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Question: "expand" power in ginac
by Clobert von Roum 16 Oct '19

16 Oct '19

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Release 1.7.8
by Richard B. Kreckel 07 Oct '19

07 Oct '19

Hi, GiNaC 1.7.8 has just been released. The changes are: * Fix pseries::evalf(), broken since 1.7.0. * Fix a corner-case bug in H_evalf(). Happy hacking! -richy. -- Richard B. Kreckel <https://in.terlu.de/~kreckel/>

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Release 1.7.7
by Richard B. Kreckel 22 Sep '19

22 Sep '19

Hi, GiNaC 1.7.7 has just been released. It fixes some bugs (see NEWS) and depends on Python3 for build (this may be important for distros). Happy hacking! -richy. -- Richard B. Kreckel <https://in.terlu.de/~kreckel/>

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trigonometric identity s2+c2=1
by Dario Pellegrini 19 Jul '19

19 Jul '19

I noticed that GiNaC does not apply the simplification: cos(x)^2+sin(x)^2 = 1 for any x in R Is there any way for me to force this substitution in my expressions? That would greatly simplify them...

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fsolve system of equation
by Dario Pellegrini 19 Jul '19

19 Jul '19

Hi all. I have a bidimensional system of two nonlinear equations (containing some trigonometric functions) which I would like to solve numerically. I spent few hours searching for an example but I could not find any. Is it possible to achieve this within the GiNaC framework?

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Re: [GiNaC-list] question about lorentz_g contract with sy_anti tensor
by Feng Feng 26 May '19

26 May '19

Thanks very much for your so quick reply. Yes, my mistake, I actually mean A^{abc}_{c} = A^{abcd} * g_{dc} So it goes back to my original post, it seems GiNaC will not produce a 0 result. Thanks again! Best Regards! Feng

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Re: [GiNaC-list] question about lorentz_g contract with sy_anti tensor
by Feng Feng 26 May '19

26 May '19

Tanks a lot! Yes and Sorry, I forgot to call simplified_indexed. Still, I get some confused with the index variance. and excuse me to let me explain my options in some details. I understand g_c^d = g_{ce} g^{ed}, and let’s come back to the ret expresion: "auto ret = indexed(A, sy_anti(0,1,2,3), ia,ib,ic,id) *lorentz_g(ic, id.toggle_variance());” In my understanding, it corresponds to A^{abcd} * g^{c}_{d}, is that right? if so, then A^{abcd} * g^{c}_{d} = A^{abcc}, so according to sy_anti, it is 0, directly from GiNaC. So if I want to contract two index in A, I mean, we need this one: A^{abc}_{c} = A^{abcd} * g^{dc} ? Thanks very much! Best Regards! Feng

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Re: [GiNaC-list] question about lorentz_g contract with sy_anti tensor
by Feng Feng 25 May '19

25 May '19

Thanks very for the reply. and I have tried the suggestion, i.e., using the following expression: auto ret = indexed(A, sy_anti(0,1,2,3), ia,ib,ic,id) * lorentz_g(ic, id.toggle_variance()); but I got something like: -delta.d~c*A~d~a~b~c, still a non-zero result. BTW, if c’s variance is not toggled in lorentz_g, then these two dummy index c’s variance is not matched, and is that allowed? Thanks again!

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question about lorentz_g contract with sy_anti tensor
by Feng Feng 23 May '19

23 May '19

Hi all, I wonder is there a way or method to simplify the expression with a sy_anti tensor contracted with a lorent_g to 0? Please have a look at the following code, I expect ret is 0, and the output looks like “A~b~a.c~c”, it is actually 0 due to the sy_anti, but I wonder, is there a way or a method to simplify those expression directly to 0? Thanks very much! using namespace GiNaC; using namespace std; int main(int argc, char** argv) { symbol a("a"), b("b"), c("c"), d("d"), A("A"); symbol D("D"); varidx ia(a,D), ib(b,D), ic(c,D), id(d,D); auto ret = indexed(A, sy_anti(0,1,2,3), ia,ib,ic,id) * lorentz_g(ic.toggle_variance(), id.toggle_variance()); cout << simplify_indexed(ret) << endl; return 0; } Best regards! Feng

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Crash with zeta({7, 3}), and performance difference with G(...)
by Vitaly Magerya 22 May '19

22 May '19

Hi, folks. I've run into a crash; here it is: (gdb) run Starting program: .../ginac/build/ginsh/ginsh ginsh - GiNaC Interactive Shell (GiNaC V1.7.6) __, _______ Copyright (C) 1999-2019 Johannes Gutenberg University Mainz, (__) * | Germany. This is free software with ABSOLUTELY NO WARRANTY. ._) i N a C | You are welcome to redistribute it under certain conditions. <-------------' For details type `warranty;'. Type ?? for a list of help topics. > evalf(zeta({7,3})); 0.008419668503096332361 > Digits=2000; 2000 > evalf(zeta({7,3})); Program received signal SIGSEGV, Segmentation fault. 0x00007ffff786c3ca in cln::operator*(cln::cl_N const&, cln::cl_N const&) () from /usr/lib/libcln.so.6 (gdb) bt #0 0x00007ffff786c3ca in cln::operator*(cln::cl_N const&, cln::cl_N const&) () from /usr/lib/libcln.so.6 #1 0x00005555556402c1 in GiNaC::(anonymous namespace)::crandall_Y_loop(cln::cl_N const&, std::vector<cln::cl_N, std::allocator<cln::cl_N> > const&) () #2 0x0000555555658ab9 in GiNaC::zeta1_evalf(GiNaC::ex const&) () #3 0x00005555555f5697 in GiNaC::function::evalf() const () #4 0x0000555555586023 in f_evalf(GiNaC::container<std::vector> const&) () #5 0x000055555558d7c7 in yyparse() () #6 0x0000555555577087 in main () (gdb) Interestingly evaluating the same zeta with G works fine: > Digits=2000; 2000 > evalf(G({0,0,0,0,0,0,1,0,0,1},1)); 0.00841966850309633242396857971467065063691... Also with Digits=500, evaluating zeta({7,3}) takes ~10 seconds, while it's only 0.6 seconds for G({0,0,0,0,0,0,1,0,0,1},1). Anyone has ideas as to why is there such difference in performance, and is there an even faster way?

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