[Jprogramming] Extended precision question

['Mike Day' via Programming]

One way to obtain high precision roots is to use, say, Newton’s solution to 
y = x^2 - a : 
Given an estimate x(n-1),  the next estimate is 
xn = x(n-1) - y/(dy/dx), in Maths notation.

In J, (J701 on this iPad),
   rt =: -:@] + (% +:)  NB. x is number whose root is required,  y an initial guess
   2 rt ^:_] 1.  NB.  Normal float
1.41421
   2 rt ^:(i.5)1x   NB. Extended to rational...
1 3r2 17r12 577r408 665857r470832 

I forget how I solved this Euler Problem!

Mike

>>> On 4/21/2022 12:43 PM, Ed Gottsman wrote:
>>> Hello.
>>> I’m working on the Project Euler “Diophantine equation” problem (#66) and using J’s extended precision facilities.  I’ve run into behavior that confuses me.  Boiled down (and overusing x: just to be sure):
>>>     x: %: x: 1 + x: *: x: 999999999
>>> 999999999
>>> That is (if my syntax is right), the square root of (one plus the square of a really large n) is n.  I’m apparently misunderstanding something about extended precision.  I’ve tried it with a variety of uses of x: but to no avail, and as I read the x: documentation…this is an odd result.
>>> 
>>> Any help would be much appreciated.
>>> (J901 on iPadOS, for which sincere kudos to Ian Clark.)
>>> Many thanks.
>>> Ed
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