Since Newton-Raphson is mentioned, I’d like to throw in (even though it might
be mentioned on the wiki page) that
VN=: 1 : 0
- u %. u D.1
)
is a kind of holy grail.
It can iteratively find roots not only of complex scalar functions, but also
complex vector and even tensor functions I believe. Of course accuracy and
initial guesses might become problematic, but I still find it quite cool for
such a simple function to be so powerful.
Best,
Louis
> On 29 Jan 2018, at 00:07, Raul Miller <[email protected]> wrote:
>
> 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
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>>
>>
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