1

Introduction to Numerical Analysis, Stoer, Chapter: Error Analysis, Page 8

if $|y|<\frac{eps}{\beta}|x|$ where $eps = 0.5\times 10^{1-t}$ then $$fl(x+y)=x+^*y=x$$ where $fl(x)=$ normalized floating point number closest to $x$ and $fl(x)=x(1+\epsilon)$ with $|\epsilon|\leq eps$.

absolute error is given by $$|fl(x)-x|\leq 0.5\times \beta^{e(1-t)} $$.

To prove above equality i have assumed that $\beta=10$ so: $$|fl(x+y)-x|=|(x+y)(1+\epsilon)-x|\leq|x+y||1+\epsilon|+|x|=|x|(3+2|\epsilon|)\leq 10^{e+1}(3+10^{1-t}) $$

but it is not true! PLEASE Help to prove it :)

SKMohammadi
  • 1,091
  • 8
  • 21
  • I think you mean $|\textrm{fl}(x+y) - x|$ where you say "absolute error is given by ...". Also, what's $e$ in $\beta^{e(1-t)}$? Should that perhapts be $\epsilon$? – fgp Apr 17 '14 at 10:47
  • yes $|\textrm{fl}(x+y) - x|$ and $e$ is exponent an for any $x$ in machine we have $|x|\in [\beta^e,\beta^{e+1})$ – SKMohammadi Apr 18 '14 at 06:54

0 Answers0