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
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>
--
Leo Butler <l_butler at users.sourceforge.net>
SDF Public Access UNIX System - http://sdf.lonestar.org