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Re: overflow question

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