Hi Stavros and Barton,
Thanks for the response.
I am using maxima 5.10 and 5.12. I implemented dividethru() from Stavros and
this gives the same result (see below).
Its not clear why trigsimp() introduces a singularity in Maxima 5.10/5.12 but
seemingly not in Maxima 5.13.99.
Note that I also get a singularity when I apply the following to q:
ratsimp(), fullratsimp(), radcan().
trigrat() produces a big expression asking whether it is pos, neg, etc.
Regards,
C. Frangos.
q : -k*Lc/(Lc*(cos(delta3)*sqrt(sin(delta3)^2*Lc^2
/(2*sin(delta3)*L2+cos(delta3)*Lc)^2
+1)
+1)
+2*sin(delta3)*L2
*sqrt(sin(delta3)^2*Lc^2/(2*sin(delta3)*L2+cos(delta3)*Lc)^2
+1))
(%i113) limit(q,delta3,0);
(%o113) -k/2
(%i114) limit(trigsimp(q),delta3,0);
Is Lc positive or negative?
positive;
(%o114) -k/2
(%i115) subst(0,delta3,q);
(%o115) -k/2
(%i116) subst(0,delta3,trigsimp(q));
Division by 0
-- an error. Quitting. To debug this try debugmode(true);
(%i117) subst(0,delta3,dividethru(q));
(%o117) -k/2
(%i118)
(%i121) subst(0,delta3,dividethru(trigsimp(q)));
Division by 0
-- an error. Quitting. To debug this try debugmode(true);
(%i122)
On Wednesday 19 December 2007 22:06, Stavros Macrakis wrote:
> Try limit(trigsimp(q),delta3,0) ... this is a removable singularity.
>
> You can confirm that they are equal by comparing the taylor series at 0:
>
> q: ...<your expression>
>
> taylor(q, delta3, 0, 5);
>
> taylor(trigsimp(q), delta3, 0, 5);
>
> You can also reorganize the expression using my divthru function:
>
> divthru(a):=block([n,d,div],
> if mapatom(a) then a elseif inpart(a,0) = "*" and
> numberp(n:inpart(a,1))
> then (div:divthru(a/n),if not atom(div) and inpart(div,0) = "+"
> then multthru(n*div) else n*div) elseif part(a,0) = "//"
> then (n:part(a,1),d:part(a,2),div:divide(n,d),div[2]/d+div[1])
> elseif part(a,0) = "-" then -divthru(-a) else a)$
>
> then subst(0,delta3,divthru(trigsimp(q))) gives -k/2.
>
> -s
>
> On Dec 19, 2007 1:46 PM, Constantine Frangos <cfrangos at telkomsa.net> wrote:
> > I have an expression q consisting of trigonometric terms, and q(delta3=0)
> > =
> > -k/2.
> >
> > When I apply trigsimp(q), the resulting expression has a singularity at
> > delta3 = 0 (see below). It appears that trigsimp() is introducing errors
> > ??
> >
> > Any assistance would be appreciated.
> >
> > Regards,
> >
> > C. Frangos.
> >
> >
> > (%i82) q;
> >
> > (%o82) -k*Lc/(Lc*(cos(delta3)*sqrt(sin(delta3)^2*Lc^2
> > /(2*sin(delta3)*L2+cos(delta3)*Lc)^2
> > +1)
> > +1)
> > +2*sin(delta3)*L2
> >
> > *sqrt(sin(delta3)^2*Lc^2/(2*sin(delta3)*L2+cos(delta3)*Lc)^2 +1))
> > (%i83) subst(0,delta3,q);
> >
> > (%o83) -k/2
> > (%i84) subst(0,delta3,trigsimp(q));
> >
> > Division by 0
> > -- an error. Quitting. To debug this try debugmode(true);
> > (%i85)
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