Apologies for the potentially obviously impossible/simple question.

I have 2 values:

$c_1 = b_1/a_1$

$c_2 = b_2/a_2$

$c_1, c_2$ are known, but $a_1,a_2,b_1,b_2$ are not (except being $> 0$ and finite)

Is there any way to calculate:

$c_t = (b_1+b_2)/(a_1+a_2)$

knowing only $c_1, c_2$?

  • $\begingroup$ Unless $c_1$ and $c_2$ are equal, $c_t$ can be anything (strictly) between $c_1$ and $c_2$. $\endgroup$ – Batominovski May 16 '17 at 10:34

It is impossible.

Take $(a_1,a_2,b_1,b_2)=(1,1,1,2)$. $c_1=1$ and $c_2=2$


Take $(a_1,a_2,b_1,b_2)=(1,2,1,4)$. $c_1=1$ and $c_2=2$


  • $\begingroup$ Those dependencies are not random. You didn't show how exactly they are not random, therefore it is not a solution (or answer to the problem) $\endgroup$ – iantonuk May 16 '17 at 11:17

That's a normal question. You can't get 1 number because these conditions define a map (a function from $R\rightarrow R$)- you need to reduce it to one parameter ${a_1}/{a_2}$ or ${b_1}/{b_2}$.

You have 2 bounds $c_1$, and $c_2$ plus two inequalities(exclusions) that $a_1\neq 0$ & $a_2\neq 0$ Your equation needs 3 bounds(it's just intuition, not a math) which means that you need 1 more bound. To figure it out just derive further:

You don't have $c_1$ & $c_2$ in equation $\implies$ you need to substitute them:

$b_1 = a_1\times c_1; b_2 = a_2\times c_2$;

$c_3 = (a_1\times c_1 + a_2\times c_2)/(a_1+a_2) = c_1(a_1 + a_2)/(a_1+a_2) + (c_2-c_1)\times a_2/(a_1+a_2) =$

$c_1 + (c_2-c_1)/(1+ a_1/a_2)$ where the only unknown part is $a_1/a_2 = t; t\neq 0 (a_1\neq 0)$ So,

The answer:

$c_3 = c_1 + (c_2-c_1)/(1 + t)$, where t belongs $(-\infty; 0)\bigcup(0;+\infty)$

In other words if $c_1, c_2$ are concrete - it's just one graph. If they are parameters ranging over $R$ you get $R^2$ space of of these hyporbolic functions. This is a specific object in Math and has all the rights, just as a number from $R$ to exist and be used. You can call it a number in a different peculiar space.

What's important - there is no way to know what you'll get in the end until you solve it. Therefore you must always solve. Such questions mathematically always have an answer.


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