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In my optimization problem there are elements consisting of a time and a value, i.e. $(t_0, v_0)$. These pairs are stored in a vector $v = [(t_0, v_0), (t_1, v_1), ... , (t_n, v_n)]$. The vector $v$ serves as an input to an optimization function $f$ that computes a cost value from the input.

Is it mathematically common to say that $V$ is the set of all possible variants of $v$ and thus define $f$ as $f : V \rightarrow \mathbb{R}$?

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  • $\begingroup$ Since no one else has answered, I'd say it looks alright to me. Though I would probably say $f:\mathbb{R}^{n \times 2} \rightarrow \mathbb{R}$, if indeed $t_i \in \mathbb{R}$ and $v_i \in \mathbb{R}$. $\endgroup$
    – Nachiket
    Jun 17 at 9:31

1 Answer 1

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That seems correct, but I would suggest two minor improvements:

  • Rearrange the pairs in a $(n+1)\times 2$ matrix $$ M = \begin{bmatrix} t_0 & v_0\\ \vdots & \vdots\\ t_n & v_n \end{bmatrix}, $$ so that you can replace $V$ with $\mathbb{R}^{(n+1)\times 2}$, as suggested by user @Nachiket in a comment.
  • Avoid using the letter $v$ for this matrix, because it conflicts with the common notation that $v$ is the vector with entries $v_1,v_2,\dots,v_n$.
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