Indeed I should’ve mentioned it; I was referring to my earlier message: 0 = (4^z) - (2^z) - 8 = _8 _1 1 p. 2^z
so 2^(x+iy) = z = (1 +- (%:33))/2. The rest should be clearer now. Thanks, Louis > On 29 Jan 2018, at 18:43, Raul Miller <[email protected]> wrote: > > 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 ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
