I learnt that if the relative error is 5*$10^{-s}$ then the number of correct digits the result has $s$. Why is this so? Can you illustrate with an example and/or a proof?
Another way to put it appears to be that $n$ correct digits means a relative error with order of $10^{-n}$. Is it a definition?
Best Answer
Let y be a real number whose expansion is given by $$ y = \pm(0.d_{1}d_{2}d_{3}...d_{k}d_{k+1}...)_{\beta}\times\beta^{e} $$ with $ d_{1}\neq 0 $ and $ m\le e\le M $ where $\beta$ = the base, $k$ = the number of digits in the base $\beta$ expansion, $m$ = the minimum exponent and $M$ = the maximum exponent, and let $ fl(y) $ be the floating point equivalent of y. When the number chopped, the floating point equivalent is given by $$ fl_{chop}(y) = \pm(0.d_{1}d_{2}d_{3}...d_{k})_{\beta}\times\beta^{e} $$ when the number is rounded, $$ fl_{round}(y)=\pm(0.d_{1}d_{2}d_{3}...d_{k})_{\beta}\times\beta^{e}\qquad for \qquad d_{k+1}<\frac{\beta}{2} $$ $$ fl_{round}(y)=\pm[(0.d_{1}d_{2}d_{3}...d_{k})_{\beta}+\beta^{-k}]\times\beta^{e}\qquad for \qquad d_{k+1} \ge \frac{\beta}{2}$$ We have $$ |fl_{chop}(y)-y|=(0.d_{k+1}d_{k+2}d_{k+3}...)_{\beta}\times\beta^{e-k}\le(1.0)_{\beta}\times\beta^{e-k}=\beta^{e-k} $$ also $$ |y|=(0.d_{1}d_{2}d_{3}...)_{\beta}\times\beta^{e}\ge(0.1)_{\beta}\times\beta^{e}=\beta^{e-1} $$ therefore $$ \frac{|fl_{chop}(y)-y|}{|y|}\leq \frac{\beta^{e-k}}{\beta^{e-1}}=\beta^{1-k} $$ By proceeding in a similar manner, it can be shown that when a number is rounded, the bounds on both the absolute and relative error due to roundoff are one-half the bounds obtained when a number is chopped. That is $$ |fl_{round}(y)-y|\le \frac{1}{2} \beta^{e-k} $$ and $$ \frac{|fl_{round}(y)-y|}{|y|}\le \frac{1}{2} \beta^{1-k} $$ Now we are ready to the following definition,
Source: A Friendly Introduction to Numerical Analysis by Brian Bradie.