Also if you multiply by x to get
2 + 4.9x = 0
You can solve using p.
p. 2 4.9
┌───┬─────────┐
│4.9│_0.408163│
└───┴─────────┘
On Sun, 25 Oct 2020, 09:47 Martin Kreuzer, <[email protected]> wrote:
> 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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>
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