Barton Willis

psi(complex float)

Isn't setting domain to complex and possibly m1pbranch to t at the top-level of matrixexp and matrixfun sufficient? Plus it avoids the extra complexity of tellrat for a bug that is really about the simplification of radicals, not about the matrixexp code. Actually, I've never used tellrat all that much and I'm not sure that I trust it. get: Setting domain to complex and m1pbranch to t at the top-level of matrixexp and matrixfun I get (%i1) load(linearalgebra)$ (%i2) fpprec : 28$ (%i3) m : %i*matrix([1,2,3],[3,2,0],[1,2,1])$...

Try this experiment: Globally set domain to complex. With the current linearalgebra, I get (%i1) load(linearalgebra)$ (%i2) display2d : false$ (%i3) domain : complex$ (%i4) seriesexp(a, k) := block([s : ident(3), t : ident(3)], for j : 1 thru k do (t : (t . a)/j, s : s + t), s)$ (%i5) mnorm(a) := lmax(map(abs, flatten(args(float(rectform(a))))))$ (%i6) m : %i*matrix([1,2,3],[3,2,0],[1,2,1])$ (%i7) fpprec : 28$ (%i9) yyy : ratsimp(matrixexp(m))$ (%i8) xxx : bfloat(seriesexp(m, 80))$ (%i10) bfloat(rectform(yyy))$...

error in matrix exponentiation

Try this experiment: Globally set domain to complex. With the current linearalgebra, I get (%i1) load(linearalgebra)$ (%i2) display2d : false$ (%i3) domain : complex$ (%i4) seriesexp(a, k) := block([s : ident(3), t : ident(3)], for j : 1 thru k do (t : (t . a)/j, s : s + t), s)$ (%i5) mnorm(a) := lmax(map(abs, flatten(args(float(rectform(a))))))$ (%i6) m : %i*matrix([1,2,3],[3,2,0],[1,2,1])$ (%i7) fpprec : 28$ (%i8) xxx : bfloat(seriesexp(m, 80))$ (%i9) yyy : ratsimp(matrixexp(m))$ (%i10) bfloat(rectform(yyy))$...

hypergeometric([-3],[-1,-5],2) signals a Lisp division-by-zero error

Fixed by Commit [7d3fce] . Closing this ticket.

hypergeometric.lisp: fix hypergeometric([-3],[-1,-5],2)