Note that if the equation really is (in traditional notation)
4^x - 2^x - 8 = 0
then it can be rewritten as
y^2 - y - 8 = 0, y = 2^x
and solved in closed form as well,
yielding a countably infinite set of solutions aligned along one (or two)
vertical lines in the complex plane.
(If I am not mi
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 d
@Skip et al …
also apologies for my sill definitions, I should have used y inside the
definitions not x (!!!), sorry if I confused the issue…
as in here for the first interpretation …
x,"0 (3 : '8+(2^y)-((2^2)^y)') x=:1.6+0.05*i.8
…/Rob
> On 29 Jan 2018, at 3:49 pm, Jose Mario Quintana
> wr
In that case,
(X=. (8 + (2 ^ ]) - (2 ^ 2) ^ ])New (^:22) 1)
1.75372489
is a root,
(8 + (2 ^ ]) - (2 ^ 2) ^ ])X
0
but, it is not the only one,
(X=. (8 + (2 ^ ]) - (2 ^ 2) ^ ])New (^:22) 0.5j_0.5)
1.24627511j4.53236014
(8 + (2 ^ ]) - (2 ^ 2) ^ ])X
8.8817842e_16j7.72083702e_15
On
Hmm...
I had originally thought about calling out the (2^2)^x interpretation
as a possibility, because rejected that, because that would be better
expressed as 4^x
But it's possible that Skip got the 1.75379 number from someone who
thought different about this.
And, to be honest, it is an ambigu
@Skip
Skip, I am a confused in your original post… your actual post read;
What is the best iterative way to solve this equation:
(-2^2^x) + (2^x) +8 =0
then later to Raul,
0 = 8 + (2^x) - 2^2^xNB. Is correct, and the answer is real
The answer i
Moreover, apparently there is at least another solution,
((-2^2^X) + (2^X) +8 ) [ X=. 2.9992934709539156j_13.597080425481581
5.19549681e_16j_2.92973749e_15
On Sun, Jan 28, 2018 at 7:28 PM, Jose Mario Quintana <
jose.mario.quint...@gmail.com> wrote:
> Are you sure?
>
>u New
> - (u %. u D
Are you sure?
u New
- (u %. u D.1)
,. (8 + (2 ^ ]) - 2 ^ 2 ^ ])New (^:(<22)) 1
1
3.44167448
3.25190632
3.03819348
2.7974808
2.53114635
2.25407823
2.00897742
1.86069674
1.82070294
1.81842281
1.81841595
1.81841595
1.81841595
1.81841595
1.81841595
1.81841595
1.81841595
1.81841595
1.8
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 belie
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 wrote:
> If Newton's method converges,
Oh, yes... I see the problem:
13 :' 8 + (2^y) - 2^2^y'
8 + (2 ^ ]) - 2 ^ 2 ^ ]
13 :' 8 + (2^y) - 2^2^y' d.1
|domain error
Instead, use:
8: + 2&^ + (2&^)@(2&^)
Thanks,
--
Raul
On Sun, Jan 28, 2018 at 5:52 PM, Skip Cave wrote:
> Raul,
>
> You had it right in the first place.
>
> 0 =
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^xNB. Is correct, and the answer is real
The answer is close to 1.75379
I wanted
@Brian
Tacit bz would certainly be harder to maintain than BZ for beginners like me
bz=: '0'"_ ` (I.@:(' '&=)@]) ` ]}~
Case in point : why is there a ~ at the end of bz?
Sometimes the answer is in front of me but I take a while to recognize it :
the gerund form of amend needs a left argumen
Raul,
You had it right in the first place.
0 = 8 + (2^x) - 2^2^xNB. 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-
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 b
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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