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
