N is at most times sensitive to its start value ...
it does work in this case (but as you stated, it will not be
neccessary in this case):
(4.9+2&%) N^:1 (_0.1)
_0.1755
(4.9+2&%) N^:2 (_0.1)
_0.275539
(4.9+2&%) N^:4 (_0.1)
_0.403614
(4.9+2&%) N^:14 (_0.1)
_0.408163
(4.9+2&%) N^:_ (_0.1)
_0.408163
-M
At 2020-10-24 15:52, you wrote:
I just realized that I don't need Newton: 2/x + 4.9 = 0? 2/x = 49/10
2 = 49/10 * x x = 2/(49/10) = 2*(10/49) = 20/49 = 0.4081632653 Skip
Cave Cave Consulting LLC On Sat, Oct 24, 2020 at 10:12 AM Skip Cave
<[email protected]> wrote: > The Newton Raphson root finder
using the new J definition of derivative is: > > N=: 1 : '- u % u
deriv_jcalculus_ 1' > > > How would I use N to solve for x in the
equation 2/x + 4.9 = 0? > > > Skip Cave > Cave Consulting LLC >
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