Mogens Kjaer wrote: ...
Padé approximation: http://list.dprg.org/archive/2007-January/030202.html
You could try this piece of code: static double atan2x(double y, double x) { double z; if(x==0) { if(y<0) return -M_PI_2; else if(y>0) return M_PI_2; else return 0; } else { z = y/x; if(z>=-1 && z<=1) { z=z*(15.0+4.0*z*z)/(15.0+9.0*z*z); if(x>0) return z; else { if(y>=0) return M_PI+z; else return z-M_PI; } } else { z=x/y; z=M_PI_2-z*(15.0+4.0*z*z)/(15.0+9.0*z*z); if(y<=0) return z-M_PI; else return z; } } } Testing it with: maxe=0; for(i=0;i<10001;i++) for(j=0;j<10001;j++) { x=0.0001*i-0.5; y=0.0001*j-0.5; a1=atan2(y,x); a2=atan2x(y,x); e=fabs(a1-a2); if(e>maxe) maxe=e; } printf("maxe: %g\n", maxe); gives maxe: 0.0062685 Timing (Xeon 5160 3 GHz) for the loop with only one function call: atan2: 7.028u atan2x: 1.586u I'm not an expert on this, so use it at your own risk :-) Mogens -- Mogens Kjaer, Carlsberg A/S, Computer Department Gamle Carlsberg Vej 10, DK-2500 Valby, Denmark Phone: +45 33 27 53 25, Fax: +45 33 27 47 08 Email: mk@xxxxxx Homepage: http://www.crc.dk -- fedora-list mailing list fedora-list@xxxxxxxxxx To unsubscribe: https://www.redhat.com/mailman/listinfo/fedora-list