= Interval dependency problem
{title2=$X-X\ne\{x-x:x\in X\}$}
= Interval dependency
{synonym}
The <interval dependency problem> is the loss of relationships between repeated occurrences of the same input. For $X=[0,1]$, interval <subtraction> gives $X-X=[-1,1]$: it permits $u-v$ with independent choices $u,v\in X$. But the <function> $x\mapsto x-x$ has actual <image of a function> $\{0\}=[0,0]$. Giving both operands the name $X$ does not tell an <interval arithmetic> calculation to enforce $u=v$.
For the <polynomial> $f(x)=x(1-x)$ on $X=[0,1]$, the <natural interval extension> is
$$
F(X)=X([1,1]-X)=[0,1][0,1]=[0,1].
$$
This allows the impossible combination $x=1$ in the first factor and $x=0$ in the second, producing the spurious upper bound $1$. In the actual <function>, both occurrences use the same $x$. <Completing the square> gives $x(1-x)=1/4-(x-1/2)^2$, which lies between $0$ and $1/4$ on $[0,1]$; every value between these bounds occurs, by <continuity> and the <intermediate value theorem>. The <maximum> $1/4$ occurs at $x=1/2$, while either endpoint gives the <minimum> $0$.
\Image[/interval-dependency-and-subdivision.png]
{title=Independent interval inputs lose the diagonal constraint; subdivision tightens the enclosure of x times one minus x}
{description=Left: the <Cartesian product> permits separate choices $u,v\in[0,1]$, while the actual <function> uses only $u=v$. Right: rectangles from four pieces contain the <polynomial> curve. Below: one, two and four pieces give <interval enclosures> with upper bounds $1$, $1/2$ and $3/8$; the <interval square range> gives the exact $1/4$.}
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