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?
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