[.]
Whereas...
u limit v
u (2 : '({~1 i.~[:(}.=}:)<.&.(1e10&*)) u y v 0.5^i.1076') v
... and, of course ;)
u=. (4 %~ ]) * 3 o. (2 %~ o.1) * 1 - ]
v=. +
Y=. 0
u lim v Y
0.1591549430837225
u limit v Y
|stack error: v
| ({~1 i.~[:(}.=}:)<.&.(100&*))u y v 0.
A numerical approach to finding limits is indeed tricky. I was coding a
tacit version of
Arie's conjunction...
(9!:11) 16 NB. Setting the print precision
( lim=. (@:(&(".@:('(0.5^i.1076)'"_ ({~ (1 i.~ (}. = }:) @:
(<.&.(1e11&*)))@:) )
(@: (&(".@:('(0.5^i.1076)'"_ (({~ (1 i.~ (}. =
}
Raul, I have modified your adverb lim as follows:
limr =: 1 : 0
lastgood =. ] {~ 0 i:~ 0 = y - y + ]
u y + lastgood (1e_6*1>.|y)*0.5^i.1000
)
% limr 0
5.35754303593134e306
NB. _
>. limr 0
1
NB. 1
<. limr 0
0
NB. 0
%@^. limr 1
4.5035996273705e15
NB. _
%@^. limr 0
_
NB. work on limit from right of a function
lim =: 1 : 'u 1e_11 * 1 + <. 1e11 * y' NB. Arne Groeneveld gave me this
idea.
NB. This version evaluates u at a value greater than y, never at y itself.
NB. It attempts to approximate the limit from the right of u at y .
% l
Different for me:
0=0.5^1022+i.5
0 0 0 0 0
and
0=0.5^1071+i.5
0 0 0 0 1
I noticed the following
0.5^1075
0
0.5^1074
4.94066e_324
and decided to use i.1076.
On 02-03-13 16:58, km wrote:
Arne, have you considered
0=0.5^1022+i.5
0 1 1 1 1
--Kip
Sent from my iPad
On
Arne, have you considered
0=0.5^1022+i.5
0 1 1 1 1
--Kip
Sent from my iPad
On Mar 1, 2013, at 6:45 AM, Aai wrote:
> Here's a conjunction with the choice of left/right approach.
>
> It's like Raul's approach except it uses a bit more values and it compares
> successive values up to a l
According to my description I have to formulate limit as follows:
limit=: 2 :'({~1 i.~[:(}.=}:)<.&.(1e11&*)) u y v 0.5^i.1076'
otherwise the values are cut off at 11 decimals.
BTW 11 is a value obtained by experimenting with
((4%~])* 3 o. 1r2p1 * 1-])
as input.
On 01-03-13 20:20, km wrote:
Thank you, Raul, especially for the "short form". I'll try at least the Wiki
article. --Kip
Sent from my iPad
On Mar 1, 2013, at 7:50 AM, Raul Miller wrote:
> For readers following along:
>
> IEEE 754 is a decent search term.
>
> Or, at least, it is for me - Google apparently customizes s
That's pretty good! Thank you, Arne. --Kip
Sent from my iPad
On Mar 1, 2013, at 6:45 AM, Aai wrote:
> Here's a conjunction with the choice of left/right approach.
>
> It's like Raul's approach except it uses a bit more values and it compares
> successive values up to a limited number of de
I think the machine epsilon concept
http://en.wikipedia.org/wiki/Machine_epsilon#Formal_definition )
is related to this thread discussion...
(%&2^:(1 (~:!.0) 1 + %&2)^:_)1 NB. 2^_52 as expected
2.22044605e_16
On Fri, Mar 1, 2013 at 8:50 AM, Raul Miller wrote:
> For readers following along:
>
For readers following along:
IEEE 754 is a decent search term.
Or, at least, it is for me - Google apparently customizes search
results so I cannot know for sure that they give good, relevant
results for everyone searching on that phrase (nor can I be sure
whether anyone I care about will get dif
Here's a conjunction with the choice of left/right approach.
It's like Raul's approach except it uses a bit more values and it
compares successive values up to a limited number of decimals.
limit=: 2 :'({~1 i.~(}.=}:))<.&.((10^11)&*) u y v 0.5^i.1076'
tests:
((4%~])* 3 o. 1r2p1 * 1-]) lim
I remember that comparisons with 0 are exact
0 = N - N + epsilon
0 0 0 0 0 0 0 1 1
which agrees with your first report. No, I am unfamiliar with the IEEE
structure, and am surprised by these results.
--Kip
Sent from my iPad
On Feb 28, 2013, at 4:21 PM, Raul Miller wrote:
> Are you fa
Are you familiar with the structure of IEEE 754 floating point numbers?
Consider, for example:
epsilon=: 2^_44
N=: 10^i:4
NB. this result reflects IEEE-754's structure
*N+epsilon-N
1 1 1 1 1 1 1 0 0
NB. this result reflects J's heuristic to deal with that structure
N=N+epsilon
Here is what I did
NB. right hand limit of a function
lim =: 1 : 0
value =. u y + (2^_44)
if. value <: - 2^40 do. __
elseif. value >: 2^40 do. _
elseif. do. value
end.
)
It does "reasonably well" but can be fooled, for example
] lim 2^40
_
Here it does better
*: lim
Here's a model implementation:
lim=: (1 :0)("0)
tests=. u ((1e_6*1>.|y)*0.5^i.1000)+y
tests {~{.I.((1 }. 0&~:) * 2 ~:/\ ])(,2:)(*!.0)2 -/\ tests
)
My assumptions are:
(1) the limit in question is relatively stable (that my choices for
epsilon are adequate)
(2) that the result of limit sh
Can you write an adverb lim so that
sin =: 1&o.
(sin % ])lim 0
1
% lim 0 NB. limit is from right
_
-@% lim 0
__
Kip Murray
Sent from my iPad
--
For information about J forums see http:/
Numerical is a bit tricky - you need to understand how many bits of
accuracy are retained by your calculation which in turn requires you
to understand what it is that you are modeling.
One approach, though [that often will work - not always, and it's
interesting to think about the failure modes] i
Trying yet again so Ceiling displays properly.
Sent from my iPad
On Feb 27, 2013, at 3:54 PM, km wrote:
> This may have been covered in another thread, but I invite you to write an
> adverb lim such that
>
> sin =: 1&o.
> (sin%])lim 0
> 1
> >. lim 0
> 1
>
> -- that is, the lim
Trying again so Ceiling displays properly.
Sent from my iPad
On Feb 27, 2013, at 3:54 PM, km wrote:
> This may have been covered in another thread, but I invite you to write an
> adverb lim such that
>
>sin =: 1&o.
>(sin%])lim 0
> 1
>>. lim 0
> 1
>
> -- that is, the limit is the
This may have been covered in another thread, but I invite you to write an
adverb lim such that
sin =: 1&o.
(sin%])lim 0
1
>. lim 0
1
-- that is, the limit is the limit from the right.
Kip Murray
Sent from my iPad
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