[Jprogramming] Derivatives

[Henry Rich]

See https://arxiv.org/pdf/math/9205211.pdf esp. p. 6

Henry Rich

On 2/9/2020 12:29 AM, Don Kelly wrote:
> Reading again, I apologize for mistaking x^x for x*x- and getting an 
> easy to get polynomial with 2 roots . Your original x^x =2*x is not 
> actually a normal polynomial.Newton could be used but  the derivative 
> is  so messy that a cut and try approach seems easier.
>
> When you do to get 0^x and get _ _ _ 1 0 0 0 what you should get is
>
> _ _ _  HUH? 0 0 0  where i am using HUH? to indicate that  for x=0 the 
> result is indeterminant.  Creep up on 0   and no matter how close to 
> 0  you get -the results are the same. Crossing 0  is a discontinuous 
> jump directly from infinity to 0
>
> 0^ - 1e1000
>
> _
>
> 0^ 1e1000
>
> 0
>
> 0^ 10^6
>
> 0 NB: the N^0 rule doesn't apply for N=0 but J isn't recognizing this
>
> Now if you use (10^6)^0 the result is 1 as N^0 rule has a non zero 
> value.**Don
>
>
> On 2020-02-06 6:23 p.m., Skip Cave wrote:
>
>> ]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 <dhky at 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 <
>>>> programming at 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 <
>>>>> henryhrich at 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 <henryhrich at 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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