I have a question about the proof of the following statement:
For each set of machine numbers $F(b, n, E_{min}, E_{max})$ with $E_{min} < E_{max}$ the following inequality holds: $\epsilon_{mach} \leq \frac{1}{2}b^{1-n}$ where $b$ is the basis, $n$ the number of digits and $\epsilon_{mach}$ the machine epsilon i.e. the biggest possible relative error.
Proof: Let $x_{min} < z < x_{max}$ be an arbitrary real number. Then $z$ can be represented as $$z = b^E\sum^\infty_{i=0} z_ib^{-i}$$ with $E_{min} \leq E \leq E_{max}$. We denote with $\tilde z \in F$ the machine number that results from $z$ after cutting of all digits after the $n$-th digit. Then $$\big|\frac{z-\tilde z}{z}\big| = \frac{|b^E\sum_{i=n}^\infty z_ib^{-i}|}{|z|} = \frac{|b^{E-n}\sum_{i=0}^\infty z_{i+n}b^{-i}|}{|z|}.$$ On one hand we have $$\sum_{i=0}^\infty z_{i+n}b^{-i} \leq \sum_{i=0}^\infty (b-1)b^{-i} = (b-1)\sum_{i=0}^\infty\big(\frac{1}{b}\big)^i = (b-1)\cdot \frac{1}{1-\frac{1}{b}} = b$$ and on the other hand we have $|z| = |b^E\sum^\infty_{i=0}z_ib^{-i}| \geq b^E$. Hence $$\frac{|z-\tilde z|}{|z|} \leq \frac{b^{E-n+1}}{b^E} = b^{1-n}.$$ Because the approximation error can be at most half as big as the considered truncation error the statement follows. $\Box$
The justification on the last line is unclear, could anyone help me parse why this follows? In particular, where is the $\frac{1}{2}$ coming from? Thank you.
