0
$\begingroup$

Assume we have several variables of the form $d_c$ which namely can be $d_1$, $d_2$, ..., $d_n$. I want to use mathematical notation to show for which index $c$ the value of $d_c$ is minimal for all variables.

I can write the minimum value of these variables like this: $$ d_{min} = \min\left\{d_1, d_2, \ldots, d_n\right\}$$

What I need now is to find $c$ so that $d_{min}=d_c$.


The context is the following: I have several classes $1\leq c \leq n$ that have a certain value assigned to their corresponding variables $d_c$. I'm interested to know for which class this value is minimal. Let's do a small example: $$d_1 = 20\qquad d_2=15\qquad d_3=8\qquad d_4=12\qquad d_5=42$$ $$ d_{min} = \min\left\{d_1, d_2, d_3, d_4, d_5\right\} = 8$$ $$d_{min}\overset{!}{=}d_c \quad \Longrightarrow \quad c=3$$

So class number $3$ has the smallest value assigned to its corresponding variable.


  • What mathematical notation can I use to find $c$?

I'm open for a totally different notation as well. Maybe this can be rewritten completely to make it much simpler and clearer to understand.

$\endgroup$
1
  • $\begingroup$ The standard notation is $c = \mbox{argmin}({\bf d})$ where ${\bf d} = (d_1,\dots,d_n)$, or $c = \mbox{argmin} \{d_1,\dots,d_n\}$, see math.stackexchange.com/a/228042/318321 $\endgroup$
    – Fnacool
    Dec 16, 2016 at 2:54

1 Answer 1

1
$\begingroup$

maybe it is not the best way but it's a way and it would work for positive numbers : $$max({i*(1-sign(d_i - d_{min}))})$$ where the sign(x) is equal to -1 if x<0, zero if x=0 and +1 if x>0

1) each minus min
2) sign of each
3) 1 minus sign of each
5) each multiply by each index
4) max of all

original problem: {20 15 8 12 42}
1) each minus min: {12 7 0 4 36}
2) sign of each: {1 1 0 1 1}
3) 1-sign: {0 0 1 0 0}
4) multiply to index: {0 0 3 0 0}
max of all: 3

$\endgroup$
1
  • 2
    $\begingroup$ This has the potential to be a nice answer. LaTeX markup would help, as would clear and detailed explanation of each step. $\endgroup$
    – The Count
    Dec 16, 2016 at 3:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .