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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