How to relate $|\delta x|$ and $|\tilde\delta x|$ in an inequality $A\le |\tilde\delta x|\le B$ where $A$ and $B$ contain $|\delta x|$?

Question

$|\delta x| = |(\tilde x - x)/x|$ (this is the absolute value of relative error)

$|\tilde\delta x| = |(\tilde x - x)/\tilde x|$ (bound for some distinct quantity)

I need to find an inequality which relates these in this way:

$A\le |\tilde\delta x|\le B$ where $A$ and $B$ contain $|\delta x|$.

I know $\tilde x = x(1 + \delta x)$ or $\delta x = \delta(x)/x = (\tilde x - x)/x$

No idea how to proceed with finding that inequality though. Thank you very much if you can show me how this is done.

Answer 1

Note that $$|\tilde\delta x| = |\delta x| \frac{|x|}{|\tilde x|}.$$ From the first equation in the question, $\left| \frac{ \tilde x}{ x}-1\right| = |\delta x| $, so by triangle inequality, we obtain $$\frac{|\tilde x|}{|x|} \le 1 + \left| \frac{\tilde x}{ x}-1\right| = 1+|\delta x|,$$ and $$ \frac{|\tilde x|}{| x|} \ge 1 - \left| \frac{\tilde x}{ x}-1\right| = 1-|\delta x|.$$ Thus $$ \frac{|\delta x|}{1+|\delta x|}\le |\tilde \delta x | \le \frac{|\delta x|}{1-|\delta x|}.$$