Numerical Analysis Questions & Answers | |
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No, I asked that you post the solution by hand and show at which step there is a problem (using LATEX).
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Yes, you are right, it should be: $$ g(y)=g(0)+(y-0)g'(0)+\frac{{y}^{2}}{2}g' '(0)+... $$ If I wrote otherwise on the board, then this is a mistake obviously so please change your notes in consequence. This is a good observation, I'll give you 2 points bonus boost.
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Oops, it's due on Thursday November 17th. Thanks for correcting this. I'll give you 2 points bonus boost.
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Well, user-specified constants are numbers that are specified by you, the user of the code. Of course, such could be any real number you wish to specify..
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Well, the number of data points $N$ can be obtained from the data shown in the tables..
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In the reminder, last row, it is written $2\le i\le N-1$, not $2\le i\le N$..
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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.
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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.
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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.
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This was explained in class. You need to exclude the reserved exponents for the special cases.
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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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Well because for the same exponent (the smallest possible one), 0.f will be just below 1.f. 1 point bonus boost.
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$\pi$ |