Question by Student 201327107 Professor, I have question about Big O notation. Sometimes you write $$(b-a)O(\delta(x^2))$$ but sometimes you write just $$O(\delta(x^2))$$ except $(b-a)$. Do these two have same meaning?
 11.25.16
I'm not sure what your question is.. The big O notation $O(\Delta x^2)$ means that the average truncation error leading term scales with $\Delta x^2$, that is all. Your question is not clear and is not well typeset either. I'll give you 0.5 point bonus boost only.
 11.26.16
 Question by Student 201029134 Professor, I got a grade and want to check my score. I'm sorry but When do I go to your office?
 12.28.16
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.
 Question by Student 201542124 Professor, I have a question about the homework. In your homepage, there is a homework#1 Today I learned about IEEE in that class. So, When should we submit the homework? And, if so, should we upload the homework in your website or write in the paper and submit in person?
 09.06.17
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.
 Question by Student 201427565 professor, I don't understand the reason why I have to use $e =11111110$ instead of $e=11111111$ when i'm finding max P. and also for the min P, why is that $e=00000001$? not just the zero?
 09.10.17
This was explained in class. You need to exclude the reserved exponents for the special cases.
 Question by Student 201627148 Professer, in roundoff error, You wrote $$x={-2g+\sqrt{4g*g+4}}/2$$ $$sqrt(4*g*g+4)=2000.001+\sqrt{8000\epsilon_{mach}}$$ But i think you forgot to put $$\sqrt{8000\epsilon_{mach}}$$ to get x. Is it possible to erase because it is too small?
 09.16.17
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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