question about Hilbert series

[David Eisenbud]

If I is an ideal in a (singly) graded ring then

hilbertSeries I

returns a "divide expression", something that
looks like a rational function n(t)/d(t). I can get
the power series expansion to order 10, say (a polynomial in the
"degrees ring" by doing

hilbertSeries(I, Order => 10)

My question: how can I conveniently compute the power series expansion
(to some given order) of

d(-t)/n(-t) ,

that is, of

((hilbertSeries i)(-t))^(-1) ?

Thanks,

David
--
David Eisenbud
Professor of Mathematics,
University of California, Berkeley
www.msri.org/~de

[Daniel R. Grayson]

The last function, "fract", in the attached file will help (after you
use "value denominator s" and "value numerator s" to get that parts of
the rational function, and use a ring map(R**Q,R,...) (where R is the
degrees ring) to substitute -t for t and to change the coefficients
from ZZ to QQ .