Numerical Analysis Questions & Answers  




I'll be in Dec. 30th, Jan 2nd, and Jan 3rd from 9am till 6pm. Note that the grades can't be changed after January 3rd.




I'm not sure yet when Assign 1 will be due. When I decide on the date, I will let you know on my website, and you will receive an email. You should write the Assignment on paper and submit it at the beginning of the class.




This was explained in class. You need to exclude the reserved exponents for the special cases.




Let's do it again step by step. $$ {\rm sqrt}(4.0*g*g+4.0)=\sqrt{4(g\pm\epsilon_{\rm mach}g)^2+4\pm 4\epsilon_{\rm mach}} $$ $$ {\rm sqrt}(4.0*g*g+4.0)=\sqrt{4g^2(1\pm\epsilon_{\rm mach})^2+4 \pm 4\epsilon_{\rm mach}} $$ For $\epsilon_{\rm mach}\ll 1$: $$ {\rm sqrt}(4.0*g*g+4.0)\approx\sqrt{4g^2(1 \pm 2\epsilon_{\rm mach})+4\pm 4\epsilon_{\rm mach}} $$ Or $$ {\rm sqrt}(4.0*g*g+4.0)\approx\sqrt{4g^2+4 \pm 8g^2 \epsilon_{\rm mach}\pm 4\epsilon_{\rm mach}} $$ But for $8g^2 \gg 4$: $$ {\rm sqrt}(4.0*g*g+4.0)\approx\sqrt{4g^2+4 \pm 8g^2 \epsilon_{\rm mach}} $$ But for $8g^2\epsilon_{\rm mach}\ll 4g^2+4$ can show that $$ {\rm sqrt}(4.0*g*g+4.0)\approx\sqrt{4g^2+4} \pm \frac{1}{2} \frac{8g^2 \epsilon_{\rm mach}}{4g^2+4}\sqrt{4g^2+4} $$ If $4g^2\gg 4$: $$ {\rm sqrt}(4.0*g*g+4.0)\approx\sqrt{4g^2+4} \pm \epsilon_{\rm mach}\sqrt{4g^2+4} $$ Thus, for $\epsilon_{\rm mach}$ much smaller than 1 and $g$ much greater than 1, a very good approximation to the error associated with the sqrt of $4g^2+4$ is $\epsilon_{\rm mach}$ times the sqrt of $4g^2+4$ (I think that was what I wrote last class). If not, make a correction. Good question. I'll give you 2 points bonus boost.



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