[Maxima-discuss] coefficients of taylor polynomials

[Barton W. <wi...@un...>]

I have been looking at the residue code (see residu.lisp).  One idea I was trying was to use a non-default value for taylor_simplifer as a way to find residues somewhat symbolically; for example:

(%i10) myresidue(1/(x^2+x+1),x,a);
Is 0 equal to a^2+a+1?
y;
(%o10) 1/(2*a+1)

(%i11) myresidue(1/(x^2+x+1),x,a);
Is 0 equal to a^2+a+1?
n;
(%o11) 0

These answers are OK.    But I'm mystified by the following:

(%i1) (display2d : false, linel : 100)$

(%i2) taylor_simplifier : lambda([s], scsimp(s, a^2=-1))$

(%i3) taylor(1/(x^2+1),x,a,4);
(%o3) -(a/(2*(x-a)))+1/4+(x-a)/(8*a^3)-(x-a)^2/16+(x-a)^3/(32*a^5)+(x-a)^4/64

It looks like the simply pole term is  -a/2, which is correct, but coeff  tells me something else

(%i4) coeff(%,x-a,-1);
(%o4) -((a^5-4*a^3+48*a)/64)

(%i5) subst(a=%i,%);
(%o5) -((53*%i)/64)

But if I do ratdisrep before finding the simple pole term, all is well:

(%i6) taylor(1/(x^2+1),x,a,4);
(%o6) -(a/(2*(x-a)))+1/4+(x-a)/(8*a^3)-(x-a)^2/16+(x-a)^3/(32*a^5)+(x-a)^4/64

(%i7) coeff(ratdisrep(%),x-a,-1);
(%o7) -(a/2)

Question: How is the output in %o4 correct?  It seems wrong to me—what is the story?