re: LF: Math Question, possible ROM error..?

To: list-fargo@lists.ticalc.org
Subject: re: LF: Math Question, possible ROM error..?
From: Shawn Prestridge <sprestridge@ti.com>
Date: Tue, 22 Jul 1997 14:33:40 -0700
Conversation-Id: <Pine.SOL.3.93.970722150134.4497A-100000@joe.math.uga.edu>
In-Reply-To: <Pine.SOL.3.93.970722150134.4497A-100000@joe.math.uga.edu>
Reply-To: list-fargo@lists.ticalc.org

You are correct if you assume that the function is real-valued.  If you allow
the expression to be complex valued, mapping the set of reals into complex,
then you can say that:

f(x) = sqrt(x-4) = i*sqrt(4-x)

As x approaches 4 for the second function, f(x) = i*sqrt(0) = i*0 =0.

>  From: Shawn Walker <swalker@joe.math.uga.edu>, on 7/22/97 2:10 PM:
>  
>  Perhaps, but as Bryan pointed out, the answer is not _mathematically_
>  correct.  The square root function is only defined on [0,oo), and as such
>  lim_{x->0} \sqrt(x) makes no sense (i.e. as Bryan pointed out, there is no
>  limit from the left)
>  
Regards,
Shawn Prestridge
Texas Instruments, Inc.
Educational & Productivity Solutions
7800 Banner Drive MS 3908
Dallas, Texas 75251
(972) 917-1698
(972) 917-7103 FAX

References:

re: LF: Math Question, possible ROM error..?

From: Shawn Walker <swalker@joe.math.uga.edu>

Prev: re: LF: Math Question, possible ROM error..?
Next: re: LF: Math Question, possible ROM error..?
Index(es):
- Main
- Thread