Numerical Analysis Questions & Answers
 Question by Student 201427127 Professor, I'll prove the question that you told us at the class before the lecture. $A < n ...ㄱ$ and $B < n+1 ...ㄴ$ And then, I'll substract ㄴ from ㄱ. the result of this is $A-B < -1$. But if $A$ is equal to $B$ or $A$ is bigger than $B$, this inequation will be false because left handside is positive number or zero. The prove is over.
 10.01.18
It's fine, but it's not clear enough. You need to provide a more clear example. 1.5 point bonus. Also, avoid naming equations with Hangul. See the mini HOWTO in the Skylounge on how to give a number to equations.
 Question by Student 201427113 Professor, I'll prove the question about inequation that you told us on class. $A < n ...ㄱ$ and $B < n+1 ...ㄴ$ And then, I'll substract ㄴ from ㄱ. the result of this is $A-B < -1$. For example, if $A = n-1 and B = n-1$ so $A - B = 0 < -1$ is false. So for generally $A$ is equal to $B$ or $A$ is bigger than $B$, this inequation will be false because left handside is positive number or zero and right handside is $-1$ so although general form is inconsistent. The prove is over.
Better. 0.5 point extra bonus. If you would have typeset this with proper equation numbers you would gotten more.
 Question by Student 201527121 1. wanted Prove $$\frac{1}{1+y}=1-y+y^2-y^3+...$$      2. Given condition $$|y|<<1$$     3. Solution $$\frac{1}{1-x}=1+x+x^2+x^3+...=\sum_{n=1}^\infty (x)^n$$ Let's substitute $$x=-y$$ $$\frac{1}{1+y}=1-y+y^2-y^3+...$$      4. Let's prove $$\frac{1}{1-x}=\sum_{n=1}^\infty (x)^n$$ $|x|<<1$ (It should be for convergence) $$S_n=\sum_{k=1}^n (x)^k$$ $$S_n=x+x^2+x^3+...+x^n$$ $$(1-x)S_n=(1-x)(x+x^2+x^3+...+x^n)$$ $$=x+x^2+...+x^n-x^2-x^3-...-x^{n+1}=x-x^{n+1}$$ $$S_n=\frac{x}{1-x}-\frac{x^{n+1}}{1-x}$$ Because of $|x|<<1$, If $n\rightarrow\infty$ then $S_n\rightarrow\frac{x}{1-x}$ $$\sum_{k=N}^n (x)^k=x^N+...+x^n=x^{N-1}\sum_{k=1}^{n-N+1} (x)^k$$ $$\sum_{k=N}^\infty x^k=\varinjlim({n}\rightarrow{\infty})\sum_{k=N}^n x^k=\varinjlim({n}\rightarrow{\infty})\cdot x^{N-1}\sum_{k=1}^{n-N+1} x^k=\frac{x^N}{1-x}$$ If $$N=0$$ then $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$      $$5. Conclusion$$ $$\frac{1}{1-x}=1+x+x^2+x^3+...=\sum_{n=1}^\infty (x)^n$$ $$x=-y$$ $$\frac{1}{1-x}=\sum_{n=1}^\infty (x)^n$$ $$\frac{1}{1+y}=1-y+y^2-y^3+...$$
This is great. I'll give you 3 points bonus. It could be made a bit shorter thus, and you shouldn't separate the proof into sections: this makes it harder to read. Also, I was looking for a simpler proof using Taylor series. If someone else can do it, another 3 bonus points are up for grabs.
 Question by Student 201427113 Professor, I'll prove $\frac{1}{1+x}=1-x+x^2-x^3...$ Taylor series is one of the power series. So $f(x)=C_0+C_1(x-a)+C_2(x-a)^2+C_3(x-a)^3+.... → A-equation$ And we have to decide what is $C_0, C_1, C_2, C_3, .....$ 1. Substitute x=a on A-equation So $f(a)=C_0$ 2. Apply differential on A-equation So $f^\prime(x)=C_1+2C_2(x-a)+3C_3(x-a)^2+....→ B-equation$ And substitute x=a on B-equation So $f^\prime(a)=C_1$ 3. Apply differential on B-equation So $f^{\prime\prime}(x)=2C_2+6C_3(x-a)+12C_4(x-a)^2....→ C-equation$ And substitute x=a on C-equation So $f^{\prime\prime}(a)=2C_2 → C_2=\frac{f^{\prime\prime}(a)}{2}$ 4. Apply differential on C-equation So $f^{\prime\prime\prime}(x)=6C_3+24C_4(x-a)....→ D-equation$ And substitute x=a on D-equation So $f^{\prime\prime\prime}(a)=6C_3 → C_3=\frac{f^{\prime\prime\prime}(a)}{6}$ 5. Apply this sequence infinitely and generalized the eqs. $C_n=\frac{f^n(a)}{n!}$ Finally substitute $C_n=\frac{f^n(a)}{n!}$ on A-equation So $f(x)=f(a)+\frac{f^\prime(a)}{1!}(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-a)^2+\frac{f^{\prime\prime\prime}(a)}{3!}(x-a)^3+...\\= \sum_{n=0}^\infty\frac{f^n(a)}{n!}(x-a)^n$ $f(x)=\frac{1}{1+x}=(1+x)^{-1} → P-equation$ $f^\prime(x)=-(1+x)^{-2} → Q-equation$ $f^{\prime\prime}(x)=2(1+x)^{-3} → R-equation$ $f^{\prime\prime\prime}(x)=-6(1+x)^{-4} → S-equation$ Apply x=a=0 on eqs. P,Q,R,S $f(0)=1\\ f^\prime(0)=-1\\ f^{\prime\prime}(0)=2\\ f^{\prime\prime\prime}(0)=-6$ So $f(x)= 1-1\times x+\frac{2}{2!}x^2+\frac{-6}{3!}x^3+...$ Prove is finished.!!
This is fine, I'll give you 2 points bonus. I would have given 3 if you would have typeset correctly your equations with proper equation numbers as specified in the mini HOWTO in the Skylounge. But, this proof is still not what I was looking for.. There's a much easier way to prove it using Taylor series (without power series). Three bonus points are still up for grabs.
 Question by Student 201527119 Proof. $$\frac{1}{1+x}=1-x+x^2-x^3+...$$ $$f(x) = f(a) + (x-a)f'(a) + \frac{(x-a)^2}{2!}f^”(a) + \frac{(x-a)^3}{3!}f”'(a) + ...$$ In $$\frac{1}{1+x}$$ a=0 So, $$f(x) = f(0) + (x-0)f'(0) + \frac{(x-0)^2}{2!}f”(0) + \frac{(x-0)^3}{3!}f”'(0) + ...$$ $$f'(x)=\frac{-1}{(1+x)^2}, f”(x)=\frac{2}{(1+x)^3}, f”'(x)=\frac{-6}{(1+x)^4}$$ $$f(0)=1,f'(0)=-1,f”(0)=2,f”'(0)=-6$$ Put them in equation. $$f(x)=\frac{1}{1+x}=1-x+x^2-x^3+...$$ End.
This is what I was looking for. I'll give you 2.5 points bonus. The only problem here is that you don't define $f(x)$ prior to doing the expansion with the Taylor series.
 Question by Jaehyuk Dear professor, I have question about A2#Q6(a). According to the question, iterative method is $x^{n+1}=x^{n}-0.05*\frac{f(x^{n})}{f{\prime}(x^{n})}-0.95*\frac{f(x^{n})(x^{n-1}-x^{n-2})}{f(x^{n-1})-f(x^{n-2})}$However, in order to use this as an iterative method, isn`t this should be like this;$x_{n+1}=x_{n}-0.05*\frac{f(x_{n})}{f{\prime}(x_{n})}-0.95*\frac{f(x_{n})(x_{n-1}-x_{n-2})}{f(x_{n-1})-f(x_{n-2})}$
 10.04.18
Well, in this case, it's clear what the notation means. It doesn't matter if it's a superscript or subscript..
 Question by Student 201527119 Proof. $$\frac{\epsilon_{(n+1)(Newton)}}{\epsilon_{(n+1)(secant)}}=|\epsilon_{0}*k|^{{2^n}-{1.618^n}}$$ $$k=|\frac{f”(r)}{2f'(r)}|$$ $$\epsilon_{(n+1)(Newton)}=k\epsilon^2$$ $$\epsilon_1=k*\epsilon_{0}^2$$ $$\epsilon_2=k*\epsilon_{1}^2=k^3*\epsilon_{0}^4$$ $$\epsilon_3=k*\epsilon_{2}^2=k^7*\epsilon_{0}^8$$ ... $$\epsilon_n=k^{2^n-1}*\epsilon_{0}^{2^n}$$ $$\epsilon_{(n+1)(secant)}=k^{\frac{1}{1.168}}\epsilon^{1.168}$$(1.168=p) $$\epsilon_1=k^{p^{-1}}*\epsilon_{0}^p$$ $$\epsilon_2=k^{p^{-1}}*\epsilon_{1}^p=k^{1+p^{-1}}*\epsilon_{0}^{p^2}$$ $$\epsilon_3=k^{p^{-1}}*\epsilon_{2}^p=k^{p+1+p^{-1}}*\epsilon_{0}^{p^3}$$ ... $$\epsilon_n=k^{\sum_{m=0}^{n-1} p^{m-1}}*\epsilon_{0}^{p^n}$$ Consequently, $$\frac{\epsilon_{n(newton)}}{\epsilon_{n(secant)}}=\frac{k^{2^n-1}*\epsilon_{0}^{2^n}}{k^{\sum_{m=0}^{n-1} p^{m-1}}*\epsilon_{0}^{p^n}}=\frac{k^{2^n}*\epsilon_{0}^{2^n}}{k*k^{\sum_{m=0}^{n-1} p^{m-1}}*\epsilon_{0}^{p^n}}$$ $$k^{\sum_{m=0}^{n-1} p^{m-1}}=k^{1+\sum_{m=0}^{n-1} p^{m-1}}$$ $$1+\sum_{m=0}^{n-1} p^{m-1}=1+p^{-1}+1+p+...+p^{n-2}$$ Using Geometric sequence sum, $$p^{-1}+1+p+...+p^{n-2}=\frac{p^n-1}{p^2-p}$$ $$p^2=p+1$$ $$So, \frac{p^n-1}{p^2-p}=p^n-1$$ =>$$\frac{k^{2^n}*\epsilon_{0}^{2^n}}{k*k^{p^n-1}*\epsilon_{0}^{p^n}}=\frac{k^{2^n}*\epsilon_{0}^{2^n}}{k^{p^n}*\epsilon_{0}^{p^n}}$$ $$Consequently, |\epsilon_{0}*k|^{{2^n}-{1.618^n}}$$ $$So, \frac{\epsilon_{(n)(Newton)}}{\epsilon_{(n)(secant)}}=|\epsilon_{0}*k|^{{2^n}-{1.618^n}}$$ $$\frac{\epsilon_{(n)(Newton)}}{\epsilon_{(n)(secant)}}=|\epsilon_{0}*\frac{f”(r)}{2f'(r)}|^{{2^n}-{1.618^n}}$$
 10.08.18
I don't understand this part: $$k^{\sum_{m=0}^{n-1} p^{m-1}}=k^{1+\sum_{m=0}^{n-1} p^{m-1}}$$ This doesn't make sense to me. Is this a typo or a misunderstanding? Please fix this and retype your proof below.
 Question by Student 201527119 Sorry, I missed k in left. $$k*k^{\sum_{m=0}^{n-1} p^{m-1}}=k^{1+\sum_{m=0}^{n-1} p^{m-1}}$$
Hm, but if that is true, then I don't understand how you obtain the following on the denominator on the LHS: $$\frac{k^{2^n}*\epsilon_{0}^{2^n}}{k*k^{p^n-1}*\epsilon_{0}^{p^n}}=\frac{k^{2^n}*\epsilon_{0}^{2^n}}{k^{p^n}*\epsilon_{0}^{p^n}}$$ Please fix this and rewrite your proof better below. And please never use the newline command in your posts (two blackslashes). This is not necessary and is a bad habit.
 Question by Student 201727142 Hello, professor. I can't understand that the bisection method in Assignment 2 #1. If the significant number is 4, should I round to the fourth decimal places? or round down to the fourth decimal place? And do I have to count 1 when I get the $x_{mid}$ in the initial range($x_{min}=\frac{\pi}{2}\leqq x \leqq x_{max}=\frac{\pi*3}{2}$)?
Obtaining convergence to 4 significant numbers means that iterating more won't change the first 4 significant numbers. Only one question per post please. 1 point bonus.
 Question by Student 201527119 $$\epsilon_{(n)(Newton)}=k^{2^{n}-1} \epsilon_{0}^{2n}$$ $$\epsilon_{(n)(secant)}=k^{\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^{n}}$$ $$\frac{\epsilon_{(n)(Newton)}}{\epsilon_{(n)(secant)}}=\frac{k^{2^{n}-1} \epsilon_{0}^{2n}}{k^{\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^{n}}}=\frac{\frac{k^{2^n} \epsilon_{0}^{2n}}{k}}{k^{\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^n}}=\frac{k^{2^n} \epsilon_{0}^{2n}}{k^1 k^{\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^n}} =\frac{k^{2^n} \epsilon_{0}^{2n}}{k^{1+\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^n}}$$ $$(k^{a}k^{b}=k^{a+b})$$ In denominator, $$k^{\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^n} =k^{1+p^{-1}+p^0+p^1+...+p^{n-2}} \epsilon_{0}^{p^n}$$ $$p^{-1}+p^0+p^1+...+p^{n-2}=\frac{\frac{p^n-1}{p}}{p-1}=\frac{p^n-1}{p^2-p}$$ $$and, p^2=p+1$$ $$so,\frac{p^n-1}{p^2-p}=p^n-1$$ $$k^{\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^n}=k^{1+p^{-1}+p^0+p^1+...+p^{n-2}} \epsilon_{0}^{p^n}=k^{1+p^n-1} \epsilon_{0}^{p^n}=k^{p^n} \epsilon_{0}^{p^n}$$ $$\frac{\epsilon_{(n)(Newton)}}{\epsilon_{(n)(secant)}}=\frac{k^{2n} \epsilon_{0}^{2n}}{k^{p^n} \epsilon_{0}^{p^n}}= \frac{{(k \epsilon_{0})}^{2^n}}{{(k \epsilon_{0})}^{p^n}}={(k \epsilon_{0})}^{2^n-p^n}$$ $$So, p=1.618, k=|\frac{f”(r)}{2f'(r)}|$$ $$\frac{\epsilon_{(n)(Newton)}}{\epsilon_{(n)(secant)}}= {(|\frac{f”(r)}{2f'(r)}| \epsilon_{0})}^{2^n-{1.618}^n}$$ I think this is much better than before one.
There is still the same problem with your proof. You claim that $$\frac{\epsilon_{(n)(Newton)}}{\epsilon_{(n)(secant)}}=\frac{k^{2^{n}-1} \epsilon_{0}^{2n}}{k^{\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^{n}}}=\frac{\frac{k^{2^n} \epsilon_{0}^{2n}}{k}}{k^{\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^n}}=\frac{k^{2^n} \epsilon_{0}^{2n}}{k^1 k^{\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^n}} =\frac{k^{2^n} \epsilon_{0}^{2n}}{k^{1+\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^n}}$$ and that $$k^{\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^n}=k^{1+p^{-1}+p^0+p^1+...+p^{n-2}} \epsilon_{0}^{p^n}=k^{1+p^n-1} \epsilon_{0}^{p^n}=k^{p^n} \epsilon_{0}^{p^n}$$ But if you substitute the latter in the former, you will not get $$\frac{\epsilon_{(n)(Newton)}}{\epsilon_{(n)(secant)}}=\frac{k^{2n} \epsilon_{0}^{2n}}{k^{p^n} \epsilon_{0}^{p^n}}= \frac{{(k \epsilon_{0})}^{2^n}}{{(k \epsilon_{0})}^{p^n}}={(k \epsilon_{0})}^{2^n-p^n}$$ Thus your proof is wrong. I gave you 3 chances already, and you didn't succeed. I can not give you one more chance. 1 point bonus for the effort. Let others now try.
 Question by Student 201727142 Hello, professor. I'm so sorry to have asked a lot of questions. But I still don't understand the second question. And do I have to count 1 when I get the $x_{mid}$ in the initial range($x_{min}=\frac{\pi}{2} \leqq x \leqq x_{max}= \frac{\pi∗3}{2}$)?
 10.09.18
I don't understand your question. Please formulate this better.
 Question by Student 201427128 Dear professor, I have question about A2#Q6(a). before that, when you solved problem using eq.Newton-Raphson by hand in the class. $x_{n+1}=x_n-\frac{f(x_n)(x_{n}-x_{n-1})}{x_n-x_{n-1}}$ $f(x)=x^{2}-2$ you guessed$~~x_0=1,~~x_{-1}=0.9$ and found $~~x_1=1.526$ And with many iterations, can get a more accurate x. I used this idea to solve A2#Q6(a). $x_{n+1}=x_n-0.05\frac{f(x_n)}{f'(x_n)}-0.95\frac{f(x_n)(x_{n-1}-x_{n-2})}{f(x_{n-1})-f(x_{n-2})}$ given initial condition$~~x_0 = 2.8$ I guessed$~~x_1 = 3.0,~~x_2 = 3.2$ And I got the answers(=5 iterations) on the site. but now I have question that if I guess more bigger x1 x2 , I get another iteration number Is there a more accurate way to guess?
 10.12.18
I'm not sure what you mean. The number of iterations depends on the initial guess. If the initial guess is the root, then the number of iterations is 1. I'll give you 0.5 point for the effort.
 Question by Student 201427128 professor, I'm sorry to repeat the same question. A2#6(a) ask about the number of iteration up to 4 significant digits. the number of iteration will different if the initial guess is different. I think it's possible that each person has a different answer. Is there a proper way to set the initial guess?
 10.13.18
I see what you mean. True, the number of iterations depends not only on the initial guess, but also on the values given to the previous guesses for the root (i.e. $x_{-1}$ and $x_{-2}$). This may affect slightly the number of iterations needed to reach convergence. But for fastest convergence generally choose values for $x_{-1}$ and $x_{-2}$ that are very close to $x_0$. But even if you choose values for $x_{-1}$ and $x_{-2}$ that are not very close to $x_0$, I won't take away points. When correcting the exams, I will look at the logic, not the answers only. Good question: 1.0 bonus point more. Lastly, please stop using the double backslash character when writing your posts (the “\\”). This makes your post very hard to read. In English, simply write what you want to say within one paragraph (one paragraph = one idea = one question). Don't break lines.
 Previous   1  ...  6 ,  7 ,  8 ,  9    Next  •  PDF 1✕1 2✕1 2✕2  •  New Question
 $\pi$