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

Reply via email to