Oh, yes... I see the problem: 13 :' 8 + (2^y) - 2^2^y' 8 + (2 ^ ]) - 2 ^ 2 ^ ] 13 :' 8 + (2^y) - 2^2^y' d.1 |domain error
Instead, use: 8: + 2&^ + (2&^)@(2&^) Thanks, -- Raul On Sun, Jan 28, 2018 at 5:52 PM, Skip Cave <[email protected]> wrote: > Raul, > > You had it right in the first place. > > 0 = 8 + (2^x) - 2^2^x NB. Is correct, and the answer is real > > The answer is close to 1.75379 > > I wanted to know how to construct the Newton Raphson method using the > iteration verb N described in the link: http://code.jsoftware. > com/wiki/NYCJUG/2010-11-09 > under "A Sampling of Solvers - Newton's Method" > > N=: 1 : '- u % u d. 1' > > Skip > > > > > > Skip Cave > Cave Consulting LLC > > On Sun, Jan 28, 2018 at 4:38 PM, Raul Miller <[email protected]> wrote: > >> Eh... I *think* you meant what would be expressed in J as: >> >> 0 = 8 + (2^x) - 2^2^x >> >> I'd probably try maybe a few hundred rounds of newton's method first, >> and see where that leads. >> >> But there's an ambiguity where the original expression (depending on >> the frame of reference of the poster) could have been intended to be: >> >> 0 = 8 + (2^x) + _2^2^x >> >> [if that is solvable, x might have to be complex] >> >> Thanks, >> >> -- >> Raul >> >> On Sun, Jan 28, 2018 at 5:25 PM, Skip Cave <[email protected]> >> wrote: >> > What is the best iterative way to solve this equation: >> > >> > (-2^2^x) + (2^x) +8 =0 >> > >> > >> > Skip Cave >> > Cave Consulting LLC >> > ---------------------------------------------------------------------- >> > 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 ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
