1
$\begingroup$

For $a_1,a_2$, $b_1,b_2$ $\in\mathbb{R}^+$, if $a_1 , then for any perturbation $\epsilon\in \mathbb{R}^+$, $r_1=\frac{a_1+\epsilon}{b_1+\epsilon}>\frac{a_1}{b_1} $ and if $a_2>b_2$, $r_2=\frac{a_2+\epsilon}{b_2+\epsilon}<\frac{a_2}{b_2} $

If $\epsilon$ is generated from a continuous function as $\epsilon={f_\epsilon(t)}$. What would be a way to characterize $r_1-r_2$ in terms of $f_\epsilon(t)?$

  • 0
    A rational number is shrinking while the other is getting larger with this form of a perturbation. $t$ is not related to $a_1,a_2$, $b_1,b_2$. I would like to see how the difference between the rationals varies w.r.t the function in one-varible, $t$. $f_\epsilon$ can be linear/monotonic or say, a higher order polynomial. I do agree that the class of functions, being referred to-with $f_\epsilon$ has not been specified. As $\epsilon$ asymptotically gets larger the ratios tend to converge to 1. How does the rate of change of $r_1$ and $r_2$ relate to the rate of change of $f_\epsilon$ ?2012-03-07

1 Answers 1

1

I don't understand the problem. If $r_1=(a_1+\epsilon)/(b_1+\epsilon)$ and $r_2=(a_2+\epsilon)/(b_2+\epsilon)$ and $\epsilon=f_{\epsilon}(t)$, then I can express $r_1-r_2$ in terms of $f_{\epsilon}(t)$ as $r_1-r_2={a_1b_2-a_2b_1+(a_1-a_2-b_1+b_2)f_{\epsilon}(t)\over(b_1+f_{\epsilon}(t))(b_2+f_{\epsilon}(t))}$ but I don't know what it means to characterize $r_1-r_2$ in terms of $f_{\epsilon}(t)$.

  • 1
    If $q-p$ is fixed, let's call it $L$, so you want $p$ to maximize $\int_p^{p+L}(r_1-r_2)\,dt$. Differentiate and set to zero, and you want to solve $r_1(p+L)-r_2(p+L)-r_1(p)+r_2(p)=0$ for $p$. I can't understand the 2nd question or the third - you are using mathematical vocabulary in ways never seen before.2012-03-07