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Re: How to find one expression in terms of another expression?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg119740] Re: How to find one expression in terms of another expression?
  • From: David Bailey <dave at removedbailey.co.uk>
  • Date: Mon, 20 Jun 2011 08:04:12 -0400 (EDT)
  • References: <itm0of$143$1@smc.vnet.net>

On 20/06/2011 00:29, 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
>
> 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 ?
>

I tried evaluating:

Solve[dT == a dt + b, dt]

It worked perfectly, so I suspect that something had a value at that 
point - say dT. BTW, if you get an error message, it usually helps to 
include that as part of your query!

In a more complicated case, if there is a chance that something already 
has a spurious value, a good trick is to change Solve to Solve1 (say) - 
evaluating that should just echo back, but if dT had a value, you would 
see that immediately.

David Bailey
http://www.dbaileyconsultancy.co.uk





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