Yeah, I guess it should be fine to just look at the sequence to see if
it's converging, or test the result.

My thought was that iterations are cheap, I just wanted a small finite
number of them.

Thanks,

-- 
Raul

On Sun, Jan 28, 2018 at 6:05 PM, Henry Rich <[email protected]> wrote:
> If Newton's method converges, you won't need a couple of hundred rounds -
> just a dozen or so.
>
> Henry Rich
>
>
> On 1/28/2018 5:52 PM, Skip Cave 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/forums.htm
>>>
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>>
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>
>
>
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