On 9/24/07, Ben Abbott <[EMAIL PROTECTED]> wrote: > > > Jean-François Mertens-3 wrote: > > > > Please redo ! Once is apparently not sufficient .. > > Or at least check that your equality above is not right! > > Correct coefficients with you sequence of denominators > > are -5i/54 , 5i/54 , 2/9 , 2/9 ... if I'm not mistaken. > > > > Thanks for the correction. > > > Jean-François Mertens-3 wrote: > > > > But this has no importance : > > > > The link documents quite well that it is an upstream problem, > > [...] > > > > Agreed. Just making sure its not just me. > -- > View this message in context: > http://www.nabble.com/trouble-with-Octave-tf4511817.html#a12870464 > Sent from the fink-users mailing list archive at Nabble.com. > > Unfortunately, it appears that 2.9.14 (at least on PowerPC), still doesn't give an answer that agrees with the analytic one:
octave:1> num = [1 0 1]; octave:2> den = [1 0 18 0 81]; octave:3> [a,p,k,e] = residue(num,den) a = 8.4492e+06 - 3.9658e+06i 8.4492e+06 + 3.9658e+06i -8.4492e+06 + 3.9658e+06i -8.4492e+06 - 3.9658e+06i p = 0.0000 + 3.0000i 0.0000 - 3.0000i -0.0000 + 3.0000i -0.0000 - 3.0000i k = [](0x0) e = 1 1 1 1 And for reference, 2.9.14 on my Debian/x86 box gives: octave2.9:1> num = [1 0 1]; octave2.9:2> den = [1 0 18 0 81]; octave2.9:3> [a,p,k,e] = residue(num,den) a = 7.4540e+03 - 9.9176e+06i 7.4540e+03 + 9.9176e+06i -7.4540e+03 + 9.9176e+06i -7.4540e+03 - 9.9176e+06i p = 0.0000 + 3.0000i 0.0000 - 3.0000i -0.0000 + 3.0000i -0.0000 - 3.0000i k = [](0x0) e = 1 1 1 1 So it appears upstream hasn't quite worked this out yet. -- Alexander K. Hansen akh AT finkproject DOT org Fink User Liaison and Documenter ------------------------------------------------------------------------- This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2005. http://clk.atdmt.com/MRT/go/vse0120000070mrt/direct/01/ _______________________________________________ Fink-users mailing list Fink-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/fink-users
