John Pryce wrote:
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.
Your recipe
(ff,df) = (gg intersection hh, max(dg,dh)) = ([0,1], saf)
is ok here but gives the wrong result if h(x) happens not to be defined
everywhere, such as when using
g(x)=|x|, h(x)=(sqrt(x))^2
My formula (corrected formulation below) was constructed in such a way
that this case is still correct, which is important for my original
example.