Re: overflow question
Alexandre Goldsztejn wrote:
On Tue, May 8, 2012 at 3:01 PM, Nate Hayes <nh@xxxxxxxxxxxxxxxxx> wrote:
Alexandre Goldsztejn wrote:
On Mon, May 7, 2012 at 4:24 PM, Nate Hayes <nh@xxxxxxxxxxxxxxxxx> wrote:
Alexandre Goldsztejn wrote:
As the discussion has eventually started, could you use my email as a
first
thread in this discussion and explain the interpretation in terms of
overflow of the computation
f(x)= 1+x^2 (1+sin(x))
f((-oo,+oo))=[1,+oo]
to the standard's list?
Yes.
The input to this function would be [-omega,+omega]. Since
X = upsilon([-omega,+omega]) = [-oo,+oo]
for FTIA we then have at Level 1a:
f(X) = [1,+oo]
which we may safely re-interpret as
omega(f(X)) = [1,+omega].
Nate
I am not sure of how this interpretation in term of overflow can be
used. In particular, is the non-existence of solution to f(x)=0 a
consequence of the overflow-interpretation of this computation?
Zero is not an element of any interval in the overflow family [1,+omega];
similarly, it is not an element of the unbounded interval
upsilon([1,+omega]) = [1,+Inf]
which represents the union of all intervals in the overflow family. Under
either interpretation, it is proof of non-existence of the solution
(c. f. Proposition 1).
Nate
As far as I understand, if I want to compute the image of a function
over an unbounded interval X, then first I should build a family of
overflown intervals Upsilon(X), then compute the image of each
interval of this family of intervals by F(Upsilon(X)), and then
compute the union of the resulting intervals Omega(F(Upsilon(Z))).
Almost, though it seems to me you might have it backwards. If X is an
ubounded interval, then Z = Omega(X) builds a family of intervals and
Upsilon(Z) takes the union of all these intervals. So in computation, if one
*starts* with family of intervals Z, then Omega(F(Upsilon(Z))) actually
means just the opposite order you mention above, i.e.,
a. compute the union X of the overflow family Z
b. compute the image F(X) of the *single* interval X
c. re-interpret the image F(X) as new family Omega(F(X))
So for FTIA we always are simply evaluating F(X) for a single interval
X \in overline-IR,
just as P1788 currently defines, i.e., that much does not change at all.
Here, F applies to an infinite set of intervals and returns an
infinite set of intervals, so formally:
F: P(IR^n) --> P(IR)
F(XX)={hull(range(F,X)):X in XX}
where P(E) is the set of all (bounded nonempty) subsets of E. Am I
correct?
Very close, but see my notes above (once one has the interval X \in
overline-IR, FTIA is defined *exactly* the same as in the current draft
standard text).
To be more homogeneous, I think Omega should return a set containing
only one interval in case there is no overflow, so that the process
above applies to both bounded and unbounded intervals, otherwise two
cases have to be handled.
Yes. This is exactly what Omega(X) does (see definition (6) in the paper in
the case X=[a,b] is a subset of the interval H=[-h,h]).
Nate