Numerical Analysis Questions & Answers  




What we learned in class is the standard Simpson's rule. There are several variations with some (marginal) advantages over the standard form.. You can read about those in the wikipedia, if you are interested.




You can come to my office in the afternoon. I'll be here tomorrow and friday, and next week from thursday.




Sure, you can come. I'm a bit busy now thus because I have to prepare slides for a conference I'll be going to next week. So if possible, come to see me after next Wednesday.




You need to use LATEX to write all mathematics. PI and SIGMA should be written using the same mathematical symbols I used in class. Check out the LATEX mini HOWTO in the Skylounge. Ask your question again below and I'll answer it if it is correctly typeset.




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.




There is a problem with your proof. When you write $$ 4(g^2+1)x^2=x^4\mp4\epsilon_{mach} g^2 x^2+4\epsilon_{mach}^2g^4 $$ You cannot say that the term $\epsilon_{mach}^2 g^4$ is negligible because $\epsilon_{mach}^2 \ll \epsilon_{mach}$. Here, $g^4$ can be much greater than $g^2$.. Thus, $\epsilon^2 g^4$ can be as large or larger than $\epsilon g^2 x^2$ and the final answer you give for $x$ is wrong. Nonetheless I'll give you 2 points bonus for the effort.




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.




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.




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.




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.




If it's not indicated, then choose the lowest possible.




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.



$\pi$ 