Question by Student 201527119 Sorry, I missed k in left. $$k*k^{\sum_{m=0}^{n-1} p^{m-1}}=k^{1+\sum_{m=0}^{n-1} p^{m-1}}$$
 10.08.18
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.
 10.09.18
 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($x_{min}=\frac{\pi}{2} \leqq x \leqq x_{max}= \frac{\pi∗3}{2}$)?
 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?
 $\pi$