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Re: Motion on interval flavors



John, p1788


My apologies for replying very late.

IMHO midrad intervals can be defined at level 1.

If  at level 1 given  m1 a    midrad interval 

we can define a reference midrad  interval value m2 for f(m1) 


an  analog  to the FTIA theorem can be asserted for midrad intervals.

Given  m3  an e effective midrad interval  value calculated for f(m1)
we must have 

     m2 \subseteq m3  .

With m2 = (r2, w2), m3  = ( r3, w3)
this gives r2 = r3 and w3 >= w2


Let me propose a tentative approach to the reference value m2.

It is based on the notion of super interval which is first defined

Since the interval set including unbounded intervals is a
lattice with respect to    <=_I order relation define by

    [a,b ]  <=_I [c,d]  iff  a<= c AND  b <= d 

we can defined SUPER INTERVALS as interval defined on the set of     intervals

A super interval S1  =  [ [a, b], [c, d] ] can be defined as 


 S1 = { [x, y]  | a <= x <= c AND b<= y <= d }

where [a,b] , [c,d]  and [x, y] are interval

It should be noted that empty is not a member of any super interval.


Given a point function  f  we can define 


f(S1) = { f([x,y]) | [x,y] \in S1}


To f(S1)  we can associate a super interval  FS1 = [[f1, f2] , [f3, f4] ]

with 
         f1 =  inf_[x,y]  inf( f([x, y]))

         f2 =  inf_[x,y]  sup (f[x, y]))

        f3 =  sup_[x,y]  inf(f[x, y]))

         f4 =  sup_[x,y]  sup f[x, y].

Let us now come back to midrad  intervals.

To a super interval S2 = [[a, b],[b,d]] such a+d = 2*b


we can associate a midrad   interval mr1  = (m, w ) of midpoint m and width w
 such that  m = b and w = d - a


For any function defined at x = b


we have f(S2)  = [[f1, f2] , [f2, f4]] 


with f2 = f(b)

f1 = inf f[a,d]

f4 = sup f[a,d]

In general the condition delta  = f1 + f4 - 2* f2 is not zero. So that the
result is not a midrad   interval 

However   a midrad super interval    F = [[f1bis, f2] , [f2, f4bis]] 


such that if delta > 0  
       f1bis = f1- delta
       f4bis = f'

and similarly
      f1bis = f1
      f4bis = f4 - delta if delta < 0

We thus have the reference a midrad reference (m1, w1) for f(S1)
with m1 =f(b) and w = f4bis - f1bis = f4 - f1 + |delta|

Since the of effective   calculation of   f(m, w)  by an expression  may need
a  width correction at each step, it is expected that   the effective 
width   will be greater that the reference width



BTW I would like to notice that   super interval might be an interesting
framework for interval extensions  since 

The set of modal interval  [b, c] structured with  usual arithmetic operations 
  is isomorphic to the the set  of super intervals of
the form      

      [[-inf, b], [c, +inf] ]  with the same operations


This suggest that all the available extension of intervals should defined by
an association to some subset of super intervalS

The flavors associated to the extensions the would be  the condition  used in
the definition of corresponding super interval set.

Starting with a super interval  defined as    [ [a, b], [c, d] ] 
we may have the flavor

modal <=> a = -inf  AND d   + inf 

bounded <=>   b # + inf AND c # - inf

number  <=>    b = c

centered <=>  a + d = b + c 

etc

Sincerely,

Dominique       











Such super intervals may be of interest for interval extension since 

  


On Sat, 23 Jun 2012 10:05:20 +0100, John Pryce wrote
> Dominique
> 
> On 22 Jun 2012, at 16:26, Dominique Lohez wrote:
> >       3)  The analog of the FTIA for midrad intervals requires the function
> > is defined but not necessarily continuous.
> 
> Can you clarify? FTIA is primarily a Level 1 concept so midrad 
> should be irrelevant. If this is a Level 2 fact, can it be stated 
> for general implicit interval types?
> 
> John Pryce


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