The way you worded it, it sounded as though f was the form you were looking
for...!
radcan is Maxima's 'specialist' in algebraic relations among exponentials,
so is the right tool here. Another approach which helps the rational
function package understand what is going on is
subst(x/2,y,factor(subst(2*y,x,f))) .
-s
On Fri, Apr 27, 2012 at 17:07, David Ronis <David.Ronis at mcgill.ca> wrote:
> Thanks for the reply Stavros, I thought I had given the orignal
> expression:
>
> f:(exp(x)+exp(x/2))/(exp(x/2)+1);
>
> I get there using ratsimp or factor on the result of a limit from
> something a bit more complicated. It's the next simplification that
> doesn't always work, as I described. Radcan works, but I think that
> factor should be able to factor something as simple as my f.
>
> David
>
>
> On Fri, 2012-04-27 at 17:00 -0400, Stavros Macrakis wrote:
> > We might be able to help if you could share your original expression.
> >
> >
> > But I don't really understand your question. You say that radcan puts
> > your expression in the form you want. So use radcan! -- Why would you
> > expect ratsimp or trigsimp or default simplification to do the same
> > thing as radcan?
> >
> >
> > As for representing exponentials as hyperbolic functions, here's an
> > approach that will often work:
> >
> >
> > (%i1) ex: %e^(a*x) + %e^(x-b)$
> >
> >
> > (%i3) subst(%e^(qq*%i*x),%e,ex);
> > (%o3) %e^(%i*a*qq*x^2)+%e^(%i*qq*x*(x-b))
> >
> >
> > (%i4) rectform(%);
> > (%o4) %i*(sin(a*qq*x^2)+sin(qq*x*(x-b)))+cos(a*qq*x^2)+cos(qq*x*(x-b))
> > (%i5) subst(-%i,qq,%);
> > (%o5) %i*(-%i*sinh(a*x^2)-%i*sinh(x*(x-b)))+cosh(a*x^2)+cosh(x*(x-b))
> > (%i6) ratsimp(%);
> > (%o6) sinh(x^2-b*x)+cosh(x^2-b*x)+sinh(a*x^2)+cosh(a*x^2)
> >
> >
> > or...:
> >
> >
> > (%i8) demoivre(%o3);
> > (%o8) %i*sin(a*qq*x^2)+cos(a*qq*x^2)+%
> > i*sin(qq*x*(x-b))+cos(qq*x*(x-b))
> > (%i9) subst(-%i,qq,%);
> > (%o9) sinh(a*x^2)+cosh(a*x^2)+sinh(x*(x-b))+cosh(x*(x-b))
> >
> >
> > Does this help in your case?
> >
> >
> > -s
> >
> >
> > -----------------------------------
> >
> > On Fri, Apr 27, 2012 at 15:32, David Ronis <David.Ronis at mcgill.ca>
> > wrote:
> > I have an expression that ultimately becomes:
> >
> > f:(exp(x)+exp(x/2))/(exp(x/2)+1);
> >
> > I've tried various simplification routines:
> >
> >
> > ratsimp(f);
> > trigsimp(f);
> > radcan(f);
> >
> >
> > The only one that works is radcan(). Is there a flag that
> > would make the other ones work? Alternately, is there a
> > function that
> > forces the exponentials to be represented as sums of
> > hyperbolic
> > functions?
> >
> >
> >
> > _______________________________________________
> > Maxima mailing list
> > Maxima at math.utexas.edu
> > http://www.math.utexas.edu/mailman/listinfo/maxima
> >
> >
>
>