Thanks,
This Vo/V0 thing was the blame.
I failed to notice it.
Sorry for bothering you with this stupid mistake.
Many thanks for your attention,
Eli
On Wed, Sep 29, 2010 at 8:07 PM, Stavros Macrakis <macrakis at alum.mit.edu>wrote:
> This doesn't work because Vo is not the same thing as V0. When you replace
> Vo with V0, f2/f3 is beta*ge*T^2/(2*V0).
>
> -s
>
> PS It would be better if you could send self-contained, reproducible code.
> In the example below, %i112 doesn't evaluate to %o112 because Fe is not
> defined.
>
> On Wed, Sep 29, 2010 at 13:44, Eli Brosh <ebrosh1 at gmail.com> wrote:
>
>> I am using wxmaxima 0.8.4 with maxima 5.20.1
>> I checked it. The example I sent really does work. But a slightly
>> different example does not:
>>
>>
>> (%i111) f1:-1/2*beta*(V/V0)^ge*T^2;
>> (%o111) -(beta*T^2*(V/V0)^ge)/2
>> (%i112) f2:diff(-Fe,V);
>> (%o112) (beta*ge*T^2*(V/V0)^ge)/(2*V)
>> (%i113) f3:(V/Vo)^(ge-1);
>> (%o113) (V/Vo)^(ge-1)
>> (%i114) f4:f2/f3;
>> (%o114) (beta*ge*T^2*(V/Vo)^(1-ge)*(V/V0)^ge)/(2*V)
>>
>> Why doesn't it simplify ?
>> I tried ratsim, trigsimp and some others.
>>
>> Thanks
>> Eli
>>
>>
>>
>>
>>
>> On Wed, Sep 29, 2010 at 6:13 PM, Stavros Macrakis <macrakis at alum.mit.edu>wrote:
>>
>>> What version of Maxima are you using? This works correctly in 5.20.1:
>>>
>>> (%i1) f1:(x/x0)^a;
>>> (%o1) (x/x0)^a
>>> (%i2) f2:(x/x0)^(a-1);
>>> (%o2) (x/x0)^(a-1)
>>> (%i3) f3:f1/f2;
>>> (%o3) x/x0
>>>
>>> For more complicated simplifications which Maxima does not perform
>>> automatically, there is a large number of functions to try for reorganizing
>>> expressions. Some that are worth trying are ratsimp, factor, factorsum,
>>> expand, multthru. To get more information on these, type, e.g.
>>>
>>> (%i4) ? ratsimp
>>> -- Function: ratsimp (<expr>)
>>> -- Function: ratsimp (<expr>, <x_1>, ..., <x_n>)
>>> Simplifies the expression <expr> and all of its subexpressions,
>>> including the arguments to non-rational functions. The result is
>>> ...
>>>
>>> On Wed, Sep 29, 2010 at 03:58, Eli Brosh <ebrosh1 at gmail.com> wrote:
>>>
>>>> Hello,
>>>> I hope people on the maxima mailing list can help me with the following.
>>>>
>>>> I defined the functions f1 and f2 as:
>>>> (%i53) f1:(x/x0)^a;
>>>> (%o53) (x/x0)^a
>>>> (%i48) f2:(x/x0)^(a-1);
>>>> (%o48) (x/x0)^(a-1)
>>>>
>>>> Now, f3 is the division:
>>>> (%i52) f3:f1/f2;
>>>> (%o52) x^a*(x/x0)^(1-a)
>>>>
>>>> How can I force maxima to simplify f3 down to (x/x0) ?
>>>> Now this is a simple example. What can I do with more complex
>>>> expressions where I do not have a good idea about the end result?
>>>>
>>>> Thanks,
>>>> Eli
>>>>
>>>>
>>>> _______________________________________________
>>>> Maxima mailing list
>>>> Maxima at math.utexas.edu
>>>> http://www.math.utexas.edu/mailman/listinfo/maxima
>>>>
>>>>
>>>
>>
>