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Re: How to find one expression in terms of another expression?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg119741] Re: How to find one expression in terms of another expression?
*From*: Bill Rowe <readnews at sbcglobal.net>
*Date*: Mon, 20 Jun 2011 08:04:23 -0400 (EDT)
On 6/19/11 at 7:29 PM, walkeystalkey at gmail.com (Jacare Omoplata)
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
The above isn't useful outside a list of assumptions. You cannot
set global properties of variables in this manner.
>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
This is what is causing the error message when you use Solve
>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?
Yes,
>I tried using Solve[dT == a dt + b, dt], but that gives an error.
The problem here is you have already given dt a value. And it is
that value that gets substituted in the equation you set up.
That is your input to Solve is exactly equivalent to
Solve[dT == a (t2-t1) + b, t2-t1]
Since t2-t1 is clearly not a variable you get the error message.
A couple of things can be done.
Use Block to localize dt, i.e.,
In[9]:= Block[{dt}, Solve[dT == a dt + b, dt]]
Out[9]= {{dt -> (-(b*c^2*Sqrt[(c^2 - u^2)/c^2]) + c^2*(-t1) +
c^2*t2 +
u*x1 - u*x2)/(a*c^2*Sqrt[(c^2 - u^2)/c^2])}}
Clear the definition of dt and use Solve as you did above. But I
suspect neither of these will give you the result you are
looking for.
I think you want something like
In[12]:= expr=Together[dT] /. t2 -> dt + t1 // Simplify
Out[12]= (c^2*dt + u*(x1 - x2))/(c^2*Sqrt[1 - u^2/c^2])
In[13]:= CoefficientList[expr, dt]
Out[13]= {(u*(x1 - x2))/(c^2*Sqrt[1 - u^2/c^2]), 1/Sqrt[1 - u^2/c^2]}
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