I haven't thought about this particularly deeply, but here are some
things that don't seem to work with
limits returning intervals.
suppose limit(f(x),x,a)= F and limit(g(x),x,a) =G and both limits
exist. then
(theorem 1)
H= limit(f(x)+g(x),x,a) = F+G.
now let f(x)=sin(x), and g(x)=1-sin(x).
limit(f(x),x,inf) ={?} [-1,1]
limit(g(x),x,inf)={?} [-1,1]
hence by theorem 1, limit(f(x)+g(x),x,inf) = {?} [-1,1]+[-1,1] = [-2,2]
unfortunately, the limit should be the same as limit(0,x,inf)=0, so the
conclusion is false. So
the hypotheses of theorem are not satisfied. That is, the individual
limits don't exist. Or 0=[-2,2].
Now it is true that 0 is inside [-2,2], so you could maybe bend the
situation and say that the answer is OK
if we change the meaning of equality for intervals. This would be
really convenient since
you could re-implement the limit function to just return [-inf,inf] for
every input and that would be OK too.:)
but then we could also compute limit(f(x)) as F=H-G, which means
[-1,1] = 0-[-1,1]. or [-1,1]=[-1,1].
This looks pretty good, but it is unfortunately questionable in the
world of intervals [-1,1]-[1,1] is not 0 but
[-2,2].
For a convenient online list of theorems about limits, most of which are
false for intervals, check
http://archives.math.utk.edu/visual.calculus/1/limits.18/index.html
{this is the first one I clicked ; it seemed ok.}
I don't know if this extended notion of limit has other ramifications,
like it breaks differentiation somehow.
It would not surprise me.
................
For amusement, I tried in Mathematica 6.0
Limit[Sin[x]^2+Sin[2*x]^2,x -> Infinity]
which you might naively think would be [0,2]. Mathematica gives [0,5].
Perhaps because it simplifies the argument to 5*sin(x)^2-4*sin(x)^4.
If it simplified that further to 25/16-(2*sin(x)^2-5/4)^2, it gives
[0,25/16] which is tight.
This tightness is a prediction of a theorem on "single use expressions".
...............
This does not address the issue of: what if we did not use intervals,
but allowed some objects like bounded(-1,1).
Then consider reconsider theorem 1, call it theorem1b
/* relate limit of f(x)+g(x) to limit of f and limit of g */
bounded_algorithm_sum(f(x),g(x),x,a) :=
block([F:bounded_algorithm(f(x),x,a), G:bounded_algorithm(g(x),x,a),
return ( bounded(bound_min(F)+bound_min(G), bound_max(F)+bound_max(G)))) $
but I'm not sure if we could say anything about inclusion. Maybe
intersection?
Anyway, this note is already too long. Comments, elaborations,
refutations, welcome.
RJF
Stavros Macrakis wrote:
> I don't think the "distance from interval" application is very common,
> and in fact interval operations on limitsets are meaningful in other
> ways.
>
> In particular, for continuous f, f( limitset(A), limitset(B) ) is a
> subset of limitset( f(A,B)), so interval arithmetic *does* make sense
> for that application.
>
> For example, limitset( sin(x) ) - limitset (max(0,sin(x))) =
> interval(-1,1) - interval(0,1) = interval(-2,1), whereas limitset
> (sin(x)-max(0,sin(x))) = interval(-1,0).
>
> -s
>
> On Tue, Sep 8, 2009 at 2:23 PM, Richard Fateman<fateman at cs.berkeley.edu> wrote:
>
>> Stavros Macrakis wrote:
>>
>>> If f(t) has a well-defined, unique limit as t approaches x, then the
>>> limit set of f(t) in the neighborhood of x is exactly that point.
>>>
>>> If it does not have a single limit point, then the limit set is an
>>> excellent characterization of its behavior in that neighborhood.
>>>
>>> Calculating s-[-1,1] as an interval calculation here isn't meaningful.
>>>
>>>
>> I agree. which is why I think that representing the limit set as an
>> interval is hazardous, because
>> once you have an interval, it has different semantics associated with it.
>> If you return
>> another object which is a "limit set" you might be safer. I'm not
>> sure where this will really bite you. That depends on how much you actually
>> do with intervals
>> that come out of limit calculations. Most people do nothing with this
>> information beyond
>> printing it out, which is probably not going to run into problems.
>>
>>
>>
>>
>>
>
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