Estimate the difference between two quotient

Question

Suppose I know that $|a-b|\leq \epsilon_1$ and $|c-d|\leq \epsilon_2$ all of them are non zero, can I estimate $$\left|\frac{a}{c}-\frac{b}{d}\right|\,?$$ I tried some thing simple $$\left|\frac{a}{c}-\frac{b}{c}+\frac{b}{c}-\frac{b}{d}\right|\leq\left|\frac{a-b}{c}\right|+|b|\left|\frac{d-c}{cd}\right|\leq \frac{\epsilon_1}{|c|}+|b|\frac{\epsilon_2}{|cd|}\,.$$ Is there another better way to do that?

Answer 1

You can do this as well: $$\left|\frac{a}{c}-\frac{b}{d}\right|=\left|\frac{a}{c}-\frac{a}{d}+\frac{a}{d}-\frac{b}{d}\right|\leq |a|\,\left|\frac{d-c}{cd}\right|+\left|\frac{a-b}{d}\right|\leq \frac{|a|}{|cd|}\epsilon_1+\frac{1}{|d|}\epsilon_2\,.$$ Thus, $$\left|\frac{a}{c}-\frac{b}{d}\right|\leq \min\Biggl\{\frac{|a|}{|cd|}\epsilon_1+\frac{1}{|d|}\epsilon_2,\frac{1}{|c|}\epsilon_1+\frac{|b|}{|cd|}\epsilon_2\Biggr\}\,.$$ I do not think that you can improve this inequality by much, as it is sharp.

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However, you can reduce everything into just $a$, $c$, $\epsilon_1$, and $\epsilon_2$, provided that $|c|>\epsilon_2$. That is, from $$|b|\leq |a|+\epsilon_1\text{ and }|d|\geq |c|-\epsilon_2\,,$$ we obtain $$\begin{align} \left|\frac{a}{c}-\frac{b}{d}\right|&\leq \min\Biggl\{\frac{|a|}{|cd|}\epsilon_1+\frac{1}{|d|}\epsilon_2,\frac{1}{|c|}\epsilon_1+\frac{|b|}{|cd|}\epsilon_2\Biggr\} \\&\leq \min\Biggl\{\frac{|a|}{|c|\big(|c|-\epsilon_2\big)}\epsilon_1+\frac{1}{\big(|c|-\epsilon_2\big)}\epsilon_2,\frac{1}{|c|}\epsilon_1+\frac{|a|+\epsilon_1}{|c|\big(|c|-\epsilon_2\big)}\epsilon_2\Biggr\}\,.\end{align}$$ This inequality is also sharp.