Thread Links Date Links
Thread Prev Thread Next Thread Index Date Prev Date Next Date Index

Re: KISS-decorations



Dear Marco, p1788

I like your paper.
 I especially appreciate
                The single  order relation between decorations
                 THe mention of isotonicity
The wide coverage of the various situations ( set operators, comparisons, FTDIA)
Our position are very close

IMHO The main weakness of our motion is
      No intensional definition of the order relation is given
So  the posssible decoration list is definitely   frozen.
In contrast this could lead to endless discussions on the choice of the possible decorations as well as on their ordering.

Dominique

PS
  BTW   Your second statement of theorem 1 is again wrong
Since f(x)=x is continuous the John's counterexample is again working. f(x) is a function defined by an expression in the John's wording. F(X) is THE extension calculated by a interval transcription of the expression.
                 Then for F(X)   the isotonicity holds.
Marco Nehmeier a écrit :
Am 29.06.2011 07:56, schrieb John Pryce:
Marco, Jürgen, P1788

On 28 Jun 2011, at 21:13, Jürgen Wolff von Gudenberg wrote:
we have prepared a third motion on decorations that clearly follows the KISS principle. Perhaps it is too simple, but we are convinced that it works better than the two current approaches. We only have 4 different decorations and one linear order.
We are eager to read your comments.
Perhaps we can prolong the discussion time again? but there is not much discussion(!)

Thank you. I like the push back towards simplicity. Arnold has more-or-less completed his new FTDIA proof and has sent me a part for checking. I think we both want to revert from the present 7-decoration idea, and get rid of boundedness. Yes, please chairman, can we extend the discussion period?

That said, I have some criticisms.

1. Theorem 1 on p2 is surely well known to be false.
    Let f(x)=x for real x. Define F: IR ->  IR by
       F(xx) = Entire if 0 not in xx, and xx otherwise.
    Then for any xx, and x in xx, we have f(x) in F(xx), so F
    is an interval extension of f.
    But xx=[1,2], yy=[0,2] give F(xx)=Entire, F(yy)=[0,2], so
    isotonicity fails.

2. Theorem 2 (1st part) is carelessly stated, in fact meaningless as
    written IMO.
    It's not "For every interval extension F ...", but "For the
particular interval *function* F that results from applying straight-
    forward interval computation ..., F(xx) encloses the range of f
    over xx" -- or better, "... encloses the range over xx of the point
    function defined by the expression f."

3. And surely Theorem 2 (2nd part) needs some continuity assumption,
    because of the effect of taking the hull. I think it's false for
    f(x)=1/x with xx=[-1,1]. And for sign(x) with xx=[0,0.5].

Further, and I trust more constructive, remarks to follow.

John



Dear John, P1788,

thank you for the quick comments.
You where right with your criticism.
We have changed the sections 1.2 and 1.4 according to your advice.

Please find attached the new version 1.2 of our motion (draft).

Best regards

Jürgen and Marco



--
Dr Dominique LOHEZ
ISEN
41, Bd Vauban
F59046 LILLE
France

Phone : +33 (0)3 20 30 40 71
Email: Dominique.Lohez@xxxxxxx