|HP-12C Trigs [Final]|
Message #1 Posted by Ex-PPC member on 18 Oct 2001, 5:29 a.m.
Chris Randle wrote:
"Just to whet your appetite, the sine routine is 29 steps and the article goes in to great detail (six A5 pages) over the implementation. I believe it's a preliminary article to implementing all scientific funtions on the 12C in case it becomes the final bastion of RPN"
First of all, I really, really appreciate your kindness in
trying to help with my question, but again, I have no
problem at all implementing a sine routine on the 12C,
nor do I need six A5 pages to tell how it's done, nor can it "whet my appetite" to know how it's done in 29 steps.
Matter of fact, my actual implementation of that sine function is just 18 bytes, far shorter than the 29 bytes you mention. To use 6 pages to describe a 29-byte sine
function would be considered a blatant "filler" in the
old PPC Journal times, so I hope for the benefit of its readers that the article goes deeper into some other worthy aspects of the question.
Anyway, implementing a sine routine is simple to the point of
being trivial. What troubles me, my actual question,
and I'll repeat it here for the last time, is:
"Can anyone point me to (show, produce, etc) an HP-12C
program that implements ALL 6 trigonometric functions
(sine, cosine, tangent, AND THEIR INVERSES) in a SINGLE
program, 99 steps or less ?"
That's it. No use for a 6-page article that shows how to
implement a sine function in 29 steps. It's either all
6 functions in 99 steps (or less) on the 12C or simply
you aren't answering my question at all.
Please don't get me wrong, I am *really* thankful
for the well-meant answers. It's just that they sidestep
the original question and we don't make any progress.
By now it's clear to me that the HP RPN programming comunity
cannot cope or does not want to cope with this topic
that I thought was very old hat, common knowledge and
now see that actually it's an unconquered challenged, at
least for this forum's participants. Well, perhaps this is
for the better and I'll succeed on my own. Thanks again to