I have sets $A = \left\{a_{1},a_{2},a_{3},\dots,a_{n} \right\}$ and $B = \left\{b_{1},b_{2},b_{3},\dots,b_{n} \right\}$. If they are time-series sorted by their indices, I can take the differences, $$ d = \frac{\Delta A}{\Delta B} = \left\{\frac{a_2-a_1}{b_2-b_1},\frac{a_3-a_2}{b_3-b_2}, \dots, \frac{a_n-a_{n-1}}{b_n-b_{n-1}}\right\} $$ and take the mean finite difference $\bar{d}$ to approximate the derivative $\frac{dA(t)}{dB(t)}$.

If they are not sorted by time (i.e., they are just sequences but not time-series), can I take all combinations of differences in the numerator and denominator, take their ratio, and finally their mean to approximate $\bar{d}$? That is, can we say: $$ \bar{d} \approx \frac{1}{n}\sum_{i\neq j}\frac{a_i-a_j}{b_i-b_j} $$

It will be awesome if someone could suggest an approximation. This is needed in my biology research. Thanks.

  • 1
    $\begingroup$ You write that you want to approximate $\overline d$, but then you put an equal sign. Which of the two did you mean? If you do just want to approximate, what exactly do you want to know about the approximation? And when you ask us to "suggest an approximation", do you mean other approximations of $\overline d$ using the sequences? Or a justification of the approximation you've provided? $\endgroup$
    – joriki
    Jul 26, 2018 at 3:29
  • $\begingroup$ Thanks joriki. I changed it to an approximate sign and I don't know if that is true. At this point it could either be a justification for my guess or any other approximation using sequences. $\endgroup$ Jul 26, 2018 at 11:25


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