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
>>> ----------------------------------------------------------------------
>>> For information about J forums see http://www.jsoftware.com/forums.htm
>> ----------------------------------------------------------------------
>> For information about J forums see http://www.jsoftware.com/forums.htm
>>
>
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