Right, I remember now you mentioned that visualization tool before. It
turns out that Skip function actually was,
f=. 8 + (2 ^ ]) - (2 ^ 2) ^ ]
instead of,
g=. 8 + (2 ^ ]) - 2 ^ 2 ^ ]
which I was entertaining earlier last night.
Coincidentally, today I was exploring some solutions for f...
Indeed I should’ve mentioned it; I was referring to my earlier message:
0 = (4^z) - (2^z) - 8
= _8 _1 1 p. 2^z
so 2^(x+iy) = z = (1 +- (%:33))/2.
The rest should be clearer now.
Thanks,
Louis
> On 29 Jan 2018, at 18:43, Raul Miller wrote:
>
> I do not know which of j expressions which could
I do not know which of j expressions which could have matched original
expression you were referring to.
Thanks,
--
Raul
On Mon, Jan 29, 2018 at 12:39 PM, Louis de Forcrand wrote:
> Yes, typed a little fast.
>
> sqrt33 = %:33.
>
> Why does it not make sense?
>
> Louis
>
>> On 29 Jan 2018, at
Yes, typed a little fast.
sqrt33 = %:33.
Why does it not make sense?
Louis
> On 29 Jan 2018, at 18:35, Raul Miller wrote:
>
> What is sqrt33 here? (I would have guessed %:33 but that does not make
> sense to me.)
>
> Thanks,
>
> --
> Raul
>
>
>> On Mon, Jan 29, 2018 at 12:31 PM, Louis de
I would recommend using D. instead of d. in general for more complicated or
explicit verbs. (lower case) d. assumes your verb is a scalar verb and tries to
perform a symbolic differentiation on tacit verbs only, while D. differentiates
numerically.
Not all verbs can be differentiated with d., b
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 Consultin
What is sqrt33 here? (I would have guessed %:33 but that does not make
sense to me.)
Thanks,
--
Raul
On Mon, Jan 29, 2018 at 12:31 PM, Louis de Forcrand wrote:
> I skipped a few steps there. With pencil and paper, I find that (using
> standard notation)
>
> 2^(x+iy) = (1 +- sqrt(33)) / 2
>
>
A rule I was working with was that if the individual terms were
differentiable, then their sums would be differentiable.
This turns out to be slightly more rigorous than what's actually required:
8: d. 1
0"0
8 d. 1
|nonce error
but
(8 + ]) d. 1
1x"0
In other words, 8: is a verb that d
I skipped a few steps there. With pencil and paper, I find that (using standard
notation)
2^(x+iy) = (1 +- sqrt(33)) / 2
Yet 2^(x+iy) = 2^x * 2^iy, and because the whole is real, 2^iy must be real.
Moreover, when y is such that 2^iy is real (when y is a multiple of Pi/log2)
then
2^iy
= exp(lo
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
:
Oops, yes, sorry, careless of me.
Thanks,
--
Raul
On Mon, Jan 29, 2018 at 4:12 AM, Rob Hodgkinson 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
Jose Mario Quintana wrote:
> but, it is not the only one,
>
> (X=. (8 + (2 ^ ]) - (2 ^ 2) ^ ])New (^:22) 0.5j_0.5)
> 1.24627511j4.53236014
There is way more than that.
If you use "phase portrait visualization tool" (e.g. from here
http://code.jsoftware.com/wiki/User:Andrew_Nikitin/Phase_port
Oops, yes, sorry, careless of me.
Thanks,
--
Raul
On Mon, Jan 29, 2018 at 4:12 AM, Rob Hodgkinson 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-Ra
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 i
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
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