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 >>> ---------------------------------------------------------------------- >>> For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
