What is the proper mathematical notation to deal with an array?
Beginning with the declaration, I am used to the following format as a programmer:
ARRAY:
X[number of elements in array];
FOR i=1 TO 100 BEGIN
X[i]=N[i];
END;
Many thanks.
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$\begingroup$ The nearest notion to "array" is either a vector in $\mathbb{R}^n$, or a function $a \colon [1, n] \rightarrow \mathbb{R}$. $\endgroup$– vonbrandApr 14, 2013 at 17:29
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$\begingroup$ Perhaps a matrix. Let $A$ be a $1 \times n$ matrix, and then you could write $a_{1,p} = b_{1, p} \forall p$ (where the idea that $p$ is a natural number less than $n$ is implied, and that $a_{i,j}$ is the element in the $i$th row and $j$th column). Honestly, I do not recommend looking for a translation from programming or mathematics for everything. Not all things have symbols in mathematics, and sometimes words are the best way to convey ideas. $\endgroup$– George V. WilliamsApr 14, 2013 at 17:33
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1 Answer
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The word "array" is not a mathematical notion in the way that "vector" or "matrix" are. The mathematical idea that comes closest is family.
Given an arbitrary index set $I$ and a "universe" $X$, a family of elements $x\in X$ is a function $$f:\quad I\to X,\qquad \iota\mapsto x_\iota$$ that produces for each $\iota\in I$ a certain element $x_\iota\in X$. Thereby one and the same element $x\in X$ may be produced several times. This function is not interesting as such, nobody cares about continuity or the like, it only serves to organize the list of $x$'s that we want to study now. When talking about this family we therefore don't mention the actual $f$; instead we denote this family by $(x_\iota)_{\iota\in I}\>$, or by $\ \bigl(x_\iota\, \bigm|\, \iota\in I\>\bigr)\ $ if typographically preferable.
The index set $I$ might have some internal structure by itself. If, e.g., the set $I$ is a cartesian product $I=[m]\times[n]$, where $[m]:=\{1,2,\ldots,m\}$, then a family $(a_\iota)_{\iota\in I}$ is called an $(m\times n)$-matrix.
answered Apr 14, 2013 at 18:44
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$\begingroup$ Thanks Christian - Set theory is familiar, so this makes sense to me. $\endgroup$ Apr 14, 2013 at 19:13