Louis mentions a different Newton Raphson verb:
VN=: 1 : 0
- u %. u D.1
)
This one works on either form of the equation:
v =: 3 : '8 + (2^y) - (2^2*y)'
vv =: (8: + (2&^) - (2&^)@(2&*))
vv VN (^:20) 1
1.75372489415532
v VN (^:20) 1
1.75372489415532
Skip
Skip Cave
Cave Consulting LLC
On Mon, Jan 29, 2018 at 11:15 AM, Skip Cave <[email protected]> wrote:
> What are the rules required to re-format a verb to make it processable by
> N?
>
> The original equation:
> (2^x) - 2^2*x = _8
>
> The reformatted equation:
> (8: + (2&^) - (2&^)@(2&*))
>
> Why the colon?
>
> Why the @ sign?
>
> Why the ampersand?
>
> Why not make a monadic verb?:
>
> v =: 3 : '8 + (2^y) - (2^2*y)'
> vv =: (8: + (2&^) - (2&^)@(2&*))
>
> v 1
>
> 6
>
>
> vv 1
>
> 6
>
>
> NB. Both verbs are the same, but:
>
>
> v N (^:20) 1
>
> |domain error
>
> | v N(^:20)1
>
>
> vv N (^:20) 1
>
> 1.75372
>
>
> ??
>
>
> Skip Cave
> Cave Consulting LLC
>
> On Mon, Jan 29, 2018 at 3:12 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
>>
>> ----------------------------------------------------------------------
>> For information about J forums see http://www.jsoftware.com/forums.htm
>>
>
>
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