Re: Unbounded intervals
Vincent Lefevre wrote:
On 2012-04-26 10:03:40 -0500, Nate Hayes wrote:
Vincent Lefevre wrote:
>On 2012-04-25 09:05:19 -0500, Nate Hayes wrote:
>>A few points:
>>
>> -- No computer (that I'm aware of) can numerically prove hardly
>>anything
>>useful about the domain of a function beyond the underlying numeric
>>limits
>>of the system; so for this reason alone, truly unbounded intervals are
>>never
>>necessary in numeric models or computations (you have never answered my
>>original question from long ago to show a counter-example of this).
>
>I disagree. If the user asks for the range of 1/[0,1], the math result
>is [1,+inf]. This is useful information, at least much more than an
>error.
You've avoided the question again.
I'm not avoiding it. This is a perfect answer to your question.
Others have given other examples.
The question is, what is an example of an interval algorithm, for
example, that proves all the zeros of a function on the domain
[MAXREAL,+Inf].
This question is pointless. If there is an unbounded interval in
the problem, then you need unbounded intervals to be able to express
the problem!
You've missed the point again: there is always some finite limit to which we
can prove anything useful. Hence the input [MAXREAL,+OVR] is perfectly
suitable input for such an algorithm.
>>for this reason it is possible to define
>>midpoint([1,+OVR]) at Level 1 in the same way P1788 is currently
>>considering to do so at Level 2.
>
>How would you define it at Level 1, as a *real number*?
At Level 2 P1788 has considered to define midpoint([1,+Inf]) =
(1+REALMAX)/2
or something similar.
No, it is REALMAX, if one really wants the midpoint to be defined.
At Level 1, midpoint([1,OVR]) could for the same reasons be
similarly be defined (1+H_f)/2.
This would be bad, even if you define it as H_f (a bit like in Level 2),
because at Level 1, you have intervals like [1,4*H_f], whose Level 1
midpoint is 2*H_f+1/2 > H_f.
No. H_f is the overflow threshold, just like REALMAX at Level 2. So, for
example
4*[1,OVR]=[4,OVR]
and midpoint([4,OVR]) would then still be H_f.... just like at Level 2.
Thus conventional inequalities (see the
discussion about midpoint) would no longer hold at Level 1.
>> -- Replacing "midpoint" with "any member of the interval" gives a
>>valid
>>mathematical definition of the Interval Newton, but such a definition
>>is
>>also then no longer an algorithm because the exact method of choosing
>>"any
>>member of the interval" is left undefined.
>
>Well, once the one who writes the algorithm choose the member in
>question, he has an algorithm. This member can potentially be a
>parameter, and it will still be an algorithm. I don't see any problem
>with that!
The problem is one cannot choose midpoint as this parameter at Level 1
without causing the Interval Newton to be undefined when unbounded
interval
is provided as input.
This is not a problem. You don't necessarily need a midpoint at
Level 1 for this algorithm.
Of course not, but if I choose midpoint the algorithm is undefined at
Level 1. That's a fact, Vincent.
You are the one who said unbounded intervals are important to ensure
algorithms are easy to define at Level 1. But this is clearly not
always the case.
The problem is just with your reasoning.
To me, it simply appears you do not like to accept facts when they are
inconvenient to your position or argument.
Nate