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
