]x=.i:3
_3 _2 _1 0 1 2 3
0^x
_ _ _ 1 0 0 0
x=.0
(x^x)=2*x
0 NB. Zero is not a root
(x^x)=2*x
0
x^x
1
2*x
0
So why does Newton Raphson show a zero root for (x^x)=2*x?
(^~ - +:) Newton 2
2
(^~ - +:) Newton 1
0 NB. Something wrong here!
(x^x)=2*x ?
Skip Cave
Cave Consulting LLC
On Thu, Feb 6, 2020 at 5:38 PM Don Kelly <d...@shaw.ca> wrote:
> Bo is correct
>
> There is a "zero rule" so n^0 =1 if n is not equal to 0. Bo is correct
>
> the only possible 0^0 is the same as 0*0
>
> Possibly the coding for ^ should take this into account.
>
>
>
> On 2020-02-05 10:14 p.m., Skip Cave wrote:
> > Bo,
> > The original equation is:
> > (x^x)=2*x
> > 2 is clearly a root. 0 is clearly not, as 0^0 = 1
> >
> >
> > Skip Cave
> >
> >
> >
> > On Wed, Feb 5, 2020 at 2:37 PM 'Bo Jacoby' via Programming <
> > programm...@jsoftware.com> wrote:
> >
> >> Skip. (x^x)-(2*x) = x*(x-2) is zero for x=0 and x=2. Two real roots.
> >> Newton Raphson finds one of these depending on the value of the initial
> >> guess.
> >> Bo.
> >> Den onsdag den 5. februar 2020 19.08.23 CET skrev Henry Rich <
> >> henryhr...@gmail.com>:
> >>
> >> 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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