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
