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Re: How to find one expression in terms of another
*To*: mathgroup at smc.vnet.net
*Subject*: [mg119744] Re: How to find one expression in terms of another
*From*: DrMajorBob <btreat1 at austin.rr.com>
*Date*: Mon, 20 Jun 2011 08:04:55 -0400 (EDT)
*References*: <201106192329.TAA01140@smc.vnet.net>
*Reply-to*: drmajorbob at yahoo.com
Again, this line does nothing:
Element[{x1, x2, t1, t2, u, c}, Reals]
It is a test, not a declaration of fact, and testing with unknowns gives
no test at all.
> 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?
That can be done for ANY dT and dt, no matter what their values, with
infinitely many solutions:
Solve[dT == a dt + b, {a, b}]
{{b -> -a dt + dT}}
If dt has no value, you can also evaluate
Solve[dT == a dt + b, dt]
{{dt -> (-b + dT)/a}}
If dt HAS a value, on the other hand, you can't solve for it (solving for
it would make no sense), so you get an error message.
Bobby
On Sun, 19 Jun 2011 18:29:04 -0500, 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 ?
>
--
DrMajorBob at yahoo.com
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