Re: A question Level 1 <---> level 2 mappings; arithmetic versus applications

["Chenyi Hu" <chu@xxxxxxx>]

When one endpoint of an interval [a,b] is unboundedould, would  left-rad and right-rad be helpful? For example, [-1, infty) could be expressed as mid=0, left-rad = 1 and right-rad = infty as [1; 0; infty]. Of course, the mid can be 2, 3, or whatever between (-1, infty). This might not be bad since it could provide user a choice in application.

Chenyi

>>> "Corliss, George" <george.corliss@xxxxxxxxxxxxx> 07/01/10 5:25 ?? >>>
On Jun 30, 2010, at 3:41 PM, Nate Hayes wrote:

> Yes.
> Possibly semi-infinite intervals can be considered invalid construction for mid-rad types, similar to how [-Inf,-Inf] is invalid construction for inf-sup. So there may be connections to the exception handling mechanism for these cases.
RIGHT. A semi-infinite interval is in a philosophical sense exceptional in inf-sup, too.

If we are really clever, we may be able to devise an exception-handling mechanism that works in the same manner for both inf-sup and mid-rad.

George

> Nate
>
> ----- Original Message ----- From: "Michel Hack" <hack@xxxxxxxxxxxxxx>
> To: "stds-1788" <stds-1788@xxxxxxxxxxxxxxxxx>
> Sent: Wednesday, June 30, 2010 3:25 PM
> Subject: Re: A question Level 1 <---> level 2 mappings; arithmetic versus applications
>
>
>> Nate Hayes wrote:
>>> For example, any given Level 1 interval [a,b] has the property:
>>> [a,b] = (m;r)
>>> where (m;r) is a mid-rad interval and m=(a+b)/2, r=(b-a)/2.
>>
>> Any *bounded* Level 1 interval. At level 1 a+b and a-b don't exist
>> when either a or b is infinite. At level 2 the only possible mid-rad
>> result is Entire, which loses almost all information unless the level 1
>> interval was Entire too (in which case a+b does not exist at level 2,
>> so a special rule would be followed).
>>
>> Michel.
>> ---Sent: 2010-06-30 20:31:49 UTC

Dr. George F. Corliss
Electrical and Computer Engineering
Marquette University
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