You especially need guardrails if you try something like this:
_2 0 1&p. Newton 1 NB. OK - square root of 2
1.41421
_2 0 1&p. Newton 1x NB. Try extended precision
C-c C-c|break NB. After waiting a while...
| _2 0 1&p.Newton 1
NB. Failure to terminate...
On Wed, Feb 5, 2020 at 7:34 AM Henry Rich <henryhr...@gmail.com> wrote:
> I misread your function.
>
> (^~ - +:) Newton 1.1
> 0.346323j1.2326e_32
> (^~ - +:) Newton 0.5
> 0.346323
>
> Still need those guardrails!
>
> Henry Rich
>
> On 2/5/2020 2:21 AM, Skip Cave wrote:
> > In "Fifty Shades of J" chapter 23, the Newton Raphson algorithm is
> > described thusly:
> >
> > Newton =: adverb : ']-u%(u D.1)'(^:_)("0)
> >
> > How would that be defined using the new derivative verbs?
> >
> > Also, what is the replacement for d.?
> >
> > How would I find the roots of (x^x)=2*x using Newton Raphson?
> >
> > Skip
> >
> > Skip Cave
> > Cave Consulting LLC
> > ----------------------------------------------------------------------
> > For information about J forums see http://www.jsoftware.com/forums.htm
>
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
>
--
Devon McCormick, CFA
Quantitative Consultant
----------------------------------------------------------------------
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