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
