Interval arithmetic

Body mass index for a person 1.80 m tall in relation to body weight m (in kilograms)

Consider the calculation of a person's body mass index (BMI). BMI is calculated as a person's body weight in kilograms divided by the square of their height in meters. Suppose a person uses a scale that has a precision of one kilogram, where intermediate values cannot be discerned, and the true weight is rounded to the nearest whole number. For example, 79.6 kg and 80.3kg are indistinguishable, as the scale can only display values to the nearest kilogram. It is unlikely that when the scale reads 80kg, the person has a weight of exactly 80.0kg. Thus, the scale displaying 80kg indicates a weight between 79.5kg and 80.5kg, or the interval ${\displaystyle [79.5,80.5)}$.

The BMI of a man who weighs 80kg and is 1.80m tall is approximately 24.7. A weight of 79.5kg and the same height yields a BMI of 24.537, while a weight of 80.5kg yields 24.846. Since the body mass is continuous and always increasing for all values within the specified weight interval, the true BMI must lie within the interval ${\displaystyle [24.537,24.846]}$. Since the entire interval is less than 25, which is the cutoff between normal and excessive weight, it can be concluded with certainty that the man is of normal weight.

The error in this example does not affect the conclusion (normal weight), but this is not generally true. If the man were slightly heavier, the BMI's range may include the cutoff value of 25. In such a case, the scale's precision would be insufficient to make a definitive conclusion.

The range of BMI examples could be reported as ${\displaystyle [24.5,24.9]}$ since this interval is a superset of the calculated interval. The range could not, however, be reported as ${\displaystyle [24.6,24.8]}$, as the interval does not contain possible BMI values.

Body mass index for different weights in relation to height L (in meters)

Height and body weight both affect the value of the BMI. Though the example above only considered variation in weight, height is also subject to uncertainty. Height measurements in meters are usually rounded to the nearest centimeter: a recorded measurement of 1.79 meters represents a height in the interval ${\displaystyle [1.785,1.795)}$. Since the BMI uniformly increases with respect to weight and decreases with respect to height, the error interval can be calculated by substituting the lowest and highest values of each interval, and then selecting the lowest and highest results as boundaries. The BMI must therefore exist in the interval

${\displaystyle {\frac {[79.5,80.5)}{[1.785,1.795)^{2}}}\subseteq [24.673,25.266].}$

In this case, the man may have normal weight or be overweight; the weight and height measurements were insufficiently precise to make a definitive conclusion.

To shorten the notation of intervals, brackets can be used.

${\displaystyle [x]\equiv [x_{1},x_{2}]}$ can be used to represent an interval. Note that in such a compact notation, ${\displaystyle [x]}$ should not be confused between a single-point interval ${\displaystyle [x_{1},x_{1}]}$ and a general interval. For the set of all intervals, we can use

${\displaystyle [\mathbb {R} ]:=\left\{\,[x_{1},x_{2}]\,|\,x_{1}\leq x_{2}{\text{ and }}x_{1},x_{2}\in \mathbb {R} \cup \{-\infty ,\infty \}\right\}}$

as an abbreviation. For a vector of intervals ${\displaystyle \left([x]_{1},\ldots ,[x]_{n}\right)\in [\mathbb {R} ]^{n}}$ we can use a bold font: ${\displaystyle [\mathbf {x} ]}$.

Values of a monotonic function

Interval functions beyond the four basic operators may also be defined.

For monotonic functions in one variable, the range of values is simple to compute. If ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ is monotonically increasing (resp. decreasing) in the interval ${\displaystyle [x_{1},x_{2}],}$ then for all ${\displaystyle y_{1},y_{2}\in [x_{1},x_{2}]}$ such that ${\displaystyle y_{1}\leq y_{2},}$ ${\displaystyle f(y_{1})<f(y_{2})}$ (resp. ${\displaystyle f(y_{2})<f(y_{1})}$).

The range corresponding to the interval ${\displaystyle [y_{1},y_{2}]\subseteq [x_{1},x_{2}]}$ can be therefore calculated by applying the function to its endpoints:

${\displaystyle f([y_{1},y_{2}])=\left[\min \left\{f(y_{1}),f(y_{2})\right\},\max \left\{f(y_{1}),f(y_{2})\right\}\right].}$

From this, the following basic features for interval functions can easily be defined:

• Exponential function: ${\displaystyle a^{[x_{1},x_{2}]}=[a^{x_{1}},a^{x_{2}}]}$ for ${\displaystyle a>1,}$
• Logarithm: ${\displaystyle \log _{a}[x_{1},x_{2}]=[\log _{a}{x_{1}},\log _{a}{x_{2}}]}$ for positive intervals ${\displaystyle [x_{1},x_{2}]}$ and ${\displaystyle a>1,}$
• Odd powers: ${\displaystyle [x_{1},x_{2}]^{n}=[x_{1}^{n},x_{2}^{n}]}$, for odd ${\displaystyle n\in \mathbb {N} .}$

For even powers, the range of values being considered is important and needs to be dealt with before doing any multiplication. For example, ${\displaystyle x^{n}}$ for ${\displaystyle x\in [-1,1]}$ should produce the interval ${\displaystyle [0,1]}$ when ${\displaystyle n=2,4,6,\ldots .}$ But if ${\displaystyle [-1,1]^{n}}$ is taken by repeating interval multiplication of form ${\displaystyle [-1,1]\cdot [-1,1]\cdot \cdots \cdot [-1,1]}$ then the result is ${\displaystyle [-1,1],}$ wider than necessary.

More generally one can say that, for piecewise monotonic functions, it is sufficient to consider the endpoints ${\displaystyle x_{1}}$, ${\displaystyle x_{2}}$of an interval, together with the so-called critical points within the interval, being those points where the monotonicity of the function changes direction. For the sine and cosine functions, the critical points are at ${\displaystyle \left({\tfrac {1}{2}}+n\right)\pi }$ or ${\displaystyle n\pi }$ for ${\displaystyle n\in \mathbb {Z} }$, respectively. Thus, only up to five points within an interval need to be considered, as the resulting interval is ${\displaystyle [-1,1]}$ if the interval includes at least two extrema. For sine and cosine, only the endpoints need full evaluation, as the critical points lead to easily pre-calculated values—namely -1, 0, and 1.

In general, it may not be easy to find such a simple description of the output interval for many functions. But it may still be possible to extend functions to interval arithmetic. If ${\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} }$ is a function from a real vector to a real number, then ${\displaystyle [f]:[\mathbb {R} ]^{n}\to [\mathbb {R} ]}$ is called an interval extension of ${\displaystyle f}$ if

${\displaystyle [f]([\mathbf {x} ])\supseteq \{f(\mathbf {y} )\mid \mathbf {y} \in [\mathbf {x} ]\}.}$

This definition of the interval extension does not give a precise result. For example, both ${\displaystyle [f]([x_{1},x_{2}])=[e^{x_{1}},e^{x_{2}}]}$ and ${\displaystyle [g]([x_{1},x_{2}])=[{-\infty },{\infty }]}$ are allowable extensions of the exponential function. Tighter extensions are desirable, though the relative costs of calculation and imprecision should be considered; in this case, ${\displaystyle [f]}$ should be chosen as it gives the tightest possible result.

Given a real expression, its natural interval extension is achieved by using the interval extensions of each of its subexpressions, functions, and operators.

The Taylor interval extension (of degree ${\displaystyle k}$ ) is a ${\displaystyle k+1}$ times differentiable function ${\displaystyle f}$ defined by

${\displaystyle [f]([\mathbf {x} ]):=f(\mathbf {y} )+\sum _{i=1}^{k}{\frac {1}{i!}}\mathrm {D} ^{i}f(\mathbf {y} )\cdot ([\mathbf {x} ]-\mathbf {y} )^{i}+[r]([\mathbf {x} ],[\mathbf {x} ],\mathbf {y} ),}$

for some ${\displaystyle \mathbf {y} \in [\mathbf {x} ]}$, where ${\displaystyle \mathrm {D} ^{i}f(\mathbf {y} )}$ is the ${\displaystyle i}$-th order differential of ${\displaystyle f}$ at the point ${\displaystyle \mathbf {y} }$ and ${\displaystyle [r]}$ is an interval extension of the Taylor remainder.

${\displaystyle r(\mathbf {x} ,\xi ,\mathbf {y} )={\frac {1}{(k+1)!}}\mathrm {D} ^{k+1}f(\xi )\cdot (\mathbf {x} -\mathbf {y} )^{k+1}.}$

Mean value form

The vector ${\displaystyle \xi }$ lies between ${\displaystyle \mathbf {x} }$ and ${\displaystyle \mathbf {y} }$ with ${\displaystyle \mathbf {x} ,\mathbf {y} \in [\mathbf {x} ]}$, ${\displaystyle \xi }$ is protected by ${\displaystyle [\mathbf {x} ]}$. Usually one chooses ${\displaystyle \mathbf {y} }$ to be the midpoint of the interval and uses the natural interval extension to assess the remainder.

The special case of the Taylor interval extension of degree ${\displaystyle k=0}$ is also referred to as the mean value form.

Outer bounds at different level of rounding

To work effectively in a real-life implementation, intervals must be compatible with floating point computing. The earlier operations were based on exact arithmetic, but in general fast numerical solution methods may not be available for it. The range of values of the function ${\displaystyle f(x,y)=x+y}$ for ${\displaystyle x\in [0.1,0.8]}$ and ${\displaystyle y\in [0.06,0.08]}$ are for example ${\displaystyle [0.16,0.88]}$. Where the same calculation is done with single-digit precision, the result would normally be ${\displaystyle [0.2,0.9]}$. But ${\displaystyle [0.2,0.9]\not \supseteq [0.16,0.88]}$, so this approach would contradict the basic principles of interval arithmetic, as a part of the domain of ${\displaystyle f([0.1,0.8],[0.06,0.08])}$ would be lost. Instead, the outward rounded solution ${\displaystyle [0.1,0.9]}$ is used.

The standard IEEE 754 for binary floating-point arithmetic also sets out procedures for the implementation of rounding. An IEEE 754 compliant system allows programmers to round to the nearest floating-point number; alternatives are rounding towards 0 (truncating), rounding toward positive infinity (i.e., up), or rounding towards negative infinity (i.e., down).

The required external rounding for interval arithmetic can thus be achieved by changing the rounding settings of the processor in the calculation of the upper limit (up) and lower limit (down). Alternatively, an appropriate small interval ${\displaystyle [\varepsilon _{1},\varepsilon _{2}]}$ can be added.

Approximate estimate of the value range

The so-called "dependency" problem is a major obstacle to the application of interval arithmetic. Although interval methods can determine the range of elementary arithmetic operations and functions very accurately, this is not always true with more complicated functions. If an interval occurs several times in a calculation using parameters, and each occurrence is taken independently, then this can lead to an unwanted expansion of the resulting intervals.

Treating each occurrence of a variable independently

As an illustration, take the function ${\displaystyle f}$ defined by ${\displaystyle f(x)=x^{2}+x.}$ The values of this function over the interval ${\displaystyle [-1,1]}$ are ${\displaystyle \left[-{\tfrac {1}{4}},2\right].}$ As the natural interval extension, it is calculated as:

${\displaystyle [-1,1]^{2}+[-1,1]=[0,1]+[-1,1]=[-1,2],}$

which is slightly larger; we have instead calculated the infimum and supremum of the function ${\displaystyle h(x,y)=x^{2}+y}$ over ${\displaystyle x,y\in [-1,1].}$ There is a better expression of ${\displaystyle f}$ in which the variable ${\displaystyle x}$ only appears once, namely by rewriting ${\displaystyle f(x)=x^{2}+x}$ as addition and squaring in the quadratic.

${\displaystyle f(x)=\left(x+{\frac {1}{2}}\right)^{2}-{\frac {1}{4}}.}$

So the suitable interval calculation is

${\displaystyle \left([-1,1]+{\frac {1}{2}}\right)^{2}-{\frac {1}{4}}=\left[-{\frac {1}{2}},{\frac {3}{2}}\right]^{2}-{\frac {1}{4}}=\left[0,{\frac {9}{4}}\right]-{\frac {1}{4}}=\left[-{\frac {1}{4}},2\right]}$

and gives the correct values.

In general, it can be shown that the exact range of values can be achieved, if each variable appears only once and if ${\displaystyle f}$ is continuous inside the box. However, not every function can be rewritten this way.

Wrapping effect

The dependency of the problem causing over-estimation of the value range can go as far as covering a large range, preventing more meaningful conclusions.

An additional increase in the range stems from the solution of areas that do not take the form of an interval vector. The solution set of the linear system

${\displaystyle {\begin{cases}x=p\\y=p\end{cases}}\qquad p\in [-1,1]}$

is precisely the line between the points ${\displaystyle (-1,-1)}$ and ${\displaystyle (1,1).}$ Using interval methods results in the unit square, ${\displaystyle [-1,1]\times [-1,1].}$ This is known as the wrapping effect.

A linear interval system consists of a matrix interval extension ${\displaystyle [\mathbf {A} ]\in [\mathbb {R} ]^{n\times m}}$ and an interval vector ${\displaystyle [\mathbf {b} ]\in [\mathbb {R} ]^{n}}$. We want the smallest cuboid ${\displaystyle [\mathbf {x} ]\in [\mathbb {R} ]^{m}}$, for all vectors ${\displaystyle \mathbf {x} \in \mathbb {R} ^{m}}$ which there is a pair ${\displaystyle (\mathbf {A} ,\mathbf {b} )}$ with ${\displaystyle \mathbf {A} \in [\mathbf {A} ]}$ and ${\displaystyle \mathbf {b} \in [\mathbf {b} ]}$ satisfying.

${\displaystyle \mathbf {A} \cdot \mathbf {x} =\mathbf {b} }$.

For quadratic systems – in other words, for ${\displaystyle n=m}$ – there can be such an interval vector ${\displaystyle [\mathbf {x} ]}$, which covers all possible solutions, found simply with the interval Gauss method. This replaces the numerical operations, in that the linear algebra method known as Gaussian elimination becomes its interval version. However, since this method uses the interval entities${\displaystyle [\mathbf {A} ]}$ and ${\displaystyle [\mathbf {b} ]}$ repeatedly in the calculation, it can produce poor results for some problems. Hence using the result of the interval-valued Gauss only provides first rough estimates, since although it contains the entire solution set, it also has a large area outside it.

A rough solution ${\displaystyle [\mathbf {x} ]}$ can often be improved by an interval version of the Gauss–Seidel method. The motivation for this is that the ${\displaystyle i}$-th row of the interval extension of the linear equation.

${\displaystyle {\begin{pmatrix}{[a_{11}]}&\cdots &{[a_{1n}]}\\\vdots &\ddots &\vdots \\{[a_{n1}]}&\cdots &{[a_{nn}]}\end{pmatrix}}\cdot {\begin{pmatrix}{x_{1}}\\\vdots \\{x_{n}}\end{pmatrix}}={\begin{pmatrix}{[b_{1}]}\\\vdots \\{[b_{n}]}\end{pmatrix}}}$

can be determined by the variable ${\displaystyle x_{i}}$ if the division ${\displaystyle 1/[a_{ii}]}$ is allowed. It is therefore simultaneously.

${\displaystyle x_{j}\in [x_{j}]}$ and ${\displaystyle x_{j}\in {\frac {[b_{i}]-\sum \limits _{k\not =j}[a_{ik}]\cdot [x_{k}]}{[a_{ii}]}}}$.

So we can now replace ${\displaystyle [x_{j}]}$ by

${\displaystyle [x_{j}]\cap {\frac {[b_{i}]-\sum \limits _{k\not =j}[a_{ik}]\cdot [x_{k}]}{[a_{ii}]}}}$,

and so the vector ${\displaystyle [\mathbf {x} ]}$ by each element. Since the procedure is more efficient for a diagonally dominant matrix, instead of the system ${\displaystyle [\mathbf {A} ]\cdot \mathbf {x} =[\mathbf {b} ]{\mbox{,}}}$ one can often try multiplying it by an appropriate rational matrix ${\displaystyle \mathbf {M} }$ with the resulting matrix equation.

${\displaystyle (\mathbf {M} \cdot [\mathbf {A} ])\cdot \mathbf {x} =\mathbf {M} \cdot [\mathbf {b} ]}$

left to solve. If one chooses, for example, ${\displaystyle \mathbf {M} =\mathbf {A} ^{-1}}$ for the central matrix ${\displaystyle \mathbf {A} \in [\mathbf {A} ]}$, then ${\displaystyle \mathbf {M} \cdot [\mathbf {A} ]}$ is outer extension of the identity matrix.

These methods only work well if the widths of the intervals occurring are sufficiently small. For wider intervals, it can be useful to use an interval-linear system on finite (albeit large) real number equivalent linear systems. If all the matrices ${\displaystyle \mathbf {A} \in [\mathbf {A} ]}$ are invertible, it is sufficient to consider all possible combinations (upper and lower) of the endpoints occurring in the intervals. The resulting problems can be resolved using conventional numerical methods. Interval arithmetic is still used to determine rounding errors.

This is only suitable for systems of smaller dimension, since with a fully occupied ${\displaystyle n\times n}$ matrix, ${\displaystyle 2^{n^{2}}}$ real matrices need to be inverted, with ${\displaystyle 2^{n}}$ vectors for the right-hand side. This approach was developed by Jiri Rohn and is still being developed.[4]

Reduction of the search area in the interval Newton step in "thick" functions.

An interval variant of Newton's method for finding the zeros in an interval vector ${\displaystyle [\mathbf {x} ]}$ can be derived from the average value extension.[5] For an unknown vector ${\displaystyle \mathbf {z} \in [\mathbf {x} ]}$ applied to ${\displaystyle \mathbf {y} \in [\mathbf {x} ]}$, gives.

${\displaystyle f(\mathbf {z} )\in f(\mathbf {y} )+[J_{f}](\mathbf {[x]} )\cdot (\mathbf {z} -\mathbf {y} )}$.

For a zero ${\displaystyle \mathbf {z} }$, that is ${\displaystyle f(z)=0}$, and thus, must satisfy.

${\displaystyle f(\mathbf {y} )+[J_{f}](\mathbf {[x]} )\cdot (\mathbf {z} -\mathbf {y} )=0}$.

This is equivalent to ${\displaystyle \mathbf {z} \in \mathbf {y} -[J_{f}](\mathbf {[x]} )^{-1}\cdot f(\mathbf {y} )}$. An outer estimate of ${\displaystyle [J_{f}](\mathbf {[x]} )^{-1}\cdot f(\mathbf {y} ))}$ can be determined using linear methods.

In each step of the interval Newton method, an approximate starting value ${\displaystyle [\mathbf {x} ]\in [\mathbb {R} ]^{n}}$ is replaced by ${\displaystyle [\mathbf {x} ]\cap \left(\mathbf {y} -[J_{f}](\mathbf {[x]} )^{-1}\cdot f(\mathbf {y} )\right)}$ and so the result can be improved. In contrast to traditional methods, the interval method approaches the result by containing the zeros. This guarantees that the result produces all zeros in the initial range. Conversely, it proves that no zeros of ${\displaystyle f}$ were in the initial range ${\displaystyle [\mathbf {x} ]}$ if a Newton step produces the empty set.

The method converges on all zeros in the starting region. Division by zero can lead to the separation of distinct zeros, though the separation may not be complete; it can be complemented by the bisection method.

As an example, consider the function ${\displaystyle f(x)=x^{2}-2}$, the starting range ${\displaystyle [x]=[-2,2]}$, and the point ${\displaystyle y=0}$. We then have ${\displaystyle J_{f}(x)=2\,x}$ and the first Newton step gives.

${\displaystyle [-2,2]\cap \left(0-{\frac {1}{2\cdot [-2,2]}}(0-2)\right)=[-2,2]\cap {\Big (}[{-\infty },{-0.5}]\cup [{0.5},{\infty }]{\Big )}=[{-2},{-0.5}]\cup [{0.5},{2}]}$.

More Newton steps are used separately on ${\displaystyle x\in [{-2},{-0.5}]}$ and ${\displaystyle [{0.5},{2}]}$. These converge to arbitrarily small intervals around ${\displaystyle -{\sqrt {2}}}$ and ${\displaystyle +{\sqrt {2}}}$.

The Interval Newton method can also be used with thick functions such as ${\displaystyle g(x)=x^{2}-[2,3]}$, which would in any case have interval results. The result then produces intervals containing ${\displaystyle \left[-{\sqrt {3}},-{\sqrt {2}}\right]\cup \left[{\sqrt {2}},{\sqrt {3}}\right]}$.

Rough estimate (turquoise) and improved estimates through "mincing" (red)

The various interval methods deliver conservative results as dependencies between the sizes of different interval extensions are not taken into account. However, the dependency problem becomes less significant for narrower intervals.

Covering an interval vector ${\displaystyle [\mathbf {x} ]}$ by smaller boxes ${\displaystyle [\mathbf {x} _{1}],\ldots ,[\mathbf {x} _{k}],}$ so that

${\displaystyle [\mathbf {x} ]=\bigcup _{i=1}^{k}[\mathbf {x} _{i}],}$

is then valid for the range of values.

${\displaystyle f([\mathbf {x} ])=\bigcup _{i=1}^{k}f([\mathbf {x} _{i}]).}$

So, for the interval extensions described above the following holds:

${\displaystyle [f]([\mathbf {x} ])\supseteq \bigcup _{i=1}^{k}[f]([\mathbf {x} _{i}]).}$

Since ${\displaystyle [f]([\mathbf {x} ])}$ is often a genuine superset of the right-hand side, this usually leads to an improved estimate.

Such a cover can be generated by the bisection method such as thick elements ${\displaystyle [x_{i1},x_{i2}]}$ of the interval vector ${\displaystyle [\mathbf {x} ]=([x_{11},x_{12}],\ldots ,[x_{n1},x_{n2}])}$ by splitting in the center into the two intervals ${\displaystyle \left[x_{i1},{\tfrac {1}{2}}(x_{i1}+x_{i2})\right]}$ and ${\displaystyle \left[{\tfrac {1}{2}}(x_{i1}+x_{i2}),x_{i2}\right].}$ If the result is still not suitable then further gradual subdivision is possible. A cover of ${\displaystyle 2^{r}}$ intervals results from ${\displaystyle r}$ divisions of vector elements, substantially increasing the computation costs.

With very wide intervals, it can be helpful to split all intervals into several subintervals with a constant (and smaller) width, a method known as mincing. This then avoids the calculations for intermediate bisection steps. Both methods are only suitable for problems of low dimension.

Interval arithmetic is used with error analysis, to control rounding errors arising from each calculation. The advantage of interval arithmetic is that after each operation there is an interval that reliably includes the true result. The distance between the interval boundaries gives the current calculation of rounding errors directly:

Error = ${\displaystyle \mathrm {abs} (a-b)}$ for a given interval ${\displaystyle [a,b]}$.

Interval analysis adds to rather than substituting for traditional methods for error reduction, such as pivoting.

Parameters for which no exact figures can be allocated often arise during the simulation of technical and physical processes. The production process of technical components allows certain tolerances, so some parameters fluctuate within intervals. In addition, many fundamental constants are not known precisely.[1]

If the behavior of such a system affected by tolerances satisfies, for example, ${\displaystyle f(\mathbf {x} ,\mathbf {p} )=0}$, for ${\displaystyle \mathbf {p} \in [\mathbf {p} ]}$ and unknown ${\displaystyle \mathbf {x} }$ then the set of possible solutions.

${\displaystyle \{\mathbf {x} \,|\,\exists \mathbf {p} \in [\mathbf {p} ],f(\mathbf {x} ,\mathbf {p} )=0\}}$,

can be found by interval methods. This provides an alternative to traditional propagation of error analysis. Unlike point methods, such as Monte Carlo simulation, interval arithmetic methodology ensures that no part of the solution area can be overlooked. However, the result is always a worst-case analysis for the distribution of error, as other probability-based distributions are not considered.

Approximation of the normal distribution by a sequence of intervals

Interval arithmetic can also be used with affiliation functions for fuzzy quantities as they are used in fuzzy logic. Apart from the strict statements ${\displaystyle x\in [x]}$ and ${\displaystyle x\not \in [x]}$, intermediate values are also possible, to which real numbers ${\displaystyle \mu \in [0,1]}$ are assigned. ${\displaystyle \mu =1}$ corresponds to definite membership while ${\displaystyle \mu =0}$ is non-membership. A distribution function assigns uncertainty, which can be understood as a further interval.

For fuzzy arithmetic[7] only a finite number of discrete affiliation stages ${\displaystyle \mu _{i}\in [0,1]}$ are considered. The form of such a distribution for an indistinct value can then be represented by a sequence of intervals.

${\displaystyle \left[x^{(1)}\right]\supset \left[x^{(2)}\right]\supset \cdots \supset \left[x^{(k)}\right].}$

The interval ${\displaystyle \left[x^{(i)}\right]}$ corresponds exactly to the fluctuation range for the stage ${\displaystyle \mu _{i}.}$

The appropriate distribution for a function ${\displaystyle f(x_{1},\ldots ,x_{n})}$ concerning indistinct values ${\displaystyle x_{1},\ldots ,x_{n}}$ and the corresponding sequences.

${\displaystyle \left[x_{1}^{(1)}\right]\supset \cdots \supset \left[x_{1}^{(k)}\right],\ldots ,\left[x_{n}^{(1)}\right]\supset \cdots \supset \left[x_{n}^{(k)}\right]}$

can be approximated by the sequence.

${\displaystyle \left[y^{(1)}\right]\supset \cdots \supset \left[y^{(k)}\right],}$

where

${\displaystyle \left[y^{(i)}\right]=f\left(\left[x_{1}^{(i)}\right],\ldots \left[x_{n}^{(i)}\right]\right)}$

and can be calculated by interval methods. The value ${\displaystyle \left[y^{(1)}\right]}$ corresponds to the result of an interval calculation.

Warwick Tucker used interval arithmetic in order to solve the 14th of Smale's problems, that is, to show that the Lorenz attractor is a strange attractor.[8] Thomas Hales used interval arithmetic in order to solve the Kepler conjecture.