May 2007 - GiNaC-list

Simplify clifford expresion: Works?
by Javier Ros Ganuza 29 May '07

29 May '07

NOTE: I'm using one week old CVS version. I'm playing with rotations in Cl3, following with my previous mail basis definition. I'm implementing the following trivial rotation ex R=(cos(theta/2)*dirac_ONE(0)-e00*e01*sin(theta/2)); cout << "subs Rotation R * e00 * R^\\dag = " << canonicalize_clifford((R * e00 * clifford_star(R)).expand()) << endl; That wonderfully (wonderfully!!!) gives: Rotation R * e00 * R^\dag = 2*e~1*sin(1/2*theta)*cos(1/2*theta)-sin(1/2*theta)^2*e~0 +e~0*cos(1/2*theta)^2 I'm trying to trigonometrically simmplify, 2*cos(1/2*theta)*e~1*sin(1/2*theta) -----to---> sin(theta) To that end: cout << "subs Rotation R * e00 * R^\\dag = " << canonicalize_clifford((R * e00 * clifford_star(R)).expand()).subs(lst(2*sin(1/2*theta)*cos(1/2*theta)==sin(theta))) << endl; that gives subs Rotation R * e00 * R^\dag = 2*e~1*sin(1/2*theta)*cos(1/2*theta)+e~(sin(theta))*cos(1/2*theta)^2-sin(1/2*theta)^2*e~(sin(theta)) Also, if I use cout << "subs Rotation R * e00 * R^\\dag = " << canonicalize_clifford((R * e00 * clifford_star(R)).expand()).subs(lst(2*sin(1/2*theta)*cos(1/2*theta)==sin(theta)),subs_options::algebraic) << endl; I get the same result subs Rotation R * e00 * R^\dag = 2*e~1*sin(1/2*theta)*cos(1/2*theta)+e~(sin(theta))*cos(1/2*theta)^2-sin(1/2*theta)^2*e~(sin(theta)) I think this is a bug. ???? Have a nice weekend! Javier

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real and imaginary parts of an expression
by Martin DRUON 28 May '07

28 May '07

hello, I would like to know how I can get the real part and the imaginary part of an expression (GiNaC::ex)... for example : #include <iostream> #include <ginac/ginac.h> using namespace GiNaC; int main(int argc, char **argv) { symbol x("x"), y("y"); // My complex number : x + iy symbol Q("Q"), a("a"); // Parameters // My expression ex f = (Q / 2 * Pi) * log(((x + I * y) - a) / ((x + I * y) + a)); std::cout << f << std::endl; // f.real() -> How I can get real part ??? // f.imag() -> How I can get imaginary part ??? return EXIT_SUCCESS; } Is it possible easily ? Thanks Martin

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How can I get a GiNaC::ex's name
by Zhongxing Xu 25 May '07

25 May '07

Hello, Say I construct an ex by: GiNaC::ex E(GiNaC::symbol("abc")); How can I get back "abc" from E as a C++ string or char*? Now I do this by print the ex to a std::ostringstream, and get the string from the ostringstream. Thanks. -- Zhongxing Xu

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Exterior algebra tools?
by Abraham D. Smith 24 May '07

24 May '07

After finding Maple inflexible and temperamental, I am interested in using GiNaC for some computations in exterior differential systems. I've read through the documentation fairly carefully, but I haven't seen any routines for working with exterior differential forms so far. Has anyone written a library to deal with alternating forms, including things like wedge-products, exterior differentiation, Lie derivatives (via Cartan's formula), and left-hook (inner product with vectors)? Thanks, Abe Smith (Duke University)

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Stuttgart.
by unknown＠example.com 19 May '07

19 May '07

Greetings. This is to ask whether there are any GiNaC users or developers in the Stuttgart (Germany) area. Thanks! Best regards, Marko Riedel -- +-------------------------------------------------------------+ | Marko Riedel, EDV Neue Arbeit gGmbH, markoriedelde(a)yahoo.de | | http://www.geocities.com/markoriedelde/index.html | +-------------------------------------------------------------+

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Symbolic computation and derivatives
by Antoine Begault 16 May '07

16 May '07

Hello, I am currently trying the functionalities of GiNaC in order to translate some Maple code to C++ code, for a Matrix-using application. But I'm facing some difficulties for which some help could be very useful. First, I need to represent a symbolic vector and then functions of this vector. So far, I've been using 'n' distincts symbols for a vector of size 'n', which is certainly not the best way, especially for what I need to do afterwards (c.f the example below). Then, I need to compute the derivative of a function with respect to the previous vector, i.e. a gradient or a Jacobian matrix. With the "n distinct symbols" solution, I would do that by derivating n times, but it would be much more efficient if the whole gradient computation could be done in only one step: do you see any better way to do so within GiNaC? Finally, it happens that this vector is itself a function of time and I need sometimes to compute derivatives of functions of this vector with respect to time (in a more efficient way than just computing the aforementioned gradient or Jacobian and remultiplyling it by the time-derivative of the vector). I need therefore to express that a symbol (the members of the vector) is in fact a function of another symbol, with respect to which I can compute derivatives. How we do this in Maple is substitute symbols q[i] by functions q[i](t), differentiate with respect to t, then substitute back q[i](t) with q [i] and diff(q[i](t),t) with qdot[i] with the following code: > R=matrix(3,3);//+ initialization > map(x->subs({seq(q[i](t)=q[i], i=1..NDDL), > seq(diff(q[i](t), t)=qdot[i], i=1..NDDL)}, > diff(subs({seq(q[i]=q[i](t), i=1..NDDL)}, x), t)), P)): Doing something similar with GiNaC requires that we deal with user defined functions, but the "function" objects don't look like they are really made on this purpose: am I understanding correctly? We would define 'n' functions q[i](t) with derivatives defined as 'n' other functions qdot [i](t), something like that, but isn't there any simpler and more effective way? Many thanks for your time and help, A. Begault. -- View this message in context: http://www.nabble.com/Symbolic-computation-and-derivatives-tf3726576.html#a… Sent from the Ginac - General mailing list archive at Nabble.com.

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Re: [GiNaC-devel] Problem with power series
by Chris.Dams＠mi.infn.it 09 May '07

09 May '07

Dear Stefan, > can you briefly tell me what the problem is ? Look at the file pseries.cpp. See: http://www.ginac.de/viewcvs/GiNaC/ginac/pseries.cpp?rev=1.93&view=markup . The method mul::pseries contains the code fragment. int real_ldegree = 0; try { real_ldegree = buf.expand().ldegree(sym-r.rhs()); } catch (std::runtime_error) {} if (real_ldegree == 0) { int orderloop = 0; do { orderloop++; real_ldegree = buf.series(r, orderloop, options).ldegree(sym); } while (real_ldegree == orderloop); } The do ... while loop looks for the lowest degree for which the series is nonzero. This is done for very factor in a product. Hence, if there is never going to occur a term in the series that is nonzero, there is a problem. I would like to see a solution where there is besides a series method also a method that returns the leading term, that may also return a result that says that no leading term was found up to a certain order. Best wishes, Chris

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Are Clifford units of different bases equal?
by Javier Ros Ganuza 09 May '07

09 May '07

Hello all, I think that I can explain myself better with this simplistic example (THE OUTPUTS OF COUTS ARE INLINED): //gcc -o main1 main1.cc -lginac -lcln #include <ginac/ginac.h> using namespace std; using namespace GiNaC; int main(){ realsymbol a1("a1"), b1("b1"), c1("c1"); varidx nu(symbol("nu", "\\nu"), 3), mu(symbol("mu", "\\mu"), 3); ex basis1 = clifford_unit(mu, diag_matrix(lst(1, 1, 1)),0), basis2 = clifford_unit(nu, diag_matrix(lst(1, 1, 1)),1); ex vector1 = lst_to_clifford(lst(a1,b1,c1), mu, basis1), vector2 = lst_to_clifford(lst(a1,b1,c1), nu, basis2); ex result=2*vector1+vector2; cout << result << endl; //GIVES: e~nu*[[a1],[b1],[c1]].nu+2*[[a1],[b1],[c1]].mu*e~mu cout << result.expand(expand_options::expand_indexed).simplify_indexed() << endl; //GIVES: 3*e~nu*[[a1],[b1],[c1]].nu cout << expand_dummy_sum(result) << endl; //GIVES: 3*a1*e~0+3*c1*e~2+3*b1*e~1 cout << canonicalize_clifford(expand_dummy_sum(result)) << endl; //GIVES: 3*a1*e~0+3*c1*e~2+3*b1*e~1 } So I understand that basis1 and basis2 are the units of two different bases, the last parameter (0 or 1) in the call of clifford_unit(...) does not affect the output (about this parameter the tutorial says: "optional parameter rl allows to distinguish different Clifford algebras (wich will commute each other))". I recognize that I don't understand the meaning of "wich will commute each other" ). I understand then that vector1 and vector2 are vectors with the same components but with different bases. But Outputs reveal: Output 1: Clifford_unit names are the same for both bases Output 2: Clifford_units are the same for both bases Output 3: Clifford_units are the same for both bases Output 4: Clifford_units are the same for both bases Is this intended behaviour? Can it be changed so that Outputs can be something like this: e_basis2~nu*[[a1],[b1],[c1]].nu+2*[[a1],[b1],[c1]].mu*e_basis1~mu e_basis2~nu*[[a1],[b1],[c1]].nu+2*[[a1],[b1],[c1]].mu*e_basis1~mu 2*a1*e_basis2~0+2*c1*e_basis2~2+2*b1*e_basis2~1+a1*e_basis1~0 +c1*e_basis1~2+b1*e_basis1~1 2*a1*e_basis2~0+2*c1*e_basis2~2+2*b1*e_basis2~1+a1*e_basis1~0 +c1*e_basis1~2+b1*e_basis1~1 ??? Another question: In expresion vector1*vector2, Does * stands for the Inner product? How can I write a expression involving the Outer product? As Vladimir may remember, this is a rework of a example that appeared in this list a couple of years ago. I have implemented a multibody dynamics kernel based in GiNaC around clases derived form GiNaC's matrix, but I want to investigate this path also. Javier PD: I'm ussing ginac-1.3.4-1.fc4 for if it is relevant.

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Re: [GiNaC-list] question about quantum operator algebra
by Chris Dams 09 May '07

09 May '07

Dear Can-Hua, > > { ... constructing the result of rewriting is done here ... > > Sorry, but I have no idea what goes here. Could you show me somewhere > to find examples? Thank you. Your function could look like the following ex rewrite_operators(const ex &x) { if (is_a<ncmul>(x)) { if (x.has(one)) { exvector v; v.reserve(x.nops()); for (int i=0; i<x.nops(); ++i) if (x.op(i) != one) v.push_back(x.op(i)); if (v.empty()) v.push_back(one); return rewrite_operators(ncmul(v)); } for (int i=0; i<x.nops()-1; ++i) if (x.op(i) == adag && x.op(i+1) == a) { exvector v; v.reserve(x.nops()); for (int j=0; j<i; ++j) v.push_back(x.op(j)); v.push_back(one+a*adag); for (int j=i+2; j<x.nops(); ++j) v.push_back(x.op(j)); return rewrite_operators(expand(ncmul(v))); } return x; } else return expand(x.map(rewrite_operators)); } Note that it is necesary to also declare an element "one" for your algebra because one cannot add numbers and noncommutative objects. Best wishes, Chris

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Re: [GiNaC-list] question about quantum operator algebra
by Canhua 08 May '07

08 May '07

> Date: Sat, 5 May 2007 11:13:49 +0200 (CEST) > From: Chris.Dams(a)mi.infn.it > Subject: Re: [GiNaC-list] question about quantum operator algebra > To: "GiNaC discussion list" <ginac-list(a)ginac.de> > You can declare those operators as noncommutative symbols. I assume that > you are using version 1.3.x. Then it goes as e.g., > > const unsigned my_tinfo = 2; > symbol a("a", return_types::noncommutative, my_tinfo); > symbol adag("adag", return_types::noncommutative, my_tinfo); > H = epsilon*adag*a; > ex commu = a*H - H*a > commu = rewrite_operators(commu); > > The global constant my_tinfo is arbitrary, but for safety it should not be > one that is already in use (see > http://www.ginac.de/viewcvs/GiNaC/ginac/tinfos.h?hideattic=0&rev=1.27.2.3&v…) > You would have to define the function rewrite_operators yourself. It could > be something like > > ex rewrite_operators(const ex &x) > { if (is_a<ncmul>(x)) > { ... constructing the result of rewriting is done here ... Sorry, but I have no idea what goes here. Could you show me somewhere to find examples? Thank you. > return result; > } > else > return x.map(rewrite_operators); > } Best regards, Can-Hua Chen

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