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 Mar 1, 2013, at 6:45 AM, Aai <[email protected]> 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 decimals.
limit=: 2 :'({~1 i.~(}.=}:))<.&.((10^11)&*) u y v 0.5^i.1076'
tests:
((4%~])* 3 o. 1r2p1 * 1-]) limit + 0
0.159155
(_4 0 1&p.%_2 1&p.) limit + 2
4
(1&o.%]) limit + 0
1
% limit + 0
_
% limit - 0
__
%@^. limit + 0
0
%@^. limit - 0
0
%@^. limit + 1
_
>. limit - 0
0
>. limit + 0
1
On 28-02-13 22:28, km wrote:
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 1000
1000000
dq =: 1 : (':'; 'y %~ (u x+y) - u x') NB. difference quotient
2&(^&3 dq)lim 0 NB. derivative of x^3 at 2 is 12
12
--Kip
Sent from my iPad
On Feb 28, 2013, at 7:19 AM, Raul Miller <[email protected]> wrote:
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 should be a consistent number.
Note that (2) means that _ and __ will typically not be returned,
since they are inconsistent numbers (but, since they are inconsistent,
it's impossible to make an entirely consistent guarantee about their
treatment).
(1&o.%]) lim 0
1
% lim 0
2.67877e306
-@% lim 0
_2.67877e306
For my purposes, these "e306" values are close enough to infinity to
be treated as such.
Note also that I'm probably being a bit too aggressive with the number
of epsilon values I'm using.
If you really want _ and __ results, you could use something like this:
lim=: (1 :0)("0)
tests=. u ((1e_6*1>.|y)*0.5^i.1000)+y
1e_3*1e3* tests {~{.I.((1 }. 0&~:) * 2 ~:/\ ])(,2:)(*!.0)2 -/\ tests
)
However, note that this is a heuristic and its ability to force the
result to be an inconsistent infinity depends on the stability u (and
also depends on the actual limit value). I place less faith in this
mechanism than in the ability of the user to recognize that the result
should be thought of as infinite (and if the user does not understand
what's going on well enough to make that determination it's hard to
imagine how this distinction could be useful).
FYI,
--
Raul
On Wed, Feb 27, 2013 at 10:55 PM, km <[email protected]> wrote:
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
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