I found out that in Mathematica this can be done by PolynomialReduce[dT, dt, {t1, t2}]. Output given below.
In[26]:= FullSimplify[PolynomialReduce[dT, dt, {t1, t2}]] Out[26]= {{1/Sqrt[1 - u^2/c^2]}, (u (x1 - x2))/(c^2 Sqrt[1 - u^2/ c^2])} But I'd rather use Sage. Does Sage have a counterpart to this Mathematica function? If not how do get the same result? On Jun 19, 11:19 am, Jacare Omoplata <walkeystal...@gmail.com> wrote: > The following are the expressions, > > sage: var('x1,t1,x2,t2,u,c',domain=RR);assume(u>0);assume(c>u); > (x1, t1, x2, t2, u, c) > sage: T1 = (t1-((u*x1)/(c^2)))/sqrt(1-((u^2)/(c^2))) > sage: T2 = (t2-((u*x2)/(c^2)))/sqrt(1-((u^2)/(c^2))) > sage: dT = T2-T1 > sage: dt = t2-t1 > > Suppose I know that dT is in this form, > > dT = a*dt + b, > > Assuming I DID NOT know that, > a = 1/sqrt(1-((u^2)/(c^2))) , > b = (u*(x2 - x1))/((c^2)*sqrt(1-((u^2)/(c^2)))) , > > is there any way I can find 'a' and 'b' using Sage? > > What if I didn't know that dT is in the form of a*dt + b, but just > knew dT in terms of x1,t1,x2,t2,u and c ? > > Can I still express dT in terms of dt using sage? -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org