I think that, solve is using a linear equation solver, and your
equations are not linear in x.
Even if solve were to use a polynomial system solver, your equations are
not polynomial in x.
How about solving for sin(x), getting a result like sin(x)=1, and then
solving that?
Should Maxima do this automatically? Maybe. It looks a little
ticklish. Since there
are an infinite number of solutions to sin(x)=1, how does
back-substitution work?
RJF
On 11/25/2012 7:57 AM, Norman Werner wrote:
> Hello,
>
> I am using maxima as a common-lisp library for a pet-project of mine.
> However I stumbled upon a problem when using solve.
>
> I try to solve an equation or a set of equation and I do introduce
> known variables by
> adding an appropriate equation to the set. Eg:
>
> solve([1=a*x,a=1],[a,x]);
> -> [[a = 1, x = 1]]
>
> Which works fine! If the equation however contains functions like exp,
> log, or sin solve won't find any solutions:
>
> solve([1=a*sin(x),a=1],[a,x]);
> -> []
> solve([1=a*log(x),a=1],[a,x]);
> -> []
> (%i125) solve([1=a*exp(x),a=1],[a,x]);
> -> []
>
> Obviously the solutions to those problems are not as obvious as they
> seem. Could someone clarify this behaviour? What's the correct way to
> solve such problems with maxima?
>
> Thanks
>
> Norman
> _______________________________________________
> Maxima mailing list
> Maxima at math.utexas.edu
> http://www.math.utexas.edu/mailman/listinfo/maxima