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From: edp@jareth.enet.dec.com (Eric Postpischil (Always mount a scratch monkey.))
Newsgroups: comp.sys.handhelds
Subject: Re: Bug in QUAD
Message-ID: <20650@shlump.nac.dec.com>
Date: 1 Mar 91 14:10:45 GMT
References: <89355@tut.cis.ohio-state.edu>
Sender: newsdaemon@shlump.nac.dec.com
Reply-To: edp@jareth.enet.dec.com (Eric Postpischil (Always mount a scratch monkey.))
Organization: Digital Equipment Corporation
Lines: 27

In article <89355@tut.cis.ohio-state.edu>, adkins@tortoise.cis.ohio-state.edu
(Brian Adkins) writes:

>      x^2
>  ----------- = .58    QUAD returns 'x=s1*76.15.../5000'  neither one of these
>  (.02 - x)^2           roots is correct.

QUAD works only on quadratic expressions; your equation contains the ratio of
two polynomials, but it is not itself a quadratic expression.

I suspect that the way QUAD works is to find the second order Taylor's
polynomial of the expression you give it.  That forces the expression into a
known form, which can then be dissected to give the quadratic coefficients.

To support this theory, observe that if you use QUAD on an expression containing
a function whose derivative is not known to the 48, you get an expression
contain "der" in it.  E.g., try 'X^2-RE(C)' 'X' QUAD.

As another test, I entered your expression and then 'x' 2 TAYLR COLCT to get
'2500*X^2=.58'.  Taking the QUAD of that produces 'x'=s1*76.1577310586/5000',
which is identical to what you got and is the correct solution for the root of
the Taylor's polynomial.


				-- edp (Eric Postpischil)
				"Always mount a scratch monkey."
				edp@jareth.enet.dec.com