I do not know which of j expressions which could have matched original expression you were referring to.
Thanks, -- Raul On Mon, Jan 29, 2018 at 12:39 PM, Louis de Forcrand <[email protected]> wrote: > Yes, typed a little fast. > > sqrt33 = %:33. > > Why does it not make sense? > > Louis > >> On 29 Jan 2018, at 18:35, Raul Miller <[email protected]> wrote: >> >> What is sqrt33 here? (I would have guessed %:33 but that does not make >> sense to me.) >> >> Thanks, >> >> -- >> Raul >> >> >>> On Mon, Jan 29, 2018 at 12:31 PM, Louis de Forcrand <[email protected]> >>> wrote: >>> I skipped a few steps there. With pencil and paper, I find that (using >>> standard notation) >>> >>> 2^(x+iy) = (1 +- sqrt(33)) / 2 >>> >>> Yet 2^(x+iy) = 2^x * 2^iy, and because the whole is real, 2^iy must be real. >>> >>> Moreover, when y is such that 2^iy is real (when y is a multiple of >>> Pi/log2) then >>> >>> 2^iy >>> = exp(log2 * i*k*Pi/log2) >>> = exp(i*k*Pi) >>> = (-1)^k >>> >>> We can see that, because 2^x is positive, one of the roots of the >>> polynomial corresponds to even k and the other to odd k. >>> >>> Thus (with J’s ^. log) >>> >>> x = 2 ^. (sqrt33 + (-1)^k)/2 >>> y = k*Pi/log2 >>> >>> for any integer k describes the (countable) set of solutions. >>> >>> They are indeed arranged in a zigzag pattern on two vertical lines, and the >>> one real solution occurs when k = 0. >>> >>> Cheers, >>> Louis >>> >>>> On 29 Jan 2018, at 08:09, Louis de Forcrand <[email protected]> wrote: >>>> >>>> Note that if the equation really is (in traditional notation) >>>> >>>> 4^x - 2^x - 8 = 0 >>>> >>>> then it can be rewritten as >>>> >>>> y^2 - y - 8 = 0, y = 2^x >>>> >>>> and solved in closed form as well, >>>> yielding a countably infinite set of solutions aligned along one (or two) >>>> vertical lines in the complex plane. >>>> (If I am not mistaken!) >>>> >>>> Louis >>>> >>>>> On 29 Jan 2018, at 07:19, Rob Hodgkinson <[email protected]> wrote: >>>>> >>>>> @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 >>>> >>>> ---------------------------------------------------------------------- >>>> 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
