Numerical Analysis Questions & Answers  


But $e=0$ indicates a special case, not an exponent. Thus, the minimum exponent $p_{\rm min}$ is not calculated using $e=0$ but using $e=1$.




I'm not sure why you're getting negative bits.. You have more or less the right logic thus (although not perfectly). Keep working more on it.




I will answer your question if the numbers are properly typeset. Answer it again below and use LATEX to write the math and numbers properly.






Yes, the relative error can be less than $\epsilon_{\rm mach}$. In class, $\epsilon_{\rm mach}$ was defined as the maximum relative error on a nondenormal float. For most nondenormal floats, the error will of course be less. 1 point bonus boost.



$\pi$ 