Re: How to find one expression in terms of another

*To*: mathgroup at smc.vnet.net*Subject*: [mg119746] Re: How to find one expression in terms of another*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Mon, 20 Jun 2011 08:05:17 -0400 (EDT)*Reply-to*: hanlonr at cox.net

Element is not used the way that you think. See documentation. $Assumptions = True; T1 = (t1 - ((u x1)/c^2))/Sqrt[1 - (u^2/c^2)]; T2 = (t2 - ((u x2)/c^2))/Sqrt[1 - (u^2/c^2)]; dT = T2 - T1; FullSimplify[dT /. t2 -> dt + t1] (c^2*dt + u*(x1 - x2))/(c^2*Sqrt[1 - u^2/c^2]) FullSimplify[dT /. t2 -> dt + t1, Element[c, Reals]] (c^2*dt + u*(x1 - x2))/(Sqrt[(c - u)*(c + u)]*Abs[c]) $Assumptions = Element[c, Reals]; FullSimplify[dT /. t2 -> dt + t1] (c^2*dt + u*(x1 - x2))/(Sqrt[(c - u)*(c + u)]*Abs[c]) Bob Hanlon ---- Jacare Omoplata <walkeystalkey at gmail.com> wrote: ============= I want to find dT in terms of dt. They are given below. In[1]:= Element[{x1, x2, t1, t2, u, c}, Reals] Out[1]= (x1 | x2 | t1 | t2 | u | c) \[Element] Reals In[3]:= T1 = (t1 - ((u x1)/c^2))/Sqrt[1 - (u^2/c^2)] Out[3]= (t1 - (u x1)/c^2)/Sqrt[1 - u^2/c^2] In[4]:= T2 = (t2 - ((u x2)/c^2))/Sqrt[1 - (u^2/c^2)] Out[4]= (t2 - (u x2)/c^2)/Sqrt[1 - u^2/c^2] In[5]:= dT = T2 - T1 Out[5]= -((t1 - (u x1)/c^2)/Sqrt[1 - u^2/c^2]) + ( t2 - (u x2)/c^2)/Sqrt[1 - u^2/c^2] In[6]:= dt = t2 - t1 Out[6]= -t1 + t2 If I knew that dT can be written in terms of dt in the form, dT = a dt + b, Can I use Mathematica to find a and b? I tried using Solve[dT == a dt + b, dt], but that gives an error. If I didn't know that dT can be expressed this way, can I still express it in terms of dt ?