Re: Revised version of Level 1 text (draft)
Arnold, P1788
Here are two examples for Arnold's specification of intersection of decorated intervals, using the (ill<emp<con<def<saf) scheme.
Example 1
---------
Suppose f(x) = |x| is represented in two (admittedly silly) ways as
g(x) = |2x-x|
h(x) = x sign(x)
(where sign(x) is -1 if x<0, 0 if x=0, 1 if x>0).
Let xx = [-1,1]. Then I get
gg = g(xx) = abs(2*xx-xx) = abs([-2,2]-[-1,1]) = abs([-3,3]) = [0,3] decorated dg = saf,
hh = h(xx) = [-1,1]*sign([-1,1] = [-1,1]*[-1,1] = [-1,1] decorated dh = def.
Each is a valid enclosure of range(f,xx) so let's improve it by taking the intersection. Arnold's recipe uses the dg>dh case and returns
(ff,df) = (hh,dh) = ([-1,1],def),
not very good. Whereas my analysis suggests we should return
(ff,df) = (gg intersection hh, max(dg,dh)) = ([0,1], saf)
which happens to be sharp in this case.
Example 2
---------
Suppose f(x) = x sqrt|x| (which is continuous and monotone increasing on the whole real line) is represented in two ways as
g(x) = sqrt(x)^3
h(x) = -(sqrt(-x)^3)
where ^ is the "pown" in Juergen's list of standard functions.
Let xx = [-4,4]. Then I get
gg = sqrt([-4,4])^3 = [0,2]^3 = [0,8], decorated dg = con,
hh = -sqrt(-[-4,4])^3 = -([0,2]^3) = -[0,8] = [-8,0], decorated dh = con.
This falls under Arnold's description below, "function...can be expressed in multiple ways as a (possibly not everywhere defined) expression". So let's take the intersection. In Arnold's recipe the "dg=dh ... else" case applies and gives
(ff,df) = ([0,0],con).
Obviously this has no relation to range(f,xx) = [-8,8].
Clearly what's wrong here is that Arnold's spec, of when intersection is appropriate, is too loosely drawn. It's appropriate when the domains of g and h are almost equal, whereas here they hardly overlap. This is a case where we should have taken the union, not the intersection.
How can we specify decorated intersection and union so that there isn't a huge element of "caveat emptor" in using them? I see no answer to that question. Arnold?
John
> The intersection operation must cater for the situation where a function
> (or an intermediate term in a function) can be expressed in multiple
> ways as a (possibly not everywhere defined) expression.
>
> For example,
> f(x)= (x_1-x_2)/(x_1+x_2) = 2/(1+q)-1, where q=x_2/x_1.
> The evaluation of the second expression gives the optimal range on
> boxes where x_1 does not contain 0. Depending on the argument, it may
> pay to evaluate both expressions and intersect the results.
> (See Appendix 2 for a worked example.)
>
> Thus if
> f(z) = g(z) if g(z) is defined and f(z) = h(z) if h(z) is defined,
> we must be able to get optimal information on the range of f on zz
> from the available information on g and h on zz by intersecting
> the decorated intervals g(zz) and h(zz). This requires that
> (gg,dg) intersect (hh,dh) = (ff,df),
> where
> dg>dh --> (ff,df) = (hh,dh),
> dh>dg --> (ff,df) = (gg,dg),
> dg=dh -->
> if gg disjoint hh --> (ff,df) = (Empty,emp),
> else --> (ff,df) = (gg intersect dh,dg).
Should that be
else --> (ff,df) = (gg intersect hh,dg).