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Given the function $Q$ that takes as input two sequences of positive integers $X = [x_1, x_2, ... x_n]$ and $Y = [y_1, y_2, ... y_n]$, $n \in [1, ... \infty]$ and returns the quotient of their sums

$$ Q(X, Y) = \frac{\sum_{i=0}^n x_i}{\sum_{i=0}^n y_i} $$

Is there any function $f(x, y)$ such that the function $S$

$$ S(X, Y) = \sum_{i=1}^{n} f(x_i, y_i) $$

respects that:

For any pairs of sequences $(A, B)$ and $(C, D)$:

If $Q(A, B) > Q(C, D)$ then $S(A, B) > S(C, D)$

If no such function $f$ can exists, does it exist if we relax it to:

If $Q(A, B) > Q(C, D)$ then $S(A, B) > S(C, D) + \epsilon$, for $\epsilon < 0$ bounded by some function of $(A, B)$ or $(C, D)$.

NOTE: The function $f$ is a function of the individual sequence elements only, i.e. $f(x_i, y_i)$ has no knowledge of the rest of the sequences $X_{j\neq i}$ and $Y_{j\neq i}$.

Federico Taschin
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  • In your first conditions, [1 for any pairs...] is equivalent to [2 for any pairs...]. In your relaxed conditions also, provided you replace $+\epsilon$ by $-\epsilon$ in 1 (wasn't it a misprint?). – Anne Bauval Aug 31 '22 at 17:44
  • Not a misprint, I wrote it in a bit of hurry and didn't realize points 2 were direct consequence of points 1. I removed them now – Federico Taschin Aug 31 '22 at 17:51
  • Then (to be a truly "relaxed" condition) $\epsilon<0$? – Anne Bauval Aug 31 '22 at 17:53
  • Yep, you're right again – Federico Taschin Aug 31 '22 at 22:50
  • Hence I repeat my suggestion to rewrite it more "reader friendly" : $S(A, B) > S(C, D) - \epsilon$ for $\epsilon>0$, instead of $S(A, B) > S(C, D) + \epsilon$ for $\epsilon<0$. It is of course equivalent, but less confusing, more classically worded. – Anne Bauval Sep 01 '22 at 17:54

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