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Newbie Newton Algorithm and extra precision digits


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 
1.41421.
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 ?

Regards,

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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