Numerical Analysis Questions & Answers | |
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Oups, I lost the $g$ on the denominator. So, keep $g$ there and add $\epsilon_{\rm mach}g$ on the denominator. Then, you'll get (in worse case scenario): $$ x=\frac{1\pm \epsilon_{\rm mach}}{\sqrt{g^2+1}+g\pm \epsilon_{\rm mach}(\sqrt{g^2+1}+|g|)} $$ You can figure out the following steps on your own following the same logic as shown in class.. Good observation: 2 points bonus.
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00000000 is zero either with the two's complement or the one's complement. I think you mean 11111111. Then, the one's complement is -0 but the two's complement is -1. You may have the rows mixed up.
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The answers given for one question are not necessarily in the correct order. There is no mistake in the answers. Think about this more carefully.
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If it's not indicated, then choose the lowest possible.
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Yes, you're on the right track. But $(0.f)_\min=0.00..001$, not 0. Thus, if $(0.f)=(0.f)_\min$, then $\epsilon_{\rm mach}=1$. But $\epsilon_{\rm mach}$ is not always 1 for denormal numbers. In fact, it is much less than 1 as $(0.f)$ becomes larger. 2 points bonus.
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$\pi$ |