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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 ?
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