Numerical Analysis Questions & Answers | |
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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.
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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.
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I don't understand your question. Please formulate this better.
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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.
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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. |
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$\pi$ |