In that case, (X=. (8 + (2 ^ ]) - (2 ^ 2) ^ ])New (^:22) 1) 1.75372489
is a root, (8 + (2 ^ ]) - (2 ^ 2) ^ ])X 0 but, it is not the only one, (X=. (8 + (2 ^ ]) - (2 ^ 2) ^ ])New (^:22) 0.5j_0.5) 1.24627511j4.53236014 (8 + (2 ^ ]) - (2 ^ 2) ^ ])X 8.8817842e_16j7.72083702e_15 On Sun, Jan 28, 2018 at 9:27 PM, Raul Miller <[email protected]> wrote: > Hmm... > > I had originally thought about calling out the (2^2)^x interpretation > as a possibility, because rejected that, because that would be better > expressed as 4^x > > But it's possible that Skip got the 1.75379 number from someone who > thought different about this. > > And, to be honest, it is an ambiguity in the original expression - > just one that I thought should be rejected outright, rather than > suggested. > > Which gets us into another issue, which is that what one person would > think is obviously right is almost always what some other person would > think is obviously wrong... (and this issue crops up all over, not > just in mathematic and/or programming contexts). > > -- > Raul > > > On Sun, Jan 28, 2018 at 9:08 PM, Rob Hodgkinson <[email protected]> wrote: > > @Skip > > > > Skip, I am a confused in your original post… your actual post read; > > > > ================================================ > > What is the best iterative way to solve this equation: > > (-2^2^x) + (2^x) +8 =0 > > then later to Raul, > > 0 = 8 + (2^x) - 2^2^x NB. Is correct, and the answer is real > > The answer is close to 1.75379 > > ================================================ > > > > However I suspect your original syntax was not J syntax (could it have > been math type syntax ?) as it differs on the J style right-to-left syntax > on the 2^2^x expression. > > > > The 2 possible interpretations are shown below and only the Excel type > syntax seems to get close to your expected answer. > > > > NB. Excel interpretation > > x,"0 (3 : '8+(2^x)-((2^2)^x)') x=:1.6+0.05*i.8 > > 1.6 1.84185 > > 1.65 1.28918 > > 1.7 0.692946 > > 1.75 0.0498772 NB. intercept seems close to your > expected value of 1.75379 > > 1.8 _0.64353 > > 1.85 _1.39104 > > 1.9 _2.19668 > > 1.95 _3.06478 > > > > NB. J style interpretation > > x,"0 (3 : '8+(2^x)-(2^2^x)') x=:1.6+0.05*i.8 > > 1.6 2.85522 > > 1.65 2.33325 > > 1.7 1.74188 > > 1.75 1.07063 > > 1.8 0.307207 NB. But in this model the intercept is > above 1.8, but this is the model that has been coded in responses to your > post ?? > > 1.85 _0.562844 > > 1.9 _1.5566 > > 1.95 _2.69431 > > > > Please clarify, thanks, Rob > > > > > >> On 29 Jan 2018, at 12:48 pm, Jose Mario Quintana < > [email protected]> wrote: > >> > >> Moreover, apparently there is at least another solution, > >> > >> ((-2^2^X) + (2^X) +8 ) [ X=. 2.9992934709539156j_13.597080425481581 > >> 5.19549681e_16j_2.92973749e_15 > >> > >> > >> On Sun, Jan 28, 2018 at 7:28 PM, Jose Mario Quintana < > >> [email protected]> wrote: > >> > >>> Are you sure? > >>> > >>> u New > >>> - (u %. u D.1) > >>> > >>> ,. (8 + (2 ^ ]) - 2 ^ 2 ^ ])New (^:(<22)) 1 > >>> 1 > >>> 3.44167448 > >>> 3.25190632 > >>> 3.03819348 > >>> 2.7974808 > >>> 2.53114635 > >>> 2.25407823 > >>> 2.00897742 > >>> 1.86069674 > >>> 1.82070294 > >>> 1.81842281 > >>> 1.81841595 > >>> 1.81841595 > >>> 1.81841595 > >>> 1.81841595 > >>> 1.81841595 > >>> 1.81841595 > >>> 1.81841595 > >>> 1.81841595 > >>> 1.81841595 > >>> 1.81841595 > >>> 1.81841595 > >>> > >>> (8 + (2 ^ ]) - 2 ^ 2 ^ ]) 1.81841595 > >>> _8.21739086e_8 > >>> > >>> (8 + (2 ^ ]) - 2 ^ 2 ^ ]) 1.75379 > >>> 1.01615682 > >>> > >>> PS. Notice that New is a tacit version (not shown) of Louis' VN > adverb. > >>> > >>> > >>> > >>> > >>> 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/forum > >>>> s.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 > > > > ---------------------------------------------------------------------- > > 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
