Numerical Analysis Questions & Answers

Question by Student 201612150 Professor, I think I found something. We can recall "Formula V" used in the last lecture: $\phi_{n+1} = \phi_{n} + \Delta tf(t_{n+1/2}, \phi_{n}+\frac{\Delta t}{2}f(t_{n},\phi_{n})+O(\Delta t^{2})) + O(\Delta t^{3})$ Looking closely, We multiply $\Delta t$ by $\phi_{n}+\frac{\Delta t}{2}f(t_{n},\phi_{n})+O(\Delta t^{2})$. Therefore, $\Delta t$ times $O(\Delta t^{2})$ is $O(\Delta t^{3})$. * Note: I figured out this from the progress to analyze global error. So due to this, the result of global error analysis is unaffected - since we multiply $\Delta t$ by not only the term $\phi_{n}+\frac{\Delta t}{2}f(t_{n},\phi_{n})$, but also error term $O(\Delta t^{2})$! Therefore, we now can sure that the modified Euler's method is of order two. Although I'm not sure if my deduction is correct, but I think this may be an answer.

12.05.17

It's not so simple because you need to show that $\Delta t f(t_{n+1/2}, \phi_{n}+\frac{\Delta t}{2}f(t_{n},\phi_{n}+O(\Delta t^2))$ scales with $O(\Delta t^3)$. Note that you can not simply take $O(\Delta t^2)$ out of $f$ as you did. This needs to be done more carefully.

Question by Student 201327139 Professor. In Chapter.7, we learned about simpson's rule and modified simpson's rule. But, when I was studying Chapter.7 and searched about simpson's rule from googling , I found about simpson's 1/3 rule and simpson's 3/8 rule. What's the difference between what we learned about and these rules?	12.06.17

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.

Question by Student 201427127 Professor. I want to check my answer sheet. How can I check?	01.02.18

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

Question by Student 201627143 Professor, can I check my answer sheet too?	01.03.18

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.

Question by Student 201427127 Professor, I can't distinct what is prod1 and prod2 at C++ programing code that you made which was PI and SIGMA at your note on blackboard.	09.03.18

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.

Question by Student 201427127 Professor, I can't distinct what is prod1 and prod2 at C++ programing code that you made which was $\pi$ and $\sum$ at your note on blackboard. Thank you. I can use LATEX now.	09.05.18

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.

Question by Student 201527121 I want to prove it by solving simple equation. $$y=\sqrt{(g^2+1)}\pm....$$ substitue small terms into x $$y=\sqrt{(g^2+1)}+x$$ $$\sqrt{(g^2+1)}+x=\sqrt{(g^2+1\pm2\epsilon_{mach} g^2)}$$ Square both sides. $$(g^2+1)+2\sqrt{g^2+1}x+x^2=(g^2+1\pm2\epsilon_{mach} g^2)$$ $$2\sqrt{g^2+1}x=-(g^2+1)+(g^2+1)\pm2\epsilon_{mach} g^2-x^2$$ Square of both sides again $$4(g^2+1)x^2=x^4\mp4\epsilon_{mach} g^2 x^2+4\epsilon_{mach}^2g^4$$ $\epsilon_{mach}^2$ is enough small to assume zero value. Rearrange function and eliminate $x^2$ and $\epsilon_{mach}^2$ $$4(g^2+1)\pm4\epsilon_{mach} g^2=x^2$$ $$x=\pm2\sqrt{g^2+1\pm\epsilon_{mach}g^2}$$ As a result, we can guess $$y=\sqrt{g^2+1}\pm2\sqrt{g^2+1\pm\epsilon_{mach}g^2}$$

09.10.18

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.

Question by Student 201427148 Professor, I want to check my Answer of the problem, which handed-out at Class 10th, Sep 2018. $$ -Given Equation. y=\sqrt{g^{2}+2\epsilon_{MACH} g^{2}+1} $$ $$ -Wanted Equation. y=\sqrt{g^{2}\pm1}+x $$ $$ -Used Equation. (a+b)^{c}=a^{c}+bca^{c-1}+...+ $$ $$ -PROBLEM : Find x. $$ 1st, I found that The Equation which must be used to solve the problem(Used Equation) is the Binomial Expansion. So, I Transform the Used Equation to Binomial Expansion like under Equation. $$ (a+b)^{c}=a^{c}+c\left(\frac{1}{1!}\right)a^{c-1}b+c(c-1)\left(\frac{1}{2!}\right)a^{c-2}b^{2}+For (a\gg b) $$ Then I Transform the Given Equation like under Equation. $$ y=\sqrt{(g^{2}+1)\pm\epsilon_{MACH}g^{2}} $$ And, Permutate each term to a, b, c. $$ a=g^{2}+1, $$ $$ b=\pm2\epsilon_{MACH}g^{2}, $$ $$ c=\frac{1}{2} $$ Before substituting the Permutated terms to Expansion, I confirmed that the condition $(a\gg b)$ is satisfied. As $g$ is larger number and $\epsilon_{MACH}$ is almost zero. Then, $a$ is more bigger than $g$ and $b$ is almost zero. So, I could use the upper Binomial Expansion. Then, I Substituted Each terms to Expanded Equation like under. $$ y=\sqrt{(g^{2}+1)\pm\epsilon_{MACH}g^{2}} $$ $$ = \sqrt{g^{2}+1}\pm{2\epsilon_{MACH}\left(\frac{1}{2}\right)\left(\frac{1}{1!}\right)\frac{1}{\sqrt{g^{2}+1}}} +\left(\pm2\epsilon_{MACH}g^{2}\right)^{2}\left(\frac{1}{2}\right)\left(\frac{-1}{2}\right)\left(\frac{1}{2!}\right) (\sqrt{g^{2}+1})^{\frac{-2}{3}}+...+ $$ In this Expanded Series, I found the pattern. The pattern is that, The 'n'th terms has $\left(\epsilon_{MACH}\right)^{n-1}$. As $\left(\epsilon_{MACH}\right)^{2}$ is almost zero,that I could ignore all terms without 1st term $\left(\sqrt{g^{2}+1}\right)$ and 2nd term $\left(\pm{2\epsilon_{MACH}\left(\frac{1}{2}\right)\left(\frac{1}{1!}\right)\frac{1}{\sqrt{g^{2}+1}}}\right) $. Finally Two Terms left from Series. $$ y\approx \sqrt{g^{2}+1}\pm{2\epsilon_{MACH}\left(\frac{1}{2}\right)\left(\frac{1}{1!}\right)\frac{1}{\sqrt{g^{2}+1}}} $$ Thus, the wanted $x$ is $$ x=\pm\frac{g^{2}}{\sqrt{g^{2}+1}}\epsilon_{MACH} $$ I'll wait your advice for my first using LATEX and my solution. Thank you.

09.11.18

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.

Question by Student 201727142 Professor, I can't understand that you teached last class about $\epsilon_x$. You teached $X=1.0/(SQRT(g^2+1)+g)$ and you said $y=\sqrt{g^2+1}\pm \epsilon_{mach} \sqrt{g^2+1}$. and then i think X=1.0/(y+g) , $X =\frac{1\pm \epsilon_{mach}}{\sqrt{g^2+1}\pm \epsilon_{mach} \sqrt{g^2+1} + g}$. But, you teached us $X =\frac{1\pm \epsilon_{mach}}{\sqrt{g^2+1}\pm \epsilon_{mach} \sqrt{g^2+1}}$. Why is (denominator's g) disappeared?

09.12.18

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.

Question by Student 201427113 Dear prof. B. Parent When i wanted to change base 2 :0000 0000 to base 10, the answer is 1 in the note during class. But i can't understand why the answer is 1 and it's two' complement is also 1. Because in case of one's complement, same base 2 0000 0000 is base10 '0' so i'm confused.	09.13.18

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.

Question by Student 201527119 Professor, I have an question about Question #6 in Assignment 01. I think Question #6 $\textrm{i)}$ and $\textrm{ii)}$ are changed, and answer is wrong too. Therefore, $\textrm{i)}$ is significand. and $\textrm{ii)}$ is exponent.	09.16.18

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.

Question by Student 201527147 Dear professor, I wonder if the range lower limit should be min pos normal or min pos denormal in question 5. Thank you.	09.18.18

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

Question by Jaehyuk Professor, I have question about Machine Precision of Float. For denormal number, $(0.f)_{min} = 0.00...0$(23'0's)=0 . Then, $\varepsilon _{mach} = \frac{0.00...01-0}{0.00...01} = \frac{2^{-23}}{2^{-23}}=1.$ Is this the right $\varepsilon_{mach}$ for float and denormal number?	09.26.18

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.

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