In short, what is the proper function notation/mapping for this type of functions?

From Wikipedia regarding maps:

In the communities surrounding programming languages that treat functions as first-class citizens, a map is often referred to as the binary higher-order function that takes a function $f$ and a list $[v_0, v_1, \dots, v_n]$ as arguments and returns $[f(v_0), f(v_1), ..., f(v_n)]$ (where $n \geq 0$).

In mathematics we use the function and mapping notation \begin{align} f&:\mathbb R^n\rightarrow \mathbb R^n \tag 1 \\ x&\mapsto (f_1(x), \dots, f_n(x)) \tag 2 \end{align} Where $x=(x_1,\dots, x_n)$ and we often write $$ f(x_1,\dots, x_n)=( f_1(x_1,\dots, x_n), \dots, f_n(x_1,\dots, x_n) ) \tag 3 $$


According Wikipedia we have the variable $v=(v_0,\dots, v_n)$ and a function of the form \begin{align} f(v_0,\dots, v_n)= (f(v_1), f(v_2), \dots, f(v_n)) \tag 4 \end{align} But this form doesn't correlate with the notation in $(1)-(3)$. What is the proper function/mapping notation for $(4)$?

  • $\begingroup$ What is the question precisely? Whether there is a standard mathematical notation for what computer scientists call map? $\endgroup$ – Andrej Bauer Dec 4 '20 at 16:43
  • $\begingroup$ The (computer science) mapping of a function on the components of a tuple corresponds to the Cartesian product (Cartesian power in this case) of $f$ with itself. If you have, for example $f:\mathbb{R}\to\mathbb{R}$, then you have $\prod_{i=1}^{n}f:\mathbb{R}^n\to\mathbb{R}^n$ defined as $(\prod_{i=1}^{n}f)(v_1,v_2,...,v_n)=(f(v_1),f(v_2),...,f(v_n))$. This notation is not used enough to use it without explicitly defining it. $\endgroup$ – plop Dec 4 '20 at 17:34
  • $\begingroup$ For example, the Wikipedia article is denoting it as $\langle f,f,...,f\rangle$, which is also not widely used either. $\endgroup$ – plop Dec 4 '20 at 17:39
  • $\begingroup$ @plop Can you explain your notation further? You have the product symbol over $i$ but no $i$ in your function, is it a typo? Do you mean $\prod_{i=1}^{n}f(v_i)=(f(v_1),f(v_2),...,f(v_n))$? $\endgroup$ – JDoeDoe Dec 8 '20 at 15:33
  • $\begingroup$ There is no missing $i$. If the $i$ is not in the factor, is because the factor is independent of $i$ and it is being multiplied (Cartesian product) with itself $n$ times. $\endgroup$ – plop Dec 8 '20 at 15:44

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