Also it is a good thing if you can plot the function for visual inspection.
t. Esa
-----Original Message-----
From: Programming <programming-boun...@forums.jsoftware.com> On Behalf Of Henry
Rich
Sent: Friday, February 7, 2020 12:18 PM
To: programm...@jsoftware.com
Subject: Re: [Jprogramming] Derivatives
Newton's method is good for final polishing of a root when you are very
close. It can do weird things when you are not close. Like here. Study
this case and see what happened.
Henry Rich
On 2/6/2020 9:23 PM, 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 <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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