What does 'best' method mean? Bisection is OK
f=. 3 : '(-2^2^y) + (2^y) +8'
f 1.818415 1.81842 1.61514e_5 _6.92891e_5
Den 16:04 mandag den 29. januar 2018 skrev Raul Miller
<[email protected]>:
Oops, yes, sorry, careless of me.
Thanks,
--
Raul
On Mon, Jan 29, 2018 at 4:12 AM, Rob Hodgkinson <[email protected]> wrote:
> Minor correction Raul (- instead of +) ?
> ... use (8: + (2&^) - (2&^)@(2&*))
>
> So Skip, just to clarify to see this solution put through Newton-Raphson;
>
> 1) Create the Newton-Raphson adverb as you stated earlier;
> N=: 1 : '- u % u d. 1’
>
> 2) Apply for a number of iterations using ^: and give it an initial value of
> 1 … fn N (^:iterations) start-value
> (8: + (2&^) - (2&^)@(2&*))N (^:20) 1
> 1.75372
>
> 9!:11]14 NB. or increase print precision
> (8: + (2&^) - (2&^)@(2&*))N (^:20) 1
> 1.7537248941553
>
> Hope that clarifies things, as without the function appropriately defined as
> Raul described, N returns a Domain Error if it can’t work with the function.
>
> Rob
>
>> On 29 Jan 2018, at 7:57 pm, Raul Miller <[email protected]> wrote:
>>
>> d. 1 wants to be able to use the chain rule for 2^2*x, and it seems
>> like the implementation was from an early version of J, and has not
>> kept up with all the more recent changes. So, you should put that
>> changed term into an f@g form.
>>
>> In other words, use (8: + (2&^) + (2&^)@(2&*))
>>
>> Thanks,
>>
>> --
>> Raul
>>
>>
>> On Mon, Jan 29, 2018 at 1:46 AM, Skip Cave <[email protected]> wrote:
>>> I see what I did wrong.
>>>
>>> The equation is: 8 + (2^x) - 2^2*x = 0
>>>
>>> The third term is (2^2*x) not (2^2^x)
>>>
>>> That should get close to the answer x=1.75372
>>>
>>> I'm mostly interested in how to formulate the code to implement the Newton
>>> Raphson solution
>>> using
>>>
>>> N=: 1 : '- u % u d. 1'
>>>
>>> In the NR code, where does the equation verb go?
>>> How does it need to be structured? Where does the iteration count limit go?
>>>
>>> Skip
>>>
>>>
>>>
>>>
>>> Skip Cave
>>> Cave Consulting LLC
>
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