Pau <vim.unix at googlemail.com> writes:
> Dear Leo,
>
> thanks for the help. In my case, I get:
>
> (%i5) solve(first(%o5)/a,a);
>
> first: argument must be a non-atomic expression; found %o5
> -- an error. To debug this try: debugmode(true);
Try:
solns : solve(TGW=TRRV,a);
solve(first(solns)/a, a);
The latter command takes the first solution from solve and divides both
sides by a, to give 1 = ..., then asks solve to solve the equation for
a. You will be ask some inconsequential question and then voila, a solution.
>
> The other option
>
> solve(log(TGW)=log(TRRV),a), logexpand=all;
>
> leads to something...
>
> But could you explain me what you did? What does first(%o5)/a mean?
>
> And why does the log help?
Your equations involve only multiplication. Take logs and you have a
linear equation in a.
>
> The reply starts with
>
> 3146 log(Rinfl) gamma 6292 log(M + m)
> (%o6) [a = expt(%e, --------------------- - ------------------
> 3146 gamma - 25168 3146 gamma - 25168
>
>
> ....
>
> What does that %e mean?
%e is the base of the natural logarithm: log(%e) ==> 1. Mathematicians
write it as just e.
>
> I suppose the answer is the exponent of the log?
>
Look at the answer more carefully.
> thanks!
>
>
>
> On 13 March 2012 14:58, Leo Butler <l_butler at users.sourceforge.net> wrote:
>> Pau <vim.unix at googlemail.com> writes:
>>
>>> Thanks... however something seems strange... it looks like maxima is
>>> trying to simplify the expression and in the end I get
>>>
>>> ? ? ? ? ? ? ? ? ? ? ? ? ? 3 - gamma
>>> ? ? ? ? ? ? ? ? ? ? 3/8 - --------- ? ? ? ? ? ? ? ? ? ? ? ? ?3 ? ? ? ?3
>>> ? ? ? ?1/4 ? ? ? ? ? ? ? ? ? ?8 ? ? ? ? ? ? 2 ? ? ? ? ? ? ? G ?M + m G ?1/4
>>> ? ? 4 6 ? ?sqrt(7) a ? ? ? ? ? ? ? ?sqrt(2 e ?+ 1) sqrt(M) (-----------)
>>> ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?c
>>> a = -----------------------------------------------------------------------]
>>> ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? gamma - 3
>>> ? ? ? ? ?5/4 ? ? ? ?4 ? ? ? ?2 ? ? ?1/4 ?Rinfl ? ? ? ? ?M 1/8 ? ? ?1/8
>>> ? ? ? ? 5 ? ?c (37 e ?+ 292 e ?+ 96) ? ?(----------------) ? ?(G M)
>>> ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? m
>>>
>>>
>>> (I hope that the alignment is right)
>>>
>>> As you can see, there's an "a" in the right hand side... that's strange
>>>
>>> I did this:
>>>
>>> (%i1) Fe : 1 + (73/24)*e^2 + (37/96)*e^4$
>>>
>>> (%i2) Na : (M/m) * (a/Rinfl)^(3-gamma)$
>>>
>>> (%i3) TGW : (5/64) * (c^5/G^3) * a^4/(m*M * (m + M)) * Fe$
>>>
>>> (%i4) TRRV : a^(3/2)/(sqrt(G*M)) * M/m * (0.28*(e^2+0.5))^2/sqrt(Na)$
>>>
>>> (%i5) solve(TGW=TRRV,a);
>>>
>>> Any idea of what might have gone wrong?
>>>
>>> thanks
>>
>> For some reason, solve gave up and returned a partial answer.
>> Here are a couple ideas:
>>
>> solve(first(%o5)/a,a);
>>
>> solve(log(TGW)=log(TRRV),a), logexpand=all;
>>
>> Both give me an answer.
>>
>> Btw, gamma is the name of a function in Maxima. Although variables and
>> functions may share the same name, this will likely lead to obscure and
>> wierd errors, so it is better to use another name.
>>
>> Leo
>>
>>>
>>> On 12 March 2012 00:56, Leo Butler <l_butler at users.sourceforge.net> wrote:
>>>> Pau <vim.unix at googlemail.com> writes:
>>>>
>>>>> Hi,
>>>>>
>>>>> I am sorry for a rather trivial question... If I have two equations like
>>>>>
>>>>> T(a,e) = a^4 ( 1 - e^3)
>>>>> H(a,e) = e^3 * constants * a^(-1)
>>>>>
>>>>> and I want to find the values of a ,e that equate them
>>>>>
>>>>> T(a,e) = H(a,e) ?---> a = XXXX f(e)
>>>>
>>>> Try
>>>> ? solve
>>>> at the Maxima command line.
>>>>
>>>> A couple notes:
>>>> -the assignment operator in Maxima is :
>>>> -the function definition operator is :=
>>>> -to solve your problem,
>>>> ?I did
>>>>
>>>> (%i1) T : a^4*(1 - e^3) $ H : c*e^3/a $ display2d : false $
>>>>
>>>> (%i4) solve(T=H,e);
>>>>
>>>> (%o4) [e = (sqrt(3)*%i-1)*a^(5/3)/(2*(c+a^5)^(1/3)),
>>>> ? ? ? e = -(sqrt(3)*%i+1)*a^(5/3)/(2*(c+a^5)^(1/3)),
>>>> ? ? ? e = a^(5/3)/(c+a^5)^(1/3)]
>>>>
>>>>>
>>>>> how could I do this in maxima? The best would be an example.
>>>>>
>>>>> Thanks for this nice piece of software. I am looking forward to
>>>>> learning how to use it in detail. Looks very promising.
>>>>>
>>>>> Cheers,
>>>>>
>>>>> Pau
>>>>> _______________________________________________
>>>>> Maxima mailing list
>>>>> Maxima at math.utexas.edu
>>>>> http://www.math.utexas.edu/mailman/listinfo/maxima
>>>>>
>>>>
>>>> --
>>>> Leo Butler ? ? ? ? ? ? ? ?<l_butler at users.sourceforge.net>
>>>> SDF Public Access UNIX System - ? http://sdf.lonestar.org
>>> _______________________________________________
>>> Maxima mailing list
>>> Maxima at math.utexas.edu
>>> http://www.math.utexas.edu/mailman/listinfo/maxima
>>>
>>>
>>
>> --
>> Leo Butler ? ? ? ? ? ? ? ?<l_butler at users.sourceforge.net>
>> SDF Public Access UNIX System - ? http://sdf.lonestar.org
>
>
--
Leo Butler <l_butler at users.sourceforge.net>
SDF Public Access UNIX System - http://sdf.lonestar.org