Dear List,
Any assistance with the following would be appreciated:
(1) The function radcan() seems to simplify the following expression correctly:
(%i98) radcan(sqrt(a^2)/sqrt(4*%pi^2*(q0-2000)^2));
(%o98) abs(a)/(2*%pi*abs(q0-2000))
(%i99)
(2) However, for the following more complicated expression, radcan() seems to simplify the denominator incorrectly:
(%i99) radcan(sqrt((sin(2*%pi*pmin*(q0-q))-sin(2*%pi*pmax*(q0-q)))^2/(4*%pi^2*(q0-q)^2)+(cos(2*%pi*pmin*(q0-q))-cos(2*%pi*pmax*(q0-q)))^2/(4*%pi^2*(q0-q)^2)));
rat: replaced 2.0 by 2/1 = 2.0
rat: replaced 2.0 by 2/1 = 2.0
rat: replaced 2.0 by 2/1 = 2.0
rat: replaced 2.0 by 2/1 = 2.0
rat: replaced 2.0 by 2/1 = 2.0
rat: replaced 2.0 by 2/1 = 2.0
rat: replaced 2.0 by 2/1 = 2.0
rat: replaced 2.0 by 2/1 = 2.0
(%o99) sqrt(sin(2*%pi*pmin*q0-2*%pi*pmin*q)^2
-2*sin(2*%pi*pmax*q0-2*%pi*pmax*q)
*sin(2*%pi*pmin*q0-2*%pi*pmin*q)
+cos(2*%pi*pmin*q0-2*%pi*pmin*q)^2
-2*cos(2*%pi*pmax*q0-2*%pi*pmax*q)
*cos(2*%pi*pmin*q0-2*%pi*pmin*q)
+sin(2*%pi*pmax*q0-2*%pi*pmax*q)^2
+cos(2*%pi*pmax*q0-2*%pi*pmax*q)^2)
/(2*%pi*q0-2*%pi*q)
(%i100)
(3) On the other hand, radcan() seems to take into account some facts specified by the
assume() statement in the example below. Do other Maxima built-in simplification functions
take specified constraints into account ?
(%i101) assume(q0>=2000);
(%o101) [q0 >= 2000]
(%i102) radcan(sqrt(a^2)/sqrt(4*%pi^2*(q0-2000)^2));
(%o102) abs(a)/(2*%pi*q0-4000*%pi)
(%i103)
Thanks very much.
Constantine Frangos.