Solutions for dependency problem in interval arithmetic (remedy for nonexistence of additive inverses)

346 Views Asked by At

Currently I read about the dependency problem of interval arithmetic. Mainly it's the problem that in the equation $X-X$ for $X$ being an interval the following is calculated: $$X-X=\{x-y:x\in X, y\in X\}$$ instead of $\{x-x:x\in X\}$. Both $X$ in $X-X$ are treated as different inputs. Thus $X-X\neq \{0\}$ so that there is no additive Inverse in interval arithmetic.

My Question: Is there a known solution for the dependency problem such that there is an additive (an multiplicative) inverse in interval arithmetic? Is there an extended version of interval arithmetic where addition is a group?

What I have found so far: I read about the centered form. For example, instead of calculating with $f(x)=x^2+x$ one can calculate with $(x+\tfrac 12)^2-\tfrac 14$ where $x$ only occurs once. The centered form helps, but does not provide an additive inverse...

1

There are 1 best solutions below

0
On

Generalized intervals at least partially answer your question. Basically, the idea is to allow intervals [a,b] to be a pair of numbers, without requiring a < b, and then to define adequate operations on them.

See for example the webpage

http://msse.gatech.edu/research/interval/WhyGeneralizedInterval.html

and the references therein.