Re: poldisc variable must be x or y ?

[Dirk Laurie <dirk.laurie@gmail.com>]

If you do poldisc(z/a+1,z) first in a fresh session, there is no problem.

However, first doing poldisc(x/a+1) defines a variable a which has
higher precedence than z and therefore is the default variable, and my
not appear to a negative power in the pol functions. poldisc(a/z+1)
works.

Op Do. 21 Feb. 2019 om 03:35 het Jacques Gélinas
<jacquesg00@hotmail.com> geskryf:
>
> $ poldisc(x/a+1)
> 1
> $ poldisc(y/a+1)
> 1
> $ poldisc(z/a+1)
>   ***   at top-level: poldisc(z/a+1)
>   ***                 ^--------------
>   *** poldisc: incorrect type in poldisc (t_RFRAC).
> $ poldisc(z/a+1,z)
>   ***   at top-level: poldisc(z/a+1,z)
>   ***                 ^----------------
>   *** poldisc: incorrect type in poldisc (t_RFRAC).
> $ version
> [2, 12, 0, 23528, "62f3ec361"]
>
> Jacques Gélinas
>