In our lectures, we've been looking at relative error and in our recent problem sheet we were asked to prove the following:
Let $\hat{x} \neq 0$ be an approximation of a non-zero quantity $x$. Consider the relative error $ e = \frac{|x- \hat{x}|}{|x|}$ and the error normalised using the approximation $\hat{x}$, $\hat{e} = \frac{|x-\hat{x}|}{|\hat{x}|}$.
Assuming that $\hat{e} < 1$, show that there exist quantities $f_1(\hat{e})$ and $f_2(\hat{e})$ such that $f_1(\hat{e}) ≤ e ≤ f_2(\hat{e})$.
I'm honestly not sure where to begin with this, especially since $f_1$ and $f_2$ are functions of $\hat{e}$. Any suggestions would be helpful.
Best Answer
We have $$e = \frac{|x-\hat{x}|}{|x|} = \frac{|x-\hat{x}|}{|\hat{x}|}\frac{|\hat{x}|}{|x|} \leq \hat{e}\left(1 + \frac{|\hat{x}-x|}{|x|}\right) = \hat{e}(1+e).$$ This implies $e(1-\hat{e}) \leq \hat{e}$ and therefore $e \leq \frac{\hat{e}}{1-\hat{e}}$ assuming $\hat{e} < 1$.
Can you show the other inequality now?