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 <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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