Won't (x^x)-(2*x) = 0 have two roots? A real one, and a complex one?
Will Newton Raphson find both?
Skip
On Wed, Feb 5, 2020 at 12:08 PM Henry Rich <henryhr...@gmail.com> wrote:
> Yeah, a rational y wouldn't ever quite satisfy 0 = _2 0 1 p. y
>
> Henry Rich
>
> On 2/5/2020 1:04 PM, Devon McCormick wrote:
> > 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
> >>> ----------------------------------------------------------------------
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> >> ----------------------------------------------------------------------
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> >>
> >
>
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