Re: Newbie Newton Algorithm and extra precision digits
- To: mathgroup at smc.vnet.net
- Subject: [mg47890] Re: Newbie Newton Algorithm and extra precision digits
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Thu, 29 Apr 2004 19:39:36 -0400 (EDT)
- Organization: Universitaet Leipzig
- References: <email@example.com>
- Reply-to: kuska at informatik.uni-leipzig.de
- Sender: owner-wri-mathgroup at wolfram.com
that what Precision returns is the precision of your computations
and not the error of your method.
Numerical errors have two origins, on are roundig errors
(that is reported by precision) and the second error source
are the errors of your numerical method. Mathematica can't find
out that you are using a Newton method, or that your program make
a numerical integration with a fixed step size. So Mathematica
gives you only the rounding errors.
Benjamin Collas wrote:
> Here is another mystery (I think there's something I dont understand
> with N).
> I'm using the Newton algorithm to compute one root of X^2-2 : with
> NestWhile I can iterate the fixed point formula.
> My stop test is (Abs[#2-#1]>10^(-6))&, and the Mathematica answer is
> The problem is when I ask Precision[%], Mathematica returns
> MachinePrecision (ie 15.9546 by default)...
> More over, when I copy-past the 1.41421 answer, what displays is
> 1.414235623730951' which means that Mathematica knows this number with
> 16 digits...
> But how could it be as my stop test is about 10^(-6). As Newton is a
> quadratic computional method, the best I could have is a 6 or 12
> precision digits.
> Can anyone explains where am I wrong ?
> Benjamin Collas
> PS : thank you Jens-Peer Kuska for your help,I'm so confused not to have
> find it out by myself...
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