@Skip et al … also apologies for my sill definitions, I should have used y inside the definitions not x (!!!), sorry if I confused the issue…
as in here for the first interpretation … x,"0 (3 : '8+(2^y)-((2^2)^y)') x=:1.6+0.05*i.8 …/Rob > On 29 Jan 2018, at 3:49 pm, Jose Mario Quintana > <[email protected]> wrote: > > 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 ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
