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
Thanks, that's interesting. It seems a little cumbersome, but it certainly
works.
It seems to do a lot of appending, but since it's continually building up
the series, it never duplicates work. It can only do odd-length output
sequences, but the even sequences are only a }: away.
On Thu, Feb 28,
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
Kip in
http://www.jsoftware.com/pipermail/programming/2013-February/031712.htmlpresented
the argument assuming that the square is perimeter-area optimal
for the rectangular polygons. The question is: can one show the implicit
assumption without calculus?
I think so: one only has to show that x=c
Here's one approach:
(, (, +:)@>:@{:)^:3]1
1 2 4 5 10 11 22
--
Raul
On Thu, Feb 28, 2013 at 10:26 AM, Johann Hibschman wrote:
> How do you generate a series produced by alternate application of two
> monads?
>
> For example, if I alternate >: and +: starting from 1, I get:
>
> 1 2 4 5 1
How do you generate a series produced by alternate application of two
monads?
For example, if I alternate >: and +: starting from 1, I get:
1 2 4 5 10 11 22 ...
I can generate every other item in the series via (+:@>:)^:(i.6) 1, and I
get close with (([: >: ])`([: +: ]))/\6#1, but the orderi
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
