Re: TI-M: TI-89 root question
[Prev][Next][Index][Thread]
Re: TI-M: TI-89 root question
> a^(1/n) =
> a^(n^(-1)) = a^n^(-1) because ^ is highest prescedence and right
> associative...(at least in standard arithmetic parsers, but I believe the
> TI-89 follows this convention too)
Don't use that last form (a^n^-1) on the calcs. There's a supposedly a bug
in most ROM versions of most z80 calcs where the precedence of powers to
powers is not evaluated correctly, and that may be taken as (a^n)^-1 rather
than a^(n^-1).
I'm not sure exactly which calcs this affects, but I remember HP listing the
HP49G's correct evaluation of this as a "feature" in which they claimed that
the HP48 line and all ti graphing calcs except the 89, 92(+), and I believe
86 (one of the z80 calcs) all suffer from this incorrect evaluation. I've
got easy access to a friends' 82/83/83+/85/86 calcs in school, so maybe I'll
test it out tomorrow and post the results here.
-Scott
Follow-Ups:
References: