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Interval arithmetic calculates with closed intervals that contain the possible values of a quantity. If all we know is $x\in X=[a,b]$, an interval enclosure for a function $f$ must contain $f(x)$ for every allowed $x$. The result says what is guaranteed, including uncertainty in the input and, with outward rounding, roundoff error in the calculation. A wide answer can still be correct; a narrow answer is useful only if it contains every possible value.

The attraction is that a finite calculation can make a statement about infinitely many real numbers at once. An interval enclosure can certify that a function stays positive throughout a whole closed interval, or bound its maximum for mathematical optimization. For example, on $X=[0,1]$, the actual image of a function for $f(x)=x(1-x)$ is $[0,1/4]$, but direct interval arithmetic gives the wider $[0,1]$. Interval subdivision and the interval square range improve this answer in different ways.

Historically, Ramon E. Moore's 1962 Stanford dissertation studied interval arithmetic and automatic error analysis in digital computing; his book Interval Analysis followed in 1966. Eldon Hansen's account documents earlier reports as well: these dates describe important steps in developing and disseminating the subject, rather than a claim of sole invention in a particular year.^[ref] Moore, R. Baker Kearfott and Michael J. Cloud later wrote Introduction to Interval Analysis (2009).^[ref]

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 isThis 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 $0$.

Interval root exclusion certifies that a function has no zero of a function on $X$ when a valid interval enclosure $F(X)$ excludes zero. For $g(x)=x(1-x)-1/3$ on $[0,1]$, the interval square range yields $G(X)=[-1/3,-1/12]$, which excludes zero. Direct interval arithmetic yields $[-1/3,2/3]$ and is inconclusive, even though the function has no zero of a function there.

Knowing only that a calculated interval enclosure contains zero does not certify a zero of a function: it may contain unattainable values. Even the positive constant function $f(x)=1$ has a valid, loose interval enclosure $[-1,2]$. A separate existence argument, such as reliable opposite endpoint signs together with continuity in the intermediate value theorem, is required; see root bracket and bisection method. Interval subdivision can resolve some inconclusive pieces, but remaining pieces must be reported as unresolved unless an existence argument succeeds. This makes interval root exclusion useful for systematically discarding regions in a search.

// Target: analysis.bigb

The interval square range treats squaring as one function of one input. For $X=[a,b]$ it is exactlyIt retains the relationship that ordinary interval multiplication loses: for $Z=[-1/2,1/2]$, the interval square range is $[0,1/4]$, whereas the independent-input product $ZZ$ is $[-1/4,1/4]$.

Consequently, completing the square for $x(1-x)$ and using the interval square range givesThis is the exact image of a function on $X=[0,1]$. Merely rewriting the expression and evaluating $[1/4,1/4]-ZZ$ gives $[0,1/2]$, so it does not achieve the same sharpening. A mathematical $Z^2$ must be interpreted explicitly here: a square range operation and an independent-input interval product have different meanings. Choosing an operation that preserves known structure can save many steps of interval subdivision.