because the original equation isf(y,y') and does not depend explicitly on x.> Can some person post the solutions for the two differential equationslis> below (in Dsolve Mathematica format)> (1) DSolve[ y[x]^2 - y'[x]^2 - 1 == 0, y[x], x]> (2) DSolve[ -y[x]^2 - y'[x]^2 - 1 == 0, y[x], x]> Thanks! (I don't have Mathematica)> For others, this is the differential equation(s)> (3) (+/-) y^2 - y'^2 - 1 = 0> I am interes in what functions satisfy (3). Thanks! ==== Subject: Re: two differential equations to solve>Can some person post the solutions for the two differential equations lis>below (in Dsolve Mathematica format)>(1) DSolve[ y[x]^2 - y'[x]^2 - 1 == 0, y[x], x]>(2) DSolve[ -y[x]^2 - y'[x]^2 - 1 == 0, y[x], x]>Thanks! (I don't have Mathematica)I don't either, but Maple comes up with the following solutions:for (1): / 2 |exp(x) | |------- + 1| _C1 | 2 | _C1 / y(x) = 1, y(x) = -1, y(x) = 1/2 -----------------, exp(x) / 2 | _C1 | |------- + 1| exp(x) | 2 | exp(x) / y(x) = 1/2 -------------------- _C1For (2): / 1 1/2 y(x) = I, y(x) = -I, y(x) = tan(x - _C1) |- -----------------| , | 2 | tan(x - _C1) + 1/ / 1 1/2 y(x) = -tan(x - _C1) |- -----------------| | 2 | tan(x - _C1) + 1/Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== Subject: Re: two differential equations to solve>>Can some person post the solutions for the two differential equations lis>>below (in Dsolve Mathematica format)>>(1) DSolve[ y[x]^2 - y'[x]^2 - 1 == 0, y[x], x]>>(2) DSolve[ -y[x]^2 - y'[x]^2 - 1 == 0, y[x], x]>>Thanks! (I don't have Mathematica) Glancing at the first equation, I spot thesolutions y[x] = +- cosh(x + C) (C constant)As Robert's y[x] = +- 1 solutions indicate, this listmay be incomplete. The second has no real solutions,but has the complex solutions y = +- i * cos(x + C)>I don't either, but Maple comes up with the following solutions:>for (1): > / 2 > |exp(x) |> |------- + 1| _C1> | 2 |> _C1 /> y(x) = 1, y(x) = -1, y(x) = 1/2 -----------------,> exp(x)> / 2 > | _C1 |> |------- + 1| exp(x)> | 2 |> exp(x) /> y(x) = 1/2 --------------------> _C1>For (2):> / 1 1/2> y(x) = I, y(x) = -I, y(x) = tan(x - _C1) |- -----------------| ,> | 2 |> tan(x - _C1) + 1/> / 1 1/2> y(x) = -tan(x - _C1) |- -----------------|> | 2 |> tan(x - _C1) + 1/>Robert Israel israel@math.ubc.ca>Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2-- John Adams served two terms as Vice President and one as President, but lostreelection. Later his son became President despite losing the popular vote.That son lost his reelection attempt badly. Now history is repeating itself.pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California ==== Subject: Re: two differential equations to solve === > Glancing at the first equation, I spot the solutions> y[x] = +- cosh(x + C) (C constant) I checked this out, and it indeed is a solution. And thank you to allwho pos in regard to my question. ==== Subject: Re: Journals and my papersDear newsgroup:There are many advantages of having functional coefficients in thequintic.Now it is possible to associate other interesting polynomials throughinitial value problem of the constraint differential equation and alsotransformations.Initial value problems means that without solving for a4(t), we knowthere is for example a solution such that a4(1)=1, a4'(1)=1, etc.(Looks like the master key is in this concept).I am sure it will have interesting qualitative results particularlyrelative to non-solvable quintics (also other polynomials)with integercoefficients.So to speak we have in some degree the general structure of solutionsin real sense.Thus there is now a dichotomy in Abel and Galois theories (relative tothe concept of radical solutions).I hope that the Cambridge University will take this great chance andpublish the paper.Thanks.Dr.Mehran Basti ==== Subject: Re: Riccati challenge to MaximaDear newsgroup:I have found in my files radical solutions of Riccati differentialequation:diff(x(t),t)+(x(t))^2=(k*t^2+(20-2*k)*t+k )/(t^2*(-1+t)^2);where k is any real number.Maple7 has complica non-radical solutions on Root sum of apolynomial of degree 8.It is interesting if someone to report the status of Maxima andMathematica relative to the above differential equation.I checked with Mathematica 5, it has a very complica solutionpossibly on Rootsum of a quartic of logarithms (it was very small insize).These are Riccati differential equations associa with classicalequations and there is a sequence of them, one can get them one byone.I wonder how the algebra group interprets the Maple solution as to itsstatus as a non-radical solution.Clearly based on what we have learned about the definition of radical,it is not certainly radical. But then how you interpret my radicalsolution.Certainly now it is possible to assign functional quintics to theRiccati.Thanks.Dr.Mehran Basti ==== Subject: Re: mupad: magnitude of a vector by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1TEK3W04526; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1TDpIi01841 by legacy.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.10 $, legacy) id i1TDpI715574 === If you don't assume anything, t is complex and the answer given byMuPAD is right. If your t is real thenassume(t, Type::Real)norm(v,2)gives you the desired result.Pekka Sorjonen ==== Subject: Re: mupad: magnitude of a vector>If you don't assume anything, t is complex and the answer given by>MuPAD is right. If your t is real then>assume(t, Type::Real)>norm(v,2)>gives you the desired result.That did it, Pekka. Thanks a lot.--john ==== Subject: Re: mupad: magnitude of a vector>If you don't assume anything, t is complex and the answer given by>MuPAD is right. If your t is real then>assume(t, Type::Real)>norm(v,2)>gives you the desired result.I noticed your post comparing mupad with maxima, so you may be theright person to ask this question-If I make a matrixr:=matrix([t,t^2])it displays as a vector of those two terms. If I then set t=2, andcall the vector again, it still shows in terms of t, but if ask forthe term r[1,1] or r[2,1], it gives 2 or 4. Why does it only evaluateon extracting the terms?--john