Numerical Analysis Questions & Answers | |
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It was written $\Pi$ — not $\pi$ — on the blackboard. I explained prod1 and prod2 again at the beginning of the class today. 0.5 point bonus for the effort.
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That's not a bad proof, but it will fail to be valid if $(\epsilon_{\rm mach}g^2)^2$ is not much smaller than $\epsilon_{\rm mach}$. For $\epsilon_{\rm mach}\approx 10^{-15}$, this will happen when $g>6000$.. Thus, your proof is valid only for not so high values of $g$. I'll give you 2 points bonus for the effort.
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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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$\pi$ |