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?
>
>
>
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